Bretscher_600928X_ISM_TTL.qxd:au_123456_ttl
1/12/09
8:40 AM
Page 1
INSTRUCTOR’S SOLUTIONS MANUAL KYLE BURKE Boston U...

Bretscher_600928X_ISM_TTL.qxd:au_123456_ttl

1/12/09

8:40 AM

Page 1

INSTRUCTOR’S SOLUTIONS MANUAL KYLE BURKE Boston University with art by

GEORGE WELCH

LINEAR ALGEBRA WITH APPLICATIONS FOURTH EDITION

Otto Bretscher Colby College

Bretscher_600928X_ISM_TTL.qxd:au_123456_ttl

1/12/09

8:40 AM

Page 2

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Prentice Hall from electronic files supplied by the author. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Prentice Hall, Upper Saddle River, NJ 07458. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-600928-3 ISBN-10: 0-13-600928-X 1 2 3 4 5 6 OPM 10 09 08

Table of Contents Chapter 1: Linear Equations ..……………………………………………………… 1 Chapter 2: Linear Transformations...………………………………………………… 48 Chapter 3: Subspaces of Rn and Their Dimensions….…………………………….... 120 Chapter 4: Linear Spaces…………………………………………………...…………171 Chapter 5: Orthogonality and Least Squares...……………………………………..... 217 Chapter 6: Determinants …………………………………………………………….. 266 Chapter 7: Eigenvalues and Eigenvectors…………………………………………… 297 Chapter 8: Symmetric Matrices and Quadratic Forms.……………..……………..... 360 Chapter 9: Linear Differential Equations.…………………………………………… 392

Section 1.1

Chapter 1 Section 1.1 x + 2y = 1 x + 2y = 1 → → 2x + 3y = 1 −2 × 1st equation −y = −1 ÷(−1) x + 2y = 1 −2 × 2nd equation x = −1 → , so that (x, y) = (−1, 1). y=1 y =1

1.1.1

1.1.2

x + 34 y = 21 = 2 ÷4 → → =3 7x + 5y = 3 −7 × 1st equation = 21 x + 43 y = 12 x = −1 → , → 3 y = 2 y = 2 − 4 × 2nd equation = − 12 ×(−4)

4x + 3y 7x + 5y

x + 34 y − 14 y

so that (x, y) = (−1, 2). 1.1.3

2x + 4y 3x + 6y

=3 =2

÷2

→

x + 2y 3x + 6y

= 23 =2

→

x + 2y 3x + 6y

=1 =3

−3 × 1st equation

→

x + 2y 0

= 32 . = − 52

→

x + 2y 0

=1 =0

So there is no solution. 1.1.4

2x + 4y 3x + 6y

=2 =3

÷2

−3 × 1st equation

This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation x + 2y = 1 gives us x = 1 − 2y = 1 − 2t. Therefore the general solution is (x, y) = (1 − 2t, t), where t is an arbitrary real number. 1.1.5

x + 23 y = 0 = 0 ÷2 → → =0 4x + 5y = 0 −4 × 1st equation =0 x =0 x + 23 y = 0 − 23 × 2nd equation , → → y =0 y =0 = 0 ÷(−1)

2x + 3y 4x + 5y

x + 23 y −y

so that (x, y) = (0, 0). x + 2y + 3z x + 2y + 3z = 8 y 1.1.6 x + 3y + 3z = 10 −I → z x + 2y + 4z = 9 −I x + 3z = 4 −3(III) x=1 y = 2 → y = 2 , so that z =1 z=1

x + 2y + 3z 1.1.7 x + 3y + 4z x + 4y + 5z

x + 2y + 3z =1 y+z = 3 −I → 2y + 2z = 4 −I

This system has no solution.

=8 −2(II) → = 2 =1 (x, y, z) = (1, 2, 1).

x+z = 1 −2(II) → y +z = 2 0 = 3 −2(II) 1

= −3 = 2 = −1

Chapter 1 x + 2y + 3z = 0 x + 2y + 3z =0 = 0 ÷(−3) → 1.1.8 4x + 5y + 6z = 0 −4(I) → −3y − 6z −6y − 11z = 0 7x + 8y + 10z = 0 −7(I) x + 2y + 3z = 0 −2(II) x−z =0 +III x =0 y + 2z = 0 → y + 2z = 0 −2(III) → y = 0 , −6y − 11z = 0 +6(II) z =0 z =0

so that (x, y, z) = (0, 0, 0).

x + 2y + 3z = 1 x + 2y + 3z = 1 1.1.9 3x + 2y + z = 1 −3(I) → −4y − 8z = −2 ÷(−4) → 7x + 2y − 3z = 1 −7(I) −12y − 24z = −6 −2(II) x−z =0 x + 2y + 3z = 1 y + 2z → y + 2z = 12 = 12 −12y − 24z = −6 +12(II) 0 =0

This system has infinitely many solutions: if we choose z = t, an arbitrary real number, then we get x = z = t and y = 21 − 2z = 12 − 2t. Therefore, the general solution is (x, y, z) = t, 21 − 2t, t , where t is an arbitrary real number. x + 2y + 3z =1 z = 2 −2(I) → y + 2z = 8 −3(I)

x + 2y + 3z 1.1.10 2x + 4y + 7z 3x + 7y + 11z

x + 2y + 3z y + 2z z

x−z = 1 −2(II) = 5 → y + 2z =0 z

so that (x, y, z) = (−9, 5, 0). 1.1.11

x − 2y 3x + 5y

=2 = 17

−3(I)

→

x − 2y 11y

= −9

=1 Swap : = 0 → II ↔ III =5

+III x = −9 = 5 −2(III) → y = 5 , z =0 =0

=2 = 11

÷11

→

x − 2y y

=2 =1

so that (x, y) = (4, 1). See Figure 1.1.

Figure 1.1: for Problem 1.1.11.

1.1.12

x − 2y 2x − 4y

=3 =6

−2(I)

x − 2y → 0

=3 =0

2

+2(II)

→

x =4 , y =1

Section 1.1 This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation x − 2y = 3 gives us x = 3 + 2y = 3 + 2t. Therefore the general solution is (x, y) = (3 + 2t, t), where t is an arbitrary real number. (See Figure 1.2.)

Figure 1.2: for Problem 1.1.12.

1.1.13

x − 2y 2x − 4y

=3 =8

−2(I)

→

x − 2y 0

=3 , which has no solutions. (See Figure 1.3.) =2

Figure 1.3: for Problem 1.1.13.

x + 5z 1.1.14 The system reduces to y − z 0 planes.

=0 = 0 , so that there is no solution; no point in space belongs to all three =1

Compare with Figure 2b. x =0 1.1.15 The system reduces to y = 0 so the unique solution is (x, y, z) = (0, 0, 0). The three planes intersect at z =0 the origin.

x + 5z 1.1.16 The system reduces to y − z 0 arbitrary number. The three planes 1.1.17

x + 2y 3x + 5y

=a =b

=0 = 0 , so the solutions are of the form (x, y, z) = (−5t, t, t), where t is an =0 intersect in a line; compare with Figure 2a.

x + 2y → −3(I) −y

=a = −3a + b

x + 2y → ÷(−1) y

3

=a = 3a − b

−2(II)

Chapter 1

x = −5a + 2b , so that (x, y) = (−5a + 2b, 3a − b). y = 3a − b

x + 2y + 3z = a x + 2y + 3z =a −2(II) 1.1.18 x + 3y + 8z = b −I → y + 5z = −a + b → x + 2y + 2z = c −I −z = −a + c x = 10a − 2b − 7c x − 7z = 3a − 2b +7(III) x − 7z = 3a − 2b y + 5z = −a + b → y + 5z = −a + b −5(III) → y = −6a + b + 5c , z =a−c z =a−c −z = −a + c ÷(−1) so that (x, y, z) = (10a − 2b − 7c, −6a + b + 5c, a − c).

1.1.19 a Note that the demand D1 for product 1 increases with the increase of price P2 ; likewise the demand D2 for product 2 increases with the increase of price P1 . This indicates that the two products are competing; some people will switch if one of the products gets more expensive.

70 − 2P1 + P2 105 + P1 − P2 which yields the unique solution P1 = 26 and P2 = 46.

b Setting D1 = S1 and D2 = S2 we obtain the system

−5P1 + P2 = −14 + 3P1 , or P1 − 3P2 = −7 + 2P2

= −84 , = 112

1.1.20 The total demand for the product of Industry A is 1000 (the consumer demand) plus 0.1b (the demand from Industry B). The output a must meet this demand: a = 1000 + 0.1b. a = 1000 + 0.1b a − 0.1b = 1000 Setting up a similar equation for Industry B we obtain the system or , b = 780 + 0.2a −0.2a + b = 780 which yields the unique solution a = 1100 and b = 1000. 1.1.21 The total demand for the products of Industry A is 310 (the consumer demand) plus 0.3b (the demand from Industry B). The output a must meet this demand: a = 310 + 0.3b. a = 310 + 0.3b a − 0.3b = 310 Setting up a similar equation for Industry B we obtain the system or , b = 100 + 0.5a −0.5a + b = 100 which yields the solution a = 400 and b = 300. 1.1.22 Since x(t) = a sin(t) + b cos(t) we can compute

dx dt

= a cos(t) − b sin(t) and 2

d2 x dt2

= −a sin(t) − b cos(t). Substituting these expressions into the equation ddt2x − dx dt − x = cos(t) and simplifying gives (b − 2a) sin(t) + (−a − 2b) cos(t) = cos(t). Comparing the coefficients of sin(t) and cos(t) on both sides of −2a + b = 0 , so that a = − 51 and b = − 25 . See Figure 1.4. the equation then yields the system −a − 2b = 1

Figure 1.4: for Problem 1.1.22.

4

Section 1.1 1.1.23 a Substituting λ = 5 yields the system 2x − y 2x − y =0 7x − y = 5x or or 0 −6x + 3y = 0 −6x + 8y = 5y

=0 . =0 There are infinitely many solutions, of the form (x, y) = 2t , t , where t is an arbitrary real number.

b Proceeding as in part (a), we find (x, y) = − 13 t, t .

c Proceedings as in part (a), we find only the solution (0, 0).

1.1.24 Let v be the speed of the boat relative to the water, and s be the speed of the stream; then the speed of the boat relative to the land is v + s downstream and v − s upstream. Using the fact that (distance) = (speed)(time), we obtain the system # " 8 = (v + s) 31 ← downstream ← upstream 8 = (v − s) 2 3

The solution is v = 18 and s = 6. x+z 1.1.25 The system reduces to y − 2z 0

=1 = −3 . =k−7

a. The system has solutions if k − 7 = 0, or k = 7. b. If k = 7 then the system has infinitely many solutions. c. If k = 7 then we can choose z = t freely and obtain the solutions (x, y, z) = (1 − t, −3 + 2t, t). x − 3z y + 2z 1.1.26 The system reduces to (k 2 − 4)z

= 1 = 1 = k−2

This system has a unique solution if k 2 − 4 6= 0, that is, if k 6= ±2.

If k = 2, then the last equation is 0 = 0, and there will be infinitely many solutions. If k = −2, then the last equation is 0 = −4, and there will be no solutions. 1.1.27 Let x = the number of male children and y = the number of female children. Then the statement “Emile has twice as many sisters as brothers” translates into y = 2(x − 1) and “Gertrude has as many brothers as sisters” translates into x = y − 1. Solving the system

−2x + y x−y

= −2 = −1

gives x = 3 and y = 4.

There are seven children in this family. 5

Chapter 1 1.1.28 The thermal equilibrium condition requires −4T1 + T2 We can rewrite this system as T1 − 4T2 + T3 T2 − 4T3 The solution is (T1 , T2 , T3 ) = (75, 100, 125).

, T2 = that T1 = T2 +200+0+0 4 = −200 = −200 = −400

T1 +T3 +200+0 , 4

and T3 =

T2 +400+0+0 . 4

1.1.29 To assure that the graph goes through the point (1, −1), we substitute t = 1 and f (t) = −1 into the equation f (t) = a + bt + ct2 to give −1 = a + b + c. a+b+c = −1 Proceeding likewise for the two other points, we obtain the system a + 2b + 4c = 3 . a + 3b + 9c = 13 The solution is a = 1, b = −5, and c = 3, and the polynomial is f (t) = 1 − 5t + 3t2 . (See Figure 1.5.)

Figure 1.5: for Problem 1.1.29. a + b + c = p 1.1.30 Proceeding as in the previous exercise, we obtain the system a + 2b + 4c = q . a + 3b + 9c = r a = 3p − 3q + r The unique solution is b = −2.5p + 4q − 1.5r . c = 0.5p − q + 0.5r

Only one polynomial of degree 2 goes through the three given points, namely, f (t) = 3p − 3q + r + (−2.5p + 4q − 1.5r)t + (0.5p − q + 0.5r)t2 .

1.1.31 f (t) is of the form at2 + bt + c. So f (1) = a(12 ) + b(1) + c = 3, and f (2) = a(22 ) + b(2) + c = 6. Also, f ′ (t) = 2at + b, meaning that f ′ (1) = 2a + b = 1. a+b+c=3 So we have a system of equations: 4a + 2b + c = 6 2a + b = 1

a=2

which reduces to b = −3 . c=4 6

Section 1.1 Thus, f (t) = 2t2 − 3t + 4 is the only solution. 1.1.32 f (t) is of the form at2 + bt + c. So, f (1) = a(12 ) + b(1) + c = 1 and f (2) = 4a + 2b + c = 0. Also, R2 R2 f (t)dt = 1 (at2 + bt + c)dt 1 = a3 t3 + 2b t2 + ct|21

= 83 a + 2b + 2c − ( a3 +

b 2

+ c)

= 73 a + 23 b + c = −1.

a+b+c=1 So we have a system of equations: 4a + 2b + c = 0 7 3 3 a + 2 b + c = −1 a=9 which reduces to b = −28 . c = 20

Thus, f (t) = 9t2 − 28t + 20 is the only solution.

1.1.33 f (t) is of the form at2 + bt + c. f (1) = a + b + c = 1, f (3) = 9a + 3b + c = 3, and f ′ (t) = 2at + b, so f ′ (2) = 4a + b = 1. a+b+c=1 Now we set up our system to be 9a + 3b + c = 3 . 4a + b = 1 a − 3c = 0 This reduces to b + 43 c = 1 . 0=0 We write everything in terms of a, revealing c = 3a and b = 1 − 4a. So, f (t) = at2 + (1 − 4a)t + 3a for an arbitrary a. 1.1.34 f (t) = at2 + bt + c, so f (1) = a + b + c = 1, f (3) = 9a + 3b + c = 3. Also, f ′ (2) = 3, so 2(2)a + b = 4a + b = 3. a+b+c=1 Thus, our system is 9a + 3b + c = 3 . 4a + b = 3 When we reduce this, however, our last equation becomes 0 = 2, meaning that this system is inconsistent.

1.1.35 f (t) = ae3t + be2t , so f (0) = a + b = 1 and f ′ (t) = 3ae3t + 2be2t , so f ′ (0) = 3a + 2b = 4. a+b=1 , Thus we obtain the system 3a + 2b = 4 a=2 . which reveals b = −1 So f (t) = 2e3t − e2t . 7

Chapter 1 1.1.36 f (t) = a cos(2t) + b sin(2t) and 3f (t) + 2f ′ (t) + f ′′ (t) = 17 cos(2t). f ′ (t) = 2b cos(2t) − 2a sin(2t) and f ′′ (t) = −4b sin(2t) − 4a cos(2t). So, 17 cos(2t) = 3(a cos(2t) + b sin(2t)) + 2(2b cos(2t) − 2a sin(2t)) + (−4b sin(2t) − 4a cos(2t)) = (−4a + 4b + 3a) cos(2t) + (−4b − 4a + 3b) sin(2t) = (−a + 4b) cos(2t) + (−4a − b) sin(2t). −a + 4b = 17 . So, our system is: −4a − b = 0 a = −1 This reduces to: . b=4 So our function is f (t) = − cos(2t) + 4 sin(2t). 1.1.37 Plugging the three points (x, y) into the equation a + bx + cy + x2 + y 2 = 0, leads to a system of linear equations for the three unknowns (a, b, c). a + 5b + 5c + 25 + 25 a + 4b + 6c + 16 + 36

= =

a + 6b + 2c + 36 + 4 =

0 0 0.

The solution is a = −20, b = −2, c = −4. −20 − 2x − 4y + x2 + y 2 = 0 is a circle of radius 5 centered at (1, 2). 1.1.38 Plug the three points into the equation ax2 + bxy + cy 2 = 1. We obtain a system of linear equations a + 2b + 4c 4a + 4b + 4c

= =

9a + 3b + c =

1 1 1.

The solution is a = 3/20, b = −9/40, c = 13/40. This is the ellipse (3/20)x2 − (9/40)xy + (13/40)y 2 = 1.

x−z

1.1.39 The given system reduces to y + 2z 0

−5a+2b 3 = 4a−b 3

=

= a − 2b + c

.

This system has solutions (in fact infinitely many) if a − 2b + c = 0. The points (a, b, c) with this property form a plane through the origin.

1.1.40 a x1 = −3 x2 = 14 + 3x1 = 14 + 3(−3) = 5 x3 = 9 − x1 − 2x2 = 9 + 3 − 10 = 2 x4 = 33 + x1 − 8x2 + 5x3 − x4 = 33 − 3 − 40 + 10 = 0, so that (x1 , x2 , x3 , x4 ) = (−3, 5, 2, 0). 8

Section 1.1 b x4 = 0 x3 = 2 − 2x4 = 2 x2 = 5 − 3x3 − 7x4 = 5 − 6 = −1 x1 = −3 − 2x2 + x3 − 4x4 = −3 + 2 + 2 = 1, so that (x1 , x2 , x3 , x4 ) = (1, −1, 2, 0).

Figure 1.6: for Problem 1.1.41a.

1.1.41 a The two lines intersect unless t = 2 (in which case both lines have slope −1). To draw a rough sketch of x(t), note that limt→∞ x(t) = limt→−∞ x(t) = −1 the line x + 2t y = t becomes almost horizontal and limt→2− x(t) = ∞, limt→2+ x(t) = −∞. Also note that x(t) is positive if t is between 0 and 2, and negative otherwise. Apply similar reasoning to y(t). (See Figures 1.6 and 1.7.)

Figure 1.7: for Problem 1.1.41a. 9

Chapter 1 b x(t) =

−t t−2 ,

and y(t) =

2t−2 t−2 .

1.1.42 We can think of the line through the points (1, 1, 1) and (3, 5, 0) as the intersection of any two planes through these two points; each of these planes will be defined by an equation of the form ax + by + cz = d. It is required that 1a + 1b + 1c = d and 3a + 5b + 0c = d. a +b +c −d = 0 Now the system reduces to 3a +5b −d = 0 a + 25 c −2d = 0 . b − 32 c +d = 0 We can choose arbitrary real numbers for c and d; then a = − 25 c + 2d and b = 23 c − d. For example, if we choose c = 2 and d = 0, then a = −5 and b = 3, leading to the equation −5x + 3y + 2z = 0. If we choose c = 0 and d = 1, then a = 2 and b = −1, giving the equation 2x − y = 1. −5x +3y +2z = 0 . We have found one possible answer: 2x −y = 1 1.1.43 To eliminate the arbitrary constant t, we can solve the last equation for t to give t = z − 2, and substitute x = 6 + 5(z − 2) x − 5z = −4 z − 2 for t in the first two equations, obtaining or . y = 4 + 3(z − 2) y − 3z = −2 This system does the job. 1.1.44 Let b = Boris’ money, m = Marina’s money, and c = cost of a chocolate bar. b + 21 m = c We are told that 1 , with solution (b, m) = (0, 2c). 2 b + m = 2c Boris has no money. 1.1.45 Let us start by reducing the system: x + 2y + 3z = 39 x + 2y + 3z x + 3y + 2z = 34 −I → y−z 3x + 2y + z = 26 −3(I) −4y − 8z

= 39 = −5 = −91

Note that the last two equations are exactly those we get when we substitute y−z = −5 x = 39 − 2y − 3z: either way, we end up with the system . −4y − 8z = −91

1.1.46 a We set up two equations here, with our variables: x1 = servings of rice, x2 = servings of yogurt. 3x1 +12x2 = 60 So our system is: . 30x1 +20x2 = 300 Solving this system reveals that x1 = 8, x2 = 3. b Again, we set up our equations:

3x1 30x1

+12x2 +20x2

P + and reduce them to find that x1 = − 15

C 25 ,

=P , =C

while x2 = 10

P 10

−

C 100 .

Section 1.2 1.1.47 Let x1 = number of one-dollarbills, x2 = the number offive-dollar bills, and x3 = the number of ten-dollar x1 + x2 + x3 = 32 , bills. Then our system looks like: x1 + 5x2 + 10x3 = 100 which reduces to give us solutions that fit: x1 = 15 + 45 x3 , x2 = 17 − 49 x3 , where x3 can be chosen freely. Now let’s keep in mind that x1 , x2 , and x3 must be positive integers and see what conditions this imposes on the variable x3 . We see that since x1 and x2 must be integers, x3 must be a multiple of 4. Furthermore, x3 must be positive, and x2 = 17 − 94 x3 must be positive as well, meaning that x3 < 68 9 . These constraints leave us with only one possibility, x3 = 4, and we can compute the corresponding values x1 = 15 + 45 x3 = 20 and x2 = 17 − 94 x3 = 8. Thus, we have 20 one-dollar bills, 8 five-dollar bills, and 4 ten-dollar bills. 1.1.48 Let x1 , x2 , x3 be the number of 20 cent, 50 cent, and 2 Euro coins, respectively. Then we need solutions to x1 +x2 +x3 = 1000 the system: .2x1 +.5x2 +2x3 = 1000 x1 −5x3 = − 5000 3 . this system reduces to: x2 +6x3 = 8000 3 5x3 − 5000 x1 3 . Unfortunately for the meter maids, there are no Our solutions are then of the form x2 = −6x3 + 8000 3 x3 x3 integer solutions to this problem. If x3 is an integer, then neither x1 nor x2 will be an integer, and no one will ever claim the Ferrari.

Section 1.2

. 1 −2.. . 3 4..

. 4 −1.. . 8 −2..

. −2.. . 8..

. 0 −10.. . 1 8..

1 5 1 1 5 −II 1 1.2.1 → → 2 2 −2(I) 0 1 −8 0 x − 10z = 13 x = 13 + 10z −→ y + 8z = −8 y = −8 − 8z x 13 + 10t y = −8 − 8t , where t is an arbitrary real number. z t 3 1.2.2 6

8 ÷3 3

→

1 6

4 3

8

.

− 13 .. . −2..

8 3

3

−6(I)

→

1

4 3

0

0

.

− 13 .. . 0..

13 −8

8 3

−13

This system has no solutions, since the last row represents the equation 0 = −13.

1.2.3 x = 4 − 2y − 3z y and z are free variables; let y = s and z = t. 4 − 2s − 3t x , where s and t are arbitrary real numbers. y = s t z 11

Chapter 1

1 1.2.4 2 3

1 0

0

. 0.. . 1.. . 0..

0

0 1.2.5 1

. 1.. . −1.. . 4..

. 1.. 1 1 . 5 −2(I) → 0 −3.. . 0 1.. 2 −3(I)

2 x = 2 −1 , so that y = −1 . 0 0 1 1 1 1 0

. 1.. . 0.. . 0.. . 1..

0

.. 1 1 1.. 3 ÷(−3) → 0 1.. . 0 1.. −1

1

0 swap : → 0 I ↔ III 0 0

1 0 1 1 0 1

1 0 0 .. +III 1 0 −1 0 . 0 1 0 0 .. 0 1 −III → 0 1 0 1 0. 0 . .. 0 0 0 1 1 0 0 1 .. 0 0 1 1. 0 −III 0 0 0 x1 x1 + x4 = 0 x2 − x4 = 0 −→ x2 x3 x3 + x4 = 0 1

0 0

0

. 0.. . 0.. . 1.. . 1..

. 1.. . −1.. . 1.. . 0..

0

0 0

0 −I

1 −II → −1 −II

−1

1

0 → 0

1 1 0

0 −1

. 0 0.. . 1 0.. . 1 1.. . 0 1..

0 0 0 0

= −x4 = x4 = −x4

x1 −t x2 t = , where t is an arbitrary real number. x3 −t x4 t

1.2.6 The system is in rref already. x1 = 3 + 7x2 − x5 x3 = 2 + 2x5 x4 = 1 − x5 Let x2 = t and x5 = r. 3 + 7t − r x1 t x2 x3 = 2 + 2r 1−r x4 r x5

1 0 1.2.7 0 0

2 0 0 1 0 1 0 0

. 3.. . 3 2.. . 4 −1.. . 0 1.. 2

0 1 0 0 → 0 0 −II 0 0

2 0 0 1 0 0 0 0

. 3.. . 3 2.. . 1 −3.. . 0 1.. 2

12

0 −2(III) 0 −3(III) → 0 0

0

0 0

0

−II → +II

Section 1.2 . 9.. 0 −9(IV ) 1 2 0 0 .. 0 11. 0 −11(IV ) 0 0 1 0 → .. +3(IV ) 0 0 0 1 1 −3. 0 .. 0 0 0 0 0 1. 0 =0 x1 = −2x2 = 0 =0 x3 −→ =0 x4 =0 =0 x5 =0 −2t x1 x2 t Let x2 = t. x3 = 0 , where t is an arbitrary 0 x4 0 x5

1 2 0 0 0 1 0 0 0 0 0 0 x1 + 2x2 x3 x4 x5

1.2.8

"

0 0

. 3.. . 8..

1 0 2 0 0 4

x2 − x5 x4 + 2x5

. 0.. . 0.. . 0.. . 1..

0

0

#

0

÷4

=0 x2 −→ x4 =0

→

"

0 1 0 0 0 0

. 2 3.. . 1 2..

0

0 0 0

real number.

0 0

#

= x5 = −2x5

−2(II) 0 1 0 0

0 0 0 1

.. −1. 0 . 2.. 0

Let x1 = r, x3 = s, x5 = t. r x1 x2 t x3 = s , where r, t and s are arbitrary real numbers. −2t x4 t x5 . . 1 2 0 0 1 −1.. 0 −1.. 2 swap : . . → 1.2.9 1 2 0 0 1 −1.. 0 I ↔ II → 0 0 0 1 2 −1.. 2 .. .. 1 2 2 0 −2 1. 2 −I 1 2 2 0 −1 1. 2 . . 1 2 0 0 1 −1.. 0 1 2 0 0 1 −1.. 0 swap : . . 2.. 2 ÷2 → 0 0 0 1 2 −1.. 2 II ↔ III → 0 0 2 0 −2 .. .. 0 0 0 1 2 −1. 2 0 0 2 0 −2 2. 2 . 1 2 0 0 1 −1.. 0 . 1.. 1 0 0 1 0 −1 .. 0 0 0 1 2 −1. 2 x1 + 2x2 + x5 − x6 = 0 x1 = −2x2 − x5 + x6 = 1 + x5 − x6 x3 − x5 + x6 = 1 −→ x3 x4 = 2 − 2x5 + x6 x4 + 2x5 − x6 =2

0

0 0

1

2

Let x2 = r, x5 = s, and x6 = t.

13

Chapter 1 −2r − s + t x1 r x2 x3 1 + s − t , where r, s and t are arbitrary real numbers. = x4 2 − 2s + t s x5 t x6

x1

1.2.10 The system reduces to

x2 x3

+ x4 − 3x4 + 2x4

x1 = 1 = 2 −→ x2 x3 = −3

= 1 − x4 = 2 + 3x4 = −3 − 2x4

x1 = 0 = 4 −→ x2 x4 = −2

= −2x3 = 4 + 3x3 . = −2

Let x4 = t. 1−t x1 x2 2 + 3t , where t is an arbitrary real number. = −3 − 2t x3 t x4

x1

x1

1.2.11 The system reduces to

+ 2x3 − 3x3

x2

x4

Let x3 = t. −2t x1 x2 4 + 3t = t x3 −2 x4

x2

1.2.12 The system reduces to x1 x2 x3 x4

+ +

3.5x5 x5

+

3x5

x3 x4

= −3.5x5 − x6 = −x5 . = 53 x6 = −3x5 − x6

Let x5 = r and x6 = t. −3.5r − t x1 −r x2 5 t x3 3 = x4 −3r − t x5 r x6 t

1.2.13 The system reduces to There are no solutions.

x y

− +

z 2z 0

= 0 = 0 . = 1 14

+

x6

− +

5 3 x6 x6

= 0 = 0 −→ = 0 = 0

Section 1.2

1.2.14 The system reduces to

x = −2 − 2y = −2 . −→ z =2 = 2

x +

x

x1 + 2x2 + 3x3

2y z

Let y = t. −2 − 2t x y = t 2 z 1.2.15 The system reduces to 1.2.16 The system reduces to

x1 x4

y z

4 2 . 1

= = =

x4

+5x5 +2x5

= =

6 −→ 7

= 6 − 2x2 − 3x3 − 5x5 . = 7 − 2x5

Let x2 = r, x3 = s, and x5 = t. 6 − 2r − 3s − 5t x1 r x2 s x3 = 7 − 2t x4 t x5

8221 = − 4340

x1

1.2.17 The system reduces to

x2

= x3

= x4

= x5

=

4695 . 434 459 − 434 8591 8680

699 434

1.2.18 a No, since the third column contains two leading ones. b Yes c No, since the third row contains a leading one, but the second row does not. d Yes 0 1 0 0 1.2.19 and 0 0 0 0 1.2.20 Four, namely

1 0 0 , 0 0 0

1 0 0 1 k (k is an arbitrary constant.) , , 0 1 0 0 0 15

Chapter 1

0 1.2.21 Four, namely 0 0 1.2.22 Seven, namely

1 0 0 1 1 k 0 0 , 0 0 , 0 0 , 0 1 (k is an arbitrary constant.) 0 0 0 0 0 0 0

0 0 0 0

0 1 1 a b 0 , , 0 0 0 0 0 0

0 c , 0 0

1 0 1 , 0 0 0

1 0 d , 0 1 e

f 0

0 0 , 0 1

1 0 . 0 1

Here, a, b, . . . , f are arbitrary constants. 1.2.23 The conditions a, b, and c for the reduced row-echelon form correspond to the properties P1, P2, and P3 given on Page 13. The Gauss-Jordan algorithm, summarized on Page 15, guarantees that those properties are satisfied. 1.2.24 Yes; each elementary row operation is reversible, that is, it can be “undone.” For example, the operation of row swapping can be undone by swapping the same rows again. The operation of dividing a row by a scalar can be reversed by multiplying the same row by the same scalar. 1.2.25 Yes; if A is transformed into B by a sequence of elementary row operations, then we can recover A from B by applying the inverse operations in the reversed order (compare with Exercise 24). 1.2.26 Yes, by Exercise 25, since rref(A) is obtained from A by a sequence of elementary row operations.

1 1.2.27 No; whatever elementary row operations you apply to 4 7 to zero.

2 3 5 6 , you cannot make the last column equal 8 9

a11 x1 + a12 x2 + · · · + a1n xn = b1 1.2.28 Suppose (c1 , c2 , . . . , cn ) is a solution of the system a21 x1 + a22 x2 + · · · + a2n xn = b2 . .........

To keep the notation simple, suppose we add k times the first equation to the second; then the second equation of the new system will be (a21 + ka11 )x1 + · · · + (a2n + ka1n )xn = b2 + kb1 . We have to verify that (c1 , c2 , . . . , cn ) is a solution of this new equation. Indeed, (a21 + ka11 )c1 + · · · + (a2n + ka1n )cn = a21 c1 + · · · + a2n cn + k(a11 c1 + · · · + a1n cn ) = b2 + kb1 .

We have shown that any solution of the “old” system is also a solution of the “new.” To see that, conversely, any solution of the new system is also a solution of the old system, note that elementary row operations are reversible (compare with Exercise 24); we can obtain the old system by subtracting k times the first equation from the second equation of the new system. 1.2.29 Since the number of oxygen atoms remains constant, we must have 2a + b = 2c + 3d. 2a + b = 2c + Considering hydrogen and nitrogen as well, we obtain the system 2b = c + a = c + a − 2d = 0 2a + b − 2c − 3d = 0 b − d = 0 . 2b − c − d = 0 , which reduces to c − d = 0 a − c − d = 0 16

3d d or d

Section 1.2 2t a b t The solutions are = . t c t d

To get the smallest positive integers, we set t = 1: 2N O2 + H2 O −→ HN O2 + HN O3 1.2.30 a a a a

Plugging the points into f (t), we obtain the system = 1 + b + c + d = 0 − b + c − d = 0 + 2b + 4c + 8d = −15

with unique solution a = 1, b = 2, c = −1, and d = −2, so that f (t) = 1 + 2t − t2 − 2t3 . (See Figure 1.8.)

Figure 1.8: for Problem 1.2.30. 1.2.31 a a a a a

Let f (t) = a + bt + ct2 + dt3 + et4 . Substituting the points in, we get + b + c + d + e = 1 + 2b + 4c + 8d + 16e = −1 + 3b + 9c + 27d + 81e = −59 − b + c − d + e = 5 − 2b + 4c − 8d + 16e = −29

This system has the unique solution a = 1, b = −5, c = 4, d = 3, and e = −2, so that f (t) = 1−5t+4t2 +3t3 −2t4 . (See Figure 1.9.)

Figure 1.9: for Problem 1.2.31.

17

Chapter 1 ′ ′′ 1.2.32 The requirement fi′ (ai ) = fi+1 (ai ) and fi′′ (ai ) = fi+1 (ai ) ensure that at each junction two different cubics fit “into” one another in a “smooth” way, since they must have the same slope and be equally curved. The requirement that f1′ (a0 ) = fn′ (an ) = 0 ensures that the track is horizontal at the beginning and at the end. How many unknowns are there? There are n pieces to be fit, and each one is a cubic of the form f (t) = p+qt+rt2 +st3 , with p, q, r, and s to be determined; therefore, there are 4n unknowns. How many equations are there?

fi (ai ) = bi fi (ai−1 ) = bi−1 ′ (ai ) fi′ (ai ) = fi+1 ′′ fi′′ (ai ) = fi+1 (ai ) f1′ (a0 ) = 0, fn′ (an ) = 0

for for for for

i = 1, 2, . . . , n i = 1, 2, . . . , n i = 1, 2, . . . , n − 1 i = 1, 2, . . . , n − 1

gives gives gives gives gives

n equations n equations n − 1 equations n − 1 equations 2 equations

Altogether, we have 4n equations; convince yourself that all these equations are linear. 1.2.33 Let f (t) = a + bt + ct2 + dt3 , so that f ′ (t) = b + 2ct + 3dt2 . Substituting a + b a + 2b b b

the + + + +

given points into f (t) and f ′ (t) we obtain the system c + d = 1 4c + 8d = 5 2c + 3d = 2 4c + 12d = 9

This system has the unique solution a = −5, b = 13, c = −10, and d = 3, so that f (t) = −5 + 13t − 10t2 + 3t3 . (See Figure 1.10.)

Figure 1.10: for Problem 1.2.33. x x 1 1.2.34 We want all vectors y in R3 such that y · 3 = x + 3y − z = 0. The endpoints of these vectors z z −1 form a plane. x −3r + t , where r and t are arbitrary real numbers. These vectors are of the form y = r z t

x1 + x2 + x3 + x4 1.2.35 We need to solve the system x1 + 2x2 + 3x3 + 4x4 x1 + 9x2 + 9x3 +7x4 x1 + 0.25x4 = 0 which reduces to x2 − 1.5x4 = 0 . x3 + 2.25x4 = 0 18

=0 = 0 , =0

Section 1.2 −0.25t x1 x 1.5t The solutions are of the form 2 = , where t is an arbitrary real number. −2.25t x3 t x4

1.2.36 Writing the equation ~b = x1~v1 + x2~v2 + x3~v3 in terms of its components, we obtain the system x1 + 2x2 + 4x3 = −8 4x1 + 5x2 + 6x3 = −1 7x1 + 8x2 + 9x3 = 9 5x1 + 3x2 + x3 = 15 The system has the unique solution x1 = 2, x2 = 3, and x3 = −4.

1.2.37 Compare with the solution of Exercise 1.1.21. x1 − 0.2x2 − 0.3x3 x1 = 0.2x2 + 0.3x3 + 320 The diagram tells us that x2 = 0.1x1 + 0.4x3 + 90 or −0.1x1 + x2 − 0.4x3 −0.2x1 − 0.5x2 + x3 x3 = 0.2x1 + 0.5x2 + 150 This system has the unique solution x1 = 500, x2 = 300, and x3 = 400.

= 320 = 90 . = 150

320 0.3 0.2 0 1.2.38 a ~v1 = 0.1 , ~v2 = 0 , ~v3 = 0.4 , ~b = 90 150 0 0.5 0.2

b Recall that xj is the output of industry Ij , and the ith component aij of ~vj is the demand of Industry Ij on industry Ij for each dollar of output of industry Ij . Therefore, the product xj aij (that is, the ith component of xj ~vj ), represents the total demand of industry Ij on Industry Ii (in dollars). c x1~v1 + · · · + xn~vn + ~b is the vector whose ith component represents the total demand on industry Ii (consumer demand and interindustry demand combined). d The ith component of the equation x1~v1 + · · · + xn~vn + ~b = ~x expresses the requirement that the output xi of industry Ii equal the total demand on that industry. 1.2.39 a These components are zero because neither manufacturing not the energy sector directly require agricultural products. b We have to solve the system x1~v1 + x2~v2 + x3~v3 + ~b = ~x or 0.707x1 = 13.2 −0.014x1 + 0.793x2 − 0.017x3 = 17.6 −0.044x1 + 0.01x2 + 0.784x3 = 1.8

The unique solution is approximately x1 = 18.67, x2 = 22.60, and x3 = 3.63.

1.2.40 We want to find m1 , m2 , m3 such that m1 + m2 + m3 = 1 and 19

Chapter 1 2 4 2 1 , that is, we have to solve the system = + m3 + m2 2 1 3 2 m1 + m2 + m3 =1 m1 + 2m2 + 4m3 = 2 . 2m1 + 3m2 + m3 = 2 1 1

m1

The unique solution is m1 = 12 , m2 = 14 , and m3 = 14 . 1 We will put 12 kg at the point and 41 kg at each of the two other vertices. 2 1.2.41 We know that −3m1 + 2m2 or −6m1 + 4m2 −3m1 + 2m2

m1~v1+ m2~v2 = m1 w ~ 1 + m2 w ~ 2 or m1 (~v1 − w ~ 1 ) + m2 (~v2 − w ~ 2 ) = ~0 =0 = 0 . =0

We can conclude that m1 = 23 m2 .

1.2.42 Let x1 , x2 , x3 , and x4 be the traffic volume at the four locations indicated in Figure 1.11.

Figure 1.11: for Problem 1.2.42. We are told that x1 + 300 x2 + 300 x3 + x4 + 100 150 + 120

the number of cars coming into each intersection x1 − x2 = 320 + x2 x2 − x3 = 400 + x3 or = 250 x3 + x4 = x1 + x4 x1 + x4 270 − t x1 x2 250 − t The solutions are of the form = . 150 − t x3 t x4 20

is the same as the number of cars coming out: = 20 = 100 = 150 = 270

Section 1.2 Since the xi must be positive integers (or zero), t must be an integer with 0 ≤ t ≤ 150. The lowest possible values are x1 = 120, x2 = 100, x3 = 0, and x4 = 0, while the highest possible values are x1 = 270, x2 = 250, x3 = 150, and x4 = 150. 1.2.43 Plugging the data into the function S(t) we obtain the system + c sin 2π·47 = 11.5 a + b cos 2π·47 365 365 a + b cos 2π·74 + c sin 2π·74 = 12 365 365 2π·273 a + b cos 2π·273 + c sin = 12 365 365

The unique solution is approximately a = 12.17, b = −1.15, and c = 0.18, so that 2πt 2πt + 0.18 sin 365 . S(t) = 12.17 − 1.15 cos 365

The longest day is about 13.3 hours. (See Figure 1.12.)

Figure 1.12: for Problem 1.2.43. x1 +x2 +x3 = 24 . 3x1 +2x2 + 21 x3 = 24 x1 −1.5x3 = −24 This system reduces to . x2 +2.5x3 = 48 1.5x3 − 24 x1 Thus, our solutions will be of the form x2 = −2.5x3 + 48 . Since all of our values must be non-negative x3 x3 lilies 0 3 integers (and x3 must be even), we find the following solutions for roses : 8 and 3 . Since Olivia daisies 16 18 loves lilies, Kyle spends his 24 dollars on 3 lilies, 3 roses and 18 daisies.

1.2.44 Kyle first must solve the following system:

1.2.45 a When k 6= 1 and k 6= 2, we can see that this will continue to reduce to a consistent system with a unique solution. b When k = 1, our bottom row reveals the inconsistency 0 = 2. c When k = 2, the second row and third row both represent the equation z = 2, meaning that the third row will be replaced with the equation 0 = 0 during further reduction. This reveals that we will have an infinite number of solutions. 21

Chapter 1

0 1.2.46 a We reduce our matrix in the following steps: 1 k

1 0

2

6

1

2k

k

0

2

1 0

0

0

1

.. . 2 1 .. → . 0 0 .. −k(I) 0 . 1 6 − 4k 2k

0 2 − 6k + 4k

2

.. . .. . .. .

2 0 1 − 2k

2

6

1

2k

−2k

2 − 6k

1 0 → 0 1

0 0

.. . .. . .. .

.. . 0 .. swap : . 2 I ↔ II → .. . 1

1 2k 2

6

0

2

2

0 1 − 2k

−2(II)

→

+2k(II)

6 − 4k 2k

2(2k − 1)(k − 1)

.. . 2 .. . . 0 .. . −(2k − 1)

We see that there will be a unique solution when the 2(2k − 1)(k − 1) term is not equal to zero, when 2k − 1 6= 0 and k − 1 6= 0, or k 6= 21 and k 6= 1. b We will have no solutions when the term 2(2k − 1)(k − 1) is equal to zero, but the term −(2k − 1) is not. This occurs only when k = 1. c We will have infinitely many solutions when the last row represents the equation 0 = 0. This occurs when 2k − 1 = 0, or k = 21 . 1.2.47 a So − 21 x1 + x2 − 21 x3 = 0 and − 21 x2 + x3 − 21 x4 = 0. After reduction of the system, we find that our solutions are all of the form −2 3 x1 −1 2 x2 = s + t . x3 0 1 x4 1 0

b Yes, from our solution in part (a), if we plug in 1 for x1 and 13 for x4 , we obtain 3t − 2s = 1 and s = 13, which leads to t = 9, and x2 = 5, x3 = 9. So we have the solution: x1 = 1, x2 = 5, x3 = 9 and x4 = 13, which is an arithmetic progression. 1.2.48 It is required that xk = 12 (xk−1 + xk+1 ), or 2xk = xk−1 + xk+1 , or xk − xk−1 = xk+1 − xk . This means that the difference of any two consecutive terms must be the same; we are looking at the finite arithmetic sequences. Thus the solutions are of the form (x1 , x2 , x3 , . . . , xn ) = (t, t + r, t + 2r, . . . , t + (n − 1)r), where t and r are arbitrary constants.

2 1 1.2.49 We begin by solving the system. Our augmented matrix begins as: 0 3 1 0

22

.. 0. C . 1.. C . 4.. C

Section 1.2 . 9 C 0 0.. 25 .. 7 . In order for x, y and z to be integers, C must be a multiple of 25. We 1 0. 25 C .. 4 0 0 1. 25 C want the smallest positive choice, so C = 25.

1 and is reduced to 0

1.2.50 f (t) = a + bt + ct2 + dt3 and we learn that f (0) = a = 3, f (1) = a + b + c + d = 2, f (2) = a + 2b + 4c + 8d = 0. Also, Z 2 1 1 8 1 f (t)dt = at + bt2 + ct3 + dt4 |20 = 2a + 2b + c + 4d = 4. 2 3 4 3 0

1

0

0

0

1 Now, we set up our matrix, 1

1

1

1

2

4

8

8 3

2 2 4 meaning that the system is inconsistent.

.. . .. . .. . .. .

3

2 . However, when we reduce this, the last line becomes 0 = 1, 0 4

In introductory calculus you may have seen the approximation formula: Z

b

a

f (t)dt ≈

b−a a+b (f (a) + 4f ( ) + f (b)), 6 2

the simplest form of Simpson’s Rule. For polynomials f (t) of degree ≤ 3, Simpson’s Rule gives the exact value of the integral. Thus, for the f (t) in our problem, Z

2

f (t)dt =

0

2 1 11 (f (0) + 4f (1) + f (2)) = (3 + 8 + 0) = . 6 3 3

Thus it is impossible to find such a cubic with Z

2

f (t)dt = 4,

0

as required. 1.2.51 The system of linear equations is c1 c1 + c2 + c4

= =

0 0

c1 + 2c2 + 4c4 c1 + c3 + c6 c1 + 2c3 + 4c6

= = =

0 0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, 0, 0, c5 , 0). This is the conic xy = 0. 23

Chapter 1 y 3

2

1

x -1

1

2

3

-1

1.2.52 The system of linear equations is c1 c1 + 2c2 + 4c4

= =

0 0

c1 + 2c3 + 4c6 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 c1 + c2 + 3c3 + c4 + 3c5 + 9c6

= = =

0 0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, −6c6 , −2c6 , 3c6 , 0, c6 ). This√is the conic −6x − 2y + 3x2 + y 2 = 0 or 3(x − 1)2 + (y − 1)2 = 4, the ellipse centered at (1, 1) with semiaxis 2/ 3 and 2. y 3

2

1

x -1

1

2

3

-1

1.2.53 The system of linear equations is c1

=

0

c1 + c2 + c4 c1 + 2c2 + 4c4 c1 + 3c2 + 9c4

= = =

0 0 0

c1 + c2 + c3 + c4 + c5 + c6

=

0

This system has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, −c5 − c6 , 0, c5 , c6 ). Setting c5 = a and c6 = b, we get the family of conics a(xy − y) + b(y 2 − y) = ay(x − 1) + by(y − 1) = 0 where a 6= 0 or b 6= 0. Each such conic is the union of the x axis with some line through the point (1, 1). Two sample solutions are shown in the accompanying figures. 24

Section 1.2 y

y 3

3

2

2

1

1

x

x -1

1

2

-1

3

1

2

3

-1

-1

1.2.54 The system of linear equations is c1

=

0

c1 + c2 + c3 + c4 + c5 + c6 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6

= =

0 0

c1 + 3c2 + 3c3 + 9c4 + 9c5 + 9c6 c1 + c2 + c4

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, c5 + c6 , −c5 − c6 , −c5 − c6 , c5 , c6 ). Setting c5 = a and c6 = b, we get the family of conics a(x − y + xy − x2 ) + b(x − y − x2 + y 2 ) = a(x − y)(1 − x) + b(x − y)(1 − x − y) = 0 where a 6= 0 or b 6= 0. Each such conic is the union of the line y = x with some line through the point (1, 0). Two sample solutions are shown in the accompanying figures. y

y

3

3

2

2

1

1

x -1

1

2

3

x -1

-1

1

2

3

-1

1.2.55 The system of linear equations is c1

=

0

c1 + c2 + c4 c1 + c3 + c6

= =

0 0

c1 + c2 + c3 + c4 + c5 + c6

=

0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, −c4 , −c6 , c4 , 0, c6 ). Setting c4 = a and c6 = b, we get the family of conics a(x2 − x) + b(y 2 − y) = ax(x − 1) + by(y − 1) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. 25

Chapter 1 y

y

1

1

x

x

1

1

1.2.56 The system of linear equations is c1 c1 + c2 + c4

= =

0 0

c1 + c3 + c6 c1 + c2 − c3 + c4 − c5 + c6

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, −c4 , −c6 , c4 , 2c6 , c6 ). Setting c4 = a and c6 = b, we get the family of conics a(x2 − x) + b(y 2 + 2xy − y) = ax(x − 1) + by(y + 2x − 1) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. y

y

2

2

1

1

x 1

-1

x

3

2

1

-1

-1

-1

-2

-2

2

3

1.2.57 The system of linear equations is c1 + 5c2 + 25c4 c1 + c2 + 2c3 + c4 + 2c5 + 4c6

= =

0 0

c1 + 2c2 + c3 + 4c4 + 2c5 + c6 c1 + 8c2 + c3 + 64c4 + 8c5 + c6

= =

0 0

c1 + 2c2 + 9c3 + 4c4 + 18c5 + 81c6

=

0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (25c6 , −10c6 , −10c6 , c6 , 0, c6 ). This is the conic 25−10x−10y +x2 +y 2 = 0, a circle of radius 5 centered at (5, 5). y 12

8 (5, 5)

4

x -4

4 -4

26

8

12

Section 1.2 1.2.58 The system of linear equations is c1 + c2 + c4

=

0

c1 + 2c2 + 4c4 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6

= =

0 0

c1 + 5c2 + 2c3 + 25c4 + 10c5 + 4c6 c1 + 5c2 + 6c3 + 25c4 + 30c5 + 36c6

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (2c6 , −3c6 , 2c6 , c6 , −2c6 , c6 ). This is the conic 2−3x+2y+x2 −2xy+y 2 = 0. y 6

4

2

x -2

2

4

6

-2

1.2.59 The system of linear equations is c1 c1 + c2 + c4

= =

0 0

c1 + 2c2 + 4c4 c1 + c3 + c6

= =

0 0

c1 + 2c3 + 4c6 c1 + c2 + c3 + c4 + c5 + c6

= =

0 0

It has the only solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, 0, 0, 0, 0). There is no conic which goes through these points. Alternatively, note that the only conic through the first five points is xy = 0, according to Exercise 51. But that conic fails to run through the point (1, 1), so that there is no conic through all six points. y 3

2

1

x -1

1

2

3

-1

1.2.60 The system of linear equations is c1 27

=

0

Chapter 1 c1 + 2c2 + 4c4

=

0

c1 + 2c3 + 4c6 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6

= =

0 0

c1 + c2 + 3c3 + c4 + 3c5 + 9c6 c1 + 4c2 + c3 + 16c4 + 4c5 + c6

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, 0, 0, 0, 0). There is no conic which goes through these points. y 4 3 2 1

x -1

1

2

3

4

-1

1.2.61 Let x1 be the cost of the environmental statistics book, x2 be the cost of the set theory textand x3 be the cost .. 1 1 0.. 178 of the educational psychology book. Then, from the problem, we deduce the augmented matrix 2 1 1.. 319 . . 0 1 1.. 147 .. 1 0 0.. 86 We can reduce this matrix to 0 1 0.. 92 , revealing that x1 = 86, x2 = 92 and x3 = 55. Thus, the . 0 0 1.. 55 environmental statistics book costs $ 86, the set theory book costs $ 92 and the educational psychology book is only priced at $ 55. grammar x1 1.2.62 Let our vectors x2 represent the numbers of the books W erther . Then we can set up the matrix LinearAlg. x3 .. 43 1 1 0 .. 64 . This system yields one solution, 21 , meaning that the grammar book costs 43 Euro, the . 1 0 1 . 98 . 55 0 1 1 .. 76 novel costs 21 Euro, and the linear algebra text costs 55 Euro.

1.2.63 The difficult part of this problem lies in setting up a system from which we can derive our matrix. We will define x1 to be the number of “liberal” students at the beginning of the class, and x2 to be the number of “conservative” students at the beginning. Thus, since there are 260 students in total, x1 + x2 = 260. We need one more equation involving x1 and x2 in order to set up a useful system. Since we know that the number of conservative students at the end of the semester is equal to the number of liberal student initially, we obtain the 6 7 6 3 x1 + 10 x2 = x1 , or − 10 x1 + 10 x2 = 0. equation 10 " # " # .. .. 1 1 . 260 to obtain 1 0. 120 . We then use . . 7 6 . − . 0 0 1.. 140 10

10

28

Section 1.2 Thus, there are initially 120 liberal students, and 140 conservative students. Since the number of liberal students initially is the same as the number of conservative students in the end, the class ends with 120 conservative students and 140 liberal students. 1.2.64 Let x1 and x2 be the initial number of students in Sections A and B, respectively. Then, since there are 55 students total, x1 + x2 = 55. Also, interpreting the change of students from the perspective of Section B, we gain 0.2x1 , lose 0.3x lose 4 students. Thus, 0.2x1 − 0.3x2 = −4.0. Our matrix becomes # 2 , and in the process, " " # .. .. . 55 , which reduces to 1 0 . 25 . This reveals that there are initially 25 students in Section 1 1 . . .2 −.3 .. −4 0 1 .. 30 A and 30 students in Section B. 1.2.65 We are told that five cows and two sheep cost ten liang, and two cows and five sheep silver. cost eight liang of 5C +2S = 10 . So, we let C be the cost of a cow, and S be the cost of a sheep. From this we derive 2C +5S = 8 C = 34 34 21 which gives the prices: 21 liang silver for a cow, and 20 This reduces to 21 liang silver for a sheep. S = 20 21 1.2.66 Letting x 1 , x2 , and x3 be the prize, incoins, of cows, sheep and pigs, respectively, we can represent the system .. .. 2 5 −13 . 1000 1 0 0 .. 1200 .. . We reduce this matrix to in a matrix: 3 −9 . 0 3 0 1 0 .. 500 . The prize of a cow, . .. 0 0 1 .. 300 . −600 −5 6 8 a sheep, and a pig is 1200, 500 and 300 coins, respectively. 1.2.67 The second measurement in the problem tells us that 4 sparrows and 1 swallow weigh as much as 1 sparrow and 5 swallows. We will immediately interpret this as 3 sparrows weighing the same as 4 swallows. The other measurement we use is that all the birds together weigh 16 liang. Setting x1 to be the weight of a sparrow, and . 3 −4.. 0 representing these two equations. x2 to be the weight of a swallow, we find the augmented matrix . 5 6.. 16 .. 32 . 1 0 19 , meaning that each sparrow weighs 32 liang, and each swallow weighs 24 liang. We reduce this to 19 19 .. 24 0 1. 19 1.2.68 This problem gives us three different combinations of horses that can pull exactly 40 dan up a hill. We condense the statements to fit our needs, saying that, One military horse and one ordinary horse can pull 40 dan, two ordinary and one weak horse can pull 40 dan and one military and three weak horses can also pull 40 dan. .. 40 .. 7 1 0 0 . 1 1 0 .. 40 With this information, we set up our matrix: 0 2 1 .. 40 , which reduces to 0 1 0 ... 120 . 7 . . 1 0 3 .. 40 0 0 1 .. 40 7

Thus, the military horses can pull dan each.

40 7

dan, the ordinary horses can pull

1.2.69 Here, let W be the depth of the well. 29

120 7

dan and the weak horses can pull

40 7

Chapter 1

2A

Then our system becomes

+B 3B

+C 4C

+D 5D

A

+E +6E

−W −W −W −W −W

= 0 = 0 = 0 . = 0 = 0

We 1 0 0 0

transform this system into an augmented matrix, then perform a prolonged reduction to reveal . 265 . . 0 0 0 0 0 − 721 . 191 . 1 0 0 0 − 721 . 0 . 191 129 76 265 148 . W , B = 721 W , C = 148 . Thus, A = 721 0 1 0 0 − 721 . 0 721 W , D = 721 W and E = 721 W . . 129 . . 0 0 0 1 0 − 721 . 76 . 0 0 0 0 1 − 721 . 0

If we choose 721 to be the depth of the well, then A = 265, B = 191, C = 148, D = 129 and E = 76. 1.2.70 We let x1 , x2 and x3 be the numbers of roosters, hens and chicks respectively. Then, since we buy a total of a hundred birds, and spend a hundred coins on them, we find the equations x1 + x2 + x3 = 100 and 5x1 + 3x2 + 13 x3 = 100. We fit these into our matrix,

which reduces to

4 1 0 −3

0 1

7 3

"

1 1

1

5 3

1 3

.. . .. .

−100

# .. . 100 , .. . 100 .

200

x1 So, x1 − 43 x3 = −100, and x2 + 37 x3 = 200. Now, we can write our solution vectors in terms of x3 : x2 = x3 4 x − 100 3 3 − 7 x3 + 200 . Since all of our values must be non-negative, x1 must be greater than or equal to zero, or 3 x3 4 x − 100 ≥ 0, which means that x3 ≥ 75. 3 3

Also, x3 must be greater than or equal to zero, meaning that − 37 x3 + 200 ≥ 0 or x3 ≤ 600 7 . Since x3 must be an integer, this forces x3 ≤ 85. Thus, we are looking for solutions where 75 ≤ x3 ≤ 85. We notice, however, that x1 and x2 are only integers when x3 is a multiple of 3. Thus, the possible values for x3 are 75, 78, 81 and 84. roosters 0 4 8 12 Now the possible solutions for hens are 25 , 18 , 11 , and 4 . chicks 75 78 81 84 1.2.71 We let x1 , x2 , x3 and x4 be the numbers of pigeons, sarasabirds, swans and peacocks respectively. We first determine the cost of each bird. Each pigeon costs 35 panas, each sarasabird costs 57 panas, the swans cost 79 panas apiece and each peacock costs 3 panas. We use these numbers to set up our system, but we must remember to make sure we are buying the proper amount of each to qualify for these deals when we find our solutions (for example, the number of sarasabirds we buy must be a multiple of 7). 30

Section 1.2

Our matrix then is

which reduces to

"

1

1

1

3 5

5 7

7 9

1 0 0 1

− 95 14 9

# . 1.. 100 . 3.. 100 . −20.. . 21..

−250 . 350

Thus, x1 = 95 x3 + 20x4 − 250 and x2 = − 14 9 x3 − 21x4 + 350. Then our solutions are of the form

5 9 x3 + 20x4 − 250 14 − 9 x3 − 21x4 + 350 . x3

x4

We determine the possible solutions by choosing combinations of x3 and x4 of the correct multiples (9 for x3 , 3 for x4 ) that give us non-negative integer solutions for x1 and x2 . Thus it is required that x1 = 59 x3 + 20x4 − 250 ≥ 0 and x2 = − 14 9 x3 − 21x4 + 350 ≥ 0. Solving for x3 we find that 225 −

27 2 x4

≥ x3 ≥ 450 − 36x4 .

To find all the solutions, we can begin by letting x4 = 0, and finding all corresponding values of x3 . Then we can increase x4 in increments of 3, and find the corresponding x3 values in each case, until we are through. For x4 = 0 we have the inequality 225 ≥ x3 ≥ 450, so that there aren’t any solutions for x3 . Likewise, there are no feasible x3 values for x4 = 3, 6 and 9, since 450 − 36x4 exceeds 100 in these cases. In the case of x4 = 12 our inequality becomes 63 ≥ x3 ≥ 18, so that x3 could be 18, 27, 36, 45, 54 or 63. In the next case, x4 = 15, we have

45 2

≥ x3 ≥ −90, so that the non-negative solutions are 0, 9 and 18.

If x4 is 18 or more, then the term 225 − 27 2 x4 becomes negative, so that there are only negative solutions for x3 . (Recall that it is required that 225 − 27 2 x4 ≥ x3 .) We have found nine solutions. If we compute the corresponding values of number of pigeons number of sarasabirds x1 = 95 x3 +20x4 −250 and x2 = − 14 9 x3 −21x4 +350, we end up with the following vectors for: number of swans number of peacocks 0 5 10 15 20 25 50 55 60 70 56 42 28 14 0 35 21 7 to be: , , , , , , , , . 18 27 36 45 54 63 0 9 18 12 12 12 12 12 12 15 15 15

"

1.2.72 We follow the outline of Exercise 70 to find the matrix 1 4 400−4x3 19

1

1

1 5

1

# .. . 100 , which reduces to 1 0 .. . 100 0 1

4 19 15 19

.. . .. .

3 . We find that our solutions are bound by 0 ≤ x3 ≤ 100. However, Thus, our solutions are of the form 1500−15x 19 x3 3 3 3 3 3 since both 400−4x = 4 100−x and 1500−15x = 15 100−x must be non-negative integers, the quantity 100−x must 19 19 19 19 19 be a non-negative integer, k, so that x3 = 100 − 19k. The condition x3 ≥ 0 now leaves us with the possibilities k = 0, 1, 2, 3, 4, 5.

31

400 19 1500 19

.

Chapter 1 20 16 12 8 4 0 ducks Thus, we find our solutions for sparrows : 0 . 15 , 30 , 45 , 60 and 75 . 5 24 43 62 81 100 roosters

1.2.73 We let x1 be the number of sheep, x2 be the number of goats, and x3 be the number of hogs. We can then 1 4 7 use the two equations 2 x1 + 3 x2 + 2 x3 = 100 and x1 + x2 + x3 = 100 to generate the following augmented matrix: . 4 7. 1 2 3 2 . 100 . 1 1 1.. 100 then reduce it to

1 0

.. 0 − 13 5 . . 18 . 1 5 .

40 . 60

40 + 13 5 s With this, we see that our solutions will be of the form 60 − 18 5 s . Now all three components of this vector s must be non-negative integers, meaning that s must be a non-negative multiple of 5 (that is, s = 0, 5, 10, . . .) such 50 that 60 − 18 5 s ≥ 0, or, s ≤ 3 . This leaves the possible solutions x3 = s = 0, 5, 10 and 15, and we can compute 18 the corresponding values of x1 = 40 + 13 5 s and x2 = 60 − 5 s in each case. 40 53 66 79 So we find the following solutions: 60 , 42 , 24 and 6 . 0 5 10 15

1.2.74 This problem is similar in nature to Exercise that example, revealing the ma 70, and we will follow # " .. .. 3 1 1 1 . 100 . We reduce this to 1 0 − 2 . −100 , which yields solutions of the form trix: . .. 5 3 2 21 .. 100 0 1 . 200 2 3 x − 100 3 2 − 5 x3 + 200 . Since all the values must be positive (there are at least one man, one woman and one child), 2 x3 we see 80, and beeven. From this, we use x3 to find our solutions: x3 must x3< < 66 that 17 14 11 8 5 2 30 , 25 , 20 , 15 , 10 and 5 . 78 76 74 72 70 68 1.2.75 Rather than setting up a huge system, here we will reason this out logically. Since there are 30 barrels, each son will get 10 of them. If we use the content of a full barrel as our unit for wine, we see that each brother will get 15 3 = 5 barrel-fulls of wine. Thus, the ten barrels received by each son will, on average, be half full, meaning that for every full barrel a son receives, he also receives an empty one. Now let x1 , x2 , and x3 be the numbers of half-full barrels received by each of the three sons. The first son, receiving x1 half-full barrels will also gain 10 − x1 other barrels, half of which must be full and half of which 1 must be empty, each equal to the quantity 10−x . Thus, x1 must be even. The same works for x2 and x3 . Since 2 x1 + x2 + x3 = 10, we have boiled down our problem to simply finding lists of three non-negative even numbers that add up to 10. We find our solutions by inspection: 4 4 4 4 6 6 6 8 8 10 0 ,2,0,4,2,0,6,4,2,0, 6 4 2 0 4 2 0 2 0 0 32

Section 1.2 0 0 0 0 0 0 2 2 2 2 2 8 , 6 , 4 , 2 , 0 , 10 , 8 , 6 , 4 , 2 and 0 . 10 8 6 4 2 0 8 6 4 2 0

As we stated before, the number of full and empty barrels is dependent on the number of half-full barrels. Thus, each solutionhere translates into exactly one solution for the overall problem. Here we list those solutions, for first son second son , using triples of the form (full barrels, half-full barrels, empty barrels) as our entries: third son (2, 6, 2) (2, 6, 2) (2, 6, 2) (1, 8, 1) (1, 8, 1) (0, 10, 0) (5, 0, 5) , (4, 2, 4) , (5, 0, 5) , (3, 4, 3) , (4, 2, 4) , (5, 0, 5) , (3, 4, 3) (4, 2, 4) (5, 0, 5) (4, 2, 4) (5, 0, 5) (5, 0, 5) (3, 4, 3) (3, 4, 3) (3, 4, 3) (3, 4, 3) (4, 2, 4) (4, 2, 4) (2, 6, 2) , (3, 4, 3) , (4, 2, 4) , (5, 0, 5) , (1, 8, 1) , (2, 6, 2) , (5, 0, 5) (4, 2, 4) (3, 4, 3) (2, 6, 2) (5, 0, 5) (4, 2, 4) (5, 0, 5) (5, 0, 5) (5, 0, 5) (4, 2, 4) (4, 2, 4) (4, 2, 4) (3, 4, 3) , (4, 2, 4) , (5, 0, 5) , (0, 10, 0) , (1, 8, 1) , (2, 6, 2) , (3, 4, 3) (4, 2, 4) (5, 0, 5) (1, 8, 1) (2, 6, 2) (3, 4, 3) (5, 0, 5) (5, 0, 5) (5, 0, 5) (3, 4, 3) , (4, 2, 4) and (5, 0, 5) . (2, 6, 2) (1, 8, 1) (0, 10, 0) 1.2.76 We let x1 be the amount of gold in the crown, x2 be the amount of bronze, x3 be the amount of tin and x4 be the amount of iron. Then, for example, since the first requirement in the problem is: “Let the gold and bronze together form two-thirds,” we will interpret this as x1 + x2 = 32 (60). We do this for all three requirements, and use the fact. that allcombined will be the total weight of the crown as our fourth. So we find the matrix 1 1 0 0 .. 23 (60) 30.5 gold . 1 0 1 0 .. 3 (60) bronze 9.5 4 , which has the solution . = . 14.5 tin 1 0 0 1 .. 3 (60) 5 5.5 iron .. 1 1 1 1 . 60 1.2.77 Let xi be the number of coins the ith merchant has. We interpret the statement of the first merchant, “If I keep the purse, I shall have twice as much money as the two of you together” as x1 + 60 = 2(x2 + x3 ), or −x1 + 2x2 + 2x3 = 60. We interpret the other statements in a similar fashion, translating this into the augmented .. −1 2 2. 60 . matrix, 3.. 60 . 3 −1 . 5 5 −1.. 60 . 0 0.. 4 . 1 0.. 12 . Thus we deduce that the first merchant has 4 . 0 0 1.. 20 coins, the second has 12, and the third is the richest, with 20 coins.

1 The reduced row echelon form of this matrix is 0

33

Chapter 1 1.2.78 For each of the three statements, we set up an equation of the form (initial amount of grass) + (grass growth) = (grass consumed by cows), or (#of f ields)x + (#of f ields)(#of days)y = (#of cows)(#of days)z. For the first statement, this produces the equation x + 2y = 6z, or x + 2y − 6z = 0. Similarly, we obtain the equations 4x + 16y − 28z = 0 and 2x + 10y − 15z = 0 for the other two statements. From this information, we . . 1 0 −5 .. 0 1 2 −6 .. 0 .. . 1 write the matrix 4 16 −28 .. 0 , which reduces to 0 1 − 2 . 0 . Thus our solutions are of the .. . 2 10 −15 . 0 0 0 0 .. 0 x 5t form y = 21 t , where t is an arbitrary positive real number. t z

Section 1.3 1.3.1 a No solution, since the last row indicates 0 = 1. b The unique solution is x = 5, y = 6. c Infinitely many solutions; the first variable can be chosen freely. 1.3.2 The rank is 3 since each row contains a leading one.

1 1 1 1.3.3 This matrix has rank 1 since its rref is 0 0 0 . 0 0 0

1 0 −1 1.3.4 This matrix has rank 2 since its rref is 0 1 2 0 0 0 1 2 7 1.3.5 a x +y = 3 1 11 b The solution of the system in part (a) is x = 3, y = 2. (See Figure 1.13.) 1.3.6 No solution, since any linear combination x~v1 + y~v2 of ~v1 and ~v2 will be parallel to ~v1 and ~v2 . 1.3.7 A unique solution, since there is only one parallelogram with sides along ~v1 and ~v2 and one vertex at the tip of ~v3 . 1.3.8 Infinitely many solution. There are at least two obvious solutions. Write ~v4 as a linear combination of ~v1 and ~v2 alone or as a linear combination of ~v3 and ~v2 alone. Therefore, this linear system has infinitely many solutions, by Theorem 1.3.1. 34

Section 1.3

Figure 1.13: for Problem 1.3.5.

1 1.3.9 4 7

2 3 1 x 5 6y = 4 8 9 9 z

1 1 1.3.10 2 · −2 = 1 · 1 + 2 · (−2) + 3 · 1 = 0 3 1 1.3.11 Undefined since the two vectors do not have the same number of components. 5 6 1.3.12 [1 2 3 4] · = 1 · 5 + 2 · 6 + 3 · 7 + 4 · 8 = 70 7 8 1.3.13

1 3

2 4

7 1 2 29 1 2 7 1 · 7 + 2 · 11 29 =7 + 11 = or = = 11 3 4 65 3 4 11 3 · 7 + 4 · 11 65

−1 2 3 1 2 3 6 1 1.3.14 2 = −1 +2 +1 = or 3 4 2 3 4 8 2 1 1 · (−1) + 2 · 2 + 3 · 1 6 = 2 · (−1) + 3 · 2 + 4 · 1 8

1 2

5 6 1.3.15 [1 2 3 4] = 5 · 1 + 6 · 2 + 7 · 3 + 4 · 8 = 70 either way. 7 8 1.3.16

0 3

1 2

−3 0 · 2 + 1 · (−3) 2 = = 0 3 · 2 + 2 · (−3) −3 35

−1 2 3 2 = 3 4 1

Chapter 1 1.3.17 Undefined, since the matrix has three columns, but the vector has only two components. 5 2 1 1 2 1 = 1 3 + 2 4 = 11 1.3.18 3 4 2 17 6 5 5 6

0 −1 1 1 1 −1 1 2 = 1 −5 + 2 1 + 3 1 = 0 0 3 −5 1 3 3

1 1 1 1.3.19 −5 1 −5

9 1.3.20 a 7 6 b

8 6 6

9 −9 18 27 36 45

158 70 1.3.21 81 123

1 0 1.3.22 By Theorem 1.3.4, the rref is 0 1 0 0

0 0 1

1 0 1.3.23 All variables are leading, that is, there is a leading one in each column of the rref: 0 0

1 0 1.3.24 By Theorem 1.3.4, rref (A) = 0 0

0 1 0 0

0 0 1 0

0 1 0 0

0 0 . 1 0

0 0 . 0 1

1.3.25 In this case, rref(A) has a row of zeros, so that rank(A) < 4; there will be a free variable. The system A~x = ~c could have infinitely many solutions (for example, when ~c = ~0) or no solutions (for example, when ~c = ~b), but it cannot have a unique solution, by Theorem 1.3.4. 1 0 1.3.26 From Example 3d we know that rank(A) = 3, so that rref(A) = 0 0

0 1 0 0

0 0 . 1 0

Since all variables are leading, the system A~x = ~c cannot have infinitely many solutions, but it could have a unique solution (for example, if ~c = ~b) or no solutions at all (compare with Example 3c). 36

Section 1.3 1 0 1.3.27 By Theorem 1.3.4, rref (A) = 0 0

0 1 0 0

0 0 1 0

0 0 . 0 1

1 0 1.3.28 There must be a leading one in each column: rref (A) = 0 0 0

1.3.29 A is

a and 0 0

a 0 0 of the form 0 b 0 0 0 c 0 0 5 5a 2 b 0 3 = 3b = 0 . 0 c −9 −9c 1 2

0 1 0 0 0

0 0 1 . 0 0

5

So a = 25 , b = 0 and c = − 19 , and A = 0 0

0 0 0 0 0 − 19

b c 5 2 e f 3 = 0 . Thus, 5a + 3b − 9c = 2, 5d + 3e − 9f = 0, and h i −9 1 matrix to have rank 1 is to make all the entries 2 in the second and third 0 0 5 columns zero, meaning that a = 52 , d = 0, and g = 15 . Thus, one possible matrix is 0 0 0 . 1 0 0 5

a 1.3.30 We must satisfy the equation d g 5g + 3h − 9i = 1. One way to force our

1.3.31 A is

a and 0 0

a b c of the form 0 d e 0 0 f 5 5a + 3b − 9c 2 b c d e 3 = 3d − 9e = 0 . −9 −9f 1 0 f

Clearly, f must equal − 91 . Then, since 3d = 9e, we can choose any non-zero value for the free variable e, and d will be 3e. So, if we choose 1 for e, then d = 3e = 3. Lastly, we must resolve 5a + 3b − 9c = 2. Here, b and c are . If we let b = c = 1. Then, a = 2−3(1)+9(1) = 58 . the free variables, and a = 2−3b+9c 5 5 8 1 1 5 So, in our example, A = 0 3 1 0 0 − 91 1.3.32 For this problem, we set up the same three equations as in Exercise 30. However, here, we must enforce that −2 −2 −2 our matrix, A, contains no zero entries. One possible solution to this problem is the matrix 3 1 2 . −1 −1 −1 37

Chapter 1 x1 x2 ... 1.3.33 The ith component of A~x is [0 0 . . . 1 . . . 0] = xi . (The 1 is in the ith position.) xi ... xn

Therefore, A~x = ~x.

c b a 1.3.34 a A~e1 = d , A~e2 = e , and A~e3 = f . k h g 1 b B~e1 = [~v1 ~v2 ~v3 ] 0 = 1~v1 + 0~v2 + 0~v3 = ~v1 . 0 Likewise, B~e2 = ~v2 and B~e3 = ~v3 .

1.3.35 Write A = [~v1 ~v2 . . . ~vi . . . ~vm ], then 0 0 ... A~ei = [~v1 ~v2 . . . ~vi . . . ~vm ] = 0~v1 + 0~v2 + · · · + 1~vi + · · · + 0~vm = ~vi = ith column of A. 1 ... 0

1 1.3.36 By Exercise 35, the ith column of A is A~ei , for i = 1, 2, 3. Therefore, A = 2 3

4 7 5 8 . 6 9

2 x1 = 2 − 2x2 1.3.37 We have to solve the system or . x3 1 x3 =1 2 − 2t x1 Let x2 = t. Then the solutions are of the form x2 = t , where t is an arbitrary real number. x3 1 x1

+

2x2

= =

1.3.38 We will illustrate our reasoning with an example. We generate the “random” 3 × 3 matrix 0.141 0.592 0.653 A = 0.589 0.793 0.238 . 0.462 0.643 0.383

Since the entries of this matrix are chosen from a large pool of numbers (in our case 1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and even less likely that the whole first column will consist of zeros). means that we will usually be able to to turn the first apply the Gauss-Jordan column This algorithm 1 4.199 4.631 0.141 0.592 0.653 1 into 0 ; this is indeed possible in our example: 0.589 0.793 0.238 −→ 0 −1.680 −2.490 . 0 −1.297 −1.757 0.462 0.643 0.383 0 38

Section 1.3 Again, it is unlikely thatany entries in the second column of the new matrix will be zero. Therefore, we can turn 0 the second column into 1 . 0 1 0 0 Likewise, we will be able to clear up the third column, so that rref(A) = 0 1 0 . 0 0 1

We summarize:

As we apply Gauss-Jordan elimination to a random matrix A (of any size), it is unlikely that we will ever encounter a zero on the diagonal. Therefore, rref(A) is likely to have all ones along the diagonal.

0 a 0 b , where a, b, and c are arbitrary. 1 c

1 0 1.3.39 We will usually get rref(A) = 0 1 0 0 1 0 1.3.40 We will usually have rref(A) = 0 0

0 1 0 0

0 0 . 1 0

(Compare with the summary to Exercise 38.)

1 1.3.41 If A~x = ~b is a “random” system, then rref(A) will usually be 0 0 solution.

0 0 1 0 , so that we will have a unique 0 1

1.3.42 If A~x = ~b is a “random” system of three equations with four unknowns, then rref(A) will usually be 1 0 0 a 0 1 0 b (by Exercise 39), so that the system will have infinitely many solutions (x4 is a free variable). 0 0 1 c . 1.3.43 If A~x = ~b is a “random” system of equations with three unknowns, then rref[A..~b] will usually be . 1 0 0 ..0 0 1 0 ...0 . , so that the system is inconsistent. 0 0 1 ..0 . 0 0 0 ..1 1.3.44 Let E = rref(A), and note that all the entries in the last row of E must be zero, by the definition of rref. If ~c is any vector in Rn whose last component isn’t zero, then the system E~x = ~c will be inconsistent. Now consider the elementary row operations that in reversed i i transform A into E, and apply the opposite h operations, h order, to the augmented matrix E ... ~c . You end up with an augmented matrix A ... ~b that represents an inconsistent system A~x = ~b, as required. 39

Chapter 1

x1 kx1 1.3.45 Write A = [~v1 ~v2 . . . ~vm ] and ~x = . . . . Then A(k~x) = [~v1 . . . ~vm ] . . . = kx1~v1 + · · · + kxm~vm and xm kxm k(A~x) = k(x1~v1 + · · · + xm~vm ) = kx1~v1 + · · · + kxm~vm . The two results agree, as claimed. 1.3.46 Since a, d, and f are all nonzero, we can divide the first row by a, the second row by d, and the third row by f to obtain 1 ab ac 0 1 e . d 0 0 1 It follows that the rank of the matrix is 3.

1.3.47 a ~x = ~0 is a solution. b This holds by part (a) and Theorem 1.3.3. c If ~x1 and ~x2 are solutions, then A~x1 = ~0 and A~x2 = ~0. Therefore, A(~x1 + ~x2 ) = A~x1 + A~x2 = ~0 + ~0 = ~0, so that ~x1 + ~x2 is a solution as well. Note that we have used Theorem 1.3.10a. d A(k~x) = k(A~x) = k~0 = ~0 We have used Theorem 1.3.10b. 1.3.48 The fact that ~x1 is a solution of A~x = ~b means that A~x1 = ~b. a. A(~x1 + ~xh ) = A~x1 + A~xh = ~b + ~0 = ~b b. A(~x2 − ~x1 ) = A~x2 − A~x1 = ~b − ~b = ~0 c. Parts (a) and (b) show that the solutions of A~x = ~b are exactly the vectors of the form ~x1 + ~xh , where ~xh is a solution of A~x = ~0; indeed if ~x2 is a solution of A~x = ~b, we can write ~x2 = ~x1 + (~x2 − ~x1 ), and ~x2 − ~x1 will be a solution of A~x = ~0, by part (b). Geometrically, the vectors of the form ~x1 + ~xh are those whose tips are on the line L in Figure 1.14; the line L runs through the tip of ~x1 and is parallel to the given line consisting of the solutions of A~x = ~0.

Figure 1.14: for Problem 1.3.48c.

40

Section 1.3 . 1.3.49 a This system has either infinitely many solutions (if the right-most column of rref[A..b] does not contain a leading one), or no solutions (if the right-most column does contain a leading one). . . b This system has either a unique solution (if rank[A..~b] = 3), or no solution (if rank[A..~b] = 4). . c The right-most column of rref[A..~b] must contain a leading one, so that the system has no solutions. d This system has infinitely many solutions, since there is one free variable. . 1.3.50 The right-most column of rref[A..~b] must contain a leading one, so that the system has no solutions. 1.3.51 For B~x to be defined, the number of columns of B, which is s, must equal the number of components of ~x, which is p, so that we must have s = p. Then B~x will be a vector in Rr ; for A(B~x) to be defined we must have m = r. Summary: We must have s = p and m = r.

0 −1 1.3.52 A(B~x) = A 1 0 0 −1 so that C = . 2 −1

x1 x2

1 0 = 1 2

−x2 x1

−x2 = 2x1 − x2

0 −1 = 2 −1

x1 , x2

x1 1.3.53 Yes; write A = [~v1 . . . ~vm ], B = [w ~1 . . . w ~ m ], and ~x = . . . . xm x1 ~ 1 ) + · · · + xm (~vm + w ~ m ) and Then (A + B)~x = [~v1 + w ~ 1 . . . ~vm + w ~ m ] . . . = x1 (~v1 + w xm x1 x1 ~ 1 + · · · + xm w ~ m. ~1 . . . w ~ m ] . . . = x1~v1 + · · · + xm~vm + x1 w A~x + B~x = [~v1 . . . ~vm ] . . . + [w xm xm The two results agree, as claimed.

1.3.54 The vectors of the form c1~v1 + c2~v2 form a plane through the origin containing ~v1 and ~v2 ; in Figure 1.15 we draw a typical vector in this plane. 7 4 1 1.3.55 We are looking for constants a and b such that a 2 + b 5 = 8 . 9 6 3 7 a + 4b = 7 The resulting system 2a + 5b = 8 has the unique solution a = −1, b = 2, so that 8 is indeed a linear 9 3a + 6b = 9 1 4 combination of the vector 2 and 5 . 3 6 41

Chapter 1

Figure 1.15: for Problem 1.3.54. −2 9 5 1 30 −5 2 6 7 −1 1.3.56 We can use technology to determine that the system 38 = x1 1 + x2 3 + x3 3 + x4 4 is 7 5 2 9 56 9 2 8 4 62 30 −1 inconsistent; therefore, the vector 38 fails to be a linear combination of the other four vectors. 56 62

2 1 on y = x2 and on y = 3x. 1 3 7 1 2 1 2 7 . = +b : a and as a linear combination of Then write 11 3 1 3 1 11 3 4 7 . + = The unique solution is a = 2, b = 3, so that the desired representation is 9 2 11 4 3 x is on the line y = 2 ; is on line y = 3x. 2 9

1.3.57 Pick a vector on each line, say

−1 2 1 3 1.3.58 We want b = k1 3 + k2 6 + k3 −3 , for some k1 , k2 and k3 . −2 4 2 c 1 1 1 Note that we can rewrite this right-hand side as k1 3 + 2k2 3 − k3 3 2 2 2 1 = (k1 + 2k2 − k3 ) 3 . It follows that k1 + 2k2 − k3 = 3, so that b = 9 and c = 6. 2 42

Section 1.3 a+b 1 1 5 2 a + 2b 1 7 1.3.59 = a + b = . a + 3b 3 1 c a + 4b 4 1 d a +b = 5 , which we quickly solve to find a = 3 and b = 2. Then, c = So we have a small system: a +2b = 7 a + 3b = 3 + 6 = 9 and d = a + 4b = 3 + 8 = 11. a 0 1 2 k2 + 2k3 0 b 0 0 0 1.3.60 We need = k1 + k2 + k3 = . From this we see that a, c and d can c 3 4 5 3k1 + 4k2 + 5k3 d 0 0 6 6k3 be any value, while b must equal zero. 1.3.61 We need to solve the system 1 1 1 c = x 2 + y 3 9 4 c2 with augmented matrix

1 2 4

The matrix reduces to

1 0 0

. 1.. . 1.. . 0..

. 1.. . 3.. . 9..

1 . c

c2 1

c−2 2

c − 5c + 6

.

This system is consistent if and only if c = 2 or c = 3. Thus the vector is a linear combination if c = 2 or c = 3. 1.3.62 We need to solve the system with augmented matrix

1 1 1 c = x a + y b c2 a2 b2

1 a

a2

The matrix reduces to

. 1.. 1 . 0 b − a.. . 0 0..

. 1.. . b.. . b2 ..

1 . c

c2

1 c−a (c − a)(c − b)

.

This system is consistent if and only if c = a or c = b. Thus the vector is a linear combination if c = a or c = b. 43

Chapter 1 1.3.63 This is the line parallel to w ~ which goes through the end point of the vector ~v . 1.3.64 This is the line segment connecting the head of the vector ~v to the head of the vector ~v + w. ~ 1.3.65 This is the full parallelogram spanned by the two vectors ~v and w. ~ 1.3.66 Write b = 1 − a and a~v + bw ~ = a~v + (1 − a)w ~ =w ~ + a(~v − w) ~ to see that this is the line segment connecting the head of the vector ~v to the head of the vector w. ~ 1.3.67 This is the full triangle with its vertices at the origin and at the heads of the vectors ~v and w. ~ 1.3.68 Writing ~u · ~v = ~u · w ~ as ~u · (~v − w) ~ = 0, we see that this is the line perpendicular to the vector ~v − w. ~

0 1 1.3.69 We write out the augmented matrix: 1 0 1 1

So x =

−a+b+c , 2

y=

a−b+c 2

and z =

a+b−c . 2

. 1.. . 1.. . 0..

.. a 1 0 0. .. b and reduce it to 0 1 0. . c 0 0 1..

−a+b+c 2 a−b+c 2 a+b−c 2

.

1.3.70 We find it useful to let s = x1 + x2 + · · · + xn . Adding up all n equations of the system, and realizing that n . Now the ith the term xi is missing from the ith equation, we see that (n − 1)s = b1 + · · · + bn , or, s = b1 +···+b n−1 n − bi . equation of the system can be written as s − xi = bi , so that xi = s − bi = b1 +···+b n−1

True or False Ch 1.TF.1 F, by Example 3a of Section 1.3 Ch 1.TF.2 T, by Definition 1.3.7 Ch 1.TF.3 T, by Theorem 1.3.4 Ch 1.TF.4 F, by Theorem 1.3.1 Ch 1.TF.5 F, by Theorem 1.3.4 Ch 1.TF.6 F; As a counter-example, consider the zero matrix. Ch 1.TF.7 T, by Theorem 1.3.8 Ch 1.TF.8 T, by Definition 1.3.9 Ch 1.TF.9 T, by Definition. Ch 1.TF.10 F; Consider the equation x + y + z = 0, repeated four times. 44

True or False Ch 1.TF.11 T; Find rref. Ch 1.TF.12 T; Find rref Ch 1.TF.13 F; Consider the 4 × 3 matrix A that contains all zeroes, except for a 1 in the lower left corner. 2 1 Ch 1.TF.14 F; Note that A = 2A for all 2 × 2 matrices A. 2 1 Ch 1.TF.15 F; The rank is 1. Ch 1.TF.16 F; The product on the left-hand side has two components.

−3 Ch 1.TF.17 T; Let A = −5 −7

0 0 , for example. 0

7 4 1 Ch 1.TF.18 T; We have 2 = 2 5 − 8 . 9 6 3 Ch 1.TF.19 T; The last component of the left-hand side is zero for all vectors ~x.

3 Ch 1.TF.20 T; A = 4

0 , for example. 0

0 Ch 1.TF.21 F; Let A = 0 0 to A all we want, we will

1 1 0 0 1 and B = 0 0 0 0 always end up with a

0 0 1 0 , for example. We can apply elementary row operations 0 0 matrix that has all zeros in the first column.

Ch 1.TF.22 T; If ~u = a~v + bw ~ and ~v = c~ p + d~q + e~r, then ~u = ac~ p + ad~q + ae~r + bw. ~ Ch 1.TF.23 F; The system x = 2, y = 3, x + y = 5 has a unique solution.

0 1 Ch 1.TF.24 F; Let A = , for example. 0 0

1 Ch 1.TF.25 F; Let A = 0 1

2 0 1 and ~b = 3 , for example. 5 1

Ch 1.TF.26 T, by Exercise 1.3.44. Ch 1.TF.27 F; Find rref to see that the rank is always 2. Ch 1.TF.28 T; Note that ~v = 1~v + 0w. ~ 45

Chapter 1

Ch 1.TF.29 F; Let ~u =

0 2 1 , for example. ,w ~= , ~v = 1 0 0

Ch 1.TF.30 T; Note that ~0 = 0~v + 0w ~

Ch 1.TF.31 1 0 0 0

0 1 0 0

1 1 . . F; If A 2 = ~0, then ~x = 2 is a solution to A..~0 . However, since rank(A) = 3, rref A..~0 = 3 3 0 0 0 .. 0 . , meaning that only ~0 is a solution to A~x = ~0. 1 0 0 0

Ch 1.TF.32 F; If ~b = ~0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. Ch 1.TF.33 T; By Example 3c of Section 1.3, the equation A~x = ~0 has the unique solution ~x = ~0. Now note that A(~v − w) ~ = ~0, so that ~v − w ~ = ~0 and ~v = w. ~ Ch 1.TF.34 T; Note that rank(A) = 4, by Theorem 1.3.4 Ch 1.TF.35 F; Let ~u =

2 1 0 , ~v = , w ~= , for example.. 0 0 1

−2t Ch 1.TF.36 T; We use rref to solve the system A~x = ~0 and find ~x = −3t , where t is an arbitrary constant. t −2 ~ = ~0, so that w ~ = 2~u + 3~v . Letting t = 1, we find [~u ~v w] ~ −3 = −2~u − 3~v + w 1

Ch 1.TF.37 F; Let A = B =

1 0 , for example. 0 1

1 0 ... 0 1 ... Ch 1.TF.38 T; Matrices A and B can both be transformed into I = ... ... ... 0 0 0 operations backwards, we can transform I into B. Thus we can first transform A

0 0 . Running the elementary 0 1 into I and then I into B.

Ch 1.TF.39 T; If ~v = a~u + bw, ~ then A~v = A(a~u + bw) ~ = A(a~u) + A(bw) ~ = aA~v + bAw. ~ Ch 1.TF.40 T; check that the three defining properties of a matrix in rref still hold. F; If ~b = ~0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. . ~ Ch 1.TF.41 T; A~x = b is inconsistent if and only if rank A..~b = rank(A)+1, since there will be an extra leading one in the last column of the augmented matrix: (See Figure 1.16.) 46

True or False

Figure 1.16: for Problem T/F 41. Ch 1.TF.42 T; The system A~x = ~b is consistent, by Example 3b, and there are, in fact, infinitely many solutions, by Theorem 1.3.3. Note that A~x = ~b is a system of three equations with four unknowns. .. .. .. ~ ~ ~ ~ Ch 1.TF.43 T; Recall that we use rref A.0 to solve the system A~x = 0. Now, rref A.0 = rref(A).0 = .. .. .. .. ~ ~ ~ ~ rref(B).0 = rref B .0 . Then, since rref(A).0 = rref(B).0 , they must have the same solutions. Ch 1.TF.44 F; Consider

1 2 . If we remove the first column, then the remaining matrix fails to be in rref. 0 0

Ch 1.TF.45 T; First we list all possible matrices rref(M ), where M is a 2 × 2 matrix, and show the corresponding solutions for M~x = ~0: rref(M solutions of M~x = ~0 ) 1 0 {~0} 0 1 −at 1 a , for an arbitrary t 0 0 t 0 1 t , for an arbitrary t 0 0 0 0 0 R2 0 0 Now, we see that if rref(A) 6= rref(B), then the systems A~x = ~0 and B~x = ~0 must have different solutions. Thus, it must be that if the two systems have the same solutions, then rref(A) = rref(B).

47

Chapter 2

Chapter 2 Section 2.1 2.1.1 Not a linear transformation, since y2 = x2 + 2 is not linear in our sense.

0 2.1.2 Linear, with matrix 0 1

2 0 0 3 0 0

2.1.3 Not linear, since y2 = x1 x3 is nonlinear. 9 3 −3 1 2 −9 2.1.4 A = 4 −9 −2 5 1 5

2.1.5 By Theorem 2.1.2, the three columns of the 2 × 3 matrix A are T (~e1 ), T (~e2 ), and T (~e3 ), so that 7 6 −13 . A= 11 9 17 1 4 1 2.1.6 Note that x1 2 + x2 5 = 2 3 6 3 2.1.7 Note that x1~v1 + · · · + xm~vm

2.1.8 Reducing the system

4 1 4 x 5 1 , so that T is indeed linear, with matrix 2 5 . x2 6 3 6

x1 = [~v1 . . . ~vm ] · · · , so that T is indeed linear, with matrix [~v1 ~v2 · · · ~vm ]. xm

x1 + 7x2 3x1 + 20x2

x1 = y1 , we obtain = y2

x2

= −20y1 = 3y1

+ −

7y2 . y2

y1 2 3 x1 2.1.9 We have to attempt to solve the equation = for x1 and x2 . Reducing the system y2 x2 6 9 x1 + 1.5x2 = 0.5y1 2x1 + 3x2 = y1 . we obtain 0 = −3y1 + y2 6x1 + 9x2 = y2 No unique solution (x1 , x2 ) can be found for a given (y1 , y2 ); the matrix is noninvertible. 1 2 y1 x1 = for x1 and x2 . Reducing the system 2.1.10 We have to attempt to solve the equation y2 4 9 x2 x1 + 2x2 = y1 x1 = 9y1 + 2y2 x1 9 −2 y1 we find that or = . 4x1 + 9x2 = y2 x2 = −4y1 + y2 x2 y2 −4 1 9 −2 . The inverse matrix is −4 1 48

Section 2.1 y1 x1 1 2 2.1.11 We have to attempt to solve the equation = for x1 and x2 . Reducing the system y2 3 9 x2 3 − 32 x1 = 3y1 − 23 y2 x1 + 2x2 = y1 . The inverse matrix is . we find that 1 3x1 + 9x2 = y2 −1 x2 = −y1 + 31 y2 3 2.1.12 Reducing the system 1 −k . 0 1

x1 + kx2 x2

= y1 = y2

we find that

x1 x2

2.1.13 a First suppose that a 6= 0. We have to attempt to solve the equation

ax1 cx1

x1

x1

+ bx2 + dx2 + (d +

= y1 =

b a x2 − bc a )x2

b a x2 ( ad−bc a )x2

y2

÷a

1 = a y1 = − ac y1

= =

1 a y1 − ac y1

x1 + → cx1 + → + y2

b a x2 dx2

= =

1 a y1

y2

− ky2 . The inverse matrix is y2

= y1 =

−c(I)

y1 y2

=

a b c d

x1 x2

for x1 and x2 .

→

+ y2

We can solve this system for x1 and x2 if (and only if) ad − bc 6= 0, as claimed. If a = 0, then we have to consider the system bx2 = y1 cx1 swap : I ↔ II cx1 + dx2 = y2

+ dx2 bx2

= = y1

y2

We can solve for x1 and x2 provided that both b and c are nonzero, that is if bc 6= 0. Since a = 0, this means that ad − bc 6= 0, as claimed. b First suppose that ad − bc 6= 0 and a 6= 0. Let D = ad − bc for simplicity. We continue our work in part (a): 1 x1 + ab x2 = a y1 a → D c ·D a x2 = − a y1 + y2 b 1 − a (II) x1 + ab x2 = a y1 → a x2 = − Dc y1 + D y2 bc x1 = ( a1 + aD )y1 − Db y2 a x2 = − Dc y1 + D y2 d − Db y2 x1 = D y1 a x2 = − Dc y1 + D y2 bc ad d Note that a1 + aD = D+bc aD = aD = D . It follows that system

a b c d

−1

=

1 ad−bc

d −c

−b , as claimed. If ad − bc 6= 0 and a = 0, then we have to solve the a 49

Chapter 2

cx1 + dx2 bx2

= y2 = y1

÷c ÷b x1 + dc x2 = 1c y2 − dc (II) x2 = 1b y1 d x1 = − bc y1 + 1c y2 1 x2 = b y1 −1 d − bc a b It follows that = 1 c d b 2.1.14 a By Exercise 13a,

b By Exercise 13b,

2 3 5 k

2 3 5 k −1

=

1 c

0

=

1 ad−bc

d −c

−b a

1 2n .

Since

k 2k−15

(recall that a = 0), as claimed.

is invertible if (and only if) 2k − 15 6= 0, or k 6= 7.5. 1 2k−15

−3 . 2

k −5

2 3 − 2k−15 2k−15 + 21 is an integer

If all entries of this inverse are integers, then or k = 7.5 +

= kn = 7.5n

=

1 2k−15

is a (nonzero) integer n, so that 2k −15 =

1 n

as well, n must be odd.

1 , where n is an odd integer. The We have shown: If all entries of the inverse are integers, then k = 7.5 + 2n −1 2 3 converse is true as well: If k is chosen in this way, then the entries of will be integers. 5 k

a b

−b a

is invertible if (and only if) a2 + b2 6= 0, which is the case unless 2.1.15 By Exercise 13a, the matrix a −b a b 1 is invertible, then its inverse is a2 +b a = b = 0. If , by Exercise 13b. 2 b a −b a 3 0 , then A~x = 3~x for all ~x in R2 , so that A represents a scaling by a factor of 3. Its inverse is a 0 3 1 0 3 1 −1 . (See Figure 2.1.) scaling by a factor of 3 : A = 0 13

2.1.16 If A =

2.1.17 If A =

−1 0

0 , then A~x = −~x for all ~x in R2 , so that A represents a reflection about the origin. −1

This transformation is its own inverse: A−1 = A. (See Figure 2.2.) 2.1.18 Compare with Exercise 16: This matrix represents a scaling by the factor of 12 ; the inverse is a scaling by 2. (See Figure 2.3.) x1 x 1 0 , so that A represents the orthogonal projection onto the ~e1 axis. (See , then A 1 = 2.1.19 If A = 0 0 0 x2 1 Figure 2.1.) This transformation is not invertible, since the equation A~x = has infinitely many solutions ~x. 0 (See Figure 2.4.)

50

Section 2.1

· ¸ 0 6

· ¸ 0 2

·

3 0 0 3

¸

· ¸ 1 0

· ¸ 3 0 Figure 2.1: for Problem 2.1.16.

· ¸ 0 2

·

−1 0 0 −1

¸

·

−1 0

¸

· ¸ 1 0

·

0 −2

¸

Figure 2.2: for Problem 2.1.17.

· ¸ 0 2

·

0.5 0 0 0.5

· ¸ 0 1

¸

·

· ¸ 1 0

0.5 0

¸

Figure 2.3: for Problem 2.1.18.

2.1.20 If A =

0 1 1 0

, then A

x1 x2

=

x2 , so that A represents the reflection about the line x2 = x1 . This x1

transformation is its own inverse: A−1 = A. (See Figure 2.5.)

2.1.21 Compare with Example 5. x 0 1 x2 , then A 1 = If A = . Note that the vectors ~x and A~x are perpendicular and have the same −1 0 x2 −x1 51

Chapter 2

· ¸ 0 2

·

1 0 0 0

¸

· ¸ 0 0 · ¸ 1 0

· ¸ 1 0

Figure 2.4: for Problem 2.1.19.

· ¸ 0 2

·

0 1 1 0

¸

· ¸ 0 1

· ¸ 1 0

· ¸ 2 0 Figure 2.5: for Problem 2.1.20.

length. If ~x is in the first quadrant, then A~x is in the fourth. Therefore, A represents the rotation through an 0 −1 angle of 90◦ in the clockwise direction. (See Figure 2.6.) The inverse A−1 = represents the rotation 1 0 ◦ through 90 in the counterclockwise direction.

· ¸ 0 2

·

0 1 −1 0

· ¸ 2 0

¸ ·

· ¸ 1 0

0 −1

¸

Figure 2.6: for Problem 2.1.21.

2.1.22 If A =

1 0 0 −1

x , then A 1 x2

=

x1 , so that A represents the reflection about the ~e1 axis. This −x2

transformation is its own inverse: A−1 = A. (See Figure 2.7.) 2.1.23 Compare with Exercise 21.

0 1 Note that A = 2 , so that A represents a rotation through an angle of 90◦ in the clockwise direction, −1 0 followed by a scaling by the factor of 2. 52

Section 2.1

· ¸ 0 2

·

1 0 0 −1

· ¸ 1 0

¸

· ¸ 1 0

·

0 −2

¸

Figure 2.7: for Problem 2.1.22.

−1

The inverse A

=

0 1 2

− 12 0

represents a rotation through an angle of 90◦ in the counterclockwise direction,

followed by a scaling by the factor of 12 . (See Figure 2.8.)

· ¸ 0 2

· · ¸ 1 0

0 2 −2 0

· ¸ 4 0

¸

·

0 −2

¸

Figure 2.8: for Problem 2.1.23. 2.1.24 Compare with Example 5. (See Figure 2.9.)

Figure 2.9: for Problem 2.1.24. 2.1.25 The matrix represents a scaling by the factor of 2. (See Figure 2.10.) 2.1.26 This matrix represents a reflection about the line x2 = x1 . (See Figure 2.11.) 2.1.27 This matrix represents a reflection about the ~e1 axis. (See Figure 2.12.) 53

Chapter 2

Figure 2.10: for Problem 2.1.25.

Figure 2.11: for Problem 2.1.26.

Figure 2.12: for Problem 2.1.27.

1 0 x x1 , then A 1 = , so that the x2 component is multiplied by 2, while the x1 component x2 0 2 2x2 remains unchanged. (See Figure 2.13.)

2.1.28 If A =

Figure 2.13: for Problem 2.1.28.

54

Section 2.1 2.1.29 This matrix represents a reflection about the origin. Compare with Exercise 17. (See Figure 2.14.)

Figure 2.14: for Problem 2.1.29.

2.1.30 If A = 2.15.)

0 0

0 x1 0 , then A = , so that A represents the projection onto the ~e2 axis. (See Figure x2 1 x2

Figure 2.15: for Problem 2.1.30.

2.1.31 The image must be reflected about the ~e2 axis, that is

x1 x2

must be transformed into −1 0 be accomplished by means of the linear transformation T (~x) = ~x. 0 1

3 0 2.1.32 Using Theorem 2.1.2, we find A = ...

0

else.

0 3 .. .

· · .. .

0 ···

−x1 : This can x2

0 0 . This matrix has 3’s on the diagonal and 0’s everywhere .. . 3

1 0 2.1.33 By Theorem 2.1.2, A = T T . (See Figure 2.16.) 0 1 √1 − √12 2 . Therefore, A = 1 1 √

√

2

2

2.1.34 As in Exercise 2.1.33, we find T (~e1 ) and T (~e2 ); then by Theorem 2.1.2, A = [T (~e1 ) T (~e2 )]. (See Figure 2.17.) 55

Chapter 2

Figure 2.16: for Problem 2.1.33.

Figure 2.17: for Problem 2.1.34.

cos θ Therefore, A = sin θ

− sin θ . cos θ

2.1.35 We want to find a matrix A = 5a + 42b 6a + 41b solving the system

a c

5c + 42d 6c + 41d

b d

such that A

= 89 = 88 . = 52 = 53

88 6 89 5 . This amounts to = and A = 53 41 52 42

(Here we really have two systems with two unknowns each.) The unique solution is a = 1, b = 2, c = 2, and d = 1, so that A =

1 2 . 2 1

2.1.36 First we draw w ~ in terms of ~v1 and ~v2 so that w ~ = c1~v1 + c2~v2 for some c1 and c2 . Then, we scale the ~v2 -component by 3, so our new vector equals c1~v1 + 3c2~v2 . 2.1.37 Since ~x = ~v + k(w ~ − ~v ), we have T (~x) = T (~v + k(w ~ − ~v )) = T (~v ) + k(T (w) ~ − T (~v )), by Theorem 2.1.3 Since k is between 0 and 1, the tip of this vector T (~x) is on the line segment connecting the tips of T (~v ) and T (w). ~ (See Figure 2.18.) 56

Section 2.1

Figure 2.18: for Problem 2.1.37.

2.1.38 T

2 = [~v1 −1

2 ~v2 ] = 2~v1 − ~v2 = 2~v1 + (−~v2 ). (See Figure 2.19.) −1

Figure 2.19: for Problem 2.1.38.

x1 x1 T (~ e ) . . . T (~ e ) 1 m 2.1.39 By Theorem 2.1.2, we have T . . . = . . . = x1 T (~e1 ) + · · · + xm T (~em ). xm xm

2.1.40 These linear transformations are of the form [y] = [a][x], or y = ax. The graph of such a function is a line through the origin. x1 , or y = ax1 + bx2 . The graph of such a function 2.1.41 These linear transformations are of the form [y] = [a b] x2 is a plane through the origin.

2.1.42 a See Figure 2.20. 1 0 b The image of the point 21 is the origin, . 0 1 2

57

Chapter 2

Figure 2.20: for Problem 2.1.42. # " 1 x1 − 2 x1 + x2 =0 0 x2 = . (See Figure 2.16.) c Solve the equation , or 0 − 21 0 1 x3 = 0 − 21 x1 + x3 2t x1 The solutions are of the form x2 = t , where t is an arbitrary real number. For example, for t = 12 , we t x3 1 find the point 12 considered in part b.These points are on the line through the origin and the observer’s eye.

− 21

1

0

1 2

x1 2 x1 2.1.43 a T (~x) = 3 · x2 = 2x1 + 3x2 + 4x3 = [2 3 4] x2 x3 4 x3 The transformation is indeed linear, with matrix [2 3 4].

v1 b If ~v = v2 , then T is linear with matrix [v1 v2 v3 ], as in part (a). v3

a a x1 x1 x1 c Let [a b c] be the matrix of T . Then T x2 = [a b c] x2 = ax1 + bx2 + cx3 = b · x2 , so that ~v = b c c x3 x3 x3 does the job.

0 v2 x3 − v3 x2 x1 v1 x1 2.1.44 T x2 = v2 × x2 = v3 x1 − v1 x3 = v3 −v2 v1 x2 − v2 x1 x3 v3 x3 0 −v3 v2 v3 0 −v1 . −v2 v1 0

58

−v3 0 v1

x1 v2 −v1 x2 , so that T is linear, with matrix x3 0

Section 2.1 2.1.45 Yes, ~z = L(T (~x)) is also linear, which we will verify using Theorem 2.1.3. Part a holds, since L(T (~v + w)) ~ = L(T (~v ) + T (w)) ~ = L(T (~v )) + L(T (w)), ~ and part b also works, because L(T (k~v )) = L(kT (~v )) = kL(T (~v )). pa + qc a 1 1 = =B =B A ra + sc c 0 0 0 0 b pb + qd T =B A =B = 1 1 d rb + sd 0 1 x1 b pb + qd So, T = + x2 T = x1 T = x2 1 0 d rb + sd

2.1.46 T

2.1.47 Write w ~ as a linear combination of ~v1 and ~v2 : w ~ = c1~v1 + c2~v2 . (See Figure 2.21.)

Figure 2.21: for Problem 2.1.47. Measurements show that we have roughly w ~ = 1.5~v1 + ~v2 . Therefore, by linearity, T (w) ~ = T (1.5~v1 + ~v2 ) = 1.5T (~v1 ) + T (~v2 ). (See Figure 2.22.)

Figure 2.22: for Problem 2.1.47. 2.1.48 Let ~x be some vector in R2 . Since ~v1 and ~v2 are not parallel, we can write ~x in terms of components of ~v1 and ~v2 . So, let c1 and c2 be scalars such that ~x = c1~v1 + c2~v2 . Then, by Theorem 2.1.3, T (~x) = T (c1~v1 + c2~v2 ) = T (c1~v1 ) + T (c2~v2 ) = c1 T (~v1 ) + c2 T (~v2 ) = c1 L(~v1 ) + c2 L(~v2 ) = L(c1~v1 + c2~v2 ) = L(~x). So T (~x) = L(~x) for all ~x in R2 .

2x1 2.1.49 a Let x1 be the number of 2 Franc coins, and x2 be the number of 5 Franc coins. Then x1 59

+5x2 +x2

= =

144 . 51

Chapter 2

From this we easily find our solution vector to be

b

total value of coins 2x1 = x1 total number of coins 2 5 So, A = . 1 1

+5x2 +x2

=

2 5 1 1

37 . 14

x1 . x2

c By Exercise 13, matrix A is invertible (since ad − bc = −3 6= 0), and A−1 = Then

− 13

part a.

1 −5 −1 2

144 51

=

− 13

144 −144

−5(51) +2(51)

=

− 31

−111 −42

=

1 ad−bc

d −b 1 = − 31 −c a −1

−5 . 2

37 , which was the vector we found in 14

p mass of the platinum alloy 2.1.50 a Let = . Using the definition density = mass/volume, or volume = s mass of the silver alloy mass/density, we can set up the system: p +s = 5, 000 , with the solution p = 2, 600 and s = 2, 400. We see that the platinum alloy makes up p s + 10 = 370 20 only 52 percent of the crown; this gold smith is a crook! p total mass 1 p+s b We seek the matrix A such that A = = p 1 s . Thus A = s total volume + 20 10 20

1 1 10

.

2 −20 p c Yes. By Exercise 13, A−1 = . Applied to the case considered in part a, we find that = −1 20 s 2, 600 5, 000 2 −20 total mass , confirming our answer in part a. = = A−1 2, 400 370 −1 20 total volume 5 5 C (F − 32) F− 9 2.1.51 a = = 9 1 1 1 5 − 160 9 . So A = 9 0 1

160 9

=

5 9

0

− 160 9 1

F . 1

5 b Using Exercise 13, we find 95 (1) − (− 160 9 )0 = 9 6= 0, so A is invertible. 9 1 160 32 9 5 = A−1 = 95 . So, F = 59 C + 32. 5 0 1 0 9

2.1.52 a A~x =

300 , meaning that the total value of our money is C$300, or, equivalently, ZAR2400. 2, 400

b From Exercise 13, we test the value ad − bc and find it to be zero. Thus A is not invertible. To determine when .. A is consistent, we begin to compute rref A.~b : 60

Section 2.2 .. .. 1 1 . b1 . b 1 8 . → .. .. −8I 8 1 . b2 0 0 . b2 − 8b1 Thus, the system is consistent only when b2 = 8b1 . This makes sense, since b2 is the total value of our money in terms of Rand, while b1 is the value in terms of Canadian dollars. Consider the example in part a. If the system A~x = ~b is consistent, then there will be infinitely many solutions ~x, representing various compositions of our portfolio in terms of Rand and Canadian dollars, all representing the same total value.

1

1 8

2.1.53 All four entries along the diagonal must be 1: they represent the process of converting a currency to itself. We also know that aij = 1/aji for all i, j because converting from one currency i to currencty jis the inverse 1 5/8 1/170 ∗ 8/5 1 ∗ 2 . Next, let’s to converting currency j to currency i. This gives us 3 more entries: 170 ∗ 1 ∗ ∗ 1/2 ∗ 1 find the entry a41 , giving the value of one Euro expressed in Pounds. Now E1 = $(8/5) = $1.60 and $1 = £(1/2) = £(0.50) so that E1 = £(1/2)(8/5) = £(4/5) = £0.80. We have found that a41 = a42 a21 = 4/5 and 1 5/8 1/170 5/4 8/5 1 ∗ 2 . Similarly, we have aij = aik akj for all indices i, j, k = 1, 2, 3, 4. This the matrix is 170 ∗ 1 ∗ 4/5 1/2 ∗ 1 gives a32 = a31 a12 = 170 ∗ 5/8 = 425/4 and a43 = a41 a13 = (4/5)(1/170) = 2/425. Using the fact that aij = a−1 ji , 1 5/8 1/170 5/4 8/5 1 4/425 2 . we can complete the matrix: 170 425/4 1 425/2 4/5 1/2 2/425 1 2.1.54 a 1: this represents converting a currency to itself. b aij is the reciprocal of aji , meaning that aij aji = 1. This represents converting on currency to another, then converting it back. c Note that aik is the conversion factor from currency k to currency i meaning that (1 unit of currency k) = (aik units of currency i) Likewise, (1 unit of currency j) = (akj units of currency k). It follows that (1 unit of currency j) = (akj aik units of currency i) = (aij units of currency i), so that aik akj = aij . d The rank of A is only 1, because every row is simply a scalar multiple of the top row. More precisely, since aij = ai1 a1j , by part c, the ith row is ai1 times the top row. When we compute the rref, every row but the top will be removed in the first step. Thus, rref(A) is a matrix with the top row of A and zeroes for all other entries.

Section 2.2 1 3 1 1 3 2.2.1 The standard L is transformed into a distorted L whose foot is the vector T = = . 0 1 2 0 1 61

Chapter 2

Meanwhile, the back becomes the vector T

2.2.2 By Theorem 2.2.3, this matrix is

2 0 3 1 0 . = = 4 2 1 2 2 ◦

cos(60 ) sin(60◦ )

◦

− sin(60 ) = cos(60◦ )

√ 1 2 √ 3 2

−

3 2 1 2

.

2.2.3 If ~x is in the unit square in R2 , then ~x = x1~e1 + x2~e2 with 0 ≤ x1 , x2 ≤ 1, so that T (~x) = T (x1~e1 + x2~e2 ) = x1 T (~e1 ) + x2 T (~e2 ). The image of the unit square is a parallelogram in R3 ; two of its sides are T (~e1 ) and T (~e2 ), and the origin is one of its vertices. (See Figure 2.23.)

Figure 2.23: for Problem 2.2.3.

2.2.4 By Theorem 2.2.4, this is a rotation√combined with a scaling. The transformation rotates 45 degrees counterclockwise, and has a scaling factor of 2. 2.2.5 Note that cos(θ) = −0.8, so that θ = arccos(−0.8) ≈ 2.498. 2 1 1 2.2.6 By Theorem 2.2.1, projL 1 = ~u · 1 ~u, where ~u is a unit vector on L. To get ~u, we normalize 1 : 2 1 1 2 1 ~u = 31 1 , so that projL 1 = 2 1

5 3

10 2 9 · 31 1 = 59 . 10 2 9

1 1 1 2.2.7 According to the discussion in the text, refL 1 = 2 ~u · 1 ~u − 1 , where ~u is a unit vector on L. To 1 1 1 11 1 1 2 2 2 9 get ~u, we normalize 1 : ~u = 31 1 , so that refL 1 = 2( 53 ) 31 1 − 1 = 19 . 11 1 1 2 2 2 9

62

Section 2.2 2.2.8 From Definition 2.2.2, we can see that this is a reflection about the line x1 = −x2 . 2.2.9 By Theorem 2.2.5, this is a vertical shear. 0.8 4 . Then = 2.2.10 By Theorem 2.2.1, projL ~x = (~u ·~x)~u, where ~u is a unit vector on L. We can choose ~u = 0.6 3 x1 0.64 0.48 0.64x1 + 0.48x2 0.8 0.8 x 0.8 x . = = = (0.8x1 + 0.6x2 ) · 1 = projL 1 0.48 0.36 x2 0.6 0.48x1 + 0.36x2 0.6 0.6 x2 x2 0.64 0.48 The matrix is A = . 0.48 0.36 1 5

2.2.11 In Exercise 10 we found the matrix A =

0.64 0.48

0.48 0.36

of the projection onto the line L. By Theorem 2.2.2, 0.28 0.96 refL ~x = 2(projL ~x) − ~x = 2A~x − ~x = (2A − I2 )~x, so that the matrix of the reflection is 2A − I2 = . 0.96 −0.28

p 2.2.12 Let ~u = (1/||w||) ~ w ~ be the unit vector in the direction of w. ~ It has the components u1 = w1 / w12 + w22 and 2 u2 = √ww2 +w 2 . On Pages 57/58, we see that the matrix representing the projection is 1

2

This can be written as

u21 u1 u2

1 w12 + w22

u1 u2 u22

w12 w1 w2

.

w1 w2 w22

,

as claimed. 2.2.13 By Theorem 2.2.2, x u1 x u1 x1 − 1 · 1 =2 refL x2 u2 x2 u2 x2 . 2 u x (2u1 − 1)x1 + 2u1 u2 x2 = 2(u1 x1 + u2 x2 ) 1 − 1 = . u2 x2 2u1 u2 x1 + (2u22 − 1)x2 2 a b 2u1 − 1 2u1 u2 The matrix is A = . Note that the sum of the diagonal entries is a + d = = 2u1 u2 2u22 − 1 c d a b 2 2 2(u1 + u2 ) − 2 = 0, since ~u is a unit vector. It follows that d = −a. Since c = b, A is of the form . Also, b −a 2 2 2 4 2 2 2 2 2 2 a + b = (2u1 − 1) + 4u1 u2 = 4u1 − 4u1 + 1 + 4u1 (1 − u1 ) = 1, as claimed.

2.2.14 a Proceeding as on Page 57/58 in the text, we find that A is the matrix whose ijth entry is ui uj : 2 u1 u2 u1 u3 u1 u22 u2 u3 A = u2 u1 un u1 un u2 u23 b The sum of the diagonal entries is u21 + u22 + u23 = 1, since ~u is a unit vector. 63

Chapter 2 2.2.15 According to the discussion on Page 60 in the text, refL (~x) = 2(~x · ~u)~u − ~x x1 u1 = 2(x1 u1 + x2 u2 + x3 u3 ) u2 − x2 x3 u3 (2u21 − 1)x1 +2u2 u1 x2 +2u1 u3 x3 2x1 u21 +2x2 u2 u1 +2x3 u3 u1 −x1 +(2u22 − 1)x2 +2u2 u3 x3 . +2x2 u22 +2x3 u3 u2 −x2 = 2u1 u2 x1 = 2x1 u1 u2 2u1 u3 x1 +2u2 u3 x2 +(2u23 − 1)x3 2x1 u1 u3 +2x2 u2 u3 +2x3 u23 −x3 (2u21 − 1) 2u2 u1 2u1 u3 (2u22 − 1) 2u2 u3 . So A = 2u1 u2 2u1 u3 2u2 u3 (2u23 − 1) 2.2.16 a See Figure 2.24.

Figure 2.24: for Problem 2.2.16a. b By Theorem 2.1.2, the matrix of T is [T (~e1 )

T (~e2 )].

Figure 2.25: for Problem 2.2.16b. 64

Section 2.2

T (~e2 ) is the unit vector in the fourth quadrant perpendicular to T (~e1 ) =

cos(2θ) , so that sin(2θ)

sin(2θ) cos(2θ) sin(2θ) . The matrix of T is therefore . − cos(2θ) sin(2θ) − cos(2θ) u1 cos θ Alternatively, we can use the result of Exercise 13, with = to find the matrix u2 sin θ 2 cos2 θ − 1 2 cos θ sin θ . 2 cos θ sin θ 2 sin2 θ − 1

T (~e2 ) =

You can use trigonometric identities to show that the two results agree. (See Figure 2.25.) 2.2.17 We want,

a b b −a

v1 v2

av1 = bv1

+bv2 −av2

v1 . = v2

Now, (a − 1)v1 + bv2 = 0 and bv1 − (a + 1)v2 , which is a system with solutions of the form is an arbitrary constant.

bt , where t (1 − a)t

b . Let’s choose t = 1, making ~v = 1−a

a−1 Similarly, we want Aw ~ = −w. ~ We perform a computation as above to reveal w ~ = b A quick check of ~v · w ~ = 0 reveals that they are indeed perpendicular.

as a possible choice.

Now, any vector ~x in R can be written in terms of components with respect to L = span(~v ) as ~x = ~x|| + ~x⊥ = c~v + dw. ~ Then, T (~x) = A~x = A(c~v + dw) ~ = A(c~v ) + A(dw) ~ = cA~v + dAw ~ = c~v − dw ~ = ~x|| − ~x⊥ = refL (~x), by Definition 2.2.2. (The vectors ~v and w ~ constructed above are both zero in the special case that a = 1 and b = 0. In that case, we can let ~v = ~e1 and w ~ = ~e2 instead.)

b 0.8 2 2.2.18 From Exercise 17, we know that the reflection is about the line parallel to ~v = = = 0.4 . 1 − a 0.4 1 x 2 So, every point on this line can be described as =k . So, y = k = 21 x, and y = 12 x is the line we are y 1 looking for.

1 0 2.2.19 T (~e1 ) = ~e1 , T (~e2 ) = ~e2 , and T (~e3 ) = ~0, so that the matrix is 0 1 0 0

0 0 −1 0 . 0 1

−1 0 0 0 . (See Figure 2.26.) 0 1

1 2.2.20 T (~e1 ) = ~e1 , T (~e2 ) = −~e2 , and T (~e3 ) = ~e3 , so that the matrix is 0 0 0 2.2.21 T (~e1 ) = ~e2 , T (~e2 ) = −~e1 , and T (~e3 ) = ~e3 , so that the matrix is 1 0 65

0 0 . 0

Chapter 2

Figure 2.26: for Problem 2.2.21. 2.2.22 Sketch the ~e1 − ~e3 plane, as viewed from the positive ~e2 axis.

Figure 2.27: for Problem 2.2.22.

cos θ Since T (~e2 ) = ~e2 , the matrix is 0 − sin θ

0 sin θ 1 0 . (See Figure 2.27.) 0 cos θ

0 2.2.23 T (~e1 ) = ~e3 , T (~e2 ) = ~e2 , and T (~e3 ) = ~e1 , so that the matrix is 0 1

0 1 1 0 . (See Figure 2.28.) 0 0

Figure 2.28: for Problem 2.2.23. 0 1 = w. ~ Since A preserves length, both ~v and w ~ must be unit = ~v and A 2.2.24 a A = [ ~v w ~ ] , so A 1 0 1 0 vectors. Furthermore, since A preserves angles and and are clearly perpendicular, ~v and w ~ must also 0 1 66

Section 2.2 be perpendicular. b Since w ~ is a unit vector perpendicular to ~v , it can be obtained by rotating ~v through 90 degrees, either in the counterclockwise or in the clockwise direction. Using the corresponding rotation matrices, we see that 0 −1 −b 0 1 b w ~= ~v = or w ~= ~v = . 1 0 a −1 0 −a c Following part b, A is either of the form reflection.

a b

−b a b , representing a rotation, or A = , representing a a b −a

1 k 1 −k represents a horizontal shear, and its inverse A−1 = represents such a 0 1 0 1 shear as well, but “the other way.”

2.2.25 The matrix A =

2.2.26 a

k 0

0 k

4 0 8 2k 2 . . So k = 4 and A = = = 0 4 −4 −k −1

b This is the orthogonal projection onto the horizontal axis, with matrix B = 4 3 −5b 0 a −b 4 3 5 . So a = 5 , b = − 5 , and C = = = c 4 5a − 35 5 b a a rotation matrix.

3 5 4 5

1 0

0 . 0

. Note that a2 + b2 = 1, as required for

d Since the x1 term is being modified, this must be a horizontal shear. 1 2 7 1 + 3k 1 1 k . . So k = 2 and D = = = Then 0 1 3 3 3 0 1 4 −5 7a + b 7 a b − 4 3 . So a = − 5 , b = 5 , and E = 35 = = e 5 7b − a 1 b −a 5 for a reflection matrix.

3 5 4 5

. Note that a2 + b2 = 1, as required

2.2.27 Matrix B clearly represents a scaling. Matrix C represents a projection, by Definition 2.2.1, with u1 = 0.6 and u2 = 0.8. Matrix E represents a shear, by Theorem 2.2.5. Matrix A represents a reflection, by Definition 2.2.2. Matrix D represents a rotation, by Definition 2.2.3. 2.2.28 a D is a scaling, being of the form

k 0

0 . k

b E is the shear, since it is the only matrix which has the proper form (Theorem 2.2.5). c C is the rotation, since it fits Theorem 2.2.3. 67

Chapter 2 d A is the projection, following the form given in Definition 2.2.1. e F is the reflection, using Definition 2.2.2. 2.2.29 To check that L is linear, we verify the two parts of Theorem 2.1.3: a) Use the hint and apply L to both sides of the equation ~x + ~y = T (L(~x) + L(~y )): L(~x + ~y ) = L(T (L(~x) + L(~y ))) = L(~x) + L(~y ) as claimed. b)L (k~x) = L (kT (L (~x))) = L (T (kL (~x))) = kL (~x) , as claimed ↑ ↑ ~x = T (L (~x)) T is linear

x1 2.2.30 Write A = [ ~v1 ~v2 ]; then A~x = [~v1 ~v2 ] = x1~v1 + x2~v2 . We must choose ~v1 and ~v2 in such a way that x2 1 , for all x1 and x2 . This is the case if (and only if) both ~v1 and x1~v1 + x2~v2 is a scalar multiple of the vector 2 1 ~v2 are scalar multiples of . 2 1 0 0 1 . , so that A = and ~v2 = For example, choose ~v1 = 2 0 0 2 x1 2.2.31 Write A = [~v1 ~v2 ~v3 ]; then A~x = [~v1 ~v2 ~v3 ] x2 = x1~v1 + x2~v2 + x3~v3 . x3

1 We must choose ~v1 , ~v2 , and ~v3 in such a way that x1~v1 + x2~v2 + x3~v3 is perpendicular to w ~ = 2 for all 3 x1 , x2 , and x3 . This is the case if (and only if) all the vectors ~v1 , ~v2 , and ~v3 are perpendicular to w, ~ that is, if ~v1 · w ~ = ~v2 · w ~ = ~v3 · w ~ = 0. −2 −2 0 0 For example, we can choose ~v1 = 1 and ~v2 = ~v3 = ~0, so that A = 1 0 0 . 0 0 0 0

2.2.32 a See Figure 2.29.

cos α b Compute D~v = sin α

− sin α cos α

cos β sin β

cos α cos β − sin α sin β = . sin α cos β + cos α sin β

Comparing this result with our finding in part (a), we get the addition theorems cos(α + β) = cos α cos β − sin α sin β sin(α + β) = sin α cos β − cos α sin β 68

Section 2.2

Figure 2.29: for Problem 2.2.32a.

Figure 2.30: for Problem 2.2.33. 2.2.33 Geometrically, we can find the representation ~v = ~v1 + ~v2 by means of a parallelogram, as shown in Figure 2.30. To show the existence and uniqueness of this representation algebraically, ~ 1 in L1 and choose a nonzero vector w x 1 a nonzero w ~ 2 in L2 . Then the system x1 w ~ 1 + x2 w ~ 2 = ~0 or [w ~1 w ~ 2] = ~0 has only the solution x1 = x2 = 0 x2 (if x1 w ~ 1 + x2 w ~ 2 = ~0 then x1 w ~ 1 = −x2 w ~ 2 is both in L1 and in L2 , so that it must be the zero vector). x Therefore, the system x1 w ~ 1 + x2 w ~ 2 = ~v or [w ~1 w ~ 2 ] 1 = ~v has a unique solution x1 , x2 for all ~v in R2 (by x2 Theorem 1.3.4). Now set ~v1 = x1 w ~ 1 and ~v2 = x2 w ~ 2 to obtain the desired representation ~v = ~v1 + ~v2 . (Compare with Exercise 1.3.57.) To show that the transformation T (~v ) = ~v1 is linear, we will verify the two parts of Theorem 2.1.3. Let ~v = ~v1 + ~v2 , w ~ =w ~1 + w ~ 2 , so that ~v + w ~ = (~v1 + w ~ 1 ) + (~v2 + w ~ 2 ) and k~v = k~v1 + k~v2 . ↑ ↑ ↑ ↑ in L1 in L2 in L1 in L2

↑ in L1

a. T (~v + w) ~ = ~v1 + w ~ 1 = T (~v ) + T (w), ~ and b. T (k~v ) = k~v1 = kT (~v ), as claimed. 69

↑ in L2

↑ ↑ in L1 in L2

Chapter 2 2.2.34 Keep in mind that the columns of the matrix of a linear transformation T from R3 to R3 are T (~e1 ), T (~e2 ), and T (~e3 ). If T is the orthogonal projection onto a line L, then T (~x) will be on L for all ~x in R3 ; in particular, the three columns of the matrix of T will be on L, and therefore pairwise parallel. This is the case only for matrix B: B represents an orthogonal projection onto a line. A reflection transforms orthogonal vectors into orthogonal vectors; therefore, the three columns of its matrix must be pairwise orthogonal. This is the case only for matrix E: E represents the reflection about a line. 2.2.35 If the vectors ~v1 and ~v2 are defined as shown in Figure 2.27, then the parallelogram P consists of all vectors of the form ~v = c1~v1 + c2~v2 , where 0 ≤ c1 , c2 ≤ 1. The image of P consists of all vectors of the form T (~v ) = T (c1~v1 + c2~v2 ) = c1 T (~v1 ) + c2 T (~v2 ). These vectors form the parallelogram shown in Figure 2.31 on the right.

Figure 2.31: for Problem 2.2.35. 2.2.36 If the vectors ~v0 , ~v1 , and ~v2 are defined as shown in Figure 2.28, then the parallelogram P consists of all vectors ~v of the form ~v = ~v0 + c1~v1 + c2~v2 , where 0 ≤ c1 , c2 ≤ 1. The image of P consists of all vectors of the form T (~v ) = T (~v0 + c1~v1 + c2~v2 ) = T (~v0 ) + c1 T (~v1 ) + c2 T (~v2 ). These vectors form the parallelogram shown in Figure 2.32 on the right.

Figure 2.32: for Problem 2.2.36.

2.2.37 a By Definition 2.2.1, a projection has a matrix of the form 70

u21 u1 u2

u1 u1 u2 , where is a unit vector. u2 u22

Section 2.2 So the trace is u21 + u22 = 1. b By Definition 2.2.2, reflection matrices look like

a b , so the trace is a − a = 0. b −a

cos θ c According to Theorem 2.2.3, a rotation matrix has the form sin θ for some θ. Thus, the trace is in the interval [−2, 2].

− sin θ , so the trace is cos θ +cos θ = 2 cos θ cos θ

1 k 1 0 , depending on whether it represents or 0 1 k 1 a vertical or horizontal shear. In both cases, however, the trace is 1 + 1 = 2.

d By Theorem 2.2.5, the matrix of a shear appears as either

2.2.38 a A =

b A=

u21 u1 u2

u1 u2 , so det(A) = u21 u22 − u1 u2 u1 u2 = 0. u22

a b , so det(A) = −a2 − b2 = −(a2 + b2 ) = −1. b −a

a −b c A= , so det(A) = a2 − (−b2 ) = a2 + b2 = 1. b a

1 k d A= 0 1

or

1 k

0 , both of which have determinant equal to 12 − 0 = 1. 1

1 1 1 1 1 1 2 . The matrix 2 2 represents an orthogonal projection (Definition 2.2.39 a Note that = 2 2 1 1 1 1 1 1 2 2 2 " √2 #2 1 1 u1 represents a projection combined with a scaling by a factor of 2. = √22 . So, 2.2.1), with ~u = 1 1 u2

2

b This lookssimilar to a shear, with the one zero off the diagonal. Since the two diagonal entries are identical, we 1 0 3 0 =3 can write , showing that this matrix represents a vertical shear combined with a scaling − 13 1 −1 3 by a factor of 3. c We are asked to write " 3 4 # k

4 k 3 2 (k)

k

3 4 4 −3

=k

"

3 k

4 k

4 k

− k3

#

, with our scaling factor k yet to be determined. This matrix,

a b

b −a

has the form of a reflection matrix . This form further requires that 1 = a2 + b2 = − k3 + ( k4 )2 , or k = 5. Thus, the matrix represents a reflection combined with a scaling by a factor of 5.

2.2.40 ~x = projP ~x + projQ ~x, as illustrated in Figure 2.33. 2.2.41 refQ ~x = −refP ~x since refQ ~x, refP ~x, and ~x all have the same length, and refQ ~x and refP ~x enclose an angle of 2α + 2β = 2(α + β) = π. (See Figure 2.34.) 71

Chapter 2

Figure 2.33: for Problem 2.2.40.

Figure 2.34: for Problem 2.2.41. 2.2.42 T (~x) = T (T (~x)) since T (~x) is on L hence the projection of T (~x) onto L is T (~x) itself. 2.2.43 Since ~y = A~x is obtained from ~x by a rotation through θ in the counterclockwise direction, ~x is obtained from ~y by a rotation through θ in the clockwise direction, that is, a rotation through −θ. (See Figure 2.35.)

Figure 2.35: for Problem 2.2.43. Therefore, the matrix of the inverse transformation is A−1 = use the formula in Exercise 2.1.13b to check this result. 2.2.44 By Exercise 1.1.13b, A−1 =

a b

−b a

−1

=

1 a2 +b2

cos(−θ) − sin(−θ) cos θ sin θ = . You can sin(−θ) cos(−θ) − sin θ cos θ

a b . −b a

If A represents a rotation through θ followed by a scaling by r, then A−1 represents a rotation through −θ followed by a scaling by 1r . (See Figure 2.36.) −1

2.2.45 By Exercise 2.1.13, A

=

1 −a2 −b2

−a −b = −b a

1 −(a2 +b2 )

a b −a −b −a −b . = = −1 b −a −b a −b a

So A−1 = A, which makes sense. Reflecting a vector twice about the same line will return it to its original state. 72

Section 2.2

Figure 2.36: for Problem 2.2.44.

2.2.46 We want to write A = k

a k b k

b k − ka

, where the matrix B =

a k b k

b k − ka

represents a reflection. It is required √ a b 1 b 2 a 2 −1 2 2 2 2 2 = k12 A = k1 B, that ( k ) + ( k ) = 1, meaning that a + b = k , or, k = a + b . Now A = a2 +b2 b −a for the reflection matrix B and the scaling factor k introduced above. In summary: If A represents a reflection combined with a scaling by k, then A−1 represents the same reflection combined with a scaling by k1 .

a b ax1 + bx2 x1 . = cx1 + dx2 c d x2 −a sin t + b cos t a cos t + b sin t − sin t cos t · = · T a. f (t) = T −c sin t + d cos t c cos t + d sin t cos t sin t

2.2.47 Write T

x1 x2

=

= (a cos t + b sin t)(−a sin t + b cos t) + (c cos t + d sin t)(−c sin t + d cos t) This function f (t) is continuous, since cos(t), sin(t), and constant functions are continuous, and sums and products of continuous functions are continuous. 0 −1 0 1 π b. f 2 = T ·T =− T ·T , since T is linear. 1 0 1 0 1 0 0 1 f (0) = T ·T =T ·T . The claim follows. 0 1 1 0 c. By part (b), the numbers f (0) and f π2 have different signs (one is positive and the other negative), or they are both zero. Since f (t) is continuous, by part (a), we can apply the intermediate value theorem. (See Figure 2.37.)

Figure 2.37: for Problem 2.2.47c. d. Note that

cos(t) sin(t)

and

− sin(t) cos(t)

are perpendicular unit vectors, for any t. If we set 73

Chapter 2 − sin(c) cos(c) , with the number c we found in part (c), then f (c) = T (~v1 ) · T (~v2 ) = 0, so , ~v2 = cos(c) sin(c) that T (~v1 ) and T (~v2 ) are perpendicular, as claimed. Note that T (~v1 ) or T (~v2 ) may be zero.

~v1 =

2.2.48 We find 0 4 − sin(t) cos(t) 0 4 · f (t) = cos(t) 5 −3 sin(t) 5 −3 =

4 sin(t) 4 cos(t) · 5 cos(t) − 3 sin(t) −5 sin(t) − 3 cos(t)

= 15(sin2 t − cos2 t) = 15(2 sin2 t − 1). See Figure 2.38.

Figure 2.38: for Problem 2.2.48. The only zero of f (t) between 0 and π2 is at c = π4 . √ # √2 " π 2 − 2 − sin( ) cos( π4 ) 4 = √ work. Note that T (~v1 ) = Therefore, ~v1 = = √2 and ~v2 = π π 2 2 sin( 4 ) cos( 4 ) 2

and T (~v2 ) =

√1 2

4 −8

2

are indeed perpendicular. See Figure 2.39.

Figure 2.39: for Problem 2.2.48.

cos(t) 2.2.49 If ~x = sin(t)

5 0 then T (~x) = 0 2

cos(t) 5 cos(t) 5 0 = = cos(t) + sin(t) . sin(t) 2 sin(t) 0 2 74

√1 2

4 2

Section 2.2 Thesevectors form an consider the characterization of an ellipse given in the footnote on Page 69, with ellipse; 0 5 . (See Figure 2.40.) and w ~2 = w ~1 = 2 0

Figure 2.40: for Problem 2.2.49. 2.2.50 Use the hint: Since the vectors on the unit circle are of the form ~v = cos(t)~v1 + sin(t)~v2 , the image of the unit circle consists of the vectors of the form T (~v ) = T (cos(t)~v1 + sin(t)~v2 ) = cos(t)T (~v1 ) + sin(t)T (~v2 ).

75

Chapter 2

Figure 2.41: for Problem 2.2.50. These vectors form an ellipse: Consider the characterization of an ellipse given in the footnote, with w ~ 1 = T (~v1 ) and w ~ 2 = T (~v2 ). The key point is that T (~v1 ) and T (~v2 ) are perpendicular. See Figure 2.41. 2.2.51 Consider the linear transformation T with matrix A = [w ~1 x x x1 ~ 1 + x2 w ~ 2. ~1 w ~ 2 ] 1 = x1 w = A 1 = [w T x2 x2 x2

w ~ 2 ], that is,

cos(t) is on the unit circle, sin(t) then T (~v ) = cos(t)w ~ 1 + sin(t)w ~ 2 is on the curve C. Therefore, C is an ellipse, by Exercise 50. (See Figure 2.42.) The curve C is the image of the unit circle under the transformation T : if ~v =

Figure 2.42: for Problem 2.2.51. 2.2.52 By definition, the vectors ~v on an ellipse E are of the form ~v = cos(t)~v1 + sin(t)~v2 , for some perpendicular vectors ~v1 and ~v2 . Then the vectors on the image C of E are of the form T (~v ) = cos(t)T (~v1 ) + sin(t)T (~v2 ). These vectors form an ellipse, by Exercise 51 (with w ~ 1 = T (~v1 ) and w ~ 2 = T (~v2 )). See Figure 2.43.

76

Section 2.3

Figure 2.43: for Problem 2.2.52.

Section 2.3

2.3.1

4 6 3 4

2.3.2

4 4 −8 −8

2.3.3 Undefined

2 2.3.4 2 7

2 0 4

a b 2.3.5 c d 0 0 2.3.6

ad − bc 0 0 ad − bc

−1 2.3.7 5 −6

1 0 3 4 −2 −4

2.3.8

0 0 0 0

2.3.9

0 0 0 0

2.3.10 [0 1] 2.3.11 [10] 77

Chapter 2 3 6 9

1 2 2.3.12 2 4 3 6 2.3.13 [h]

−2 2 2 2.3.14 A2 = , BC = [14 8 2], BD = [6], C 2 = 4 2 2 10 5 DE = 5 , EB = [5 10 15], E 2 = [25] 5

3 3 , 3

1 2 0 −2 −2 1 −2 , CD = 3 , DB = 1 2 1 2 6 4 −2

"

" # " # #" # " # # " # " # " 1 0 0 0 0 1 1 0 1 0 + + 1 0 [ 4 ] [ 3 ] 0 1 0 0 2 2 0 = " # 0 1" #0 2.3.15 = 2 0 1 0 +[ 4 ][ 3 ] [ 1 3 ] +[ 4 ][ 4 ] [1 3] 19 16 [ 19 ] [ 16 ] 2 0 "

1 0 2.3.16 " 0 0

0 1# 0 0

1 3 " 1 3

# "

2 1 + 4# " 0 2 1 + 4 0 # "

0 0 1 #" 0 0 0 1 0 #"

# " 0 1 0 # " 0 0 0 0 0

0 2 1 #" 4 0 2 0 4 #"

2.3.17 We must find all S such that SA = AS, or

3 1 + 5# " 0 3 1 + 5 0 # "

a b c d

0 1 1 #" 3 0 1 1 3 #"

" # 2 1 4# = " 3 2 0 0 4

1 1 0 = 0 0 2

0 2

2 4# 0 0

3 7 " 1 3

# "

# 1 5 3 9# = 2 0 4 0

2 4 0 0

3 7 1 3

5 9 2 4

a b . c d

a 2b a = c 2d 2c

b , meaning that b = 2b and c = 2c, so b and c must be zero. 2d 1 0 a 0 . ) commute with We see that all diagonal matrices (those of the form 0 2 0 d So

a b 1 2 1 2 a b a b . = . Now we want 2.3.18 As in Exercise 2.3.17, we let A = c d 0 1 0 1 c d c d a + 2c b + 2d a 2a + b , revealing that c = 0 (since a + 2c = a) and a = d (since b + 2d = 2a + b). = So, c d c 2c + d a b . Thus B is any matrix of the form 0 a

a b 0 −2 0 −2 a b a b . = . We want c d 2 0 2 0 c d c d 2b −2a −2c −2d Thus, = , meaning that c = −b and d = a. 2d −2c 2a 2b 0 −2 a b . commute with We see that all matrices of the form 2 0 −b a

2.3.19 Again, let A =

78

Section 2.3

2.3.20 Following the form of Exercise 17, we let A =

Now we want

a b c d

2 3 2 = −3 2 −3

3 2

a b . c d

a b . c d

2a − 3b 3a + 2b 2a + 3c 2b + 3d So, = , revealing that a = d (since 3a + 2b = 2b + 3d) and −b = 2c − 3d 3c + 2d −3a + 2c −3b + 2d c (since 2a + 3c = 2a − 3b). a b Thus B is any matrix of the form . −b a 2.3.21 Now we want

a b c d

2 1 2 a b = . −1 2 −1 c d

1 2

a + 2c b + 2d a + 2b 2a − b . So a + 2b = a + 2c, or c = b, and 2a − b = b + 2d, revealing = Thus, 2a − c 2b − d c + 2d 2c − d d = a − b. (The other two equations are redundant.) 1 2 a b . commute with All matrices of the form 2 −1 b a−b

a b a b 1 1 1 1 a b 2.3.22 As in Exercise 17, we let A = . Now we want = . c d c d 1 1 1 1 c d So,

a+c a+b a+b = a+c c+d c+d

b+d , revealing that a = d (since a + b = b + d) and b = c (since a + c = a + b). b+d

Thus B is any matrix of the form

2.3.23 We want

a c

b d

a b . b a

a b 1 3 1 3 . = c d 2 6 2 6

a + 2b 3a + 6b a + 3c b + 3d Then, = . So a + 2b = a + 3c, or c = 23 b, and 3a + 6b = b + 3d, revealing c + 2d 3c + 6d 2a + 6c 2b + 6d d = a + 35 b. The other two equations are redundant. Thus all matrices of the form

a b 5 2 b a + 3 3b

commute with

a b 2.3.24 Following the form of Exercise 2.3.17, we let A = d e g h

a b Now we want d e g h

c 2 0 f 0 3 i 0 0

2 0 0 0 = 0 3 0 0 4

0 a b 0d e 4 g h 79

1 2

3 . 6

c f . i

c f . i

Chapter 2 2a 2a 3b 4c So, 2d 3e 4f = 3d 4g 2g 3h 4i chosen freely.

2b 3e 4h

2c 3f , which forces b, c, d, f, g and h to be zero. a, e and i, however, can be 4i

a Thus B is any matrix of the form 0 0

0 0 e 0. 0 i

0 2 0 = 0 2 0

a 2.3.25 Now we want d g 2a 3b 2c or, 2d 3e 2f = 2g 3h 2i must all be zero.

b c 2 0 e f 0 3 h i 0 0 2a 2b 2c 3d 3e 3f . 2g 2h 2i

a 0 Thus all matrices of the form 0 e g 0

0 0 a b 3 0d e 0 2 g h

c f , i

So, 3b = 2b, 2d = 3d, 3f = 2f and 3h = 2h, meaning that b, d, f and h

2 0 0 c 0 commute with 0 3 0 . 0 0 2 i

a 2.3.26 Following the form of Exercise 2.3.17, we let A = d g a 2 0 0 2 0 0 a b c Then we want d e f 0 2 0 = 0 2 0 d g 0 0 3 0 0 3 g h i 2a 2b 2c 2a 2b 3c So, 2d 2e 3f = 2d 2e 2f . Thus c, f, g and h 3g 3h 3i 2g 2h 3i a b 0 d e 0. 0 0 i

c f . i b c e f . h i

b e h

must be zero, leaving B to be any matrix of the form

2.3.27 We will prove that A(C + D) = AC + AD, repeatedly using Theorem 1.3.10a: A(~x + ~y ) = A~x + A~y . Write B = [~v1 . . . ~vm ] and C = [w ~1 . . . w ~ m ]. Then A(C + D) = A[~v1 + w ~ 1 · · · ~vm + w ~ m ] = [A~v1 + Aw ~ 1 · · · A~vm + Aw ~ m ], and AC + AD = A[~v1 · · · ~vm ] + A[w ~1 · · · w ~ m ] = [A~v1 + Aw ~ 1 · · · A~vm + Aw ~ m ]. The results agree. 2.3.28 The ijth entries of the three matrices are p X

h=1

.

(kaih )bhj ,

p X

aih (kbhj ), and k

h=1

p X

h=1

80

aih bhj

!

Section 2.3 The three results agree. 2.3.29 a Dα Dβ and Dβ Dα are the same transformation, namely, a rotation through α + β.

cos α − sin α b Dα Dβ = sin α cos α

cos β − sin β sin β cos β

=

cos α cos β − sin α sin β sin α cos β + cos α sin β

=

cos(α + β) sin(α + β)

− cos α sin β − sin α cos β − sin α sin β + cos α cos β

− sin(α + β) cos(α + β)

Dβ Dα yields the same answer. 2.3.30 a See Figure 2.44.

Figure 2.44: for Problem 2.4.30. The vectors ~x and T (~x) have the same length (since reflections leave the length unchanged), and they enclose an angle of 2(α + β) = 2 · 30◦ = 60◦ b Based on the answer in part (a), we conclude that T is a rotation through 60◦ .

c The matrix of T is

cos(60◦ ) sin(60◦ )

1 − sin(60◦ ) = √2 ◦ 3 cos(60 )

√

−

2

3 2 1 2

.

w ~1 ~ w 2.3.31 Write A in terms of its rows: A = 2 (suppose A is n × m). ··· w ~n

81

Chapter 2 We can think of this as a partition into n w ~ 1B w ~1 ~ B ~ w w 1 × m matrices. Now AB = 2 B = 2 (a product of partitioned matrices). ··· ··· w ~ nB w ~n

We see that the ith row of AB is the product of the ith row of A and the matrix B.

a b 1 0 1 0 a b 1 0 1 0 a b 2.3.32 Let X = . Then we want X = X, or = , or cd 0 0 0 0 c d 0 0 0 0 c d a 0 a b 0 1 0 1 a 0 0 1 = , meaning that b = c = 0. Also, we want X = X, or = c 0 0 0 0 0 0 0 0 d 0 0 0 1 a 0 0 a 0 d a 0 , or = so a = d. Thus, X = = aI2 must be a multiple of the identity 0 0 0 d 0 0 0 0 0 a matrix. (X will then commute with any 2 × 2 matrix M , since XM = aM = M X.) 2.3.33 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between A = −I2 and I2 . The matrix A describes a reflection about the origin. Alternatively one can say A represents a rotation by 180◦ = π. Since A2 is the −1 0 identity, A1000 is the identity and A1001 = A = . 0 −1 2.3.34 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between A and I2 . The matrix a reflection A describes 1 0 2 1000 1001 . about the x axis. Because A is the identity, A is the identity and A =A= 0 −1 2.3.35 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between A and I2 . The matrix A describes a reflection 0 1 2 1000 1001 about the diagonal x = y. Because A is the identity, A is the identity and A =A= . 1 0

1 0

1 0 −2 1

1 3 ,A = 0

2.3.36 A = and A = . The power An represents a horizontal shear along the 1 1001 x-axis. The shear strength increases linearly in n. We have A1001 = . 0 1 2

2 1

3 1

4

1 0

4 1

1 0 1 0 , A3 = and A4 = . The power An represents a vertical shear along −3 1 −4 1 1 0 the y axis. The shear magnitude increases linearly in n. We have A1001 = . −1001 1

2.3.37 A2 =

−1 0 , A3 = −A, A4 = I2 . The matrix A represents the rotation through π/2 in the 0 −1 counterclockwise Since A4 is the identity matrix, we know that A1000 is the identity matrix and direction. 0 −1 A1001 = A = . 1 0

2.3.38 A2 =

−1 1 0 1 , A4 = −I2 . The matrix A describes a rotation by π/4 in the , A3 = √12 −1 1 −1 0 clockwise direction. Because A8 is the identity matrix, we know that A1000 is the identity matrix and A1001 =

2.3.39 A2 =

82

Section 2.3 √ A = (1/ 2)

1 −1

1 1

.

√ −1 3 √ , A3 = I2 , A4 = A. The matrix A describes a rotation by 120◦ = 2π/3 in the − 3 −1 3 999 counterclockwise direction. Because is the identity matrix and √ A is the identity matrix, we know that A 3 −1 √ . A1001 = A2 = A−1 = 21 − 3 −1

2.3.40 A2 =

1 2

2.3.41 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between I2 for even n and A for odd n. Therefore A1001 = A. The matrix represents a reflection about a line. 2.3.42 An = A. The matrix A represents a projection on the line x = y spanned by the vector 1 1 A1001 = A = (1/2) . 1 1 2.3.43 An example is A =

1 0

0 −1

−1 √ 3

0 −1 1 0

√ − 3 , the rotation through 2π/3. See Problem 2.3.40. −1

2.3.46 For example, A =

1 1

1 1

1 , the orthogonal projection onto the line spanned by . 1

2.3.47 For example, A =

1 2

1 1

1 1

1 . , the orthogonal projection onto the line spanned by 1

. We have

.

1 2

2.3.48 For example, the shear A =

1 1

, representing the reflection about the horizontal axis.

2.3.44 A rotation by π/2 given by the matrix A =

2.3.45 For example, A = (1/2)

1 1/10 0 1

.

1 0 −1 0 represents the reflection about the x-axis, while F A = represents the reflection 0 −1 0 1 about the y-axis. (See Figure 2.45.)

2.3.49 AF =

0 0 1 represents a reflection about the line x = y, while GC = −1 1 0 about the line x = −y. (See Figure 2.46.)

2.3.50 CG =

2.3.51 F J = JF = Figure 2.47.)

−1 −1 1 −1

−1 0

represents a reflection

both represent a rotation through 3π/4 combined with a scaling by

83

√ 2. (See

Chapter 2

A

F

AF

A

F

FA

Figure 2.45: for Problem 2.3.49. 0.2 −1.4 . Since H represents a rotation and J represents a rotation through π/4 combined 2.3.52 JH = HJ = 1.4 0.2 √ with a scaling √ by 2, the products in either order will be the same, representing a rotation combined with a scaling by 2. (See Figure 2.48.)

0 −1 0 1 2.3.53 CD = represents the rotation through π/2, while DC = represents the rotation 1 0 −1 0 through −π/2. (See Figure 2.49.)

−0.6 −0.8 2.3.54 BE = represents the rotation through the angle θ = arccos(−0.6) ≈ 2.21, while EB = 0.8 −0.6 −0.6 0.8 represents the rotation through −θ. (See Figure 2.50.) −0.8 −0.6 2.3.55 We need to solve the matrix equation

1 2

2 4

a b c d 84

=

0 0 0 0

,

Section 2.3

C

G

CG

C

G

GC

Figure 2.46: for Problem 2.3.50. which amounts to solving the system a + 2c = 0, 2a + 4c = 0, b + 2d = 0 and 2b + 4d = 0. The solutions are of −2c −2d , where c, d are arbitrary constants. the form a = −2c and b = −2d. Thus X = c d 2.3.56 Proceeding as in Exercise 55, we find X = 2.3.57 We need to solve the matrix equation

1 3

2 5

, where b and d are arbitrary.

−2b b −2d d

a b c d

=

1 0 0 1

,

which amounts to solving the system a + 2c = 1, 3a + 5c = 0, b + 2d = 0 and 3b + 5d = 1. The solution is −5 2 X= . 3 −1 2.3.58 Proceeding as in Exercise 57, we find X =

−5 2 3 −1 85

.

Chapter 2

F

J

and FJ

and F

J

JF

Figure 2.47: for Problem 2.3.51. 2.3.59 The matrix equation

a c

b d

2 1 4 2

=

1 0 0 1

has no solutions, since we have the inconsistent equations 2a + 4b = 1 and a + 2b = 0. 2.3.60 Proceeding as in Exercise 59, we find that this equation has no solutions. 2.3.61 We need to solve the matrix equation

1 2 0 1

3 2

a b c d = 1 0 e f

0 1

,

which amounts to solving the system a+ 2c + 3e = 0, c + 2e = 0, b + 2d + 3f = 0 and d + 2f = 1. The solutions e+1 f −2 are of the form X = −2e 1 − 2f , where e, f are arbitrary constants. e f 86

Section 2.3

37 ◦

H

J

and HJ

and H

J

37 ◦

JH

Figure 2.48: for Problem 2.3.52.

e − 5/3 2.3.62 Proceeding as in Exercise 61, we find X = −2e + 4/3 e

f + 2/3 −2f − 1/3 , where e, f are arbitrary constants. f

2.3.63 The matrix equation

1 4 2 5 a b d e 3 6

c f

1 0 = 0 1 0 0

0 0 1

has no solutions, since we have the inconsistent equations a + 4d = 1, 2a + 5d = 0, and 3a + 6d = 0. 2.3.64 The matrix equation

1 0 2 1 a b d e 3 2

c f

1 0 = 0 1 0 0

0 0 1

has no solutions, since we have the inconsistent equations a = 1, 2a + d = 0 and 3a + 2d = 0. 87

Chapter 2

C

D

CD

C

D

DC

Figure 2.49: for Problem 2.3.53.

2 2.3.65 With X = , we have to solve X = 0 can be arbitrary. The general solution is X = 0 a b 0 c

a2 ab + bc 0 c2 b . 0

=

0 0 0 0

. This means a = 0, c = 0 and b

0 0 c 0 then the diagonal entries of X 3 will be a3 , c3 , and f 3 . Since we want X 3 = 0, we must e f 0 0 0 have a = c = f = 0. If X = b 0 0 , then a direct computation shows that X 3 = 0. Thus the solutions d e 0 0 0 0 are of the form X = b 0 0 , where b, d, e are arbitrary. d e 0

a 2.3.66 If X = b d

2.3.67 For a horizontal shear, A =

1 k 0 1

0 2 , we have (A − I2 ) = 0 88

k 0

2

=

0 0 0 0

. Note that A~x − ~x =

Section 2.4

B

E

BE

B

E

EB

Figure 2.50: for Problem 2.3.54. A2 ~x − A~x for all vectors ~x, as illustrated in the accompanying figure. This equation means that A2 ~x − 2A~x + ~x = (A − I2 )2 ~x = ~0. Analogous results hold for vertical shears. x

A2 x

2.3.68 Let ~v1 , . . . , ~vn be the columns of the matrix X. Solving the matrix equation AX = In amounts to solving the linear systems A~vi = ~ei for i = 1, . . . , n. Since A is a n × m matrix of rank n, all these systems are consistent, so that the matrix equation AX = In does have at least one solution. If n < m, then each of the systems A~vi = ~ei has infinitely many solutions, so that the matrix equation AX = In has infinitely many solutions as well. See the examples in Exercices 2.3.57,2.3.61 and 2.3.62. 2.3.69 Let ~v1 , . . . , ~vn be the columns of the matrix X. Solving the matrix equation AX = In amounts to solving the linear systems A~vi = ~ei for i = 1, . . . , n. Since A is an n × n matrix of rank n, all these systems have a unique solution, by Theorem 1.3.4, so that the matrix equation AX = In has a unique solution as well.

89

Chapter 2

Section 2.4 .. 2.4.1 rref 2 3.. 5 8.. "

1

0

0

1

.. 2.4.2 rref 1 1.. 1 1..

1

0

0

1

.. . 0 2 2.4.3 rref . 1 1..

1

0

0

1

"

"

#

. 1 0.. = . 0 1..

#

. 1 1.. = . 0 0..

#

. 1 0.. = . 0 1..

8

1 2.4.5 rref 1 1

2 2 1 3 1 = 0 1 3 0

−3 , so that 2

−5

0

1 , so that −1

1 − 21

1

1 2

0

2.4.4 Use Theorem 2.4.5; the inverse is

1 2 3 0 2 = 0 0 0 3

0 1

0 0 2.4.8 Use Theorem 2.4.5; the inverse is 0 1 1 0 1 2.4.9 rref 1 1

1 1 1 1 1 = 0 1 1 0

−1

fails to be invertible.

−1

=

=

8 −3 . −5 2

− 21 1 2

1 0

.

.

1 −2 . 1

2 0 0 1 , so that the matrix fails to be invertible, by Theorem 2.4.3. 0 0

0 2 1 1

1 2 1 −2 1 2

−1

1 −2 1 2.4.6 Use Theorem 2.4.5; the inverse is 0 0 0 1 2.4.7 rref 0 0

1 1

3 8

0 4 1 −1 , so that the matrix fails to be invertible, by Theorem 2.4.3. 0 0

1 1

, so that

3 2 1 2 − 32

2 5

1 0 . 0

1 1 0 0 , so that the matrix fails to be invertible, by Theorem 2.4.3. 0 0

3 2.4.10 Use Theorem 2.4.5; the inverse is −3 1

1 2.4.11 Use Theorem 2.4.5; the inverse is 0 0

−3 1 5 −2 . −2 1 0 −1 1 0 . 0 1 90

Section 2.4 5 0 2.4.12 Use Theorem 2.4.5; the inverse is −2 0

−20 −1 6 3

1 −2 2.4.13 Use Theorem 2.4.5; the inverse is 1 0

0 0 1 0 −2 1 1 −2

0 0 . 0 1

3 −1 2.4.14 Use Theorem 2.4.5; the inverse is 0 0

−5 0 2 0 0 5 0 −2

0 0 . −2 1

−6 9 2.4.15 Use Theorem 2.4.5; the inverse is −5 1

9 −5 −1 −5 −5 9 2 −3

1 2 −3 1

−2 −7 0 0 1 2 0 1

2.4.16 Solving for x1 and x2 in terms of y1 and y2 we find that x1 x2

= −8y1 + 5y2 = 5y1 − 3y2

2.4.17 We make an attempt to solve for x1 and x2 in terms of y1 and y2 : x1 + 2x2 = y1 x1 + 2x2 = y1 −−−→ . 0 = −4y1 + y2 4x1 + 8x2 = y2 −4(I) This system has no solutions (x1 , x2 ) for some (y1 , y2 ), and infinitely many solutions for others; the transformation fails to be invertible. 2.4.18 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 we find that x1 x2 x3

= y3 = y1 = y2

2.4.19 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 , we find that x1 x2 x3

= 3y1 − 52 y2 + 21 y3

= −3y1 + 4y2 − y3 = y1 − 32 y2 + 21 y3

2.4.20 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 we find that x1 x2 x3

= −8y1 − 15y2 + 12y3 = 4y1 + 6y2 − 5y3 = −y1 − y2 + y3

2.4.21 f (x) = x2 fails to be invertible, since the equation f (x) = x2 = 1 has two solutions, x = ±1. 91

Chapter 2 2.4.22 f (x) = 2x fails to be invertible, since the equation f (x) = 2x = 0 has no solution x. 2.4.23 Note that f ′ (x) = 3x2 + 1 is always positive; this implies that the function f (x) = x3 + x is increasing throughout. Therefore, the equation f (x) = b has at most one solution x for all b. (See Figure 2.51.) Now observe that limx→∞ f (x) = ∞ and limx→−∞ f (x) = −∞; this implies that the equation f (x) = b has at least one solution x for a given b (for a careful proof, use the intermediate value theorem; compare with Exercise 2.2.47c).

Figure 2.51: for Problem 2.3.23. 2.4.24 We can write f (x) = x3 − x = x(x2 − 1) = x(x − 1)(x + 1). The equation f (x) = 0 has three solutions, x = 0, 1, −1, so that f (x) fails to be invertible. 2.4.25 Invertible, with inverse

x1 x2

=

√ 3 y 1 y2

2.4.26 Invertible, with inverse

x1 x2

=

√ 3 y2 − y1 y1

2.4.27 This transformation fails to be invertible, since the equation

x1 + x2 x1 x2

22 13 −16 −3 2.4.28 We are asked to find the inverse of the matrix A = 8 9 5 4 1 −2 9 − 25 5 −22 60 −2 −1 We find that A = . 4 −9 41 −112 −9 17 80 222

1 1 −II → 0 k−1 0 k 2 − 1 −3(II)

0 2−k 1 k−1 0 k 2 − 3k + 2 92

0 has no solution. = 1

8 3 −2 −2 . 7 2 3 1

T −1 is the transformation from R4 to R4 with matrix A−1 .

2.4.29 Use Theorem 2.4.3: 1 1 1 1 1 1 2 k −I → 0 1 1 4 k 2 −I 0 3

Section 2.4 The matrix is invertible if (and only if) k 2 − 3k + 2 = (k − 2)(k − 1) 6= 0, in which case we can further reduce it to I3 . Therefore, the matrix is invertible if k 6= 1 and k 6= 2. 2.4.30 Use Theorem 2.4.3: −1 0 1 b −−−−→ −1 0 c I ↔ II 0 −b −b −c 0

0 1 −c

−−−−→ 1 c ÷(−1) 0 b −b 0

1 0 −c −→ 0 1 b 0 0 0 +b(I) + c(II)

0 1 −c

This matrix fails to be invertible, regarless of the values of b and c.

−c b 0

2.4.31 Use Theorem 2.4.3; first assume that a 6= 0.

0 −a −b

a 0 −c

0

− ac

1

0

a b ÷a → 0 −c − bc 0 a

1

1

0 0

−a b swap : c → 0 I ↔ II −b 0

−c

1 0 c ÷(−a) → 0 a b −c 0 −b − ac

b a − bc a

Now consider the case when a = 0: −b −c 0 0 b swap : 0 → 0 0 0 c I ↔ III 0 0 −b −c 0

0 It follows that the matrix −a −b

+c(II)

1

→ 0 0

− ac

0

→ b +b(I) 0

a −c

0 − ac b a

1 0

0

0 c : The second entry on the diagonal of rref will be 0. b

a b 0 c fails to be invertible, regardless of the values of a, b, and c. −c 0

2.4.32 Use Theorem 2.4.9. a b If A = is a matrix such that ad − bc = 1 and A−1 = A, then c d A−1 =

1 ad−bc

d −c

−b d = a −c

−b a b = , so that b = 0, c = 0, and a = d. a c d

The condition ad − bc = a2 = 1 now implies that a = d = 1 or a = d = −1. This leaves only two matrices A, namely, I2 and −I2 . Check that these two matrices do indeed satisfy the given requirements. 2.4.33 Use Theorem 2.4.9. 1 The requirement A−1 = A means that − a2 +b 2

a2 + b2 = 1.

−a −b 93

−b a

=

a b

b . This is the case if (and only if) −a

Chapter 2 2.4.34 a By Theorem 2.4.3, A is invertible if (and only if) a, b, and c are all nonzero. 1 0 0 a 0 1b 0 . 0

0

In this case, A−1 =

1 c

b In general, a diagonal matrix is invertible if (and only if) all of its diagonal entries are nonzero. 2.4.35 a A is invertible if (and only if) all its diagonal entries, a, d, and f , are nonzero. b As in part (a): if all the diagonal entries are nonzero. . c Yes, A−1 will be upper triangular as well; as you construct rref[A..In ], you will perform only the following row operations: • divide rows by scalars • subtract a multiple of the jth row from the ith row, where j > i. Applying these operations to In , you end up with an upper triangular matrix. d As in part (b): if all diagonal entries are nonzero. 2.4.36 If a matrix A can be transformed into B by elementary row operations, then A is invertible if (and only if) B is invertible. The claim now follows from Exercise 35, where we show that a triangular matrix is invertible if (and only if) its diagonal entries are nonzero. 2.4.37 Make an attempt to solve the linear equation ~y = (cA)~x = c(A~x) for ~x: A~x = 1c ~y , so that ~x = A−1 1c ~y = 1c A−1 ~y . This shows that cA is indeed invertible, with (cA)−1 = 1c A−1 .

2.4.38 Use Theorem 2.4.9; A−1 =

1 −1

−1 −k 1 = 0 1 0

k (= A). −1

2.4.39 Suppose the ijth entry of M is k, and all other entries are as in the identity matrix. Then we can find . rref[M ..In ] by subtracting k times the jth row from the ith row. Therefore, M is indeed invertible, and M −1 differs from the identity matrix only at the ijth entry; that entry is −k. (See Figure 2.52.) 2.4.40 If you apply an elementary row operation to a matrix with two equal columns, then the resulting matrix will also have two equal columns. Therefore, rref(A) has two equal columns, so that rref(A) 6= In . Now use Theorem 2.4.3. 2.4.41 a Invertible: the transformation is its own inverse. b Not invertible: the equation T (~x) = ~b has infinitely many solutions if ~b is on the plane, and none otherwise. 94

Section 2.4

Figure 2.52: for Problem 2.3.39. c Invertible: The inverse is a scaling by

1 5

(that is, a contraction by 5). If ~y = 5~x, then ~x = 51 ~y .

d Invertible: The inverse is a rotation about the same axis through the same angle in the opposite direction. 2.4.42 Permutation matrices are invertible since they row reduce to In in an obvious way, just by row swaps. The . . inverse of a permutation matrix A is also a permutation matrix since rref[A..In ] = [In ..A−1 ] is obtained from . [A..In ] by a sequence of row swaps. 2.4.43 We make an attempt to solve the equation ~y = A(B~x) for ~x: B~x = A−1 ~y , so that ~x = B −1 (A−1 ~y ). 1 0 2.4.44 a rref(M4 ) = 0 0

0 −1 −2 1 2 3 , so that rank(M4 ) = 2. 0 0 0 0 0 0

b To simplify the notation, we introduce the row vectors ~v = [1 1 . . . 1] and w ~ = [0 n 2n . . . (n − 1)n] with n components. ~v + w ~ ~ −2(I) 2~v + w Then we can write Mn in terms of its rows as Mn = . ··· ... n~v + w ~ −n(I) ~v + w ~ −w ~ Applying the Gauss-Jordan algorithm to the first column we get −2w ~ . ... −(n − 1)w ~ All the rows below the second are scalar multiples of the second; therefore, rank(Mn ) = 2. c By part (b), the matrix Mn is invertible only if n = 1 or n = 2. 2.4.45 a Each of the three row divisions requires three multiplicative operations, and each of the six row subtractions requires three multiplicative operations as well; altogether, we have 3 · 3 + 6 · 3 = 9 · 3 = 33 = 27 operations. 95

Chapter 2 . b Suppose we have already taken care of the first m columns: [A..In ] has been reduced the matrix in Figure 2.53.

Figure 2.53: for Problem 2.3.45b. Here, the stars represent arbitrary entries. Suppose the (m+1)th entry on the diagonal is k. Dividing the (m+1)th row by k requires n operations: n−m−1 to the left of the dotted line not counting the computation kk = 1 , and m + 1 to the right of the dotted line including k1 . Now the matrix has the form shown in Figure 2.54.

Figure 2.54: for Problem 2.4.45b. Eliminating each of the other n − 1 components of the (m + 1)th column now requires n multiplicative operations (n − m − 1 to the left of the dotted line, and m + 1 to the right). Altogether, it requires n + (n − 1)n = n2 operations to process the mth column. To process all n columns requires n · n2 = n3 operations. c The inversion of a 12 × 12 matrix requires 123 = 43 33 = 64 · 33 operations, that is, 64 times as much as the inversion of a 3 × 3 matrix. If the inversion of a 3 × 3 matrix takes one second, then the inversion of a 12 × 12 matrix takes 64 seconds. 2.4.46 Computing A−1~b requires n3 + n2 operations: First, we need n3 operations to find A−1 (see Exercise 45b) and then n2 operations to compute A−1~b (n multiplications for each component). . How many operations are required to perform Gauss-Jordan eliminations on [A..~b]? Let us count these operations “column by column.” If m columns of the coefficient matrix are left, then processing the next column requires nm operations (compare with Exercise 45b). To process all the columns requires = n · n + n(n − 1) + · · · + n · 2 + n · 1 = n(n + n − 1 + · · · + 2 + 1) = n n(n+1) 2 96

n3 +n2 2

operations.

Section 2.4 only half of what was required to compute A−1~b. We mention in passing that one can reduce the number of operations further (by about 50% for large matrices) by performing the steps of the row reduction in a different order. 2.4.47 Let f (x) = x2 ; the equation f (x) = 0 has the unique solution x = 0. 2.4.48 Consider the linear system A~x = ~0. The equation A~x = ~0 implies that BA~x = ~0, so ~x = ~0 since BA = Im . Thus the system A~x = ~0 has the unique solution ~x = ~0. This implies m ≤ n, by Theorem 1.3,3. Likewise the linear system B~y = ~0 has the unique solution ~y = ~0, implying that n ≤ m. It follows that n = m, as claimed.

0.293 0 0 0.707 2.4.49 a A = 0.014 0.207 0.017 , I3 − A = −0.014 0.044 0.01 0.216 −0.044 1.41 0 0 0.0274 (I3 − A)−1 = 0.0267 1.26 0.0797 0.0161 1.28

0 −0.017 0.784

0 0.793 −0.01

1.41 1 b We have ~b = 0 , so that ~x = (I3 − A)−1~e1 = first column of (I3 − A)−1 ≈ 0.0267 . 0.0797 0 c As illustrated in part (b), the ith column of (I3 − A)−1 gives the output vector required to satisfy a consumer demand of 1 unit on industry i, in the absence of any other consumer demands. In particular, the ith diagonal entry of (I3 − A)−1 gives the output of industry i required to satisfy this demand. Since industry i has to satisfy the consumer demand of 1 as well as the interindustry demand, its total output will be at least 1. d Suppose the consumer demand increases from ~b to ~b + ~e2 (that is, the demand on manufacturing increases by one unit). Then the output must change from (I3 − A)−1~b to (I3 − A)−1 (~v + ~e2 ) = (I3 − A)−1~b + (I3 − A)−1~e2 = (I3 − A)−1~b+ (second column of (I3 − A)−1 ). The components of the second column of (I3 −A)−1 tells us by how much each industry has to increase its output. e The ijth entry of (In − A)−1 gives the required increase of the output xi of industry i to satisfy an increase of the consumer demand bj on industry j by one unit. In the language of multivariable calculus, this quantity is ∂xi ∂bj . 2.4.50 Recall that 1 + k + k 2 + · · · =

1 1−k .

The top left entry of I3 − A is I − k, and the top left entry of (I3 − A)−1 will therefore be

1 − k ∗ ∗

0 0 ∗ ∗ ∗ ∗

.. . 1 .. . 0 .. . 0

0 0 ÷(1 − k) −→ 1 0 0 1

1 0

∗ ∗ ∗ ∗

. 0 .. . ∗ .. . ∗ ..

1 1−k

0 0

→ . . . (first row will remain unchanged). 97

0 0

1 0 0 1

1 1−k ,

as claimed:

Chapter 2 In terms of economics, we can explain this fact as follows: The top left entry of (I3 − A)−1 is the output of industry 1 (Agriculture) required to satisfy a consumer demand of 1 unit on industry 1. Producting this one unit to satisfy the consumer demand will generate an extra demand of k = 0.293 units on industry 1. Producting these k units in turn will generate an extra demand of k · k = k 2 units, and so forth. We are faced with an infinite series of (ever smaller) demands, 1 + k + k 2 + · · · . 2.4.51 a Since rank(A)< n, the matrix E =rref(A) will not have a leading one in the last row, and all entries in the last row of E will be zero. 0 0 . . Let ~c = . . Then the last equation of the system E~x = ~c reads 0 = 1, so this system is inconsistent. 0 1 .. ~ Now, we can “rebuild” b from ~c by performing the reverse row-operations in the opposite order on E .~c until . we reach A..~b . Since E~x = ~c is inconsistent, A~x = ~b is inconsistent as well. b Since rank(A)≤ min(n, m), and m < n, rank(A) < n also. Thus, by part a, there is a ~b such that A~x = ~b is inconsistent. 0 0 . 0 0 2.4.52 Let ~b = . Then A..~b = 1 0 0 1 an inconsistency in the third row. 2.4.53 a A − λI2 =

.. . .. . .. . .. .

1 2 2 4 3 6 4 8

0

1 0

. 0 . We find that rref A..~b = 0 1 0 0 1 0 0

0

. 0 .. 0 . 2 .. 0 , which has . 0 .. 1 . 0 .. 0

1 . 5−λ

3−λ 3

This fails to be invertible when (3 − λ)(5 − λ) − 3 = 0, or 15 − 8λ + λ2 − 3 = 0, or 12 − 8λ + λ2 = 0 or (6 − λ)(2 − λ) = 0. So λ = 6 or λ = 2. b For λ = 6, A − λI2 =

−3 1 . 3 −1

The system (A − 6I2 )~x = ~0 has the solutions

1 t , for example. , where t is an arbitrary constant. Pick ~x = 3 3t

1 1 For λ = 2, A − λI2 = . 3 3 98

Section 2.4 The system (A − 2I2 )~x = ~0 has the solutions example. c For λ = 6, A~x =

3 3

For λ = 2, A~x =

3 3

1 t , for , where t is an arbitrary constant. Pick ~x = −1 −t

1 6 1 . =6 = 3 18 3 1 1 2 1 = =2 . 5 −1 −2 −1 1 5

1−λ 10 2.4.54 A − λI2 = . This fails to be invertible when det(A − λI2 ) = 0, −3 12 − λ so 0 = (1 − λ)(12 − λ) + 30 = 12 − 13λ + λ2 + 30 = λ2 − 13λ + 42 = (λ − 6)(λ − 7). In order for this to be zero, λ must be 6 or 7. −5 10 . We solve the system (A − 6I2 ) ~x = ~0 and find that the solutions are of the If λ = 6, then A − 6I2 = −3 6 2t 2 form ~x = . For example, when t = 1, we find ~x = . t 1 −6 10 If λ = 7, then A − 7I2 = . Here we solve the system (A − 7I2 ) ~x = ~0, this time finding that our −3 5 5 5t . . For example, for t = 1, we find ~x = solutions are of the form ~x = 3 3t 1/2 0 . The linear transformation defined by A is a 0 1/2 −1 scaling by a factor 2 and A defines a scaling by 1/2. The determinant of A is the area of the square spanned 2 0 by ~v = and w ~= . The angle θ from ~v to w ~ is π/2. (See Figure 2.55.) 0 2

2.4.55 The determinant of A is equal to 4 and A−1 =

w=

0 2 θ= π 2

v=

2 0

Figure 2.55: for Problem 2.4.55.

−3−1 0 2.4.56 The determinant of A is 9. The matrix is invertible with inverse A = . The linear 0 −3−1 transformation defined by A is a reflection about the origin combined with a scaling by a factor 3. The inverse −1

99

Chapter 2 defines a reflection about combined with the origin a scaling by a factor 1/3. The determinant is the area of the −3 0 square spanned by ~v = and w ~= . The angle θ from ~v to w ~ is π/2. (See Figure 2.56.) 0 −3

v=

−3 0 θ= π 2

w=

0 −3

Figure 2.56: for Problem 2.4.56. 2.4.57 The determinant of A is −1. Matrix A is invertible, with A−1 = A. Matrices A and A−1 define reflection cos(α/2) . The absolute value of the determinant of A is the area of the about the line spanned by the ~v = sin(α/2) cos(α) sin(α) unit square spanned by ~v = and w ~ = . The angle θ from ~v to w ~ is −π/2. (See Figure sin(α) − cos(α) 2.57.)

v=

cos α sin α θ=− π 2

w=

sin α − cos α

Figure 2.57: for Problem 2.4.57.

cos(α) sin(α) 2.4.58 The determinant of A is 1. The matrix is invertible with inverse A = . The linear − sin(α) cos(α) transformation defined by A is a rotation by angle α in the counterclockwise direction. The inverse represents a rotation by the angle α in the clockwise direction. The determinant of A is the area of the unit square spanned −1

100

Section 2.4

by ~v =

cos(α) sin(α)

and ~v =

θ= π 2

w=

v=

− sin(α) cos(α)

. The angle θ from ~v to w ~ is π/2. (See Figure 2.58.)

cos α sin α

α

− sin α cos α

Figure 2.58: for Problem 2.4.58.

0.6 0.8 . The matrix A −0.8 0.6 represents the rotation through the angle α = arccos(0.6). Its inverse represents a rotation by thesame angle 0.6 in the clockwise direction. The determinant of A is the area of the unit square spanned by ~v = and 0.8 −0.8 . The angle θ from ~v to w ~ is π/2. (See Figure 2.59.) w ~= 0.6

2.4.59 The determinant of A is 1. The matrix A is invertible with inverse A−1 =

θ= π 2

w=

v=

0.6 0.8

−0.8 0.6

Figure 2.59: for Problem 2.4.59. −1 −1 2.4.60 The determinant of A is −1. The matrix A is invertible with inverse A = A. Matrices A and A define cos(α/2) , where α = arccos(−0.8). The absolute value of the the reflection about the line spanned by ~v = sin(α/2) −0.8 0.6 determinant of A is the area of the unit square spanned by ~v = and w ~= . The angle θ from v 0.6 0.8

101

Chapter 2 to w is −π/2. (See Figure 2.60.)

θ=− π 2

v=

w=

0.6 0.8

−0.8 0.6

Figure 2.60: for Problem 2.4.60.

1 −1 2.4.61 The determinant of A is 2 and A = . The matrix A represents a rotation through the angle 1 1 √ √ −π/4 combined with scaling by 2. describes through a rotation π/4 and scaling by 1/ 2. The determinant of √ 1 1 A is the area of the square spanned by ~v = and w ~ = with side length 2. The angle θ from ~v to −1 1 w ~ is π/2. (See Figure 2.61.) −1

w=

1 2

1 1 θ= π 2

v=

1 −1

Figure 2.61: for Problem 2.4.61.

2.4.62 The determinant of A is 25. The matrix A is a rotation dilation matrix with scaling factor 5 and rotation by 3 −4 is a rotation dilation too an angle arccos(0.6) in the clockwise direction. The inverse A−1 = (1/25) 4 3 with a scaling factor 1/5 angle arccos(0.6). The determinant of A is the area of the parallelogram and rotation 4 3 with side length 5. The angle from ~v to w ~ is π/2. (See Figure 2.62.) and w ~= spanned by ~v = 3 −4 102

Section 2.4

w=

4 3

θ= π 2

v=

3 −4

Figure 2.62: for Problem 2.4.62.

−3 4 1 A. The matrix A represents a reflection 2.4.63 The determinant of A is −25 and A = (1/25) = 25 4 3 −1 about a line combined with a scaling by 5 whilc A represents a reflection about the same line combined with −3 a scaling by 1/5. The absolute value of the determinant of A is the area of the square spanned by ~v = 4 4 and w ~= with side length 5. The angle from ~v to w ~ is −π/2. (See Figure 2.63.) 3 −1

v=

−3 4

θ=− π 2

w=

4 3

Figure 2.63: for Problem 2.4.63.

−1

1 1 0 1

. Both A and A−1 represent horizontal shears. The −1 1 . The angle from and w ~ = determinant of A is the area of the parallelogram spanned by ~v = 1 0 ~v to w ~ is 3π/4. (See Figure 2.64.)

2.4.64 The determinant of A is 1 and A

−1

2.4.65 The determinant of A is 1 and A

=

=

1 −1

0 1

. Both A and A−1 represent vertical shears. The determinant 103

Chapter 2

w=

−1 1

θ= 3π 4

v=

1 0

Figure 2.64: for Problem 2.4.64.

of A is the area of the parallelogram spanned by ~v = Figure 2.65.)

w=

0 1

θ= π 4

v=

1 1

and w ~=

0 1

. The angle from ~v to w ~ is π/4. (See

1 1

Figure 2.65: for Problem 2.4.65. 2.4.66 We can write AB(AB)−1 = A(B(AB)−1 ) = In and (AB)−1 AB = ((AB)−1 A)B = In . By Theorem 2.4.8, A and B are invertible. 2.4.67 Not necessarily true; (A + B)2 = (A + B)(A + B) = A2 + AB + BA + B 2 6= A2 + 2AB + B 2 if AB 6= BA. 2.4.68 True; apply Theorem 2.4.7 to B = A. 2.4.69 Not necessarily true; consider the case A = In and B = −In . 104

Section 2.4 2.4.70 Not necessarily true; (A − B)(A + B) = A2 + AB − BA − B 2 6= A2 − B 2 if AB 6= BA. 2.4.71 True; ABB −1 A−1 = AIn A−1 = AA−1 = In . 2.4.72 Not necessarily true; the equation ABA−1 = B is equivalent to AB = BA (multiply by A from the right), which is not true in general. 2.4.73 True; (ABA−1 )3 = ABA−1 ABA−1 ABA−1 = AB 3 A−1 . 2.4.74 True; (In + A)(In + A−1 ) = In2 + A + A−1 + AA−1 = 2In + A + A−1 . 2.4.75 True; (A−1 B)−1 = B −1 (A−1 )−1 = B −1 A (use Theorem 2.4.7).

1 2 2 1 2 2.4.76 We want A such that A = , so that A = 2 5 1 3 1

1 3

1 2

2 5

−1

8 −3 = . −1 1

2.4.77 We want A such that A~vi = w ~ i , for i = 1, 2, . . . , m, or A[~v1 ~v2 . . . ~vm ] = [w ~1 w ~2 . . . w ~ m ], or AS = B. Multiplying by S −1 from the right we find the unique solution A = BS −1 .

1 2 2 5

3 1 2.4.79 Use the result of Exercise 2.4.77, with S = 1 2 9 3 . A = BS −1 = 15 −2 16

2.4.78 Use the result of Exercise 2.4.77, with S =

A = BS −1

−13 − 8 −3

33 = 21 9

T

T

T

7 and B = 5 3

1 2 ; 3

6 3 and B = ; 2 6

T

2.4.80 P0 −→ P1 , P1 −→ P3 , P2 −→ P2 , P3 −→ P0 L

L

L

L

P0 −→ P0 , P1 −→ P2 , P2 −→ P1 , P3 −→ P3 a. T −1 is the rotation about the axis through 0 and P2 that transforms P3 into P1 . b. L−1 = L c. T 2 = T −1 (See part (a).) d. P0 −→ P1

T ◦L

P0 −→ P2

P1 −→ P2 P2 −→ P3 P3 −→ P0

P1 −→ P3 P2 −→ P1 P3 −→ P0

e.

L◦T

The transformations T ◦ L and L ◦ T are not the same.

105

Chapter 2 P0 P1 P2 P3

L◦T ◦L

−→ P2 −→ P1 −→ P3 −→ P0

This is the rotation about the axis through 0 and P1 that sends P0 to P2 . 2.4.81 Let A be the matrix of T and C the matrix of L. We want that AP0 = P1 , AP1 = P3 , and AP2 = P2 . We 1 −1 −1 1 1 −1 1 . 1 and B = −1 −1 can use the result of Exercise 77, with S = 1 −1 −1 1 −1 1 −1 −1 0 0 1 0 0 . Then A = BS −1 = −1 0 −1 0 0 1 0 Using an analogous approach, we find that C = 1 0 0 . 0 0 1 a b c 2.4.82 a EA = d − 3a e − 3b f − 3c g h k

The matrix EA is obtained from A by an elementary row operation: subtract three times the first row from the second.

a 1 b EA = 4 d g

b 1 4e h

c 1 4f k

The matrix EA is obtained from A by dividing the second row of A by 4 (an elementary row operation).

1 0 c If we set E = 0 0 0 1

1 0 0 1 then 0 0 0 1 0

a b a b c 0 1d e f = g h d e g h k 0

c k , as desired. f

d An elementary n × n matrix E has the same form as In except that either • eij = k(6= 0) for some i 6= j [as in part (a)], or • eii = k(6= 0, 1) for some i [as in part (b)], or • eij = eji = 1, eii = ejj = 0 for some i 6= j [as in part (c)].

2.4.83 Let E be an elementary n × n matrix (obtained from In by a certain elementary row operation), and let F be the elementary matrix obtained from In by the reversed row operation. Our work in Exercise 2.4.82 [parts (a) through (c)] shows that EF = In , so that E is indeed invertible, and E −1 = F is an elementary matrix as well. 2.4.84 a The matrix rref(A) is obtained from A by performing a sequence of p elementary row operations. By Exercise 2.4.82 [parts (a) through (c)] each of these operations can be represented by the left multiplication with an elementary matrix, so that rref(A) = E1 E2 . . . Ep A. 106

Section 2.4

b

A=

0 1

2 3

3 2

3 1

0 1

swap rows 1 and 2, represented by

0 1

1 0

↓

1 0

÷2

, represented by

1 0

0 1 2

↓

1 0

−3(II)

, represented by

1 −3 0 1

1 0

0

↓ rref(A) =

1 0

1 0 1 Therefore, rref(A) = = 0 1 0

−3 1

1 2

0 1

1 0

0 1

2 = E1 E2 E3 A. 3

2.4.85 a Let S = E1 E2 . . . Ep in Exercise 2.4.84a. By Exercise 2.4.83, the elementary matrices Ei are invertible: now use Theorem 2.4.7 repeatedly to see that S is invertible. b A=

1 2 4 8

2 4 4 8

÷2

, represented by

, represented by

−4(I) 1 2 rref(A) = 0 0

Therefore, rref(A) =

S=

1 −4

0 1

1 2

0

1

0 1

1 0 −4 1

2

0

1 2 1 = 0 0 −4

1 0 2 = 1 −2

0 1

1

2

0

0 1

2 4

4 = E1 E2 A = SA, where 8

0 . 1

(There are other correct answers.) 2.4.86 a By Exercise 2.4.84a, In = rref(A) = E1 E2 . . . Ep A, for some elementary matrices E1 , . . . , Ep . By Exercise 2.4.83, the Ei are invertible and their inverses are elementary as well. Therefore, A = (E1 E2 . . . Ep )−1 = Ep−1 . . . E2−1 E1−1 expresses A as a product of elementary matrices. b We can use out work in Exercise 2.4.84 b: −1 −1 −1 1 3 1 0 1 0 0 1 0 1 0 1 1 −3 1 0 1 −3 0 2 = = = 0 1 0 2 0 21 1 0 1 0 1 0 0 1 0 12 0 1 1 3 107

Chapter 2

2.4.87

k 0

1 0

0 1

1 0 1 k represents a vertical shear, represents a horizontal shear, k 1 0 1 0 represents a “scaling in ~e1 direction” (leaving the ~e2 component unchanged), 1 0 represents a “scaling in ~e2 direction” (leaving the ~e1 component unchanged), and k 1 1 . represents the reflection about the line spanned by 1 0

2.4.88 Performing a sequence of p elementary row operations on a matrix A amounts to multiplying A with E1 E2 . . . Ep from the left, where the Ei are elementary matrices. If In = E1 E2 . . . Ep A, then E1 E2 . . . Ep = A−1 , so that a. E1 E2 . . . Ep AB = B, and b. E1 E2 . . . Ep In = A−1 . 2.4.89 Let A and B be two lower triangular n×n matrices. We need to show that the ijth entry of AB is 0 whenever i < j. This entry is the dot product of the ith row of A and the jth column of B, 0 .. . 0 [ai1 ai2 . . . aii 0 . . . 0] · b , which is indeed 0 if i < j. jj .. . bnj

1 2.4.90 a 2 2 ↓

2 3 1 0 6 7 −2I , represented by 0 1 2 4 −2I −2 0

1 0 0

2 3 1 2 1 represented by 0 −2 −2 +II 0

2 3 2 1 , so that 0 −1 1 1 0 0 2 3 2 1 = 0 1 0 0 −2 0 1 1 0 −1

1 0 0 1 0 0

↓

↑ U

↑ E3

0 0 1 0 1 1

1 0 0 1 0 −2 0 0 1

↑ E2

0 1 0 0 0 −2 1 0 1 0 0 1

1 2 0 0 1 02 6 2 2 0 1

↑ A

↑ E1

108

3 7 4

Section 2.4 1 b A = (E3 E2 E1 )−1 U = E1−1 E2−1 E3−1 U = 2 0

↑ M1

c Let L = M1 M2 M3 in part (b);

1 2 Then 2 6 2 2

↑ A

1 0 3 1 7 = 2 2 −1 4

↑ L

1 0 0 0 1 1 00 0 −1 0 1

1 0 0 1 00 2 0 1

↑ M2

↑ M3

1 2 3 0 1 00 2 0 0 −1 1 ↑ U

0 0 1 0 . −1 1

1 we compute L = 2 2 1 2 3 0 1 00 2 0 0 −1 1 ↑ U

1 d We can use the matrix L we found in part (c), but U needs to be modified. Let D = 0 0 (Take the 1 Then 2 2

diagonal entries of the matrix U in part (c)). 2 3 1 0 0 1 0 0 1 2 6 7 = 2 1 00 2 00 1 0 0 2 4 2 −1 1 0 0 −1 ↑ A

↑ L

↑ D

3 1 2

1

↑ U

.

2.4.91 a Write the system L~y = ~b in components: y1 = −3 = 14 −3y1 + y2 , so that y1 = −3, y2 = 14 + 3y1 = 5, y1 + 2y2 + y3 = 9 −y1 + 8y2 − 5y3 + y4 = 33 y3 = 9 − y1 − 2y2 = 2, and y4 = 33 + y1 − 8y2 + 5y3 = 0: −3 5 ~y = . 2 0

1 −1 b Proceeding as in part (a) we find that ~x = . 2 0

a 0 2.4.92 We try to find matrices L = b c

d and U = 0 109

e f

such that

0 0 2 0 . 0 −1

Chapter 2

d a 0 0 1 = 0 b c 1 0

e f

=

ad ae . bd be + cf

Note that the equations ad = 0, ae = 1, and bd = 1 cannot be solved simultaneously: If ad = 0 then a or d is 0 so that ae or bd is zero. 0 1 Therefore, the matrix does not have an LU factorization. 1 0 (m) L(m) 0 U U2 and U = . L3 L4 0 U4 (m) (m) L U L(m) U2 Then A = LU = , so that A(m) = L(m) U (m) , as claimed. L3 U (m) L3 U2 + L4 U4

2.4.93 a Write L =

b By Exercise 2.4.66, the matrices L and U are both invertible. By Exercise 2.4.35, the diagonal entries of L and U are all nonzero. For any m, the matrices L(m) and U (m) are triangular, with nonzero diagonal entries, so that they are invertible. By Theorem 2.4.7, the matrix A(m) = L(m) U (m) is invertible as well.

A(n−1) c Using the hint, we write A = w ~

~v k

L′ = ~x

0 t

U′ 0

~y . s

We are looking for a column vector ~y , a row vector ~x, and scalars t and s satisfying these equations. The following equations need to be satisfied: ~v = L′ ~y , w ~ = ~xU ′ , and k = ~x~y + ts. We find that ~y = (L′ )−1~v , ~x = w(U ~ ′ )−1 , and ts = k − w(U ~ ′ )−1 (L′ )−1~v . We can choose, for example, s = 1 and t = k − w(U ~ ′ )−1 (L′ )−1~v , proving that A does indeed have an LU factorization. Alternatively, one can show that if all principal submatrices are invertible then no row swaps are required in the Gauss-Jordan Algorithm. In this case, we can find an LU -factorization as outlined in Exercise 2.4.90. 2.4.94 a If A = LU is an LU factorization, then the diagonal entries of L and U are nonzero (compare with Exercise 2.4.93). Let D1 and D2 be the diagonal matrices whose diagonal entries are the same as those of L and U , respectively. Then A = (LD1−1 )(D1 D2 )(D2−1 U ) is the desired factorization ↑ new L

↑ D

↑ new U

(verify that LD1−1 and D2−1 U are of the required form). b If A = L1 D1 U1 = L2 D2 U2 and A is invertible, then L1 , D1 , U1 , L2 , D2 , U2 are all invertible, so that we can −1 multiply the above equation by D2−1 L−1 from the right: 2 from the left and by U1 D2−1 L2−1 L1 D1 = U2 U1−1 . Since products and inverses of upper triangular matrices are upper triangular (and likewise for lower triangular −1 is both upper and lower triangular, that is, it is diagonal. Since matrices), the matrix D2−1 L−1 2 L1 D1 = U2 U1 the diagonal entries of U2 and U1 are all 1, so are the diagonal entries of U2 U1−1 , that is U2 U1−1 = In , and thus U2 = U1 . 110

Section 2.4 −1 Now L1 D1 = L2 D2 , so that L−1 is diagonal. As above, we have in fact L−1 2 L1 = D2 D1 2 L1 = In and therefore L2 = L1 .

2.4.95 Suppose A11 is a p×p matrix and A22 is a q ×q matrix. For B to be the inverse of A we must have AB = Ip+q . Let us partition B the same way as A: B11 B12 B= , where B11 is p × p and B22 is q × q. B21 B22 I 0 A11 B11 A11 B12 B11 B12 A11 0 means that = p = Then AB = 0 Iq A22 B21 A22 B22 B21 B22 0 A22 A11 B11 = Ip , A22 B22 = Iq , A11 B12 = 0, A22 B21 = 0. −1 This implies that A11 and A22 are invertible, and B11 = A−1 11 , B22 = A22 .

This in turn implies that B12 = 0 and B21 = 0. We summarize: A is invertible if (and only if) both A11 and A22 are invertible; in this case −1 0 A11 −1 A = . 0 A−1 22 2.4.96 This exercise is very similar to Example 7 in the text. We outline the solution: I 0 A11 0 B11 B12 = p means that B21 B22 0 Iq A21 A22 A11 B11 = Iq , A11 B12 = 0, A21 B11 + A22 B21 = 0, A21 B12 + A22 B22 = Iq . −1 This implies that A11 is invertible, and B11 = A−1 11 . Multiplying the second equation with A11 , we conclude that −1 B12 = 0. Then the last equation simplifies to A22 B22 = Iq , so that B22 = A22 . −1 −1 Finally, B21 = −A−1 22 A21 B11 = −A22 A21 A11 .

We summarize: A is invertible if (and only if) both A11 and A22 are invertible. In this case, 0 A−1 −1 11 . A = −1 A−1 −A−1 22 22 A21 A11 2.4.97 Suppose A11 is a p × p matrix. Since A11

I is invertible, rref(A) = p 0

A12 0

∗ , so that rref(A23 )

rank(A) = p + rank(A23 ) = rank(A11 ) + rank(A23 ). 2.4.98 Try to find a matrix B =

I AB = n w ~

~v 1

X ~y

X ~y

~x t

~x X + ~v~y = t wX ~ + ~y

(where X is n × n) such that ~x + t~v I = n w~ ~x + t 0

0 . 1

We want X + ~v~y = In , ~x + t~v = ~0, wX ~ + ~y = ~0, and w~ ~ x + t = 1. Substituting ~x = −t~v into the last equation we find −tw~ ~ v + t = 1 or t(1 − w~ ~ v ) = 1. 111

Chapter 2 v~y into the This equation can be solved only if w~ ~ v 6= 1, in which case t = 1−1w~ ~ v . Now substituting X = In − ~ 1 ~ third equation, we find w ~ − w~ ~ v~y + ~y = 0 or ~y = − 1−w~ ~ = −tw. ~ ~v w In + t~v w ~ −t~v −1 We summarize: A is invertible if (and only if) w~ ~ v 6= 1. In this case, A = , where t = 1−1w~ ~v . −tw ~ t . The same result can be found (perhaps more easily) by working with rref[A..In+1 ], rather than partitioned matrices. 2.4.99 Multiplying both sides with A−1 we find that A = In : The identity matrix is the only invertible matrix with this property. 2.4.100 Suppose the entries of A are all a, where a 6= 0. Then the entries of A2 are all na2 . The equation na2 = a 1 1 · · · n1 n n 1 1 ··· 1 n n n 1 . is satisfied if a = n . Thus the solution is A = .. . 1 1 1 ··· n n n 2.4.101 The ijth entry of AB is n X

aik bkj .

k=1

Then

n X

k=1

aik bkj ≤

n X

sbkj = s

k=1

n X

bkj

k=1

!

≤ sr.

↑ ↑ since aik ≤ s this is ≤ r, as it is the j th column sum of B. 2.4.102 a We proceed by induction on m. Since the column sums of A are ≤ r, the entries of A1 = A are also ≤ r1 = r, so that the claim holds for m = 1. Suppose the claim holds for some fixed m. Now write Am+1 = Am A; since the entries of Am are ≤ rm and the column sums of A are ≤ r, we can conclude that the entries of Am+1 are ≤ rm r = rm+1 , by Exercise 101. b For a fixed i and j, let bm be the ijth entry of Am . In part (a) we have seen that 0 ≤ bm ≤ rm . Note that limm→∞ rm = 0 (since r < 1), so that limm→∞ bm = 0 as well (this follows from what some calculus texts call the “squeeze theorem”). c For a fixed i and j, let cm be the ijth entry of the matrix In + A + A2 + · · · + Am . By part (a), cm ≤ 1 + r + r2 + · · · + rm <

1 1−r .

Since the cm form an increasing bounded sequence, limm→∞ cm exists (this is a fundamental fact of calculus). 112

Section 2.4 d (In − A)(In + A + A2 + · · · + Am ) = In + A + A2 + · · · Am − A − A2 − · · · − Am − Am+1 = In − Am+1 Now let m go to infinity; use parts (b) and (c). (In − A)(In + A + A2 + · · · + Am + · · ·) = In , so that (In − A)−1 = In + A + A2 + · · · + Am + · · ·. 2.4.103 a The components of the jth column of the technology matrix A give the demands industry Jj makes on the other industries, per unit output of Jj . The fact that the jth column sum is less than 1 means that industry Jj adds value to the products it produces. b A productive economy can satisfy any consumer demand ~b, since the equation (In − A)~x = ~b can be solved for the output vector ~x : ~x = (In − A)−1~b (compare with Exercise 2.4.49). c The output ~x required to satisfy a consumer demand ~b is ~x = (In − A)−1~b = (In + A + A2 + · · · + Am + · · ·) ~b = ~b + A~b + A2~b + · · · + Am~b + · · ·. To interpret the terms in this series, keep in mind that whatever output ~v the industries produce generates an interindustry demand of A~v . The industries first need to satisfy the consumer demand, ~b. Producing the output ~b will generate an interindustry demand, A~b. Producing A~b in turn generates an extra interindustry demand, A(A~b) = A2~b, and so forth. For a simple example, see Exercise 2.4.50; also read the discussion of “chains of interindustry demands” in the footnote to Exercise 2.4.49. 2.4.104 a We write our three equations below: I L S

1 = 13 R + 31 G + 13 B 3 =R−G , so that the matrix is P = 1 = − 12 R − 12 G + B − 21

R R 1 b G is transformed into G , with matrix A = 0 B 0 0

c This matrix is P A =

1 3

1 3

1

−1

− 21

− 12

1 3

−1 − 12

1 3

0 . 1

0 0 1 0 . 0 0

0 0 (we apply first A, then P .) 0

d See Figure 2.66. A “diagram chase” shows that M = P AP −1 = 2.4.105 a A−1

0 = 1 0

1 0 0 1 0 0 and B −1 = 0 0 0 1 1 0

0 1 . 0 113

2 3

0 −1

0 − 29

1 0

0 . 1 3

Chapter 2

Figure 2.66: for Problem 2.4.104d. Matrix A−1 transforms a wife’s clan into her husband’s clan, and B −1 transforms a child’s clan into the mother’s clan. b B 2 transforms a women’s clan into the clan of a child of her daughter. c AB transforms a woman’s clan into the clan of her daughter-in-law (her son’s wife), while BA transforms a man’s clan into the clan of his children. The two transformations are different. (See Figure 2.67.)

Figure 2.67: for Problem 2.4.105c. d The matrices for the four given diagrams 0 0 1 0 BAB −1 = 1 0 0 , B(BA)−1 = 0 0 1 0 1 e Yes; since BAB −1 = A−1 clan.

0 0 = 1 0 0 1

(in the same order) are BB −1 = I3 , 1 0 0 1 , BA(BA)−1 = I3 . 0 0

1 0 , in the second case in part (d) the cousin belongs to Bueya’s husband’s 0

2.4.106 a We need 8 multiplications: 2 to compute each of the four entries of the product. b We need n multiplications to compute each of the mp entries of the product, mnp multiplications altogether. 2.4.107 g(f (x)) = x, for all x, so that g ◦ f is the identity, but f (g(x)) = 114

x x+1

if x is even . if x is odd

True or False

2.4.108 a The formula

1 − Rk y = −k n

L + R − kLR 1 − kL

x m

is given, which implies that

y = (1 − Rk)x + (L + R − kLR)m. In order for y to be independent of x it is required that 1 − Rk = 0, or k =

1 R

= 40 (diopters).

1 k

then equals R, which is the distance between the plane of the lens and the plane on which parallel incoming rays focus at a point; thus the term “focal length” for k1 . b Now we want y to be independent of the slope m (it must depend on x alone). In view of the formula above, 1 1 10 L+R = + = 40 + ≈ 43.3 (diopters). this is the case if L + R − kLR = 0, or k = LR R L 3 c Here the transformation is y 1 0 1 D 1 = n −k1 1 0 1 −k1

0 1

x 1 − k1 D = m k1 k2 D − k1 − k2

D 1 − k2 D

x . m

We want the slope n of the outgoing rays to depend on the slope m of the incoming rays alone, and not on x; 1 1 2 this forces k1 k2 D − k1 − k2 = 0, or, D = kk11+k k2 = k1 + k2 , the sum of the focal lengths of the two lenses. See Figure 2.68.

Figure 2.68: for Problem 2.4.108c.

True or False Ch 2.TF.1 T, by Theorem 2.4.3. Ch 2.TF.2 T; Let A = B in Theorem 2.4.7. Ch 2.TF.3 F, by Theorem 2.3.3. Ch 2.TF.4 T, by Theorem 2.4.8. Ch 2.TF.5 F; Matrix AB will be 3 × 5, by Definition 2.3.1b. 0 0 . A linear transformation transforms ~0 into ~0. = Ch 2.TF.6 F; Note that T 1 0 115

Chapter 2 Ch 2.TF.7 T, by Theorem 2.2.4. Ch 2.TF.8 T, by Theorem 2.4.6. Ch 2.TF.9 T; The matrix is

1 −1 . −1 1

Ch 2.TF.10 F; The columns of a rotation matrix are unit vectors; see Theorem 2.2.3. Ch 2.TF.11 F; Note that det(A) = (k − 2)2 + 9 is always positive, so that A is invertible for all values of k. Ch 2.TF.12 T; Note that the columns are unit vectors, since (−0.6)2 + (±0.8)2 = 1. The matrix has the form presented in Theorem 2.2.3. Ch 2.TF.13 F; Consider A = I2 (or any other invertible 2 × 2 matrix).

1 2 Ch 2.TF.14 T; Note that A = 3 4

−1

1 1

1 1

5 6 7 8

−1

is the unique solution.

Ch 2.TF.15 F, by Theorem 2.4.9. Note that the determinant is 0. Ch 2.TF.16 T, by Theorem 2.4.3.

1 Ch 2.TF.17 T; The shear matrix A = 0 Ch 2.TF.18 T; Simplify to see that T

1 2

1

works.

x 4y 0 4 x = = . y −12x −12 0 y

Ch 2.TF.19 T; The equation det(A) = k 2 − 6k + 10 = 0 has no real solution. Ch 2.TF.20 T; The matrix fails to be invertible for k = 5 and k = −1, since the determinant det A = k 2 − 4k − 5 = (k − 5)(k + 1) is 0 for these values of k. Ch 2.TF.21 T; The product is det(A)I2 . Ch 2.TF.22 T; Writing an upper triangular matrix A = 0 b , where b is any nonzero constant. that A = 0 0 Ch 2.TF.23 T; Note that the matrix of 4) works.

0 −1 1 0

a b 0 c

and solving the equation A2 =

0 0

0 0

we find

represents a rotation through π/2. Thus n = 4 (or any multiple

−1

Ch 2.TF.24 F; If a matrix A is invertible, then so is A

. But

116

1 1 1 1

fails to be invertible.

True or False Ch 2.TF.25 F; If matrix A has two identical rows, then so does AB, for any matrix B. Thus AB cannot be In , so that A fails to be invertible. Ch 2.TF.26 T, by Theorem 2.4.8. Note that A−1 = A in this case.

1 1 Ch 2.TF.27 F; For any 2 × 2 matrix A, the two columns of A 1 1 Ch 2.TF.28 T; One solution is A =

1 0

will be identical.

1 . 0

Ch 2.TF.29 F; A reflection matrix is of the form

a b , where a2 + b2 = 1. Here, a2 + b2 = 1 + 1 = 2. b −a

Ch 2.TF.30 T; Just multiply it out.

0 Ch 2.TF.31 F; Consider matrix 0 1

0 1 1 0 , for example. 0 0

Ch 2.TF.32 T; Apply Theorem 2.4.8 to the equation (A2 )−1 AA = In , with B = (A2 )−1 A. Ch 2.TF.33 F; Consider the matrix A that represents a rotation through the angle 2π/17. Ch 2.TF.34 F; Consider the reflection matrix A =

1 0 . 0 −1

Ch 2.TF.35 T; We have (5A)−1 = 15 A−1 . Ch 2.TF.36 T; The equation A~ei = B~ei means that the ith columns of A and B are identical. This observation applies to all the columns. Ch 2.TF.37 T; Note that A2 B = AAB = ABA = BAA = BA2 . Ch 2.TF.38 T; Multiply both sides of the equation A2 = A with A−1 . Ch 2.TF.39 F; Consider A = I2 and B = −I2 . Ch 2.TF.40 T; Since A~x is on the line onto which we project, the vector A~x remains unchanged when we project again: A(A~x) = A~x, or A2 ~x = A~x, for all ~x. Thus A2 = A. Ch 2.TF.41 T; If you reflect twice in a row (about the same line), you will get the original vector back: A(A~x) = ~x, or, A2 ~x = ~x = I2 ~x. Thus A2 = I2 and A−1 = A. 0 1 1 1 , for example. ,w ~= , ~v = Ch 2.TF.42 F; Let A = 1 0 0 1

117

Chapter 2 1 0 1 0 0 , B = 0 1 , for example. Ch 2.TF.43 T; Let A = 0 1 0 0 0

Ch 2.TF.44 F; By Theorem 1.3.3, there is a nonzero vector ~x such that B~x = ~0, so that AB~x = ~0 as well. But I3 ~x = ~x 6= ~0, so that AB 6= I3 . Ch 2.TF.45 T; We can rewrite the given equation as A2 + 3A = −4I3 and − 41 (A + 3I3 )A = I3 . By Theorem 2.4.8, the matrix A is invertible, with A−1 = − 14 (A + 3I3 ). Ch 2.TF.46 T; Note that (In + A)(In − A) = In2 − A2 = In , so that (In + A)−1 = In − A. Ch 2.TF.47 F; A and C can be two matrices which fail to commute, and B could be In , which commutes with anything. Ch 2.TF.48 F; Consider T (~x) = 2~x, ~v = ~e1 , and w ~ = ~e2 . Ch 2.TF.49 F; Since there are only eight entries that are not 1, there will be at least two rows that contain only ones. Having two identical rows, the matrix fails to be invertible. Ch 2.TF.50 F; Let A = B =

0 0 , for example. 0 1

0 1 a b S fails to be diagonal, for an arbitrary invertible matrix S = Ch 2.TF.51 F; We will show that S −1 . 0 0 c d 0 1 c d d −b cd d2 1 1 S = ad−bc Now, S −1 = ad−bc . Since c and d cannot both be zero (as S 2 0 0 0 0 −c a −c −cd must be invertible), at least one of the off-diagonal entries (−c2 and d2 ) is nonzero, proving the claim. Ch 2.TF.52 T; Consider an ~x such that A2 ~x = ~b, and let ~x0 = A~x. Then A~x0 = A(A~x) = A2 ~x = ~b, as required.

a b d −b −a −b 1 . Now we want A−1 = −A, or ad−bc = . This holds if c d −c a −c −d ad − bc = 1 and d = −a. These equations have many solutions: for example, a = d = 0, b = 1, c = −1. More 2 generally, we can choose an arbitrary a and an arbitrary nonzero b. Then, d = −a and c = − 1+a b .

Ch 2.TF.53 T; Let A =

2 a b a + bc Ch 2.TF.54 F; Consider a 2×2 matrix A = . We make an attempt to solve the equation A2 = c d ac + cd 2 a + bc b(a + d) 1 0 = . Now the equation b(a + d) = 0 implies that b = 0 or d = −a. c(a + d) d2 + bc 0 −1

ab + bd = cb + d2

If b = 0, then the equation d2 + bc = −1 cannot be solved. If d = −a, then the two diagonal entries of A2 , a2 + bc and d2 + bc, will be equal, so that the equations a2 + bc = 1 and d2 + bc = −1 cannot be solved simultaneously. 1 0 2 cannot be solved. In summary, the equation A = 0 −1 118

True or False u21 u1 u2 u1 , where is a u1 u2 u22 u2 unit vector. Thus, a2 + b2 + c2 + d2 = u41 + (u1 u2 )2 + (u1 u2 )2 + u42 = u41 + 2(u1 u2 )2 + u42 = (u21 + u22 )2 = 12 = 1.

Ch 2.TF.55 T; Recall from Definition 2.2.1 that a projection matrix has the form

Ch 2.TF.56 T; We observe that the systems AB~x = 0 and B~x = 0 have the same solutions (multiply with A−1 and A, respectively, to obtain one system from the other). Then, by True or False Exercise 45 in Chapter 1, rref(AB) =rref(B).

119

Chapter 3

Chapter 3 Section 3.1 3.1.1 Find all ~x such that A~x = ~0: " # " # . . 1 2.. 0 −→ 1 0.. 0 , so that x = x = 0. 1 2 . . 3 4.. 0 0 1.. 0 ker(A) = {~0}. 3.1.2 Find all ~x such that A~x = ~0: " # . .. 3. 3t . 0 1 2 3. 0 −→ 2 , so that x1 = − 2 . . x2 t 6 9.. 0 0 0.. 0 Setting t = 2 we find ker(A) = span

−3 . 2

3.1.3 Find all ~x such that A~x = ~0; note that all ~x in R2 satisfy the equation, so that ker(A) = R2 = span(~e1 , ~e2 ). 3.1.4 Find all ~x such that A~x = ~0, or x1 + 2x2 + 3x2 = 0. −2t − 3r −2 −3 x1 = t 1 + r 0 , so that t The solutions are of the form x2 = r 0 1 x2 −2 −3 ker(A) = span 1 , 0 . 0 1 3.1.5 Find all ~x such that A~x = ~0. . .. 1 0 −1.. 1 1 1 . 0 . . 2.. 1 2 3.. 0 −→ 0 1 . . 1 3 5.. 0 0 0 0..

0

x1 0 ; x2 0

1 ker(A) = span −2 . 1

3.1.6 Find all ~x such that A~x = ~0. .. .. 1 1 1.. 1 1 1.. 0 1 1 1.. 0 −→ 0 0 0.. . . 0 0 0.. 1 1 1.. 0

t x = x3 1 ; x2 = −2t = −2x3 t x3

0 0 ; x1 + x2 + x3 = 0 0

120

Section 3.1 −1 −1 −r − t x1 x2 = r = r 1 + t 0 1 0 t x3

−1 −1 ker(A) = span 1 , 0 . 1 0

3.1.7 Find all ~x such that A~x = ~0. Since rref(A) = I3 we have ker(A) = {~0}.

1 3.1.8 Find all ~x such that A~x = ~0. Solving this system yields ker(A) = span −2 . 1 3.1.9 Find all ~x such that A~x = ~0. Solving this system yields ker(A) = {~0}. 1 −2 3.1.10 Solving the system A~x = ~0 we find that ker(A) = span . 1 0

−2 3 3.1.11 Solving the system A~x = ~0 we find that ker(A) = span . 1 0

1 −2 1 0 3.1.12 Solving the system A~x = ~0 we find that ker(A) = span 0 , −1 . 0 1 0 0 0 −3 −2 1 0 0 0 −2 0 3.1.13 Solving the system A~x = ~0 we find that ker(A) = span , . , 0 −1 0 0 1 0 1 0 0

3.1.14 By Theorem 3.1.3, the image of A is the span of the column vectors of A: 1 1 1 2 im(A) = span , . 1 3 1 4 3.1.15 By Theorem 3.1.3, the image of A is the span of the columns of A: 121

Chapter 3

im(A) = span

1 1 1 1 . , , , 4 3 2 1

Since any two of these vectors span all of R2 already, we can write 1 1 . , im(A) = span 2 1 3.1.16 By Theorem 3.1.3, the image of A is the span of the column vectors of A: 3 2 1 im(A) = span 1 , 2 , 3 . 3 2 1

Since these three vectors are parallel, we need only one of them to span the image: 1 im(A) = span 1 . 1

2 1 = R2 (the whole plane). , 3.1.17 By Theorem 3.1.3, im(A) = span 4 3 1 4 1 (a line in R2 ). = span , 3.1.18 By Theorem 3.1.3, im(A) = span 3 12 3 3.1.19 Since the four column vectors of A are parallel, we have im(A) = span

1 , a line in R2 . −2

1 3.1.20 Since the three column vectors of A are parallel, we have im(A) = span 1 , a line in R3 . 1 3 7 4 3.1.21 By Theorem 3.1.3, im(A) = span 1 , 9 , 2 . 8 6 5 We We 4 1

must simply find out can detemine this by . .. 1 0.. 7. 3 . . 9.. 2 −→ 0 1.. . . 0 0.. 5 6.. 8 R3 .

how many of the column vectors are not redundant to determine a basis of the image. taking the rref of the matrix: 0 0 , which shows us that all three column vectors are independent: the span is all of 1

3.1.22 Compare with the solution to Exercise 21. . .. 2 1 0.. 2 1 . 3 . . 3 4.. 2 −→ 0 1.. −1 .. . 6 5. 7 0 0.. 0 122

Section 3.1 This computation shows that the third column vector of A, ~v3 , is a linear combination of the first two, Thus, only the first two vectors are independent, and the image is a plane in R3 . 3.1.23 im(T ) = R2 and ker(T ) = {~0}, since T is invertible (see Summary 3.1.8). 3.1.24 im(T ) is the plane x + 2y + 3z = 0, and ker(T ) is the line perpendicular to this plane, spanned by the vector 1 2 (compare with Examples 5 and 9). 3 3.1.25 im(T ) = R2 and ker(T ) = {~0}, since T is invertible (see Summary 3.1.8). 3.1.26 Since limt→∞ f (t) = ∞ and limt→−∞ f (t) = −∞, we have im(f ) = R. A careful proof involves the intermediate value theorem (see Exercise 2.2.47c),

Figure 3.1: for Problem 3.1.26. Any horizontal line intersects this graph at least once (compare with Example 3 and see Figure 3.1). 3.1.27 Let f (x) = x3 − x = x(x2 − 1) = x(x − 1)(x + 1). Then im(f ) = R, since lim f (x) = ∞ and

x→∞

lim f (x) = −∞

x→−∞

but the function fails to be invertible since the equation f (x) = 0 has three solutions, x = 0, 1, and −1. 3.1.28 This ellipse can be obtained from the unit circle by means of the linear transformation with matrix as shown in Figure 3.2 (compare with Exercise 2.2.49). 1 0 cos(t) cos(t) We obtain the parametrization = for the ellipse. 0 2 sin(t) 2 sin(t) We can check that x2 +

y2 4

= cos2 (t) +

4 sin2 (t) 4

= 1.

sin(φ) cos(θ) φ = sin(φ) sin(θ) 3.1.29 Use spherical coordinates (see any good text on multivariable calculus): f θ cos(φ) 123

1 0 , 0 2

Chapter 3

Figure 3.2: for Problem 3.1.28. 1 does the job. There are many other possible answers: any nonzero 2 × n matrix 5 1 A whose column vectors are scalar multiples of vector . 5

3.1.30 By Theorem 3.1.3, A =

−2 −3 3.1.31 The plane x + 3y + 2z = 0 is spanned by the two vectors 0 and 1 , for example. Therefore, 1 0 −2 −3 A= 0 1 does the job. There are many other correct answers. 1 0 7 3.1.32 By Theorem 3.1.3, A = 6 does the job. There are many other correct answers: any nonzero 3 × n matrix 5 7 A whose column vectors are scalar multiples of 6 . 5 x 3.1.33 The plane is the kernel of the linear transformation T y = x + 2y + 3z from R3 to R. z 3.1.34 To describe a subset of R3 as a kernel means to describe it as an intersection of planes (think about it). By inspection, the given line is the intersection of the planes x+y 2x + z

= =

0 0.

and

x x+y from R3 to R2 . This means that the line is the kernel of the linear transformation T y = 2x + z z 3.1.35 ker(T ) = {~x : T (~x) = ~v · ~x = 0} = the plane with normal vector ~v . im(T ) = R, since for every real number k there is a vector ~x such that T (~x) = k, for example, ~x = 3.1.36 ker(T ) = {~x : T (~x) = ~v × ~x = ~0} = the line spanned by ~v 124

k v. ~ v ·~ v~

Section 3.1 (see Theorem A.10d in the Appendix) im(T ) = the plane with normal vector ~v By Definition A.9, T (~x) = ~v × ~x is in this plane, for all ~x in R3 . Conversely, for every vector w ~ in this plane ~ (verify this!). there is an ~x in R3 such that T (~x) = w, ~ namely ~x = − ~v1·~v T (w)

0 1 3.1.37 A = 0 0 0 0

0 0 0 1 0 0 1 , A2 = 0 0 0 , A3 = 0 0 0 0 0 0 0 0

0 0 , so that 0

ker(A) = span(~e1 ), ker(A2 ) = span(~e1 , ~e2 ), ker(A3 ) = R3 , and im(A) = span(~e1 , ~e2 ), im(A2 ) = span(~e1 ), im(A3 ) = {~0}.

3.1.38 a If a vector ~x is in ker(Ak ), that is, Ak ~x = ~0, then ~x is also in ker(Ak+1 ), since Ak+1 ~x = AAk ~x = A~0 = ~0. Therefore, ker(A) ⊆ ker(A2 ) ⊆ ker(A3 ) ⊆ . . . Exercise 37 shows that these kernels need not be equal. b If a vector ~y is in im(Ak+1 ), that is, ~y = Ak+1 ~x for some ~x, then ~y is also in im(Ak ), since we can write ~y = Ak (A~x). Therefore, im(A) ⊇ im(A2 ) ⊇ im(A3 ) ⊇ . . .. Exercise 37 shows that these images need not be equal. 3.1.39 a If a vector ~x is in ker(B), that is, B~x = ~0, then ~x is also in ker(AB), since AB(~x) = A(B~x) = A~0 = ~0: ker(B) ⊆ ker(AB). Exercise 37 (with A = B) illustrates that these kernels need not be equal. b If a vector ~y is in im(AB), that is, ~y = AB~x for some ~x, then ~y is also in im(A), since we can write ~y = A(B~x): im(AB) ⊆ im(A). Exercise 37 (with A = B) illustrates that these images need not be equal. 3.1.40 For any ~x in Rm , the vector B~x is in im(B) = ker(A), so that AB~x = ~0. If we apply this fact to ~x = ~e1 , ~e2 , . . . , ~em , we find that all the columns of the matrix AB are zero, so that AB = 0. 4 −4 3.1.41 a rref(A) = 1 3 , so that ker(A) = span . 3 0 0 0.36 3 im(A) = span = span . 0.48 4 Note that im(A) and ker(A) are perpendicular lines. b A2 = A If ~v is in im(A), with ~v = A~x, then A~v = A2 ~x = A~x = ~v . 125

Chapter 3

Figure 3.3: for Problem 3.1.41c. c Any vector ~v in R2 can be written uniquely as ~v = ~v1 + ~v2 , where ~v1 is in im(A) and ~v2 is in ker(A). (See Figure 3.3.) Then A~v = A~v1 + A~v2 = ~v1 (A~v1 =~v1 by part b, A~v2 = ~0 since ~v2 is in ker(A)), so that A represents the 3 orthogonal projection onto im(A) = span . 4 y1 y 3.1.42 Using the hint, we see that the vector ~y = 2 is in the image of A if y3 y4

y1

y2

−3y3 −2y3

+2y4 +y4

=0 = 0.

and

This means that im(A) is the kernel of the matrix

1 0

0 −3 2 . 1 −2 1

3.1.43 Using our work in Exercise 42 as a guide, we come up with the following procedure to express the image of an n × m matrix A as the kernel of a matrix B: If rank(A) = n, let B be the n × n zero matrix. If r = rank(A) < n, let B be the (n − r) × n matrix obtained by omitting the first r rows and the first m columns . of rref[A..I ]. n

3.1.44 a Yes; by construction of the echelon form, the systems A~x = ~0 and B~x = ~0 have the same solutions (it is the whole point of Gaussian elimination not to change the solutions of a system). b No; as a counterexample, consider A = im(B) = span(~e1 ).

0 1

1 0 0 , with , with im(A) = span(~e2 ), but B = rref(A) = 0 0 0

3.1.45 As we solve the system A~x = ~0, we obtain r leading variables and m − r free variables. The “general vector” in ker(A) can be written as a linear combination of m − r vectors, with the free variables as coefficients. (See Example 11, where m − r = 5 − 3 = 2.) 126

Section 3.1 3.1.46 If rank(A) = r, then im(A) = span(~e1 , . . . , ~er ). See Figure 3.4.

Figure 3.4: for Problem 3.1.46. 3.1.47 im(T ) = L2 and ker(T ) = L1 . 3.1.48 a w ~ = A~x, for some ~x, so that Aw ~ = A2 ~x = A~x = w. ~ b If rank(A) = 2, then A is invertible, and the equation A2 = A implies that A = I2 (multiply by A−1 ). 0 0 If rank(A) = 0 then A = . 0 0 c First note that im(A) and ker(A) are lines (there is one nonleading variable).

Figure 3.5: for Problem 3.1.48c. By definition of a projection, we need to verify that ~x − A~x is in ker(A). This is indeed the case, since A(~x − A~x) = A~x − A2 ~x = A~x − A~x = ~0 (we are told that A2 = A). See Figure 3.5. 3.1.49 If ~v and w ~ are in ker(T ), then T (~v + w) ~ = T (~v ) + T (w) ~ = ~0 + ~0 = ~0, so that ~v + w ~ is in ker(T ) as well. If ~v is in ker(T ) and k is an arbitrary scalar, then T (k~v ) = kT (~v ) = k~0 = ~0, so that k~v is in ker(T ) as well. 3.1.50 From Exercise 38 we know that ker(A3 ) ⊆ ker(A4 ). Conversely, if ~x is in ker(A4 ), then A4 ~x = A3 (A~x) = ~0, so that A~x is in ker(A3 ) = ker(A2 ), which implies that A2 (A~x) = A3 ~x = ~0, that is, ~x is in ker(A3 ). We have shown that ker(A3 ) = ker(A4 ). 3.1.51 We need to find all ~x such that AB~x = ~0. If AB~x = ~0, then B~x is in ker(A) = {~0}, so that B~x = ~0. Since ker(B) = {~0}, we can conclude that ~x = ~0. It follows that ker(AB) = {~0}. 127

Chapter 3 A~x A , we can conclude that C~x = ~0 if (and only if) both A~x = ~0 and B~x = ~0. It ~x = B~x B follows that ker(C) is the intersection of ker(A) and ker(B): ker(C) = ker(A) ∩ ker(B).

3.1.52 Since C~x =

3.1.53 a Using the equation 1 + 1 = 0 (or −1 = 1), we can write the general vector ~x in ker(H) as x1 p+r+s x2 p + q + s x3 p + q + r ~x = x4 = p q x5 x6 r x7 s 1 0 1 1 1 1 0 1 1 1 1 0 = p1 + q0 + r0 + s0 0 1 0 0 0 0 1 0 0 0 0 1 ↑ ~v1

↑ ~v2

↑ ~v3

↑ ~v4

b ker(H) = span(~v1 , ~v2 , ~v3 , ~v4 ) by part (a), and im(M ) = span(~v1 , ~v2 , ~v3 , ~v4 ) by Theorem 3.1.3, so that im(M ) = ker(H). M~x is in im(M ) = ker(H), so that H(M~x) = ~0. 3.1.54 a If no error occurred, then w ~ = ~v = M~u, and H w ~ = H(M~u) = ~0, by Exercise 53b. If an error occurred in the ith component, then w ~ = ~v + ~ei = M~u + ~ei , so that Hw ~ = H(M~u) + H~ei = ith column of H. Since the columns of H are all different, this method allows us to find out where an error occurred. 1 b Hw ~ = 1 = seventh column of H: an error occurred in the seventh component of ~v . 0 1 0 0 1 1 Therefore ~v = w ~ + ~e7 = 0 and ~u = . 0 1 1 0 1

Section 3.2 3.2.1 Not a subspace, since W does not contain the zero vector. 128

Section 3.2 −1 1 3.2.2 Not a subspace, since W contains the vector ~v = 2 but not the vector (−1)~v = −2 . −3 3

1 2 3.2.3 W = im 4 5 7 8

3 6 is a subspace of R3 , by Theorem 3.2.2. 9

3.2.4 span(~v1 , . . . , ~vm ) = im[~v1 . . . ~vm ] is a subspace of Rn , by Theorem 3.2.2. 3.2.5 We have subspaces {~0}, R3 , and all lines and planes (through the origin). To prove this, mimic the reasoning in Example 2. 3.2.6 a Yes! • The zero vector is in V ∩ W , since ~0 is in both V and W . • If ~x and ~y are in V ∩ W , then both ~x and ~y are in V , so that ~x + ~y is in V as well, since V is a subspace of Rn . Likewise, ~x + ~y is in W , so that ~x + ~y is in V ∩ W . • If ~x is in V ∩ W and k is an arbitrary scalar, then k~x is in both V and W , since they are subspaces of Rn . Therefore, k~x is in V ∩ W . b No; as a counterexample consider V = span(~e1 ) and W = span(~e2 ) in R2 . 3.2.7 Yes; we need to show that W contains the zero vector. We are told that W is nonempty, so that it contains some vector ~v . Since W is closed under scalar multiplication, it will contain the vector 0~v = ~0, as claimed. 1 2 3 1 2 3.2.8 We need to solve the system c1 + c2 + c3 = 2 3 4 2 3 t c1 The general solution is c2 = −2t . t c3

c1 3 0 c2 = . 4 0 c3

1 2 3 0 Picking t = 1 we find the nontrivial relation 1 −2 +1 = . 2 3 4 0

3.2.9 These vectors are linearly dependent, since ~vm = 0~v1 + 0~v2 + · · · + 0~vm−1 . 3.2.10 Linearly dependent, since

6 2 6 is redundant. . Thus, the vector =3 3 1 3

3.2.11 Linearly independent, since the two vectors are not parallel, and therefore not redundant. 7 0 . Thus, the vector ~0 is redundant. =0 3.2.12 Linearly dependent, since 11 0 129

Chapter 3

3.2.13 Linearly dependent, since the second vector is redundant

1 1 3.2.14 Linearly independent, since ref 0 2 0 0

1 1 . =1 2 2

1 2 = I3 (use Theorem 3.2.6). 3

3.2.15 Linearly dependent. By Theorem 3.2.8, since we have three vectors in R2 , at least one must be redundant. We can perform a straightforward computation to reveal that ~v3 = −~v1 + 2~v2 . 6 3 1 6 1 3 3.2.16 Certainly 2 is not a multiple of 1 , so it is not redundant. However, 5 = 3 1 + 1 2 , so 5 4 1 1 4 1 1 is redundant. Thus, these vectors are linearly dependent.

1 1 1 3.2.17 Linearly independent. The first two vectors are clearly not redundant, and since rref 1 2 3 = I3 , the 1 3 6 last vector is also not redundant. Thus, the three vectors turn out to be linearly independent. 1 1 3.2.18 Linearly dependent, since rref 1 1 be redundant.

1 2 3 4

1 1 4 0 = 0 7 0 10

1 0 −2 1 3 4 . So, we find that the vector turns out to 7 0 0 10 0 0

1 0 3.2.19 Linearly dependent. First we see that is not redundant, because it is first, and non-zero. However, 0 0 2 1 0 0 = 2 , so it is redundant. 0 0 0 0 0 0 3 1 0 0 1 0 4 0 1 0 and are clearly not redundant, but = 3 + 4 + 5 , so it is redundant. 0 1 5 0 0 1 0 0 0 0 0 0 0 3.2.20 0 is redundant, simply because it is the zero vector. 0 1 0 is our first non-zero vector, and thus, is not redundant. 0 1 3 0 = 3 0 and is redundant. 0 0 130

Section 3.2 1 0 1 is not a multiple of 0 and is not redundant. 0 0 0 1 4 5 = 4 0 + 5 1 and is redundant. 0 0 0 0 1 6 Similarly, 7 = 6 0 + 7 1 and is also redundant. 0 0 0 0 1 0 However, by inspection, 0 is not a linear combination of 0 and 1 , meaning that this last vector is not 0 0 1 redundant. Thus, the seven vectors are linearly dependent. 3.2.21 Certainly, sincethe second vector equals the first, the second is redundant. So ~v1 = ~v2 , 1~v1 − 1~v2 = ~0, 1 is in ker(A). revealing that −1 3.2.22

1 3 3 1 1 3 3 is in the kernel of = ~0. Thus, −1 . So, 3 =3 is redundant, because 2 −1 6 2 2 6 6

3.2.23 The first column is ~0, so it is redundant. 1~v1 = ~0, so

3 . 6

1 is in ker(A). 0

6 6 1 3 1 3 6 3 3.2.24 5 is redundant, because 5 = 3 1 + 1 2 . Thus, 3 1 + 1 2 − 1 5 = ~0 and 1 is in 4 4 1 1 1 1 4 −1 1 3 6 the kernel of 1 2 5 . 1 1 4

1 3.2.25 The third column equals the first, so it is redundant and ~v1 = ~v3 , or 1~v1 + 0~v2 − 1~v3 = ~0. Thus, 0 is in −1 ker(A). 2 1 0 3.2.26 3 = 2 0 + 3 1 , 0 0 0 2 1 3 is in the kernel of 0 −1 0 0

2 1 0 2 0 so 3 is redundant. Now, 2 0 + 3 1 − 1 3 + 0 0 = ~0, revealing that 0 0 0 0 1 0 2 0 1 3 0. 0 0 1

1 1 3.2.27 A basis of im(A) is 1 , 2 , by Theorem 3.2.4. 3 1 131

Chapter 3 1 1 1 3.2.28 The three column vectors are linearly independent, since rref 1 2 5 = I3 . 1 3 7

Therefore, the three columns form a basis of im(A)(= R3 ): 1 1 1 1 , 2 , 5 . 1 3 7 Another sensible choice for a basis of im(A) is ~e1 , ~e2 , ~e3 .

3.2.29 The three column vectors of A span all of R2 , so that im(A) = R2 . We can choose any two of the columns of A to form a basis of im(A); another sensible choice is ~e1 , ~e2 . 3.2.30 im(A) = span(~e1 , ~e2 ) We can choose ~e1 , ~e2 as a basis of im(A). 3.2.31 The two column vectors of the given matrix A are linearly independent (they are not parallel), so that they form a basis of im(A). 3.2.32 By inspection, the first, third and sixth columns are redundant. Thus, a basis of the image consists of the 1 0 0 0 1 0 remaining column vectors: , , . 0 0 1 0 0 0 3.2.33 im(A) = span(~e1 , ~e2 , ~e3 ), so that ~e1 , ~e2 , ~e3 is a basis of im(A). 1 2 3.2.34 The fact that is in ker(A) means that 3 4 1 1 2 2 A = [~v1 ~v2 ~v3 ~v4 ] = ~v1 + 2~v2 + 3~v3 + 4~v4 = ~0, so that ~v4 = − 14 ~v1 − 12 ~v2 − 43 ~v3 . 3 3 4 4 3.2.35 If ~vi is a linear combination of the other vectors in the list, ~vi = c1~v1 + · · · + ci−1~vi−1 + ci+1~vi+1 + · · · + cn~vn , then we can subtract ~vi from both sides to generate a nontrivial relation (the coefficient of ~vi will be -1). Conversely, if there is a nontrivial relation c1~v1 + · · · + ci~vi + · · · + cn~vn = ~0, with ci 6= 0, then we can solve for vector ~vi and thus express ~vi as a linear combination of the other vectors in the list. 3.2.36 Yes; we know that there is a nontrivial relation c1~v1 + c2~v2 + · · · + cm~vm = ~0. Now apply the transformation T to the vectors on both sides, and use linearity: T (c1~v1 + c2~v2 + · · · + cm~vm ) = T (~0), so that c1 T (~v1 ) + c2 T (~v2 ) + · · · + cm T (~vm ) = ~0. 132

Section 3.2 This is a nontrivial relation among the vectors T (~v1 ), . . . , T (~vm ), so that these vectors are linearly dependent, as claimed. 3.2.37 No; as a counterexample, consider the extreme case when T is the zero transformation, that is, T (~x) = ~0 for all ~x. Then the vectors T (~v1 ), . . . , T (~vm ) will all be zero, so that they are linearly dependent. 3.2.38 a Using the terminology introduced in the exercise, we need to show that any vector ~v in V is a linear combination of ~v1 , . . . , ~vm . Choose a specific vector ~v in V . Since we can find no more than m linearly independent vectors in V , the m + 1 vectors ~v1 , . . . , ~vm , ~v will be linearly dependent. Since the vectors ~v1 , . . . , ~vm are independent, ~v must be redundant, meaning that ~v is a linear combination of ~v1 , . . . , ~vm , as claimed. b With the terminology introduced in part a, we can let V = im [ ~v1

· · · ~vm ] .

3.2.39 Yes; the vectors are linearly independent. The vectors in the list ~v1 , . . . , ~vm are linearly independent (and therefore non-redundant), and ~v is non-redundant since it fails to be in the span of ~v1 , . . . , ~vm . 3.2.40 Yes; by Theorem 3.2.8, ker(A) = {~0} and ker(B) = {~0}. Then ker(AB) = {~0} by Exercise 3.1.51, so that the columns of AB are linearly independent, by Theorem 3.2.8. 3.2.41 To show that the columns of B are linearly independent, we show that ker(B) = {~0}. Indeed, if B~x = ~0, then AB~x = A~0 = ~0, so that ~x = ~0 (since AB = Im ). By Theorem 3.2.8, rank(B) = # columns = m, so that m ≤ n and in fact m < n (we are told that m 6= n). This implies that the rank of the m × n matrix A is less than n, so that the columns of A are linearly dependent (by Theorem 3.2.8). 3.2.42 We can use the hint and form the dot product of ~vi and both sides of the relation c1~v1 + · · · + ci~vi + · · · + cm~vm = ~0: (c1~v1 + · · · + ci~vi + · · · + cm~vm ) · ~vi = ~0 · ~vi , so that c1 (~v1 · ~vi ) + · · · + ci (~vi · ~vi ) + · · · + cm (~vm · ~vi ) = 0. Since ~vi is perpendicular to all the other ~vj , we will have ~vi · ~vj = 0 whenever j 6= i; since ~vi is a unit vector, we will have ~vi · ~vi = 1. Therefore, the equation above simplifies to ci = 0. Since this reasoning applies to all i = 1, . . . , m, we have only the trivial relation among the vectors ~v1 , ~v2 , . . . , ~vm , so that these vectors are linearly independent, as claimed. 3.2.43 Consider a linear relation c1~v1 + c2 (~v1 +~v2 ) + c3 (~v1 +~v2 +~v3 ) = ~0, or, (c1 + c2 + c3 )~v1 + (c2 + c3 )~v2 + c3~v3 = ~0. Since there is only the trivial relation among the vectors ~v1 , ~v2 , ~v3 , we must have c1 + c2 + c3 = c2 + c3 = c3 = 0, so that c3 = 0 and then c2 = 0 and then c1 = 0, as claimed. 3.2.44 Yes; this is a special case of Exercise 40 (recall that ker(A) = {~0}, by Theorem 3.1.7b). 3.2.45 Yes; if A is invertible, then ker(A) = {~0}, so that the columns of A are linearly independent, by Theorem 3.2.8. 3.2.46 Solve the system

x1

+

2x2 x3

+ +

3x4 4x4

+ +

The solutions are of the form 133

5x5 6x5

=0 . =0

Chapter 3 −5 −3 −2 −2s − 3t − 5r x1 s 0 0 1 x2 x3 = −4t − 6r = s 0 + t −4 + r −6 . 0 1 0 t x4 1 0 0 r x5 −2 −3 −5 1 0 0 The vectors 0 , −4 , −6 span the kernel, by construction, and they are linearly independent, by 0 1 0 0 0 1 Theorem 3.2.5. Therefore, the three vectors form a basis of the kernel.

1 0 3.2.47 By Theorem 3.2.8, the rank of A is 3. Thus, rref(A) = 0 0

0 1 0 0

0 0 . 1 0

x1 3.2.48 We can write 3x1 + 4x2 + 5x3 = [3 4 5] x2 = 0, so that V = ker[3 4 5]. x3 0 4 To express V as an image, choose a basis of V , for example, −3 , 5 . −4 0 4 0 Then, V = im −3 5 . 0 −4

There are other solutions.

1 3.2.49 L = im 1 1

To write L as a kernel, think of L as the intersection of the planes x = y and y = z, that is, as the solution set x − y =0 of the system . y − z =0 1 −1 0 Therefore, L = ker . 0 1 −1

There are other solutions. 3.2.50 The verification of the three properties listed in Definition 3.2.1 is straightforward. Alternatively, we can choose a basis ~v1 , . . . , ~vp of V and a basis w ~ 1, . . . , w ~ q of W (see Exercise 38a) and show that V + W = span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ) (compare with Exercise 4). Indeed, if ~v + w ~ is in V +W , then ~v is a linear combination of ~v1 , . . . , ~vp and w ~ is a linear combination of w ~ 1, . . . , w ~q, so that ~v + w ~ is a linear combination of ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q . Conversely, if ~x is in span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ), then ~x = (c1~v1 + · · · + cp~vp ) + (d1 w ~ 1 + · · · + dq w ~ q ), so that ~x is in V + W . 134

Section 3.2 If V and W are distinct lines in R3 (spanned by ~v and w, ~ respectively), then V + W is the plane spanned by ~v and w. ~ 3.2.51 a Consider a relation c1~v1 + · · · + cp~vp + d1 w ~ 1 + · · · + dq w ~ q = ~0. Then the vector c1~v1 + · · · + cp~vp = −d1 w ~ 1 − · · · − dq w ~ q is both in V and in W , so that this vector is ~0 : ~ ~ c1~v1 + · · · + cp~vp = 0 and d1 w ~ 1 + · · · + dq w ~ q = 0. Now the ci are all zero (since the ~vi are linearly independent) and the dj are zero (since the w ~ j are linearly independent). Since there is only the trivial relation among the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q , they are linearly independent. b In Exercise 50 we show that V + W = span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ), and in part (a) we show that these vectors are linearly independent. a b d 1 0 0 0 c e 0 1 0 3.2.52 If a, c and f are nonzero, then rref = , and the three vectors are linearly independent, 0 0 f 0 0 1 0 0 0 0 0 0 by Theorem 3.2.6. If at least one of the constants a, c or f is zero, then at least one column of rref will not contain a leading one, so that the three vectors are linearly dependent.

3.2.53 The zero vector is in V ⊥ , since ~0 · ~v = 0 for all ~v in V . If w ~ 1 and w ~ 2 are both in V ⊥ , then (w ~1 + w ~ 2 ) · ~v = w ~ 1 · ~v + w ~ 2 · ~v = 0 + 0 = 0 for all ~v in V , so that w ~1 + w ~ 2 is in ⊥ V as well. If w ~ is in V ⊥ and k is an arbitrary constant, then (k w) ~ · ~v = k(w ~ · ~v ) = k0 = 0 for all ~v in V , so that k w ~ is in V ⊥ as well. 1 x x 3.2.54 We need to find all vectors y in R3 such that y · 2 = x + 2y + 3z = 0. 3 z z −3 −2 −2s − 3t x = s 1 + t 0 . s These vectors have the form y = 1 0 t z −3 −2 Therefore, 1 , 0 is a basis of L⊥ . 1 0 1 x1 x2 2 3.2.55 We need to find all vectors ~x in R5 such that x3 · 3 = x1 + 2x2 + 3x3 + 4x4 + 5x5 = 0. 4 x4 5 x5

These vectors are of the form 135

Chapter 3 −5 −4 −3 −2 −2a − 3b − 4c − 5d x1 a 0 1 0 0 x2 b = a 0 + b 1 + c 0 + d 0 . x3 = 0 1 0 0 c x4 1 0 0 0 d x5

The four vectors to the right form a basis of L⊥ ; they span L⊥ , by construction, and they are linearly independent, by Theorem 3.2.5. 3.2.56 Consider a linear relation c1~v1 + c2~v2 + c3~v3 + c4~v4 = ~0 among the four given vectors. The last component of the vector on the left hand side is c3 , so that c3 = 0. Now the fifth component on the left is c1 , so that c1 = 0. The third component is now c4 , so c4 = 0. It follows that c2 = 0 as well. We have shown that there is only the trivial relation among the given vectors, so that they are linearly independent, regardless of the values of the constants a, b . . . , m. 3.2.57 We will begin to go through the possibilties for j until we see a pattern: 1 0 0 j = 1: Yes, because 0 is in ker(A) (the first column is ~0). 0 0 0 j = 2: No, this would just be a multiple of the second column, and only ~0 if the jth component is zero. 0 2 −1 j = 3: Yes, since 0 is in ker(A). 0 0 0

At this point, we realize that we are choosing the redundant columns. Thus, j can also be 6 and 7, because 0 0 0 3 0 0 4 , and 0 are in ker(A). 0 5 0 −1 0 1 3.2.58 This occurs for each column, j, that is redundant. If ~x is in the kernel, and the j th component of ~x is the last non-zero component, then x1~v1 + · · · + xj ~vj + xj+1~vj+1 + · · · + xm~vm = ~0, but xj+1 = · · · = xm = 0, so x1~v1 + · · · + xj ~vj = ~0. 136

Section 3.3 Thus, since xj 6= 0, ~vj = −

x1 ~ v1 +···+xj−1 ~ vj−1 xj

and ~vj is redundant. Conversely, if ~vj is redundant, with ~vj = c1 .. . cj−1 c1~v1 + · · · + cj−1~vj−1 , then the vector ~x = −1 is in the kernel of A. The last non-zero component of ~x is the 0 . .. 0 j th , as required.

Section 3.3 3.3.1 Clearly the second column is just three time the first, and thus is redundant. Applying the notion of Kyle Numbers, we see: 3 1 2

−1 3 3 3 , so is in the ker(A). No other vectors belong in our list, so a basis of the kernel is , −1 −1 6 1 and a basis of the image is . 2 3.3.2 Using Kyle Numbers, we see that the second column is redundant: 4 1 2

−1 4 4 , 4 , so is in the ker(A). No other vectors belong in our list, so a basis of the kernel is −1 −1 8 1 . and a basis of the image is 2 3.3.3 The two columns here are independent, so there redundant vectors. Thus, ∅ is a basis of the kernel, are no 1 2 and the two columns form a basis of the image: , . 3 4 3.3.4 The first column is redundant. We use the following Kyle Numbers: 1 0 0

0 1 1 1 , so is in the ker(A). No other vectors belong in our list, so a basis of the kernel is , and 0 0 2 1 a basis of the image is . 2 3.3.5 The first two vectors are non-redundant, but the third is a multiple of the first. We see: 3 0 −1 3 1 −2 3 , so a basis of the kernel is 0 , and a basis of the image consists of the non-redundant 2 4 6 −1 −2 1 . , columns, or 4 2 137

Chapter 3 3.3.6 The first two vectors are non-redundant, but the third is a combination of the first two: 1 2 −1 1 1 1 1 1 3 , so a basis of the kernel is , and a basis of the image is 2 , . 2 1 2 1 4 −1 3.3.7 We immediately see fitting Kyle numbers for one relation: 2 −1 1 2 1 2 zero above

0 3 . Now, since the second column is redundant, we remove it from further inspection and keep a 4 it:

0 2 3 . However, in this case, there are no more redundant vectors. Thus, a basis of the kernel is 2 4 2 1 3 −1 , and a basis of the image is , . 1 4 0

1 1

3.3.8 Here the second column is redundant, with Kyle Numbers as: 3 1 2 3

1 1 −3 3 . This reveals a basis of our kernel as and a basis of the image to be 2 . −6 1 3 −9

3.3.9 The second column is redundant, and we can choose Kyle numbers as follows: 2 1 1 1

−1 2 2 2

0 2 1 , but the third column is non-redundant. Thus, a basis of the kernel is −1 , while a basis 2 0 3 1 1 of the image is 1 , 2 . 3 1 1 3.3.10 The first column is redundant, and 0 is in the kernel: 0 1 0 0 0

0 1 1 1

0 1 1 0 , while . No other columns are redundant, however, meaning that a basis of the kernel is 2 0 3 1 1 a basis of the image is 1 , 2 . 1 3

3.3.11 Here itis clear third the that only column is redundant, since it is equal to the first. Thus, a basis of the 1 1 0 kernel is 0 , and 0 , 1 is a basis of the image. −1 0 1 138

Section 3.3 3.3.12 The third vector is the only redundant vector here, shown by: 0 1 1 1

1 0 0 1

−1 0 0 . No other columns are redundant, however, meaning that a basis of the kernel is 1 , while 0 −1 1 0 1 a basis of the image is 1 , 0 . 1 1 3.3.13 Here we first see

2 [1

−1 0 3 , then 2 3] [1

0 −1 , 2 3]

3 2 so both the second and third columns are redundant, and a basis of the kernel is −1 , 0 . This leaves −1 0 ([ 1 ]) to be a basis of the image.

3.3.14 The first and the third columns are redundant, as the Kyle Numbers show us: 0 1 0 2 −1 1 0 0 , so that a basis of the kernel is 0 , 2 . This leaves ([ 1 ]) to be a basis , then 2] [0 1 [0 1 2] −1 0 of the image. 3.3.15 We quickly find that the third column is redundant, with the Kyle numbers 2 1 0 1 0

2 0 1 0 1

−1 2 2 2 2

0 1 0 1 0

0 0 1 0 1

0 2 2 2 2

0 0 0 , then see that the fourth column is also redundant, 0 0 1 0 0. 0 0

0 1 0 2 0 1 2 0 Thus, a basis of our kernel is , , while , is a basis of our image. 0 1 0 −1 1 0 1 0

3.3.16 The third column is redundant, as we find with 3 1 0 0 0

2 1 1 1 1

−1 5 2 2 2

0 1 2 . The fourth column, however, fails to be redundant. 3 4 139

Chapter 3 1 1 1 3 0 1 2 2 Thus, a basis of our kernel is , while , , is a basis of our image. 3 1 0 −1 4 1 0 0

3.3.17 For this problem, we again successively use Kyle Numbers to find our kernel, investigating the columns from left to right. We initially see that the first column is redunant: 0 0 2 −1 3 , then the third column: 0 1 2 4 0 0 0 0 0 1 0 2 3 Thus, 0 , −1 , 0 is a basis of the kernel, and 4 0 0 −1 0 0 1 0 0

0 1 0

0 2 0

0 0 1

0 0 1

0 0 3 3 , followed by the fifth column: 0 1 0 0 4

0 2 0

1 0 , is a basis of the image. 0 1

3.3.18 This matrix is already in rref, and we see that there are two columns without leading ones. These will be our redundant columns. Thus we see 2 1 0 0

1 −2 0 0

0 0 1 0

0 −1 5 0

0 −1 0 1 0 , and 0 1 0

0 −2 0 0

5 0 1 0

−1 −1 5 0

0 0 . 0 1

−1 2 0 0 1 1 0 Then 0 , 5 is a basis of the kernel, and 0 , 1 , 0 is a basis of the image. 1 0 0 −1 0 0 0 3.3.19 We see that the third column is redundant, and choose Kyle numbers as follows: 5 1 0 0 0

3 1 0 0 0

4 0 1 0 0 2 0 1 0 0

−1 5 4 0 0

0 3 2 0 0

0 0 0 , then we see that the fourth column is also redundant, 1 0

0 −1 0 5 3 0 4 2 0. 0 0 1 0 0 0 3 5 1 0 0 4 2 0 1 0 Thus, −1 , 0 is a basis of the kernel, and , , is a basis of the image. 0 0 1 −1 0 0 0 0 0 0 3.3.20 Although this matrix is not quite in rref, we can still quickly see that columns 2, 3, and 5 are the redundant columns: 140

4 0 1

−1 3 . 4

Section 3.3 0 5 −3 1 3 , 0 0 0 0 0 −12 5 0 1 0 0 So, 0 , −1 , 0 is 3 0 0 −1 0 0 0 1 0 0 0

1 0 0 0 0

0 5 0 0 0

0 3 1 0 0

0 −1 0 5 0 0 0 0 0 0

0 3 1 0 0

0 −3 3 , 0 0

−12 1 0 0 0

0 0 0 0 0

0 5 0 0 0

3 3 1 0 0

−1 −3 3 . 0 0

1 3 0 1 a basis of the kernel, and , is a basis of the image. 0 0 0 0

−3 1 0 −3 1 3 9 3.3.21 rref 4 5 8 = 0 1 4 , which we can use to “spot” a vector in the kernel: 4 . Since the third −1 0 0 0 7 6 3 column is the only redundant one, thisforms abasis of the kernel, and implies that the third column of A is also 1 3 redundant. Thus, a basis of im(A) is 4 , 5 . 7 6

2 4 8 1 0 −6 3.3.22 rref 4 5 1 = 0 1 5 . It is clear that the third vector is redundant, and we quickly see that the 0 0 0 7 9 3 −6 −6 vector 5 is in the kernel. Since this is the only redundant column, 5 is a basis of the kernel. Thus, −1 −1 2 4 a basis of im(A) is 4 , 5 . 7 9 1 0 3.3.23 rref 3 0

0 2 4 1 1 −3 −1 0 = 4 −6 8 0 −1 3 1 0

0 1 0 0

2 4 −3 −1 . 0 0 0 0

Using the method of Exercises 17 and 19, we find the kernel: 4 −1 0 −1 0 4 4 1 0 2 −1 , then 0 1 −3 −1 . 0 0 0 0 0 0 0 0 0 0 4 2 −3 −1 So a basis of ker(A) is , . The non-redundant column vectors of A form a basis of im(A): −1 0 0 −1 0 1 0 1 . , 4 3 −1 0 2 1 0 0 0

−3 0 1 0 0

−1 2 −3 0 0

141

Chapter 3 4 3 3.3.24 rref 2 1 2 −1 0 , 0 0

1 3 3.3.25 rref 1 2

8 6 4 2

1 1 1 3

1 1 6 2 5 0 = 0 9 10 0 2 0

2 0 0 0

0 1 0 0

0 0 1 0

0 0 . Here our kernel is the span of only one vector: 0 1

4 1 1 6 3 1 2 5 while a basis of the image of A is , , , . 2 1 9 10 1 3 2 0 2 6 2 4

3 9 4 9

2 6 1 1

1 1 3 0 = 2 0 2 0

2 0 0 0

0 1 0 0

5 −1 0 0

5 0 0 1 0 , then 0 1 0 0 0 5 2 −1 0 So a basis of ker(A) is 0 , −1 and −1 0 0 0

2 1 kernel: 0 0 0

−1 2 0 0 0

0 0 1 0 0

0 5 −1 0 0

0 0 . We will emulate Exercise 23 to find the 1 0 0 −1 2 0 0 1 0 0 0 0

−1 5 −1 0 0

0 0 0. 1 0

1 3 1 3 9 3 a basis of im(A) is , , . 1 4 2 2 9 2

3.3.26 a We notice that each of the six matrices has two identical In matrices C and L, the second column columns. 0 is identical to the third, so that ker(C) = ker(L) = span 1 . In matrices H, T, X and Y, the first column is −1 1 identical to the third, so that ker(H) = ker(T ) = ker(X) = ker(Y ) = span 0 . Thus, only L has the same −1 kernel as C. b We observe that each of the six matrices in the list has two identical rows. For example, the first and the y1 last row of matrix C are identical, so that any vector y2 in im(C) will satisfy the equation y1 = y3 . We y3 y1 y1 can conclude that im(C) = im(H) = im(X) = y2 : y1 = y3 , im(L) = y2 : y1 = y2 , and im(T ) = y3 y3 y1 im(Y ) = y2 : y2 = y3 . y3 c Our discussion in part b shows that the answer is matrix L.

3.3.27 Form a 4 × 4 matrix A with the given vectors as its columns. We find that rref(A) = I4 , so that the vectors 142

Section 3.3 do indeed form a basis of R4 , by Summary 3.3.10. 3.3.28 Form a 4 × 4 matrix A with the given vectors as its columns. The matrix A reduces to 1 0 0 0

0 1 0 0

0 0 1 0

2 3 . 4 k − 29

This matrix can be reduced further to I4 if (and only if) k − 29 6= 0, that is, if k 6= 29. By Summary 3.3.10, the four given vectors form a basis of R4 unless k = 29. 3.3.29 x1 = − 23 x2 − 12 x3 ; let x2 = s and x3 = t. Then the solutions are of the form 3 1 3 − 2 s − 21 t x1 −2 −2 x2 = = s 1 + t 0 . s x3 0 1 t

−1 −3 Multiplying the two vectors by 2 to simplify, we obtain the basis 2 , 0 . 2 0

1 −1 −2 2 0 0 3.3.30 Proceeding as in Exercise 29, we find the basis , , . 0 1 0 0 0 1 1 −2 −4 1 0 0 3.3.31 Proceeding as in Exercise 29, we can find the following basis of V : , , . 0 1 0 0 0 1 Now let A be the 4 × 3 matrix with these three vectors as its columns. Then im(A) = V by Theorem 3.1.3, and ker(A) = {~0} by Theorem 3.2.8, so that A does the job. 1 −2 −4 0 0 1 A= . 0 1 0 0 0 1

0 1 x1 x1 x2 1 x2 0 4 3.3.32 We need to find all vectors ~x in R such that · = 0 and · = 0. 2 −1 x3 x3 3 x4 1 x4 x1 − x3 + x4 = 0 , which in turn amounts to finding the This amounts to solving the system x + 2x3 + 3x4 = 0 2 1 0 −1 1 kernel of . 0 1 2 3

143

Chapter 3 −1 1 −2 −3 Using Kyle Numbers, we find the basis . , 0 1 1 0

3.3.33 We can write V = ker(A), where A is the 1 × n matrix A = [c1 c2 · · · cn ]. Since at least one of the ci is nonzero, the rank of A is 1, so that dim(V ) = dim(ker(A)) = n − rank(A) = n − 1, by Theorem 3.3.7. A “hyperplane” in R2 is a line, and a “hyperplane” in R3 is just a plane. 3.3.34 We can write V = ker(A), where A is the n × m matrix with entries aij . Note that rank(A) ≤ n. Therefore, dim(V ) = dim(ker(A)) = m − rank(A) ≥ m − n, by Theorem 3.3.7. 3.3.35 We need to find all vectors ~x in Rn such that ~v · ~x = 0, or v1 x1 + v2 x2 + · · · + vn xn = 0, where the vi are the components of the vector ~v . These vectors form a hyperplane in Rn (see Exercise 33), so that the dimension of the space is n − 1. 3.3.36 No; if im(A) = ker(A) for an n × n matrix A, then n = dim(ker(A)) + dim(im(A)) = 2 dim(im(A)), so that n is an even number. 3.3.37 Since 1 0 A= 0 0

dim(ker(A)) = 5 − rank(A), any 4 × 5 matrix with rank 2 will do; for example, 0 0 0 0 1 0 0 0 . 0 0 0 0 0 0 0 0

3.3.38 a The rank of a 3 × 5 matrix A is 0,1,2, or 3, so that dim(ker(A)) = 5 − rank(A) is 2,3,4, or 5. b The rank of a 7 × 4 matrix A is at most 4, so that dim(im(A)) = rank(A) is 0,1,2,3, or 4. 3.3.39 Note that ker(C) 6= {~0}, by Theorem 3.1.7a, and ker(C) ⊆ ker(A). Therefore, ker(A) 6= {~0}, so that A is not invertible. 3.3.40 Substituting a point (x, y) = (xi , yi ) into the equation c1 + c2 x + c3 y + c4 x2 + c5 xy + c6 y 2 = 0 produces a linear equation in the six unknows c1 , ..., c6 . Fitting a conic through m points Pi (xi , yi ), for i = 1, ..., m, amounts to solving a system of m homogenous linear equations in six unknowns. This system can be written in matrix form as A~x = ~0, where A is an m × 6 matrix. The ith row of A is 1 xi yi x2i xi yi yi2 .

3.3.41 The kernel of a 4 × 6 matrix is at least two-dimensional. Since every one-dimensional subspace of this kernel defines a conic through the four given points, there will be infinitely many such conics. 3.3.42 The kernel of a 5 × 6 matrix is at least one-dimensional, so that there is at least one conic passing through five given points. Exercise 1.2.52 provides an example with exactly one solution, while there are infinitely many solutions in Exercise 1.2.54. 3.3.43 Here, “anything can happen”: 144

Section 3.3 n o 1. If ker A = ~0 , then there is no solution, as in Exercise 1.2.59

2. If dim ker A = 1, then there is a unique solution. For example, the only conic through the six points (0, 0) , (1, 0) , (2, 0) , (3, 0) , (0, 1) , (0, 2) is xy = 0. 3. If dim ker A > 1, then there are infinitely many solutions. For example, any conic consisting of the x axis and some other line runs through the six points (0, 0) , (1, 0) , (2, 0) , (3, 0) , (4, 0) , (5, 0).

3.3.44 To run through the points (0, 0) , (1, 0) , (2, 0), and (3, 0), the cubic must satisfy the equations c1 = 0, c1 + c2 + c4 + c7 = 0, c1 + 2c2 + 4c4 + 8c7 = 0, and c1 + 3c2 + 9c4 + 27c7 = 0. This means that c1 = c2 = c4 = c7 = 0. Likewise, the cubic runs through the points (0, 0) , (0, 1) , (0, 2) and (0, 3) if (and only if) c1 = c3 = c6 = c10 = 0. Therefore, to run through the first seven given points, the equation must be of the form c5 xy + c8 x2 y + c9 xy 2 = 0. The last point, (1, 1) imposes the condition c5 + c8 + c9 = 0, or c5 = −c8 − c9 . Setting c8 = a and c9 = b, we find the family of cubics (−a − b) xy + ax2 y + bxy 2 = xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Such a cubic is the union of the x axis, the y axis, and some line through the point (1, 1). Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x 1

-1

2

1

-1

4

3

2

3

4

-1

-1

3.3.45 To run through the points (0, 0) , (1, 0) , (2, 0), and (3, 0), the cubic must satisfy the equations c1 = 0, c1 + c2 + c4 + c7 = 0, c1 + 2c2 + 4c4 + 8c7 = 0, and c1 + 3c2 + 9c4 + 27c7 = 0. This means that c1 = c2 = c4 = c7 = 0. Likewise, the cubic runs through the points (0, 0) , (0, 1) , (0, 2) and (0, 3) if (and only if) c1 = c3 = c6 = c10 = 0. Therefore, to run through the first seven given points, the equation must be of the form c5 xy + c8 x2 y + c9 xy 2 = 0 . The last two point, (1, 1) and (2, 2), impose the conditions c5 + c8 + c9 = 0 and 4c5 + 8c8 + 8c9 = 0 , implying that c5 = 0 and c8 = −c9 . Letting c9 = 1, we find that there is one such cubic, −x2 y + xy 2 = xy (y − x) = 0, the union of the x axis, the y axis, and the line y = x. y 4

3

2

1

x 1

-1

2

3

4

-1

3.3.46 Here we add the point (4, 0) to the list of points in Exercise 44 . In Exercise 44 we found the family of cubics xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Since all these cubics run through the point (4, 0), we find the same solutions here: xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. 145

Chapter 3 y

y 4

4

3

3

2

2

1

1

x

x 1

-1

2

1

-1

4

3

2

3

4

-1

-1

3.3.47 Here we add the point (2, 1) to the list of points in Exercise 45 . In Exercise 45 we found the cubic xy (y − x) = 0. Since this cubic fails to run through the point (2, 1), there are no solutions here. y 4

3

2

1

x 1

-1

2

4

3

-1

3.3.48 Here we add the point (3, 3) to the list of points in Exercise 45 . In Exercise 45 we found the cubic xy (y − x) = 0. Since this cubic runs though the point (3, 3), we get the same solution here, xy (y − x) = 0. y 4

3

2

1

x 1

-1

2

4

3

-1

3.3.49 Here we add the point (0, 4) to the list of points in Exercise 46 . In Exercise 46 we found the family of cubics xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Since all these cubics run through the point (0, 4), we find the same solutions here: xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Each such cubic is the union of the x axis, the y axis, and some line through the point (1,1). Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x 1

-1

2

3

1

-1

4

2

3

4

-1

-1

3.3.50 To run through the points (0, 0) , (1, 0) , and (2, 0), the cubic must satisfy the equations c1 = 0, c1 +c2 +c4 + c7 = 0, and c1 + 2c2 + 4c4 + 8c7 = 0. This means that c1 = 0, c2 = 2c7 and c4 = −3c7 . Likewise, the cubic runs 146

Section 3.3 through the points (0, 0) , (0, 1) , and (0, 2) if (and only if) c1 = 0, c3 = 2c10 and c6 = −3c10 . Therefore, to run through the points (0, 0), (1, 0), (2,0), (0, 1), (0, 2) the cubic must be of the form c5 xy + c7 x3 − 3x2 + 2x + c8 x2 y + c9 xy 2 + c10 y 3 − 3y 2 + 2y = 0. The three additional points, (1, 1) , (2, 1) , and (1, 2) impose the conditions c5 + c8 + c9 = 0, 2c5 + 4c8 + 2c9 = 0 and 2c5 + 2c8 + 4c 9 = 0, implying that c5 = c8 = c9 = 0. Letting c7 = a and c10 = b, we find the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x -1

1

2

3

1

-1

4

2

3

4

-1

-1

3.3.51 Here we add the point (3, 2) to the list of points in Exercise 50 . In Exercise 50 we found the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. The point (3, 2) now imposed the condition a = 0, so that the only solution here is y 3 − 3y 2 + 2y = 0, the union of the lines y = 0, y = 1, and y = 2. y 4

3

2

1

x 1

-1

2

3

4

-1

3.3.52 Here we add the point (2, 2) to the list of points in Exercise 50 . In Exercise 50 we found the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0or b 6= 0. Since all these cubics run through the point (2, 2), we find the same solutions here: a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x -1

1

2

3

1

-1

4

2

3

4

-1

-1

3.3.53 Here we add the point (3, 3) to the list of points in Exercise 52 . In Exercise 52 we found the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. The point (3, 3) now imposes the condition a = −b, so that the only cubic here (with b = 1) is −2x + 2y + 3x2 − 3y 2 − x3 + y 3 = 0, or x (x − 1) (x − 2) = y (y − 1) (y − 2). 147

Chapter 3 y 4

3

2

1

x 1

-1

2

3

4

-1

3.3.54 Substituting a point (x, y) = (xi , yi ) into the equation c1 + c2 x + ... + c9 xy 2 + c10 y 3 = 0 produces a linear equation in the ten unknows c1 , ..., c10 . Fitting a cubic through m points Pi (xi , yi ), for i = 1, ..., m, amounts to solving a system of m homogenous linear equations in ten unknowns. This system can be written in matrix form as A~x = ~0, where A is an m × 10 matrix. The ith row of A is 1 xi yi x2i xi yi yi2 x3i x2i yi xi yi2 yi3 . 3.3.55 The kernel of a 8 × 10 matrix is at least two-dimensional. Since every one-dimensional subspace of this kernel defines a cubic through the eight given points, there will be infinitely many such cubics. 3.3.56 The kernel of a 9 × 10 matrix is at least one-dimensional, so that there is at least one cubic passing through nine given points. Exercise 51 provides an example with exactly one solution, while there are infinitely many solutions in Exercise 52. 3.3.57 Here, “anything can happen”: n o If ker A = ~0 , then there is no solution, as in Exercise 47

If dim ker A = 1, then there is a unique solution, as in Exercise 48 If dim ker A > 1, then there are infinitely many solutions, as in Exercise 49

3.3.58 It is not true that “Any nine distinct points determine a unique cubic,” as illustrated in Exercises 46 and 52, although “most” sets of nine distinct points do determine a unique cubic. 3.3.59 Here we add the point P to the list of points in Exercise 44 . In Exercise 44 we found the family of cubics xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. As long as we choose a point P on the x or on the y axis (with y = 0 or x = 0), all these cubics will run through P, so that there are infinitely many cubics through the nine given points. (Another, “cheap” solution is P = (1, 1)). However, if P is neither (1, 1) nor located on one of the axes, then the only cubic through the nine points is the union of the two axes and the line through (1, 1) and P.

148

Section 3.3 3.3.60 We can choose a basis ~v1 , . . . , ~vp in V , where p = dim(V ). Then ~v1 , . . . , ~vp are linearly independent vectors in W , so that dim(V ) = p ≤ dim(W ), by Theorem 3.3.4a, as claimed. 3.3.61 We can choose a basis ~v1 , . . . , ~vp of V , where p = dim(V ) = dim(W ). Then ~v1 , . . . , ~vp is a basis of W as well, by Theorem 3.3.4c, so that V = W = span(~v1 , . . . , ~vp ), as claimed. 3.3.62 Consider a basis ~v1 , . . . , ~vn of V . Since the ~vi are n linearly independent vectors in Rn , they form a basis of Rn (by parts (vii) and (ix) of Summary 3.3.10), so that V = span(~v1 , . . . , ~vn ) = Rn , as claimed. (Note that Exercise 3.3.62 is a special case of Exercise 3.3.61.) 3.3.63 dim(V + W ) = dim(V ) + dim(W ), by Exercise 3.2.51b. 3.3.64 Suppose that V ∩ W = {~0} and dim(V ) + dim(W ) = n. Choose a basis ~v1 , . . . , ~vp of V and a basis w ~ 1, . . . , w ~ q in W ; note that p + q = n. By Exercise 3.2.51b, the n vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q in Rn are linearly independent, so that they form a basis of Rn (by parts (vii) and (ix) of Summary 3.3.10). By Theorem 3.2.10, any vector ~x can be written uniquely as ~x = (c1~v1 + · · · + cp~vp ) + (d1 w ~ 1 + · · · + dq w ~ q ), with ~v = c1~v1 + · · · + cp~vp in V and w ~ = d1 w ~ 1 + · · · + dq w ~ q in W , which gives the desired representation. Conversely, suppose V and W are complements. Let us first show that V ∩ W = {~0} in this case. Indeed, if ~x is in V ∩ W , then we can write ~x = ~x + ~0 = ~0 + ~x ↑ ↑ in in V W

↑ ↑ in in V W

Since this representation is unique (by definition of complements), we must have ~x = ~0, so that V ∩ W = {~0}. By definition of complements, we have Rn = V + W , so that n = dim(V + W ) = dim(V ) + dim(W ), by Exercise 63. 3.3.65 Note that im(A) = span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ) = V , since the w ~ j alone span V . To find a basis of V = im(A), we omit the redundant vectors from the list ~v1 , . . . , ~vp , w ~ 1, . . . w ~ q , by Theorem 3.2.4. Since the vectors ~v1 , . . . , ~vp are linearly independent, none of them are redundant, so that our basis of V contains all vectors ~v1 , . . . , ~vp and some of the vectors from the list w ~ 1, . . . , w ~q.

3.3.66 Use Exercise 65

1 2 Now rref 3 4

1 4 6 8

1 0 0 0

1 1 4 2 with ~v1 = , ~v2 = , and 6 3 8 4 1 0 2 0 0 0 0 0 0 1 −1 0 0 1 0 0 = 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1

w ~ i = ~ei for i = 1, 2, 3, 4.

− 14

1 4 − 12 − 34

.

0 0 1 1 2 4 1 0 Picking the non-redundant columns gives the basis , , , . 1 0 6 3 0 0 8 4 149

Chapter 3 3.3.67 Using the terminology suggested in the hint, we need to show that ~u1 , . . . , ~um , ~v1 , . . . , ~vp , w ~ 1, . . . , w ~q is a basis of V + W . Then dim(V + W ) + dim(V ∩ W ) = (m + p + q) + m = (m + p) + (m + q) = dim(V ) + dim(W ), as claimed. Any vector ~x in V + W can be written as ~x = ~v + w, ~ where ~v is in V and w ~ is in W . Since ~v is a linear combination of the ~ui and the ~vj , and w ~ is a linear combination of the ~ui and w ~ j , ~x will be a linear combination of the ~ui , ~vj , and w ~ k ; this shows that the vectors ~u1 , . . . , ~um , ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q span V + W . To show linear independence, consider the relation a1 ~u1 + · · · + am ~um + b1~v1 + · · · + bp~vp + c1 w ~ 1 + · · · + cq w ~ q = ~0. Then the vector a1 ~u1 + · · · + am ~um + b1~v1 + · · · + bp~vp = −c1 w ~ 1 − · · · − cq w ~ q is in V ∩ W , so that it can be expressed uniquely as a linear combination of ~u1 , . . . , ~um alone; this implies that the bi are all zero. Now our relation simplifies to a1 ~u1 + · · · + am ~um + c1 w ~ 1 + · · · + cq w ~ q = ~0, which implies that the ai and the cj are zero as well (since the vectors ~u1 , . . . , ~um , w ~ 1, . . . , w ~ q are linearly independent). 3.3.68 By Exercise 3.3.67, dim(V ∩ W ) = dim(V ) + dim(W ) − dim(V + W ) = 13 − dim(V + W ). The dimension of V + W is at least 7 (since W ⊆ V + W ) and at most 10 (since V + W ⊆ R10 ); therefore the dimension of V ∩ W is at least 3 and at most 6. 3.3.69 The nonzero rows of E span the row space, and they are linearly independent (consider the leading ones), so that they form a basis of the row space: [0 1 0 2 0], [0 0 1 3 0], [0 0 0 0 1]. 3.3.70 As in Exercise 3.3.69, we observe that the nonzero rows of E form a basis of the row space, so that dim(row space of E) = rank(E). 3.3.71 a All elementary row operations leave the row space unchanged, so that A and rref(A) have the same row space. b By part (a) and Exercise 3.3.70, dim(row space of A) = dim(row space of rref(A)) = rank(rref(A)) = rank(A). 1 0 3.3.72 rref(A) = 0 0

0 −1 1 2 0 0 0 0

−2 3 0 0

By Exercises 3.3.70 and 3.3.71a, [1 0 − 1 − 2], [0 1 2 3] is a basis of the row space of A.

3.3.73 Using the terminology suggested in the hint, we observe that the vectors ~v , A~v , . . . , An~v are linearly dependent (by Theorem 3.2.8), so that there is a nontrivial relation c0~v + c1 A~v + · · · + cn An~v = ~0. We can rewrite this relation in the form (c0 In + c1 A + · · · + cn An )~v = ~0.

The nonzero vector ~v is in the kernel of the matrix c0 In + c1 A + · · · + cn An , so that this matrix fails to be invertible. 1 3.3.74 We can use the approach outlined in Exercise 3.3.73, with ~v = , say. 0 −3 1 −3 −4 1 1 2 . = , and A ~v = , A~v = Then ~v = 4 0 4 −3 2 0 We find the relation 5~v − 2A~v + A2~v = ~0, so that the matrix 5I2 − 2A + A2 does the job. 150

Section 3.3 3.3.75 If rank(A) = n, then the n non-redundant columns of A form a basis of im(A) = Rn , so that the matrix formed by the non-redundant columns is invertible (by Summary 3.3.10). Conversely, if A has an invertible n × n submatrix B, then the columns of B form a basis of Rn (again by Summary 3.3.10), so that im(A) = Rn and therefore rank(A) = dim(im(A)) = n. 3.3.76 Using the terminology suggested in the Exercise, we multiply the relation c0~v + c1 A~v + · · · + cm−1 Am−1~v = ~0 with Am−1 and obtain c0 Am−1~v = ~0 (all other terms vanish since Am = 0). Since the vector Am−1~v is nonzero (by construction), the scalar c0 must be zero, and our relation simplifies to c1 A~v + c2 A2~v + · · · + cm−1 Am−1~v = ~0. Now we multiply both sides with Am−2 and obtain c1 Am−1~v = ~0, so that c1 = 0 as above. Continuing like this we conclude that all the ci must be zero, as claimed. 3.3.77 As in Exercise 76, let m be the smallest positive integer such that Am = 0. In Exercise 76 we construct m linearly independent vectors ~v , A~v , . . . , Am−1~v in Rn ; now m ≤ n by Theorem 3.2.8. Therefore An = Am An−m = 0An−m = 0, as claimed. 3.3.78 If the vectors w ~ 1, . . . , w ~ q span an m-dimensional space V (with basis ~v1 , . . . , ~vm ), then m ≤ q by Theorem 3.3.1 (since the vectors ~vi are linearly independent). 3.3.79 Prove Theorem 3.3.4d: If m vectors ~v1 , . . . , ~vm span an m-dimensional space V , then they form a basis of V . We need to show that the vectors ~vi are linearly independent. We will argue indirectly, assuming that the vectors are linearly dependent; this means that at least one of the vectors ~vi is redundant, say ~vp . But then V = span(~v1 , . . . , ~vp , . . . , ~vm ) = span(~v1 , . . . , ~vp−1 , ~vp+1 , . . . , ~vm ), contradicting Theorem 3.3.4b. 3.3.80 im(A) is the plane onto which we project, so that rank(A) = dim(im(A)) = 2. 3.3.81 a Note that rank(B) ≤ 2, so that dim(ker(B)) = 5 − rank(B) ≥ 3 and dim(ker(AB)) ≥ 3 since ker(B) ⊆ ker(AB). Since ker(AB) is an subspace of R5 , dim(ker(AB)) could be 3,4, or 5. It is easy to give an example 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 1 , then AB = for each case; for example, if A = and and B = 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 dim(ker(AB)) = 3. b Since dim(im(AB)) = 5 − dim(ker(AB)), the possible values of dim(im(AB)) are 0,1, and 2, by part a. 3.3.82 Write A = [~v1 . . . ~vm ] and B = [w ~1 . . . w ~ m ], so that A + B = [~v1 + w ~ 1 · · · ~vm + w ~ m ]. Any linear combination of the columns of A + B, ~y = c1 (~v1 + w ~ 1 ) + · · · + cm (~vm + w ~ m ), can be written as ~ 1 + · · · + cm w ~ m) ~y = (c1~v + · · · + cm~vm ) + (c1 w {z } | {z } | in im(A)

in im(B)

so that im(A + B) ⊆ im(A) + im(B) (see Exercise 3.2.50). Since dim(V + W ) ≤ dim(V ) + dim(W ), by Exercise 3.3.67, we can conclude that rank(A+B) = dim(im(A+B)) ≤ dim(im(A))+dim(im(B)) = rank(A)+rank(B). Summary: rank(A + B) ≤ rank(A) + rank(B). 151

Chapter 3 3.3.83 a By Exercise 3.1.39b, im(AB) ⊆ im(A), and therefore rank(AB) ≤ rank(A). b Write B = [~v1 · · · ~vm ] and AB = [A~v1 · · · A~vm ]. If r = rank(B), then the r non-redundant columns of B will span im(B), and the corresponding r columns of AB will span im(AB), by linearity of A. By Theorem 3.3.4b, rank(AB) = dim(im(AB)) ≤ r = rank(B). Summary: rank(AB) ≤ rank(A), and rank(AB) ≤ rank(B). 3.3.84 Same answer as Exercise 3.3.85. 3.3.85 Let ~v1 , . . . , ~v6 be the columns of matrix A. Following the hint, we observe that ~v5 = 4~v1 + 5~v2 + 6~v4 , which gives the relation 4~v1 + 5~v2 + 6~v4 − ~v5 = ~0. Thus the vector 4 5 0 ~x = 6 −1 0 is in the kernel of matrix A. Since ~x fails to be in the kernel of matrix B, the two kernels are different, as claimed.

3.3.86 We will freely use the terminology introduced in the hint. First we need to show that at ileast one of the h ~ ~ column vectors ~ak and bk fails to contain a leading 1. If rank[~a1 · · · ~ak−1 ] = rank b1 · · · ~bk−1 = r, and if ~ak contains a leading 1, then ~ak is the standard vector ~er+1 ; likewise for ~bk . Since ~ak and ~bk are different vectors, they cannot both contain a leading 1. Without loss of generality, we can assume that ~ak fails to contain a leading 1, so that ~ak is redundant: We can write ~ak = c1~a1 + · · · + ck−1~ak−1 . Then the vector c1 .. . ck−1 ~x = −1 is in the kernel of A. We will show that ~x fails to be in the kernel of matrix B, so that ker(A) 6= 0 . .. 0 ker(B), as claimed. Indeed, B~x = c1~b1 + · · · + ck−1~bk−1 − ~bk = c1~a1 + · · · + ck−1~ak−1 − ~bk = ~ak − ~bk 6= ~0. We have used the fact that the first k − 1 columns of B are identical to those of A, while ~bk 6= ~ak . 3.3.87 Exercise 86 shows that if two matrices A and B of the same size are both in rref and have the same kernel, then A = B. Apply this fact to A and B = rref(M ).

Section 3.4 2 0 1 2 . , so [~x]B = +3 =2 3.4.1 3 1 0 3 3.4.2

0 3 1 −4 = −4 +3 . , so [~x]B = −4 1 3 0 152

Section 3.4

3.4.3

0 31 23 31 . , so [~x]B = +1 =0 1 37 29 37

3.4.4

23 = 29

3.4.5

−4 5 2 7 . , so [~x]B = +3 = −4 3 12 5 16

1 2

1 61 46 , so [~x]B = 2 . +0 0 67 58

This may not be as obvious as Exercises 1 and 3, but we can find our coefficients simply by reducing the matrix 2 5 .. 7 . . 5 12 16 3.4.6

−4 1 5 11 = 11 −3 , so [~x]B = . 4 2 6 −3

We arrive at this solution by reducing the matrix

1 2

5 .. −4 . . 6 4

0 1 3 3.4.7 We need to find the scalars c1 and c2 such that 1 = c1 −1 + c2 1 . Solving a linear system gives −1 0 −4 c1 3 c1 = 3, c2 = 4. Thus [~x]B = = . c2 4

2 1 2 3.4.8 We need to find the scalars c1 and c2 such that 3 = c1 1 + c2 0 . Attempting to solve the linear 1 0 4 system reveals an inconsistency; ~x is not in the span of ~v1 and ~v2 . 3.4.9 We can solve this by inspection: Note that our first coefficient must be 3 because of the first terms of the vectors. Also, the second coefficient must be 2 due to the last terms. 3 0 3 However, 3~v1 + 2~v2 = 3 + −2 = 1 . Thus, we reason that ~x is not in the span of ~v1 and ~v2 . 4 4 0 1 0 .3 We can also see this by attempting to solve 1 −1 .. 3 , which turns out to be inconsistent. Thus, ~x is not in 0 2 4 V. 3 3.4.10 Proceeding as in Example 1, we find [~x]B = . 1 3.4.11 Proceeding as in Example 1, we find [~x]B =

1 2 1 2

.

−3 . 3.4.12 Proceeding as in Example 1, we find [~x]B = 5

153

Chapter 3 3.4.13 Here, we quickly see that since x1 = 1 = 1c1 + 0c2 + 0c3 , c1 must equal 1. We find c2 = −1 similarly, since x2 = 1 = 2(1) + 1c2 + 0c3 . Finally, now that x3 = 3(1) + 2(−1) + 1c3 , c3 must be zero. 1 0 0 1 1 So 1 = 1 2 − 1 1 + 0 0 , and [~x]B = −1 . 0 1 2 3 1 3.4.14 We proceed by inspection here, noting that we need c1 = 3, then see that c2 must be 4. Finally, c3 must be 6. 3 0 0 1 3 Thus, 7 = 3 1 + 4 1 + 6 0 , and [~x]B = 4 . 6 1 1 1 13 1 1.1 4 .. 0 to 0 0 8 0

1 1 3.4.15 This may be a bit too difficult to do by inspection. Instead we reduce 2 3 1 4

0 0. 8 1 0 .. −12 , 0 1 5

8 revealing that ~x = 8~v1 − 12~v2 + 5~v3 , and [~x]B = −12 . 5

1 1 3.4.16 We reduce 1 2 1 3

1 0 1.7 3 .. 1 to 0 1 0 0 6 3

0 . 21 0 .. −22 , 1 8

21 revealing that ~x = 21~v1 − 22~v2 + 8~v3 , and [~x]B = −22 . 8 3.4.17 By inspection, we see that in order for ~x to be in V , ~x = 1~v1 +1~v2 −1~v3 (by paying attention to the first, second, and fourth terms). Now we need to verify that the third terms “work out”. So, 1(2)+1(3)−1(4) = 5−4 = 1 = x3 .

1 Thus ~x is in V , and [~x]B = 1 . −1 3.4.18 Here, ~x is not in V, as we find an inconsistency while attempting to solve the system.

1 3.4.19 a S = 1 Then B = S

−1

1 , and we find the inverse S −1 to be equal to −1 AS =

1 2

1 1

1 −1

0 1

1 0

1 1 = 1 −1

1 2

b Our commutative diagram: 154

1 1 −1 1

1 2

1 1 . 1 −1

1 1 = 1 −1

1 2

2 0

0 1 = −2 0

0 . −1

Section 3.4 1 1 + c2 ~x = c1 −1 1 y

c [~x]B = 1 c2 So,

a b c d

c1 c2

1 1 T (~x) = A~x = c1 A + c2 A −1 1 1 −1 1 1 = c1 + c2 = c1 − c2 1 1 1 −1 y c1 −−−→ B [T (~x)]B = −c2 −−→ T

c B = [[T (~v1 )]B [T (~v2 )]B ] =

3.4.20 a S =

0 1

1 0

1 1 B

0 1 1 0

1 −1

−1

AS =

1 2

1 1

1 −1

1 1

1 1

1 1 = 1 −1

1 2

=

B

1 , and we find the inverse S −1 to be equal to −1

1 1

Then B = S

0 . −1

c1 1 = , and we quickly find B = −c2 0

2 2 0 0

1 2

1 −1 1 0 = . 1 B 1 B 0 −1

1 1 . 1 −1

1 1 = 1 −1

1 2

0 2 0 = . 0 0 0

4 0

b Our commutative diagram: 1 1 + c2 A T (~x) = A~x = c1 A −1 1 2 0 1 1 = c1 + c2 = 2c +0 2 0 1 1 −1 y 2c1 −−−→ B [T (~x)]B = 0

1 1 + c2 ~x = c1 −1 1 y

[~x]B =

So,

a b c d

c1 c2

c1 c2

−−→ T

2c1 2 0 . = , and we quickly find B = 0 0 0

c B = [[T (~v1 )]B [T (~v2 )]B ] =

3.4.21 a S =

1 3

Then B = S

−1

1 1

1 1

1 1 B

1 1 1 1

1 −1

B

−2 , and we find the inverse S −1 to be equal to 1 AS =

1 7

1 −3

2 1

1 3

2 6

1 −2 = 3 1

1 7

b Our commutative diagram: 155

7 14 0 0

1 7

2 0 2 0 = = . 2 B 0 B 0 0

1 2 . −3 1

1 −2 = 3 1

1 7

49 0

0 7 0 = . 0 0 0

Chapter 3 −2 1 + c2 ~x = c1 1 3

1 −2 T (~x) = A~x = c1 A + c2 A 3 1 7 0 1 = c1 + c2 = 7c1 21 0 3 y y 7c1 c1 −−−→ B [T (~x)]B = [~x]B = 0 c2 c1 a b 7c1 7 0 . So, = , and we quickly find B = c d c2 0 0 0 −−→ T

c B = [[T (~v1 )]B [T (~v2 )]B ] = =

7 21

B

1 2 3 6 0 . 0

0 7 = 0 B 0

1 3 B

1 2 3 6

−2 1

B

1 2 −2 1 −1 . , and we find the inverse S to be equal to 5 −2 1 1 1 2 5 10 −3 4 1 −2 1 −2 Then B = S −1 AS = 15 = = 15 −2 1 10 −5 4 3 2 1 2 1

1 3.4.22 a S = 2

1 5

25 0

b Our commutative diagram: −2 1 −−→ T + c2 ~x = c1 1 2

−2 1 + c2 A T (~x) = A~x = c1 A 1 2 5 10 1 −2 = c1 + c2 = 5c1 − 5c2 10 −5 2 1 y y 5c1 c1 −−−→ B [T (~x)]B = [~x]B = −5c2 c2 a b c1 5c1 5 0 So, = , and we quickly find B = . c2 c d −5c2 0 −5

c B = [[T (~v1 )]B [T (~v2 )]B ] = =

5 10

B

10 −5

B

=

5 0

−3 4

4 3

1 2 B

−3 4 4 3

−2 1

B

0 . −5

2 −1 1 −1 . , and we find the inverse S to be equal to −1 1 2 2 0 1 1 4 −2 1 1 5 −3 2 −1 −1 . = = Then B = S AS = 0 −1 1 2 1 −1 1 2 6 −4 −1 1

1 3.4.23 a S = 1

156

0 5 0 = . −25 0 −5

Section 3.4 b Our commutative diagram: 1 1 −−→ + c2 ~x = c1 T 2 1 y

c [~x]B = 1 c2 So,

a b c d

c1 c2

1 1 T (~x) = A~x = c1 A + c2 A 1 2 1 1 −1 2 − 1c2 = 2c1 + c2 = c1 2 1 −2 2 y 2c1 −−−→ B [T (~x)]B = −c2

2c1 2 = , and we quickly find B = −c2 0

c B = [[T (~v1 )]B [T (~v2 )]B ] = 2 = 2 B 3.4.24 a S =

−1 −2

5 6

−3 −4

1 1 B

5 −3 6 −4

1 2 B

2 0 = . 0 −1

B

3 −5 5 . , and we find the inverse S −1 to be equal to −1 2 3

2 1

Then B = S

0 . −1

−1

3 −5 AS = −1 2

13 6

−20 −9

2 5 9 −15 2 = 1 3 −1 2 1

5 3 = 3 0

0 . 1

b Our commutative diagram: 5 2 −−→ T + c2 ~x = c1 3 1 y

[~x]B =

So,

a b c d

c1 c2

c1 c2

5 2 + c2 A T (~x) = A~x = c1 A 3 1 6 5 2 5 = c1 + c2 = 3c + 1c2 3 3 1 1 3 y 3c1 −−−→ B [T (~x)]B = c2

=

3c1 3 0 . , and we quickly find B = c2 0 1

c B = [[T (~v1 )]B [T (~v2 )]B ] = 6 = 3 B

13 6

−20 −9

2 1 B

13 6

−20 −9

5 3 B

5 3 0 . = 3 B 0 1

3.4.25 We will use the commutative diagram method here (though any method suffices). 157

Chapter 3 1 1 + c2 ~x = c1 2 1

1 2 1 2 1 1 T (~x) = A~x = c1 +c 3 4 1 2 3 4 2 3 5 = c1 + c2 7 11 1 1 1 1 +6 + c2 −1 +4 = c1 −1 1 2 1 2 1 1 + (4c1 + 6c2 ) = (−c1 − c2 ) 2 1 y y c1 −c1 − c2 −−−→ [~x]B = B [T (~x)]B = c2 4c1 + 6c2 c −c1 − c2 −1 −1 B 1 = , so B = . c2 4c1 + 6c2 4 6 −−→ T

3.4.26 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B ] 1 0 1 1 0 1 = 1 B 2 3 2 B 2 3 1 2 6 4 = . = 5 B 8 B −4 −3 3.4.27 We use a commutative diagram: 2 0 1 −−→ ~x = c1 1 + c2 2 + c3 0 T −2 1 1

y

c1 [~x]B = c2 c3 9 9c1 c1 B c2 = 0 , so B = 0 0 0 c3

T (~x) = A~x

1 0 2 = c1 A 1 + c2 A 2 + c3 A 0 1 1 −2 2 18 = c1 9 + ~0 + ~0 = 9c1 1 −2 −18 y 9c1 −−−→ B [T (~x)]B = 0 0 0 0 0 0 . 0 0

3.4.28 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] 158

Section 3.4

5 −4 −2 5 2 −4 = −4 5 −2 2 −2 −2 8 −2 1 B 0 9 0 0 −9 9 = 0 = 0 0 B 0 B −18 B 0

−4 −2 1 5 −2 −1 −2 8 0 B 0 0 9 0 . 0 9

5 −4 −2 0 −4 5 −2 1 −2 −2 8 −2 B

3.4.29 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] −1 1 0 1 −1 0 = 0 −2 2 1 3 −9 6 1 3 B 0 1 2 0 2 6 = 0 = 0 0 B 3 B 12 B 0

1 0 1 −2 2 2 −9 6 3 B 0 0 1 0 . 0 2

0 1 2 3 6 6 B

1 −1 0 2 4 1 B

−1 1 0 −2 3 −9

3.4.30 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] 0 2 1 0 2 −1 2 −1 = 2 −1 0 1 4 −4 1 4 −4 1 B 0 0 1 1 0 0 = 0 −1 −1 = 1 0 B −2 B 1 B 0 0

0 −1 0 1 2 1 B 0 0 . 0

0 2 2 −1 4 −4

3.4.31 We can use a commutative diagram to see how this works: ~x = c1~v1 + c2~v2 + c3~v3 y

c1 [~x]B = c2 c3 c3 c1 B c2 = 0 , −c1 c3

−−→ T

T (~x) = ~v2 × ~x = c1 (~v2 × ~v1 ) + c2 (~v2 × ~v2 ) + c3 (~v2 × ~v3 ) = c1 (−~v3 ) + c2 (~0) +c3 (~v1 ) = c3~v1 − c1~v3 y c3 −−−→ B [T (~x)]B = 0 −c1 0 0 1 so B = 0 0 0 . −1 0 0

3.4.32 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

= [ [~v1 × ~v3 ]B

[~v2 × ~v3 ]B

[T (~v3 )]B ] [~v3 × ~v3 ]B ] = [ [−~v2 ]B

[~v1 ]B ~0 ], since all three are perpendicular unit vectors. 159

Chapter 3

0 So, B = −1 0

1 0 0 0 . 0 0

3.4.33 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

[T (~v3 )]B ]

= [ [(~v2 · ~v1 )~v2 ]B [(~v2 · ~v2 )~v2 ]B 0 0 0 So, B = 0 1 0 . 0 0 0

[(~v2 · ~v3 )~v2 ]B ] = [ ~0

[1~v2 ]B ~0 ], since all three are perpendicular unit vectors.

3.4.34 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

[T (~v3 )]B ]

= [ [~v1 − 2(~v3 · ~v1 )~v3 ]B [~v2 − 2(~v3 · ~v2 )~v3 ]B [~v3 − 2(~v3 · ~v3 )~v3 ]B ] = [ [~v1 ]B [~v2 ]B [−~v3 ]B ]. 1 0 0 So, B = 0 1 0 . This is the reflection about the plane spanned by ~v1 and ~v2 . 0 0 −1 3.4.35 Using another commutative diagram: −−→ T ~x = c1~v1 + c2~v2 + c3~v3 T (~x) = c1 T (~v1 ) + c2 T (~v2 ) + c3 T (~v3 ) = c1 (~v1 − 2(~v1 · ~v1 )~v2 ) + c2 (~v2 − 2(~v1 · ~v2 )~v2 )+ c3 (~v3 − 2(~v1 · ~v3 )~v2 ) = c1 (~v1 − 2~v2 ) + c2 (~v2 − ~0) + c3 (~v3 − ~0) = c1~v1 + (−2c1+ c2 )~v2 + c3~v3 y y c1 c1 − − − → B [T (~x)]B = −2c1 + c2 [~x]B = c2 c3 c3 1 0 0 So B = −2 1 0 . This is a shear along the second term. 0 0 1 3.4.36 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

= [ [~v1 × ~v1 + (~v1 · ~v1 )~v1 ]B = [ [~v1 ]B

[T (~v3 )]B ] [~v1 × ~v2 + (~v1 · ~v2 )~v1 ]B

[~v1 × ~v3 + (~v1 · ~v3 )~v1 ]B ]

[~v3 ]B

[−~v2 ]B ]. 1 0 0 So, B = 0 0 −1 . This is a 90-degree rotation about the line spanned by ~v1 . The rotation is counterclockwise 0 1 0 when looking from the positive ~v1 direction. 160

Section 3.4 3.4.37 We want a basis B = (~v1 , ~v2 ) such that T (~v1 ) = a~ v1 a the B-matrix of T will be B = [ [T (~v1 )]B [T (~v2 )]B ] = 0 that T (~v ) = ~v = 1~v for vectors parallel to the line L onto perpendicular L. Thus, we can pick a basis where ~v1 is to 1 −2 B= , . 2 1

and T (~v2 ) = b~v2 for some scalars a and b. Then 0 , which is a diagonal matrix as required. Note b which we project, and T (w) ~ = ~0 = 0w ~ for vectors parallel to L and ~v2 is perpendicular, for example,

3.4.38 We want a basis B = (~v1 , ~v2 ) such that T (~v1 ) = T (~v2 ) = b~v2 for some scalars a and b. Then the a~v1 and a 0 , which is a diagonal matrix as required. Note that B-matrix of T will be B = [ [T (~v1 )]B [T (~v2 )]B ] = 0 b T (~v ) = ~v = 1~v for vectors parallel to the line L about which we reflect, and T (w) ~ = −w ~ = (−1)w ~ for vectors perpendicular to L. Thus, we can pick a basis where ~ v is parallel to L and ~ v is perpendicular, for example, 1 2 −3 2 . , B= 2 3 3.4.39 Using the same approach as in Exercise 37, we want a basis, ~v1 , ~v2 , ~v3 such that T (~v1 ) = a~v1 , T (~v2 ) = b~v2 and 1 T (~v3 ) = c~v3 . First we see that if ~v1 = 2 , then T (~v1 ) = ~v1 . Next we notice that if ~v2 and ~v3 are perpendicular 3 −3 −2 to ~v1 , then T (~v2 ) = −~v2 and T (~v3 ) = −~v3 . So we can pick ~v2 = 1 and ~v3 = 0 , for example. 1 0 3.4.40 From Exercise 37, we see that we want one of our basis vectors to parallel be to the line, while the others 1 1 1 must be perpendicular the line. We can easily find such a basis: B = 1 , −1 , 0 . 1 0 −1 3.4.41 We will use the same approach as in Exercises 37 and 39. Any basis with 2vectors and one in the plane 0 −1 perpendicular to it will work nicely here! So, let ~v1 , ~v2 be in the plane. ~v1 can be 3 , and ~v2 = −2 (note 0 1 3 that these must be independent). Then ~v3 should be perpendicular to the plane. We will use ~v3 = 1 —the 2 3 coefficient vector. This is perpendicular to the plane because all vectors perpendicular to 1 lie in the plane. 2

−1 0 3 So, our basis is: 3 , −2 , 1 . 0 1 2 3.4.42 From Exercise 38, we deduce that one of our vectors should be perpendicular to this plane, while two should 1 fall inside it. Finding the perpendicular is not difficult: we simply take the coefficient vector: −2 . Then 2 161

Chapter 3 0 2 we add two linearly independent vectors on the plane, 1 , 1 , for instance. These three vectors form one 1 0 possible basis. 4 −2 −1 3.4.43 By definition of coordinates (Definition 3.4.1), ~x = 2 0 + (−3) 1 = −3 . 2 0 1

11 5 8 3.4.44 By definition of coordinates, ~x = 2 4 + (−1) 2 = 6 . −1 −1 −1

3.4.45 If ~v1 , ~v2 is a basis with the desired property, then ~x = 2~v1 + 3~v2 , or ~v2 = 13 ~x − 32 ~v1 . Thus we can make ~v 1 any 3 vector in the plane that is not parallel to ~x, and then let ~v2 = 31 ~x − 32 ~v1 . For example, if we choose ~v1 = 2 , 0 −4 then ~v2 = 13 −4 . −1 3.4.46 As in Exercise 3.4.45, we canmake~v1 any vectorin the plane that is not parallel to ~x, and then let ~v2 = 2~v1 −~x. 1 1 For example, if we choose ~v1 = 0 , then ~v2 = 1 . −3 −1 3.4.47 By Theorem 3.4.4, we have A = SBS

3.4.48 [~x]B =

−1 2

−1

0 1 = 1 0

a b c d

0 1 1 0

−1

d = b

means that ~x = −~v + 2w. ~ See Figure 3.6

Figure 3.6: for Problem 3.4.48. −1 . 3.4.49 ~u + ~v = −w, ~ so that w ~ = −~u − ~v , i.e., [w] ~B= −1

1 2 −−→ −−→ −−→ −−→ . , OQ = ~v + 2w, ~ so that [OQ]B = 3.4.50 a OP = w ~ + 2~v , so that [OP ]B = 2 1 −−→ b OR = 3~v + 2w. ~ See Figure 3.7. 162

c . a

Section 3.4

Figure 3.7: for Problem 3.4.50. c If the tip of ~u is a vertex, then so is the tip of ~u + 3~v and also the tip of ~u + 3w ~ (draw a sketch!). We know that −→ the tip P of 2~v + w ~ is a vertex (see part a.). Therefore, the tip S of OS = 17~v + 13w ~ = (2~v + w) ~ + 5(3~v ) + 4(3w) ~ is a vertex as well. 3.4.51 Let B = (~v1 , ~v2 , · · · , ~vm ). Then, let ~x = a1~v1 + a2~v2 + · · · + am~vm and ~y = b1~v1 + b2~v2 + · · · + bm~vm . Then [~x + ~y ]B = [a1~v1 + a2~v2 + · · · + am~vm + b1~v1 + b2~v2 + · · · + bm~vm ]B = [(a1 + b1 )~v1 + (a2 + b2 )~v2 + · · · + (am + bm )~vm ]B b1 a1 a1 + b1 a2 + b2 a2 b2 = . + . = [~x]B + [~y ]B = .. .. .. . am

am + bm

bm

3.4.52 Yes; T (~x) = [~x]B = S −1 ~x, so that T is “given by a matrix.” (See Definition 2.1.1.)

1 3 3.4.53 By Definition 3.4.1, we have ~x = S[~x]B = 2 4

7 40 = . 11 58

3.4.54 Let Q be the matrix whose columns are the vectors of the basis T . Then [[~v1 ]T . . . [~vn ]T ] = [Q−1~v1 . . . Q−1~vn ] = Q−1 [~v1 . . . ~vn ] is an invertible matrix, so that the vectors [~v1 ]T . . . [~vn ]T form a basis of Rn . 1 1 1 1 3 1 1 [~x]B = [~x]R , so that [~x]B and ~x = 3.4.55 By Definition 3.4.1, we have ~x = 2 1 2 2 4 1" 2 # −1 1 − 1 1 3 1 1 2 [~x]R = [~x]B , i.e., P = . 1 2 4 1 2 0 2 | {z }

3 [~x]R and 4

P

1 3 3 2 3.4.56 Let S = [~v1~v2 ] where ~v1 , ~v2 is the desired basis. Then by Theorem 3.4.1, =S and =S , 2 5 4 3 −1 3 2 1 3 1 3 3 2 12 −7 12 −7 i.e. S = . Hence S = = . The desired basis is , . 5 3 2 4 2 4 5 3 14 −8 14 −8 3.4.57 If we can find a basis B = (~v1 , ~v2 , ~v3 ) such that the B-matrix of A is 1 0 0 1 0 0 B = 0 1 0 , then A must be similar to 0 1 0 . Because of the entries in the matrix B, it is required 0 0 −1 0 0 −1 that A~v1 = ~v1 , A~v2 = ~v2 and A~v3 = −~v3 . So, all we need for our basis is to pick independent ~v1 , ~v2 in the plane, and ~v3 perpendicular to the plane. 3.4.58 a Consider a linear relation c1 A2~v + c2 A~v + c3~v = ~0. 163

Chapter 3 Multiplying A2 with the vectors on both sides and using that A3~v = ~0 and A4~v = ~0, we find that c3 A2~v = ~0 and therefore c3 = 0, since A2~v 6= ~0. Therefore, our relation simplifies to c1 A2~v + c2 A~v = ~0. Multiplying A with the vectors on both sides we find that c2 A2~v = ~0 and therefore c2 = 0. Then c1 = 0 as well. We have shown that there is only the trivial relation among the vectors A2~v , A~v , and ~v , so that these three vectors from a basis of R3 , as claimed. 0 b T (A2~v ) = A3~v = ~0 so [T (A2~v )]B = 0 . 0 1 T (A~v ) = A2~v so [T (A~v )]B = 0 . 0 0 T (~v ) = A~v so [T (~v )]B = 1 . 0

0 Hence, by Theorem 3.4.3, the desired matrix is 0 0

1 0 0 1 . 0 0

x y x y 2 1 = , or z t z t 0 3 −y y 2x x + 3y 2x 2y , where y and t are arbitrary constants. . The solutions are of the form S = = 0 t 2z z + 3t 3z 3t 2 0 2 1 Since there are invertible solutions S (for example, let y = t = 1), the matrices and are indeed 0 3 0 3 similar.

3.4.59 First we find the matrices S =

x y z t

such that

2 0

0 3

1 0 x y x y 0 1 x y 3.4.60 First we find the matrices S = such that = , or, = z t z t 1 0 −z −t 0 −1 y x y y . The solutions are of the form S = , where y and t are arbitrary constants. Since there are t z −t t 0 1 1 0 are indeed similar. and invertible solutions S (for example, let y = t = 1), the matrices 1 0 0 −1 x z

y t

x y x y 3.4.61 We seek a basis ~v1 = , ~v2 = such that the matrix S = [~v1 ~v2 ] = satisfies the equation z t z t 3z z 3t −2 4− 2 1 1 x y x y −5 −9 . . Solving the ensuing linear system gives S = = 0 1 z t z t 4 7 z t −9 0 We need to choose z and t so that S will be invertible. For example, if we let z = 6 and t = 1, then S = , 6 1 0 −9 . , ~v2 = so that ~v1 = 1 6 164

Section 3.4 x y y x satisfies the equation such that the matrix S = [~v1 ~v2 ] = , ~v2 = 3.4.62 We seek a basis ~v1 = t zz t z −t 5 0 x y x y 1 2 . We need to choose . Solving the ensuing linear system gives S = 2 = z t 0 −1 z t z t 4 3 1 −1 both z and t nonzero to make S invertible. For example, if we let z = 2 and t = 1, then S = , so that 2 1 1 −1 ~v1 = , ~v2 = . 2 1 px − qz py − qt p q x y x y p −q x y = , or, = such that 3.4.63 First we find the matrices S = qx + pz qy + pt −q p z t z t q p z t px − qy qx + py −t z . If q 6= 0, then the solutions are of the form S = , where z and t are arbitrary conpz − qt qz + pt z t p q p −q and stants. Since there are invertible solutions S (for example, let z = t = 1), the matrices −q p q p are indeed similar. (If q = 0, then the two matrices are equal.)

3.4.64 If b and c are both zero, then the given matrices are equal, so that they are similar, by Theorem 3.4.6.a. Let’s now assume that at least one of the scalars b and c is nonzero; reversing the roles of b and c if necessary, we can assume that c 6= 0. x y a b x y x y a c Let’s find the matrices S = such that = , or z t c d z t z t b d ax + by cx + dy ax + bz ay + bt . The solutions are of the form = az + bt cz + dt cx + bz cy + dt " # (a−d)z+b z c S= , where z and t are arbitrary constants. Since there are invertible solutions S (for example, z t a b a c let z = 1, t = 0), the matrices and are indeed similar. c d b d 3.4.65 a If S = In , then S −1 AS = A. b If S −1 AS = B, then SBS −1 = A. If we let R = S −1 , then R−1 BR = A, showing that B is similar to A. 3.4.66 We build B “column-by-column”: 2 a + b2 − a ab + b − ba a−1 b = T B= T b2 + a2 − a B ba − b − ab B b 1−a B B b 1−a 1 0 = . = 1−a B −b B 0 −1 b b Thus, this matrix represents the reflection about the line spanned by . Note that the two vectors 1−a 1−a a−1 are perpendicular. and b 165

Chapter 3

3.4.67 The matrix we seek is

2 a a + bc 1 a 0 T T = = c B ac + cd B 0 B c B 1

3.4.68 Using Exercise 67 as a guide, consider the basis

bc − ad . a+d

1 a 1 1 1 , = , and let S = . 0 c 3 0 3

1 2 3 −2 3.4.69 The matrix of the transformation T (~x) = A~x with respect to the basis , is D = = 2 1 6 B −1 B 3 0 1 2 . Thus S −1 AS = D for S = . 0 −1 2 1

a b , then 0 c

3.4.70 Suppose such a basis ~v1 , ~v2 exists. If B = [[T (~v1 )]B [T (~v2 )]B ] is upper triangular, of the form a , so that T (~v1 ) = a~v1 , that is, T (~v1 ) is parallel to ~v1 . But this is impossible, since T is a rotation [T (~v1 )]B = 0 π through 2 . 3.4.71 a Note that AS = SB. If ~x is in ker(B), then A(S~x) = SB~x = S~0 = ~0, so that S~x is in ker(A), as claimed.

b We use the hint and observe that nullity (B) = dim(ker B) = p ≤ dim(ker A) = nullity(A), since S~v1 , . . . , S~vp are p linearly independent vectors in ker(A). Reversing the roles of A and B shows that, conversely, nullity(A) ≤ nullity(B), so that the equation nullity(A) = nullity(B) holds, as claimed. 3.4.72 If A and B are similar n × n matrices, then rank(A) = n − nullity(A) = n − nullity(B) = rank(B), by Exercise 71 and the rank nullity theorem (Theorem 3.3.7). 0 0.6 3.4.73 a By inspection, we can find an orthonormal basis ~v1 = ~v , ~v2 , ~v3 of R3 , ~v1 = ~v = 0.8 , ~v2 = 0 , 1 0 0.8 ~v3 = −0.6 . 0

Figure 3.8: for Problem 3.4.73b.

b Now T (~v1 ) = ~v1 , T (~v2 )= 1 basis ~v1 , ~v2 , ~v3 is B = 0 0

~v3 and T B of T with respect to the 3.8), so that the matrix (~v3 ) = −~v2 (see Figure 0 0 0.36 0.48 0.8 0 −1 .Then A = SBS −1 = 0.48 0.64 −0.6 . 1 0 −0.8 0.6 0 166

Section 3.4 0 −1 −1 1 1 3.4.74 a ~v0 + ~v1 + ~v2 + ~v3 = 1 + −1 + 1 + −1 = 0 . 0 1 −1 −1 1 −1 b If B is the basis ~v1 , ~v2 , ~v3 , then ~v0 + ~v1 + ~v2 + ~v3 = ~0 (by part a) so ~v0 = −~v1 − ~v2 − ~v3 , i.e. [~v0 ]B = −1 . −1

c T (~v2 ) = T (−~v0 − ~v1 − ~v3 ) = −T (~v0 ) − T (~v1 ) − T (~v3 ) = −~v3 − ~v0 − ~v1 = ~v2 Hence, T is a rotation through 120◦ about the line spanned by ~v2 . Its matrix, B, is given by [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] where −1 T (~v1 ) = ~v0 = −~v1 − ~v2 − ~v3 so [T (~v1 )]B = −1 −1 0 T (~v2 ) = ~v2 so [T (~v2 )]B = 1 0 1 T (~v3 ) = ~v1 so [T (~v3 )]B = 0 0 −1 0 1 and B = −1 1 0 . 1 0 0

B 3 = I3 since if the tetrahedron rotates through 120◦ three times, it returns to the original position.

3.4.75 B = S

−1

3.4.76 B = S

−1

0 AS, where S = A = 1

0 −1 −1 . Thus B = A = 1 0 0

cos(t) AS, where S = A = sin(t)

− sin(t) cos(t) . Thus B = A = cos(t) sin(t)

− sin(t) . cos(t)

3.4.77 Let S be the n × n matrix whose columns are ~en , ~en−1 , . . . , ~e1 . Note that S has all 1’s on “the other diagonal” and 0’s elsewhere: 1 if i + j = n + 1 sij = 0 otherwise Also, S −1 = S. Now, B = S −1 AS = SAS; the entries of B are bij = si,n+1−i an+1−i,n+1−j sn+1−j,j = an+1−i,n+1−j . Answer: bij = an+1−i,n+1−j B is obtained from A by reversing the order of the rows and of the columns. 167

Chapter 3 3.4.78 Note first that the diagonal entry sij of S gives the unit price of good i. If aij tells us how many dollars’ worth of good i are required to produce one dollar’s worth of good j, then aij sjj tells us how many dollars’ worth of good i are required to produce one unit of good j, and s−1 ii aij sjj is the number of units of good i required to produce one unit of good j. Thus bij = s−1 a s , and B = S −1 AS. ij jj ii

True or False Ch 3.TF.1 T, by Summary 3.3.10. Ch 3.TF.2 F, by Theorem 3.3.7. Ch 3.TF.3 T, by Summary 3.3.10. Ch 3.TF.4 F; The identity matrix is similar only to itself. Ch 3.TF.5 T; We have the nontrivial relation 3~u + 3~v + 3w ~ = ~0. Ch 3.TF.6 F; The columns could be ~e1 , ~e2 , ~e3 , ~e4 in R5 , for example. Ch 3.TF.7 T, by Theorem 3.3.2. Ch 3.TF.8 F; The nullity is 6 − 4 = 2, by Theorem 3.3.7. Ch 3.TF.9 F; It’s a subspace of R3 . Ch 3.TF.10 T; by Definition 3.1.2. Ch 3.TF.11 T, by Definition 3.2.3. Ch 3.TF.12 T, by Definition 3.2.1. Ch 3.TF.13 T; Check that

1 0

0 −1

1 −1

1 1 1 0 = 1 −1 1 1

1 . 0

Ch 3.TF.14 T, by Theorem 3.3.9. Ch 3.TF.15 T, by Theorem 3.2.8. Ch 3.TF.16 T, by Summary 3.3.10. Ch 3.TF.17 F; The number n may exceed 4. Ch 3.TF.18 T, by Definition 3.2.1 (V is closed under linear combinations) Ch 3.TF.19 T, by Theorem 3.4.6, parts b and c. 168

True or False

Ch 3.TF.20 F; Let V = span

1 in R2 , for example. 1

Ch 3.TF.21 T, since AB~v = A~0 = ~0. Ch 3.TF.22 T, by Definition 3.2.3. Ch 3.TF.23 F; Suppose ~v2 = 2~v1 . Then T (~v2 ) = 2T (~v1 ) = 2~e1 cannot be ~e2 . Ch 3.TF.24 F; Consider ~u = ~e1 , ~v = 2~e1 , and w ~ = ~e2 . Ch 3.TF.25 T, since A−1 (AB)A = BA. Ch 3.TF.26 T, since both kernels consist of the zero vector alone. Ch 3.TF.27 F; There is no invertible matrix S as required in the definition of similarity. Ch 3.TF.28 F; Five vectors in R4 must be dependent, by Theorem 3.2.8. Ch 3.TF.29 T, by Definition 3.2.1 (all vectors in R3 are linear combinations of ~e1 , ~e2 , ~e3 ). Ch 3.TF.30 T; Use a basis with one vector on the line and the other perpendicular to it. Ch 3.TF.31 F; Note that R2 isn’t even a subset of R3 . A vector in R2 , with two components, does not belong to R3 . Ch 3.TF.32 T; If B = S −1 AS, then B + 7In = S −1 (A + 7In )S. Ch 3.TF.33 T; for any ~v1 , ~v2 , ~v3 of V also k~v1 , ~v2 , ~v3 is a basis too, for any nonzero scalar k. Ch 3.TF.34 F; The identity matrix is similar only to itself. Ch 3.TF.35 F; Consider

0 1 0 1

1 0 0 1 1 , but = 0 0 0 0 0

1 0

0 0

0 1 = 0 1

2 . 0

Ch 3.TF.36 F; Let A = I2 , B = −I2 and ~v = ~e1 , for example. Ch 3.TF.37 F; Let V = span(~e1 ) and W = span(~e2 ) in R2 , for example. Ch 3.TF.38 T; If A~v = Aw, ~ then A(~v − w) ~ = ~0, so that ~v − w ~ = ~0 and ~v = w. ~ Ch 3.TF.39 T; Consider the linear transformation with matrix [w ~1 . . . w ~ n ][~v1 . . . ~vn ]−1 . Ch 3.TF.40 F; Suppose A were similar to B. Then A4 = I2 were similar to B 4 = −I2 , by Example 7 of Section 3.4. But this isn’t the case: I2 is similar only to itself. 169

Chapter 3

Ch 3.TF.41 T; Let A =

0 1 , for example, with ker(A) = im(A) = span(~e1 ). 0 0

Ch 3.TF.42 F; Consider In and 2In , for example. Ch 3.TF.43 T; Matrix B = S −1 AS is invertible, being the product of invertible matrices. Ch 3.TF.44 T; Note that im(A) is a subspace of ker(A), so that dim(im A) = rank(A) ≤ dim(ker A) = 10 − rank(A). Ch 3.TF.45 T; Pick three vectors ~v1 , ~v2 , ~v3 that span V . Then V = im[~v1 ~v2 ~v3 ]. Ch 3.TF.46 T; Check that

0 0

1 0

is similar to

0 0

2 . 0

Ch 3.TF.47 T; Pick a vector ~v that is neither on the line nor perpendicular to it. Then the matrix of the linear 0 1 transformation T (~x) = R~x with respect to the basis ~v , R~v is , since R(R~v ) = ~v . 1 0 Ch 3.TF.48 F; If B = S −1 AS, then B = (2S)−1 A(2S) as well. Ch 3.TF.49 T; Note that A(B − C) = 0, so that all the columns of matrix B − C are in the kernel of A. Thus B − C = 0 and B = C, as claimed. Ch 3.TF.50 T; Suppose ~v is in both ker(A) and im(A), so that ~v = Aw ~ for some vector w. ~ Then ~0 = A~v = A2 w ~= Aw ~ = ~v , as claimed. Ch 3.TF.51 F; Suppose such a matrix A exists. Then there is a vector ~v in R2 such that A2~v 6= ~0 but A3~v = ~0. As in Exercise 3.4.58a we can show that vectors ~v , A~v , A2~v are linearly independent, a contradiction (we are looking at three vectors in R2 ). Ch 3.TF.52 T; The ith column ~ai of A, being in the image of A, is also in the image of B, so that ~ai = B~ci for some ~ci in Rm . If we let C = [~c1 · · · ~cm ] , then BC = [B~c1 · · · B~cm ] = [~a1 · · · ~am ] = A, as required. Ch 3.TF.53 F; Think about this problem in terms of “building” such an invertible matrix column by column. If we wish the matrix to be invertible, then the first column can be any column other than ~0 (7 choices). Then the second column can be any column other than ~0 or the first column (6 choices). For the third column, we have at most 5 choices (not ~0 or the first or second columns, as well as possibly some other columns). For some choices of the first two columns there will be other columns we have to exclude (the sum or difference of the first two), but not for others. Thus, in total, fewer than 7 × 6 × 5 = 210 matrices are invertible, out of a total 29 = 512 matrices. Thus, most are not invertible.

170

Section 4.1

Chapter 4 Section 4.1 4.1.1 Not a subspace since it does not contain the neutral element, that is, the function f (t) = 0, for all t. 4.1.2 This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V . • If f and g are in V (so that f (2) = g(2) = 0), then (f + g)(2) = f (2) + g(2) = 0 + 0 = 0, so that f + g is in V . • If f is in V (so that f (2) = 0), and k is any constant, then (kf )(2) = kf (2) = 0, so that kf is in V . A polynomial f (t) = a + bt + ct2 is in V if f (2) = a + 2b + 4c = 0, or a = −2b − 4c. The general element of V is of the form f (t) = (−2b − 4c) + bt + ct2 = b(t − 2) + c(t2 − 4), so that t − 2, t2 − 4 is a basis of V . 4.1.3 This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V since f ′ (1) = f (2) = 0. • If f and g are in V (so that f ′ (1) = f (2) and g ′ (1) = g(2)), then (f + g)′ (1) = (f ′ + g ′ )(1) = f ′ (1) + g ′ (1) = f (2) + g(2) = (f + g)(2), so that f + g is in V . • If f is in V (so that f ′ (1) = f (2)) and k is any constant, then (kf )′ (1) = (kf ′ )(1) = kf ′ (1) = kf (2) = (kf )(2), so that kf is in V . If f (t) = a + bt + ct2 then f ′ (t) = b + 2ct, and f is in V if f ′ (1) = b + 2c = a + 2b + 4c = f (2), or a + b + 2c = 0. The general element of V is of the form f (t) = (−b − 2c) + bt + ct2 = b(t − 1) + c(t2 − 2), so that t − 1, t2 − 2 is a basis of V . 4.1.4 This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V since • If f and g are in V • If f is in V

so that

so that

Z

1

f=

0

Z

0

1

Z

0

1

Z

1

0 dt = 0.

0

Z Z 1 (f + g) = g = 0 then 0

Z

0

1

f+

1

kf = k

Z

1

g = 0, so that f + g is in V .

0

0

0

Z f = 0 and k is any constant, then

If f (t) = a + bt + ct2 then

1

Z

1

f = 0, so that kf is in V .

0

1 b c b c b c f (t)dt = at + t2 + t3 = a + + = 0 if a = − − . 2 3 0 2 3 2 3

b 1 1 1 c 1 2 2 The general element of V is f (t) = − − + bt + ct = b t − +c t − , so that t − , t2 − is a 2 3 2 3 2 3 basis of V . 171

Chapter 4 4.1.5 If p(t) = a + bt + ct2 then p(−t) = a − bt + ct2 and −p(−t) = −a + bt − ct2 . Comparing coefficients we see that p(t) = −p(−t) for all t if (and only if) a = c = 0. The general element of the subset is of the form p(t) = bt. These polynomials form a subspace of P2 , with basis t. 4.1.6 Not a subspace, since I3 and −I3 are invertible, but their sum is not. 4.1.7 a The set V of diagonal 3 × 3 matrices is a subspace of R3×3 :

0 0 0 b The zero matrix 0 0 0 is in V , 0 0 0 p 0 0 a 0 0 c If A = 0 b 0 and B = 0 q 0 are in V , then so is their sum 0 0 r 0 0 c a+p 0 0 b+q 0 . A+B = 0 0 0 c+r

a 0 0 ka 0 d If A = 0 b 0 is in V , then so is kA = 0 kb 0 0 c 0 0

0 0 , for all constants k. kc

4.1.8 This is a subspace; the justification is analogous to Exercise 7. 4.1.9 Not a subspace; consider multiplication with a negative scalar. I3 belongs to the set, but −I3 doesn’t. 1 4.1.10 a Let ~v = 2 . Let V be the set of all 3 × 3 matrices A such that A~v = ~0. Then V is a subspace of R3×3 : 3 b The zero matrix 0 is in V , since 0~v = ~0. c If A and B are in V , then so is A + B, since (A + B)~v = A~v + B~v = ~0 + ~0 = ~0. d If A is in V , then so is kA for all scalars k, since (kA)~v = k(A~v ) = k~0 = ~0. 4.1.11 Not a subspace: I3 is in rref, but the scalar multiple 2I3 isn’t. 4.1.12 Yes, the set W of all arithmetic sequences is a subspace of V . Use the fact that a sequence (x0 , x1 , x2 , . . .) is arithmetic if xn = x0 + kn for some constant k. • The sequence (0, 0, 0, . . .) is an arithmetic sequence, with k = 0. 172

Section 4.1 • If (xn ) and (yn ) are arithmetic sequences (with xn = x0 + pn and yn = y0 + qn), then xn + yn = x0 + y0 + (p + q)n, so that (xn + yn ) is an arithmetic sequence as well. • If (xn ) is an arithmetic sequence (with xn = x0 + pn) and k is an arbitrary constant, then kxn = kx0 + (kp)n, so that (kxn ) is an arithmetic sequence as well. 4.1.13 Not a subspace: (1, 2, 4, 8, . . .) and (1, 1, 1, 1, . . .) are both geometric sequences, but their sum (2, 3, 5, 9, . . .) is not, since the ratios of consecutive terms fail to be equal, for example, 32 6= 35 . 4.1.14 Yes • (0, 0, 0, . . . , 0, . . .) converges to 0. • If limn→∞ xn = 0 and limn→∞ yn = 0, then limn→∞ (xn + yn ) = limn→∞ xn + limn→∞ yn = 0. • If limn→∞ xn = 0 and k is any constant, then limn→∞ (kxn ) = k limn→∞ xn = 0. 4.1.15 The set W of all square-summable sequences is a subspace of V : • The sequence (0, 0, 0, . . .) is in W . • Suppose (xn ) and (yn ) are in W . Note that the inequality (xn +P yn )2 ≤ 2x2n + 2yn2 holds n, since P∞ for all P ∞ ∞ 2 2 2 2 2 2 2xn + 2yn − (xn + yn ) = xn + yn − 2xn yn = (xn − yn ) ≥ 0. Thus n=1 (xn + yn )2 ≤ 2 n=1 x2n + 2 n=1 yn2 converges, so that the sequence (xn + yn ) is in W as well. P∞ • If (xn ) is in W so that n=1 x2n converges , then (kxn ) is in W as well, for any constant k, since ∞ X

(kxn )2 = k 2

n=1

∞ X

x2n

n=1

will converge. 0 0 1 1 0 a b 4.1.16 c d = a 0 0 + b 0 0 + c 1 0 0 0 0 0 e f 0 0 0 0 1 1 0 The matrices 0 0 , 0 0 , 1 0 , 0 0 0 0 0 0 0 0 6.

0 0 0 0 0 0 0 0 + d0 1 + e0 0 + f 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 , 0 0 , 0 0 form a basis of R3×2 , so that dim(R3×2 ) = 0 1 1 0 0

4.1.17 Let Eij be the n × m matrix with a 1 as its ijth entry, and zeros everywhere else. Any A in Rn×m can be written as the sum of all aij Eij , and the Eij are linearly independent, so that they form a basis of Rn×m . Thus dim(Rn×m ) = nm. 4.1.18 Any f in Pn can be written as a linear combination of 1, t, t2 , . . . , tn , by definition of Pn . Also, 1, t, . . . , tn are linearly independent; to see this consider a relation c0 +c1 t+· · ·+cn tn = 0; since the polynomial c0 +c1 t+· · ·+cn tn has more than n zeros, we must have c0 = c1 = · · · = cn = 0, as claimed. Thus, dim(Pn ) = n + 1. 173

Chapter 4 0 0 i 1 a + bi +d +c +b =a i 1 0 0 c + di 0 0 i 1 form a basis of C2 as a real linear space, so that dim(C2 ) = 4. , , , The vectors i 1 0 0

4.1.19

4.1.20 We use Summary 4.1.6. We have a = −d, so that the general element of the subspace is 0 1 0 0 −1 0 b +c +d . 0 0 1 0 0 1 0 1 0 0 −1 0 Thus , , is a basis of the subspace; the dimensions is 3. 0 0 1 0 0 1

−d c

b d

=

4.1.21 Use Summary 4.1.6. The general element of the subspace is 0 0 1 0 0 0 1 0 a 0 is a basis of the subspace; the dimension is 2. , . Thus +d =a 0 1 0 0 0 1 0 0 0 d 4.1.22 Using Exercise 21 as a guide, we find the basis E11 , E22 , . . . , Enn , where Eii is the n × n matrix with all 0 entries, except for a 1 at the ith place on the diagonal. The dimension of this space is n. 4.1.23 Proceeding as in Exercise 21, we find the basis

0 ; the dimension is 3. 1

0 0 , 0 0

0 1 0 , 1 0 0

4.1.24 Proceeding as in Exercise 21, we find the basis E11 , E12 , E13 , E22 , E23 , E33 . Here Eij is the 3 × 3 matrix with all 0 entries, except for a 1 in the ith component of the jth column; the dimension is 6. 4.1.25 A polynomial f (t) = a + bt + ct2 is in this subspace if f (1) = a + b + c = 0, or a = −b − c. The polynomials in the subspace are of the form f (t) = (−b − c) + bt + ct2 = b(t − 1) + c(t2 − 1), so that t − 1, t2 − 1 is a basis of the subspace, whose dimension is 2. 4.1.26 Denote the subspace by W . A polynomial f (t) = a + bt + ct2 + dt3 is in W if f (1) = a + b + c + d = 0 and 1 Z 1 b c d 2 f (t)dt = at + t2 + t3 + t4 = 2a + c = 0. 2 3 4 −1 3 −1 − 31 c " # " # 2 a+b+c+d =0 − c − d =0 a + 13 c 3 . The system reduces to , with general solution 2a + 23 c b + 23 c + d = 0 =0 c d The polynomials in W are of the form f (t) = − 31 c − 32 c + d t + ct2 + dt3 = c t2 − 23 t − 31 + d(t3 − t), so that t2 − 23 t − 13 , t3 − t is a basis of W , and dim(W ) = 2.

4.1.27

a b c d

is in the subspace if

a b c d

1 0 0 2

= 174

a 2b c 2d

equals

1 0 0 2

a b c d

=

a b , which is 2c 2d

Section 4.1

the case if b = c = 0. The matrices in the subspace are of the form 0 0 1 0 is a basis, and the dimension is 2. , 0 1 0 0

a 0 0 d

=a

0 1 0 +d 0 0 0

0 , so that 1

a b a b 1 1 a a+b 1 1 a b a+c b+d 4.1.28 is in the subspace if = equals = c d c d 0 1 c c+d 0 1 c d c d which is the case if c = 0 and a = d. 0 1 1 0 0 1 1 0 a b is a , , so that +b = a The matrices in the subspace are of the form 0 0 0 1 0 0 0 1 0 a basis, and the dimension is 2.

a b c d

a b c d

1 4.1.29 We are looking for the matrices such that 1 −b is required that a = −b and c = −d. The general element is −d −1 1 0 0 , is a basis, and the dimension is 2. 0 0 −1 1

1 a+b a+b 0 0 = = . It 1 c+d c+d 0 0 b −1 1 0 0 = b +d . Thus d 0 0 −1 1

0 0 a + 2c b + 2d a b . It = = 0 0 3a + 6c 3b + 6d c d −2c −2d −2 0 0 −2 is required that a = −2c and b = −2d. Thus the general element is =c +d . c d 1 0 0 1

4.1.30 We are looking for the matrices

Thus

a b c d

such that

1 2 3 6

0 −2 −2 0 is a basis, and the dimension is 2. , 0 1 1 0

a b 0 1 a 4.1.31 We are looking for the matrices such that c d 1 0 c a −b c d . It is required that a = c and b = −d. = or, c −d a b The general element is is 2.

1 c −d =c 1 c d

0 0 +d 0 0

1 −1 . Thus 1 1

b a b 1 = d c d 0

0 −1 0 is a basis, and the dimension , 0 1 0

2 a b a b 1 1 a b = such that 4.1.32 We are looking for the matrices 0 c d c d 1 1 c d a+c b+d 2a 0 or, = . It is required that a = c and b = −d. a+c b+d 2c 0

0 , −1

0 , 0

a b 1 0 0 1 1 0 0 1 The general element is =a +b . Thus , is a basis, and the dimension a −b 1 0 0 −1 1 0 0 −1 is 2.

a b 1 1 a b a b 4.1.33 Let S = . Then = , meaning c d 1 1 c d c d 175

Chapter 4 b . So a + c = a, b + d = b, a + c = c and b + d = d. These imply, respectively, that d b = 0. 0 0 , and the basis is ∅. Thus, S can only equal 0 0 a a+c b+d = c a+c b+d c = 0, d = 0, a = 0 and

a b 3 2 a b a b . We want = , meaning c d 4 5 c d c d 3a + 2c 3b + 2d a b = . So 3a + 2c = a, 3b + 2d = b, 4a + 5c = c and 4b + 5d = d. These imply that 4a + 5c 4b + 5d c d a = −c and b = −d. 0 1 1 0 0 1 1 0 a b is a basis, and the , . Thus +b =a So the general element is 0 −1 −1 0 0 −1 −1 0 −a −b dimension is 2. a b c 4.1.35 Let A = d e f . We want AB = BA, or g h i a b c 2 0 0 2 0 0 a b c d e f 0 3 0 = 0 3 0 d e f , or g h i 0 0 4 0 0 4 g h i 2a 2b 2c 2a 3b 4c 2d 3e 4f = 3d 3e 3f . 4g 4h 4i 2g 3h 4i 4.1.34 Let S =

We note that b, c, d, f, g, h must be zero, but a, e, and i are chosen freely. So, our space, V , consists of all matrices a 0 0 of the form 0 e 0 0 0 i 0 0 0 0 0 0 1 0 0 = a 0 0 0 + e 0 1 0 + i 0 0 0 . 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 Thus, 0 0 0 , 0 1 0 , 0 0 0 is a basis of V , and dim(V ) = 3. 0 0 1 0 0 0 0 0 0

a b c 4.1.36 Let A = d e f . We want g h i a b c 2 0 0 2 0 d e f 0 3 0 = 0 3 g h i 0 0 3 0 0 2a 2b 2c 2a 3b 3c 2d 3e 3f = 3d 3e 3f . 3g 3h 3i 2g 3h 3i

AB = BA, or 0 a b 0d e 3 g h

c f , or i

176

Section 4.1 We note that b, c, d, g must be zero, but a, e, f, h and i are chosen freely. So, our space, V , consists of all matrices a 0 0 of the form 0 e f 0 h i

1 = a0 0

0 0 0 0 0 0 + e0 1 0 0 0 0

1 Thus, 0 0

0 0 0 0 0 0,0 1 0 0 0 0

0 0 0 + f 0 0 0

0 0 0 0,0 0 0 0 0

0 0 0 0 0 1 + h0 0 0 0 0 1

0 0 0 + i0 0 0

0 0 0 0 is a basis of V , and dim(V ) = 5. 0 1

0 0 0 0 0,0 0 1 0

0 0 1,0 0 0

0 0 0 0 . 0 1

4.1.37 If all diagonal entries of B are different, then dim(V ) = 3, as in Exercise 35. If two of the diagonal entries are equal, then dim(V ) = 5, as in Example 36. If all three entries are equal, then B is a scalar multiple of I3 , and will commute with all 3x3 matrices, so that dim(V ) = dim(R3×3 ) = 9. So, in summary, dim(V ) could be 3, 5 or 9. 4.1.38 We look at similar cases mentioned in Exercise 37, and see that the different possibilities occur when all four entries are different (dim(V ) = 4), when exactly two are the same, but the other two are different (dim(V ) = 6), when exactly three are the same (dim(V ) = 10), when all four are the same (dim(V ) = 16) and when two of the terms of B are equal, and the other two diagonal terms of B are also equal, but different from the first pair (dim(V ) = 8). a11 0 4.1.39 An upper-triangular matrix has the form: ...

0

1 0 0 0 = a11 ... ... 0 0

0 0 0 1 + a22 ... ... 0 0

··· ··· .. . ··· ··· ··· .. . ···

0 1 0 0 0 0 + a12 .. ... ... . 0 0 0

··· ··· .. . ···

0 0 0 0 + · · · + a2n .. ... .

0

0

a12 a22 .. .

··· ··· .. .

0

···

a1n a2n .. .

ann

0 0 0 0 + · · · + a1n .. ... .

0

0

0 ··· 0 ··· .. . . . . 0 ···

0 ··· 0 ··· .. . . . . 0 ···

1 0 .. . 0

0 0 0 0 0 1 + · · · + ann . . .. .. .. . 0 0 0

So the dimension of this space is n + (n − 1) + · · · + 1 =

··· ··· .. . ···

0 0 . .. .

1

n(n+1) . 2

4.1.40 Let V be the space of all n × n matrices A such that A~v = ~0. We look at some possibilities for ~v . If ~v = ~0, then any matrix A will work, and dim(V ) = n2 . Now assume that ~v 6= 0, and suppose that the ith component vi is nonzero. If we denote the columns of A by w ~ 1, . . . , w ~ n , then the condition A~v = ~0 means that v1 w ~ 1 + · · · + vi w ~ i + · · · + vn w ~ n = ~0. We can solve this relation for w ~ i and express w ~ i in terms of the other n − 1 column vectors of A. This means that we can choose n − 1 columns (or n(n − 1) entries) of A freely; the column w ~ i is then determined by these choices. Thus dim(V ) = n(n − 1) in this case. In summary, we have dim(V ) = n2 if ~v = ~0, and dim(V ) = n(n − 1) otherwise. 177

Chapter 4 4.1.41 Let V be the space of all matrices A such that BA = 0. Note that A is in V if (and only if) all the columns of A are in the kernel of B. Since the columns of A can be chosen independently, it is plausible that dim(V ) = 3dim(ker(B)). We show this more clearly by investigating the different possibilities for ker(B). In the case where ker(B) 6= {~0}, let ~v1 , . . . , ~vn be a basis of the kernel of B, and note that n = nullity(B) can be either 1, 2 or 3. Then, the general element of V is of the form: [ a1~v1 + . . . + an~vn b1~v1 + . . . + bn~vn c1~v1 . . . + cn~vn ], which has the 3n arbitrary constants a1 , . . . , an , b1 , . . . , bn , c1 , . . . , cn . Using Summary 4.1.6, we can construct a basis of V with 3n elements, proving our claim that dim(V ) = 3dim(ker(B)). Thus, dim(V ) can be 0, 3, 6 or 9. 4.1.42 Let B be a matrix such that dim(ker(B))= k. Then, it is required that the columns of A contain only vectors in the kernel of B. Thus, each column of A can be written as: c1~v1 + c2~v2 + · · · + ck~vk , where the vectors ~vi form as basis of the kernel of B. Thus, each of the n columns in A involves k arbitrary constants, and matrix A involves nk arbitrary constants overall. The space of matrices A has dimension nk, where k is an integer in the range [0, n]. 1 0 . Using the terminology introduced in the 4.1.43 Let V be the space of all matrices S such that AS = S 0 −1 1 0 hint, we want A [ ~v w ~ ] = [ ~v w ~] or [ A~v Aw ~ ] = [ ~v −w ~ ]. Thus, ~v must be parallel to L, and w ~ 0 −1 must be perpendicular to L. If ~v1 is a nonzero vector parallel to L, and ~v2 a nonzero vector perpendicular to L, then the general element of V is of the form S = [ ~v w ~ ] = [ a~v1 b~v2 ] = a [ ~v1 0 ] + b [ 0 ~v2 ] , where a and b are arbitrary constants. We see that [ ~v1 0 ] , [ 0 ~v2 ] is a basis of V and dim(V ) = 2.

1 0 0 4.1.44 Let V be the space of all matrices S such that AS = S 0 1 0 . Let’s denote the column vectors of S 0 0 0 1 0 0 by ~u, ~v and w. ~ The condition AS = S 0 1 0 means that A~u = ~u, A~v = ~v and Aw ~ = ~0. This in turn means 0 0 0 that the vectors ~u and ~v have to be on the plane V, while w ~ is perpendicular to V . If we choose a basis ~v1 , ~v2 of V and a nonzero vector ~v3 perpendicular to V, then we can write ~u = a~v1 + b~v2 , ~v = c~v1 + d~v2 , w ~ = e~v3 , and

S = [~u ~v w] ~ = [a~v1 + b~v2 c~v1 + d~v2 e~v3 ] = a[~v1 ~0 ~0] + b[~v2 ~0 ~0] + c[~0 ~v1 ~0] + d[~0 ~v2 ~0] + e[~0 ~0~v3 ]. Thus dim(V ) = 5; the five matrices in the linear combination above form a basis of V .

4.1.45 Let

a d g 0 0 0

b e h a d g

a b A = d e g h 0 1 c f 0 0 0 0 i b d e = g h 0

c f . We want i 0 1 0 1 = 0 0 0 0 0 e f h i . 0 0

AB = BA, or a b 0 1d e g h 0

c f , or i

1 a b c So d, g, h = 0, a = e = i and f = b. Our space V consists of all matrices of the form 0 a b = a 0 0 0 0 a

178

0 0 1 0 + 0 1

Section 4.1 0 0 1 0 1 0 b 0 0 1 + c 0 0 0 . 0 0 0 0 0 0 0 0 0 1 0 1 0 0 Thus, 0 1 0 , 0 0 1 , 0 0 0 0 0 0 0 0 0 1

1 0 is a basis of V and dim(V ) = 3. 0

4.1.46 The arithmetic sequences are of the form (a, a + k, a + 2k, a + 3k, . . .) = a(1, 1, 1, 1, . . .) + k(0, 1, 2, 3, . . .), so that the sequences (1, 1, 1, . . .) (all 1’s) and (0, 1, 2, 3, . . .) (the nth entry is n) form a basis of this space, whose dimension is 2. 4.1.47 We show that the set of all even functions is a subspace of F (R, R): • If f (t) = 0 for all t, then f (−t) = f (t) = 0 for all t. • If f and g are even (that is, f (−t) = f (t) and g(−t) = g(t) for all t), then (f + g)(−t) = f (−t) + g(−t) = f (t) + g(t) = (f + g)(t), so that f + g is even as well. • If f is even and k is any constant, then (kf )(−t) = kf (−t) = kf (t) = (kf )(t), so that kf is even as well. An analogous proof shows that the odd functions form a subspace of F (R, R). 4.1.48 a f is even if f (−t) = f (t) for all t. Comparing coefficients we find that b = d = 0, so that f (t) is of the form f (t) = a + ct2 + et4 , with basis 1, t2 , t4 . The dimension is 3. b f is odd if f (−t) = −f (t), which is the case if a = c = e = 0. The odd polynomials are of the form f (t) = bt+dt3 , with basis t, t3 and dimension 2. 4.1.49 We show that L(Rm , Rn ) is a subspace of F (Rm , Rn ): • The zero transformation T (~x) = ~0 (for all ~x) is linear, represented by the zero matrix. • If T and S are linear transformations from Rm to Rn (with T (~x) = A~x and S(~x) = B~x for some n × m matrices A and B), then (T + S)(~x) = T (~x) + S(~x) = A~x + B~x = (A + B)~x, so that T + S is linear as well, given by the matrix A + B. • If T is a linear transformation from Rm to Rn (with T (~x) = A~x) and k is any constant, then (kT )(~x) = kT (~x) = kA~x = (kA)~x, so that kT is linear as well, with matrix kA. 4.1.50 Using Example 18 as a guide, we first look for solutions of the form f (x) = ekx . It is required that f ′′ (x) + 8f ′ (x) − 20f (x) = k 2 ekx + 8kekx − 20ekx = 0 for all x, or k 2 + 8k − 20 = (k − 2)(k + 10) = 0. Thus k = 2 or k = −10. By Theorem 4.1.5, the solutions of the differential equation are of the form f (x) = c1 e2x + c2 e−10x , where c1 and c2 are arbitrary constants. 4.1.51 Using Example 18 as a guide, we first look for solutions of the form f (x) = ekx . It is required that f ′′ (x) − 7f ′ (x) + 12f (x) = k 2 ekx − 7kekx + 12ekx = 0 for all x, or k 2 − 7k + 12 = (k − 3)(k − 4) = 0. Thus k = 3 or k = 4. By Theorem 4.1.7, the solutions of the differential equation are of the form f (x) = c1 e3x + c2 e4x , where c1 and c2 are arbitrary constants. 179

Chapter 4 4.1.52 We have to find constants a and b such that the functions e−x and e−5x are solutions of the differential equation f ′′ (x) + af ′ (x) + bf (x) = 0. Thus it is required that e−x − ae−x + be−x = 0, or 1 − a + b = 0, and also that 25 − 5a + b = 0. The solution of this system of two equations in two unknowns is a = 6, b = 5, so that the desired differential equation is f ′′ (x) + 6f ′ (x) + 5f (x) = 0. 4.1.53 Let B = (f1 , . . . , fn ) be a basis of V and suppose that the elements g1 , . . . , gm in V are linearly independent. In the proof of Theorem 4.1.5, we show that the coordinate vectors [g1 ]B , . . . , [gm ]B in Rn are linearly independent, so that m ≤ n by Theorem 3.2.8. 4.1.54 We can adapt the answer to Exercise 3.2.38a. Let m be the largest number of linearly independent elements we can find in W ; note that m ≤ n, by Exercise 53. Choose linearly independent elements f1 , . . . , fm in W . We claim that the elements f1 , . . . , fm span W . Indeed, if f is any element of W, then the m + 1 elements f1 , . . . , fm , f are linearly dependent, so that f is redundant: f is a linear combination of f1 , . . . , fm . It follows that f1 , . . . , fm is a basis of W, so that dim(W ) = m ≤ n =dim(V ), as claimed. 4.1.55 We will argue indirectly, assuming that F (R, R) is n-dimensional for some n. Now, the n + 1 polynomials 1, x, x2 , . . . , xn in F (R, R) are linearly independent, contradicting the fact that we can find at most n linearly independent elements in an n-dimensional space (see Exercise 53). We conclude that F (R, R) is infinite dimensional, as claimed. 4.1.56 Argue indirectly and assume that the space V of infinite sequences is finite-dimensional, with dim(V ) = n. According to the solution to Exercise 57, there can be at most n linearly independent elements in V . But here is our contradiction: It is easy to give n + 1 linearly independent infinite sequences, namely, (1, 0, 0, 0, . . .), (0, 1, 0, 0, . . .), (0, 0, 1, 0, . . .), . . . , (0, 0, 0, . . . , 0, 1, 0, . . .); in the last sequence the 1 is in the (n + 1)th place. 4.1.57 We can construct a basis of V by omitting the redundant elements from the list g1 , . . . , gm . It follows that V is finite-dimensional, and, in fact, dim(V ) ≤ m, since our basis is a “sub-list” of the original list g1 , . . . , gm . 4.1.58 a Let g(x) be a function in V . Thus, g ′′ (x) = −g(x). Now, if f (x) = g(x)2 +g ′ (x)2 , then f ′ (x) = 2(g(x))g ′ (x)+ 2(g ′ (x))g ′′ (x) = 2(g(x))g ′ (x) − 2(g(x))g ′ (x) = 0. So, f (x) = g(x)2 + g ′ (x)2 is a constant function. b Let g(x) be a function in V such that g(0) = g ′ (0) = 0. From part a we know that g(x)2 + g ′ (x)2 = k, a constant. Now g(0)2 + g ′ (0)2 = 02 + 02 = 0, so that k = 0. The equation g(x)2 + g ′ (x)2 = 0 means that g(x) = g ′ (x) = 0 for all x, as claimed. c First we note that g(x) = f (x) − f (0) cos(x) − f ′ (0) sin(x) is in V, since the functions f (x), cos(x) and sin(x) are all in V, and V is a subspace of F (R, R). Note that g(0) = f (0) − f (0) cos(0) − f ′ (0) sin(0) = f (0) − f (0) = 0. Also, g ′ (x) = f ′ (x) + f (0) sin(x) − f ′ (0) cos(x), so that g ′ (0) = f ′ (0) − f ′ (0) = 0. By part b, we can conclude that g(x) = 0 for all x, so that f (x) = f (0) cos(x) + f ′ (0) sin(x), as claimed.

Section 4.2 4.2.1 Fails to be linear, since T (A + B) = A + B + I2 doesn’t equal T (A) + T (B) = A + I2 + B + I2 = A + B + 2I2 . 180

Section 4.2 4.2.2 Linear, since T (A + B) = 7(A + B) = 7A + 7B equals T (A) + T (B) = 7A + 7B, and T (kA) = 7kA equals kT (A) = k(7A) = 7kA. Yes, T is an isomorphism, with T −1 (A) = 71 A.

a b p q a+p b+q + =T =a+p+d+s c d r s c+r d+s a b p q a b ka kb equals T +T = a + d + p + s, and T k =T = ka + kd c d r s c d kc kd a b equals kT = k(a + d) = ka + kd. c d

4.2.3 Linear, since T

No, T fails to be an isomorphism, since 4 = dim(R2×2 ) 6= dim(R) = 1; see Theorem 4.2.4b. 4.2.4 Fails to be linear, since T (2I2 ) = det(2I2 ) = 4 does not equal 2T (I2 ) = 2 det(I2 ) = 2. 4.2.5 Fails to be linear, since T (2I2 ) = 4I2 does not equal 2T (I2 ) = 2I2 . 4.2.6 Let P =

1 3

2 . Transformation T is linear, since 6

T (A + B) = (A + B)P = AP + BP equals T (A) + T (B) = AP + BP , and T (kA) = (kA)P = kAP equals kT (A) = kAP . No, T isn’t an isomorphism, since ker(T ) 6= {0}; the matrix A = rem 4.2.4a).

3 −1 0 0

is in the kernel of T (see Theo-

1 2 1 2 1 2 (M + N ) = M+ N = T (M ) + T (N ). Also, T (kM ) = 3 4 3 4 3 4

4.2.7 Linear, since T (M + N ) = 1 2 1 2 (kM ) = k M = kT (M ). 3 4 3 4

−1 1 2 1 2 M for M to find the inverse M = N= This is also an isomorphism. Solve the equation N = 3 4 3 4 2 1 −4 N. 2 3 −1

1 1 2 2 (kM ) = k M 4.2.8 Not linear, since T (kM ) = (kM ) 3 3 4

2 M = k 2 T (M ) 6= kT (M ), in general. 4

4.2.9 Linear, since T (A + B) = S −1 (A + B)S = S −1 AS + S −1 BS equals T (A) + T (B) = S −1 AS + S −1 BS, and T (kA) = S −1 (kA)S = kS −1 AS equals kT (A) = kS −1 AS. 181

Chapter 4 Yes, T is an isomorphism. Solve the equation B = S −1 AS for A to find the inverse A = SBS −1 . 4.2.10 Linear, since T (A + B) = P (A + B)P −1 = P AP −1 + P BP −1 equals T (A) + T (B) = P AP −1 + P BP −1 , and T (kA) = P (kA)P −1 = kP AP −1 equals kT (A) = kP AP −1 . Yes, T is an isomorphism. Solve the equation B = P AP −1 for A to find the inverse A = P −1 BP . 4.2.11 Linear, since T (M + N ) = P (M + N )Q = (P M + P N )Q = P M Q + P N Q = T (M ) + T (N ). Also, T (kM ) = P (kM )Q = kP M Q = kT (M ). This is also an isomorphism. Solve the equation N = P M Q for M to find the inverse M = P −1 N Q−1 .

2 4.2.12 Linear. Let A = 4

3 . Then T (c + d) = (c + d)A = cA + dA equals T (c) + T (d) = cA + dA, and 5

T (kc) = kcA equals kT (c) = kcA. No, T isn’t an isomorphism, since domain and codomain have different dimensions.

1 4.2.13 Let Q = 0

2 . Transformation T is linear, since 1

T (M + N ) = (M + N )Q − Q(M + N ) = M Q + N Q − QM − QN equals T (M ) + T (N ) = M Q − QM + N Q − QN , and T (kM ) = (kM )Q − Q(kM ) = kM Q − kQM equals kT (M ) = k(M Q − QM ) = kM Q − kQM . No, T isn’t an isomorphism, since A = I2 is in ker(T ); see Theorem 4.2.4a.

2 4.2.14 Let Q = 5

3 . Transformation T is linear, since 7

T (M + N ) = Q(M + N ) − (M + N )Q = QM + QN − M Q − N Q equals T (M ) + T (N ) = QM − M Q + QN − N Q, and T (kM ) = Q(kM ) − (kM )Q = kQM − kM Q equals kT (M ) = k(QM − M Q) = kQM − kM Q. No, T isn’t an isomorphism, since A = I2 is in ker(T ); see Theorem 4.2.4a. 2 0 4 0 2 0 2 4.2.15 Linear, since T (M + N ) = (M + N ) − (M + N ) = M+ 0 3 0 5 0 3 0 4 0 2 0 4 0 2 0 4 0 = T (M ) + T (N ). N −N + M −M = N 0 5 0 3 0 5 0 3 0 5 2 0 4 0 2 0 4 0 Also, T (kM ) = (kM ) − (kM ) =k M − kM = kT (M ). 0 3 0 5 0 3 0 5 182

0 4 0 N −M − 3 0 5

Section 4.2 This is also an isomorphism. We first of V and W are show that the kernel then equal, the dimensions see that 4a 5b 2a 2b 4 0 a b a b 2 0 a b = − = − = contains only zero. Let T (M ) = T 4c 5d 3c 3d 0 5 c d c d 0 3 c d −2a −3b . Clearly the only time this matrix will equal zero is when a, b, c, d = 0. Thus, ker(T ) = { 0}. −c −2d 3 0 2 0 2 0 3 0 2 0 M− − +N (M + N ) = M − 4.2.16 Linear, since T (M + N ) = (M + N ) 0 4 0 3 0 3 0 3 0 4 3 0 2 0 3 0 2 0 3 0 N =M − M +N − N = T (M ) + T (N ). 0 4 0 3 0 4 0 3 0 4 2 0 3 0 2 0 3 0 Also, T (kM ) = (kM ) − (kM ) = k M − M = kT (M ). 0 3 0 4 0 3 0 4 2 0 a b a b − = To determine whether T is an isomorphism, we find the kernel of T . Now, T (M ) = T 0 3 c d c d 3 0 a b 2a 3b 3a 3b −a 0 a b = − = . We see that M = is in the kernel of T if 0 4 c d 2c 3d 4c 4d −2c −d c d 0 b a = c = d = 0. Thus, the kernel consists of all matrices , where b is arbitrary. Since the kernel is nonzero, 0 0 T fails to be an isomorphism.

4.2.17 T is linear: T ((a + ib) + (c + id)) = T (a + c + i(b + d)) = a + c = T (a + ib) + T (c + id), and T (k(x + iy)) = T (kx + iky) = kx = kT (x + iy). However, T (5i) = 0, so T fails to be an isomorphism. 4.2.18 This transformation fails to be linear, since 2T (3) = 2(9) = 18 6= T (6) = 36. 4.2.19 T is linear: if w and z are complex numbers, then T (w + z) = i(w + z) = iw + iz = T (w) + T (z) and T (kz) = i(kz) = k(iz) = kT (z). T is also an isomorphism. Solve the equation w = iz for z to find that z = 1i w = −iw. We have used the fact that i2 = −1, so that 1i = −i. 4.2.20 Linear, since T ((x + iy) + (z + it)) = T (x + z + i(y + t)) = x + z − i(y + t) equals T (x + iy) + T (z + it) = x − iy + z − it = x + z − i(y + t) and T (k(x + iy)) = T (kx + iky) = kx − iky equals kT (x + iy) = k(x − iy) = kx − iky. Yes, T is an isomorphism; it’s its own inverse, since T (T (x + iy)) = T (x − iy) = x + iy. 4.2.21 Linear, since T ((x + iy) + (z + it)) = T (x + z + i(y + t)) = y + t + i(x + z) equals T (x + iy) + T (z + it) = y + ix + t + iz = y + t + i(x + z), and T (k(x + iy)) = T (kx + iky) = ky + ikx equals kT (x + iy) = k(y + ix) = ky + ikx. Yes, T is an isomorphism; it’s its own inverse, since T (T (x + iy)) = T (y + ix) = x + iy. 183

Chapter 4

4.2.22 Linear, since T (f + g) =

Z

3

(f + g) =

Z

3

−2

f+

Z

3

Z

3

f+

−2

−2

T (f ) + T (g) =

Z

3

g, and T (kf ) =

Z

3

kf = k

Z

3

f equals kT (f ) = k

−2

−2

−2

g equals

−2

Z

3

f.

−2

No, T isn’t an isomorphism, since domain and codomain have different dimensions. 4.2.23 T is linear, because T (f (t) + g(t)) = f (7) + g(7) = T (f (t)) + T (g(t)), and T (kf (t)) = kf (7) = kT (f (t)). However, T cannot be an isomorphism, because the dimensions of the domain and codomain fail to be equal. 4.2.24 T is not linear. 2T (t2 ) = 2(2(t2 )) = 4t2 6= T (2t2 ) = 4(2t2 ) = 8t2 . 4.2.25 Linear, since T (f + g) = (f + g)′′ + 4(f + g)′ = f ′′ + g ′′ + 4f ′ + 4g ′ equals T (f ) + T (g) = f ′′ + 4f ′ + g ′′ + 4g ′ , and T (kf ) = (kf )′′ + 4(kf )′ = kf ′′ + 4kf ′ equals kT (f ) = k(f ′′ + 4f ′ ) = kf ′′ + 4kf ′ . No, it isn’t an isomorphism, since the constant function f (x) = 1 is in ker(T ). 4.2.26 Linear, since T (f (t) + g(t)) = f (−t) + g(−t) equals T (f (t)) + T (g(t)) = f (−t) + g(−t), and T (kf (t)) = kf (−t) equals kT (f (t)) = kf (−t). Yes, T is an isomorphism; it’s its own inverse, since T (T (f (t))) = T (f (−t)) = f (t). 4.2.27 Linear, since T (f (t) + g(t)) = f (2t) + g(2t) equals T (f (t)) + T (g(t)) = f (2t) + g(2t), and T (kf (t)) = kf (2t) equals kT (f (t)) = kf (2t). Yes, T is an isomorphism; the inverse is T −1 (g(t)) = g 2t .

4.2.28 T is linear, since T (f (t) + g(t)) = f (2t) + g(2t) − f (t) − g(t) = f (2t) − f (t) + g(2t) − g(t) = T (f (t)) + T (g(t)), and T (kf (t)) = kf (2t) − kf (t) = k(f (2t) − f (t)) = kT (f (t)). T is not an isomorphism, however, since T (3) = 3 − 3 = 0. 4.2.29 Linear, because T (f (t) + g(t)) = f ′ (t) + g ′ (t) = T (f (t)) + T (g(t)) and T (kf (t)) = kf ′ (t) = kT (f (t)). However, since T (5) = 0, the kernel is nonzero, and T fails to be an isomorphism. 4.2.30 Linear, since T (f (t) + g(t)) = t(f ′ (t) + g ′ (t)) = t(f ′ (t)) + t(g ′ (t)) equals T (f (t)) + T (g(t)) = t(f ′ (t)) + t(g ′ (t)), and T (kf (t)) = t(kf ′ (t)) = kt(f ′ (t)) equals kT (f (t)) = kt(f ′ (t)). No, T isn’t an isomorphism, since the constant function f (t) = 1 is in ker(T ).

f (0) + g(0) f (1) + g(1) f (0) f (1) g(0) 4.2.31 T is linear: T (f (t) + g(t)) = = + f (2) + g(2) f (3) + g(3) f (2) f (3) g(2) kf (0) kf (1) f (0) f (1) = T (f (t)) + T (g(t)), and T (kf (t)) = =k = kT (f (t)). kf (2) kf (3) f (2) f (3) 184

g(1) g(3)

Section 4.2 However, the dimensions here are different, so that T fails to be an isomorphism. 4.2.32 T is not linear: T (2) = 0 + t2 6= 2T (1) = 2(0 + t2 ). 4.2.33 Linear, since T ((x0 , x1 , x2 , . . .) + (y0 , y1 , y2 , . . .)) = T (x0 + y0 , x1 + y1 , x2 + y2 , . . .) = (x0 + y0 , x2 + y2 , . . .) equals T (x0 , x1 , x2 , . . .) + T (y0 , y1 , y2 , . . .) = (x0 , x2 , . . .) + (y0 , y2 , . . .) = (x0 + y0 , x2 + y2 , . . .), and T (k(x0 , x1 , x2 , . . .)) = T (kx0 , kx1 , kx2 , . . .) = (kx0 , kx2 , . . .) equals kT (x0 , x1 , x2 , . . .) = k(x0 , x2 , . . .) = (kx0 , kx2 , . . .). No, T isn’t an isomorphism, since (0, 1, 0, 0, 0, . . .) is in ker(T ). 4.2.34 Linear, since T ((x0 , x1 , x2 , . . .) + (y0 , y1 , y2 , . . .)) = T (x0 + y0 , x1 + y1 , x2 + y2 , . . .) = (0, x0 + y0 , x1 + y1 , x2 + y2 , . . .) equals T (x0 , x1 , x2 , . . .) + T (y0 , y1 , y2 , . . .) = (0, x0 , x1 , x2 , . . .) + (0, y0 , y1 , y2 , . . .) = (0, x0 + y0 , x1 + y1 , x2 + y2 , . . .), and T (k(x0 , x1 , x2 , . . .)) = T (kx0 , kx1 , kx2 , . . .) = (0, kx0 , kx1 , kx2 , . . .) equals kT (x0 , x1 , x2 , . . .) = k(0, x0 , x1 , x2 , . . .) = (0, kx0 , kx1 , kx2 , . . .). No, T isn’t an isomorphism, since (1, 0, 0, 0, . . .) isn’t in im(T ). 4.2.35 Linear, since T (f (t) + g(t)) = (f (0) + g(0), f ′ (0) + g ′ (0), · · ·) = (f (0), f ′ (0), · · ·) + (g(0), g ′ (0), · · ·) = T (f (t)) + T (g(t)) and T (kf (t)) = (kf (0), kf ′ (0), · · ·) = k(f (0), f ′ (0), · · ·) = kT (f (t)). T fails to be an isomorphism. Note that the sequences in the image of T have only finitely many nonzero entries, so that a sequence like (1, 1, 1, 1, . . .), with all 1’s, fails to be in the image of T . Now use Theorem 4.2.4a. 4.2.36 T is linear: T (f (t)+g(t)) = (f (0)+g(0), f (1)+g(1), ...) = (f (0), f (1), ...)+(g(0), g(1), ...) = T (f (t))+T (g(t)), and T (kf (t)) = (kf (0), kf (1), ...) = k(f (0), f (1), ...) = kT (f (t)). We will show that the image of T isn’t all of V , so that T fails to be an isomorphism. More specifically, we claim that the sequence (1, 0, 0, 0, . . .), a 1 followed by all 0’s, fails to be in the image. We make an attempt to find a polynomial f (t) such that T (f (t)) = (f (0), f (1), f (2), . . .) = (1, 0, 0, . . .). This polynomial f (t) is required to have infinitely many zeros, at t = 1, 2, 3, . . . , so that f (t) must be the zero polynomial, and the equation f (0) = 1 isn’t satisfied. Thus there is no polynomial f (t) such that T (f (t)) = (1, 0, 0, 0, . . .). 4.2.37 Linear, since T (f +g) = f +g+f ′ +g ′ = f +f ′ +g+g ′ = T (f )+T (g) and T (kf ) = kf +kf ′ = k(f +f ′ ) = kT (f ). However, T (e−x ) = e−x − e−x = 0, so T fails to be an isomorphism. 4.2.38 Linear, just as in Exercise 37. T is not an isomorphism, since T (sin(x)) = sin(x) − sin(x) = 0. 185

Chapter 4 4.2.39 Linear; the proof is analogous to Exercise 25. No, T isn’t an isomorphism, since the kernel of T is two-dimensional, by Theorem 4.1.7. 4.2.40 Same answer as in Exercise 39. 4.2.41 Not linear, because T (f (t) + g(t)) = f (t) + g(t) + f ′′ (t) + g ′′ (t) + sin(t) does not equal T (f (t)) + T (g(t)) = f (t) + f ′′ (t) + sin(t) + g(t) + g ′′ (t) + sin(t).

f (7) + g(7) 4.2.42 Linear, since T (f (t) + g(t)) = equals T (f (t)) + T (g(t)) = f (11) + g(11) kf (7) f (7) + g(7) g(7) f (7) equals , and T (kf (t)) = = + kf (11) f (11) + g(11) g(11) f (11) f (7) kf (7) kT (f (t)) = k = . f (11) kf (11) Not an isomorphism, since domain and codomain have different dimensions. g(5) f (5) f (5) + g(5) 4.2.43 Linear, since T (f (t) + g(t)) = f (7) + g(7) = f (7) + g(7) = T (f (t)) + T (g(t)), and T (kf (t)) = g(11) f (11) f (11) + g(11) f (5) kf (5) kf (7) = k f (7) = kT (f (t)). f (11) kf (11)

T is an isomorphism; the proof is analogous to Example 6b.

g(1) f (1) f (1) + g(1) 4.2.44 Linear, since T (f (t) + g(t)) = f ′ (2) + g ′ (2) = f ′ (2) + g ′ (2) = T (f (t)) + T (g(t)), and g(3) f (3) f (3) + g(3) f (1) kf (1) T (kf (t)) = kf ′ (2) = k f ′ (2) = kT (f (t)). f (3) kf (3)

Not an isomorphism, since T ((t − 1)(t − 3)) = T (t2 − 4t + 3) = ~0.

4.2.45 Linear, since T (f (t) + g(t)) = t(f (t) + g(t)) = tf (t) + tg(t) = T (f (t)) + T (g(t)) and T (kf (t)) = t(kf (t)) = ktf (t) = kT (f (t)). This is not an isomorphism, since the constant function f (t) = 1 isn’t in the image. 4.2.46 Linear, since T (f (t) + g(t)) = (t − 1)(f (t) + g(t)) = (t − 1)f (t) + (t − 1)g(t) = T (f (t)) + T (g(t)) and T (kf (t)) = (t − 1)(kf (t)) = k(t − 1)f (t) = kT (f (t)). This is not an isomorphism, since the constant function f (t) = 1 isn’t in the image. Rt Rt Rt 4.2.47 T is linear, because T (f (t) + g(t)) = 0 (f (x) + g(x))dx = 0 f (x)dx + 0 g(x)dx = T (f (t)) + T (g(t)). Also, Rt Rt T (kf (t)) = 0 kf (x)dx = k 0 f (x)dx = kT (f (t)). 186

Section 4.2 However, there is no polynomial f (t) such that T (f (t)) =

Rt 0

f (x) = 6. Thus, T is not an isomorphism.

4.2.48 Linear, since T (f (t)+g(t)) = (f +g)′ (t) = f ′ (t)+g ′ (t) = T (f (t))+T (g(t)) and T (kf (t)) = kf ′ (t) = kT (f (t)). This is not an isomorphism, however, since T (5) = 0. 4.2.49 Linear, since T (f (t) + g(t)) = f (t2 ) + g(t2 ) = T (f (t)) + T (g(t)) and T (kf (t)) = kf (t2 ) = kT (f (t)). However, there is no f (t) in P such that T (f (t)) = f (t2 ) = t. Thus, the image of T fails to be all of P , and T fails to be an isomorphism. (t)−g(t) = 4.2.50 Linear. T (f (t)+g(t)) = f (t+2)+g(t+2)−f 2 kf (t+2)−kf (t) f (t+2)−f (t) =k = kT (f (t)). 2 2

This is not an isomorphism, however, since T (5) =

f (t+2)−f (t) + g(t+2)−g(t) 2 2

5−5 2

= T (f (t))+T (g(t)), and T (kf (t) =

= 0.

x y 1 2 that commute with , that is, z t 0 1 x + 2z y + 2t x 2x + y x y 1 2 1 2 x y . = , or, = z t z 2z + t z t 0 1 0 1 z t

4.2.51 We need to find the matrices M =

It follows that z = 0 and x = t, so that the kernel of T consists of all matrices of the form (i.e., the dimension of the kernel of T ) is 2. 4.2.52 We need to find the matrices A =

x z

y t

t y . The nullity 0 t

such that

0 0 = . It is required that x = −3y and z = −3t, so that the kernel 0 0 −3y y of T consists of all matrices of the form . −3t t x y z t

1 2 3 6

=

x + 3y z + 3t

2x + 6y 2z + 6t

The nullity (i.e., the dimension of the kernel of T ) is 2. 4.2.53 Note that T (a + bt + ct2 ) = 2c + 4b + 8ct. Thus the kernel consists of all constant polynomials f (t) = a(when b = c = 0), and the nullity is 1. The image consists of all linear polynomials f (t) = p + qt, and the rank is 2. 4.2.54 Use calculus to see that T (a + bt + ct2 ) = 5a + 52 b +

35 3 c.

The image is all of

R, so that the rank is 1. The kernel consists of all polynomials of the form f (t) = − 21 b − 37 c + bt + ct2 , and thus the nullity is 2. 4.2.55 The kernel consists of all infinite sequences (x0 , x1 , x2 , x3 , . . .) such that T (x0 , x1 , x2 , x3 , x4 , . . .) = (x0 , x2 , x4 , . . .) = (0, 0, 0, . . .), that is, all terms xk with even k must be 0. Thus the kernel consists of all sequences of the form (0, x1 , 0, x3 , 0, . . .). The image consists of all infinite sequences (y0 , y1 , y2 , . . .), since (y0 , y1 , y2 , . . .) = 187

Chapter 4 T (y0 , 0, y1 , 0, y2 , 0, . . .) for example. 4.2.56 Note that T (a + bt + ct2 ) = bt + 2ct2 . Thus the kernel consists of all constant polynomials f (t) = a (when b = c = 0), and the nullity is 1. The image consists of all polynomials of the form f (t) = pt + qt2 , and the rank is 2. 4.2.57 The kernel consists of the solutions of the differential equation f ′′ (t) − 5f ′ (t) + 6f (t) = 0. Using the approach outlined in Example 18 of Section 4.1 (involving a trial solution f (t) = ekt ), we find the general solution f (t) = c1 e2t + c2 e3t ; thus the nullity is 2. 4.2.58 The kernel consists of all infinite sequences such that T (x0 , x1 , x2 , . . .) = (0, x0 , x1 , x2 , . . .) = (0, 0, 0, 0, . . .), that is, all terms xk must be 0. Thus the kernel consists of the zero sequence (0, 0, 0, . . .) alone. The image consists of all infinite sequences of the form (0, x0 , x1 , x2 , . . .). 4.2.59 To find the kernel, we solve the equation T (f (t)) = T (a + bt + ct2 ) = a + 7b + 49c = 0. It follows that a = −7b − 49c, and the general element of the kernel is (−7b − 49c) + bt + ct2 = b(−7 + t) + c(−49 + t2 ). Then a basis of the kernel is −7 + t, −49 + t2 , and the nullity of T is 2. Now the rank of T must be 1, and the image is all of R. a + 7b + 49c a + 7b + 49c = . To find the kernel, solve the linear system 4.2.60 Note that T (a+bt+ct ) = a + 11b + 121c a + 11b + 121c 0 . The solution is a = 77c, b = −18c, so that the kernel consists of all polynomials of the form f (t) = c(77 − 0 18t + t2 ) = c(t − 11)(t − 7). You can also see directly that the quadratic polynomials f (t) with f (7) = f (11) = 0 are of this form. The nullity is 1. The image consists of all of R2 , so that the rank is 2. 2

4.2.61 The kernel consists of all polynomials f (t) such that t(f (t)) = 0 for all t, that is, the zero polynomial f (t) = 0 alone. The image consists of all polynomials g(t) that can be written as g(t) = t(f (t)), meaning that we can factor out a t. These are the polynomials with constant term 0, of the form g(t) = a1 t + a2 t2 + · · · + an tn . 4.2.62 The image of this transformation consists of all polynomials, since any polynomial is the derivative of another. The kernel of this transformation consists of all constant functions, or the span of the function f (t) = 1. 4.2.63 This is impossible, since dim(P3 ) = 4 and dim(R3 ) = 3. See Theorem 4.2.4b.

a b 4.2.64 Consider T (a + bt + ct + dt ) = , for example. c d 2

3

4.2.65 a First, we need to show that T (A + B) = T (A) + T (B) for all n × m matrices A and B, that is (T (A + B))(~v ) = (T (A) + T (B))(~v ) for all ~v in Rm . Indeed (T (A + B))(~v ) = (A + B)~v = A~v + B~v equals (T (A) + T (B))(~v ) = T (A)(~v ) + T (B)(~v ) = A~v + B~v . Also (T (kA))(~v ) = (kA)~v = kA~v equals (kT (A))(~v ) = k(T (A))(~v ) = kA~v . 188

Section 4.2 b The kernel of T consists of all n × m matrices A such that T (A) = 0, that is (T (A))(~v ) = A~v = ~0 for all ~v in Rm . This holds for the zero matrix only. Thus ker(T ) = {0}. c This is true by definition of a linear transformation (Definition 2.1.1). d Note that T gives an isomorphism from Rn×m to L(Rm , Rn ), by parts a, b, and c. Since dim(Rn×m ) = nm, by Exercise 17 of Seciton 4.1, we have dim(L(Rm , Rn )) = nm, by Theorem 4.2.4b. 4.2.66 The kernel of T consists of all smooth functions f (t) such that T (f (t)) = f (t) − f ′ (t) = 0, or f ′ (t) = f (t). As you may recall from a discussion of exponential functions in calculus, those are the functions of the form f (t) = Cet , where C is a constant. Thus the nullity of T is 1. 4.2.67 To show that T is linear, proceed as in Exercise 15. Now let M =

a b . c d

3 0 a b a b 2 3 − Then T (M ) = 0 k c d c d 0 4 2a + 3c 2b + 3d 3a kb −a + 3c (2 − k)b + 3d = − = . 4c 4d 3c kd c (4 − k)d a b The matrix is in the kernel of T if a = c = 0, (2 − k)b + 3d = 0 and (4 − k)d = 0. c d

If k is neither 2 nor 4, then the equation (4 − k)d = 0 implies that d = 0, and (2 − k)b + 3d = 0 then implies that b = 0. Thus the kernel of T is 0, and T is an isomorphism. If k is 2, then bis arbitrary (while a = c = d = 0), and T fails to be an isomorphism. A nonzero matrix in 0 1 kernel is in this case. 0 0 0 2 If k is 4, then b is arbitrary, and d = 3 b, so that, again, the kernel contains nonzero matrices for example, 0 and T fails to be an isomorphism.

the

3 2

,

In summary, T is an isomorphism if (and only if) k is neither 2 nor 4.

a b a b 5 0 4.2.68 We find the constants k which make the kernel of T nonzero. Let M = . Then T (M ) = − c d c d 0 1 0 0 0 0 3a −b 2a 2b 5a b a b 2 0 . = . We see that if k = 1, T = − = 0 0 0 1 (5 − k)c (1 − k)d kc kd 5c d c d 0 k 0 0 0 0 Similarly, if k = 5, T = . Thus, for k = 1, 5, the kernel of T is non-zero, and T is not an 1 0 0 0 isomorphism. For all other values of k, however, T is an isomorphism. 189

Chapter 4 4.2.69 No, since A is similar to B, there exists an invertible S such that AS = SB. Now T (S) = AS − SB = 0, so that the nonzero matrix S is in the kernel of T . 4.2.70 Since the dimensions of the domain and codomain are equal, it suffices to examine when the kernel is zero. Now f (t) is in the kernel of T if f (c0 ) = f (c1 ) = · · · = f (cn ) = 0. Recall that a nonzero polynomial of degree ≤ n has at most n zeros. If the n + 1 numbers c0 , c1 , . . . , cn are all different, then the only polynomial in Pn with f (c0 ) = f (c1 ) = . . . . = f (cn ) = 0 is the zero polynomial, so that the kernel of T is zero, and T is an isomorphism. However, if some of the numbers are equal, for example, cp = cq , then there will be nonzero polynomials in the kernel (for example, the product of all (t − ci ) where i 6= q), so that T fails to be an isomorphism. 4.2.71 Exercise 70 tells us that T is an isomorphism so long as c0 , c1 , · · · , cn are all different. This condition is met f (2) f (3) here, so T (f (t)) = f (5) is an isomorphism from P4 to R5 . Thus, there is exactly one polynomial f (t) with f (7) f (11) the required properties. 4.2.72 We satisfy all the requirements of Definition 4.1.2. Clearly 0 is an element of Zn . Let h = f + g, where f and g are elements of Zn . Then h(0) = f (0) + g(0) = 0 + 0 = 0. Also, if f is in Zn , then kf (0) = k(0) = 0. We notice that the space Zn has the basis, t, t2 , . . . , tn . Thus, the dimension of Zn is n. 4.2.73 Yes. T is an isomorphism; the inverse transformation is D(f (t)) = f ′ (t) = df dt , the derivative. We will check that the composite of T with D is the identity, in either order. Indeed Z d d t D(T (f (t))) = (T (f (t)) = f (x)dx = f (t) dt dt 0 by what is sometimes called the first fundamental theorem. And Z t f ′ (x)dx = f (t) − f (0) = f (t) for all f (t) in Zn T (D(f (t))) = T (f ′ (t)) = 0

by the ”second” fundamental theorem. 4.2.74 Let T (f (t)) = f ′ (t). This is an isomorphism, since the spaces have the same dimension, and the kernel consists only of the zero polynomial. 4.2.75 To see that 0 + 0 = 0, apply the formula f + 0 = f to f = 0 (see Definition 4.1.1). Then 0 = k0 − k0 = k(0 + 0) − k0 = k0 + k0 − k0 = k0. 4.2.76 0W = T (0V ) − T (0V ) = T (0V + 0V ) − T (0V ) = T (0V ) + T (0V ) − T (0V ) = T (0V ) We have used the equation 0V + 0V = 0V derived in Exercise 41. 4.2.77 If T and L are linear, then (L ◦ T )(f + g) = L(T (f + g)) = L(T (f ) + T (g)) = L(T (f )) + L(T (g)) 190

Section 4.2 = (L ◦ T )(f ) + (L ◦ T )(g), and (L ◦ T )(kf ) = L(T (kf )) = L(kT (f )) = kL(T (f )) = k(L ◦ T )(f ), so that L ◦ T is linear as well. If T and L are isomorphism, then L ◦ T is an isomorphism as well, since the composite of invertible functions is invertible. (See Figure 4.1.)

Figure 4.1: for Problem 4.2.77.

4.2.78 a Check all the conditions in Definition 4.1.1. A basis is 2. b T (x ⊕ y) = T (xy) = ln(xy) = ln(x) + ln(y) = T (x) + T (y) and T (k ⊙ x) = T (xk ) = ln(xk ) = k ln(x) = kT (x). The inverse of T is L(y) = ey , so that T is indeed an isomorphism. 4.2.79 Yes; let T be an invertible function from R2 to R. On R2 we can define the “exotic” operations ~v ⊕ w ~ = T −1 (T (~v ) + T (w)) ~ and k ⊙ ~v = T −1 (kT (~v )) (check that the conditions of Definition 4.1.1 hold). Then T is a linear transformation from R2 (with these exotic operations) to R (with the usual operations): T (~v ⊕ w) ~ = T (T −1 (T (~v ) + T (w))) ~ = T (~v ) + T (w), ~ and T (k ⊙ v) = kT (~v ). Now T is an isomorphism, so that the dimension of R2 (with our exotic operations) equals the dimension of the “usual” R, which is 1. 4.2.80 If a real linear space V has more than one element, then it is infinite; to see this, note that the scalar multiples kf of a nonzero f in V are all distinct (think about it!). Since there is more than one student in your class, but the number of students is finite, we cannot make the set X into a real linear space. 4.2.81 a ker(T ) is a subspace of V , so by Exercise 4.1.54, ker(T ) must also be finite-dimensional. Also, im(T ) is finite-dimensional, because it is finitely-generated by some elements T (f1 ), T (f2 ), · · · , T (fk ), where f1 , f2 , · · · , fk is a basis of V . b Following the hint: T (c1 u1 + · · · + cr ur + d1 v1 + · · · + dn vn ) = T (0), so c1 T (u1 ) + · · · + cr T (ur ) + d1 T (v1 ) + · · · + dn T (vn ) = c1 w1 + · · · + cr wr + 0 + · · · + 0 = 0. So, since w1 , · · · , wr is a basis and must be linearly independent, c1 , · · · cr must all be zero. Now, d1 v1 + · · · + dn vn = 0, but v1 , · · · , vn is also a basis, so d1 , · · · dn also must all equal zero. Thus, the elements u1 , · · · , ur , v1 , · · · , vn are linearly independent. c The hint guides us right along here. Let v be in V . Then T (v) is in im(T ), so T (v) = d1 w1 + · · · + dr wr . Now, T (v − d1 u1 − · · · − dr ur ) = T (v) − d1 T (u1 ) − · · · − dr T (ur ) = d1 w1 + · · · + dr wr − d1 w1 − · · · − dr wr = 0. Thus, v − d1 u1 − · · · − dr ur is in the kernel of T , and there are some c1 , · · · , cn such that v − d1 u1 − · · · − dr ur = c1 v1 + · · · + cn vn . Thus, v = d1 u1 + · · · + dr ur + c1 v1 + · · · + cn vn . Thus the elements u1 , · · · , ur , v1 , · · · , vn span V. 191

Chapter 4 4.2.82 Consider a basis v1 , ..., vn of ker T and a basis w1 , ..., wr of imT . Consider elements u1 , ..., ur in V such that T (ui ) = wi for i = 1, ..., r. In Exercise 81, parts b and c, we prove that the elements v1 , ..., vn , u1 , ..., ur form a basis of V, proving our claim. 4.2.83 The transformation T induces a transformation T˜ from ker (L ◦ T ) to ker L, with ker T˜ = ker T . Since ker L is assumed to be finite-dimensional, its subspace imT˜ will be finite-dimensional as well, with dim imT˜ ≤ dim ker L. Thus we can apply the rank-nullity-theorem as presented in Exercise 82 to T˜, finding that ker (L ◦ T ) is finitedimensional, with dim ker(L ◦ T ) = dim ker T˜ + dim imT˜ ≤ dim ker T + dim ker L, as claimed. 4.2.84 Using the terminology and the results introduced in Exercise 83, we observe that imT˜ = ker L. Indeed, if w is in ker L, then w = T (v) for some v in V, since imT = W . But this v will be in ker (L ◦ T ) since L (T (v)) = L(w) = 0. Thus w is in imT˜. Therefore dim ker(L◦T ) = dim ker T˜ +dim imT˜ = dim ker T +dim ker L, as claimed.

Section 4.3 2 4.3.1 Let B bethe of the given polynomials with respect to B t, t . Then the coordinates of P2 : 1, standard basis 7 9 3 3 9 7 are [f ]B = 3 , [g]B = 9 , [h]B = 2 . Finding rref 3 9 2 = I3 , we conclude that [f ]B , [g]B , [h]B are 1 4 1 1 4 1 linearly independent, hence so are f, g, h, since the coordinate transformation is an isomorphism.

0 0 0 0 0 1 1 0 of R2×2 . Then the coordinates of the given matrices , , , 0 1 1 0 0 0 0 0 1 1 2 1 1 1 1 1 2 2 2 3 3 1 4 4 = , = , = , = . Finding with respect to B are 1 3 5 1 1 B 6 3 4 B 5 7 B 6 8 B 1 4 7 8 1 1 2 1 1 0 0 −1 1 1 2 1 1 2 3 4 0 1 0 4 1 2 3 4 rref = 6= I4 , we conclude that the four vectors , , , are 1 3 5 6 0 0 1 −1 1 3 5 6 1 4 7 8 0 0 0 0 1 4 7 8 1 4 1 1 1 2 2 3 linearly dependent, and so are the four given matrices. In fact =− +4 − . 6 8 1 1 3 4 5 7

4.3.2 Let B be the basis

1 2 4.3.3 We proceed as in Exercise 1. Since rref 9 1

1 7 0 7

1 8 1 5

1 8 = I4 , the four given polynomials do form a basis of P3 . 4 8

4.3.4 Consider the coordinate vectors of the 3 given polynomials with respect to the standard basis of P2 : 1, t, t2 . 2k 0 1 1,1,2 + k 1 1 0 1 0 2k 1 0 2k Since matrix 1 1 2 + k reduces to 0 1 2 − k , these three vectors form a basis of R3 unless k = 1. 0 0 k−1 0 1 1 192

Section 4.3 Therefore, the three polynomials f (t), tf (t), g(t) form a basis of P2 unless k = 1. 4.3.5 Use a diagram: a b + 2c a b a b 1 2 −−→ = T 0 3c 0 c 0 c 0 3 y y a a − − → b b + 2c A c 3c 1 a a The matrix A that transforms b into b + 2c is A = 0 0 3c c A is invertible, so T is an isomorphism.

0 0 1 2 . 0 3

4.3.6 Use Theorem 4.3.2. to construct matrix B column by column: 1 0 0 1 0 1 B= T T T 0 0 B 0 0 B 0 1 B =

1 0

0 0

B

0 0

1 0

B

0 0

3 3

B

1 0 = 0 1 0 0

0 0 . 3

This matrix is invertible, so T is an isomorphism. 4.3.7 Use Theorem 4.3.2 to construct B column by column: 1 0 0 1 1 0 0 0 0 0 0 4 B= T T T = 0 1 B 0 0 B 0 −1 B 0 0 B 0 0 B 0 0 B 0 0 1 0 0 0 = 0 0 4 . 0 , 1 is a basis of the kernel of B, and 1 is a basis of the image of B. This implies that 0 0 0 0 0 0 0 1 0 1 1 0 is a basis of the image of T . Thus, T fails to be an is a basis of the kernel of T , and , 0 0 0 0 0 1 isomorphism.

4.3.8 Use a diagram: 0 2a − 2c a b −−→ T 0 0 c 0 y y 0 a − − → 2a − 2c b A 0 c

193

Chapter 4 0 0 0 0 a The matrix that transforms b into 2a − 2c is A = 2 0 −2 . 0 0 0 0 c 0 1 0 We see that a basis of the kernel of A is 1 , 0 and a basis of the image of A is 2 . Thus, a basis of 0 1 0 0 1 0 2 the kernel of T is , I2 while a basis of the image is . Thus, the rank of T is 1 and T is not an 0 0 0 0 isomorphism. 4.3.9 Use a diagram: a b a 2b −−→ T 0 c 0 c y y a a −−→ b 2b A c c

a a 1 The matrix A that transforms b into 2b is A = 0 c c 0 A is invertible, so T is an isomorphism.

4.3.10 Use a diagram: a b −−→ 1 3 T 3 0 0 c y a −−→ b A c 1 The matrix is A = 2 0

−2 1

0 3 0

0 −2 . 1

0 0 2 0 . 0 1

1 2 a 2a + 3b − 2c = 0 3 0 c y a 2a + 3b − 2c c

a b 0 c

Since A is invertible, T is an isomorphism.

4.3.11 Use Theorem 4.3.2. to construct matrix B column by column: 1 −1 0 1 0 1 1 −1 0 1 0 3 B= T T T = 0 0 B 0 1 B 0 0 B 0 0 B 0 1 B 0 0 B 1 0 0 = 0 1 0 . 0 0 3 A is invertible, so T is an isomorphism.

194

Section 4.3 4.3.12 Use a diagram as in Definition 4.3.1: a b 2a 3b −−→ T cd 2c 3d y y a 2a −−→ b 3b A c 2c d 3d 2 0 0 0 0 3 0 0 The matrix is A = . 0 0 2 0 0 0 0 3

Since A is invertible, T is an isomorphism.

4.3.13 Use a diagram: a b a+c b+d −−→ T cd 2a + 2c 2b + 2d y y a a+c −−→ b b+d A c 2a + 2c d 2b + 2d 1 0 1 0 0 1 0 1 The matrix is A = . 2 0 2 0 0 2 0 2 0 1 0 1 0 1 0 1 We find to be a basis of the kernel of A, and , to be a basis of the image of A. Thus, a , 0 2 0 −1 2 0 −1 0 1 0 0 1 1 0 0 1 basis of the kernel of T is , , a basis of the image of T is , , and T fails to be −1 0 0 −1 2 0 0 2 an isomorphism. 4.3.14 Use Theorem 4.3.2 to construct matrix B column by column: 0 0 0 0 0 0 0 0 3 0 0 3 0 0 0 0 B= = 0 0 B 0 0 B 6 0 B 0 6 B 0 0 3 0 0 0 0 3

1 0 0 We find bases of the kernel and image of B, and use them to find that a basis of the kernel of T is , −1 0 0 0 1 1 0 . Thus the rank of T is 2, and T is not an isomorphism. , and a basis of the image is 0 2 2 0 195

1 , −1

Chapter 4 4.3.15 We use a diagram again: x+ iy y x y

−−→ T

x− iy y x −−→ A −y 1 0 . Thus A = 0 −1 A is invertible, so T is an isomorphism. 4.3.16 B = [[1 − i]B [1 + i]B ] =

0 1 . Since B is invertible, T must be an isomorphism. 1 0

4.3.17 Another diagram: −−→ x+ T −y + ix iy y y −y x −−→ A x y 0 −1 A= . 1 0 A is invertible, so T is an isomorphism. −−→ x+ T 2x − 3y +i(3x + 2y) iy y 4.3.18 y . x 2x − 3y −−→ A y 3x + 2y 2 −3 . Since A is invertible, T must be an isomorphism. A= 3 2 4.3.19 We use a diagram to show our work: −−→ x+ T px − qy +i(qx + py) iy y y x px − qy −−→ A y qx + py p −q A = . If p = q = 0, then T (z) = 0 for all z, so that the kernel is all of C, while the image is {0}. q p Otherwise, T is an isomorphism. 196

Section 4.3 −−→ T a + bt+ ct2 b +2ct y y 4.3.20 a b −−→ b A 2c c 0 0 1 0 1 1 0 Thus A = 0 0 2 . We see that a basis of the kernel of A is 0 and a basis of the image of A is 0 , 2 . 0 0 0 0 0 0 Thus, a basis of the kernel of T is 1, while a basis of the image of T is 1, 2t. Thus, the rank of T is 2, and T is not an isomorphism. 4.3.21 We use a diagram to show our work: −−→ a + bt+ ct2 T b − 3a + (2c− 3b)t − 3ct2 y y a b − 3a −−→ b 2c − 3b A c −3c −3 1 0 2 . Thus A = 0 −3 0 0 −3 A is invertible, so T is an isomorphism.

−−→ a + bt+ ct2 T 4b + 2c + 8ct y y 4.3.22 a 4b + 2c −−→ b A 8c c 0 0 4 2 1 4 2 Thus A = 0 0 8 . We find a basis of the kernel of A to be 0 , a basis of the image of A to be 0 , 8 . 0 0 0 0 0 0 From this we see that a basis of kernel of T is 1, while a basis of the image of T is 4, 2 + 8t. Thus, T is not an isomorphism, since the rank of T is only 2. 4.3.23 A diagram shows our work: −−→ a + bt+ ct2 T a + 3b + 9c y y a + 3b + 9c a − − → b 0 A 0 c 1 3 9 Thus A = 0 0 0 . 0 0 0 197

Chapter 4 1 9 3 We find −1 , 0 to be a basis of the kernel of A, and 0 to be a basis of the image of A. Thus, a basis 0 −1 0 of the kernel of T is 3 − t, 9 − t2 , a basis of the image of T is 1 (the image is R), and T fails to be an isomorphism.

4.3.24 B = [[1]B [0]B respectively.

1 0 [0]B ] = 0 0 0 0

1 0 0 0 0 . A basis of the kernel and image of B are obvious, 1 , 0 and 0 0 1 0 0

So we find a basis of the kernel of T to be t − 3, (t − 3)2 , and a basis of the image of T to be 1. Thus T has a rank of 1, and is not an isomorphism. 4.3.25 We use the following diagram: −−→ a + bt+ ct2 T a − bt+ ct2 y y a a −−→ −b b A c c 1 0 0 Thus A = 0 −1 0 . 0 0 1

A is invertible, so T is an isomorphism.

−−→ a + bt+ ct2 T a + 2bt+ 4ct2 y y 4.3.26 a a −−→ 2b b A 4c c 1 0 0 Thus A = 0 2 0 . Since A is invertible, T must be an isomorphism. 0 0 4 4.3.27 We use a diagram: a + b(2t − 1) + c(2t − 1)2 −−→ a + bt + ct2 T = a − b + c +(2b − 4c)t + 4ct2 y y a a−b+c −−→ b 2b − 4c A c 4c 1 −1 1 Thus A = 0 2 −4 . 0 0 4 198

Section 4.3 A is invertible, so T is an isomorphism. 1 0 0 4.3.28 B = [1]B [2t − 2]B 4(t − 1)2 B = 0 2 0 . Since B is invertible, T will be an isomorphism. 0 0 4 4.3.29 This diagram shows our work: a + bt+ ct2 y a b c

−−→ T

2a + 2b+ 8c/3 y

2a + 2b + 8c/3 −−→ 0 A 0

2 2 Thus A = 0 0 0 0

8/3 0 . 0

4 1 1 3 We find −1 , 0 to be a basis of the kernel of A, and 0 to be a basis of the image of A. Thus, a basis 0 0 −1 of the kernel of T is 1 − t, 34 − t2 , a basis of the image of T is 1 (the image is R), and T fails to be an isomorphism.

4.3.30

a + bt + ct y a b c

2

−−→ T

a+b(t+h)+c(t+h)2 −a−bt−ct2 h

= b +ch + 2ct y b + ch 2c 0

−−→ A

0 1 h 1 1 h Thus A = 0 0 2 . A basis of the kernel of A is 0 , while a basis of the image of A is 0 , 2 . Thus, 0 0 0 0 0 0 a basis of the kernel of T is 1 while a basis of the image of T is 1, h + 2t. So the rank of T is 2, and it is not an isomorphism.

Recall that in calculus

f (t+h)−f (t) h

is called the difference quotient of function f (t).

Geometrically, it represents the slope of the secant through points (t, f (t)) and (t + h, f (t + h)). The limit of the difference quotient, as h approaches 0, is the derivative f ′ (t). Compare with Exercise 20. 199

Chapter 4 a+b(t+h)+c(t+h)2 −a−b(t−h)−c(t−h)2

−−→ 2h a + bt + ct2 T = b+ 2ct y y 4.3.31 a b −−→ b 2c A c 0 0 1 0 Thus A = 0 0 2 . 0 0 0 0 1 1 We find 0 to be a basis of the kernel of A, and 0 , 2 to be a basis of the image of A. Thus, a basis of 0 0 0 the kernel of T is 1, a basis of the image of T is 1, 2t, and T fails to be an isomorphism. (t−h) Geometrically, f (t+h)−f is the slope of the secant through points (t − h, f (t − h)) and (t + h, f (t + h)). This 2h exercise shows that for a quadratic polynomial f (t) this secant is parallel to the tangent at the midpoint (t, f (t)), a fact that was known to Archimedes. Draw a sketch!

Thus T (f (t)) =

f (t+h)−f (t−h) 2h

= f ′ (t) in this case. Compare with Exercise 20.

a + b + c + (b + 2c)(t − 1) −−→ a + bt + ct2 T = a − c+ (b + 2c)t y y 4.3.32 a−c a − − → b + 2c b A 0 c −1 1 0 −1 2 . We find a basis of the kernel of A to be 2 , while a basis of the image of A is Thus A = 0 1 −1 0 0 0 0 1 0 , 1 . Thus, a basis of the kernel of T is −1 + 2t − t2 and a basis of the image of T is 1, t. The rank of T 0 0 is only 2, and T is not an isomorphism. Note that T (f (t)) = f (1) + f ′ (1)(t − 1) is the equation of the tangent to the graph of f (t) at the point (1, f (1)).

4.3.33 B = [[1]B [t − 1]B

1 [0]B ] = 0 0

0 0 1 0 . 0 0

0 1 0 We find 0 to be a basis of the kernel of B, and 0 , 1 to be a basis of the image of B. Thus, a basis of 0 0 1 the kernel of T is (t − 1)2 , a basis of the image of T is 1, t − 1, and T fails to be an isomorphism. 200

Section 4.3

4.3.34 A =

0 0 1 1 0 T T T 1 0 0 A 0 0 A

0 0

A

0 T 0

0 1

A

0 0 0 0 0 −3 0 0 = = . 0 0 3 0 A A A A 0 0 0 0 0 0 0 1 0 0 −3 0 We find , to be a basis of the kernel of A, and , to be a basis of the image of A. Thus, a 0 0 3 0 0 0 1 0 1 0 0 0 0 −3 0 0 basis of the kernel of T is , , a basis of the image of T is , , meaning that T has 0 0 0 1 0 0 3 0 a rank of 2 and T fails to be an isomorphism.

0 0

0 0

0 −3 0 0

0 0 3 0

0 0 0 0

4.3.35 Again, we use a diagram to show our work: a b c d−a 1 0 0 1 0 0 −−→ M= T T (M ) = =c + (d − a) −c 0 −c 0 0 0 0 0 1 c d y y a c b −−→ d − a [M ]A = A [T (M )]A = c 0 d −c 0 0 1 0 −1 0 0 1 Thus, A = . 0 0 0 0 0 0 −1 0 0 1 1 0 By inspection, a basis of the kernel and image of this matrix are , and 0 0 0 1 0 1 1 0 0 1 −1 0 , , , , respectively. Thus, we see that a basis of ker(T ) is 0 1 0 0 0 0 0 −1 0 −1 1 0 and a basis of im(T ) is , . Thus, T fails to be an isomorphism. 0 0 0 −1 4.3.36 We will build A column-by-column: 1 0 0 1 0 0 0 0 A= T T T T 0 0 A 0 0 A 1 0 A 0 1 A =

0 1

−1 0

A

−1 0 0 1

A

1 0 0 −1

A

0 1 −1 0

A

0 −1 1 0 −1 0 = 1 0 0 0 1 −1

201

0 1 . −1 0

Chapter 4 −1 0 0 1 −1 0 0 1 We find to be a basis of the image of A. , to be a basis of the kernel of A, and , 0 1 −1 0 1 0 0 −1 1 0 0 1 0 −1 −1 0 Thus, a basis of the kernel of T is , , a basis of the image of T is , , the 0 −1 −1 0 1 0 0 1 rank of T is 2 and T fails to be an isomorphism.

4.3.37 We will construct our matrix B column-by-column: −2 0 0 0 2 2 2 −2 0 2 0 0 B = 0− −0 [0 − 0]B [0 − 0]B = . −2 −2 B 0 0 0 0 2 −2 B 0 0 0 0 0 −2 0 0 0 2 0 0 We find , to be a basis of the kernel of B, and , to be a basis of the image of B. Thus, a 0 0 0 1 0 0 1 0 2 −2 −2 −2 0 1 , and T fails to be , , and a basis of the image of T is basis of the kernel of T is I2 , 2 −2 2 2 1 0 an isomorphism.

4.3.38 We will construct our matrix B column-by-column: 0 0 0 0 −2 0 0 2 0 −2 0 0 B = [0]B [0]B = . 2 0 B 0 2 B 0 0 2 0 0 0 0 0 1 0 0 0 0 0 −2 0 We find , to be a basis of the kernel of B, and , to be a basis of the image of B. From this 0 0 0 2 0 1 0 0 1 0 0 1 −2 0 0 2 we conclude that a basis of ker(T ) is , , a basis of im(T ) is , and the rank of 1 0 0 −1 2 0 0 2 T is 2. So, T fails to be an isomorphism. 4.3.39 We use a diagram to show our work: a b c−a d−b −−→ M= T T (M ) = a− c b − d c d y y Let a c−a b −−→ d−b [M ]A = A [T (M )]A = c a−c d b−d −1 0 1 0 1 0 −1 0 Then our A-matrix of T is: . 1 0 −1 0 0 1 0 −1 202

Section 4.3 0 −1 0 1 0 −1 0 1 Thus, , is a basis of the kernel, and is a basis of the image. From this we conclude , 0 1 0 1 1 0 1 0 1 0 0 1 −1 0 0 −1 that a basis of ker(T ) is , and a basis of im(T ) is , . So, T fails to be an 1 0 0 1 1 0 0 1 isomorphism. 4.3.40 We will construct our matrix A column-by-column: −4 0 2 0 −4 0 0 2 2 0 0 2 0 2 0 2 A= = . 4 0 A 0 4 A −2 0 A 0 4 A 4 0 −2 0 0 4 0 4 1 0 −4 0 2 0 1 0 2 We find , to be a basis of the kernel of A, and , to be a basis of the image of A. 1 0 4 0 0 −1 0 4 1 0 2 −4 0 0 1 0 2 . , and a basis of im(T ) is , From this we conclude that a basis of ker(T ) is 0 4 4 0 0 −1 1 0 So, the rank of T is 2 and T fails to be an isomorphism.

4.3.41 a In Exercise 5 we consider the standard basis A basis B consisting of 1 0 0 1 0 1 , , . 0 0 0 0 0 1 1 1 0 0 1 0 1 Now SB→A = = 0 0 0 A 0 0 A 0 1 A 0

1 b Check that AS = SB = 0 0 −1 c SA→B = SB→A

1 = 0 0

of U 2×2 , and in Exercise 6 we work with the alternative

0 1 0

0 1 . 1

0 0 1 3 . 0 3

0 0 1 −1 . 0 1

4.3.42 a If A is the standard basis considered in Exercise 8 and B is the basis in Exercise 7, then 1 0 1 1 0 0 1 1 0 0 . = 0 1 S= 0 1 A 0 0 A 0 −1 A 1 0 −1

0 b Check that AS = SB = 0 0

0 0 0 4 . 0 0 203

Chapter 4 1

2

−1 c SA→B = SB→A =0 1 2

0 1 0

1 2

0 . − 21

4.3.43 a If A is the standard basis considered in Exercise 10 and B is the basis in Exercise 11, then 1 0 0 1 −1 0 1 0 1 SB→A = = −1 1 1 . 0 0 A 0 1 A 0 0 A 0 1 0 0 0 1 3 . 1 0

1 b Check that AS = SB = −1 0

1 = 0 1

−1 c SA→B = SB→A

0 0 0 1 . 1 −1

4.3.44 a If A is the standard basis considered in Exercise 13 and B is the basis in Exercise 14, then

1 0 2 0

0 0 b Check that AS = SB = 0 0

0 0 0 0

3 0 6 0

S=

1 −1

0 0

A

0 1 0 −1

A

A

0 1 0 2

A

1 0 1 0 = −1 0 0 −1

1 0 2 0

0 1 . 0 2

0 3 . 0 6

4.3.45 a If A is the standard basis considered in Exercise 15 and B is the basis in Exercise 16, then 1 1 . SB→A = [[1 + i]A [1 − i]A ] = 1 −1 b In Exercises 16 and 15 we found B = 1 1 . −1 1 −1 c SA→B = SB→A =

1 2

0 1 1 0

and A =

1 0

0 , respectively. Now check that AS = SB = −1

1 1 . 1 −1

4.3.46 a If A is the standard basis considered in Exercise 23 and B is the basis in Exercise 24, then 1 −3 9 S = [1]A [−3 + t]A 9 − 6t + t2 A = 0 1 −6 . 0 0 1 204

Section 4.3

1 b Check that AS = SB = 0 0 c SA→B =

−1 SB→A

1 = 0 0

0 0 0 0 . 0 0

3 −9 1 6. 0 1

4.3.47 a If A is the standard basis considered in Exercise 27 and B is the basis in Exercise 28, then 1 −1 1 SB→A = [1]A [−1 + t]A 1 − 2t + t2 A = 0 1 −2 . 0 0 1 b In Exercises 28

1 AS = SB = 0 0 −1 c SA→B = SB→A

1 0 and 27 we found B = 0 2 0 0 −2 4 2 −8 . 0 4

1 = 0 0

1 0 0 and A = 0 0 4

−1 1 2 −4 , respectively. Now check that 0 4

1 1 1 2 . 0 1

4.3.48 Use a diagram: −−→ a cos(t) + b sin(t) T b cos(t) − a sin(t) y y a b −−−→ B b −a 0 1 Thus B = . −1 0 4.3.49 B = [[2 cos(t) − 2 sin(t)]B [2 cos(t) + 2 sin(t)]B ] = invertible. 4.3.50 B = [[(b − 1) cos(t) − a sin(t)]B

2 −2

2 . Yes, T is an isomorphism, since matrix B is 2

b−1 [a cos(t) + (b − 1) sin(t)]B ] = −a

a . b−1

4.3.51 Note that cos(t − π/2) = sin(t) and sin(t − π/2) = − cos(t). Thus 0 −1 . Yes, T is an isomorphism. B = [[cos(t − π/2)]B [sin(t − π/2)]B ] = [[sin(t)]B [− cos(t)]B ] = 1 0 4.3.52 Recall that cos(t − δ) = cos(δ) cos(t) + sin(δ) sin(t) and 205

Chapter 4 sin(t − δ) = cos(δ) sin(t) − sin(δ) cos(t). Also, cos(π/4) = sin(π/4) =

√

2/2.

Thus B = [[cos(t − π/4)]B [sin(t − π/4)]B ] =

hh √

2 2

√

cos(t) +

2 2

i sin(t)

B

h

√

−

2 2

√

cos(t) +

2 2

i i sin(t) = B

√ 2 2

1 1

−1 1

4.3.53 Recall that cos(t − θ) = cos(θ) cos(t) + sin(θ) sin(t) and sin(t − θ) = cos(θ) sin(t) − sin(θ) cos(t). Thus B = [[cos(t − θ)]B [sin(t − θ)]B ] cos(θ) − sin(θ) . Yes, T is an isomor= [[cos(θ) cos(t) + sin(θ) sin(t)]B [− sin(θ) cos(t) + cos(θ) sin(t)]B ] = sin(θ) cos(θ) phism. Note that B is a rotation matrix. 5 1 4.3.54 Note that the two basis vectors 1 and −4 are perpendicular. 1 −1 0 1 0 5 1 0 . T −4 = 0 . Now B = 1 0 = 0 0 −1 B 0 B 0 1

1 1 Thus T 1 = 1 and −1 −1

1 1 4.3.55 Let ~u = √16 −2 be the unit vector in the direction of −2 . 1 1 1 1 1 1 1 Now T 1 = 16 −2 · 1 −2 = − 31 −2 and −1 1 −1 1 1 1 5 1 1 5 T −4 = 16 −2 · −4 −2 = 73 −2 , by Theorem 2.2.1. 1 1 1 1 1 1 1 5 1 2 1 1 −2 Also note that −2 = − 3 1 + 3 −4 , so that −2 . =3 1 1 −1 1 1 B 1 1 2 −14 7 1 1 =9 − 3 −2 −2 Now B = . 3 −1 7 1 1 B B

1 1 1 −5 5 1 5 14 4.3.56 T 1 = 2 × 1 = 4 and T −4 = 2 × −4 = 14 . Thus −1 3 −1 −1 1 3 1 −14 14 −5 0 14 B= = . 14 4 −1 0 −1 B −14 B 206

Section 4.3 1 3 5 5 1 −6 1 4.3.57 T 1 = 3 = − 1 − −4 and T −4 = 3 = 3 1 . −1 −3 1 1 −1 0 −1 −6 3 −1 3 Thus B = 3 3 = . −1 0 0 B −3 B

1 3 1 5 0 3 0 3 0 4.3.58 T 1 = 3 = 3 1 and T −4 = 0 . Thus B = 3 0 = . 0 0 −1 −3 −1 1 0 −3 B 0 B 4.3.59 T (f ) = t · f is linear and ker(T ) = {0}, but T is not an isomorphism since the constant function 1 is not in the image of T .

4.3.60 a We will use Theorem 4.3.3. SB→A

−1 b Here, we know SA→B = SB→A =

3 0 = 1 0 3 A

1 = 2 2 A

1 . 1

1 −1 . 0 1

c Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A 4.3.61 a We will use Theorem 4.3.3. SB→A = reflection combined with a scaling. b Here, we know SA→B =

−1 SB→A

=

1 −9−16

3 −4

5 −10

A

10 5

A

−4 3 1 = − 25 −3 −4

c Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A

=

−3 4

−4 = −3

1 25

4 3

−3 4

= 5

"

− 35

4 5

4 5

3 5

#

, and is thus a

4 . 3

4.3.62 a Finding this basis is equivalent to finding of[ 1 −22 ] that does not contain any of the kernel a basis 4 2 4 2 zeroes. We can quickly spot the vectors 2 and 1 , so B = 2 , 1 , for example. −1 1 −1 1 b From Theorem 4.3.3, SB→A

c SA→B =

−1 SB→A

=

1 −3

4 1 = 2 1 −1 A

2 = 2 1 A

−1 −1 = −1 2

1 3

1 1

1 . −2

d Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A

207

1 . −1

Chapter 4 4.3.63 a Finding this basis to finding is equivalent a basis of the kernel of [ 1 2 0 2 0 vectors 0 and 2 , so B = 0 , 2 , for example. 1 3 1 3 b From Theorem 4.3.3, SB→A

−1 c SA→B = SB→A =

1 −3+1

2 = 0 1 A

0 2 = −1 1 3 A

3 1 3 = − 12 −1 −1 −1

1 = −1

1 2

−1 . 3

−3 −1 . 1 1

d Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A

4.3.64 a We find A column-by-column: A = [I2 ]B [P ]B P

3 −2 ]. We can quickly spot the

2 B

1 1 = 0 2 1 3

1 8. 9

1 1 −3 −3 b We find a basis of the kernel of A to be 4 , and a basis of the image of A to be 0 , 2 . Thus, 0 1 3 −1 is a basis of the kernel of T , and I2 , P is a basis of the image of T.

2 2 a b a + bc 4.3.65 a P 2 = = c d ac + cd bc − ad . [P 2 ]B = a+d

ab + bd bc + d2

=

a2 + bc (a + d)b (a + d)c bc + d2

= (a + d)

4 −1

a b 1 0 + (bc − ad) . So c d 0 1

b We will do this column-by-column: B = [[T (I2 )]B [T (P )]B ] = [P ]B P 2 B

0 = 1

bc − ad , by part a. T is an isomorphism if bc − ad 6= 0, that is, if P is invertible. a+d

0 0 a+d c Assume that bc − ad = 0. Then the B-matrix of T is B = . So, ker(B) = span and −1 d 1 a+ d −b 0 , a basis of im(T ) is P , and the rank of . Thus a basis of ker(T ) is (a + d)I2 − P = im(B) =span −c a 1 T is 1. 0 −1 0 4.3.66 a B = T (x21 ) B [T (x1 x2 )]B T (x22 ) B = 2 0 1

1 b Note that rref(B) = 0 0

0 −2 . 0

−1 0 0 −1 1 0 . Now im(B) = span 2 , 0 and 1 0 0 0 208

Section 4.3 1 ker(B) = span 0 . Thus 2x1 x2 , −x21 + x22 is a basis of im(T ) and x21 + x22 is a basis of ker(T ). 1 4.3.67 a

[T (cos t)]B [T (sin t)]B

= [0]B = ~0 = [0]B = ~0

0 −2 [T (t cos(t))]B = [−2 sin(t)]B = 0 0

2 0 [T (t sin(t))]B = [2 cos(t)]B = 0 0 0 0 so M = 0 0

0 0 0 0

0 −2 0 0

2 0 . 0 0

1 p 0 q b The equation T (f ) = cos(t) corresponds to M~x = , with solutions ~x = , where p and q are arbitrary. 0 0 1 0 2 1 Thus f (t) = p cos(t) + q sin(t) + 2 t sin(t). In Figure 4.2 we graph f (t) for p = q = 0; note that p cos(t) + q sin(t) is just a sinusoidal function 2π

t 2 1 t sin (t) 2

π –π

π

2π

3π

t

–π –2π

– 2t

Figure 4.2: for Problem 4.3.67b.

4.3.68 a The sequence (0, 0, 0, . . .) is in W . • If the sequences (xn ) and (yn ) are in W (that is, xn+2 = xn+1 + 6xn and yn+2 = yn+1 + 6yn for all n), then xn+2 + yn+2 = xn+1 + yn+1 + 6(xn + yn ) so that the sequence (xn + yn ) is in W as well. • If the sequence (xn ) is in W and k is any constant, then kxn+2 = kxn+1 + 6kxn , so that the sequence (kxn ) is in W as well. b A sequence in W is determined by its first two components (a and b, say), which we can choose freely. All the later components can then be expressed in terms of a and b, since xn+2 = xn+1 + 6xn : (a, b, b + 6a, 6a + 7b, 42a + 13b, . . .) = a(1, 0, 6, 6, 42, . . .) + b(0, 1, 1, 7, 13, . . .). The two sequences (1, 0, 6, 6, 42, . . .) and (0, 1, 1, 7, 13, . . .) form a basis of W , so that dim(W ) = 2. 209

Chapter 4 c It is required that cn+2 = cn+1 + 6cn for all n ≥ 0 or c2 = c + 6 or c2 − c − 6 = 0 or (c − 3)(c + 2) = 0 The solution are c1 = 3 and c2 = −2. The geometric sequences (1, 3, 9, 27, 81, . . .) and (1, −2, 4, −8, 16, . . .) are in W . d Yes, the two sequences we found in part c do the job, since dim(W ) = 2. e (x0 , x1 , x2 , x3 , x4 , . . .) = (0, 1, 1, 7, 13, . . .). We are looking for constants p and q such that (x0 , x1 , x2 , x3 , . . .) = (0, 1, 1, 7, . . .) = p(1, 3, 9, 27, . . .) + q(1, −2, 4, −8, . . .). 0=p+q , so that p = To find p and q, it suffices to consider the first two components: 1 = 3p − 2q Thus xn = 51 3n − 15 (−2)n . f (a ) 1 1 f1 (a2 ) 4.3.69 As the hint suggests, we find the kernel of M = .. . f (a ) c1 1 1 c2 f1 (a2 ) If M ... = .. .

cn

f2 (a1 ) f2 (a2 ) .. .

··· ···

f1 (an ) f2 (an ) · · ·

f2 (a1 ) f2 (a2 ) .. .

··· ···

1 5

and q = − 15 .

fn (a1 ) fn (a2 ) .. . .

f1 (an ) f2 (an ) · · · fn (an ) c f (a ) + · · · + c f (a ) fn (a1 ) c1 0 1 1 1 n n 1 fn (a2 ) c2 c1 f1 (a2 ) + · · · + cn fn (a2 ) 0 = . , .. .. .. ... = . . 0 cn c1 f1 (an ) + · · · + cn fn (an ) fn (an )

then the polynomial c1 f1 + · · · + cn fn in Pn−1 has at least n zeros, namely, a1 , a2 , · · · , an . It follows that c1 f1 + c2 f2 + · · · + cn fn = 0 and therefore c1 = c2 = · · · = cn = 0 since the fi are linearly independent. We have shown that ker(M ) = {~0}, so that M is invertible.

4.3.70 Yes; suppose dim(V ) = dim(W ) = n, and let B and C be bases of V and W , respectively. Then the coordinate transformation TB and TC define isomorphisms from V to Rn and from W to Rn , respectively. TC−1 ◦ TB is an isomorphism from V to W (see Exercise 4.2.77). 4.3.71 We need to show that there are constants w1 , . . . , wn such that Z 1 f1 w1 f1 (a1 ) + w2 f1 (a2 ) + · · · + wn f1 (an ) = −1 1

w1 f2 (a1 ) + w2 f2 (a2 ) + · · · + wn f2 (an ) = .. .

.. .

w1 fn (a1 ) + w2 fn (a2 ) + · · · + wn fn (an ) =

Z

f2

−1

Z

1

fn .

−1

210

True or False In order to be able to use the a somewhat unusual way, as f1 (a1 ) f1 (a2 ) [ w1 w2 . . . wn ] ··· f1 (an ) |

matrix M introduced in Exercise 69, as the hint suggests, we write this system in f2 (a1 ) f2 (a2 ) ··· f2 (an ) {z M

· · · fn (a1 ) Z 1 · · · fn (a2 ) f1 = ··· ··· −1 · · · fn (an ) }

Z

1

f2

−1

···

Z

1

−1

fn

Since M is invertible (Exercise 69), we have the unique solution Z 1 Z 1 Z 1 [ w1 w2 . . . wn ] = fn M −1 . f1 f2 · · · −1

−1

−1

Now if f is any polynomial in Pn−1 and f =

n X

cj fj , then

j=1

Z

1

f=

−1

n X j=1

cj

Z

1

fj =

−1

n n n n n X X X X X cj wi cj fj (ai ) = wi fj (ai ) = wi f (ai ), j=1

i=1

i=1

i=1

j=1

as claimed. 4.3.72 If we work with the basis f1 (t) = 1, f2 (t) = t, and f3 (t) = t2 of P2 , then we have to solve the system w1 + w2 + w3 =2 −w1 + w2 = 0 (see Exercise 71). The solution is w1 = w3 = 13 and w2 = 34 , so that w1 + w2 = 23 1

4 1 1 f (−1) + f (0) + f (1) for f in P2 . This is what you get when you apply Simpson’s Rule (with two 3 3 3 −1 subintervals) to f ; note that Simpson’s Rule gives the exact value of the integral in this case. Z

f =

True or False Ch 4.TF.1 F; A basis of R2×3 is

0 0 0 0

1 0

0 0 0 , 0 0 0

1 0 0 0 , 0 0 0 0

1 0 0 , 0 1 0

0 0 , 0 0

0 0 , 1 0

0 , so it has a dimension of 6. 1

Ch 4.TF.2 T; check with Definition 4.1.3c. Ch 4.TF.3 T; The linear transformation T (ax + b) = a + ib is an isomorphism from P1 to C, with the inverse T −1 (a + ib) = ax + b. Ch 4.TF.4 T; by Theorem 4.2.4c. Ch 4.TF.5 T; This fits all properties of Definition 4.1.2. 211

Chapter 4

Ch 4.TF.6 F; The transformation T could be: T (f ) = dimension of the kernel would be 7.

0 0 , in which case the kernel would be all of P6 and the 0 0

Ch 4.TF.7 T; We are looking at P6 , with a basis 1, t, t2 , t3 , t4 , t5 , t6 , which has seven elements. Ch 4.TF.8 True, we can check both requirements of Definition 4.2.1. Ch 4.TF.9 T; check the three properties listed in Definition 4.1.2. Ch 4.TF.10 T; by Definition 4.2.1. Ch 4.TF.11 F; T (sin(x)) = sin(x) − sin(x) = 0. Ch 4.TF.12 F; T (f ) =

0 0 0 0

is not an isomorphism.

1 1 −1 . . Now im(A) = ker(A) = span Ch 4.TF.13 F; Let V =R , A = 1 1 −1 2

Ch 4.TF.14 T; the dimensions of both spaces are the same: 10. Ch 4.TF.15 F; dim(P3 )= 4, so the three given polynomials cannot span P3 . Ch 4.TF.16 T; We can construct a basis of V by omitting the redundant elements from a list of ten elements that span V . Thus dim(V ) ≤ 10. 0 1 0 + det Ch 4.TF.17 F; det 0 0 0

1 0 = 0 6= det 0 1

0 = 1. 1

Ch 4.TF.18 F; For any matrix A, the space of matrices commuting with A is at least two-dimensional. Indeed, if A is a scalar multiple of I2 , then A commutes with all 2 × 2 matrices, and if A fails to be a scalar multiple of I2 , then A commutes with the linearly independent matrices A and I2 . Ch 4.TF.19 F; t3 , t3 + t2 , t3 + t, t3 + 1 is a basis of P3 . Ch 4.TF.20 T; If T is linear and invertible, then T −1 will be linear and invertible as well. Ch 4.TF.21 F; T (f (t)) = f ′ (t) is not an isomorphism. Ch 4.TF.22 T; We need only show that either the new list contains no redundant elements, or spans the whole space. The latter is slightly easier to show. Since f1 , f2 , f3 form a basis of V , it suffices to show that these three elements are in the span of f1 , f1 + f2 , f1 + f2 + f3 . This is simple to demonstrate: f2 = (f1 + f2 ) − f1 , and f3 = (f1 + f2 + f3 ) − (f1 + f2 ). Ch 4.TF.23 T; We show that none of the polynomials is redundant; let’s call them f (x), g(x) and h(x). Now g(x) isn’t a multiple of f (x) since f (b) = 0, but g(b) 6= 0. Likewise, h(x) isn’t a linear combination of f (x) and g(x) since f (c) = g(c) = 0, but h(c) 6= 0. 212

True or False Ch 4.TF.24 T; Make the substitution 4t − 3 = s to see that the inverse is T −1 (g(s)) = g( s+3 4 ). Ch 4.TF.25 F; T

1 = (a + 2c) 3

a b 1 = c d 3 0 0 + (b + 2d) 0 0

a b a + 2c b + 2d = c d 3a + 6c 3b + 6d 0 1 0 1 and . So the image is the span of 0 3 0 3

2 6

1 , and rank(T )= 2. 3

Ch 4.TF.26 T; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . Thus f = f1 does the job. Ch 4.TF.27 T; If T (f (t)) = f (t2 ) = 0, f (t) must also be zero. Ch 4.TF.28 T; The inverse is T −1 (N ) = S −1 N S −1 .

1 1 1 1 a b a b 1 1 Ch 4.TF.29 T; Let A = . Then we want = , or 0 0 0 0 c d c d 0 0 0 1 0 a a a+c b+d and . Thus, c = 0 and a = b + d. So our space is the span of = 1 0 1 c c 0 0 Ch 4.TF.30 T; Let our basis be clearly none are redundant.

1 0

0 . −1

0 1 0 1 1 0 0 . Each matrix here is invertible, and also , , , −1 0 1 0 0 −1 1

Ch 4.TF.31 F; P2 is a subspace of P , and P is infinite dimensional.

a b Ch 4.TF.32 T; Let T d e g h are exactly as required.

c b f = 0 i

Ch 4.TF.33 F; The space spanned by

1 0

0 0

c . We can easily see that the kernel and image of this transformation f

and

0 1 0 0

contains no invertible matrices.

Ch 4.TF.34 F; This is the change of basis matrix from B to A. The change of basis matrix we are looking for is: 1 −1 . 0 1

1 2 1 Ch 4.TF.35 F; Let B = (f, g) and C = (g, f ). The fact that is the B- matrix of T implies that [T (f )]B = , 3 4 3 3 3 . This , meaning that the second column of the C-matrix of T is or T (f ) = f + 3g. But then [T (f )]C = 1 1 2 1 shows that the matrix fails to be the C-matrix of T . 4 3 Ch 4.TF.36 T; The image of T is Pn−1 , so that rank(T ) = dim (imT ) = dim(Pn−1 ) = n. 213

Chapter 4 Ch 4.TF.37 T; because the matrix is invertible. Ch 4.TF.38 T; The dimension of P9 is 10, and the dimension of R3×4 is 12. Thus, any 10-dimensional subspace of R3×4 will be acceptable. For example, we can consider the space of all 3 × 4 matrices A with a11 = a12 = 0. Ch 4.TF.39 T; let W1 be {~0}. Then any other subspace W2 unioned with W1 will simply be W2 again, which we know is a subspace. Ch 4.TF.40 T; Let T (a0 +a1 t+a2 t2 +· · ·+a5 t5 +· · ·) = a0 +a1 t+a2 t2 +· · ·+a5 t5 . The image of this transformation is clearly all of P5 , and T satisfies the requirements of Definition 4.2.1. Ch 4.TF.41 T; there will be no redundant elements in this list. Ch 4.TF.42 F; The kernel of T consists of all constant functions. Ch 4.TF.43 T; We apply the rank-nullity theorem: dim(W ) = dim(im(T )) = dim(P4 ) − dim(ker(T )) = 5 − dim(ker(T )) ≤ 5. Ch 4.TF.44 F; We can construct as many linearly independent elements in ker(T ) as we want, for example, the 1 , for all positive integers n. polynomials f (t) = tn − n+1 Ch 4.TF.45 T; 0 is in our set, and if f and g are in our set, then T (f + g) = T (f ) + T (g) = f + g so that f + g is in V as well. Also, if f is in V and k is an arbitrary scalar, then T (kf ) = kT (f ) = kf , so kf is in V as well. Ch 4.TF.46 F; Consider the transformation that drops the constant term: T (c0 + c1 t + · · · + c6 t6 ) = c1 t + · · · + c6 t6 . Ch 4.TF.47 T; Let P = I2 , Q = −I2 . Then T (M ) = I2 M − M (−I2 ) = 2M , which is an isomorphism. Ch 4.TF.48 F; We use dimension arithmetic here to show that this cannot happen. Any transformation T from P6 to C must have a kernel of at least 5 dimensions, since P6 is 7-dimensional and C is only a 2-dimensional space. Thus, any such kernel cannot be isomorphic to R2×2 , which is a 4-dimensional space. Ch 4.TF.49 F; If f = −f1 , then 0 is a member of the list! Ch 4.TF.50 T; Consider the space of all matrices of the form

a −b , for example. b a

Ch 4.TF.51 T; note that dim(P11 ) = 12 = dim(R3×4 ). The linear spaces P11 and R3×4 are both isomorphic to R12 , via the coordinate transformation, and thus they are isomorphic to each other. Ch 4.TF.52 F; Consider the linear transformation T (f (t)) = f (t) from P2 to P , for example. Ch 4.TF.53 T; We use the rank-nullity theorem: dim(V ) =dim(im(T ))+dim(ker(T ))=dim(im(T ))≤ dim(R2×2 ) = 4. Ch 4.TF.54 T; Using the fundamental theorem of calculus, we can write g(t) = T (f (t)) = 3f (3t + 4). Make the substitution 3t + 4 = s to see that the inverse is T −1 (g(s)) = g((s − 4)/3)/3. 214

True or False Ch 4.TF.55 T; Using a coordinate transformation, it suffices to show this for R4 . For every real number k, we define the three dimensional subspace Vk of R4 consisting of all vectors ~x such that x4 = kx3 . If c is different from k, 0 0 4 then Vc and Vk will be different subspaces of R , since Vk contains the vector , but Vc does not. Thus we 1 k have generated infinitely many distinct three-dimensional subspaces Vk of R4 , one for every real number k. Ch 4.TF.56 T; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . We are looking for a nonzero f = af1 + bf2 such that T (f ) = 4f . Now T (f ) = aT (f1 ) + bT (f2 ) = 3af1 + 5bf1 + 4bf2 = (3a + 5b)f1 + 4bf2 must be equal to 4f = 4af1 + 4bf2 . Thus it is required that 3a + 5b = 4a, or a = 5b. For example, f = 5f1 + f2 does the job. Ch 4.TF.57 T; This is logically equivalent to the following statement: If the domain of T is finite dimensional, then so is the image of T . Compare with Exercises 4.2.81a and 4.1.57. Ch 4.TF.58 F; If A is a scalar multiple of I2 , then all2 × 2 matrices commute with A, so that the space of a b commuting matrices is 4 - dimensional. If A = fails to be a scalar multiple of I2 , consider the equation c d a b x y x y a b = , which amounts to the system cy − bz = 0, bx + (d − a)y − bt = 0, cx + c d z t z t c d (d − a)z − ct = 0. If b 6= 0, then the first two equations are independent; if c 6= 0, then the first and the third equation are independent; and if a 6= d, then the second and the third equation are independent. Thus the rank of the system is at least two and the solution space is at most two-dimensional. (The solution space is in fact two -dimensional, since A and I2 are independent solutions.) Ch 4.TF.59 T; If A = 0, then we are done. If rank(A) = 1, then the image h of itheh linear i transformation T (M ) = AM 2×2 2×2 ~ ~ from R to R is two dimensional (if ~v is a basis of im(A), then ~v 0 , 0 ~v is a basis of im(T )). Since the three matrices AB = T (B), AC = T (C) , and AD = T (D) are all in im(T ), they must be linearly dependent.

Ch 4.TF.60 F; Consider two distinct three-dimensional subspaces W1 and W2 of P4 . Since the spaces W1 and W2 are distinct, neither of them is a subspace of the other, so that we can find a polynomial f1 that is in W1 but not in W2 as well as an f2 that is in W2 but not in W1 . Then f1 and f2 are both in the union of W1 and W2 , but f1 + f2 isn’t. Ch 4.TF.61 T; Pick the first redundant element fk in the list. Since the elements f1 , . . . , fk−1 are linearly independent, the representation of fk as a linear combination of the preceding elements will be unique. Ch 4.TF.62 F; T (I3 ) = P − P = 0, and T can never be an isomorphism. Ch 4.TF.63 T; Let W = span(f1 , f2 , f3 , f4 , f5 ) = span(f2 , f4 , f5 , f1 , f3 ). If we omit the two redundant elements from the first list, f1 , f2 , f3 , f4 , f5 , we end up with a basis of W with three elements, so that dim(W ) = 3. If we omit the redundant elements from the second list, f2 , f4 , f5 , f1 , f3 , we end up with a (possibly different) basis of W , but that basis must consist of 3 elements as well. Thus there must be two redundant elements in the second list. Ch 4.TF.64 F; The dimensions of the kernel and image would have to be equal, and both add up to the dimension of P6 , which is the odd number 7. 215

Chapter 4 Ch 4.TF.65 T; Consider the proof of the rank nullity theorem outlined in Exercise 4.2.81. In the proof, we use bases of ker(T ) and im(T ) to construct a basis of the domain. Ch 4.TF.66 F; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . We are looking for a nonzero f = af1 + bf2 such that T (f ) = 5f . Now T (f ) = aT (f1 ) + bT (f2 ) = 3af1 + 5bf1 + 4bf2 = (3a + 5b)f1 + 4bf2 must be equal to 5f = 5af1 + 5bf2 . Thus it is required that 3a + 5b = 5a and 4b = 5b, implying that a = b = 0. We are unable to find a nonzero f with the desired property. x y in W can be described by a z t single linear equation ax + by + cz + dt = 0 , where at least one of the coefficients is nonzero. Suppose x is the leading variable (meaning that a 6= 0), and y, z and t are the free variables. We can choose y = z = 1 and t = 0, ∗ 1 in W will be invertible. We represent x by a star, since its value does not affect and the resulting matrix 1 0 the invertibility. If y is the leading variable and the other three are the free variables, then we canconstruct the 1 0 1 ∗ . Finally, for in W . If z is the leading variable, we have the invertible matrix invertible matrix ∗ 1 0 1 0 1 the leading variable t we have . 1 ∗

Ch 4.TF.67 T; Consider a 3-dimensional subspace W of R2×2 . The matrices

216

Section 5.1

Chapter 5 Section 5.1 5.1.1 k~v k = 5.1.2 k~v k = 5.1.3 k~v k =

√

72 + 112 =

√

22 + 32 + 42 =

√

22 + 32 + 42 + 52 =

√

49 + 121 =

√ 170 ≈ 13.04

√ √ 4 + 9 + 16 = 29 ≈ 5.39 √ √ 4 + 9 + 16 + 25 = 54 ≈ 7.35

·~ v √ √7+11 √18 5.1.4 θ = arccos k~u~ukk~ v k = arccos 2 170 = arccos 340 ≈ 0.219 (radians) 2+6+12 ·~ v √ √ 5.1.5 θ = arccos k~u~ukk~ v k = arccos 14 29 ≈ 0.122 (radians) ·~ v 2−3+8−10 √ √ 5.1.6 θ = arccos k~u~ukk~ ≈ 1.700 (radians) v k = arccos 10 54

5.1.7 Use the fact that ~u · ~v = k~ukk~v k cos θ, so that the angle is acute if ~u · ~v > 0, and obtuse if ~u · ~v < 0. Since ~u · ~v = 10 − 12 = −2, the angle is obtuse. 5.1.8 Since ~u · ~v = 4 − 24 + 20 = 0, the two vectors enclose a right angle. 5.1.9 Since ~u · ~v = 3 − 4 + 5 − 3 = 1, the angle is acute (see Exercise 7). 5.1.10 ~u · ~v = 2 + 3k + 4 = 6 + 3k. The two vectors enclose a right angle if ~u · ~v = 6 + 3k = 0, that is, if k = −2. ·~ v √1 5.1.11 a θn = arccos k~u~ukk~ v k = arccos n

θ2 = arccos √12 =

π 4 (=

45◦ )

θ3 = arccos √13 ≈ 0.955 (radians) θ4 = arccos 12 =

π 3 (=

60◦ )

b Since y = arccos(x) is a continuous function, lim θn = arccos lim √1n = arccos(0) = π2 (= 90◦ ) n→∞

n→∞

5.1.12 k~v + wk ~ 2 = (~v + w) ~ · (~v + w) ~ (by hint) = k~v k2 + kwk ~ 2 + 2(~v · w) ~ (by definition of length) ≤ k~v k2 + kwk ~ 2 + 2k~v kkwk ~ (by Cauchy-Schwarz) = (k~v k + kwk) ~ 2 , so that k~v + wk ~ 2 ≤ (k~v k + kwk) ~ 2 Taking square roots of both sides, we find that k~v + wk ~ ≤ k~v k + kwk, ~ as claimed. 217

Chapter 5 kF~2 k = 20 cos θ2 . It is required that kF~2 + F~3 k = 16, so that 20 cos θ2 = 16, or θ = 2 arccos(0.8) ≈ 74◦ .

5.1.13 Figure 5.1 shows that kF~2 + F~3 k = 2 cos

θ 2

Figure 5.1: for Problem 5.1.13. 5.1.14 The horizontal components of F~1 and F~2 are −kF~1 k sin β and kF~2 k sin α, respectively (the horizontal component of F~3 is zero). Since the system is at rest, the horizontal components must add up to 0, so that −kF~1 k sin β + kF~2 k sin α = 0 or ~ kF~1 k sin β = kF~2 k sin α or kF1 k = sin α . ~2 k kF

To find

EA , EB

sin β

note that EA = ED tan α and EB = ED tan β so that

α and β are two distinct acute angles, it follows that

EA EB

6=

~1 k kF ~2 k , kF

EA EB

=

tan α tan β

=

sin α sin β

·

cos β cos α

=

~1 k cos β kF ~2 k cos α . kF

Since

so that Leonardo was mistaken.

5.1.15 The subspace consists of all vectors ~x in R4 such that 1 x1 x2 2 ~x · ~v = · = x1 + 2x2 + 3x3 + 4x4 = 0. 3 x3 4 x4 −2r −3s −4t −2 −3 −4 r 1 0 0 These are vectors of the form = r + s + t . s 0 1 0 t 0 0 1 The three vectors to the right form a basis.

5.1.16 You may be able to find the solutions by educated guessing. Here is the systematic approach: we first find all vectors ~x that are orthogonal to ~v1 , ~v2 , and ~v3 , then we identify the unit vectors among them. Finding the vectors ~x with ~x · ~v1 = ~x · ~v2 = ~x · ~v3 = 0 amounts to solving the system x1 + x2 + x3 + x4 = 0 x1 + x2 − x3 − x4 = 0 x1 − x2 + x3 − x4 = 0 we can omit all the coefficients 12 . t x1 x2 −t The solutions are of the form ~x = = . −t x3 t x4

218

Section 5.1 1 2

Since k~xk = 2|t|, we have a unit vector if t = 1 2 − 12 −1 2 1 2

and

− 21

or t = − 21 . Thus there are two possible choices for ~v4 :

1 2 1 2 − 21

.

5.1.17 The orthogonal complement W ⊥ of W consists of the vectors ~x in R4 such that 5 1 x1 x1 2 x x 2 6 2 · = 0 and · = 0. 7 3 x3 x3 8 x4 4 x4 x1 + 2x2 + 3x3 + 4x4 = 0 . Finding these vectors amounts to solving the system 5x1 + 6x2 + 7x3 + 8x4 = 0 The solutions are of the form s + 2t 1 2 x1 −2 −3 x2 −2s − 3t = = s + t . x3 s 1 0 x4 t 0 1 The two vectors to the right form a basis of W ⊥ . 5.1.18 a k~xk2 = 1 + k~xk =

√2 3

1 4

+

1 16

+

1 64

≈ 1.155.

+ ··· =

1 1− 41

=

4 3

use the formula for a geometric series, with a =

1 4

, so that

b If we let ~u = (1, 0, 0, . . .) and ~v = 1, 12 , 41 , · · · , then √

·~ v θ = arccos k~uu~kk~ v k = arccos

1

2 √ 3

= arccos

3 2

=

π 6 (=

30◦ ).

c ~x = 1, √12 , √13 , · · · , √1n , · · · does the job, since the harmonic series 1 + introductory calculus classes).

1 2

+

1 3

+ · · · diverges (a fact discussed in

√

d If we let ~v = (1, 0, 0, . . .), ~x = 1, 21 , 41 , · · · and ~u = projL~v = (~u · ~v )~u =

5.1.19 See Figure 5.2.

3 4

~ x k~ xk

=

1, 21 , 41 , · · · .

3 2

1, 12 , 41 , · · · then

5.1.20 On the line L spanned by ~x we want to find the vector m~x closest to ~y (that is, we want km~x − ~y k to be minimal). We want m~x − ~y to be perpendicular to L (that is, to ~x), which means that ~x · (m~x − ~y ) = 0 or ·~ y 4182.9 m(~x · ~x) − ~x · ~y = 0 or m = ~x~x·~ x ≈ 198.532 ≈ 0.106. Recall that the correlation coefficient r is r =

~ x·~ y k~ xkk~ yk ,

so that m =

219

k~ yk k~ xk r.

See Figure 5.3.

Chapter 5

Figure 5.2: for Problem 5.1.19.

Figure 5.3: for Problem 5.1.20. 5.1.21 Call the three given vectors ~v1 , ~v2 , and ~v3 . Since ~v2 is required to be a unit vector, we must have b = g = 0. Now ~v1 · ~v2 = d must be zero, so that d = 0. Likewise, ~v2 · ~v3 = e must be zero, so that e = 0.

√

Since ~v3 must be a unit vector, we have k~v3 k2 = c2 +

1 4

= 1, so that c = ±

3 2 .

√

Since we are asked to find just one solution, let us pick c = √

The condition ~v1 · ~v3 = 0 now implies that

3 2 a

3 2 .

√ + 12 f = 0, or f = − 3a.

Finally, it is required that k~v1 k2 = a2 + f 2 = a2 + 3a2 = 4a2 = 1, so that a = ± 12 . √

Let us pick a = 12 , so that f = −

3 2 .

220

Section 5.1 Summary:

√ 3 0 2 ~v1 = √0 , ~v2 = 1 , ~v3 = 0 1 0 − 3 1 2

2

2

There are other solutions; some components will have different signs.

5.1.22 Let W = {~x in Rn : ~x · ~vi = 0 for all i = 1, . . . , m}. We are asked to show that V ⊥ = W , that is, any ~x in V ⊥ is in W , and vice versa. If ~x is in V ⊥ , then ~x · ~v = 0 for all ~v in V ; in particular, x · ~vi = 0 for all i (since the ~vi are in V ), so that ~x is in W. Conversely, consider a vector ~x in W . To show that ~x is in V ⊥ , we have to verify that ~x · ~v = 0 for all ~v in V . Pick a particular ~v in V . Since the ~vi span V , we can write ~v = c1~v1 + · · · + cm~vm , for some scalars ci . Then ~x · ~v = c1 (~x · ~v1 ) + · · · + cm (~x · ~vm ) = 0, as claimed. 5.1.23 We will follow the hint. Let ~v be a vector in V . Then ~v · ~x = 0 for all ~x in V ⊥ . Since (V ⊥ )⊥ contains all vectors ~y such that ~y · ~x = 0, ~v is in (V ⊥ )⊥ . So V is a subspace of (V ⊥ )⊥ . Then, by Theorem 5.1.8c, dim (V ) + dim(V ⊥ ) = n and dim(V ⊥ ) + dim((V ⊥ )⊥ ) = n, so dim (V ) + dim(V ⊥ ) = dim(V ⊥ ) + dim((V ⊥ )⊥ ) and dim (V ) = dim((V ⊥ )⊥ ). Since V is a subspace of (V ⊥ )⊥ , it follows that V = (V ⊥ )⊥ , by Exercise 3.3.61. 5.1.24 Write T (~x) = projV (~x) for simplicity. To prove the linearity of T we will use the definition of a projection: T (~x) is in V , and ~x − T (~x) is in V ⊥ . To show that T (~x + ~y ) = T (~x) + T (~y ), note that T (~x) + T (~y ) is in V (since V is a subspace), and ~x + ~y − (T (~x) + T (~y )) = (~x − T (~x)) + (~y − T (~y )) is in V ⊥ (since V ⊥ is a subspace, by Theorem 5.1.8a). To show that T (k~x) = kT (~x), note that kT (~x) is in V (since V is a subspace), and k~x − kT (~x) = k(~x − T (~x)) is in V ⊥ (since V ⊥ is a subspace). 5.1.25 a kk~v k2 = (k~v ) · (k~v ) = k 2 (~v · ~v ) = k 2 k~v k2 Now take square roots of both sides; note that is negative). kk~v k = |k|k~v k, as claimed.

b k~uk = k~v1k ~v =

1 vk k~ v k k~

√ k 2 = |k|, the absolute value of k (think about the case when k

= 1, as claimed.

↑

by part a 5.1.26 The two given vectors spanning the subspace are orthogonal, but they are not unit vectors: both have length 7. To obtain an orthonormal basis ~u1 , ~u2 of the subspace, we divide by 7: 2 3 ~u1 = 17 3 , ~u2 = 17 −6 . 6 2 221

Chapter 5 49 Now we can use Theorem 5.1.5, with ~x = 49 : 49 19 3 2 projV ~x = (~u1 · ~x)~u1 + (~u2 · ~x)~u2 = 11 3 − −6 = 39 . 64 2 6

5.1.27 Since the two given vectors in the subspace are orthogonal, we have the orthonormal basis 2 −2 2 2 ~u1 = 31 , ~u2 = 13 . 1 0 0 1 Now we can use Theorem 5.1.5, with ~x = 9~e1 : projV ~x = (~u1 · ~x)~u1 + (~u2 · ~x)~u2 2 −2 8 2 2 0 = 2 − 2 = . 1 0 2 0 1 −2 5.1.28 Since the three given vectors in the subspace are orthogonal, we have the orthonormal basis 1 1 1 1 1 −1 ~u1 = 12 , ~u2 = 12 , ~u3 = 21 . 1 −1 −1 1 −1 1 3 1 Now we can use Theorem 5.1.5, with ~x = ~e1 : projV ~x = (~u1 · ~x)~u1 + (~u2 · ~x)~u2 + (~u3 · ~x)~u3 = 14 . −1 1

5.1.29 By the Pythagorean theorem (Theorem 5.1.9), k~xk2

= k7~u1 − 3~u2 + 2~u3 + ~u4 − ~u5 k2 = k7~u1 k2 + k3~u2 k2 + k2~u3 k2 + k~u4 k2 + k~u5 k2 . = 49 + 9 + 4 + 1 + 1 = 64, so that k~xk = 8.

5.1.30 Since ~y = projV ~x, the vector ~x − ~y is orthogonal to ~y , by definition of an orthogonal projection (see Theorem 5.1.4): (~x − ~y ) · ~y = 0 or ~x · ~y − k~y k2 = 0 or ~x · ~y = k~y k2 . See Figure 5.4.

Figure 5.4: for Problem 5.1.30.

222

Section 5.1 5.1.31 If V = span(~u1 , . . . , ~um ), then projV ~x = (~u1 · ~x)~u1 + · · · + (~um · ~x)~um , by Theorem 5.1.5, and kprojV ~xk2 = (~u1 ·~x)2 +· · ·+(~um ·~x)2 = p, by the Pythagorean theorem (Theorem 5.1.9). Therefore p ≤ k~xk2 , by Theorem 5.1.10. The two quantities are equal if (and only if) ~x is in V . 5.1.32 By Theorem 2.4.9a, the matrix G is invertible if (and only if) (~v1 · ~v1 )(~v2 · ~v2 ) − (~v1 · ~v2 )2 = k~v1 k2 k~v2 k2 − (~v1 · ~v2 )2 6= 0. The Cauchy-Schwarz inequality (Theorem 5.1.11) tells us that k~v1 k2 k~v2 k2 − (~v1 · ~v2 )2 ≥ 0; equality holds if (and only if) ~v1 and ~v2 are parallel (that is, linearly dependent). 1 x1 5.1.33 Let ~x = · · · be a vector in Rn whose components add up to 1, that is, x1 + · · · + xn = 1. Let ~y = · · · 1 xn (all n components are √ 1). The Cauchy-Schwarz inequality (Theorem 5.1.11) tells us that |~x · ~y | ≤ k~xkk~y k, or, |x1 + · · · + xn | ≤ k~xk n, or k~xk ≥ √1n . By Theorem 5.1.11, the equation k~xk = √1n holds if (and only if) the vectors ~x and ~y are parallel, that is, x1 = x2 = · · · = xn = n1 . Thus the vector of minimal length is 1 n ~x = · · · all components are n1 .

1 n

Figure 5.5 illustrates the case n = 2. x2

1 → = x X

1 2 1 2

1

x1

x1 + x2 = 1

Figure 5.5: for Problem 5.1.33.

1 5.1.34 Let ~x be a unit vector in Rn , that is, k~xk = 1. Let ~y = . . . (all n components are 1). The Cauchy-Schwarz 1 √ √ inequality (Theorem 5.1.11) tells us that |~x · ~y | ≤ k~xkk~y k, or, |x1 + . . . + xn | ≤ k~xk n = n. By Theorem 5.1.11, k √ the equation x1 +. . .+xn = n holds if ~x = k~y for positive k. Thus ~x must be a unit vector of the form ~x = . . . k 1 √ n for some positive k. It is required that nk 2 = 1, or, k = √1n . Thus ~x = . . . all components are √1n .

√1 n

Figure 5.6 illustrates the case n = 2. 223

Chapter 5 x2 1 √2 1 √2

→ x = X

x1

1

x1 + x2 =√ 2

Figure 5.6: for Problem 5.1.34. 1 x 5.1.35 Applying the Cauchy-Schwarz inequality to ~u = y and ~v = 2 gives |~u · ~v | ≤ k~ukk~v k, or |x + 2y + 3z| ≤ 3 z √ √ 14. The minimal value x + 2y + 3z = − 14 is attained when ~u = k~v for negative k. Thus ~u must be a unit − √114 k √2 vector of the form ~u = 2k , for negative k. It is required that 14k 2 = 1, or, k = − √114 . Thus ~u = − 14 . 3k − √3 14

a 0.2 5.1.36 Let ~x = b and ~y = 0.3 . It is required that ~x · ~y = 0.2a + 0.3b + 0.5c = 76. Our goal is to minimize c 0.5 2 2 2 quantity ~x · ~x = a + b + c . The Cauchy-Schwarz inequality (squared) tells us that (~x · ~y )2 ≤ k~xk2 k~y k2 , or 762 2 2 2 762 ≤ (a2 + b2 + c2 )(0.22 + 0.32 + 0.52 ) or a2+ b2 + c2 ≥ 0.38 . The quantity a + b + c is minimal when 0.2k a 762 a2 + b2 + c2 = 0.38 . This is the case when ~x = b = 0.3k for some positive constant k. It is required that 0.5k c 0.2a + 0.3b + 0.5c = (0.2)2 k + (0.3)2 k + (0.5)2 k = 0.38k = 76, so that k = 200. Thus a = 40, b = 60, c = 100: The student must study 40 hours for the first exam, 60 hours for the second, and 100 hours for the third. 5.1.37 Using Definition 2.2.2 as a guide, we find that ref V ~x = 2(projV~x) − ~x = 2(~u1 · ~x)~u1 + 2(~u2 · ~x)~u2 − ~x. 5.1.38 Since ~v1 and ~v2 are unit vectors, the condition ~v1 · ~v2 = k~v1 kk~v2 k cos(α) = cos(α) = 12 implies that ~v1 and ~v2 enclose an angle of 60◦ = π3 . The vectors ~v1 and ~v3 enclose an angle of 60◦ as well. In the case n = 2 there are two possible scenarios: either ~v2 = ~v3 , or ~v2 and ~v3 enclose an angle of 120◦ . Therefore, either ~v2 · ~v3 = 1 or ~v2 · ~v3 = cos(120◦ ) = − 21 . In the case n = 3, the vectors ~v2 and ~v3 could enclose any angle between 0◦ (if ~v2 = ~v3 ) and 120◦ , as illustrated in Figure 5.7. We have − 12 ≤ ~v2 · ~v3 ≤ 1. √ 3 cos θ 2 0 0 √ √ 3 For example, consider ~v1 = 0 , ~v2 = 23 , ~v3 = sin θ 2 1 1 1 2

2

Note that ~v2 · ~v3 = claimed.

3 4

sin θ +

1 4

could be anything between − 12 (when sin θ = −1) and 1 (when sin θ = 1), as 224

Section 5.1

Figure 5.7: for Problem 5.1.38. If n exceeds three, we can consider the orthogonal projection w ~ of ~v3 onto the plane E spanned by ~v1 and ~v2 . ~ = (~v1 · w)~ ~ v1 = 21 ~v1 , and since kwk Since proj~v1 w ~ ≤ k~v3 k = 1, (by Theorem 5.1.10), the tip of w ~ will be on the line segment in Figure 5.8. Note that the angle φ enclosed by the vectors ~v2 and w ~ is between 0◦ and 120◦ , so that cos φ is between − 12 and 1. Therefore, ~v2 · ~v3 = ~v2 · w ~ = kwk ~ cos φ is between − 12 and 1. This implies that ∠(~v2 , ~v3 ) is between 0◦ and 120◦ as well. To see that all these values are attained, add (n − 3) zeros to the three vectors ~v1 , ~v2 , ~v3 in R3 given above. v1

w v2

φ

Figure 5.8: for Problem 5.1.38.

5.1.39 No! By definition of a projection, the vector ~x − projL ~x is perpendicular to projL ~x, so that (~x − projL ~x) · (projL ~x) = ~x · projL ~x − kprojL ~xk2 = 0 and ~x · projL ~x = kprojL ~xk2 ≥ 0. (See Figure 5.9.)

Figure 5.9: for Problem 5.1.39.

5.1.40 ||~v2 || =

√ √ ~v2 · ~v2 = a22 = 3.

·~ v3 20 √ a23 √ 5.1.41 θ =arccos( ||~v~v22||||~ v3 || ) =arccos( a22 a33 ) =arccos( 21 ) ≈ 0.31 radians.

5.1.42 ||~v1 + ~v2 || =

p √ √ (~v1 + ~v2 ) · (~v1 + ~v2 ) = a11 + 2a12 + a22 = 22. 225

Chapter 5 ~ v2 u is an orthonormal basis 3 . Then, ~ ~ v2 ~ v2 1 ( 3 · ~v1 ) 3 = 3 (~v2 · ~v1 ) ~v32 = 31 (a12 ) ~v32 = 95 ~v2 .

5.1.43 Let ~u = (~u · ~v1 )~u =

~ v2 ||~ v2 ||

=

for span(~v2 ). Using Theorem 5.1.5, proj~v2 (~v1 ) =

5.1.44 One method to solve this is to take ~v = ~v2 − proj~v3 ~v2 = ~v2 −

20 v3 . 49 ~

5.1.45 Write the projection as a linear combination of ~v2 and ~v3 , c2~v2 + c3~v3 . Now you want ~v1 − c2~v2 − c3~v3 to be perpendicular to V , that is, perpendicular to both ~v2 and ~v3 . Using dot products, this boils down to two linear 25 1 equation in two unknowns, 9c2 + 20c3 = 5, and 20c2 + 49c3 = 11, with the solution c2 = 41 and c3 = − 41 . Thus 1 25 the answer is 41 ~v2 − 41 ~v3 . 5.1.46 Write the projection as a linear combination of ~v1 and ~v2 : c1~v1 + c2~v2 . Now we want ~v3 − c1~v1 + c2~v2 to be perpendicular to V , that is, perpendicular to both ~v1 and ~v2 . Using dot products, this boils down to two linear equations in two unknowns, 11 = 3c1 + 5c2 and 20 = 5c1 + 9c2 , with the solution c1 = − 12 , c2 = 52 . Thus, the answer is − 21 ~v1 + 52 ~v2 .

Section 5.2 In Exercises 1–14, we will refer to the given vectors as ~v1 , . . . , ~vm , where m = 1, 2, or 3.

2 5.2.1 ~u1 = k~v11 k ~v1 = 31 1 −2 6 5.2.2 ~u1 = k~v11 k ~v1 = 71 3 2 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

Note that ~u1 · ~v2 = 0. 4 5.2.3 ~u1 = k~v11 k ~v1 = 51 0 3 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

4 5.2.4 ~u1 = 51 0 and ~u2 = 3

2 = 17 −6 3

3 = 15 0 −4

3 1 0 as in Exercise 3. 5 −4

0 Since ~v3 is orthogonal to ~u1 and ~u2 , ~u3 = k~v13 k ~v3 = −1 . 0 226

Section 5.2 2 5.2.5 ~u1 = k~v11 k ~v1 = 31 2 1 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 = √118 −1 = 4

−1 1 √ −1 3 2 4

1 5.2.6 ~u1 = k~v11 k ~v1 = 0 = ~e1 0 ~u2 =

~ v2⊥ k~ v2⊥ k

~u3 =

~ v3⊥ k~ v3⊥ k

0 u1 ·~ v2 )~ u1 1 = ~e2 = = k~~vv22 −(~ −(~ u1 ·~ v2 )~ u1 k 0 0 u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 0 = ~e3 = k~~vv33 −(~ = −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k 1

2 5.2.7 Note that ~v1 and ~v2 are orthogonal, so that ~u1 = k~v11 k ~v1 = 13 2 and ~u2 = 1 1 2 ~ v⊥ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 √1 −4 = 1 −2 . ~u3 = k~v3⊥ k = k~~vv33 −(~ −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k = 3 36 3 4 2 5 1 4 1 5.2.8 ~u1 = k~v1 k ~v1 = 7 2 2 −2 2 ~u2 = k~v12 k ~v2 = 71 5 −4 1 1 1 1 5.2.9 ~u1 = k~v1 k ~v1 = 2 1 1 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 1 7 = 10 −7 1

1 1 1 1 5.2.10 ~u1 = k~v1 k ~v1 = 2 1 1 227

−2 1 v2 = 31 1 . Then k~ v2 k ~ 2

Chapter 5

~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

1 −1 = 12 1 −1

4 0 5.2.11 ~u1 = k~v11 k ~v1 = 51 0 3 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−3 1 2 = √225 = 14 4

−3 1 2 15 14 4

2 1 3 5.2.12 ~u1 = 7 0 6 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

0 −2 = 13 2 1

1 1 1 1 5.2.13 ~u1 = k~v1 k ~v1 = 2 1 1 ~u2 =

~u3 =

~ v2⊥ k~ v2⊥ k

~ v3⊥ k~ v3⊥ k

=

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

=

1 2 −1 2 1 −2 1 2

~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k

=

1 2 1 2 −1 2 − 12

1 7 1 5.2.14 ~u1 = k~v11 k ~v1 = 10 1 7 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 0 = √12 1 0

228

Section 5.2

~u3 =

~ v3⊥ k~ v3⊥ k

=

0 1 = √12 0 −1

~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k

In Exercises 15–28, we will use the results of Exercises 1–14 (note that Exercise k, where k = 1, . . . , 14, gives the QR factorization of the matrix in Exercise (k + 14)). We can set Q = [~u1 . . . ~um ]; the entries of R are r11 r22 r33 rij

= k~v1 k = k~v2⊥ k = k~v2 − (~u1 · ~v2 )~u1 k = k~v3⊥ k = k~v3 − (~u1 · ~v3 )~u1 − (~u2 · ~v3 )~u2 k = ~ui · ~vj , where i < j.

2 5.2.15 Q = 13 1 , R = [3] −2

2 7 0 −6 , R = 0 7 3

3 5 5 0,R = 0 35 −4

5 5 0 3 0 0 −5 , R = 0 35 0 0 0 2 −4 0

6 5.2.16 Q = 17 5 2 4 5.2.17 Q = 15 0 5 4 5.2.18 Q = 15 0 5

5.2.19 Q =

1 3

2

2 1

− √12

1 − √12 , R = 3 0

√1 2

√4 2

5.2.20 Q = I3 , R = [ ~v1

~v2

2 3 ~v3 ] = 0 4 0 0

2 5.2.21 Q = 31 2 1

−2 1 3 0 1 −2 , R = 0 3 2 2 0 0

5 4 5.2.22 Q = 71 2 2

−2 2 7 7 , R = 5 0 7 −4

5 6 7 12 −12 6

229

Chapter 5 0.5 0.5 5.2.23 Q = 0.5 0.5

−0.1 2 0.7 , R = 0 −0.7 0.1

1 1 1 5.2.24 Q = 2 1 1

1 −1 2 10 , R = 1 0 2 −1

4 10

12 −3 2 5 1 0 5.2.25 Q = 15 , R = 0 14 0 9 4

2 7 3 7

5.2.26 Q = 0 6 7

0

− 23 7 ,R= 2 0 3

1 1 5.2.27 Q = 21 1 1 5.2.28 Q = 5.2.29 ~u1 =

1 10 7 10 1 10 7 10

10 15

14 3

1 3

1 1 2 1 −1 1 , R = 0 1 −1 −1 0 0 1 −1

− √12 0

1 v1 k~ v1 k ~

0

10 , R = 0 0 0

√1 2

√1 2

− √12

0

=

1 5

−3 4

~u2 =

~ v2⊥ k~ v2⊥ k

=

1 −2 1 10 10 √ 2 √0 0 2 ~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

=

1 5

4 . (See Figure 5.10.) 3

Figure 5.10: for Problem 5.2.29.

230

Section 5.2 5.2.30 See Figure 5.11.

Figure 5.11: for Problem 5.2.30. 1 5.2.31 ~u1 = k~v11 k ~v1 = 0 = ~e1 0

0 b b ~v2⊥ = ~v2 − projV1 ~v2 = c − 0 = c , so that ~u2 = 0 0 0

Here V1 = span(~e1 ) = x axis. d d 0 ~v3⊥ = ~v3 − projV2 ~v3 = e − e = 0 , so that ~u3 = f 0 f Here V2 = span(~e1 , ~e2 ) = x-y plane. (See Figure 5.12.)

~ v2⊥ k~ v2⊥ k

~ v3⊥ k~ v3⊥ k

0 = 1 = ~e2 0 0 = 0 = ~e3 . 1

Figure 5.12: for Problem 5.2.31. −1 −1 5.2.32 A basis of the plane is ~v1 = 1 , ~v2 = 0 . 1 0

231

Chapter 5 Now apply the Gram-Schmidt process. ~u1

~u2

1 v1 k~ v1 k ~

=

=

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 = √12 1 0 −1 = √16 −1 2

Your solution may be different if you start with a different basis ~v1 , ~v2 of the plane.

1 0 0 5.2.33 rref(A) = 0 1 1

1 0

−1 0 0 −1 A basis of ker(A) is ~v1 = , ~v2 = . 0 1 1 0

−1 0 Since ~v1 and ~v2 are orthogonal already, we obtain ~u1 = √12 , ~u2 = 0 1

5.2.34 rref(A) =

1 0 −1 0 1 2

−2 3

0 −1 √1 . 2 1 0

1 2 −2 −3 A basis of ker(A) is ~v1 = , ~v2 = . 1 0 0 1

We apply the Gram-Schmidt process and obtain

~u1

~u2

1 5.2.35 rref(A) = 0 0

0 1 0

1 3 1 3

0

=

=

1 v1 k~ v1 k ~

~ v2⊥ k~ v2⊥ k

1 −2 = √16 1 0 2 −1 = √130 −4 3

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

The non-redundant columns of A give us a basis of im(A): 232

Section 5.2 2 1 ~v1 = 2 , ~v2 = 1 −2 2

1 Since ~v1 and ~v2 are orthogonal already, we obtain ~u1 = 13 2 , ~u2 = 2

1 1 1 1 −1 5.2.36 Write M = 2 1 −1 1 1

↑ Q0

1 2 3 −1 0 −4 1 0 0 −1

2 1 1 . 3 −2

5 6 7

↑ R0

This is almost the QR factorization of M : the matrix Q0 has orthonormal columns and R0 is upper triangular; the only problem is the entry −4 on the diagonal of R0 . Keeping in mind how matrices are multiplied, we can change all the signs in the second column of Q0 and in the second row of R0 to fix this problem: 1 −1 1 2 3 5 1 1 −1 M = 12 0 4 −6 1 1 −1 0 0 7 1 −1 1 ↑ Q

↑ R

1 1 1 −1 5.2.37 Write M = 21 1 −1 1 1

↑ Q0

1 1 3 −1 1 0 1 −1 0 −1 −1 0

↑ R0

4 5 0 0

Note that the last two columns of Q0 and the last two rows of R0 have no effect on the product Q0 R0 ; if we drop them, we have the QR factorization of M : 1 1 3 4 1 −1 M = 12 1 −1 0 5 1 1 ↑ Q

↑ R

5.2.38 Since ~v1 = 2~e3 , ~v2 = −3~e1 and ~v3 0 −1 ~v1 0 0 Q = k~v1 k k~~vv22 k k~~vv33 k = 1 0 0 0

= 4~e4 are orthogonal, we have 0 k~v1 k 0 0 2 0 0 k~v2 k 0 = 0 3 and R = 0 0 0 0 k~v3 k 0 0 1 233

0 0. 4

Chapter 5 1 5.2.39 ~u1 = √114 2 , ~u2 = 3

1 √1 1 , ~ u3 = ~u1 × ~u2 = 3 −1

5.2.40 If ~v1 , . . . , ~vn are the columns of A, then Q =

−5 √1 4 42 −1

~ v1 k~ v1 k

~ vn k~ vn k

···

(See Exercise 38 as an example.)

and R =

k~v1 k 0

0 ..

. k~vn k

.

5.2.41 If all diagonal entries of A are positive, then we have Q = In and R = A. A small modification is necessary if A has negative entries on the diagonal: if aii < 0 we let rij = −aij for all j, and we let qii = −1; if aii > 0 we let rij = aij and qii = 1. Furthermore, qij = 0 if i 6= j (that is, Q is diagonal). −1 2 3 −1 0 0 1 −2 −3 For example, 0 4 5 = 0 1 0 0 4 5 0 0 −6 0 0 −1 0 0 6 ↑ A

↑ Q

↑ R

5.2.42 We have r11 = k~v1 k and r22 = k~v2⊥ k = k~v2 − projL~v2 k, so that r11 r22 is the area of the parallelogram defined by ~v1 and ~v2 . See Figure 5.13.

Figure 5.13: for Problem 5.2.42. 5.2.43 Partition the matrices Q and R in the QR factorization of A as follows: R1 R2 [ A1 A2 ] = A = QR = [ Q1 Q2 ] = [ Q1 R1 Q1 R2 + Q2 R3 ] , 0 R3 where Q1 is n × m1 , Q2 is n × m2 , R1 is m1 × m1 , and R3 is m2 × m2 . Then, A1 = Q1 R1 is the QR factorization of A1 : note that the columns of A1 are orthonormal, and R1 is upper triangular with positive diagonal entries. 5.2.44 No! If m exceeds n, then there is no n × m matrix Q with orthonormal columns (if the columns of a matrix are orthonormal, then they are linearly independent). 5.2.45 Yes. Let A = [ ~v1 with ~um = k~v1m k ~vm .

· · · ~vm ]. The idea is to perform the Gram-Schmidt process in reversed order, starting

Then we can express ~vj as a linear combination of ~uj , . . . , ~um , so that [ ~v1 for some lower triangular matrix L, with 234

· · · ~vj

· · · ~vm ] = [ ~u1

· · · ~uj

· · · ~um ] L

Section 5.3

~vj = [ ~u1

· · · ~uj

l1j ··· · · · ~um ] ljj = ljj ~uj + · · · + lmj ~um . ··· lmj

Section 5.3 5.3.1 Not orthogonal, the column vectors fail to be perpendicular to each other. 5.3.2 This matrix is orthogonal. Check that the column vectors are unit vectors, and that they are perpendicular to each other. 5.3.3 This matrix is orthogonal. Check that the column vectors are unit vectors, and that they are perpendicular to each other. 5.3.4 Not orthogonal, the first and third column vectors fail to be perpendicular to each other. 5.3.5 3A will not be orthogonal, because the length of the column vectors will be 3 instead of 1, and they will fail to be unit vectors. 5.3.6 −B will certainly be orthogonal, since the columns will be perpendicular unit vectors. 5.3.7 AB is orthogonal by Theorem 5.3.4a. 5.3.8 A + B will not necessarily be orthogonal, because the columns may not be unit vectors. For example, if A = B = In , then A + B = 2In , which is not orthogonal. 5.3.9 B −1 is orthogonal by Theorem 5.3.4b. 5.3.10 This matrix will be orthogonal, by Theorem 5.3.4. 5.3.11 AT is orthogonal. AT = A−1 , by Theorem 5.3.7, and A−1 is orthogonal by Theorem 5.3.4b. 5.3.13 3A is symmetric, since (3A)T = 3AT = 3A. 5.3.14 −B is symmetric, since (−B)T = −B T = −B. 5.3.15 AB is not necessarily symmetric, since (AB)T = B T AT = BA, which is not necessarily the same as AB. (Here we used Theorem 5.3.9a.) 5.3.16 A + B is symmetric, since (A + B)T = AT + B T = A + B. 5.3.17 B −1 is symmetric, because (B −1 )T = (B T )−1 = B −1 . In the first step we have used 5.3.9b. 5.3.18 A10 is symmetric, since (A10 )T = (AT )10 = A10 . 235

Chapter 5 5.3.19 This matrix is symmetric. First note that (A2 )T = (AT )2 = A2 for a symmetric matrix A. Now we can use the linearity of the transpose, (2In + 3A − 4A2 )T = 2InT + 3AT − (4A2 )T = 2In + 3A − 4(AT )2 = 2In + 3A − 4A2 . 5.3.20 AB 2 A is symmetric, since (AB 2 A)T = (ABBA)T = (BA)T (AB)T = AT B T B T AT = AB 2 A. 5.3.21 Symmetric. (AT A)T = AT (AT )T = AT A. 5.3.22 BB T is symmetric: (BB T )T = (B T )T B T = BB T . 5.3.23 Not necessarily symmetric. (A − AT )T = AT − A = −(A − AT ). 5.3.24 Not necessarily symmetric. (AT BA)T = AT (AT B)T = AT B T A. 5.3.25 Symmetric, because (AT B T BA)T = AT B T (B T )T (AT )T = AT B T BA. 5.3.26 Symmetric, since (B(A + AT )B T )T = ((A + AT )B T )T B T = B(A + AT )T B T = B(AT + A)T B T = B((AT )T + AT )B T = B(A + AT )B T . 5.3.27 Using Theorems 5.3.6 and 5.3.9a, we find that (A~v ) · w ~ = (A~v )T w ~ = ~v T AT w ~ = ~v · (AT w), ~ as claimed. 5.3.28 Write L(~x) = A~x; by Definition 5.3.1, A is an orthogonal n × n matrix, so that AT A = In , by Theorem 5.3.7. Now L(~v ) · L(w) ~ = (A~v ) · (Aw) ~ = (A~v )T Aw ~ = ~v T AT Aw ~ = ~v T In w ~ = ~v T w ~ = ~v · w, ~ as claimed. Note that we have used Theorems 5.3.6 and 5.3.9a. 5.3.29 We will use the fact that L preserves length (by Definition 5.3.1) and the dot product (by Exercise 28): L(~ v )·L(w) ~ ·w ~ = arccos k~v~vkk v , w). ~ ∠(L(~v ), L(w)) ~ = arccos kL(~ v )kkL(w)k ~ wk ~ = ∠(~

5.3.30 If L(~x) = ~0, then kL(~x)k = k~xk = 0, so that ~x = ~0. Therefore, ker(L) = {~0}. By Theorem 3.3.7, dim(im(L)) = m − dim(ker(L)) = m. Since Rn has an m-dimensional subspace (namely, im(L)), the inequality m ≤ n holds. The transformation L preserves right angles (the proof of Theorem 5.3.2 applies), so that the columns of A are orthonormal (since they are L(~e1 ), . . . , L(~em )). Therefore, we have AT A = Im (the proof of Theorem 5.3.7 applies). Since the vectors ~v1 , . . . , ~vm form an orthonormal basis projection onto im(A), by Theorem 5.3.10. 1 A simple example of such a transformation is L(~x) = 0 0

of im(A), the matrix AAT represents the orthogonal 0 x1 x 1 ~x, that is, L 1 = x2 . x2 0 0

5.3.31 Yes! If A is orthogonal, then so is AT , by Exercise 11. Since the columns of AT are orthogonal, so are the rows of A. 236

Section 5.3 1 0 5.3.32 a No! As a counterexample, consider A = 0 1 (see Exercise 30). 0 0

b Yes! More generally, if A and B are n × n matrices such that BA = In , then AB = In , by Theorem 2.4.8c. cos(φ) , for some φ. Then ~v2 will be one 5.3.33 Write A = [ ~v1 ~v2 ]. The unit vector ~v1 can be expressed as ~v = sin(φ) − sin(φ) sin(φ) of the two unit vectors orthogonal to ~v1 : ~v2 = or ~v2 = . (See Figure 5.7.) cos(φ) − cos(φ)

cos(φ) − sin(φ) cos(φ) sin(φ) Therefore, an orthogonal 2 × 2 matrix is either of the form A = or A = , sin(φ) cos(φ) sin(φ) − cos(φ) representing a rotation or a reflection. Compare with Exercise 2.2.24. See Figure 5.14.

Figure 5.14: for Problem 5.3.33.

a b 5.3.34 Since the first two columns are orthogonal to the third, we have c = d = 0. Then is an e f cos(φ) − sin(φ) 0 gonal 2 × 2 matrix; By Exercise 33, the 3 × 3 matrix A is either of the form A = 0 sin(φ) cos(φ) cos(φ) sin(φ) 0 0 1 . A= 0 sin(φ) − cos(φ) 0 5.3.35 Let us first think about the inverse L = T −1 of T . 2 3

Write L(~x) = A~x = [ ~v1

~v2

~v3 ] ~x. It is required that L(~e3 ) = ~v3 = 23 . 1 3

237

ortho 0 1 or 0

Chapter 5

Furthermore, the vectors ~v1 , ~v2 , ~v3 must form an orthonormal basis of R3 . By inspection, we find ~v1 =

− 13

− 23 1 3 2 3

.

−1 2 2 2 ~x. −2 1

−2 Then ~v2 = ~v1 × ~v3 = 23 does the job. In summary, we have L(~x) = 13 1 2 − 23

Since the matrix of L is orthogonal, the matrix of T = L−1 is the transpose of the matrix of L: −2 1 2 T (~x) = 31 −1 2 −2 ~x. 2 2 1 There are many other answers (since there are many choices for the vector ~v1 above). 2

5.3.36 Let the third column be the cross product of the first two: A =

3 2 3 1 3

√1 2 − √12

0

√1 18 √1 18 − √418

There is another solution, with the signs in the last column reversed.

.

2 3 −3 2 5.3.37 No, since the vectors 3 and 2 are orthogonal, whereas 0 and −3 are not (see Theorem 5.3.2). 0 2 0 0

0 5.3.38 a The general form of a skew-symmetric 3 × 3 matrix is A = −a −b 2 −a − b2 −bc ac −a2 − c2 −ab , a symmetric matrix. A2 = −bc ac −ab −b2 − c2

a b 0 c , with −c 0

b By Theorem 5.3.9.a, (A2 )T = (AT )2 = (−A)2 = A2 , so that A2 is symmetric. 5.3.39 By Theorem 5.3.10, the matrix of the projection is ~u~uT ; the ij th entry of this matrix is ui uj . 0.5 −0.1 0.5 0.7 5.3.40 An orthonormal basis of W is ~u1 = , ~u2 = (see Exercise 5.2.9). 0.5 −0.7 0.5 0.1

By Theorem 5.3.10, the matrix of the projection onto W is QQT , where Q = [ ~u1 26 18 1 QQT = 100 32 24

18 74 −24 32

32 −24 74 18

24 32 18 26 238

~u2 ].

Section 5.3 1 . 1 5.3.41 A unit vector on the line is ~u = √n .. . 1

The matrix of the orthogonal projection is ~u~uT , the n × n matrix whose entries are all cise 39).

1 n

(compare with Exer-

5.3.42 a Suppose we are projecting onto a subspace W of Rn . Since A~x is in W already, the orthogonal projection of A~x onto W is just A~x itself: A(A~x) = A~x, or A2 ~x = A~x. Since this equation holds for all ~x, we have A2 = A. b A = QQT , for some matrix Q with orthonormal columns ~u1 , . . . , ~um . Note that QT Q = Im , since the ij th entry of QT Q is ~ui · ~uj . Then A2 = QQT QQT = Q(QT Q)QT = QIm QT = QQT = A. 5.3.43 Examine how A acts on ~u, and on a vector ~v orthogonal to ~u: A~u = (2~u~uT − I3 )~u = 2~u~uT ~u − ~u = ~u, since ~uT ~u = ~u · ~u = k~uk2 = 1. A~v = (2~u~uT − I3 )~v = 2~u~uT ~v − ~v = −~v , since ~uT ~v = ~u · ~v = 0. Since A leaves the vectors in L = span(~u) unchanged and reverses the vectors in V = L⊥ , it represents the reflection about L. Note that B = −A, so that B reverses the vectors in L and leaves the vectors in V unchanged; that is, B represents the reflection about V . 5.3.44 Note that AT is an m × n matrix. By Theorems 3.3.7 and 5.3.9c we have dim(ker(AT )) = n − rank(AT ) = n − rank(A). By Theorem 3.3.6, dim(im(A)) = rank(A), so that dim(im(A)) + dim(ker(AT )) = n. 5.3.45 Note that AT is an m × n matrix. By Theorems 3.3.7 and 5.3.9c, we have dim(ker(A)) = m − rank(A) and dim(ker(AT )) = n − rank(AT ) = n − rank(A), so that dim(ker(A)) = dim(ker(AT )) if (and only if) A is a square matrix. 5.3.46 By Theorem 5.2.2, the columns ~u1 , . . . , ~um of Q are orthonormal. Therefore, QT Q = Im , since the ij th entry of QT Q is ~ui · ~uj . If we multiply the equation M = QR by QT from the left then QT M = QT QR = R, as claimed. 5.3.47 By Theorem 5.2.2, the columns ~u1 , . . . , ~um of Q are orthonormal. Therefore, QT Q = Im , since the ij th entry of QT Q is ~ui · ~uj . By Theorem 5.3.9a, we now have AT A = (QR)T QR = RT QT QR = RT R. 5.3.48 As suggested, we consider the QR factorization AT = P R 239

Chapter 5 of AT , where P is orthogonal and R is upper triangular with positive diagonal entries. By Theorem 5.3.9a, A = (P R)T = RT P T . Note that L = RT is lower triangular and Q = P T is orthogonal. 5.3.49 Yes! By Exercise 5.2.45, we can write AT = P L, where P is orthogonal and L is lower triangular. By Theorem 5.3.9a, A = (P L)T = LT P T . Note that R = LT is upper triangular, and Q = P T is orthogonal (by Exercise 11). 5.3.50 a If an n×n matrix A is orthogonal and upper triangular, then A−1 is both lower triangular (since A−1 = AT ) and upper triangular (being the inverse of an upper triangular matrix; compare with Exercise 2.4.35c). Therefore, A−1 = AT is a diagonal matrix, and so is A itself. Since A is orthogonal with positive diagonal entries, all the diagonal entries must be 1, so that A = In . −1 b Using the terminology suggested in the hint, we observe that Q−1 2 Q1 is orthogonal (by Theorem 5.3.4) and R2 R1 −1 −1 is upper triangular with positive diagonal entries. By part a, the matrix Q2 Q1 = R2 R1 is In , so that Q1 = Q2 and R1 = R2 , as claimed.

5.3.51 a Using the terminology suggested in the hint, we observe that Im = QT1 Q1 = (Q2 S)T Q2 S = S T QT2 Q2 S = S T S, so that S is orthogonal, by Theorem 5.3.7. b Using the terminology suggested in the hint, we observe that R2 R1−1 is both orthogonal (let S = R2 R1−1 in part a) and upper triangular, with positive diagonal entries. By Exercise 50a, we have R2 R1−1 = Im , so that R1 = R2 . Then Q1 = Q2 R2 R1−1 = Q2 , as claimed.

5.3.52 Applying the strategy outlined in Summary 4.1.6 to the

1 0 0

0 0 0 0 0,1 0 0 0

1 0 0 0 0 0,0 0 0 0 1 0

5.3.53 Applying the strategy

0 1 0 0 basis −1 0 0 , 0 0 0 0 −1

1 0 0 0,0 1 0 0 0

0 0 0 · 0 0 0

0 0 1

a b c general element b d e of V , we find the basis c e f 0 0 0 0 1 , 0 0 0 , so that dim(V ) = 6. 0 0 0 1

0 outlined in Summary 4.1.6 to the general element −b −c 0 1 0 0 0 0 0,0 0 1 , so that dim(V ) = 3. 0 0 0 −1 0

b c 0 e of V , we find the −e 0

5.3.54 To write the general form of a skew-symmetric n × n matrix A, we can place arbitrary constants above the diagonal, the opposite entries below the diagonal (aij = −aji ), and zeros on the diagonal (since aii = −aii ). See Exercise 53 for the case n = 3. Thus the dimension of the space equals the number of entries above the diagonal of an n × n matrix. In Exercise 55 we will see that there are (n2 − n)/2 such entries. Thus dim(V ) = (n2 − n)/2. 5.3.55 To write the general form of a symmetric n × n matrix A, we can place arbitrary constants on and above the diagonal, and then write the corresponding entries below the diagonal (aij = aji ). See Exercise 52 for the 240

Section 5.3 case n = 3. Thus the dimension of the space equals the number of entries on and above the diagonal of an n × n matrix. Now there are n2 entries in the matrix, n2 − n off the diagonal, and half of them, (n2 − n)/2, above the diagonal. Since there are n entries on the diagonal, we have dim(V ) = (n2 − n)/2 + n = (n2 + n)/2. 5.3.56 Yes and yes (see Exercise 57). 5.3.57 Yes, L is linear, since L(A + B) = (A + B)T = AT + B T = L(A) + L(B) and L(kA) = (kA)T = kAT = kL(A). Yes, L is an isomorphism; the inverse is the transformation R(A) = AT from Rn×m to Rm×n . 5.3.58 Adapting the solution of Exercise 59, we see that the kernel consists of all skew-symmetric matrices, and the image consists of all symmetric matrices. 5.3.59 The kernel consists of all matrixes A such that L(A) = symmetric matrices.

1 2 (A

− AT ) = 0, that is, AT = A; those are the

Following the hint, let’s apply L to a skew-symmetric matrix A, with AT = −A. Then L(A) = (1/2)(A − AT ) = (1/2)2A = A, so that A is in the image of L. Conversely, if A is any 2 × 2 matrix, then L(A) will be skewsymmetric, since (L(A))T = (1/2)(A − AT )T = (1/2)(AT − A) = −L(A). In conclusion: The kernel of L consists of all symmetric matrices, and the image consists of all skew-symmetric matrices. 1 0 5.3.60 Using Theorem 4.3.2, we find the matrix 0 0

0 1 0 0

0 0 1 0

0 0 . 0 −1

5.3.61 Note that the first three matrices of the given basis B are symmetric, so that L(A) = A − AT = 0, and the coordinate vector [L(A)]B is ~0 for all three of them. The last matrix of the basis is skew-symmetric, so that 0 0 0 0 0 0 0 0 L(A) = 2A, and [L(A)]B = 2~e4 . Using Theorem 4.3.2, we find that the B-matrix of L is . 0 0 0 0 0 0 0 2

1 0 5.3.62 By Theorem 5.3.9a, AT = 1 1 −1 2

0 1 0 0 0 −1 1 0 0

0 1 3 2 0 0 1 0 . 2 0 0 1

5.3.63 By Exercise 2.4.94b, the given LDU factorization of A is unique. By Theorem 5.3.9a, A = AT = (LDU )T = U T DT LT = U T DLT is another way to write the LDU factorization of A (since U T is lower triangular and LT is upper triangular). By the uniqueness of the LDU factorization, we have U = LT (and L = U T ), as claimed. A −B T C −DT A + C −B T − DT 5.3.64 a + = AT D CT B+D AT + C T B T A + C −(B + D) = is of the required form. B+D (A + C)T

A b k B

−B T AT

kA = kB

−kB T kAT

kA = kB

−(kB)T (kA)T

is of the required form. 241

Chapter 5 p −q −r −s p s −r q c The general element of H is M = , with four arbitrary constants, r,s,p, and q. Thus r −s p q s r −q p dim(H) = 4; use the strategy outlined in Summary 4.1.6 to construct a basis.

d

A −B T C −DT B AT D CT is of the required form.

=

AC − B T D BC + AT D

−ADT − B T C T −BDT + AT C T

=

AC − B T D BC + AT D

−(BC + AT D)T (AC − B T D)T

Note that A, B, C, D, and their transposes are rotation-dilation matrices, so that they all commute. e

A B

−B T AT

T

=

AT −B

BT (AT )T

is of the required form.

p f Note that the columns ~v1 , ~v2 , ~v3 , ~v4 or M are orthogonal, and they all have length p2 + q 2 + r2 + s2 . Now M T M is the 4 × 4 matrix whose ij th entry is ~vi · ~vj , so that M T M = (p2 + q 2 + r2 + s2 )I4 . g If M 6= 0, then k = p2 + q 2 + r2 + s2 > 0, and M −1 =

1 T p2 +q 2 +r 2 +s2 M .

1 T kM

M = I4 , so that M is invertible, with

By parts b and e, M −1 is in H as well. 0 −1 0 1 h No! A = 0 0 0 0

0 0 0 −1

0 0 0 0 0 0 and B = 1 1 0 0 0 1

−1 0 0 −1 do not commute (AB = −BA). 0 0 0 0

a b a b 10 must be an orthogonal ; it is required that a, b, c and d be integers. Now A = 10 c d c d 10 10 a 2 c 2 matrix, implying that ( 10 ) + ( 10 ) = 1, or a2 + c2 = 100. Checking the squares of all integers from 1 to 9, we see that there are only two ways to write 100 as a sum of two positive perfect squares: 100 = 36 + 64 = 64 + 36. Since a and c are required to be positive, we have either a = 6 and c = 8 or a = 8 and c = 6. In each case we have two options for the second column of A, namely, the two unit vectors perpendicular to the first column vector. Thus we end up with four solutions: .6 −.8 .6 .8 .8 −.6 .8 .6 A= , , or . .8 .6 .8 −.6 .6 .8 .6 −.8

5.3.65 Write 10A =

5.3.66 One approach is to take one of the solutions from Exercise 65, say, the rotation matrix B = 0.28 −0.96 and then let A = B 2 = . Matrix A is orthogonal by Theorems 5.3.4a. 0.96 0.28

0.8 0.6

−0.6 , 0.8

5.3.67 a We need to show that AT A~c = AT ~x, or, equivalently, that AT (~x − A~c) = ~0. But AT (~x − A~c) = AT (~x − c1~v1 − · · · − cm~vm ) is the vector whose ith component is (~vi )T (~x − c1~v1 − · · · − cm~vm ) = ~vi · (~x − c1~v1 − · · · − cm~vm ), which we know to be zero. 242

Section 5.4 b The system AT A~c = AT ~x has a unique solution ~c for a given ~x, since ~c is the coordinate vector of projV ~x with respect to the basis ~v1 , . . . , ~vm . Thus the coefficient matrix AT A must be invertible, so that we can solve for ~c and write ~c = (AT A)−1 AT ~x. Then projV ~x = c1~v1 + · · · + cm~vm = A~c = A(AT A)−1 AT ~x. 5.3.68 If A = QR, then A(AT A)−1 AT = QR(RT QT QR)−1 RT QT = QR(RT R)−1 RT QT = QRR−1 (RT )−1 RT QT = QQT , as in Theorem 5.3.10. The equation QT Q = Im holds since the columns of Q are orthonormal.

Section 5.4 T

5.4.1 A basis of ker(A ) is

−3 . (See Figure 5.15.) 2

Figure 5.15: for Problem 5.4.1.

1 5.4.2 A basis of ker(AT ) is −2 . im(A) is the plane perpendicular to this line. 1 5.4.3 We will first show that the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q span Rn . Any vector ~v in Rn can be written as ⊥ ~v = ~vk + ~v⊥ , where ~vk is in V and ~v⊥ is in V (by definition of orthogonal projection, Theorem 5.1.4). Now ~vk is a linear combination of ~v1 , . . . , ~vp , and ~v⊥ is a linear combination of w ~ 1, . . . , w ~ q , showing that the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q span Rn . Note that p + q = n, by Theorem 5.1.8c; therefore, the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q form a basis of Rn , by Theorem 3.3.4d. 5.4.4 By Theorem 5.4.1, the equation (im B)⊥ = ker(B T ) holds for any matrix B. Now let B = AT . Then (im(AT ))⊥ = ker(A). Taking transposes of both sides and using Theorem 5.1.8d we obtain im(AT ) = (kerA)⊥ , as claimed. 243

Chapter 5

5.4.5 V = ker(A), where A =

1 1

1 1 1 . 2 5 4

Then V ⊥ = (kerA)⊥ = im(AT ), by Exercise 4. The two columns of AT form a basis of V ⊥ : 1 1 1 2 , 5 1 4 1 5.4.6 Yes! For any matrix A, im(A)

= (ker(AT ))⊥ ↑

Theorem 5.4.1

= ↑

(ker(AAT ))⊥

=

(ker(AAT )T )⊥

Theorem 5.4.2a

= ↑

im(AAT ).

Theorems 5.4.1 and 5.1.8d.

5.4.7 im(A) and ker(A) are orthogonal complements by Theorem 5.4.1: (imA)⊥ = ker(AT ) = ker(A) 5.4.8 a By Theorem 5.4.6, L+ (~y ) = (AT A)−1 AT ~y . The transformation L+ is linear since it is “given by a matrix,” by Definition 2.1.1. b If L (and therefore A) is invertible, then L+ (~y ) = A−1 (AT )−1 AT ~y = A−1 ~y = L−1 ~y , so that L+ = L−1 . c L+ (L(~x)) =

the unique least-squares solution ~u of L(~u) = L(~x) = ~x.

d L(L+ (~y )) = A(AT A)−1 AT ~y = projV ~y , where V = im(A), by Theorem 5.4.7.

1 e Here A = 0 0

0 1 0 1 . Then L+ (~y ) = (AT A)−1 AT ~y = 0 1 0

0 ~y . 0

5.4.9 ~x0 is the shortest of all the vectors in S. (See Figure 5.16.) 5.4.10 a If ~x is an arbitrary solution of the system A~x = ~b, let ~xh = projV ~x, where V = ker(A), and ~x0 = ~x − projV ~x. Note that ~b = A~x = A(~xh + ~x0 ) = A~xh + A~x0 = A~x0 , since ~xh is in ker(A). b If ~x0 and ~x1 are two solutions of the system A~x = ~b, both from (kerA)⊥ , then ~x1 − ~x0 is in the subspace (kerA)⊥ as well. Also, A(~x1 − ~x0 ) = A~x1 − A~x0 = ~b − ~b = ~0, so that ~x1 − ~x0 is in ker(A). By Theorem 5.1.8b, it follows that ~x1 − ~x0 = ~0, or ~x1 = ~x0 , as claimed. c Write ~x1 = ~xh + ~x0 as in part a; note that ~xh is orthogonal to ~x0 . The claim now follows from the Pythagorean Theorem (Theorem 5.1.9). 5.4.11 a Note that L+ (~y ) = AT (AAT )−1 ~y ; indeed, this vector is in im(AT ) = (kerA)⊥ , and it is a solution of L(~x) = A~x = ~y . 244

Section 5.4

Figure 5.16: for Problem 5.4.9. b L(L+ (~y )) = ~y , by definition of L+ . c L+ (L(~x)) = AT (AAT )−1 A~x = projV ~x, where V = im(AT ) = (kerA)⊥ , by Theorem 5.4.7. d im(L+ ) = im(AT ), by part c, and ker(L+ ) = {~0} (if ~y is in ker(L+ ), then ~y = L(L+ (~y )) = L(~0) = ~0, by part b). e Let A =

1 0

1 0 0 0 ; then the matrix of L+ is AT (AAT )−1 = AT = 0 1 . 1 0 0 0

5.4.12 By Theorem 5.4.5, the least-squares solutions of the linear system A~x = ~b are the exact solutions of the (consistent) system AT A~x = AT ~b. The minimal solution of this normal equation (in the sense of Exercise 10) is called the minimal least-squares solution of the system A~x = ~b. Equivalently, the minimal least-squares solution of A~x = ~b can be defined as the minimal solution of the consistent system A~x = projV ~b, where V = im(A). 5.4.13 a Suppose that L+ (~y1 ) = ~x1 and L+ (~y2 ) = ~x2 ; this means that ~x1 and ~x2 are both in (kerA)⊥ = im(AT ), AT A~x1 = AT ~y1 , and AT A~x2 = AT ~y2 . Then ~x1 + ~x2 is in im(AT ) as well, and AT A(~x1 + ~x2 ) = AT (~y1 + ~y2 ), so that L+ (~y1 + ~y2 ) = ~x1 + ~x2 . The verification of the property L+ (k~y ) = kL+ (~y ) is analogous. b L+ (L(~x)) is the orthogonal projection of ~x onto (kerA)⊥ = im(AT ). c L(L+ (~y )) is the orthogonal projection of ~y onto im(A) = (kerAT ))⊥ . d im(L+ ) = im(AT ) and ker(L+ ) = ker(AT ), by parts b and c. 245

Chapter 5

e L+

y1 y2

y1

1

2

2

= 0 , so that the matrix of L+ is 0 0 0

0 0 . 0

Figure 5.17: for Problem 5.4.14. 5.4.14 L+ (w ~ 1 ) is the minimal solution of the system L(~x) = w ~ 1 . The line S in Figure 5.17 shows all solutions of the system L(~x) = w ~ 1 (compare with Exercise 9). The minimal solution, L+ (w ~ 1 ), is perpendicular to ker(L). L+ (w ~ 2 ) = L+ (projim(L) w ~ 2 ) = L+ (~0) = ~0 L+ (w ~ 3 ) = L+ (projim(L) w ~ 3 ) ≈ L+ (0.55w ~ 1 ) = 0.55L+ (w ~ 1) 5.4.15 Note that (AT A)−1 AT A = In ; let B = (AT A)−1 AT . 5.4.16 If A is an m × n matrix, then dim(imA)⊥

= m − dim(imA) ↑

Theorem 5.1.8c and dim(ker(AT ))

= m − rank(A) ↑ Theorem 3.3.6

= m − rank(AT ). ↑

Theorem 3.3.7 It follows that rank(A) = rank(AT ), as claimed. 5.4.17 Yes! By Theorem 5.4.2, ker(A) = ker(AT A). Taking dimensions of both sides and using Theorem 3.3.7, we find that n − rank(A) = n − rank(AT A); the claim follows. 5.4.18 Yes! By Exercise 17, rank(A) = rank(AT A). Substituting AT for A in Exercise 17 and using Theorem 5.3.9c, we find that rank(A) = rank(AT ) = rank(AAT ). The claim follows. ∗

T

−1

5.4.19 ~x = (A A)

1 , by Theorem 5.4.6. A b= 1 T~

246

Section 5.4 −1 2 and ~b − A~x∗ = 1 . 5.4.20 Using Theorem 5.4.6, we find ~x∗ = 2 1 Note that ~b − A~x∗ is perpendicular to the two columns of A.

5.4.21 Using Theorem 5.4.6, we find ~x∗ =

∗

5.4.22 Using Theorem 5.4.6, we find ~x = exact solution; the error k~b − A~x∗ k is 0.

−1 2

−12 and ~b − A~x∗ = 36 , so that k~b − A~x∗ k = 42. −18

and ~b − A~x∗ = ~0. This system is in fact consistent and ~x∗ is the

3 −2

5.4.23 Using Theorem 5.4.6, we find ~x∗ = ~0; here ~b is perpendicular to im(A). 5.4.24 Using Theorem 5.4.6, we find ~x∗ = [2]. 5 15 x1 5 5.4.25 In this case, the normal equation AT A~x = AT ~b is = , which simplifies to x1 + 3x2 = 1, x 15 45 15 2 1 − 3t , where t is an arbitrary constant. or x1 = 1 − 3x2 . The solutions are of the form ~x∗ = t

66 5.4.26 Here, the normal equation AT A~x = AT ~b is 78 90 where t is an arbitrary constant.

78 93 108

t − 67 1 90 x1 108 x2 = 2 , with solutions ~x∗ = 1 − 2t , 3 126 x3 t

5.4.27 The least-squares solutions of the system SA~x = S~b are the exact solutions of the normal equation (SA)T SA~x = (SA)T S~b. Note that S T S = In , since S is orthogonal; therefore, the normal equation simplifies as follows: (SA)T SA~x = AT S T SA~x = AT A~x and (SA)T S~b = AT S T S~b = AT ~b, so that the normal equation is AT A~x = AT ~b, the same as the normal equation of the A~x = ~b. Therefore, the systems A~x = ~b and SA~x = S~b have the same system 7 . least-squares solution, ~x∗ = 11 5.4.28 The least-squares solutions of the system A~x = ~un are the exact solutions of A~x = projim(A) ~un . Note that ~un is orthogonal to im(A), so that projim(A) ~un = ~0, and the unique least-squares solution is ~x∗ = ~0. 5.4.29 By Theorem 5.4.6, ~x∗ = (AT A)−1 AT ~b =

1+ε 1

1 1+ε

−1

1+ε = 1+ε

1 2+ε

1 1+ε ≈ 21 , where ε = 10−20 . 1+ε 2

If we use a hand-held calculator, due to roundoff errors we find the normal equation infinitely many solutions. 247

1 1 1 1

x1 x2

=

1 , with 1

Chapter 5 5.4.30 We attempt to solve the system c0 + 0c1 = 0 1 0 0 c c0 + 0c1 = 1 , or 1 0 0 = 1 . c1 c0 + 1c1 = 1 1 1 1 This system cannot be solved exactly; the least-squares solution is points best is f ∗ (t) =

1 2

+ 12 t.

c∗0 c∗1

1 = 21 . The line that fits the data 2

Figure 5.18: for Problem 5.4.30. The line goes through the point (1, 1) and “splits the difference” between (0, 0) and (0, 1). See Figure 5.18. 5.4.31 We want

c0 c1

such that 3 1 0 3 = c0 + 0c1 c 3 = c0 + 1c1 or 1 1 1 = 3 . c2 6 6 = c0 + 1c1 1 1

−1 ∗ 1 0 1 0 1 1 1 c 1 Since ker 1 1 = {~0}, 0 = 1 1 c1 0 1 1 0 1 1 1 1 −1 3 3 2 12 = = 3 so f ∗ (t) = 3 + 23 t. (See Figure 5.19.) 2 2 9 2

3 1 1 3 1 1 6

Figure 5.19: for Problem 5.4.31. c0 5.4.32 We want c1 of f (t) = c0 + c1 t + c2 t2 such that c2

248

Section 5.4 1 27 = c0 + 0c1 + 0c2 0 = c0 + 1c1 + 1c2 1 or 1 0 = c0 + 2c1 + 4c2 1 0 = c0 + 3c1 + 9c2

0 1 2 3

27 0 c0 1 0 c1 = 0 4 c2 0 9

If we call the coefficient matrix A, we notice that ker(A) = {~0} so ∗ 27 c0 25.65 0 c1 = (AT A)−1 AT = −28.35 so f ∗ (t) = 25.65 − 28.35t + 6.75t2 . 0 c2 6.75 0

c0 5.4.33 We want c1 such that c2

0 1 c cos(1) 1 1 c2 = . 2 cos(2) c3 3 cos(3) ∗ c1 Since the coefficient matrix has kernel {~0}, we compute c2 using Theorem 5.4.6, obtaining c3 1 0 = c0 + sin(0)c1 + cos(0)c2 1 = c0 + sin(1)c1 + cos(1)c2 1 or 2 = c0 + sin(2)c1 + cos(2)c2 1 1 3 = c0 + sin(3)c1 + cos(3)c2

0 sin(1) sin(2) sin(3)

∗ 1.5 c0 c1 ≈ 0.1 so f ∗ (t) ≈ 1.5 + 0.1 sin t − 1.41 cos t. −1.41 c2

c0 c1 5.4.34 We want c2 such that c3 c4

1 1 1 1 1 1 1

0 sin(0.5) sin(1) sin(1.5) sin(2) sin(2.5) sin(3)

1 cos(0.5) cos(1) cos(1.5) cos(2) cos(2.5) cos(3)

0 sin(1) sin(2) sin(3) sin(4) sin(5) sin(6)

0 1 cos(1) c0 0.5 cos(2) c1 1 cos(3) c2 = 1.5 cos(4) c3 2 2.5 c4 cos(5) 3 cos(6)

Since the columns of the coefficient matrix are linearly independent, its kernel is {~0}. We can use Theorem 5.4.6 1.5 c0 c1 0.109 to compute c2 ≈ −1.537 so f ∗ (t) ≈ 1.5 + 0.109 sin(t) − 1.537 cos(t) + 0.303 sin(2t) + 0.043 cos(2t). 0.303 c3 0.043 c4

5.4.35 a The ij th entry of ATn An is the dot product of the ith row of ATn and the jth column of An , i.e. 249

Chapter 5 n X

n n X sin ai ATn An = i=1 X n cos ai

i=1 n X

n X

sin ai

i=1

sin2 ai sin ai cos ai

i=1

2π Z 2π T b lim 2π A A = sin t dt n n n n→∞ 0 Z 2π cos t dt 0

Z

i=1

i=1

Z

2π

sin t dt 2π

sin2 t dt

0

Z

2π

Z

cos t dt

2π 2π 0 = sin t cos t dt 0 0 Z 2π cos2 t dt 0

0

Z

g(ai )

i=1 n X g(ai ) sin ai sin ai cos ai and ATn~b = . i=1 i=1 n n X X 2 g(ai ) cos ai cos ai

i=1 n X

i=1

n X

X n

cos ai

2π

sin t cos t dt

Z

0 π 0

0 0 π

0

0

2π

g(t) dt 0 Z 2π T~ and lim 2π b = A g(t) sin t dt n n n→∞ 0 Z 2π g(t) cos t dt 0

( Here

2π n

=

2π n→∞ n

∆t so lim

n X

cos(ti )

=

i=1

lim

n→∞

All other limits are obtained similarly. )

2π cn c c p = lim pn = 0 n→∞ 0 qn q

0 π 0

where c, p, q are given above.

Z

n X

cos(ti )∆t

=

Z

2π

cos t dt for instance.

0

i=1

Z 2π 1 g(t) dt g(t) dt 2π 0 −1 0 0 1 Z 2π Z 2π 0 g(t) sin t dt = g(t) sin t dt and f (t) = c+p sin t+q cos t, π 0 0 π Z 2π Z 2π 1 g(t) cos t dt g(t) cos t dt π 0 0 2π

a 5.4.36 We want b such that c

2π 2π a + b sin 32 + c cos 32 366 366 2π 2π a + b sin 77 + c cos 77 366 366 2π 2π 121 + c cos 121 a + b sin 366 366 250

=

10

=

12

=

14

Section 5.4

a + b sin

1

sin

1 Using A = 1

sin

sin

1

sin

2π 366 32 2π 366 77 2π 366 121 2π 366 152

cos cos cos cos

2π 2π 152 + c cos 152 366 366

2π 366 32 2π 366 77 2π 366 121 2π 366 152

∗ 10 a 12 and ~b = , we compute b 14 c 15

c0 c1

∗

= (AT A)−1 AT ~b ≈

2π 366 t

− 2.899 cos

2π 366 t

.

35 log 35 46 c0 log 46 = 59 c1 log 77 69 log 133

↑

↑

~b

A so

15

12.26 = (AT A)−1 AT ~b ≈ 0.431 and f ∗ (t) ≈ 12.26 + 0.431 sin −2.899 5.4.37 a We want c0 , c1 such that c0 + c1 (35) = log 35 1 c0 + c1 (46) = log 46 1 or c0 + c1 (59) = log 77 1 c0 + c1 (69) = log 133 1

=

0.915 0.017

so log(d) ≈ 0.915 + 0.017t.

b d ≈ 100.915 · 100.017t ≈ 8.22 · 100.017t c If t = 88 then d ≈ 258. Since the Airbus has only 93 displays, new technologies must have rendered the old trends obsolete. c0 5.4.38 We want c1 such that c2

110 2 1 12 0 c0 180 5 1 c1 = 120 . 160 c2 11 1 160 6 0 ∗ 125 c0 5 , so that w∗ = 125 + 5h − 25g. The least-squares solution is c1 = −25 c2 1 110 = c0 + 2c1 + c2 180 = c0 + 12c1 + 0c2 1 120 = c0 + 5c1 + c2 or 1 1 160 = c0 + 11c1 + c2 160 = c0 + 6c1 + 0c2 1

For a general population, we expect c0 and c1 to be positive, since c0 gives the weight of a 5′ male, and increased height should contribute positively to the weight. We expect c2 to be negative, since females tend to be lighter than males of equal height. 5.4.39 a We want

c0 c1

such that 251

Chapter 5 log(250) = c0 + c1 log(600, 000) log(60) = c0 + c1 log(200, 000) log(25) = c0 + c1 log(60, 000) log(12) = c0 + c1 log(10, 000) log(5) = c0 + c1 log(2500) The least-squares solution to the above system is

c0 c1

∗

≈

−1.616 0.664

so log z ≈ −1.616 + 0.664 log g.

b Exponentiating both sides of the answer to a, we get z ≈ 10−1.616 · g 0.664 ≈ 0.0242 · g 0.664 . c This model is close since 5.4.40 First we look for

c0 c1

√

g = g 0.5 .

such that log D = c0 + c1 log a.

Proceeding as in Exercise 39, we get

c0 c1

∗

0 ≈ , i.e. log D ≈ 1.5 log a, hence D ≈ 101.5 log a = a1.5 . 1.5

Note that the formula D = a1.5 is Kepler’s third law of planetary motion. 5.4.41 a We want log 533 log 1, 823 log 4, 974 log 7, 933

c0 c1

such that log D = c0 + c1 t (t in years since 1975), i.e.

= c0 + c1 (0) = c0 + c1 (10) = c0 + c1 (20) = c0 + c1 (30)

The least-squares solution to the system is

c0 c1

∗

≈

2.8 , i.e. log D ≈ 2.8 + 0.04 or D ≈ 102.8 · 100.040t . 0.040

b In the year 2015, we have t = 40 and D ≈ 24, 200. The formula predicts a debt of about US$24 trillion. 5.4.42 Clearly, L is a linear transformation. We will use Theorem 4.2.4a and show that ker(L) = {~0} and im(L) = im(A). Now ker(L) = ker(A)∩im(AT ) = {~0}, by Theorems 5.4.1 and 5.1.8b. Also, im(L) = {A~v : ~v in im(AT )} = im(AAT ) = im(A), by Exercise 6.

Section 5.5 5.5.1 Since f is nonzero, there is a c on [a, b] such that f (c) = d = 6 0. By continuity, there is an ε > 0 such that for all x on the open interval (c − ε, c + ε) where f is defined (see any good Calculus text). |f (x)| > |d| 2 252

Section 5.5

Figure 5.20: for Problem 5.5.1.

We assume that c − ε ≥ a and c + ε ≤ b and leave the other cases to the reader. Then hf, f i = Z c+ε 2 d2 ε d dt = > 0, as claimed. (See Figure 5.20.) 2 2 c−ε

Z

a

b

(f (t))2 dt ≥

5.5.2 We perform the following operations: hf, g + hi = hg + h, f i ↑ Definition 5.5.1a

= hg, f i + hh, f i ↑

Definition 5.5.1b

= hf, gi + hf, hi ↑

Definition 5.5.1a

5.5.3 a Note that h~x, ~y i = (S~x)T S~y = S~x · S~y . We will check the four parts of Definition 5.5.1 α. h~x, ~y i = S~x · S~y = S~y · S~x = h~y , ~xi β. h~x + ~y , ~zi = S(~x + ~y ) · S~z = (S~x + S~y ) · S~z = (S~x · S~z) + (S~y · S~z) = h~x, ~zi + h~y , ~zi γ.

hc~x, ~y i = S(c~x) · S~y = c(S~x) · S~y = ch~x, ~y i

δ. If ~x 6= 0, then h~x, ~xi = S~x · S~x = kS~xk2 is positive if S~x 6= ~0, that is, if ~x is not in the kernel of S. It is required that S~x 6= ~0 whenever ~x 6= ~0, that is, ker(S) = {~0}. Answer: S must be invertible. b It is required that h~x, ~y i = (S~x)T S~y = ~xT S T S~y equal ~x · ~y = ~xT ~y for all ~x and ~y . This is the case if and only if S T S = In , that is, S is orthogonal. 5.5.4 a For column vectors ~v , w, ~ we have h~v , wi ~ = trace(~v T w) ~ = trace(~v · w) ~ = ~v · w, ~ the dot product. b For row vectors ~v , w, ~ the ij th entry of ~v T w ~ is vi wj , so that h~v , wi ~ = trace(~v T w) ~ = product.

m X i=1

vi wi = ~v · w, ~ again the dot

5.5.5 a hhA, Bii = tr(AB T ) = tr((AB T )T ) = tr(BAT ) = hhB, Aii In the second step we have used the fact that tr(M ) = tr(M T ), for any square matrix M . 253

Chapter 5 b hhA + B, Cii = tr((A + B)C T ) = tr(AC T + BC T ) = tr(AC T ) + tr(BC T ) = hhA, Cii + hhB, Cii c hhcA, Bii = tr(cAB T ) = ctr(AB T ) = chhA, Bii d hhA, Aii = tr(AAT ) =

n X i=1

k~vi k2 > 0 if A 6= 0, where ~vi is the ith row of A.

We have shown that hh·, ·ii does indeed define an inner product. 5.5.6 a The ii th entry of PQ is

m X

pik qki , so that tr(P Q) =

n m X X

pik qki .

i=1 k=1

k=1

Likewise, tr(QP ) =

m n X X

qik pki .

i=1 k=1

Reversing the roles of i and k and the order of summation we see that tr(P Q) = tr(QP ), as claimed. b Using part a and the fact that tr(M ) = tr(M T ) for any square matrix M , we find that hA, Bi = tr(AT B) = tr(BAT ) = tr((BAT )T ) = tr(AB T ) = hhA, Bii 5.5.7 Axioms a, b, and c hold for any choice of k (check this!). Also, it is required that hhv, vii = khv, vi be positive for nonzero v. Since hv, vi is positive, this is the case if (and only if) k is positive. 5.5.8 By parts b and c of Definition 5.5.1, we have T (u + v) = hu + v, wi = hu, wi + hv, wi = T (u) + T (v) and T (cv) = hcv, wi = chv, wi = cT (v), so that T is linear. If w = 0, then im(T ) = {0} and ker(T ) = V . If w 6= 0, then im(T ) = R and ker(T ) consists of all v perpendicular to w. 5.5.9 If f is even and g is odd, then fg is odd, so that hf, gi =

Z

1

f g = 0.

−1

5.5.10 A function g(t) = a + bt + ct2 is orthogonal to f (t) = t if Z 1 2 1 (at + bt2 + ct3 ) dt = [ a2 t2 + 3b t3 + 4c t4 ]−1 = b = 0, that is, if b = 0. hf, gi = 3 −1 Thus, the functions 1 and t2 form a basis of the space of all functions in P2 orthogonal to f (t) = t. To find an Z 1 orthonormal basis g1 (t), g2 (t), we apply Gram-Schmidt. Now k1k = 12 1 dt = 1, so that we can let g1 (t) = 1. Then g2 (t) =

t2 −h1,t2 i1 kt2 −h1,t2 i1k

=

t2 − 13 kt2 − 13 k

−1

√

=

5 2 2 (3t

− 1)

√

Answer : g1 (t) = 1, g2 (t) =

5 2 2 (3t

− 1)

5.5.11 hf, gi = hcos(t), cos(t + δ)i = hcos(t), cos(t) cos(δ) − sin(t) sin(δ)i = cos(δ)hcos(t), cos(t)i − sin(δ) hcos(t), sin(t)i = cos(δ), by Theorem 5.5.4. Also, hg, gi = 1 (left to reader). 254

Section 5.5 Thus, ∠(f, g) = arccos

5.5.12 By Theorem 5.5.5Z E D a0 = |t|, √12 = √12π

hf,gi kf kkgk

= arccos(cos δ) = δ.

π

π 1 |t| dt = √ , bk = h|t|, sin(kt)i = π 2 −π

function.

Z

π

−π

|t| sin(kt) dt = 0, since the integrand is an odd

π

Z 2 π |t| cos(kt) dt = ck = h|t|, cos(kt)i = t cos(kt) dt π 0 −π R π π = π2 k1 t sin(kt) 0 − 0 k1 sin(kt) dt 0 if k is even = π2 k12 [cos(kt)]π0 = πk2 2 (cos(kπ) − 1) = − k24π if k is odd In summary: 1 π

a0 bk ck

=0 =0 0 = − k24π

Z

if k is even if k is odd

5.5.13 The sequence (a0 , b1 , c1 , b2 , c2 , . . .) is “square-summable” by Theorem 5.5.6, so that it is in ℓ2 . Also, k(a0 , b1 , c1 , b2 , c2 , . . .)k a20 + b21 + c21 + b22 + c22 + · · · = kf k2 , by Theorem 5.5.6, so that the two norms are equal. 5.5.14 a This is not an inner product since there are nonzero polynomials f (t) in P2 with f (1) = f (2) = 0, so that hf, f i = (f (1))2 + (f (2))2 = 0. (For example, let f (t) = (t − 1)(t − 2).) b This is an inner product. We leave it to the reader to check axioms a to c. As for d: A nonzero polynomial f in P2 has at most two zeros, so that f (1) 6= 0 or f (2) 6= 0 or f (3) 6= 0, and hf, f i = (f (1))2 + (f (2))2 + (f (3))2 > 0. 1 0 0 1 , = , = c, by part a of Definition 5.5.1, so that b = c. Check 0 1 1 0 that if b = c then h~v , wi ~ = hw, ~ ~v i for all ~v , w ~ in R2 . Check that parts (b) and (c) of Definition 5.5.1 are satisfied for of constants c,d. Next, we need to worry about part allvalues (d) of that definition. It is required that x1 x1 x1 2 2 , = x1 + 2bx1 x2 + dx2 be positive for all nonzero . We can complete the square and write x2 x2 x2 x1 if (and only if) x21 + 2bx1 x2 + dx22 = (x1 + bx2 )2 + (d − b2 )x22 ; this quantity is positive for all nonzero x2 d − b2 > 0, or, d > b2 . In summary, the function is an inner product if (and only if) b = c and d > b2 .

5.5.15 First note that b =

5.5.16 a We start with the standard basis 1, t and use the Gram-Schmidt process to construct an orthonormal basis g1 (t), g2 (t). s Z 1 √ t− 1 t−h1,ti1 dt = 1, so that we can let g1 (t) = 1. Then g2 (t) = kt−h1,ti1k = kt− 21 k = 3(2t − 1). k1k = 2

0

Summary: g1 (t) = 1 and g2 (t) =

√ 3(2t − 1) is an orthonormal basis.

b We are looking for projP1 (t2 ) = hg1 (t), t2 ig1 (t) + hg2 (t), t2 ig2 (t), by Theorem 5.5.3. 255

Chapter 5

We find that hg1 (t), t2 i =

Z

1

t2 dt =

0

1) = t − 16 . See Figure 5.21.

√ 1 and hg2 (t), t2 i = 3 3

Z

0

1

(2t3 − t2 ) dt =

√

3 , so that projP1 t2 = 6

1 3

+ 21 (2t −

Figure 5.21: for Problem 5.5.16b. 5.5.17 We leave it to the reader to check that the first three axioms are satisfied for any such T . As for axiom d: It is required that hv, vi = T (v) · T (v) = kT (v)k2 be positive for any nonzero v, that is, T (v) 6= ~0. This means that the kernel of T must be {0}. c1 c2 5.5.18 Let the orthonormal basis be f1 , . . . , fn and f = c1 f1 + · · · + cn fn ; then [f ]B = ... . The Pythagorean

Theorem tells us that kf k2 = c21 + · · · + c2n = k[f ]B k2 , so that kf k = k[f ]B k.

cn

x1 y p q x1 y ,w ~ = 1 , and A = , then , 1 x2 y2 x2 r s y2 p q y1 = [x1 x2 ] = px1 y1 + qx1 y2 + rx2 y1 + sx2 y2 . Note that in Exercise 15 we considered the special case y2 r s 1 1 p = 1. First it is required that p = , be positive. 0 0 i h x1 y Now we can write , 1 = p x1 y1 + pq x1 y2 + pr x2 y1 + ps x2 y2 and use our work in Exercise 15 with b = pq , c = pr , d x2 y2 2 to see that the conditions q = r and q < ps must hold. In summary, the function is an inner product if (and p q only if) the entries of matrix A = satisfy the conditions p > 0, q = r and det(A) = ps − q 2 > 0. r s

5.5.19 If we write ~v =

5.5.20 a −2 . 1

x1 1 1 = [x1 x2 ] , 2 x2 0

2 8

1 = x1 + 2x2 = 0 when x1 = −2x2 . This is the line spanned by vector 0

1 −2 b Since vectors and are orthogonal, we merely have to multiply each of them with the recipro0 1 256

Section 5.5

2

1 1 2 1 1

cal of its norm. Now = 1, so that is a unit vector, and

= [1 0] 2 8 0 0 0

−2 −1 1 −2 1 2

is an orthonormal basis. = 4, so that [ −2 1 ] 1

1 = 2. Thus 0 , 1 2 8 2

−2 2

1 =

5.5.21 Note that h~v , wi ~ = (~v + w) ~ · (~v + w) ~ − ~v · ~v − w ~ ·w ~ = 2(~v · w), ~ double the dot product. Thus it’s an inner product, by Exercise 7. 5.5.22 Apply the Cauchy-Schwarz inequality to f (t) and g(t) = 1; note that kgk = 1: |hf, gi| ≤ kf kkgk = kf k or hf, gi2 ≤ kf k2 or

Z

1

f (t) dt

0

2

≤

R1 0

(f (t))2 dt.

5.5.23 We start with the standard basis 1, t of P1 and use the Gram-Schmidt process to construct and orthonormal basis g1 (t), g2 (t). q t− 1 t−h1,ti1 = kt− 21 k = 2t − 1. k1k = 12 (1 · 1) + 1 · 1 = 1, so that we can let g1 (t) = 1. Then g2 (t) = kt−(1,t)1k 2

Summary: g1 (t) = 1 and g2 (t) = 2t − 1 is an orthonormal basis.

5.5.24 a hf, g + hi = hf, gi + hf, hi = 0 + 8 = 8 b kg + hk =

p

hg + h, g + hi =

p

hg, gi + 2hg, hi + hh, hi =

√ √ 1 + 6 + 50 = 57

c Since hf, gi = 0, kgk = 1, and kf k = 2, we know that f2 , g is an orthonormal basis of span (f, g). E D Now projE h = f2 , h f2 + hg, hig = 14 hf, hif + hg, hig = 2f + 3g. d From part c we know that 12 f, g are orthonormal, so we apply Theorem 5.2.1 to obtain the third polynomial in an orthonormal basis of span(f, g, h): h−projE h kh−projE hk

=

h−2f −3g kh−2f −3gk

Orthonormal basis:

=

h−2f −3g 5

2 1 2 f, g, − 5 f

= − 25 f − 53 g + 15 h

− 53 g + 15 h

5.5.25 Using the inner product defined in Example 2, we find that q q p 2 k~xk = h~x, ~xi = 1 + 41 + 91 + · · · + n12 + · · · = π6 = √π6 (see the text right after Theorem 5.5.6). 5.5.26 a0 = bk =

1 π

Z

√1 2π

Z

π

f (t) dt = 0

−π

π

f (t) sin(kt) dt =

−π

=

2 − kπ [cos(kt)]π0

=

0 4 πk

1 π

Z −

0

sin(kt) dt +

−π

Z

π

sin(kt) dt

0

if k is even if k is odd 257

=

2 π

Z

0

π

sin(kt) dt

Chapter 5

Figure 5.22: for Problem 5.5.26.

ck =

1 π

Z

π

f (t) cos(kt) dt = 0, since the integrand is odd.

−π

f1 (t) = f2 (t) =

4 π

sin(t). See Figure 5.22.

Figure 5.23: for Problem 5.5.26.

f3 (t) = f4 (t) =

4 π

sin(t) +

4 3π

sin(3t). See Figure 5.23.

f5 (t) = f6 (t) =

4 π

sin(t) +

4 3π

sin(3t) +

4 5π

sin(5t). See Figure 5.24. 258

Section 5.5

Figure 5.24: for Problem 5.5.26.

bk =

1 π

ck =

1 π

π

1 f (t) dt = √ . 2 −π Z π Z 0 1 π 1 π f (t) sin(kt) dt = sin(kt) dt = − [cos(kt)]0 = 2 π 0 kπ −π kπ Z π Z 1 π 1 f (t) cos(kt) dt = [sin(kt)]π0 = 0 cos(kt) dt = π kπ −π 0

5.5.27 a0 =

√1 2π

Z

+

2 π

sin(t), f3 (t) = f4 (t) =

5.5.28 kf k = hf, f i =

1 π

Z

f1 (t) = f2 (t) =

1 2

1 2

+

2 π

sin(t) +

2 3π

if k is even if k is odd

sin(3t)

.. . 2

π

1 (f (t)) dt = π −π 2

Now Theorem 5.5.6 tells us that

5.5.29 kf k2 = hf, f i =

1 π

Z

1 dt = 2

−π

16 16 16 π 2 + 9π 2 + 25π 2 +· · ·

π

(f (t))2 dt =

−π

Now Theorem 5.5.6 tells us that

1 2

+

X 1 1 1 1 =1+ + + + ··· = k2 9 25 49

k odd

π

Z

1 π

4 π2

1 2 4 π2

Z

π

+

4 9π 2

=

16 π2

X 1 k2

k odd

!

= 2, or

X 1 1 1 1 π2 = 1+ + + +· · · = . k2 9 25 49 8

k odd

1 dt = 1

0

=

+

4 25π 2

+ · · · = 1, or

π2 . 8

1 2

+

4 π2

X 1 k2

k odd

!

= 1 or

5.5.30 There is an invertible linear transformation T (~x) = A~x from R2 to R2 that transforms E into the unit circle. −1 (If E is parametrized by cos(t)w ~ 1 + sin(t)w ~ 2 , let A = [ w ~1 w ~ 2 ] . Compare with Exercises 2.2.50 and 2.2.51.) 259

Chapter 5 Now ~x is on E if kT (~x)k = 1, or T (~x) · T (~x) = 1. This means that the inner product h~x, ~y i = T (~x) · T (~y ) does the job (see Exercise 17). (There are other, very different approaches to this problem.) √1 , f1 (t) 2

5.5.31 An orthonormal basis of P2 of the desired form is f0 (t) = with Exercise 10), and the zeros of f2 (t) are a1,2 = Next we find the weights w1 , w2 such that

Z

± √13 .

1

f (t) dt =

−1

2 X

=

q

3 2 t, f2 (t)

=

1 2

q

5 2 2 (3t

− 1) (compare

wi f (ai ) for all f in P1 . We need to make sure that

i=1

the equation holds for 1 and t. # " 2 = w1 + w2 , with solution w1 = w2 = 1. 0 = √13 w1 − √13 w2 1

1 1 √ + f −√ holds for all polynomials f in 3 3 −1 P1 . We can check that it holds for t2 and t3 as well, that is, it holds in fact for all cubic polynomials. Z 1 Z 1 2 2 3 3 2 t2 dt = equals √13 + − √13 = 32 , and t3 dt = 0 equals √13 + − √13 = 0. 3 −1 −1

It follows that the equation

5.5.32 a.htn , tm i = b. ktn k =

p

1 2

R1

Z

f (t) dt = f (a1 ) + f (a2 ) = f

tn+m dt = −1

htn , tn i =

q

1 2n+1

1 2

h

tn+m+1 n+m+1

i1

=

−1

1 n+m+1

if n + m is even if n + m is odd

0

by part a.

c. It helps to observe that 1 and t2 are orthogonal to t and t3 . Accordingly, many terms in the Gram-Schmidt formulas vanish.

√ √ t2 − t2 , 1 1 5 1 t g0 (t) = = 1, g1 (t) = = 3 t, g2 (t) = 2 = 3t2 − 1 2 k1k ktk kt − ht , 1i 1k 2

√ √ √ t3 − t3 , 3 t 3 t t3 − 3t/5 7

√ √

= 3 g3 (t) = = 5t3 − 3t

t3 − t3 , 3 t 3 t kt − 3t/5k 2

d. The first few Legendre Polynomials are

g0 (t) g1 (1)

= 1,

g1 (t) g1 (1)

= t,

g2 (t) g2 (1)

= 3t2 − 1 /2,

g3 (t) g3 (1)

= 5t3 − 3t /2.

1 e. Since f (t) = 1+t 2 is an even function, we need to consider g0 (t) and g2 (t) only. The solution is hf (t), g0 (t)i g0 (t)+

15 3π 2 hf (t), g2 (t)i g2 (t) = hf (t), 1i 1 + 45 f (t), 3t2 − 1 3t2 − 1 = π4 + 85 (6 − 2π) 3t2 − 1 = 15 4 (3 − π) t − 4 + 2 . 1

f (t) = g(t) =

15 4 (3

− π)t2 −

-1

15 4

+ 32 π 1

260

1 1 + t2

t

Section 5.5 5.5.33 a The first property of an inner product, hf, gi = hg, f i, follows from the fact that f (t)g(t) = g(t)f (t). The property hf + h, gi = hf, gi + hh, gi follows from the sum rule for integrals, and the property hcf, gi = c hf, gi follows from the constant multiple rule for integrals. Proceed as in Exercise 1 to show that hf, f i > 0 for all nonzero f. b If f (t) = 1 then kf k =

qR qR p b b 2 dt = hf, f i = w(t) · 1 w(t)dt = 1. a a

√ R1 5.5.34 a. Note that the graph of 1 − t2 is the upper half of the unit circle centered at the origin. Thus −1 w(t) dt = R √ 2 1 1 − t2 dt = π2 (half the area of the unit circle) = 1 π −1 p b. If f (t) = 1 then kf k = hf, f i = 1 by part a and Exercise 33b. R1 √ c. If n + m is odd then htn , tm i = π2 −1 1 − t2 tn+m dt = 0 since the integrand is an odd function.

d. It follows from the definition of this inner product that ht, ti = 1, t2 = 1/4 and t2 , t2 = t, t3 = 1/8 so p

2 p √ that ktk = 1/4 = 1/2 and t = 1/8 = 2/4.

e. It helps to observe that 1 and t2 are orthogonal to t and t3 . Accordingly, many terms in the Gram-Schmidt formulas vanish. The first few Chebyshev Polynomials of the Second Kind are

t2 − t2 , 1 1 1 t t2 − 1/4 p g0 (t) = = 1, g1 (t) = = 2 t, g2 (t) = 2 = = 4t2 − 1 k1k ktk kt − ht2 , 1i 1k 1/16 g3 (t) =

t3 − t3 , 2 t 2 t t3 − t/2 = p = 8t3 − 4t. 3 3 kt − ht , 2 ti 2 tk 1/64

f. Since f (t) = t4 is an even we need to consider g0 (t) and g2 (t) only. The solution is hf (t), g0 (t)i g0 (t) + function,

5 1 − 18 4t2 − 1 = 43 t2 − 16 . hf (t), g2 (t)i g2 (t) = t4 , 1 1 + t4 , 4t2 − 1 4t2 − 1 = 81 + 16 f (t) = t4

1

g(t) = 34 t2 −

-1

1

1 16

t

q R p p p 1 5.5.35 a. ktk32 = 12 −1 t2 dt = 1/3 and ktk34 = ht, ti34 = h1, t2 i34 = 1/2, so that ktk32 > ktk34 . Note that

2 1, t 34 = 1/4 is given in Exercise 34. 1

w32 (t) =

w34 (t) =

1 2

2 π

√ 1 − t2

-1

t

1

1

w32 (t) t2 = 12 t2 w34 (t) t2 =

-1

2 2 πt

√ 1 − t2

1

261

t

Chapter 5 q R q √ p √ √ 1 1 2 ) dt = = (1 − t b. For f (t) = 1 − t2 we have kf k32 = 2/3 and kf k 1 − t2 , 1 − t2 34 = 34 2 −1 p p p h1, 1 − t2 i34 = 1 − 1/4 = 3/4. 1

w32 (t)

w34 (t)

√ √ 1 − t2 = 21 1 − t2

√ 1 − t2 =

2 π (1

-1

− t2 ) 1

t

True or False Ch 5.TF.1 T, since (A + B)T = AT + B T = A + B Ch 5.TF.2 T, by Theorem 5.3.4 Ch 5.TF.3 F. Consider

1 1

1 . 1

Ch 5.TF.4 T. First note that AT = A−1 , by Theorem 2.4.8. Thus A is orthogonal, by Theorem 5.3.7. Ch 5.TF.5 F. The correct formula is projL (~x) = (~x · ~u)~u, by Definition 2.2.1. Ch 5.TF.6 T, since (7A)T = 7AT = 7A. Ch 5.TF.7 F. Consider T (~x) =

1 ~x. 0

1 0

Ch 5.TF.8 T, by Theorem 5.3.9.b Ch 5.TF.9 T, by Theorem 5.3.4a Ch 5.TF.10 F. We have (AB)T = B T AT , by Theorem 5.3.9a. Ch 5.TF.11 T. If A is orthogonal, then AT = A−1 , and A−1 is orthogonal by Theorem 5.3.4b. Ch 5.TF.12 F. Consider

0 1 1 0

0 −1 1 0 . = 1 0 0 −1

0 Ch 5.TF.13 F. Consider A = B = 0

1 1 0 0 T T . Then AB = isn’t equal to B A = 0 0 0 0

0 . 1

Ch 5.TF.14 that the columns of A be orthonormal (Theorem 5.3.10). As a counterexample, consider F. It is required 4 0 2 . , with AAT = A= 0 0 0 262

True or False Ch 5.TF.15 T, since (ABBA)T = AT B T B T AT = ABBA, by Theorem 5.3.9a Ch 5.TF.16 T, since AT B T = (BA)T = (AB)T = B T AT , by Theorem 5.3.9a Ch 5.TF.17 F. dim(V ) + dim(V ⊥ ) = 5, by Theorem 5.1.8c. Thus one of the dimensions is even and the other odd. Ch 5.TF.18 T. Consider the QR factorization (Theorem 5.2.2) Ch 5.TF.19 F. The Pythagorean Theorem holds for orthogonal vectors ~x, ~y only (Theorem 5.1.9) a c a b Ch 5.TF.20 T. det = ad − bc = det . b d c d Ch 5.TF.21 F. As a counterexample, consider

1 0

0 −1

and

0 1

−1 . 0

Ch 5.TF.22 T, by Theorem 5.4.1. Ch 5.TF.23 T, by Theorem 5.4.2a. Ch 5.TF.24 F. Consider A =

2 0 , or any other symmetric matrix that fails to be orthogonal. 0 2

−1 0 −1 0 Ch 5.TF.25 F. det = −1 − 0 = −1, yet is orthogonal. 0 1 0 1 Ch 5.TF.26 T.

1

2 (A

T − AT ) = 12 (A − AT )T = 12 (AT − A) = − 21 (A − AT ) .

Ch 5.TF.27 T, since the columns are unit vectors.

Ch 5.TF.28 T. Use the Gram-Schmidt process to construct such a basis (Theorem 5.2.1) Ch 5.TF.29 F. The columns fail to be unit vectors (use Theorem 5.3.3b) Ch 5.TF.30 T, by definition of an orthogonal projection (Theorem 5.1.4). 1 0 cos θ − sin θ 1 + cos θ − sin θ Ch 5.TF.31 T. Try A = and B = , so that A + B = . It is required sin θ cos θ sin θ 1 + cos θ 0 1 1 + cos θ − sin θ that and be unit vectors, meaning that 1 + 2 cos θ + cos2 θ + sin2 θ = 2 + 2 cos θ = 1, or sin θ 1 + cos θ √ 3 1 √ − − 1 0 2 is a solution. and B = √2 cos θ = − 12 , and sin θ = ± 23 . Thus A = 3 0 1 −1 2

Ch 5.TF.32 F. Consider A =

0 1 2

− 21

0

2

, for example, representing a rotation combined with a scaling. 263

Chapter 5

Ch 5.TF.33 F. Consider A =

−1 1 . 0 1

Ch 5.TF.34 T. By Definition 5.1.12, quantity cos(θ) =

~ v ·w ~ k~ v kkwk ~

is positive, so that θ is an acute angle.

Ch 5.TF.35 T. In Theorem 5.4.1, let A = B T to see that (im(B T ))⊥ = ker(B). Now take the orthogonal complements of both sides and use Theorem 5.1.8d. Ch 5.TF.36 T, since (AT A)T = AT (AT )T = AT A, by Theorem 5.3.9a. Ch 5.TF.37 F. Verify that matrices A =

Ch 5.TF.38 F. Consider B = 5.4.2.

1 0

0 −1

and B =

1 0

1 −1

are similar.

0 1 . The correct formula im(B) = im(BB T ) follows from Theorems 5.4.1 and 0 0

Ch 5.TF.39 T. We know that AT = A and S −1 = S T . Now (S −1 AS)T = S T AT (S −1 )T = S −1 AS, by Theorem 5.3.9a. Ch 5.TF.40 T. By Theorem 5.4.2, we have ker(A) = ker(AT A). Replacing A by AT in this formula, we find that ker(AT ) = ker(AAT ). Now ker(A) = ker(AT A) = ker(AAT ) = ker(AT ). Ch 5.TF.41 T. We attempt to write A = S + Q, where S is symmetric and Q is skew-symmetric. Then AT = S T + QT = S − Q. Adding the equations A = S + Q and AT = S − Q together gives 2S = A + AT and S = 12 (A + AT ). Similarly we find Q = 21 (A − AT ). Check that the decomposition A = S + Q = ( 12 (A + AT )) + ( 21 (A − AT )) does the job.

x1 1 Ch 5.TF.42 T. Apply the Cauchy-Schwarz inequality (squared), (~x · ~y )2 ≤ k~xk2 k~y k2 , to ~x = . . . and ~y = . . . xn 1 (all n entries are 1). 2 2 x y x + y 2 xz + yt x + yz xy + yt T 2 Ch 5.TF.43 T. Let A = . We know that AA = A , or = . We z t xz + yt z 2 + t2 zx + tz yz + t2 2 2 2 need to show that y = z. If y 6= 0, this follows from the equation x + y = x + yz; if z 6= 0, it follows from z 2 + t2 = yz + t2 ; if both y and z are zero, we are all set.

Ch 5.TF.44 T, since ~x · (projV ~x) = (projV ~x + (~x − projV ~x)) · projV ~x = kprojV ~xk2 ≥ 0. Note that ~x − projV ~x is orthogonal to projV ~x, by the definition of a projection.

~ Ch 5.TF.45 T. Note that 1 = A k~x1k ~x = k~x1k Ax

= Definition 5.3.1.

1 xk k~ xk kA~

for all nonzero ~x, so that kA~xk = k~xk. See

a b a−x b is a symmetric matrix, then A − xI2 = . This matrix fails to be b c b c−x invertible if (and only if) det(A − xI2 ) = (a − x)(c − x) − b2 = 0. We use the quadratic formula to find the (real)

Ch 5.TF.46 T. If A =

264

True or False

solutions x = or zero.

a+c±

√

(a+c)2 −4ac+4b2 2

Ch 5.TF.47 T; one basis is:

1 0

=

1 0 , 0 1

a+c±

√

(a−c)2 +4b2 . 2

Note that the discriminant (a − c)2 + 4b2 is positive

0 −1 0 1 0 . , , 1 0 1 0 −1

Ch 5.TF.48 F; A direct computation or a geometrical argument shows that Q = reflection, not a rotation.

√1 5

1 2

2 , representing a −1

Ch 5.TF.49 F; dim(R3×3 )= 9, dim(R2×2 )= 4, so dim(ker(L))≥ 5, but the space of all 3×3 skew-symmetric matrices has dimension of 3. 0 0 0 0 0 −1 0 −1 0 A basis is 1 0 0 , 0 0 0 , 0 0 −1 . 0 1 0 1 0 0 0 0 0 Ch 5.TF.50 T; Consider an orthonormal basis ~v1 , ~v2 of V, and a unit vector ~v3 perpendicular to V, and form the 1 0 0 i h orthogonal matrix S = [~v1 ~v2 ~v3 ]. Now AS = ~v1 ~v2 ~0 = S 0 1 0 . Since S is orthogonal, we have 0 0 0 1 0 0 S T AS = S −1 AS = 0 1 0 , a diagonal matrix. 0 0 0

265

Chapter 6

Chapter 6 Section 6.1 1 2 = 6 − 6 = 0. 6.1.1 Fails to be invertible; since det 3 6 2 3 6.1.2 Invertible; since det = 10 − 12 = −2. 4 5 3 5 6.1.3 Invertible; since det = 33 − 35 = −2. 7 11 1 4 6.1.4 Fails to be invertible; since det = 8 − 8 = 0. 2 8

2 5 6.1.5 Invertible; since det 0 11 0 0

6 0 6.1.6 Invertible; since det 5 4 3 2

7 7 = 2 · 11 · 5 + 0 + 0 − 0 − 0 − 0 = 110. 5 0 0 = 6 · 4 · 1 + 0 + 0 − 0 − 0 − 0 = 24. 1

6.1.7 This matrix is clearly not invertible, so the determinant must be zero. 6.1.8 This matrix fails to be invertible, since the det(A) = 0.

0 1 6.1.9 Invertible; since det 7 8 6 5

1 6.1.10 Invertible; since det 1 1

2 3 = 0 + 3 · 6 + 2 · 7 · 5 − 7 · 4 − 2 · 8 · 6 = −36. 4 1 1 2 3 = 1 · 2 · 6 + 1 · 3 · 1 + 1 · 1 · 3 − 3 · 3 · 1 − 2 · 1 · 1 − 6 · 1 · 1 = 1. 3 6

k 6.1.11 det 3

2 6 0 when 4k 6= 6, or k 6= 32 . = 4

1 6.1.12 det k

k 6 0 when k 2 6= 4, or k 6= 2, −2. = 4

k 6.1.13 det 0 0

3 2 0

5 6 = 8k, so k 6= 0 will ensure that this matrix is invertible. 4 266

Section 6.1

0 k 1

0 0 = 0, so the matrix will never be invertible, no matter which k is chosen. 0

k 3 6

1 4 = 6k − 3. This matrix is invertible when k 6= 12 . 7

2 k 7

3 5 = 60 + 84 + 8k − 18k − 35 − 64 = 45 − 10k. So this matrix is invertible when k 6= 4.5. 8

1 k k2

1 −1 = 2k 2 − 2 = 2(k 2 − 1) = 2(k − 1)(k + 1). So k cannot be 1 or -1. 1

1 2k 7

k 5 = 30 + 21k − 18k 2 = −3(k − 2)(6k + 5). So k cannot be 2 or − 56 . 5

1 k k

4 6.1.14 det 3 2 0 6.1.15 det 2 5 1 6.1.16 det 4 6 1 6.1.17 det 1 1 0 6.1.18 det 3 9 1 6.1.19 det 1 k

k k = −k 3 + 2k 2 − k = −k(k − 1)2 . So k cannot be 0 or 1. k

1 k 1 6.1.20 det 1 k + 1 k + 2 = (k + 1)(2k + 4) + k(k + 2) + (k + 2) − (k + 1) − k(2k + 4) − (k + 2)(k + 2) = 1 k + 2 2k + 4 (k + 1)(3k + 6) − (3k 2 + 9k + 5) = 1. Thus, A will always be invertible, no matter the value of k, meaning that k can have any value.

k 6.1.21 det 1 1

1 k 1

cos k 6.1.22 det 0 sin k

1 1 = k 3 − 3k + 2 = (k − 1)2 (k + 2). So k cannot be -2 or 1. k 1 − sin k 2 0 = 2 cos2 k + 2 sin2 k = 2. So k can have any value. 0 cos k

6.1.23 det(A − λI2 ) = det

1−λ 0

2 = (1 − λ)(4 − λ) = 0 if λ is 1 or 4. 4−λ

6.1.24 det(A − λI2 ) = det

2−λ 1

0 = (2 − λ)(−λ) = 0 if λ is 2 or 0. 0−λ

4−λ 6.1.25 det(A − λI2 ) = det 4

2 = (4 − λ)(6 − λ) − 8 = (λ − 8)(λ − 2) = 0 if λ is 2 or 8. 6−λ 267

Chapter 6

6.1.26 det(A − λI2 ) = det

4−λ 2

2 = (4 − λ)(7 − λ) − 4 = (λ − 8)(λ − 3) = 0 if λ is 3 or 8. 7−λ

6.1.27 A−λI3 is a lower triangular matrix with the diagonal entries (2−λ), (3−λ) and (4−λ). Now, det(A−λI3 ) = (2 − λ)(3 − λ)(4 − λ) = 0 if λ is 2, 3 or 4. 6.1.28 A−λI3 is an upper triangular matrix with the diagonal entries (2−λ), (3−λ) and (5−λ). Now, det(A−λI3 ) = (2 − λ)(3 − λ)(5 − λ) = 0 if λ is 2, 3 or 5. 3−λ 5 6 4−λ 2 = (3 − λ)(λ − 8)(λ − 3) = 0 if λ is 3 or 8. 6.1.29 det(A − λI3 ) = det 0 0 2 7−λ

4−λ 2 0 6−λ 0 = (4 − λ)(6 − λ)(3 − λ) − 8(3 − λ) 6.1.30 det(A − λI3 ) = det 4 5 2 3−λ

= (3 − λ)(8 − λ)(2 − λ) = 0 if λ is 3, 8 or 2.

6.1.31 This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 24. 6.1.32 This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 210. 6.1.33 The determinant of this block matrix is det 6.1.5. 6.1.34 The determinant of this block matrix is det 6.1.5.

1 8

4 3

2 7

5 6

det

det

2 3 7 5

1 4 2 3

= (7 − 16) (10 − 21) = 99,by Theorem

= (24 − 15) (3 − 8) = −45, by Theorem

6.1.35 There are two patterns with a nonzero product, (a12 , a23 , a31 , a44 ) = (3, 2, 6, 4), with two inversions, and (a12 , a23 , a34 , a41 ) = (3, 2, 3, 7), with 3 inversions. Thus det A = 3 · 2 · 6 · 4 − 3 · 2 · 3 · 7 = 18. 6.1.36 There is one pattern with a nonzero product, containing all the 1’s, with six inversions. Thus det A = 1. 6.1.37 The determinant block matrix is of this 5 6 7 5 4 det det 0 1 2 = (35 − 24) (5 · 1 · 1) = 55, by Theorem 6.1.5. 6 7 0 0 1 6.1.38 The 1 det 3 2

determinant of this block matrix is 2 3 6 5 0 4 det = (2 · 4 · 2 + 3 · 3 · 1 − 1 · 4 · 1 − 2 · 3 · 2) (6 · 6 − 5 · 5) = 99, by Theorem 6.1.5. 5 6 1 2

6.1.39 There is only one pattern with a nonzero product, containing all the nonzero entries of the matrix, with eight inversions. Thus det A = 1 · 2 · 4 · 3 · 5 = 120. 268

Section 6.1 6.1.40 There is only one pattern with a nonzero product, containing all the nonzero entries of the matrix, with seven inversions. Thus det A = −3 · 2 · 4 · 1 · 5 = −120. 6.1.41 There are two patterns with a nonzero product, (a15 , a24 , a32 , a41 , a53 ) = (2, 2, 3, 2, 3), with eight inversions, and (a13 , a24 , a32 , a41 , a55 ) = (1, 2, 3, 2, 4), with five inversions. Thus det A = 2 · 2 · 3 · 2 · 3 − 1 · 2 · 3 · 2 · 4 = 24. 6.1.42 There is only one pattern with a nonzero product, (a13 , a24 , a32 , a45 , a51 ) = (2, 2, 9, 5, 3), with six inversions. Thus det A = 2 · 2 · 9 · 5 · 3 = 540. 6.1.43 For each pattern P in A, consider the corresponding pattern Popp in −A, with all the n entries being opposites. Then prod (Popp ) = (−1)n prod(P ) and sgn (Popp ) = sgn (P ), so that det(−A) = (−1)n det A. 6.1.44 For each pattern P in A, consider the corresponding pattern Pm in kA, with all the n entries being multiplied by the scalar k. Then prod (Pm ) = k n prod(P ) and sgn (Pm ) = sgn (P ), so that det(kA) = k n det A. 6.1.45 If A =

6.1.46 Let A =

a c a b = ad − cb = det(A). It turns out that det(AT ) = det(A). , then det(AT ) = det b d c d

a1 a3

a2 . If a1 a4 − a2 a3 6= 0, then A−1 = a4

By Exercise 44, det(A−1 ) =

1 det(A)

2

(a1 a4 − a2 a3 ) =

1 det(A)

1 det(A)

a4 −a3

2

−a2 . a1

· det(A) so det(A−1 ) =

1 det(A) .

6.1.47 We have det(A) = (ah − cf )k + bef + cdg − aeg − bdh. Thus matrix A is invertible for all k if (and only if) the coefficient (ah − cf ) of k is 0, while the sum bef + cdg − aeg − bdh is nonzero. A numerical example is a = c = d = f = h = g = 1 and b = e = 2, but there are infinitely many other solutions as well. 0 0 0 1 0 0 1 0 so det(A) = det(B) = det(C) = det(D) = 0 ,D = ,C = ,B = 0 0 1 0 0 0 0 1 A B hence det(A) det(D) − det(B) det(C) = 0 but det = −1. C D

6.1.48 Consider A =

6.1.49 The kernel of T consists of all vectors ~x such that the matrix [~x ~v w] ~ fails to be invertible. This is the case if ~x is a linear combination of ~v and w ~ as discussed on Pages 249 and 250. Thus ker(T ) = span(~v , w). ~ The image of T isn’t {0}, since T (~v × w) ~ 6= 0, for example. Being a subspace of R, the image must be all of R. 6.1.50 Theorem 6.1.1 tells us that det[~u ~v w] ~ = ~u · (~v × w) ~ = k~uk cos(θ)k~v × wk ~ = k~uk cos(θ)k~v k sin(α)kwk ~ = cos(θ) sin(α), where θ is the angle enclosed by vectors ~u and ~v × w, ~ and α is the angle between ~v and w. ~ Thus det[~u ~v w] ~ can be any number on the closed interval [−1, 1]. 6.1.51 Let aii be the first entry on the diagonal that fails to belong to the pattern. The pattern must contain an entry in the ith row to the right of aii , above the diagonal, and also an entry in the ith column below aii , below the diagonal.

~ ~v 6.1.52 By Definition 6.1.1, we have det ~v × w

2 w ~ = (~v × w) ~ · (~v × w) ~ = k~v × wk ~ .

269

Chapter 6 6.1.53 There is one pattern with a nonzero product, containing all the 1’s. We have n2 inversions, since each of the 1’s in the lower left block forms an inversion with each of the 1’s in the upper right block. Thus det A = n2 n (−1) = (−1) . 5

6.1.54 The pattern containing all the 1000’s has 4 inversions so it contributes (1000) = 1015 to the determinant. There are 5! − 1 = 119 other patterns with at most 3 entries being 1000, the others being ≤ 9. Thus the product 3 2 associated with each of those patterns is less than(1000) (10) = 1011 . Now det A > 1015 − 119 · 1011 > 0. 6.1.55 By Exercise 2.4.93, a square matrix admits an LU factorization if (and only if) all its principal submatrices are invertible. Now 7 4 2 7 4 A(1) = [7], A(2) = , A(3) = A = 5 3 1 , 5 3 3 1 4 with det(A(1) ) = 7, det(A(2) ) = 1, det(A(3) ) = 1. Since all principal submatrices turn out to be invertible, the matrix A does indeed admit an LU factorization. 6.1.56 There is only one pattern with a nonzero product, containing all the 1’s. The number of inversions is Pn−1 . This number is even if either n or n − 1 is divisible by 4, that (n − 1) + (n − 2) + ... + 2 + 1 = k=1 k = (n−1)n 2 is, for n = 4, 5, 8, 9, 12, 13, .... a. det M4 = det M5 = 1, b. det Mn = (−1)

n(n−1)/2

det M2 = det M3 = det M6 = det M7 = −1.

6.1.57 In a permutation matrix P, there is only one pattern with a nonzero product, containing all the 1’s. Depending on the number of inversions in that pattern, we have det P = 1 or det P = −1. 6.1.58 a If a, b, c, d are prime numbers, then ad 6= bc, since the prime factorization of a positive integer is distinct a b 6= 0: No matrix of the required form exists. unique. Thus det c d b We are looking for a noninvertible matrix A = [~u ~v w] ~ whose entries are nine distinct prime numbers. The last column vector, w, ~ must be redundant; to keep things simple, we will make w ~ = ~u + 2~v . Now we have to pick six distinct prime entries for the first two columns, ~u and ~v , suchthat the entries ~ = ~u + 2~v are prime as well. of w 7 2 11 This can be done in many different ways; one solution is A = 17 3 23 . 19 5 29 6.1.59 F

a b c d

=

a c

b · = ab + cd d

a. Yes, F is linear in both columns. To prove linearity in the second column, observe that ab + cd is a linear combination of the variables b and d, with the constant coefficients a and c. An analogous argument proves linearity in the first column. b. No, since ab + cd fails to be a linear combination of a and b. c. No, F

w ~ ~v

=F

~v

w ~

= ~v · w: ~ swapping the columns leaves F unchanged. 270

Section 6.1 6.1.60 The functions in parts a, b, and d are linear in both columns; the functions in parts a, c, and d are linear in both rows; and F (A) = − det A in part d is alternating on the columns. For example, the function F (A) = cd in part b is linear in the first column since cd is a linear combination of the entries a and c in the first column. However, F (A) = cd fails to be linear in the cd fails second row since to be a linear combination of the entries c a b b a and d in the second row. Furthermore, F =F = cd, showing that F fails to be alternating c d d c on the columns. 6.1.61 The function F (A) = bf g is linear in all three columns and in all three rows since the product bf g contains exactly one factor from each row and from each column; it is the product associated with a pattern. For example, F is linear in the second row since bf g is a scalar multiple of f and combination d, e, f of the entries thus a linear 1 0 0 0 1 0 in the second row. F fails to be alternating; for example, F 0 0 1 = 1 but F 0 0 1 = 0 after 0 1 0 1 0 0 swapping the first two columns. 6.1.62 If A =

a a c c

a a , then D(A) = D c c

−D = |{z} swap columns

a a c c

= −D(A), so that D(A) = 0.

a b a b a 0 1 1 1 0 = D + D = ab D + ad D = ad. In Step 1, we 0 d |{z} 0 0 0 d |{z} 0 0 0 1 |{z} Step 2 Step 3 Step1 b b 0 a 1 write = + and use linearity in the second column. In Step 2, we write =a etc. d 0 d 0 0 and use linearity in both columns. In Step 3, we use Exercise 62 for the first summand and the given property D (I2 ) = 1 for the second summand.

6.1.63 D

a a 0 6.1.64 Writing = + and repeatedly using linearity in the columns, we find 0 c c a b a b 0 b 0 b 0 0 D =D +D = ad + D +D c d 0 d c d |{z} c 0 c d Step 2 0 1 0 0 1 0 = ad + bc D + cd D = ad − bc D = ad − bc = det A 1 0 1 1 |{z} 0 1 Step 4 a b = ad from Exercise See the analogous computations in Exercise 63. In Step 2, we are using the result D 0 d 63. In Step 4, we swap the columns of the matrix in the second summand and we apply Exercise 62 to the third summand. 6.1.65 Freely using the linearity and alternating properties in the columns, and omitting terms with two equal columns (see Exercise 62), we find 0 b c 0 b c 1 b c a b c D d e f = a D 0 e f +d D 1 e f +g D 0 e f 0 h i 0 h g h i 1 h i i 0 0 c 0 1 c 1 0 c 1 0 c = ae D 0 1 f + ah D 0 0 f + db D 1 0 f + dh D 1 0 f 0 1 i 0 0 i 0 1 i 0 0 i 271

Chapter 6

0 +gb D 0 1

1 0 0 0 1 0 + ahf D 0 0 0 1 0 1

1 0 0 c 1 c 0 f + ge D 0 1 f = aei D 0 0 1 0 i 0 i

0 1 0 dbi D 1 0 0 + dhc D 0 0 1 = aei − ahf − dbi + dhc + gbf

0 1 0

0 1 1 0 0 0

0 0 1 0 0 0 1 1 0 0 + gbf D 0 0 1 + gec D 0 1 1 0 0 0 1 0 − gec = det A

In the penultimate step, we perform up to two column swaps on each of the six matrices to make them I3 , recalling that D (I3 ) = 1.

a c

b d

1 1 0 0

1 0 0 1

+ + ad F = ab F 6.1.66 a. Proceeding as in Exercises 63 and 64, we can see that F 0 1 0 0 cb F + cd F for all functions F in V. Thus the functions ab, ad, cb, and cd span V . Apply 1 0 1 1 1 0 1 1 , , these functions to the matrices 0 1 0 0 0 1 0 0 , and to see that the four functions are linearly independent. Thus the functions ab, ad, cb, 1 0 1 1 and cd form a basis of V , and dim V = 4. a b 1 0 1 0 b. In Exercise 64 we see that the formula F =F (ad − bc) = F (det A) holds for all c d 0 1 0 1 functions F in W. Thus W is one-dimensional, with det A as a basis.

Section 6.2

1 1 1 1 6.2.1 A = 1 3 3 −I → B = 0 2 2 5 −2I 0

1 1 2 2 . Now det(A) = det(B) = 6, by Algorithm 6.2.5b. 0 3

1 2 3 1 6.2.2 A = 1 6 8 −I → B = 0 −2 −4 0 +2I 0 1 1 6.2.3 A = 1 2

3 6 3 6

2 4 0 4

4 1 8 −I 0 →B= 0 −I 0 12 −2I 0

1 −1 2 1 −1 2 6.2.4 A = 2 1 14 −2 6 10

1 0 0 0

−1 2 1 3 3 10 4 14

2 3 4 5 . Now det(A) = det(B) = 24, by Algorithm 6.2.5b. 0 6 3 2 4 3 2 4 . Now det(A) = det(B) = −24, by Algorithm 6.2.5b. 0 −2 −4 0 0 4

−2 6 +I → 10 −2I 33 +2I

−2 4 → 14 −3II 29 −4II 272

Section 6.2 1 0 0 0

−1 1 0 0

2 3 1 2

1 −2 4 0 →B= 0 2 0 13 −2III

−1 1 0 0

2 3 1 0

1 0 6.2.5 After three row swaps, we end up with B = 0 0 −24.

1 1 1 1 1 1 4 4 −I 6.2.6 A = → 1 −1 2 −2 −I 1 −1 8 −8 −I 1 1 1 1 0 0 3 3 swap: → 0 −2 1 −3 II ↔ III 0 −2 7 −9 1 1 1 1 0 −2 1 −3 ÷ − 2 → 0 0 3 3 0 −2 7 −9 1 1 1 1 1 3 0 1 −2 2 → 0 0 3 3 0 −2 7 −9 +2II 1 1 1 1 1 1 1 3 0 1 0 1 −2 2 → B= 0 0 3 3 0 0 0 0 6 −6 −2III 0 0

−2 4 . Now det(A) = det(B) = 9, by Algorithm 6.2.5b. 2 9 2 2 0 0

3 3 3 0

4 4 . Now, by Algorithm 6.2.5b, det(A) = (−1)3 det(B) = 4 4

1 − 21 3 0

1 3 2

3 −12

det(A) = − 21 (−1) det(B) = −72, by Algorithm 6.2.5b.

6.2.7 After two row swaps, we end up with an upper triangular matrix B with all 1’s along the diagonal. Now det(A) = (−1)2 det(B) = 1, by Algorithm 6.2.5b. 6.2.8 After four row swaps, we end up with an upper triangular matrix B with all 1’s along the diagonal, except for a 2 in the bottom right corner. Now det(A) = (−1)4 det(B) = 2, by Algorithm 6.2.5b. 6.2.9 If we subtract the first row from every other row, then we have an upper triangular matrix B, with diagonal entries 1, 1, 2, 3 and 4. Then det(A) = det(B) = 24 by Algorithm 6.2.5b. 1 1 6.2.10 A = 1 1 1

1 2 3 4 5

1 3 6 10 15

1 4 10 20 35

1 5 15 35 70

−I −I → −I −I 273

Chapter 6

1 0 0 0 0

1 1 1 2 2 5 3 9 4 14

1 0 0 0 0

1 1 0 0 0

1 2 1 3 6

1 3 3 10 22

1 0 0 0 0

1 1 0 0 0

1 2 1 0 0

1 1 1 3 4 0 3 6 → B = 0 1 4 0 4 17 −4IV 0

1 3 9 19 34

1 4 14 −2II → 34 −3II 69 −4II 1 4 → 6 22 −3III 53 −6III 1 1 0 0 0

1 2 1 0 0

1 3 3 1 0

1 4 6 . 4 1

Now det(A) = det(B) = 1 by Algorithm 6.2.5b. 6.2.11 By Theorem 6.2.3a, the desired determinant is (−9)(8) = −72. 6.2.12 By Theorem 6.2.3b, the desired determinant is −8. 6.2.13 By Theorem 6.2.3b, applied twice, since there are two row swaps, the desired determinant is (−1)(−1)(8) = 8. 6.2.14 By Theorem 6.2.3c, the desired determinant is 8. 6.2.15 By Theorem 6.2.3c, the desired determinant is 8. 6.2.16 This determinant is 0, since the first row is twice the last.

3 0 2 6 , so that det(T ) = det(A) = 8. 0 2

−2 4 3 −12 , so that det(T ) = det(A) = 27. 0 9

1 6.2.19 The standard matrix of T is A = 0 0

0 0 −1 0 , so that det(T ) = det(A) = −1. 0 1

1 0 6.2.20 The standard matrix of L is M = 0 0

0 0 1 0

2 6.2.17 The standard matrix of T is A = 0 0 1 6.2.18 The standard matrix of T is A = 0 0

0 1 0 0

0 0 , so that det(L) = det(M ) = −1. 0 1 274

Section 6.2 1 0 6.2.21 The standard matrix of T is A = 0 0

0 −1 0 0

0 0 0 0 , so that det(T ) = det(A) = 1. 1 0 0 −1

6.2.22 Using Exercises 19 and 21 as a guide, we observe that the standard matrix A of T is diagonal, of size (n + 1) × (n + 1), with diagonal entries (−1)0 , (−1)1 , (−1)2 , . . . , (−1)n . Thus det(T ) = det(A) = (−1)1+2+···+n = (−1)n(n+1)/2 . 6.2.23 Consider the matrix M of T with respect to a basis consisting of n(n+1)/2 symmetric matrices and n(n−1)/2 skew-symmetric matrices (see Exercises 54 and 55 or Section 5.3). Matrix M will be diagonal, with n(n + 1)/2 entries 1 and n(n − 1)/2 entries -1 on the diagonal. Thus, det(T ) = det(M ) = (−1)n(n−1)/2 . 6.2.24 The standard matrix of T is A =

2 3

2 6.2.25 The standard matrix of T is A = 0 0

−3 , so that det(T ) = det(A) = 13. 2 0 0 2 3 , so that det(T ) = det(A) = 16. 0 4

6.2.26 The matrix of T with respect to the basis

1 0 0 , 0 0 1

1 0 , 0 0

det(A) = −16.

0 1

2 4 is A = 2 4 0 4

0 2 , so that det(T ) = 6

−b a 6.2.27 The matrix of T with respect to the basis cos(x), sin(x) is A = , so that det(T ) = det(A) = a2 +b2 . −a −b −3 −2 −6 −10 , so that det(T ) = det(A) = 14. 0 is A = 1 , 6.2.28 The matrix of T with respect to the basis 5 6 1 0

6.2.29 Expand down the first column, realizing that all but the first contribution are zero, since a21 = 0 and Ai1 has two equal rows for all i > 2. Therefore, det(Pn ) = det(Pn−1 ). Since det(P1 ) = 1, we can conclude that det(Pn ) = 1, for all n. 1 1 1 6.2.30 a f (t) = det a b t = (ab2 − a2 b) + (a2 − b2 )t + (b − a)t2 so f (t) is a quadratic function of t. The a2 b2 t2 2 coefficient of t is (b − a).

b In the cases t = a and t = b the matrix has two identical columns. It follows that f (t) = k(t − a)(t − b) with k = coefficient of t2 = (b − a). c The matrix is invertible for the values of t for which f (t) 6= 0, i.e., for t 6= a, t 6= b.

1 6.2.31 a If n = 1, then A = a0

1 , so det(A) = a1 − a0 (and the product formula holds). a1 275

Chapter 6 b Expanding the given determinant down the right-most column, we see that the coefficient k of tn is the n − 1 Vandermonde determinant which we assume is Y (ai − aj ). n−1≥i>j

Now f (a0 ) = f (a1 ) = · · · = f (an−1 ) = 0, since in each case the given matrix has two identical columns, hence its determinant equals zero. Therefore Y f (t) = (ai − aj ) (t − a0 )(t − a1 ) · · · (t − an−1 ) n−1≥i>j

and

det(A) = f (an ) =

Y

n≥i>j

(ai − aj ),

as required. 6.2.32 By Exercise 31, we need to compute

Q

i>j

Q

i>j

(ai − aj ) where a0 = 1, a1 = 2, a2 = 3, a3 = 4, a4 = 5 so

(ai − aj ) = (2 − 1)(3 − 1)(3 − 2)(4 − 1)(4 − 2)(4 − 3)(5 − 1)(5 − 2)(5 − 3)(5 − 4) = 288.

th

6.2.33 Think of the i

1 ai 2 a column of the given matrix as ai .i . .

1 a1 2 a be written as (a1 a2 · · · an ) det .1 . .

a1n−1 (see Exercise 6.2.31), and we get

1 a2 a22 .. .

··· ··· ···

1 an a2n .. .

a2n−1

···

ann−1

n Y

i=1

ai

Y

i>j

ain−1

, so, by Theorem 6.2.3a, the determinant can

. The new determinant is a Vandermonde determinant (ai − aj ).

6.2.34 a The hint pretty much gives it away. Since the columns of matrix B have [ In M ] = B − M = 0, and M = B, as claimed. −In

B −In

are in the kernel of [ In

. . b If B = A−1 we get rref[A..In ] = [In ..A−1 ] which tells us how to compute A−1 (see Theorem 2.4.5).

6.2.35

x1 x2

1 must satisfy det x1 x2

1 a1 a2

1 b1 = 0, i.e., must satisfy the linear equation b2 276

M ], we

Section 6.2 (a1 b2 − a2 b1 ) − x1 (b2 − a2 ) + x2 (b1 − a1 ) = 0. b x1 a1 x1 = 1 satisfy this equation, since the matrix has two identical columns and = We can see that b2 x2 a2 x2 in these cases. 6.2.36 Expanding down the first column we see that the equation has the form A − Bx1 + Cx2 − D(x21 + x22 ) = 0. If D 6= 0 this equation defines it is a line. From Exercise 35 acircle; otherwise a1 b1 c1 we know that D = 0 if and only if the three given points , , are collinear. Note that the circle a2 b2 c2 or line runs through the three given points. 6.2.37 Applying Theorem 6.2.6 to the equation AA−1 = In we see that det(A) det(A−1 ) = 1. The only way the product of the two integers det(A) and det(A−1 ) can be 1 is that they are both 1 or both -1. Therefore, det(A) = 1 or det(A) = −1. 6.2.38 det(AT A) = det(AT ) det(A) = [det(A)]2 = 9 ↑

↑

Theorem 6.2.6

Theorem 6.2.1 2

6.2.39 det(AT A) = det(AT ) det(A) = [det(A)] > 0 ↑

↑

Theorem 6.2.6

Theorem 6.2.1

6.2.40 By Exercise 38, det(AT A) = [det(A)]2 . Since A is orthogonal, AT A = In so that 1 = det(In ) = det (AT A) = [det(A)]2 and det(A) = ±1. 6.2.41 det(A) = det(AT ) = det(−A) = (−1)n (det A) = − det(A), so that det(A) = 0. We have used Theorems 6.2.1 and 6.2.3a. T 6.2.42 det AT A |{z} = det (QR) QR = det RT QT QR |{z} = det RT Im R Step 1 Step 3 Qm 2 2 T T = det R R |{z} = det R det(R) |{z} = (det R) |{z} = ( i=1 rii ) > 0. Step 5

Step 6

Step 7

In Step 1 we are using the definition of matrix A. Equation 3 holds since the column vectors of matrix Q are orthonormal. In Steps 5 and 6 we are using Theorems 6.2.6 and 6.2.1, respectively. Finally, Equation 7 holds since matrix R is triangular. T

6.2.43 det(A A) = det

~v T w ~T

~v · ~v [~v w] ~ = det ~v · w ~

~v · w ~ k~v k2 = det w ~ ·w ~ ~v · w ~

~v · w ~ kwk ~ 2

= k~v k2 kwk ~ 2 − (~v · w) ~ 2 ≥ 0 by the Cauchy-Schwarz inequality (Theorem 5.1.11). 6.2.44 a We claim that ~v2 ×~v3 ×· · ·×~vn 6= ~0 if and only if the vectors ~v2 , . . . , ~vn are linearly independent. If the vectors ~v2 , . . . , ~vn are linearly independent, then we can find a basis ~x, ~v2 , . . . , ~vn of Rn (any vector ~x that is not in span (~v2 , . . . , ~vn ) will do). Then ~x·(~v2 ×· · ·×~vn ) = det[~x ~v2 · · · ~vn ] 6= 0, so that ~v2 ×· · ·×~vn 6= ~0. Conversely, suppose that 277

Chapter 6 ~v2 ×~v3 ×· · ·×~vn 6= 0; say the ith component of this vector is nonzero. Then 0 6= ~ei ·(~v2 ×· · ·×~vn ) = det[~ei ~v2 · · · ~vn ], so that the vectors ~v2 , . . . , ~vn are linearly independent (being columns of an invertible matrix).

1 b ith component of ~e2 × ~e3 × · · · × ~en = det ~ei 1

1 ~e2 1

so ~e2 × ~e3 × · · · × ~en = ~e1 .

1 c ~vi · (~v2 × ~v3 × · · · × ~vn ) = det ~vi 1

1 ~v2 1

1 ~v3 1

1 n 1 · · · ~en = 0 1

if i = 1 if i > 1

1 · · · ~vn = 0 1

for any 2 ≤ i ≤ n since the above matrix has two identical columns.

d Compare the 1 1 det ~ei ~v2 1 1

ith components of the two vectors: 1 1 1 1 1 ~v3 · · · ~vn and det ~ei ~v3 ~v2 1 1 1 1 1

1 · · · ~vn · 1

The two determinants differ by a factor of −1 by Theorem 6.2.3b, so that ~v2 × ~v3 × · · · × ~vn = −~v3 × ~v2 × · · · × ~vn .

e det[~v2 × ~v3 × · · · × ~vn ~v2 ~v3 · · · ~vn ] = (~v2 × ~v3 × · · · × ~vn ) · (~v2 × ~v3 × · · · × ~vn ) = k~v2 × · · · × ~vn k2 f In Definition 6.1.1 we saw that the “old” cross product satisfies the defining equation of the “new” cross product: ~x · (~v2 × ~v3 ) = det [ ~x ~v2 ~v3 ]. 6.2.45 f (x) is a linear function, so f ′ (x) 1 0 that the coefficient of x is − det 0 0

a 3 6.2.46 a det b 3 c 3

is 2 2 0 0

the 3 3 3 0

coefficient of x (the slope). Expanding down the first column, we see 4 4 = −24, so f ′ (x) = −24. 4 4

a 1 d d e = 3 det b 1 e = 21. We have used Theorem 6.2.3. c 1 f f

a 1 a 2(1) d a 2(1) + 1 d a 3 d b det b 5 e = det b 2(2) + 1 e = det b 2(2) e +det b 1 c 1 c 2(3) f c 2(3) + 1 f c 7 f 2 · 11 + 7 = 29. We have used Theorem 6.2.2 and Theorem 6.2.3a.

6.2.47 Yes! For example, T

a 1 a 1 d d e = 2 det b 2 e +det b 1 c 1 c 3 f f

x b = dx + by is given by the matrix [d b], so that T is linear in the first column. y d

6.2.48 Since ~v2 , . . . , ~vn are linearly independent, T (~x) = 0 only if ~x is a linear combination of the ~vi ‘s, (otherwise the matrix [~x ~v2 · · · ~vn ] is invertible, and T (~x) 6= 0). Hence, the kernel of T is the span of ~v2 , . . . , ~vn , an (n − 1)dimensional subspace of Rn . The image of T is the real line R (since it must be 1-dimensional). 278

d e= f

Section 6.2 1 1 1 6.2.49 For example, we can start with an upper triangular matrix B with det(B) = 13, such as B = 0 1 1 . 0 0 13 Adding the first row of B to both the second and the third to make all entries nonzero, we end up with 1 1 1 A = 1 2 2 . Note that det(A) = det(B) = 13. 1 1 14

6.2.50 There are many ways to do this problem; here is one possible approach: Subtracting the second to last row from the last, we can make the last row into [0

0

···

0

1 ].

Now expanding along the last row we see that det(Mn ) = det(Mn−1 ). Since det(M1 ) = 1 we can conclude that det(Mn ) = 1 for all n. 6.2.51 Notice that it takes n row swaps (swap row i with n + i for each i between 1 and n) to turn A into I2n . So, det(A) = (−1)n det(I2n ) = (−1)n . 6.2.52 a We build B column-by-column: d −b −b d 0 1 . = = T T −c a a −c 1 0 b det(A) = ad − bc = det(B). The two determinants are equal.

d −b a b da − bc 0 c BA = = = (ad − bc)I2 . −c a c d 0 −cb + ad ad − bc 0 d −b a b = (ad − bc)I2 also. = AB = 0 −cb + da −c a c d a d −b a b a = = . We note that B + c2 d Any vector ~u in the image of A will be of the form c1 c −c a c d c ad − bc b = ~0. The same is true of B . Thus, anything in the image of A will be in the kernel of B. Since −ca + ac d both matrices have a rank of 1, the dimensions of the kernel and image of each will be exactly 1. So, it must be that im(A) = ker(B). −b a b −b d = . However, we see that +c2 Also, any vector ~u in the image of B will be of the form c1 a c d a −c −ab + ba d = ~0. The same is true for A . Thus, by the same reasoning as above, the image of B will equal −bc + ad −c the kernel of A. e A−1 =

1 ad−bc

d −b = −c a

1 ad−bc B.

6.2.53 a See Exercise 37. 279

Chapter 6

b If A =

a b , then A−1 = c d

1 det(A)

6.2.54 f (t) = (det(A + tB))2 − 1 =

d −b −c a

has integer entries.

a + tb1 det 1 a3 + tb3

a2 + tb2 a4 + tb4

2

− 1 assuming A =

a1 a3

b a2 ,B= 1 b3 a4

b2 . b4

Then the determinant above is a polynomial of degree ≤2 so its square is a polynomial of degree ≤4. Hence f (t) is a polynomial of degree ≤4. Since A, A + B, A + 2B, A + 3B, A + 4B are invertible and their inverses have integer entries, by Exercise 53a, it follows that their determinants are ±1. Hence f (0) = f (1) = f (2) = f (3) = f (4) = 0. Since f is a polynomial of degree ≤ 4 with at least 5 roots, it follows that f (t) = 0 for all t, in particular for t = 5, so det(A + 5B) = ±1. Hence A + 5B is an invertible 2 × 2 matrix whose inverse has integer entries by Exercise 53b. 6.2.55 We start out with a preliminary remark: If a square matrix A has two equal rows, then D(A) = 0. Indeed, if we swap the two equal rows and call the resulting matrix B, then B = A, so that D(A) = D(B) = −D(A), by property b, and D(A) = 0 as claimed. Next we need to understand how the elementary row operations affect D. Properties a and b tell us about how row multiplications and row swaps, but we still need to think about row additions. We will show that if B is obtained from A by adding k times the ith row to the j th , then D(B) = D(A). Let’s label the row vectors of A by ~v1 , . . . ~vn . By linearity of D in the j th row (property c) we have .. .. . . ~ v ~ v i i . . .. .. D(B) = D = D(A) + kD = D(A). ~ v + k~ v ~ v j i i .. .. . .

Note that in the last step we have used the preliminary remark. Now, using the terminology introduced on Page 265, we can write D(A) = (−1)s k1 k2 · · · kr D(rref A).

Next we observe that D(rref A) = det(rref A) for all square matrices A. Indeed, if A is invertible, then rref(A) = In , and D(In ) = 1 = det(In ) by property c of function D. If A fails to be invertible, then D(rref A) = 0 = det(rref A) by linearity in the last row. It follows that D(A) = (−1)s k1 k2 · · · kr D(rref A) = (−1)s k1 k2 · · · kr det(rref A) = det(A) for all square matrices, as claimed. 6.2.56 a We show first that D is linear in the i th row. ~v1 M ~v1 . . . . . . ~xM ~x . . 1 D .. = det M det .. ~vn M ~vn ↑ ↑ AM A 280

Section 6.2 The entries in the ith row of AM are linear combinations of the components xi of the vector ~x, while the other entries of AM are constants. Therefore, ) is a linear combination of the xi (expand along the ith row). det(AM ~v1 .. . Since det1M is a constant, we have D ~x = c1 x1 + c2 x2 + · · · + cn xn for some constants ci , as claimed. . .. ~vn c The property D(In ) = 1 is obvious. det(AM ) det(M )

It now follows from Exercise 41 that det(A) = D(A) = det(AM ) = det(A) det(M ).

and therefore

6.2.57 Note that matrix A1 is invertible, since det(A1 ) 6= 0. Now ~y ~y T = [A1 A2 ] = A1 ~y + A2 ~x = ~0 when A1 ~y = −A2 ~x, or, ~x ~x ~y = −A−1 x. This shows that for every ~x there is a unique ~y (that is, ~y is a function of ~x); furthermore, this 1 A2 ~ function is linear, with matrix M = −A−1 1 A2 .

1 2 1 2 6.2.58 Using the approach of Exercise 57, we have A1 = , A2 = , 3 7 4 3 1 −8 y1 1 −8 x1 and M = −A−1 A = . The function is = . 2 1 y2 −1 3 x2 −1 3 Alternatively, we can solve the linear system y1 + 2y2 + x1 + 2x2 3y1 + 7y2 + 4x1 + 3x2

=0 =0

Gaussian Elimination gives y1 − x1 + 8x2 = 0

and

y2 + x1 − 3x2 = 0

0 6.2.59 det 0

y1 = x1 − 8x2 y2 = −x1 + 3x2

0 1 0 = 0 = det , but these matrices fail to be similar. 0 0 0

6.2.60 We argue using induction on n. The base case (n = 2) is discussed in the text. Now we assume that B is obtained from the n × n matrix A by adding k times the pth row to the q th row. We will evaluate the determinant of B by expanding across the ith row (where i is neither p nor q). det(B) = Σnj=1 (−1)i+j bij det(Bij ) = Σnj=1 (−1)i+j aij det(Bij ) = Σnj=1 (−1)i+j aij det(Aij ) = det(A) 281

Chapter 6 Note that the (n − 1) × (n − 1) matrix Bij is obtained from Aij by adding k times some row to another row. Now, det(Bij ) = det(Aij ) by induction hypothesis. A B A B In 0 = 6.2.61 We follow the hint: −CA + AC −CB + AD C D −C A A B In 0 A B = . So, det = det(A) det(AD − CB). 0 AD − CB −C A C D A B = det(A) det(AD − CB), which leads to Thus, det(In ) det(A) det C D A B det = det(AD − CB), since det(A) 6= 0. C D

A B A B In 0 6.2.62 a We compute = . Since the matrix is invert−1 C D 0 −CA−1 B −CA +D In A B A B ible (its determinant is 1),the product will have the same rank as , namely, n. 0 −CA−1 B + D C D With A being invertible, this implies that −CA−1 B + D = 0, or CA−1 B = D, as claimed. In −CA−1

0 In

b Take determinants on both sides of the equation D = CA−1 B from part (a) to find that det(D) = det(C)(det A)−1 det(B), or det(A) det(D) − det(B) det(C) = 0, proving the claim. 6.2.63 Let Mn be the number of multiplications required to compute the determinant of an n × n matrix by Laplace expansion. We will use induction on n to prove that Mn > n!, for n ≥ 3. In the lowest applicable case, n = 3, we can check that M3 = 9 and 3! = 6. Now let’s do the induction step. If A is an n × n matrix, then det(A) = a11 det(A11 ) + · · · + (−1)n+1 an1 det(An1 ), by Theorem 6.2.10. We need to compute n determinants of (n − 1) × (n − 1) matrices, and then do n extra multiplications ai1 det(Ai1 ), so that Mn = nMn−1 + n. If n > 3, then Mn−1 > (n − 1)!, by induction hypothesis, so that Mn > n(n − 1)! + n > n!, as claimed. 6.2.64 To compute det(A) for an n × n matrix A by Laplace expansion, det(A) = a11 det(A11 ) − a21 det(A21 ) + · · · + (−1)n+1 an1 det(An1 ), we first need to compute the n minors, which requires nLn−1 operations; then we compute the n products ai1 det(Ai1 ); and finally we have to do n − 1 additions. Altogether, Ln = nLn−1 + n + (n − 1) = nLn−1 + 2n − 1. Now we can prove the formula

Ln n!

=1+1+

1 2!

+

1 3!

+ ··· +

1 (n−1)!

−

1 n!

by induction on n.

For n = 2, the formula gives L22 = 1 + 1 − 21 , or L2 = 3, which is correct: We have to perform 2 multiplications and 1 addition to compute the determinant of a 2 × 2 matrix. nLn−1 +2n−1 Ln−1 Ln = (n−1)! n! = n! Ln−1 2 1 Applying the induction hypothesis to the first summand, we find that Ln!n = (n−1)! + (n−1)! − n! 1 1 1 2 1 1 1 1 1 3! + · · · + (n−2)! − (n−1)! + (n−1)! − n! = 1 + 1 + 2! + 3! + · · · + (n−1)! − n! , as claimed.

For an n × n matrix A we can use the recursive formula derived above to see that 1 − n! . 1 + 2! +

2 (n−1)!

1+1

Now recall from the theory of Taylor series in calculus that e = e1 = 1 + 1 + 1 1 1 1 Thus Ln = (1 + 1 + 2! + 3! + · · · + (n−1)! − n! )n! < en!, as claimed. 282

1 2!

+

1 3!

+ ··· +

1 (n−1)!

+

1 n!

+ =

+ · · ·.

Section 6.3

6.2.65 a. Using Laplace expansion along the first row, we find dn = det (Mn ) = det (Mn−1 ) − det det (Mn−1 ) + det (Mn−2 ) = dn−1 + dn−2 . b. d1 = 1, d2 = 2, d3 = 3, d4 = 5, ..., d10 = 89 c. Since dn > 0 for all positive integers n, the matrix Mn is always invertible.

6.2.66 a. Using Laplace expansion along the first row, we find dn = det (Mn ) = det (Mn−1 ) − det

−1 ∗ 0 Mn−2

=

=

1 ∗ 0 Mn−2

det (Mn−1 ) − det (Mn−2 ) = dn−1 − dn−2 . b. d1 = 1, d2 = 0, d3 = −1, d4 = −1, d5 = 0, d6 = 1, d7 = 1, d8 = 0. c. Since d4 = −d1 and d5 = −d2 , the formula dn+3 = −dn holds for all positive integers n. (Give a formal proof by induction.) Now dn+6 = −dn+3 = − (−dn ) = dn for all positive integers n, meaning that the sequence dn has a period of six. 6.2.67 Let k be the number of pattern entries to the left and above aij . Then the number of pattern entries to the right and above aij is i − 1 − k, since there are i − 1 rows above aij , each of which contains exactly one pattern entry. Likewise, the number of pattern entries to the left and below aij is j − 1 − k. Now the number #(inversions involving aij ) of inversions involving aij is (i − 1 − k) + (j − 1 − k) = i + j − 2 − 2k. It follows that (−1) i+j−2−2k i+j = (−1) = (−1) . This means that the number of inversions involving aij is even if (and only if) i + j is even, as claimed. 6.2.68 Using Exercise 67 and the terminology introduced in the proof of Theorem 6.2.10, we have #(inversions in P ) #(inversions involving aij ) #(inversions not involving aij ) sgnP = (−1) = (−1) (−1) = (−1)

i+j

(−1)

#(inversions in Pij )

= (−1)

i+j

sgn (Pij ) .

Section 6.3 6.3.1 By Theorem 2.4.10, the area equals | det 6.3.2 By Theorem 2.4.10, Area =

6.3.3 Area of triangle = 21 | det

1 2

6 −2

3 8 | = | − 50| = 50. 7 2

3 8 det = 12 |−50| = 25 7 2 1 | = 13 (See Figure 6.1.) 4

b1 − a1 c1 − a1 6.3.4 Note that area of triangle = det . (See Figure 6.2.) b2 − a2 c2 − a2 a1 b1 − a1 c1 − a1 a1 b1 c1 On the other hand, det a2 b2 c2 = det a2 b2 − a2 c2 − a2 , by subtracting the first column from the 1 0 0 1 1 1 second and third. b1 − a1 c1 − a1 , by expanding across the bottom row. This, in turn, equals det b2 − a2 c2 − a2 a1 b1 c1 Therefore, area of triangle = 12 det a2 b2 c2 . 1 1 1 1 2

283

Chapter 6

Figure 6.1: for Problem 6.3.3.

Figure 6.2: for Problem 6.3.4. 6.3.5 The volume of the tetrahedron T0 defined by ~e1 , ~e2 , ~e3 is 13 (base)(height) = 16 . Here we are using the formula for the volume of a pyramid. (See Figure 6.3.)

Figure 6.3: for Problem 6.3.5. The tetrahedron T defined by ~v1 , ~v2 , ~v3 can be obtained by applying the linear transformation with matrix [~v1 ~v2 ~v3 ] to T0 . Now we have vol(T ) = | det[~v1 ~v2 ~v3 ]|vol(T0 ) = 16 | det[~v1 ~v2 ~v3 ]| = 16 V (~v1 , ~v2 , ~v3 ). ↑

↑

Theorem 6.3.7 and Page 283

6.3.6 From Exercise 5 we know that

area of triangle =

1 2

a1 det a2 1

b1 b2 1

Theorem 6.3.4 and Page 280

a1 b1 c1 volume of tetrahedron = 61 det a2 b2 c2 , and Exercise 4 tells us that 1 1 1 c1 c2 , so that area of tetrahedron = 31 (area of triangle). 1

284

Section 6.3 We can see this result more directly if we think of the tetrahedron as an inverted pyramid whose base is the triangle and whose height is 1. (See Figure 6.4.)

Figure 6.4: for Problem 6.3.6. c1 b1 a1 The three vertices of the shaded triangle are a2 , b2 , c2 . 1 1 1

6.3.7 Area =

1 2

10 det 11

−2 13

+

1 2

8 det 2

10 11

= 110. (See Figure 6.5.)

Figure 6.5: for Problem 6.3.7. 6.3.8 We need to show that both sides of the equation in Theorem 6.3.3 give zero. | det(A)| = 0 since A is not invertible. On the other hand, since A is not invertible, the ~vi will be linearly k dependent, i.e., one of the ~vi will be redundant. This implies that ~vi = ~vi and ~vi⊥ = ~0, so that the right-hand side of the equation is 0, as claimed. 6.3.9 Using linearity in the second column, we find that det det

~v1

~v2⊥ . Thus the two determinants are equal.

285

~v1

~v2

= det

h

~v1

i i h k ~v2 + ~v2⊥ = det ~v1 ~v2k + det ~v1 {z } | noninvertible

~v

Chapter 6 6.3.10 | det(A)| ≤ k~v1 kk~v2 k · · · k~vn k since | det(A)| = k~v1 kk~v2⊥ k · · · k~vn⊥ k and k~vi k ≥ k~vi⊥ k. The equality holds if k~vi k = k~vi⊥ k for all i, that is, if the ~vi ‘s are mutually perpendicular. 6.3.11 The matrix of the transformation T with respect to the basis ~v1 , ~v2 is B = det(B) = 12, by Theorem 6.2.7.

3 0 , so that det(A) = 0 4

6.3.12 Denote the columns by ~v1 , ~v2 , ~v3 , ~v4 . By Theorem 6.3.3 and Exercise 6.3.8 we know that | det(A)| ≤ k~v1 kk~v2 k k~v3√ kk~v4 k; equality holds if the columns are orthogonal. Since the entries of the ~vi are 0, 1, and −1, we have k~vi k ≤ 1 + 1 + 1 + 1 = 2. Therefore, | det A| ≤ 16. To build an example where det(A) = 16 we want all 1’s and −1’s as entries, and the columns need to be 1 1 1 1 1 −1 −1 1 orthogonal. A little experimentation produces A = (there are other solutions). Note that 1 −1 −1 1 1 −1 1 −1 we need to check that det(A) = 16 (and not −16). 6.3.13 By Theorem 6.3.6, the v u u 1 u 1 1 1 1 1 udet t 1 2 3 4 1 1

desired 2-volume is 1 s 2 4 = det 3 10 4

√ 10 = 20. 30

6.3.14 By Theorem 6.3.6, the desired 3-volume is v u 1 1 1 v u u 1 1 1 1 0 0 0 u u √ 0 1 2 u udet 4 10 = 6. 1 1 1 1 = tdet 1 t 0 1 3 1 10 30 1 2 3 4 0 1 4 6.3.15 If ~v1 , ~v2 , . . . , ~vm are linearly dependent and if A = [~v1 · · · ~vm ], then det(AT A) = 0 since AT A and A have equal and nonzero kernels (by Theorem 5.4.2), hence AT A fails to be invertible. On the other hand, since the ~vi are linearly dependent, at least one of them will be redundant. For such a k redundant ~vi , we will have ~vi = p ~vi and ~vi⊥ = ~0, so that V (~v1 , . . . , ~vm ) = 0, by Definition 6.3.5. This discussion shows that V (~v1 , . . . , ~vm ) = 0 = det(AT A) if the vectors ~v1 , . . . , ~vm are linearly dependent.

6.3.16 False

If T is given by A = 2I3 then | det(A)| = 8. But if Ω is the square defined by ~e1 , ~e2 in R3 (of area 1), then T (Ω) is the square defined by 2~e1 , 2~e2 and the area of T (Ω) is 4. 6.3.17 a Let w ~ = ~v1 ×~v2 ×~v3 . Note that w ~ is orthogonal to ~v1 , ~v2 and ~v3 , by Exercise 6.2.44c. Then V (~v1 , ~v2 , ~v3 , w) ~ = V (~v1 , ~v2 , ~v3 )kw ~ ⊥ k = V (~v1 , ~v2 , ~v3 )kwk. ~ ↑ by Definition 6.3.5. 286

Section 6.3 b By Exercise 6.2.44e, V (~v1 , ~v2 , ~v3 , ~v1 × ~v2 × ~v3 )

= | det [ ~v1 ~v2 ~v3 ~v1 × ~v2 × ~v3 ] | . = | det [ ~v1 × ~v2 × ~v3 ~v1 ~v2 ~v3 ] | = k~v1 × ~v2 × ~v3 k2

c By parts a and b, V (~v1 , ~v2 , ~v3 ) = k~v1 × ~v2 × ~v3 k. If the vectors ~v1 , ~v2 , ~v3 are linearly dependent, then both sides of the equation are 0, by Exercise 15 and Exercise 6.2.44a.

Figure 6.6: for Problem 6.3.18a. p 0 cos(t) p · cos(t) p 0 6.3.18 a (See Figure 6.6.) = , the ellipse with semi-axis ± and ± . 0 q sin(t) q · sin(t) 0 q (area of the ellipse) = | det(A)|(area of the unit circle) = pqπ b By Theorem 6.3.7, | det(A)| =

area of the ellipse = area of the unit circle

abπ π

= ab so | det(A)| = ab.

c The unit circle consists of all vectors of the form ~x = cos(t) √12 consisting of all vectors √ 1 √ 1 T (~x) = cos(t) 2 2 + sin(t) 2 1 −1 | {z } | {z } . (See Figure 6.7.) semi-major axis semi-minor axis

1 1 + sin(t) √12 ; its image is the ellipse 1 −1

6.3.19 det[~v1 ~v2 ~v3 ] = ~v1 ·(~v2 ×~v3 ) = k~v1 kk~v2 ×~v3 k cos θ where θ is the angle between ~v1 and ~v2 ×~v3 so det[~v1 ~v2 ~v3 ] > 0 if and only if cos θ > 0, i.e., if and only if θ is acute (0 ≤ θ ≤ π2 ). (See Figure 6.8.) 6.3.20 By Exercise 19, ~v1 , ~v2 , ~v3 constitute a positively oriented basis if and only if det[~v1 ~v2 ~v3 ] > 0. Assume that ~v1 , ~v2 , ~v3 is such a basis. We want to show that A~v1 , A~v2 , A~v3 is positively oriented if and only if det(A) > 0. We have det[A~v1 A~v2 A~v3 ] = det(A[~v1 ~v2 ~v3 ]) = det(A) det[~v1 ~v2 ~v3 ] so since det [~v1 ~v2 ~v3 ] > 0 by assumption, det [A~v1 A~v2 A~v3 ] > 0 if and only if det(A) > 0. Hence A is orientation preserving if and only if det(A) > 0. 6.3.21 a Reverses 287

Chapter 6

Figure 6.7: for Problem 6.3.18c. v2 × v3

v1 θ

v3

v2

Figure 6.8: for Problem 6.3.19. Consider ~v2 and ~v3 in the plane (not parallel), and let ~v1 = ~v2 × ~v3 ; then ~v1 , ~v2 , ~v3 is a positively oriented basis, but T (~v1 ) = −~v1 , T (~v2 ) = ~v2 , T (~v3 ) = ~v3 is negatively oriented. b Preserves Consider ~v2 and ~v3 orthogonal to the line (not parallel), and let ~v1 = ~v2 ×~v3 ; then ~v1 , ~v2 , ~v3 is a positively oriented basis, and T (~v1 ) = ~v1 , T (~v2 ) = −~v2 , T (~v3 ) = −~v3 is positively oriented as well. c Reverses The standard basis ~e1 , ~e2 , ~e3 is positively oriented, but T (~e1 ) = −~e1 , T (~e2 ) = −~e2 , T (~e3 ) = −~e3 is negatively oriented. 6.3.22 Here A = det

x=

"

1 7 3 11 5

#

1 3 7 , so by Theorem 6.3.8 , det(A) = 5, ~b = 3 4 11 "

det

= −2, y =

3 4 5

1 3

#

= 1. 288

Section 6.3

6.3.23 Here A =

1 5 −3 , so by Theorem 6.3.8 , det(A) = 17, ~b = 0 −6 7

1 −3 det 0 7 x1 = 17 2 3 6.3.24 Here A = 0 4 6 0 8 3 −1

x=

det

3 0 4 5 0 7

146

5 1 det −6 0 7 = , x2 = 17 17

=

6 . 17

8 0 5 , det(A) = 146, ~b = 3 , so by Theorem 6.3.8, −1 7

= 1, y =

2

8 0 3 5 6 −1 7

det 0

146

= 2, z =

2 3 4 6 0

det 0

146

8 3 −1

= −1.

6.3.25 By Theorem 6.3.9, the ij th entry of adj(A) is given by (−1)i+j det(Aji ), so since 1 0 1 0 1 0 3 2 = 1, and for i = 1, j = 2 we get (−1) det A = 0 1 0 for i = 1, j = 1, we get (−1) det 0 0 1 2 0 1 and so forth. 1 0 −1 Completing this process gives adj(A) = 0 −1 0 , hence by Theorem 6.3.9, −2 0 1 1 0 −1 −1 0 1 1 1 A−1 = det(A) adj(A) = −1 0 −1 0 = 0 1 0 . −2 0 1 2 0 −1

1 = 0, 1

1 adj(A), so if det(A) = 1, A−1 = adj(A). If A has integer entries then 6.3.26 By Theorem 6.3.9, A−1 = det(A) i+j (−1) det(Aji ) will be an integer for all 1 ≤ i, j ≤ n, hence adj(A) will have integer entries. Therefore, A−1 will also have integer entries.

x = det

1 0

a −b 1 , det(A) = a2 + b2 , ~b = , we get b a 0 −b a 1 a −b 1 1 = , y = det a2 +b2 a2 +b2 a2 +b2 = a2 +b2 , a b 0

6.3.27 By Theorem 6.3.8, using A =

so x is positive, y is negative (since a, b > 0), and x decreases as b increases.

s 6.3.28 Here A = m det

Y =

I◦ + G Ms − M ◦

−sh−ma

"

det

r=

"

◦ a I +G ~ , det(A) = −sh − ma, b = so, by Theorem 6.3.8 −h Ms + M ◦ a −h

s I◦ + G m Ms − M ◦ −sh−ma

#

#

=

=

−h(I ◦ +G)−a(Ms −M ◦ ) −sh−ma

s(Ms −M ◦ )−m(I ◦ +G) −sh−ma

=

=

h(I ◦ +G)+a(Ms −M ◦ ) , sh+ma

m(I ◦ +G)−s(Ms −M ◦ ) . sh+ma

289

Chapter 6 6.3.29 By Theorem 6.3.8,

dx1 =

det

dy1 =

det

dp =

det

0 0 −R2 de2 −R1 α R2

−R1 α R2

R1 1−α −R2 D

0 0 −R2 de2 D

R1 1−α −R2 D

−(1 − α) −(1 − α)2 2 − (1−α) α

−(1 − α) −(1 − α)2 2 − (1−α) α

=

R1 R2 (1−α)2 de2 −R2 (1−α)2 de2 D

0 0 −R2 de2

=

R2 de2 (R1 (1−α)2 +α(1−α)) D

>0

=

R1 R2 de2 D

> 0.

0 0 6 0. −5 3

−6 0 6.3.31 Using the procedure outlined in Exercise 25, we find adj(A) = −3 5 4 −5

1 −2 . 1

0 0 6.3.32 Using the procedure outlined in Exercise 25, we find that adj(A) = 0 −1

0 0 −1 −1 0 0 . 0 −1 0 0 0 0

18 6.3.30 Using the procedure outlined in Exercise 25, we find adj(A) = −12 −2

24 0 0 0 0 12 0 0 6.3.33 Using the procedure outlined in Exercise 25, we find that adj(A) = . Note that the matrix 0 0 8 0 0 0 0 6 adj(A) is diagonal, and the ith diagonal entry of adj(A) is the product of all ajj where j 6= i.

6.3.34 For an invertible n × n matrix A, Aadj(A) = A(det(A)A−1 ) = det(A)AA−1 = det(A)In . The same is true for adj(A)A. 6.3.35 det(adj(A)) = det(det(A)A−1 ). Taking the product det(A)A−1 amounts to multiplying each row of A−1 by 1 det(A), so that det(adj(A)) = (det A)n det(A−1 ) = (det A)n det(A) = (det A)n−1 . 6.3.36 adj(adjA) = adj(det(A)A−1 ) = det(det(A)A−1 )(det(A)A−1 )−1 = (det A)n det(A−1 )(det(A)A−1 )−1 1 (A−1 )−1 = (det A)n−1 (det(A)A−1 )−1 = (det A)n−1 det(A)

= (det A)n−2 A. 6.3.37 adj(A−1 ) = det(A−1 )(A−1 )−1 = (det A)−1 (A−1 )−1 = (adjA)−1 . 290

Section 6.3 6.3.38 adj(AB) = det(AB)(AB)−1 = det(A)(det(B)B −1 )A−1 = det(B)B −1 (det(A)A−1 ) = adj(B)adj(A). 6.3.39 Yes, let S be an invertible matrix such that AS = SB, or SB −1 = A−1 S. Multiplying both sides by det(A) = det(B), we find that S(det(B)B −1 ) = (det(A)A−1 )S, or, S(adjB) = (adjA)S, as claimed. 6.3.40 The ij th entry of the matrix B of T is (ith component of T (~ej )) = det(A~ej ,i ). Expand down the i′ th column to see that this is the ij th entry of adj(A). Thus B = adj(A). See Theorem 6.3.9. 6.3.41 If A has a nonzero minor det(Aij ), then the n − 1 columns of the invertible matrix Aij will be independent, so that the n − 1 columns of A, minus the j th , will be independent as well. Thus, the rank of A (the dimension of the image) is at least n − 1. Conversely, if rank(A) ≥ n−1, then we can find n−1 independent columns of A. The n×(n−1) matrix consisting of those n − 1 columns will have rank n − 1, so that there will be exactly one redundant row (compare with Exercises 3.3.69 through 3.3.71). Omitting this redundant row produces an invertible (n − 1) × (n − 1) submatrix of A, giving us a nonzero minor of A. 6.3.42 By Theorem 6.3.9, adj(A) = 0 if (and only if) all the minors Aji of A are zero. By Exercise 41, this is the case if (and only if) rank(A) ≤ n − 2. 6.3.43 A direct computation shows that A(adjA) = (adjA)A = (det A)(In ) for all square matrices. Thus we have A(adjA) = (adjA)A = 0 for noninvertible matrices, as claimed. Let’s write B = adj(A), and let’s verify the equation AB = (det A)(In ) for the diagonal entries; the verification for the off-diagonal entries is analogous. The ith diagonal entry of AB is n ith X aij bji . [ith row of A] column = ai1 b1i + · · · + ain bni = j=1 of B Since B is the adjunct of A, bji = (−1)j+i det(Aij ). So, our summation equals n X

aij (−1)i+j det(Aij )

j=1

which is our formula for Laplace expansion across the ith row, and equals det(A), proving our claim for the diagonal entries. 6.3.44 The equation A(adjA) = 0 from Exercise 43 means that im(adjA) is a subspace of ker(A). Thus rank(adjA) = dim(im(adjA)) ≤ dim(kerA) = n − rank(A) = n − (n − 1) = 1, implying that rank(adjA) ≤ 1. Since adj(A) 6= 0, by Exercise 42, we can conclude that rank(adjA) = 1. 291

Chapter 6 d −b a c a b . So, a = d and b = −c. Thus, the = . We want AT = adj(A), or b d c d −c a a b . equation AT = adj(A) holds for all matrices of the form −b a

6.3.45 Let A =

6.3.46 In the simple case when f (x, y) = 1 we have

Z

f (x, y)dA =

Ω2

Z

g(u, v)dA =

Ω1

Z

dA = area of Ω1 = 1, so that

Ω1

Z

Ω2

Z

Ω2

dA = area of Ω2 = | det M | and

f (x, y)dA = | det M | ·

Z

g(u, v)dA.

Ω1

This formula holds, in fact, for any continuous function f (x, y); see an introductory text in multivariable calculus for a justification.

x1 x2 is the area of the triangle OP1 P2 , where O denotes the origin. This is likewise y1 y2 true for one-half the second matrix. See Theorem 2.4.10. However, because of the reversal in orientation, x3 x4 1 is negative the area of the triangle OP3 P4 ; likewise for the last matrix. Finally, note that the 2 det y y4 3 area of the quadrilateral P1 P2 P3 P4 is equal to:

6.3.47 Note that

1 2

det

the area of triangle OP1 P2 + area of triangle OP2 P3 − area of triangle OP3 P4 − area of triangle OP4 P1 . 6.3.48 In what follows, we will freely use the fact that an invertible linear transformation L from R2 to R2 maps an ellipse into an ellipse (see Exercise 2.2.52). Now consider a linear transformation L that transforms our 3-4-5 right triangle R into an equilateral triangle T . 0 4 0 , and the vertices of the equilateral , , If we place the vertices of the right triangle R at the points 3 0 0 "1 # 1 1 2 0 2 3 1 √ , , triangle T at , then the transformation L has the matrix A = , with det(A) = 2√ . 3 0 0 3 0 √1 3

According to the hint, L will map the largest ellipse E inscribed √ into R into the circle C inscribed into T . The Figure 6.9 illustrates that the radius of C is tan(π/6) = 1/ 3, so that the area of C is π/3. Using the interpretation of the determinant as an expansion fact, we find that (area of C) = (det A)(area of E), or (area of C = √ 2π of E) = area ≈ 3.6 det(A) 3 6.3.49 We will use the terminology introduced in√ the solution of Exercise 48 throughout. Note that the trans√ 3 2 −2/ √ formation L−1 , with matrix A−1 = , maps the circle C (with radius 1/ 3) into the ellipse E. 0 3 cos θ of C, and find the maximal value M and the minimal value m of Now consider a radial vector ~v = √13 sin θ 2 25 8 1 4 (sin θ)(cos θ) = 18 (sin 2θ) (we are taking the square to facilitate kA−1~v k = 34 + 19 sin2 θ − 3√ − 18 (cos 2θ) − 3√ 3 √3 √ the computations). Then M and mq will be the lengths of the semi-axis of E. The function above is sinusoidal √ √ √ 25+ 193 193 16 1 and m = 25−18 193 , so that the with average value 25 18 and amplitude 182 + 27 = 18 . Thus M = 18 length of the semi-major axis of E is 292

True or False

1 2

[ 03 [

1 3 1 √3

A= 0 R

[ √13 [

L

2

–1

L

T 1

A

–1

=

2

−2 √3

0

√3 C

1

2

3

[ 40 [

π 6

r= 1 √3

[ 20 [

Figure 6.9: for Problem 6.3.48 and Problem 6.3.49. q √ √ M = 25+18 193 ≈ 1.47, and for the semi-minor axis we get q √ √ m = 25−18 193 ≈ 0.79.

True or False Ch 6.TF.1 T, by Definition 6.1.1 Ch 6.TF.2 F; We have det(4A) = 44 det(A), by Theorem 6.2.3a. Ch 6.TF.3 F; Let A = B = I5 , for example Ch 6.TF.4 T; We have det(−A) = (−1)6 det(A) = det(A), by Theorem 6.2.3a. Ch 6.TF.5 F; In fact, det(A) = 0, since A fails to be invertible Ch 6.TF.6 F; The matrix A fails to be invertible if det(A) = 0 by Theorem 6.2.4. Ch 6.TF.7 T, by Theorem 6.2.3a, applied to the columns. Ch 6.TF.8 T, by Theorem 6.2.6. Ch 6.TF.9 T, By theorem 6.1.4, a diagonal matrix is triangular as well. Ch 6.TF.10 T, by Theorem 6.2.3b. Ch 6.TF.11 T. Without computing its exact value, we will show that the determinant is positive. The pattern that contains all the entries 100 has a product of 1004 = 108 , with two inversions. Each of the other 4! − 1 = 23 patterns contains at most two entries 100, with the other entries being less than 10, so that the product of each of these patterns is less than 1002 · 102 = 106 . Thus the determinant in more than 108 − 23 · 106 > 0, so that the matrix in invertible. 293

Chapter 6 Ch 6.TF.12 F; The correct formula is det(A−1 ) =

1 , det(AT )

by Theorems 6.2.1 and 6.2.8.

Ch 6.TF.13 T; The matrix A is invertible. Ch 6.TF.14 T; Any nonzero noninvertible matrix A will do. Ch 6.TF.15 T, by Theorem 6.2.7. Ch 6.TF.16 F, by Theorem 6.3.1. The determinant can be −1. Ch 6.TF.17 T, by Theorem 6.2.6. Ch 6.TF.18 F; The second and the fourth column are linearly dependent. Ch 6.TF.19 T; The determinant is 0 for k = −1 or k = −2, so that the matrix is invertible for all positive k. Ch 6.TF.20 F. There is only one pattern with a nonzero product, containing all the 1’s. Since there are three inversions in this pattern, det A = −1.

1 1 Ch 6.TF.21 T; Let A = 1 1

1 1 1 −1 −1 1 −1 −1

0

8 Ch 6.TF.22 F; Let A = 0

1 2

1 −1 . The column vectors of A are orthogonal and they all have length 2. −1 1

1 , for example. and ~v = 0

Ch 6.TF.23 F; In fact, det(A) = det[~u ~v w] ~ = − det[~v ~u w] ~ = −~v · (~u × w). ~ We have used Theorem 6.2.3b and Definition 6.1.1.

1 0 Ch 6.TF.24 T; Let A = 0 1

1 and B = 0

1 Ch 6.TF.25 F; Note that det 0 1

0 , for example. −1

1 0 1 1 = 2. 0 1

Ch 6.TF.26 T, by Theorem 6.3.9. Ch 6.TF.27 T, by Theorem 6.3.3, since k~vi⊥ k ≤ k~vi k = 1 for all column vectors ~vi . Ch 6.TF.28 T; We have det(A) = det(rref A) = 0. 3 2 , for example. See Theorem 6.2.10. Ch 6.TF.29 F; Let A = 5 3

Ch 6.TF.30 F; Let A = 2I2 , for example 294

True or False Ch 6.TF.31 F; Note that det(S −1 AS) = det(A) but det(2A) = 23 (det A) = 8(det A). Ch 6.TF.32 F; Note that det(S T AS) = (det S)2 (det A) and det(−A) = −(det A) have opposite signs. Ch 6.TF.33 F; Let A = 2I2 , for example. 0 1 Ch 6.TF.34 F; Let A = 0 0

−1 0 0 0

0 0 0 1

0 0 , for example. −1 0

Ch 6.TF.35 F; Let A = I2 and B = −I2 , for example. Ch 6.TF.36 T; Note that det(B) = − det(A) < det(A), so that det(A) > 0. Ch 6.TF.37 T; Let’s do Laplace expansion along the first row, for example (see Theorem 6.2.10). Then det(A) =

n X j=1

(−1)1+j a1j det(A1j ) 6= 0. Thus det(A1j ) 6= 0 for at least one j, so that A1j is invertible.

Ch 6.TF.38 T; Note that det(A) and det(A−1 ) are both integers, and (det A)(det A−1 ) = 1. This leaves only the possibilities det(A) = det(A−1 ) = 1 and det(A) = det(A−1 ) = −1. Ch 6.TF.39 T, since adj(A) = (det A)(A−1 ), by Theorem 6.3.9. Ch 6.TF.40 F; Note that det(A2 ) = (det A)2 cannot be negative, but det(−I3 ) = −1. Ch 6.TF.41 T; The product associated with the diagonal pattern is odd, while the products associated with all other patterns are even. Thus the determinant of A is odd, so that A is invertible, as claimed.

2 Ch 6.TF.42 F; Let A = 1 1

1 1 2 1 , for example 5 2

a b 0 Ch 6.TF.43 T; Let A = . If a 6= 0, let B = c d 0 1 0 and if d 6= 0, let B = . 0 0

0 0 0 0 1 ; if b 6= 0, let B = ; if c 6= 0, let B = , 1 1 0 0 0

Ch 6.TF.44 T; Use Gaussian elimination for the first column only to transform A into a matrix of the form 1 ±1 ±1 ±1 ∗ ∗ 0 ∗ B= 0 ∗ ∗ ∗ 0 ∗ ∗ ∗ Note that det(B) = det(A) or det(B) = −(det A). The stars in matrix B all represent numbers (±1) ± (±1), so that they are 2, 0, or −2. Thus the determinant of the 3 × 3 matrix M containing the stars is divisible by 8, 295

Chapter 6 since each of the 6 terms in Sarrus’ rule is 8, 0 or -8. Now perform Laplace expansion down the first column of B to see that det(M ) = det(B) = +/ − det(A). Ch 6.TF.45 T; A(adjA) = A(det(A)A−1 ) = det(A)In = det(A)A−1 A = adj(A)A.

1 Ch 6.TF.46 T; Laplace expansion along the second row gives det(A) = −k det 8 0 constant C (we need not compute that C = −259). Thus A is invertible except 37 be 259 35 = 5 = 7.4).

1 Ch 6.TF.47 F; A = 0 0 AB 6= BA.

2 9 0 for

4 7 + C = 35k + C, for some 5 k = −C 35 (which turns out to

−1 0 0 0 0 0 −1 and B = 0 −1 0 are both orthogonal and det(A) = det(B) = 1. However, 0 0 1 1 0

296

Section 7.1

Chapter 7 Section 7.1 7.1.1 If ~v is an eigenvector of A, then A~v = λ~v . Hence A3~v = A2 (A~v ) = A2 (λ~v ) = A(Aλ~v ) = A(λA~v ) = A(λ2~v ) = λ2 A~v = λ3~v , so ~v is an eigenvector of A3 with eigenvalue λ3 . 7.1.2 We know A~v = λ~v so ~v = A−1 A~v = A−1 λ~v = λA−1~v , so ~v = λA−1~v or A−1~v = λ1 ~v . Hence ~v is an eigenvector of A−1 with eigenvalue

1 λ.

7.1.3 We know A~v = λ~v , so (A + 2In )~v = A~v + 2In~v = λ~v + 2~v = (λ + 2)~v , hence ~v is an eigenvector of (A + 2In ) with eigenvalue λ + 2. 7.1.4 We know A~v = λ~v , so 7A~v = 7λ~v , hence ~v is an eigenvector of 7A with eigenvalue 7λ. 7.1.5 Assume A~v = λ~v and B~v = β~v for some eigenvalues λ, β. Then (A + B)~v = A~v + B~v = λ~v + β~v = (λ + β)~v so ~v is an eigenvector of A + B with eigenvalue λ + β. 7.1.6 Yes. If A~v = λ~v and B~v = µ~v , then AB~v = A(µ~v ) = µ(A~v ) = µλ~v 7.1.7 We know A~v = λ~v so (A − λIn )~v = A~v − λIn~v = λ~v − λ~v = ~0 so a nonzero vector ~v is in the kernel of (A − λIn ) so ker(A − λIn ) 6= {~0} and A − λIn is not invertible. a b a b 1 1 a 5 7.1.8 We want all such that =5 hence = , i.e. the desired matrices must have c d c d 0 0 c 0 5 b the form . 0 d

a b c d

1 1 a λ =λ for any λ. Hence = , i.e., the desired matrices must have the form 0 0 c 0

7.1.9 We want λ b , they must be upper triangular. 0 d 7.1.10 We want

a c

a b c d

b d

5 − 2b 1 1 , i.e. the desired matrices must have the form =5 10 − 2d 2 2

2 −2 = . So, 2a + 3b = −2 and 2c + 3d = −3. Thus, b = 3 −3 a −2−2a 3 will fit. all matrices of the form c −3−2c 3

7.1.11 We want

7.1.12 Solving solving

2 3

t v1 2 0 v v1 = 2 1 we get = (with t 6= 0) and 3 4 v2 v2 v2 − 32 t v1 0 v1 v1 0 =4 we get (with t 6= 0). = 4 v2 v2 v2 t

297

−2−2a , 3

b . d and d =

−3−2c . 3

So

Chapter 7

7.1.13 Solving

−6 6 −15 13

v1 v2

=4

3 t v1 v1 (with t 6= 0). , we get = 5 v2 v2 t

7.1.14 We want to find all 4 × 4 matrices A such that A~e2 = λ~e2 , i.e. the second column of A must be of the form a 0 c d 0 e λ f g λ . , so A = h 0 i j 0 k 0 l m 0 7.1.15 Any vector on L is unaffected by the reflection, so that a nonzero vector on L is an eigenvector with eigenvalue 1. Any vector on L⊥ is flipped about L, so that a nonzero vector on L⊥ is an eigenvector with eigenvalue −1. Picking a nonzero vector from L and one from L⊥ , we obtain a basis consisting of eigenvectors. 7.1.16 Rotation by 180◦ is a flip about the origin so every nonzero vector is an eigenvector with the eigenvalue −1. Any basis for R2 consists of eigenvectors. 7.1.17 No (real) eigenvalues 7.1.18 Any nonzero vector in the plane is unchanged, hence is an eigenvector with the eigenvalue 1. Since any nonzero vector in V ⊥ is flipped about the origin, it is an eigenvector with eigenvalue −1. Pick any two non-collinear vectors from V and one from V ⊥ to form a basis consisting of eigenvectors. 7.1.19 Any nonzero vector in L is an eigenvector with eigenvalue 1, and any nonzero vector in the plane L⊥ is an eigenvector with eigenvalue 0. Form a basis consisting of eigenvectors by picking any nonzero vector in L and any two nonparallel vectors in L⊥ . 7.1.20 Any nonzero vector along the ~e3 -axis is unchanged, hence is an eigenvector with eigenvalue 1. No other (real) eigenvalues can be found. 7.1.21 Any nonzero vector in R3 is an eigenvector with eigenvalue 5. Any basis for R3 consists of eigenvectors. 7.1.22 Any nonzero scalar multiple of ~v is an eigenvector with eigenvalue 1. 7.1.23 a Since S = [~v1 · · · ~vn ], S −1~vi = S −1 (S~ei ) = ~ei . b i th column of S −1 AS = S −1 AS~ei = S −1 A~vi (by definition of S) = S −1 λi~vi (since ~vi is an eigenvector) = λi S −1~vi = λi~ei (by part a)

λ1 0 hence S −1 AS = ... 0

0 λ2

0 ··· 0 ···

0

0 ···

0 0 .

λn

298

Section 7.1 7.1.24 See Figure 7.1.

Figure 7.1: for Problem 7.1.24. 7.1.25 See Figure 7.2.

Figure 7.2: for Problem 7.1.25. 7.1.26 See Figure 7.3.

Figure 7.3: for Problem 7.1.26. 7.1.27 See Figure 7.4. 7.1.28 See Figure 7.5. 7.1.29 See Figure 7.6. 299

Chapter 7

Figure 7.4: for Problem 7.1.27.

Figure 7.5: for Problem 7.1.28.

Figure 7.6: for Problem 7.1.29. 7.1.30 Since the matrix is diagonal, ~e1 and ~e2 are eigenvectors. See Figure 7.7. 7.1.31 See Figure 7.8. 7.1.32 Since the matrix is diagonal, ~e1 and ~e2 are eigenvectors. See Figure 7.9. 1 t −1 , hence we know that the eigenvalues are 2 and 6 with corresponding +6 7.1.33 We are given that ~x(t) = 2 1 1 1 −1 −1 1 = respectively (see Theorem 7.1.3), so we want a matrix A such that A and eigenvectors 1 1 1 1 −1 1 −1 4 −2 2 −6 . . Multiplying on the right by , we get A = 1 1 −2 4 2 6 t

300

Section 7.1

Figure 7.7: for Problem 7.1.30.

Figure 7.8: for Problem 7.1.31.

Figure 7.9: for Problem 7.1.32. 7.1.34 (A2 + 2A + 3In )~v = A2~v + 2A~v + 3In~v = 42~v + 2 · 4~v + 3~v = (16 + 8 + 3)~v = 27~v so ~v is an eigenvector of A2 + 2A + 3In with eigenvalue 27. 7.1.35 Let λ be an eigenvalue of S −1 AS. Then for some nonzero vector ~v , S −1 AS~v = λ~v , i.e., AS~v = Sλ~v = λS~v so λ is an eigenvalue of A with eigenvector S~v . Conversely, if α is an eigenvalue of A with eigenvector w, ~ then Aw ~ = αw, ~ for some nonzero w. ~ Therefore, S −1 AS(S −1 w) ~ = S −1 Aw ~ = S −1 αw ~ = αS −1 w, ~ so S −1 w ~ is an eigenvector of S −1 AS with eigenvalue α. 3 15 1 10 3 1 15 10 7.1.36 We want A such that A = and A = , i.e. A = , so 1 5 2 20 1 2 5 20 301

Chapter 7

15 A= 5

10 20

3 1 1 2

−1

4 = −2

3 . 11

0.6 0.8 7.1.37 a A = 5 is a scalar multiple of an orthogonal matrix. By Theorem 7.1.2, the possible eigenvalues 0.8 −0.6 of the orthogonal matrix are ±1, so that the possible eigenvalues of A are ±5. In part b we see that both are indeed eigenvalues. b Solve A~v = ±5~v to get ~v1 =

1 1 2 1 −3 −1 = −2 = 2 −1 . The associated eigenvalue is 2. 2 −1 −2 −1

4 1 7.1.38 −5 0 −1 −1 7.1.39 a c So,

2 −1 , ~v2 = . 1 2

a We want c 0 1 =a d 0 0 1 0 , 1 0 0

b 0 0 0 =λ = . So b = 0, and d = λ (for any λ). Thus, we need matrices of the form d 1 1 λ 0 0 0 0 0 +c +d . 0 1 0 0 1 0 0 0 is a basis of V , and dim(V )= 3. , 0 1 0

7.1.40 We need all matrices A such that

a c

b d

1 1 λ =λ = . −3 −3 −3λ

Thus, a − 3b = λ and c − 3d = −3λ. Thus, c − 3d = −3(a − 3b) = −3a + 9b, or c = −3a + 9b + 3d. a b 1 0 0 1 0 0 So A must be of the form = a +b +d . Thus, a basis of V is −3 0 9 0 3 1 −3a + 9b + 3d d 0 0 0 1 1 0 , and the dimension of V is 3. , , 3 1 9 0 −3 0 7.1.41 We want 2λ2 = c + 2d.

a b c d

1 1 a , and = λ1 1 1 c

b d

1 1 . So, a + b = λ1 = c + d and a + 2b = λ2 and = λ2 2 2

So (a+2b)−(a+b) = λ2 −λ1 = b, a = λ1 −b = 2λ1 −λ = 2λ2−λ1 = d, c = λ1 −d = 2λ1 −2λ2 . (c+2d)−(c+d) 2 . Also, −1 1 2 −1 2λ1 − λ2 λ2 − λ1 . + λ2 = λ1 So A must be of the form: −2 2 2 −1 2λ1 − 2λ2 2λ2 − λ1 −1 1 2 −1 , and dim(V )= 2. , So a basis of V is −2 2 2 −1 1 7.1.42 We will do this in a slightly simpler manner than Exercise 40. Since A 0 is simply the first column of A, 0 the first column must be a multiple of ~e1 . Similarly, the third column must be a multiple of ~e3 . There are no 302

Section 7.1

other restrictions on the

0 0 0 0 + c0 1 0 0 0 1 Thus, a basis of V is 0 0 5.

0 1 b0 0 0 0

1 0 a b 0 form of A, meaning it can be any matrix of the form 0 c 0 = a 0 0 0 0 0 d e 0 0 0 0 0 0 0 0 + d 0 0 0 + e 0 0 0 . 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 , 0 0 0 , 0 1 0 , 0 0 0 , 0 0 0 and the dimension 0 0 1 0 1 0 0 0 0 0 0 0 0 0

7.1.43 A = AIn = A[ ~e1 . . . ~en ] = [ λ1~e1 . . . can be any diagonal matrix, and dim(V ) = n.

0 0 + 0

of V is

λn~en ], where the eigenvalues λ1 , . . . , λn are arbitrary. Thus A

7.1.44 We see that each of the columns 1 through m of A will have to be a multiple of its respective vector ~ei . Thus, there will be m free variables in the first m columns. The remaining n − m columns will each have n free variables. Thus, in total, the dimension of V is m + (n − m)n = m + n2 − nm. 7.1.45 Consider a vector w ~ that is not parallel to ~v . We want A[~v w] ~ = [λ~v a~v + bw], ~ where λ, a and b are arbitrary constants. Thus the matrices A in V are of the form A = [λ~v a~v + bw][~ ~ v w] ~ −1 . Using Summary 4.1.6, we see that [~v ~0][~v w] ~ −1 , [~0 ~v ][~v w] ~ −1 , [~0 w][~ ~ v w] ~ −1 is a basis of V , so that dim(V ) = 3. 7.1.46 a We need all matrices A such that

a b c d

k 1 1 . = =k 2k 2 2

Thus, a + 2b = k and c+ 2d = 2k. So, c + 2d = 2a + 4b, or c = −2d + 2a + 4b and A must be of the form a b 1 0 0 1 0 0 1 0 0 1 0 0 =a +b +d . So a basis of V is , , , and −2d + 2a + 4b d 2 0 4 0 −2 1 2 0 4 0 −2 1 the dimension of V is 3. 1 is a basis of the image of T by definition of V , so that the rank of T is 1. The kernel of T consists b Clearly 2 a b a b 1 of all matrices such that = ~0, or a + 2b = 0, c + 2d = 0. These are the matrices of the form c d c d 2 −2b b −2 1 0 0 −2 1 0 0 =b +d . Thus a basis of the kernel of T is , . −2d d 0 0 −2 1 0 0 −2 1 a b c Let’s find the kernel of L first. In part (a) we saw that the matrices in V are of the form A = . −2d + 2a + 4b d a b 1 A matrix A in V is in the kernel of L if = ~0, or a + 3b = 0, 2a + 4b + d = 0. This −2d + 2a + 4b d 3 −3b b = system simplifies to a = −3b and d = 2b, so that the matrices in the kernel of L are of the form −6b 2b −3 1 −3 1 forms a basis of the kernel of L. By the rank-nullity theorem, the rank of L . The matrix b −6 2 −6 2 is dim(V )−dim(ker L) = 3 − 1 = 2, and the image of L is all of R2 . 7.1.47 Suppose V is a one-dimensional A-invariant subspace of Rn , and ~v is a non-zero vector in V . Then A~v will be in V, so that A~v = λ~v for some λ, and ~v is an eigenvector of A. Conversely, if ~v is any eigenvector of A, then 303

Chapter 7 V = span(~v ) will be a one-dimensional A-invariant subspace. Thus the one-dimensional A-invariant subspaces V are of the form V = span(~v ), where ~v is an eigenvector of A. 7.1.48 a Since span(~e1 ) is an A-invariant subspace of R3 , it must be that ~e1 is an eigenvector of A, as revealed in a Exercise 47. Thus, the first column of A must be of the form 0 . Since span(~e1 , ~e2 ) is also an A-invariant 0 b subspace, it must be that A~e2 is in span(~e1 , ~e2 ). Thus, the second column of A must have the form c . The 0 1 1 1 third column may be any vector in R3 . Thus, we can choose A = 0 1 1 to maximize the number of non-zero 0 0 1 entries. b We see, from our construction a of all matrices of the form 0 0

above, that upper-triangular matrices fit this description. This space, V consists b d c e and has a dimension of 6. 0 f

100 7.1.49 The eigenvalues of the system are λ1 = 1.1, and λ2 = 0.9 and corresponding eigenvectors are ~v1 = 300 100 200 , we can see that ~x0 = 3~v1 − ~v2 . Therefore, by Theorem 7.1.3, , respectively. So if ~x0 = and ~v2 = 100 800 200 100 , i.e. c(t) = 300(1.1)t − 200(0.9)t and r(t) = 900(1.1)t − 100(0.9)t . − (0.9)t we have ~x(t) = 3(1.1)t 100 300 4 100 , and we see that A~v (0) = 7.1.50 a ~v (0) = 1 100 100 100 . = 2t At 100 100

−2 1

100 100

=

200 200

= 2

100 . So, ~v (t) = At~v (0) = 100

So c(t) = r(t) = 100(2)t . 4 200 , and we see that A~v (0) = b ~v (0) = 1 100 t 200 . 3 100

−2 1

200 600 200 t t 200 = . So, ~v (t) = A ~v (0) = A =3 = 100 100 300 100

So c(t) = 200(3)t and r(t) = 100(3)t . 200 100 600 . So, + . We can write this in terms of the previous eigenvectors as ~v (0) = 4 c ~v (0) = 100 100 500 200 100 200 100 . ~v (t) = At~v (0) = At 4 + (3)t = 4(2)t + At 100 100 100 100

So c(t) = 400(2)t + 200(3)t and r(t) = 400(2)t + 100(3)t . 304

Section 7.1 0 100 , and we see that A~v (0) = 7.1.51 a ~v (0) = −1.5 200 100 100 . = (1.5)t At 200 200

.75 2.25

100 150 100 . So, ~v (t) = At~v (0) = = 1.5 = 200 300 200

So c(t) = 100(1.5)t and r(t) = 200(1.5)t . 100 0 b ~v (0) = , and we see that A~v (0) = 100 −1.5 t 100 t 100 A = (.75) . 100 100

.75 2.25

100 100

=

75 75

= .75

100 . So, ~v (t) = At~v (0) = 100

So c(t) = 100(.75)t and r(t) = 100(.75)t .

500 100 100 c ~v (0) = . We can write this in terms of the previous eigenvectors as ~v (0) = 3 +2 . So, 700 100 200 100 100 100 100 . + 2(1.5)t = 3(.75)t + At 2 ~v (t) = At~v (0) = At 3 200 100 200 100 So c(t) = 300(.75)t + 200(1.5)t and r(t) = 300(.75)t + 400(1.5)t . 7.1.52 a and

0.978 0.004

0.978 0.004

g b ~x0 = 0 l0

−0.006 −1 −0.99 −1 = = 0.99 , 0.992 2 1.98 2 3 2.94 3 −0.006 . The eigenvalues are λ1 = 0.99 and λ2 = 0.98. = 0.98 = −1 −0.98 −1 0.992

100 −1 3 3 t t −1 = = 20 + 40 , hence + 40(0.98) so ~x(t) = 20(0.99) 0 2 −1 2 −1

g(t) = −20(0.99)t + 120(0.98)t and h(t) = 40(0.99)t − 40(0.98)t .

Figure 7.10: for Problem 7.1.52b. h(t) first rises, then falls back to zero. g(t) falls a little below zero, then goes back up to zero. See Figure 7.10. c We set g(t) = −20(0.99)t + 120(0.98)t = 0. Solving for t we get that g(t) = 0 for t ≈ 176 minutes. (After t = 176, g(t) < 0). 305

Chapter 7 1 1 1 6 7.1.53 a ~v (0) = 1 = 3 1 + 2 −1 + 0 . −1 0 1 2 1 1 1 So, ~v (t) = At~v (0) = At 3 1 + 2 −1 + 0 −1 0 1 1 1 1 1 1 1 = 3At 1 + 2At −1 + At 0 = 3λt1 1 + 2λt2 −1 + λt3 0 1 −1 0 0 1 −1 1 1 1 = 3 1 + 2(− 12 )t −1 + (− 21 )t 0 . 1 0 −1 So a(t) = 3 + 3(− 12 )t , b(t) = 3 − 2(− 12 )t and c(t) = 3 − (− 21 )t .

b a(365) = 3 + 3(− 12 )365 = 3 − c(365) = 3 − (− 12 )365 = 3 +

3 2365 ,

1 2365 .

b(365) = 3 − 2(− 12 )365 = 3 +

1 2364

and

So, Benjamin will have the most gold.

7.1.54 a We are given that n(t + 1) = 2a(t) a(t + 1) = n(t) + a(t), so that the matrix is A = 1 0 b A = 1 1

0 2 . 1 1

1 2 1 2 0 2 2 −2 2 = =2 and A = = =− , hence 2 and −1 are 1 2 1 1 −1 1 −1 1 −1 1 2 the eigenvalues associated with and respectively. 1 −1 2 1

2 2 n0 1 1 1 1 t 1 1 1 t + 3 (−1) (by Theorem 7.1.3), +3 , and ~x(t) = 3 2 c We are given ~x0 = = so ~x0 = 3 1 −1 1 −1 a0 0 hence n(t) = 31 2t + 23 (−1)t and a(t) = 31 2t − 13 (−1)t .

Section 7.2 7.2.1 λ1 = 1, λ2 = 3 by Theorem 7.2.2. 7.2.2 λ1 = 2 (Algebraic multiplicity 2) λ2 = 1 (Algebraic multiplicity 2), by Theorem 7.2.2.

5−λ 7.2.3 det(A − λI2 ) = det 2

−4 = λ2 − 4λ + 3 = (λ − 1)(λ − 3) = 0 so λ1 = 1, λ2 = 3. −1 − λ 306

Section 7.2 4 = −λ(4 − λ) + 4 = (λ − 2)2 = 0 so λ = 2 with algebraic multiplicity 2. 4−λ

7.2.4 det(A − λI2 ) = det

−λ −1

7.2.5 det(A − λI2 ) = det

11 − λ 6

1−λ 7.2.6 det(A − λI2 ) = det 3

−15 = λ2 − 4λ + 13 so det(A − λI2 ) = 0 for no real λ. −7 − λ 2 = λ2 − 5λ − 2 = 0 so λ1,2 = 4−λ

√ 5± 33 . 2

7.2.7 λ = 1 with algebraic multiplicity 3, by Theorem 7.2.2. 7.2.8 fA (λ) = −λ2 (λ + 3) so λ1 = 0 (Algebraic multiplicity 2) λ2 = −3. 7.2.9 fA (λ) = −(λ − 2)2 (λ − 1) so λ1 = 2 (Algebraic multiplicity 2) λ2 = 1. 7.2.10 fA (λ) = (1 + λ)2 (1 − λ) so λ1 = −1 (Algebraic multiplicity 2), λ2 = 1. 7.2.11 fA (λ) = −λ3 − λ2 − λ − 1 = −(λ + 1)(λ2 + 1) = 0 λ = −1 (Algebraic multiplicity 1). 7.2.12 fA (λ) = λ(λ + 1)(λ − 1)2 so λ1 = 0, λ2 = −1, λ3 = 1 (Algebraic multiplicity 2). 7.2.13 fA (λ) = −λ3 + 1 = −(λ − 1)(λ2 + λ + 1) so λ = 1 (Algebraic multiplicity 1). 7.2.14 fA (λ) = det(B − λI2 ) det(D − λI2 ) (see Theorem 6.1.5). The eigenvalues of A are the eigenvalues of B and D. The eigenvalues of C are irrelevant. 2

7.2.15 fA (λ) = λ − 2λ + (1 − k) = 0 if λ1,2 =

2±

√

4−4(1−k) 2

=1±

√ k

The matrix A has 2 distinct real eigenvalues when k > 0, no real eigenvalues when k < 0. 7.2.16 fA (λ) = λ2 − (a + c)λ + (ac − b2 ) The discriminant of this quadratic equation is (a + c)2 − 4(ac − b2 ) = a2 + 2ac + c2 − 4ac + 4b2 = (a − c)2 + 4b2 ; this quantity is always positive (since b 6= 0). There will always be two distinct real eigenvalues. √ 7.2.17 fA (λ) = λ2 − a2 − b2 = 0 so λ1,2 = ± a2 + b2 . The matrix A represents a reflection about a line followed by a scaling by 307

√

a2 + b2 , hence the eigenvalues.

Chapter 7 7.2.18 fA (λ) = λ2 − 2aλ + a2 − b2 so λ1,2 =

+2a±

√

4a2 −4(a2 −b2 ) 2

= a ± b.

Hence the eigenvalues are a ± b. 7.2.19 True, since fA (λ) = λ2 − tr(A)λ + det(A) and the discriminant [tr(A)]2 − 4 det(A) is positive if det(A) is negative. 7.2.20 The characteristic polynomial of A is fA (λ) = (λ − λ1 )(λ − λ2 ) = λ2 − (λ1 + λ2 )λ + λ1 λ2 . But from Theorem 7.2.4 we know that fA (λ) = λ2 − tr(A)λ + det(A). Comparing the coefficient of λ, we see that λ1 + λ2 = tr(A), as claimed. 7.2.21 If A has n eigenvalues, then fA (λ) = (λ1 − λ)(λ2 − λ) · · · (λn − λ). Then fA (λ) = (−λ)n + (λ1 + λ2 + · · · + λn )(−λ)n−1 + · · · + (λ1 λ2 · · · λn ). But, by Theorem 7.2.5, the coefficient of (−λ)n−1 is tr(A). So, tr(A) = λ1 + · · · + λn . 7.2.22 By Theorem 6.2.1, fA (λ) = det(A−λIn ) = det(A−λIn )T = det(AT −λIn ) = fAT (λ). Since the characteristic polynomials of A and AT are identical, the two matrices have the same eigenvalues, with the same algebraic multiplicities. 7.2.23 fB (λ) = det(B − λIn ) = det(S −1 AS − λIn ) = det(S −1 AS − λS −1 In S) = det(S −1 (A − λIn )S) = det(S −1 ) det(A − λIn ) det(S) = (det S)−1 det(A − λIn ) det(S) = det(A − λIn ) = fA (λ) Hence, since fA (λ) = fB (λ), A and B have the same eigenvalues. 7.2.24 λ1 = 0.25, λ2 = 1 b ab + cb (a + c)b b b = = = since a + c = b + d = 1; therefore, is an eigenvector with c cb + cd (b + d)c c c eigenvalue λ1 = 1. 1 a−b 1 1 Also, A = = (a−b) since a−b = −(c−d); therefore, is an eigenvector with eigenvalue −1 c−d −1 −1 λ2 = a − b. Note that |a − b| < 1; a possible phase portrait is shown in Figure 7.11.

7.2.25 A

7.2.26 Here

1 0.25 b with λ2 = a − b = 0.25. See Figure 7.12. with λ1 = 1 and = −1 0.5 c

7.2.27 a We know ~v1 = x1 (t) =

1 3

+

2 3

1 1 1 then ~x0 = 31 ~v1 + 23 ~v2 , so by Theorem 7.1.3, , λ2 = 14 . If ~x0 = , λ1 = 1 and ~v2 = 0 −1 2

1 t 4

t − 23 41 . 0 If ~x0 = then ~x0 = 13 ~v1 − 31 ~v2 , so by Theorem 7.1.3, 1 x2 (t) =

2 3

308

Section 7.2

Figure 7.11: for Problem 7.2.25.

Figure 7.12: for Problem 7.2.26. x1 (t) =

1 3

−

1 3

x2 (t) =

2 3

+

1 3

t

b A approaches

1 t 4

1 t 4 . 1 3

See Figure 7.13.

1 1 , as t → ∞. See part c for a justification. 2 2

c Let us think about the first column of At , which is At~e1 . We can use Theorem 7.1.3 to compute At~e1 . b 1 1 and c2 = Start by writing ~e1 = c1 + c2 ; a straightforward computation shows that c1 = b+c c −1 309

c b+c .

Chapter 7

Figure 7.13: for Problem 7.2.27a.

Now At~e1 =

1 b+c

b + c

c t b+c (λ2 )

1 , where λ2 = a − b. −1

Since |λ2 | < 1, the second summand goes to zero, so that lim (At~e1 ) = t→∞

Likewise, lim (At~e2 ) = t→∞

1 b+c

b , so that lim At = t→∞ c

1 b+c

b c

1 b+c

b . c

b . c

7.2.28 a w(t + 1) = 0.8w(t) + 0.1m(t) m(t + 1) = 0.2w(t) + 0.9m(t)

0.8 so A = 0.2

0.1 0.9

which is a regular transition matrix since its columns sum to 1 and its entries are positive.

b The eigenvectors of A are

~x0 =

0.1 0.2

or

1 1 with λ2 = 0.7. with λ1 = 1, and −1 2

1 1 1 1 1200 or + 800(0.7)t so ~x(t) = 400 + 800 = 400 −1 2 −1 2 0

w(t) = 400 + 800(0.7)t m(t) = 800 − 800(0.7)t . c As t → ∞, w(t) → 400 so Wipfs won’t have to close the store. 7.2.29 The ith entry of A~e is [ai1 ai2 · · · ain ]~e = to the eigenvector ~e.

n X

aij = 1, so A~e = ~e and λ = 1 is an eigenvalue of A, corresponding

j=1

310

Section 7.2 7.2.30 a Let vi be the largest component of the vector ~v , that is, vi ≥ vj for j = 1, . . . , n. Then the ith component of n n n X X X aij vi = vi aij vi = aij vj ≤ A ~v is λvi = j=1

j=1

↑

vj ≤ v i

j=1

n X

↑

aij = 1

j=1

We can conclude that λvi ≤ vi , and therefore λ ≤ 1, as claimed. Also note that if ~v is not a multiple of the n n X X aij vi and therefore aij vj < eigenvector ~e discussed in Exercise 29, then vj < vi for some index j, so that j=1

j=1

λ < 1.

b Let vi be the component of ~v with the largest absolute . . , n. Then the is, |vi | ≥ |vj | for j = 1, 2, . value, that n n n n X X X X aij |vi | = aij |vi | = aij |vj | ≤ aij vj ≤ absolute value of the ith component of A~v is |λkvi | = j=1

j=1

j=1

j=1

|vi | so that |λkvi | ≤ |vi | and |λ| ≤ 1, as claimed.

7.2.31 Since A and AT have the same eigenvalues (by Exercise 22), Exercise 29 states that λ = 1 is an eigenvalue of A,and Exercise 30 says that |λ| ≤ 1 for all eigenvalues λ. Vector ~e need not be an eigenvector of A; consider 0.9 0.9 A= . 0.1 0.1 7.2.32 fA (λ) = −λ3 + 3λ + k. The eigenvalues of A are the solutions of the equation −λ3 + 3λ + k = 0, or, λ3 − 3λ = k. Following the hint, we graph the function g(λ) = λ3 − 3λ as shown in Figure 7.14. We use the derivative f ′ (λ) = 3λ2 − 3 to see that g(λ) has a global minimum at (1, −2) and a global maximum at (−1, 2). To count the eigenvalues of A, we need to find out how many times the horizontal line y = k intersects the graph of g(λ). In Figure 7.14, we see that there are three solutions if k satisfies the inequality 2 > k > −2, two solutions if k = 2 or k = −2, and one solution if |k| > 2.

(−1, 2)

g(λ) = λ3 − 3λ (1, − 2)

Figure 7.14: for Problem 7.2.32. 7.2.33 a fA (λ) = det(A − λI3 ) = −λ3 + cλ2 + bλ + a 311

Chapter 7

0 b By part a, we have c = 17, b = −5 and a = π, so M = 0 π

1 0 0 1 . −5 17

7.2.34 Consider the possible graphs of fA (λ) assuming that it has 2 distinct real roots.

(1 – λ) (–2 – λ) (λ2 + 1)

Figure 7.15: for Problem 7.2.34. 1 0 Algebraic multiplicity of each eigenvalue is 1. Example: 0 0

0 −2 0 0

0 0 0 1

0 0 . See Figure 7.15. −1 0

(–2 – λ)2 (1 – λ)2

Figure 7.16: for Problem 7.2.34. −2 0 Algebraic multiplicity of each eigenvalue is 2. Example: 0 0

Algebraic multiplicity of λ1 is 1, and of λ2 is 3. −2 0 0 0 0 1 0 0 Example: . See Figure 7.17. 0 0 1 0 0 0 0 1 0 −1 0 1 7.2.35 A = 0 0 0 0

0 0 0 0 , with fA (λ) = (λ2 + 1)2 0 −1 1 0 312

0 −2 0 0

0 0 1 0

0 0 . See Figure 7.16. 0 1

Section 7.2

(– 2 – λ) (1 – λ)3

Figure 7.17: for Problem 7.2.34.

7.2.36 Let A =

B

0 B

0

..

. B

where B = 0 1

−1 , fA (λ) = (λ2 + 1)n . 0

7.2.37 We can write fA (λ) = (λ − λ0 )2 g(λ), for some polynomial g. The product rule for derivatives tells us that fA′ (λ) = 2(λ − λ0 )g(λ) + (λ − λ0 )2 g ′ (λ), so that fA′ (λ0 ) = 0, as claimed. 7.2.38 By Theorem 7.2.4, the characteristic polynomial of A is fA (λ) = λ2 − 5λ − 14 = (λ − 7)(λ + 2), so that the eigenvalues are 7 and -2. ae + bg − − − e f a b = ae + bg + cf + dh. =tr 7.2.39 tr(AB) =tr − − − cf + dh g h c d e f a b ea + f c − − − tr(BA) =tr =tr = ea + f c + gb + hd. So they are equal. g h c d − − − gb + hd 7.2.40 Let the entries of A be aij and the entries of B be bij . Now, tr(AB) = (a11 b11 + a12 b21 + · · · + a1n bn1 ) + (a21 b12 + · · · + a2n bn2 ) + · · · + (an1 b1n + · · · +ann bnn ). This is the sum of all products of the form aij bji . We see that tr(BA) = (b11 a11 + · · · + b1n an1 )+· · ·+(bn1 a1n + · · · + bnn ann ) , which also is the sum of all products of the form bji aij = aij bji . Thus, tr(AB) = tr(BA). 7.2.41 So there exists an invertible S such that B = S −1 AS, and tr(B) =tr(S −1 AS) =tr((S −1 A)S). By Exercise 40, this equals tr(S(S −1 A)) =tr(A). 7.2.42 tr (A + B)2 = tr(A2 + AB + BA + B 2 ) = tr(A2 ) + tr(AB) + tr(BA) + tr(B 2 ).. By Exercise 40, tr(AB) = tr(BA). Thus, tr (A + B)2 = tr(A2 ) + 2tr(BA) + tr(B 2 ) = tr(A2 ) + tr(B 2 ), since BA = 0.

7.2.43 tr(AB − BA) =tr(AB)−tr(BA) =tr(AB)−tr(AB) = 0, but tr(In ) = n, so no such A, B exist. We have used Exercise 40.

7.2.44 No, there are no such matrices A and B. We will argue indirectly, assuming that invertible matrices A and 313

Chapter 7 B with AB − BA = A do exist. Then AB = BA + A = (B + In )A, and ABA−1 = B + In . Using Exercise 41, we see that tr(B) = tr(ABA−1 ) = tr(B + In ) = tr(B) + n, a contradiction. 7.2.45 fA (λ) = λ2 −tr(A)λ+det(A) = λ2 − 2λ + (−3 − 4k). We want fA (5) = 25 − 10 − 3 − 4k = 0, or, 12 − 4k = 0, or k = 3. 7.2.46 a λ21 + λ22 = (λ1 + λ2 )2 − 2λ1 λ2 = (trA)2 − 2 det(A) = (a + d)2 − 2(ad − bc) = a2 + d2 + 2bc. b Based on part (a), we need to show that a2 + d2 + 2bc ≤ a2 + b2 + c2 + d2 , or 2bc ≤ b2 + c2 , or 0 ≤ (b − c)2 . But the last inequality is obvious. c By parts (a) and (b), the equality λ21 + λ22 = a2 + b2 + c2 + d2 holds if (and only if) 0 = (b − c)2 , or b = c. Thus equality holds for symmetric matrices A. 7.2.47 Let M = [ ~v1

~v2 ]. We want A[ ~v1

~v2 ] = [ ~v1

must be nonzero, 2 or 3 must be an eigenvalue of A.

2 0 , or, [ A~v1 ~v2 ] 0 3

A~v2 ] = [ 2~v1

3~v2 ]. Since ~v1 or ~v2

7.2.48 Let S = [~v1 ~v2 ]. Then AS = [A~v1 A~v2 ] and SD = [2~v1 3~v2 ], so that ~v1 must be an eigenvector with eigenvalue 2, and ~v2 must be an eigenvector with eigenvalue 3. Thus, both 2 and 3 must be eigenvalues of A. 7.2.49 As in problem 47, such an M will exist if A has an eigenvalue 2, 3 or 4. 7.2.50 a If f (x) = x3 + 6x − 20 then f ′ (x) = 3x2 + 6 so f ′ (x) > 0 for all x, i.e. f is always increasing, hence has only one real root. b If v 3 − u3 = 20 and vu = 2 then (v − u)3 + 6(v − u) = v 3 − 3v 2 u + 3vu2 − u3 + 6(v − u) = v 3 − u3 − 3vu(v − u) + 6(v − u) = 20 − 6(v − u) + 6(v − u) = 20 Hence x = v − u satisfies the equation x3 + 6x = 20. c The second equation tells us that u =

2 v

or u3 =

8 v3 .

Substituting into the first equation we find that

8 v3

= 20, or, (v 3 )2 − 8 = 20v 3 or (v 3 )2 − 20v 3 − 8 = 0, with solutions p √ √ √ √ 3 v 3 = 20± 400+32 = 10 ± 108 = 10 ± 6 3 and v = 10 ± 108. 2 p √ √ 3 Now u3 = v 3 − 20 = −10 ± 108 and u = −10 ± 108. v3 −

d Let v =

r 3

q 2

+

q

q 2 2

+

p 3 3

Then v 3 − u3 = q and vu = Since x = v − u we have

and u =

q 3

q 2 2

+

r 3

− 2q +

p 3 3

−

q

q 2 2

q 2 2

+

= p3 .

p 3 3 .

x3 + px = v 3 − 3v 2 u + 3vu2 − u3 + p(v − u) = v 3 − u3 − 3vu(v − u) + p(v − u) 314

Section 7.3 = q − p(v − u) + p(v − u) = q, as claimed. 2 3 If p is negative, the expression 2q + p3 may be negative. Also, the equation x3 + px = q may have more than one solution in this case. e Setting x = t −

a 3

we get t −

a 3 3

+a t−

a 2 3

+b t−

a 3

+ c = 0 or

t3 − at2 + at2 + (linear and constant terms) = 0 or t3 + (linear and constant terms) = 0, as claimed (bring the constant terms to the right-hand side).

Section 7.3

8 1 −2 8 4 = span , E9 = ker = span 2 0 0 0 1

0 7.3.1 λ1 = 7, λ2 = 9, E7 = ker 0 1 4 Eigenbasis: , 0 1 7.3.2 λ1 = 2, λ2 = 0, E2 = span

1 −1 , E0 = span 1 1

−1 1 , Eigenbasis: 1 1 1 3 , E9 = span 7.3.3 λ1 = 4, λ2 = 9, E4 = span 1 −2 1 3 , Eigenbasis: 1 −2

−1 7.3.4 λ1 = λ2 = 1, E1 = span 1

No eigenbasis 7.3.5 No real eigenvalues as fA (λ) = λ2 − 2λ + 2. 7.3.6 λ1,2 =

√ 7± 57 2

Eigenbasis:

3 3 3 3 ≈ , ≈ −2.27 λ2 − 2 5.27 λ1 − 2

7.3.7 λ1 = 1, λ2 = 2, λ3 = 3, eigenbasis: ~e1 , ~e2 , ~e3 1 1 1 7.3.8 λ1 = 1, λ2 = 2, λ3 = 3, eigenbasis: 0 , 1 , 2 0 0 1 315

Chapter 7 −1 0 1 7.3.9 λ1 = λ2 = 1, λ3 = 0, eigenbasis: 0 , 1 , 0 1 0 0 1 0 7.3.10 λ1 = λ2 = 1, λ3 = 0, E1 = span 0 , E0 = span 0 0 1 No eigenbasis

1 1 1 7.3.11 λ1 = λ2 = 0, λ3 = 3, eigenbasis: −1 , 0 , 1 1 −1 0

1 7.3.12 λ1 = λ2 = λ3 = 1, E1 = span 0 , no eigenbasis 0 1 1 0 7.3.13 λ1 = 0, λ2 = 1, λ3 = −1, eigenbasis: 1 , −3 , −1 2 1 0 0 1 0 7.3.14 λ1 = 0, λ1 = λ3 = 1, eigenbasis: 1 , −5 , 2 0 0 1 0 7.3.15 λ1 = 0, λ2 = λ3 = 1, E0 = span 1 . We can use Kyle Numbers to see that 0 1 −2 E1 = ker −3 −4

−1 0 −1 0

2 1 1 = span −1 . 1 2 2

There is no eigenbasis since the eigenvalue 1 has algebraic multiplicity 2, but the geometric multiplicity is only 1.

1 7.3.16 λ1 = 0 (no other real eigenvalues), with eigenvector −1 1 No real eigenbasis

7.3.17 λ1 = λ2 = 0, λ3 = λ4 = 1 0 0 0 1 0 −1 1 0 with eigenbasis , , , 0 0 1 0 1 0 0 0 316

Section 7.3 7.3.18 λ1 = λ2 = 0, λ3 = λ4 = 1, E0 = span(~e1 , ~e3 ), E1 = span(~e2 ) No eigenbasis 7.3.19 Since 1 is the only eigenvalue, with algebraic multiplicity 3, there exists an eigenbasis for A 0 the geometric multiplicity of the eigenvalue 1 is 3 as well, that is, if E1 = R3 . Now E1 = ker 0 0 (and only if) a = b = c = 0.

if (and only if) a b 0 c is R3 if 0 0

If a = b = c = 0 then E1 is 3-dimensional with eigenbasis ~e1 , ~e2 , ~e3 . If a 6= 0 and c 6= 0 then E1 is 1-dimensional and otherwise E1 is 2-dimensional. The geometric multiplicity of the eigenvalue 1 is dim(E1 ). 0 a 0 0 a b 7.3.20 For λ1 = 1, E1 = ker 0 0 c = ker 0 0 1 so if a = 0 then E1 is 2-dimensional, otherwise it is 0 0 0 0 0 1 1-dimensional. 1 −a −b 1 −c so E2 is 1-dimensional. For λ2 = 2, E2 = ker 0 0 0 0

Hence, there is an eigenbasis if a = 0.

1 4 2 2 1 1 , i.e. A = = 2 and A = 7.3.21 We want A such that A 2 6 3 3 2 2 −1 1 4 1 2 5 −2 . = 2 6 6 −2 2 3

2 3

=

1 4 2 6

so A =

The answer is unique.

7 0 7.3.22 We want A such that A~e1 = 7~e1 and A~e2 = 7~e2 hence A = . 0 7 7.3.23 λ1 = λ2 = 1 and E1 = span(~e1 ), hence there is no eigenbasis. The matrix represents a shear parallel to the x-axis. 2 2 a b a b , or 2a + b = 2, 2c + d = 1. This condition is satisfied = . First we want 7.3.24 Let A = 1 1 c d c d a 2 − 2a by all matrices of the form A = . Next, we want there to be no other eigenvalue, besides 1, so that c 1 − 2c 1 must have an algebraic multiplicity of 2. We want the characteristic polynomial to be (λ−1)2 = λ2−2λ+1, so that the trace must be 2, and a+(1−2c) = 2, 1 + 2c −4c . or, a = 1 + 2c. Thus we want a matrix of the form A = c 1 − 2c 2 instead of E1 = R2 . This means that we must exclude the case Finally, we have to make sure the E1 = span 1 1 + 2c −4c A = I2 . In order to ensure this, we state simply that A = , where c is any nonzero constant. c 1 − 2c 317

Chapter 7

−λ 1 7.3.25 If λ is an eigenvalue of A, then Eλ = ker(A − λI3 ) = ker 0 −λ a b

0 1 . c−λ

The second and third columns of the above matrix aren’t parallel, hence Eλ is always 1-dimensional, i.e., the geometric multiplicity of λ is 1.

7.3.26 Note that fA (0) = det(A − 0I6 ) = det(A) is negative. Since lim fA (λ) = ∞, there must be a positive root, λ→∞

by the Intermediate Value Theorem (see Exercise 2.2.47c). Therefore, the matrix A has a positive eigenvalue. See Figure 7.18.

Figure 7.18: for Problem 7.3.26. 7.3.27 By Theorem 7.2.4, we have fA (λ) = λ2 − 5λ + 6 = (λ − 3)(λ − 2) so λ1 = 2, λ2 = 3. 7.3.28 Since Jn (k) is triangular, its 0 0 . Ek = ker(Jn (k) − kIn ) = ker .. . . . 0

eigenvalues are its diagonal entries, hence its only eigenvalue is k. Moreover, 1 0 ··· 0 .. 0 1 . .. .. = span (~e1 ). . 0 . .. .. 1 . . 0 0 0

The geometric multiplicity of k is 1 while its algebraic multiplicity is n.

7.3.29 Note that r is the number of nonzero diagonal entries of A, since the nonzero columns of A form a basis of im(A). Therefore, there are n − r zeros on the diagonal, so that the algebraic multiplicity of the eigenvalue 0 is n − r. It is true for any n × n matrix A that the geometric multiplicity of the eigenvalue 0 is dim(ker(A)) = n − rank(A) = n − r. 7.3.30 Since A is triangular, fA (λ) = (a11 − λ)(a22 − λ) · · · (amm − λ)(0 − λ)n−m . Hence the algebraic multiplicity of λ = 0 is (n − m). Also note that the rank of A is at least m, since the first m columns of A are linearly independent. Therefore, the geometric multiplicity of the eigenvalue 0 is dim(ker(A)) = n − rank(A) ≤ n − m. 7.3.31 They must be the same. For if they are not, by Theorem 7.3.7, the geometric multiplicities would not add up to n. 318

Section 7.3 7.3.32 Recall that a matrix and its transpose have the same rank (Theorem 5.3.9c). The geometric multiplicity of λ as an eigenvalue of A is dim(ker(A − λIn )) = n − rank(A − λIn ). The geometric multiplicity of λ as an eigenvalue of AT is dim(ker(AT − λIn )) = dim(ker(A − λIn )T ) = n − rank(A − λIn )T = n − rank(A − λIn ). We can see that the two multiplicities are the same. 7.3.33 If S −1 AS = B, then S −1 (A − λIn )S = S −1 (AS − λS) = S −1 AS − λS −1 S = B − λIn . 7.3.34 a If ~x is in the kernel of B, then AS~x = SB~x = S~0 = ~0, so that S~x is in ker(A). b T is clearly linear, and the transformation R(~x) = S −1 ~x is the inverse of T (if ~x is in the kernel of B, then S −1 ~x is in the kernel of A, by part (a)). c The equation nullity(A) = nullity(B) follows from part (b); the equation rank(A) = rank(B) then follows from the rank-nullity theorem (Theorem 3.3.7). 7.3.35 No, since the two matrices have different eigenvalues (see Theorem 7.3.6c). 7.3.36 No, since the two matrices have different traces (see Theorem 7.3.6.d) 7.3.37 a A~v · w ~ = (A~v )T w ~ = (~v T AT )w ~ = (~v T A)w ~ = v T (Aw) ~ = ~v · Aw ~ ↑ A symmetric b Assume A~v = λ~v and Aw ~ = αw ~ for λ 6= α, then (A~v ) · w ~ = (λ~v ) · w ~ = λ(~v · w), ~ and ~v · Aw ~ = ~v · αw ~ = α(~v · w). ~ By part a, λ(~v · w) ~ = α(~v · w) ~ i.e., (λ − α)(~v · w) ~ = 0. Since λ 6= α, it must be that ~v · w ~ = 0, i.e., ~v and w ~ are perpendicular. 7.3.38 Note that fA (0) = det(A − 0I3 ) = det(A) = 1. Since lim fA (λ) = −∞, the polynomial fA (λ) must have a positive root λ0 , by the Intermediate Value Theorem λ→∞

in single variable calculus. In other words, the matrix A will have a positive eigenvalue λ0 . Since A is orthogonal, this eigenvalue λ0 will be 1, by Theorem 7.1.2. This means that there is a nonzero vector ~v in R3 such that A~v = 1~v = ~v , as claimed. See Figure 7.19. 7.3.39 a There are two eigenvalues, λ1 = 1 (with E1 = V ) and λ2 = 0 (with E0 = V ⊥ ). Now geometric multiplicity(1) = dim(E1 ) = dim(V ) = m, and geometric multiplicity(0) = dim(E0 ) = dim(V ⊥ ) = n − dim(V ) = n − m. Since geometric multiplicity(λ) ≤ algebraic multiplicity(λ), by Theorem 7.3.7, and the algebraic multiplicities cannot add up to more than n, the geometric and algebraic multiplicities of the eigenvalues are the same here. 319

Chapter 7

fA (λ) 1 λ0 = 1

Figure 7.19: for Problem 7.3.38. b Analogous to part a: E1 = V and E−1 = V ⊥ . geometric multiplicity(1) = algebraic multiplicity(1) = dim(V ) = m, and geometric multiplicity(−1) = algebraic multiplicity(−1) = dim(V ⊥ ) = n − m.

a b 7.3.40 The matrix of the dynamical system is A = b a

Hence, λ1,2 = a ± b, and the respective eigenvectors are Since ~x(0) =

3 1 2

=

7 4

1 − 1

5 4

so fA (λ) = (a − λ)2 − b2 . 1 −1 and . 1 1

1 −1 5 7 t 1 t + 4 (a − b) . , by Theorem 7.1.3, ~x(t) = 4 (a + b) 1 −1 1

Note that a − b is between 0 and 1, so that the second summand in the formula above goes to ~0 as t goes to infinity. Qualitatively different outcomes occur depending on whether a + b exceeds 1, equals 1, or is less than 1. See Figure 7.20.

Figure 7.20: for Problem 7.3.40. 1 2 9 7.3.41 The eigenvalues of A are 1.2, −0.8, −0.4 with eigenvectors 6 , −2 , −2 . 2 1 2 320

Section 7.3 2 9 1 2 9 Since ~x0 = 50 6 + 50 −2 + 50 −2 we have ~x(t) = 50(1.2)t 6 + 50(−0.8)t −2 + 1 2 2 1 2 1 50(−0.4)t −2 , so, as t goes to infinity, j(t) : n(t) : a(t) approaches the proportion 9 : 6 : 2. 2 50 1 0.8 10 C(t + 1) , . A has eigenvectors ,A = = 1 0 0 1 1 C(0) 0 0 1 50 corresponding to λ1 = 0.8 and λ2 = 1. Since = , and = −50 + , we have C(t) = 1 1 1 0 1 t −50(0.9) + 50, hence in the long run, there will be 50 spectators. The graph of C(t) looks similar to the graph in Figure 7.21.

7.3.42 C(t + 1) = 0.8C(t) + 10 so if A

C(t) 1

Figure 7.21: for Problem 7.3.42.

0 1 7.3.43 a A = 12 1 0 1 1

1 1 0

7 7.6660156 b After 10 rounds, we have A10 11 ≈ 7.6699219 . 5 7.6640625 7.66666666667 7 After 50 rounds, we have A50 11 ≈ 7.66666666667 . 7.66666666667 5 c The eigenvalues of A are 1 and − 21 with −1 0 1 E1 = span 1 and E− 12 = span 1 , −1 2 −1 1 1 −1 0 t t so ~x(t) = 1 + c30 1 + − 21 1 + − 21 c30 −1 . 1 2 −1 After 1001 rounds, Alberich will be ahead of Brunnhilde by

1 1001 2

, so that Carl needs to beat Alberich to 1001 win the game. A straightforward computation shows that c(1001) − a(1001) = 21 (1 − c0 ); Carl wins if this quantity is positive, which is the case if c0 is less than 1. Alternatively, observe that the ranking of the players is reversed in each round: whoever is first will be last after the next round. Since the total number of rounds is odd (1001), Carl wants to be last initially to win the game; he wants to choose a smaller number than both Alberich and Brunnhilde. 321

Chapter 7 7.3.44 a a11 = 0.7 means that only 70% of the pollutant present in Lake Sils at a given time is still there a week later; some is carried down to Lake Silvaplana by the river Inn, and some is absorbed or evaporates. The other diagonal entries can be interpreted analogously: a21 = 0.1 means that 10% of the pollutant present in Lake Sils at any given time can be found in Lake Silvaplana a week later, carried down by the river Inn. The significance of the coefficient a32 = 0.2 is analogous; a31 = 0 means that no pollutant is carried down from Lake Sils to Lake St. Moritz in just one week. The matrix is lower triangular since no pollutant is carried from Lake Silvaplana to Lake Sils. The river Inn would have to flow the other way. b The eigenvalues of A are 0.8, 0.6, 0.7, with corresponding eigenvectors 1 0 0 0 , 1 , 1 . −2 −1 1 100 0 0 1 ~x(0) = 0 = 100 0 − 100 1 + 100 1 , 0 1 −1 −2 1 0 0 so ~x(t) = 100(0.8)t 0 − 100(0.6)t 1 + 100(0.7)t 1 or −2 −1 1 x1 (t) = 100(0.7)t

x2 (t) = 100(0.7)t − 100(0.6)t x3 (t) = 100(0.8)t + 100(0.6)t − 200(0.7)t . See Figure 7.22.

Figure 7.22: for Problem 7.3.44b. Using calculus, we find that the function x2 (t) = 100(0.7)t − 100(0.6)t reaches its maximum at t ≈ 2.33. Keep in mind, however, that our model holds for integer t only.

0.1 7.3.45 a A = 0.4 A ~b b B= 0 1

1 0.2 ~ ,b = 2 0.3

322

Section 7.3

c The eigenvalues of A are 0.5 and −0.1 with associated eigenvectors

1 1 . and −1 2

~v ~v ~v The eigenvalues of B are 0.5, −0.1, and 1. If A~v = λ~v then B =λ so is an eigenvector of B. 0 0 0 2 −(A − I2 )−1~b Furthermore, 4 is an eigenvector of B corresponding to the eigenvalue 1. Note that this vector is . 1 1 2 1 1 x1 (0) d Write ~y (0) = x2 (0) = c1 2 + c2 −1 + c3 4 . 1 0 0 1

Note that c3 = 1.

2 2 1 1 2 t→∞ t→∞ t t Now ~y (t) = c1 (0.5) 2 + c2 (−0.1) −1 + 4 −→ 4 so that ~x(t) −→ . 4 1 1 0 0 7.3.46 a T1 (t + 1) = 0.6T1 (t) + 0.1T2 (t) + 20 T2 (t + 1) = 0.1T1 (t) + 0.6T2 (t) + 0.1T3 (t) + 20 T3 (t + 1) = 0.1T2 (t) + 0.6T3 (t) + 40 20 0.6 0.1 0 so A = 0.1 0.6 0.1 , ~b = 20 40 0 0.1 0.6 b B=

A ~b 0 1

0 70.86 0 93.95 c ~y (10) = B 10 ≈ 0 120.56 1 1 0 74.989 0 99.985 ~y (30) = B 30 ≈ 0 124.989 1 1 75 100 ~y (t) seems to approach as t → ∞ 125 1

d The eigenvalues of A are λ1 ≈ 0.45858, λ2 = 0.6, λ3 ≈ 0.74142 so the eigenvalues of B are λ1 ≈ 0.45858, λ2 = 0.6, λ3 ≈ 0.74142, λ4 = 1. 323

Chapter 7 ~v1 ~v ~v If ~v1 , ~v2 , ~v3 are eigenvectors of A (with A~vi = λi~vi ), then , 2 , 3 are corresponding eigenvectors 0 0 0 75 100 of B. Furthermore, is an eigenvector if B with eigenvalue 1. Since λ1 , λ2 , λ3 are all less than 1, 125 1 75 lim ~x(t) = 100 , as in Exercise 45. t→∞ 125 1 r(t) 2 7.3.47 a If ~x(t) = p(t) , then ~x(t + 1) = A~x(t) with A = 12 w(t) 0

1 4 1 2 1 4

0 1 2 1 2

.

1 1 1 The eigenvalues of A are 0, 12 , 1 with eigenvectors −2 , 0 , 2 . 1 −1 1 1 1 1 1 1 t Since ~x(0) = 0 = 14 −2 + 21 0 + 41 2 , ~x(t) = 12 21 0 + 1 −1 1 −1 0

1 1 2 for t > 0. 4 1

b As t → ∞ the ratio is 1 : 2 : 1 (since the first term of ~x(t) drops out). 7.3.48 a We are told that a(t + 1) = a(t) + j(t)

1 1 j(t + 1) = a(t), so that A = . 1 0 λ2 λ1 . and 1 1 λ λ 1 λ1 λ2 = √15 Since ~x(0) = − √15 we have ~x(t) = √15 (λ1 )t 1 − √15 (λ2 )t 2 , i.e. 1 1 0 1 1

b fA (λ) = λ(λ − 1) − 1 = λ2 − λ − 1 so λ1,2 =

a(t) =

√1 ((λ1 )t+1 5

j(t) =

√1 ((λ1 )t 5

c As t → ∞,

a(t) j(t)

√ 1± 5 2

with eigenvectors

− (λ2 )t+1 )

− (λ2 )t ). → λ1 =

√ 1+ 5 2 ,

since |λ2 | < 1.

7.3.49 This “random” matrix A = [~0 ~v2 · · · ~vn ] is unlikely to have any zeros above the diagonal. In this case, the columns ~v2 , . . . , ~vn will be linearly independent (none of them is redundant), so that rank(A) = n − 1 and geometric multiplicity(0) = dim(ker(A)) = n − rank(A) = 1. Alternatively, you can argue in terms of rref(A). √ 1 −1 7.3.50 a fA (λ) = (2 − λ)2 − 3 so λ1,2 = 2 ± 3 (or approximately 3.73 and 0.27) with eigenvectors √ and √ . 3 3 See Figure 7.23. 324

Section 7.3

Figure 7.23: for Problem 7.3.50a.

b The trajectory starting at

0 0 = is above the line Eλ1 , so that At 1 1

(second column of At ) has a slope of more than

√ √ 3, for all t. Applying this to t = 6 gives the estimate 3 <

1351 780 .

1 t 1 Likewise, the trajectory starting at = is below Eλ1 , so that A 1 1 (sum of the two columns of √ At ) has a slope of less than 3. Applying this to t = 4 gives

265 153

<

√ 3.

c det(A6 ) = (det A)6 = 1 and det(A6 ) = 13512 − 780 · 2340, so that 13512 − 780 · 2340 = 1. Dividing both sides by 1351 · 780 we obtain Now note that Therefore

1351 780

2340 1351

−

d The slope of A6

is the slope of A6

√ 3<

1351 780

−

2340 1351

1351 780

−

=

1 780·1351

< 10−6 .

√ 1 , which is less than 3. 0

< 10−6 , as claimed.

√ 2131 1 is less than 3, i.e. = 3691 1

2340 1351

−λ 7.3.51 fA (λ) = det(A − λI3 ) = det 1 0

3691 2131

<

√

3.

0 a −λ b = −λ3 + cλ2 + bλ + a. 1 c−λ

n 7.3.52 The characteristic polynomial is fA (λ) = (−1) λn − an−1 λn−1 − ... − a1 λ − a0 . We can prove this by induction, using Laplace expansion along the first row: n+1 fA (λ) = det (A − λIn ) = −λ det (A − λIn )11 + (−1) a0 det (A − λIn )1n n−1 n+1 = −λ (−1) λn−1 − an−1 λn−2 − ... − a2 λ − a1 + (−1) a0 n n n−1 = (−1) λ − an−1 λ − ... − a1 λ − a0 . Note that (A − λIn )1n is upper triangular with all 1’s on the diagonal, so that det (A − λIn )1n = 1. 325

Chapter 7

7.3.53 a B =

0 1 0 0 0

0 a ∗ 0 b ∗ 1 c ∗ 0 0 ∗ 0 0 ∗

∗ ∗ ∗ ∗ ∗

.

0 0 a b Let B11 = 1 0 b and B22 be the diagonal blocks of the matrix B we found in part (a). Since A is similar 0 1 c to B, we have fA (λ) = fB (λ) = fB22 (λ) fB11 (λ) = fB22 (λ) −λ3 + cλ2 + bλ + a , by Exercise 51. Now we observe that h (λ) = fB22 (λ)is a quadratic polynomial, completing our proof. c From part (b) we know that fA (A) = h(A) −A3 + cA2 + bA + aI5 . The given equation A3~v = a~v + bA~v + cA2~v can be written as −A3 + cA2 + bA + aI5 ~v = ~0, implying that fA (A) ~v = h(A) −A3 + cA2 + bA + aI5 ~v = ~0, as claimed.

7.3.54 a. B11

=

0 1 0 ... 0 0

0 0 1 ... 0 0

0 0 0 ... 0 0

... ... ... ... ... ...

0 a0 0 a1 0 a2 ... ... 0 am−2 1 am−1

, B21 = 0.

b. We find fA (λ) = fB (λ) = fB22 (λ) fB11 (λ) as in part (b) of Exercise 53. Now matrix B11 has the form discussed in Exercise 52, where we show that m that det (B11 ) = (−1) λm − am−1 λm−1 − ... − a1 λ − a0 . It follows m fA (λ) = (−1) fB22 (λ) λm − am−1 λm−1 − ... − a1 λ − a0 , as claimed. c. Using the same approach as in Exercise 53, part (c ), we can write the given equation for Am~v as m m−1 ~ A − am−1 A − ... − a1 A − a0 In ~v = 0. Thus m fA (A) ~v = (−1) fB22 (A) Am − am−1 Am−1 − ... − a1 A − a0 In ~v = ~0, as claimed. d. Our work in parts (a) through (c) shows that fA (A) ~v = ~0 for all ~v in Rn , meaning that fA (A) = 0, the zero matrix.

Section 7.4 7.4.1 Matrix A is diagonal already, so it’s certainly diagonalizable. Let S = I2 . 7.4.2 Diagonalizable. The eigenvalues are 2,3, with associated eigenvectors 2 0 −1 . S AS = D = 0 3 326

1 1 1 1 , . If we let S = , then 0 1 0 1

Section 7.4

7.4.3 Diagonalizable. The eigenvalues are 0,3, with associated eigenvectors 0 0 . then S −1 AS = D = 0 3 7.4.4 Diagonalizable. The eigenvalues are 0,7, with associated eigenvectors 0 0 then S −1 AS = D = . 0 7

−1 1 1 −1 , . If we let S = , 1 2 2 1

−2 1 −2 1 , . If we let S = , 1 3 1 3

7.4.5 Fails to be diagonalizable. There is only one eigenvalue, 1, with a one-dimensional eigenspace. 7.4.6 Fails to be diagonalizable. There is only one eigenvalue, 2, with a one-dimensional eigenspace. 4 −1 4 −1 7.4.7 Diagonalizable. The eigenvalues are 2,−3, with associated eigenvectors , . If we let S = , 1 1 1 1 2 0 then S −1 AS = D = . 0 −3 7.4.8 Diagonalizable. The eigenvalues are 4,−2, with associated eigenvectors 4 0 −1 . then S AS = D = 0 −2

1 −1 −1 1 , . If we let S = , 1 1 1 1

7.4.9 Fails to be diagonalizable. There is only one eigenvalue, 1, with a one-dimensional eigenspace. 7.4.10 Fails to be diagonalizable. There is only one eigenvalue, 1, with a one-dimensional eigenspace. 7.4.11 Fails to be diagonalizable. The eigenvalues are 1,2,1, and the eigenspace E1 = ker(A − I3 ) = span(~e1 ) is only one-dimensional. 1 0 −1 7.4.12 Diagonalizable. The eigenvalues are 2,1,1, with associated eigenvectors 0 , 1 , 0 . If we let S = 0 0 1 2 0 0 1 0 −1 0 1 0 , then S −1 AS = D = 0 1 0 . 0 0 1 0 0 1 7.4.13 Diagonalizable. The eigenvalues are

1 0 0

1 0 1 1 1 2 , then S −1 AS = D = 0 2 0 0 0 1

1 1 1 1,2,3, with associated eigenvectors 0 , 1 , 2 . If we let S = 0 0 1 0 0 . 3

0 −1 1 7.4.14 Diagonalizable. The eigenvalues are 3,2,1, with associated eigenvectors 0 , 1 , −1 . If we let 1 0 0 327

Chapter 7

1 −1 1 S = 0 0 0 7.4.15

2 1 0

3 0 0 −1 , then S −1 AS = D = 0 2 0 0 1

2 1 0 Diagonalizable. The eigenvalues are 1, −1, 1, with associated eigenvectors 1 , 1 , 0 . If we let S = 0 0 1 1 0 0 1 0 1 0 , then S −1 AS = D = 0 −1 0 . 0 0 1 0 1

7.4.16 Diagonalizable. The eigenvalues are

2 0 1

0 0 . 1

1 0 3 0 0 1 , then S −1 AS = D = 0 2 1 0 0 0

2 1 0 3,2,1, with associated eigenvectors 0 , 0 , 1 . If we let S = 1 1 0 0 0 . 1

7.4.17 Diagonalizable. 1 1 1 1 1 The eigenvalues are 0,3,0, with associated eigenvectors −1 , 1 , 0 . If we let S = −1 1 0 1 −1 1 0 0 0 0 S −1 AS = D = 0 3 0 . 0 0 0

1 0 , then −1

−1 1 0 7.4.18 Diagonalizable. The eigenvalues are 0,2,1, with associated eigenvectors 0 , 0 , 1 . If we let S = 1 1 0 0 0 0 −1 1 0 0 0 1 , then S −1 AS = D = 0 2 0 . 0 0 1 1 1 0 7.4.19 Fails to be diagonalizable. The eigenvalues are 1,0,1, and the eigenspace E1 = ker(A − I3 ) = span(~e1 ) is only one-dimensional. 0 1 −1 7.4.20 Diagonalizable. The eigenvalues are 1,2,0, with associated eigenvectors 1 , 2 , 0 . If we let S = 0 1 1 0 1 −1 1 0 0 1 2 0 , then S −1 AS = D = 0 2 0 . 0 1 1 0 0 0 7.4.21 Diagonalizable for all values of a, since there are always two distinct eigenvalues, 1 and 2. See Theorem 7.4.3. 7.4.22 Diagonalizable except if b = 1 and a 6= 0. (In that case we have only one eigenvalue, 1, with a one-dimensional eigenspace.). 328

Section 7.4 7.4.23 Diagonalizable for positive a. The characteristic polynomial is (λ − 1)2 − a, so that the eigenvalues are √ λ = 1 ± a. If a is positive, then we have two distinct real eigenvalues, so that the matrix is diagonalizable. If a is negative, then there are no real eigenvalues. If a is 0, then 1 is the only eigenvalue, with a one-dimensional eigenspace. 7.4.24 Diagonalizable for all values polynomial is λ2 − (a + c)λ + ac − b2 , so that √ √ of a, b, and c. The characteristic a+c± (a−c)2 +4b2 a+c± (a+c)2 −4(ac−b2 ) = . Note that the expression whose square root the eigenvalues are λ = 2 2 we take (the “discriminant”) is always positive or 0, since it is the sum of two squares. If the discriminant is positive, then we have two distinct real eigenvalues, and everything is fine. The discriminant is 0 only if a = c and b = 0. In that case the matrix is diagonal already, and certainly diagonalizable as well. 7.4.25 Diagonalizable for all values of a, b, and c, since we have three distinct eigenvalues, 1, 2, and 3. 7.4.26 The eigenvalues are 1, 2, 1, and thematrix is diagonalizable if (and only if) the eigenspace E1 is two0 a b 0 1 c dimensional. Now E1 = ker(A − I3 ) = ker 0 1 c = ker 0 0 b − ac is two-dimensional if (and only 0 0 0 0 0 0 if) b − ac = 0. Thus the matrix is diagonalizable if and only if b = ac. 7.4.27 Diagonalizable only if a = b = c = 0. Since 1 is the only eigenvalue, it is required that E1 = R3 , that is, the matrix must be the identity matrix. 7.4.28 Diagonalizable for positive values of a. The characteristic polynomial is −λ3 + aλ = −λ(λ2 − a). If a is √ positive, then we have three distinct real eigenvalues, 0, ± a, so that the matrix will be diagonalizable. If a is negative or 0, then 0 is the only real eigenvalue, and the matrix fails to be diagonalizable. 7.4.29 Not diagonalizable for any a. The characteristic polynomial is −λ3 + a, so that there is only one real √ eigenvalue, 3 a, for all a. Since the corresponding eigenspace isn’t all of R3 , the matrix fails to be diagonalizable.

0 0 a 7.4.30 First we observe that all the eigenspaces of A = 1 0 3 are one-dimensional, regardless of the value of 0 1 0 1 0 ∗ a, since rref(A − λI3 ) is of the form 0 1 ∗ for all λ. Thus A is diagonalizable if and only if there are three 0 0 ∗ distinct real eigenvalues. The characteristic polynomial of A is −λ3 + 3λ + a. Thus the eigenvalues of A are the solutions of the equation λ3 − 3λ = a. See Figure 7.24 with the function f (λ) = λ3 − 3λ; using calculus, we find the local maximum f (−1) = 2 and the local minimum f (1) = −2. To count the distinct eigenvalues of A, we have to examine how many times the horizontal line y = a intersects the graph of f (λ). The answer is three if |a| < 2, two if a = ±2, and one if |a| > 2. Thus A is diagonalizable if and only if |a| < 2, that is, −2 < a < 2. 1 2 are −1 and 5, with associated 7.4.31 In Example 2 of Section 7.3 we see that the eigenvalues of A = 4 3 1 1 1 1 −1 0 −1 eigenvectors and . If we let S = , then S AS = D = . −1 2 −1 2 0 5

329

Chapter 7

Thus A = SDS

−1

t

t

and A = SD S

−1

=

1 3

=

1 3

1 −1

1 2

(−1)t 0

2(−1)t + 5t 2(5t ) − 2(−1)t

0 55

2 1

−1 1

(−1)t+1 + 5t 2(5t ) + (−1)t

1 2 4 −2 . If we let S = and are 3 and 2, with associated eigenvectors 7.4.32 The eigenvalues of A = 1 1 1 t 1 2 1 3 0 2 1 3 0 1 −1 , then S −1 AS = D = . Thus A = SDS −1 and At = SDt S −1 = = 0 2 1 1 0 2t −1 2 1 t1 2(3 ) − 2t 2t+1 − 2(3t ) . 3t − 2t 2t+1 − 3t

1 2 −2 1 7.4.33 The eigenvalues of A = are 0 and 7, with associated eigenvectors and . If we let S = 3 6 1 3 0 0 −3 1 0 0 −2 1 1 −2 1 −1 t t −1 −1 = . Thus A = SDS and A = SD S = 7 , then S AS = D = 0 7t 1 2 1 3 0 7 1 t3 t 7 2(7 ) 1 t−1 A. We can find the same result more directly by observing that A2 = 7A. 7 3(7t ) 6(7t ) = 7

−1 1 −1 1 7.4.34 The eigenvalues of A are 1/4 and 1, with associated eigenvectors and . If we let S = , 1 2 1 2 −2 1 (1/4)t 0 −1 1 1/4 0 = . Thus A = SDS −1 and At = SDt S −1 = 13 then S −1 AS = D = 1 1 0 1 1 2 0 1 t 1 − (1/4)t 1 1 + 2(1/4) · 3 2 − 2(1/4)t 2 + (1/4)t −1 6 7.4.35 Matrix has the eigenvalues 3 and 2. If ~v and w ~ are associated eigenvectors, and if we let S = [~v w], ~ −2 6 3 0 −1 6 3 0 −1 6 . is indeed similar to , so that matrix S= then S −1 0 2 −2 6 0 2 −2 6

−1 −2 3 to the diagonal matrix 0 Theorem 3.4.6.

7.4.36 Yes. The matrices

2 both have the eigenvalues 3 and 2, so that each of them is similar 4 0 −1 6 1 2 , by Algorithm 7.4.4. Thus is similar to , by parts b and c of 2 −2 6 −1 4 1 6 and −1 6

7.4.37 Yes. Matrices A and B have the same characteristic polynomial, λ2 − 7λ + 7, so that they have the same √ λ1 0 7± 21 two distinct real eigenvalues λ1,2 = 2 . Thus both A and B are similar to the diagonal matrix , by 0 λ2 Algorithm 7.4.4. Therefore A is similar to B, by parts b and c of Theorem 3.4.6.

2 0 7.4.38 No. As a counterexample, consider A = 0 2

2 1 and B = . 0 2

7.4.39 The eigenfunctions with eigenvalue λ are the nonzero functions f (x) such that T (f (x)) = f ′ (x) − f (x) = λf (x), or f ′ (x) = (λ + 1)f (x). From calculus we recall that those are the exponential functions of the form 330

Section 7.4 f (x) = Ce(λ+1)x , where C is a nonzero constant. Thus all real numbers are eigenvalues of T , and the eigenspace Eλ is one-dimensional, spanned by e(λ+1)x . 7.4.40 The eigenfunctions with eigenvalue λ are the nonzero functions f (x) such that T (f (x)) = 5f ′ (x) − 3f (x) = λf (x), or f ′ (x) = λ+3 5 f (x). From calculus we recall that those are the exponential functions of the form (λ+3)x/5 f (x) = Ce , where C is a nonzero constant. Thus all real numbers are eigenvalues of T , and the eigenspace Eλ is one-dimensional, spanned by e(λ+3)x/5 . 7.4.41 The nonzero symmetric matrices are eigenmatrices with eigenvalue 2, since L(A) = A + AT = 2A in this T case. The nonzero skew-symmetric matrices have 0, eigenvalue since L(A) = A+ A = A − A = 0. Yes, L 1 0 0 1 0 0 0 1 is diagonalizable, since we have the eigenbasis , , , (three symmetric matrices, 0 0 1 0 0 1 −1 0 and one skew-symmetric one). 7.4.42 The nonzero symmetric matrices are eigenmatrices with eigenvalue 0, since L(A) = A − AT = A − A = 0 in T this case. The nonzero skew-symmetric matrices have eigenvalue 2, since L(A) = A− A = A + A = 2A. Yes, L 1 0 0 1 0 0 0 1 is diagonalizable, since we have the eigenbasis , , , (three symmetric matrices, 0 0 1 0 0 1 −1 0 and one skew-symmetric one). 7.4.43 The nonzero real numbers are “eigenvectors” with eigenvalue 1, and the nonzero imaginary numbers (of the form iy) are “eigenvectors” with eigenvalue −1. Yes, T is diagonalizable, since we have the eigenbasis 1,i. 7.4.44 The nonzero sequence (x0 , x1 , x2 , . . .) is an eigensequence with eigenvalue λ if T (x0 , x1 , x2 , . . .) = (x2 , x3 , x4 , . . .) = λ(x0 , x1 , x2 , . . .) = (λx0 , λx1 , λx2 , . . .). This means that x2 = λx0 , x3 = λx1 , . . . , xn+2 = λxn , . . .. These are the sequences of the form (a, b, λa, λb, λ2 a, λ2 b, . . .), where at least one of the first two terms, a and b, is nonzero. Thus all real numbers λ are eigenvalues of T , and the eigenspace Eλ is two-dimensional, with basis (1, 0, λ, 0, λ2 , 0, . . .), (0, 1, 0, λ, 0, λ2 , . . .). 7.4.45 The nonzero sequence (x0 , x1 , x2 , . . .) is an eigensequence with eigenvalue λ if T (x0 , x1 , x2 , . . .) = (0, x0 , x1 , x2 , . . .) = λ(x0 , x1 , x2 , . . .) = (λx0 , λx1 , λx2 , . . .). This means that 0 = λx0 , x0 = λx1 , x1 = λx2 , . . . , xn = λxn+1 , . . . . If λ is nonzero, then these equations imply that x0 = λ1 0 = 0, x1 = 1 1 λ x0 = 0, x2 = λ x1 = 0, . . . , so that there are no eigensequences in this case. If λ = 0, then we have x0 = λx1 = 0, x1 = λx2 = 0, x2 = λx3 = 0, . . . , so that there aren’t any eigensequences either. In summary: There are no eigenvalues and eigensequences for T . 7.4.46 The nonzero sequence (x0 , x1 , x2 , . . .) is an eigensequence with eigenvalue λ if T (x0 , x1 , x2 , . . .) = (x0 , x2 , x4 , . . .) = λ(x0 , x1 , x2 , . . .) = (λx0 , λx1 , λx2 , . . .). This means that x0 = λx0 , x2 = λx1 , x4 = λx2 , . . . , x2n = λxn , . . . . For each λ, there are lots of eigensequences: we can choose the terms xk for odd k freely and then fix the xk for even k according to the formula x2n = λxn . For example, eigenspace E3 consists of the sequences of the form (x0 = 0, x1 , x2 = 3x1 , x3 , x4 = 9x1 , x5 , x6 = 3x3 , x7 , x8 = 27x1 , x9 , . . .), where x1 , x3 , x5 , x7 , x9 , . . . are arbitrary. Note that all the eigenspaces are infinite-dimensional. The condition x0 = λx0 implies that x0 = 0, except for λ = 1, in which case x0 is arbitrary. 7.4.47 The nonzero even functions, of the form f (x) = a+cx2 , are eigenfunctions with eigenvalue 1, and the nonzero odd functions, of the form f (x) = bx, have eigenvalue −1. Yes, T is diagonalizable, since the standard basis, 331

Chapter 7 1, x, x2 , is an eigenbasis for T . 7.4.48 Apply T to the standard basis: T (1) = 1, T (x) = 2x, and T (x2 ) = (2x)2 = 4x2 . This gives the eigenvalues 1, 2, and 4, with corresponding eigenfunctions 1, x, x2 . Yes, T is diagonalizable, since the standard basis is an eigenbasis for T . 1 −1 1 3 −6 . The eigenvalues of B are 7.4.49 The matrix of T with respect to the standard basis 1, x, x2 is B = 0 0 0 9 1 −1 1 1, 3, 9, with corresponding eigenvectors 0 , 2 , −4 . The eigenvalues of T are 1,3,9, with corresponding 4 0 0 eigenfunctions 1, 2x − 1, 4x2 − 4x + 1 = (2x − 1)2 . Yes, T is diagonalizable, since the functions 1, 2x − 1, (2x − 1)2 from an eigenbasis.

1 −3 9 7.4.50 The matrix of T with respect to the standard basis 1, x, x2 is B = 0 1 −6 . The only eigenvalue of B 0 0 1 1 is 1, with corresponding eigenvector 0 . The only eigenvalue of T is 1 as well, with corresponding eigenfunction 0 f (x) = 1. T fails to be diagonalizable, since there is only one eigenvalue, with a one-dimensional eigenspace. 7.4.51 The nonzero constant functions f (x) = b are the eigenfunctions with eigenvalue 0. If f (x) is a polynomial of degree ≥ 1, then the degree of f (x) exceeds the degree of f ′ (x) by 1 (by the power rule of calculus), so that f ′ (x) cannot be a scalar multiple of f (x). Thus 0 is the only eigenvalue of T , and the eigenspace E0 consists of the constant functions. 7.4.52 Let f (x) = a0 + a1 x + a2 x2 + · · · + an xn , with an 6= 0, be an eigenfunction of T with eigenvalue λ. Then T (f (x)) = x(a1 + 2a2 x + · · · + nan xn−1 ) = a1 x + 2a2 x2 + · · · + nan xn = λ(a0 + a1 x + a2 x2 + · · · + an xn ) = λa0 + λa1 x + λa2 x2 + · · · + λan xn . This means that λa0 = 0, λa1 = a1 , λa2 = 2a2 , . . . , λan = nan . Since we assumed that an 6= 0, we can conclude that λ = n. Now it follows that a0 = a1 = · · · = an−1 = 0, so that the eigenfunctions with eigenvalue n are the nonzero scalar multiples of xn , of the form f (x) = an xn . This makes good sense, since T (xn ) = x(nxn−1 ) = nxn . In summary: The eigenvalues are the integers n = 0, 1, 2, . . ., and the eigenspace En is span (xn ) 7.4.53 Suppose basis D consists of f1 , . . . , fn . We are told that the D-matrix D of T is diagonal; let λ1 , λ2 , . . . , λn be the diagonal entries of D. By Theorem 4.3.3., we know that [T (fi )]D = (ith column of D) = λi~ei , for i = 1, 2, . . . , n, so that T (fi ) = λi fi , by definition of coordinates. Thus f1 , . . . , fn is an eigenbasis for T , as claimed. 7.4.54 Note that A2 = 0, but B 2 6= 0. Since A2 fails to be similar to B 2 , matrix A isn’t similar to B (see Example 7 of Section 3.4). 7.4.55 Let A =

0 1 0 0

and B =

1 0

0 , for example. 0

AB 7.4.56 The hint shows that matrix M = B

0 0

0 is similar to N = B 332

0 ; thus matrices M and N have the BA

Section 7.4 AB − λIn 0 = (−λ)n det(AB − B −λIn λIn ) = (−λ)n fAB (λ). To understnd the second equality, consider Theorem 6.1.5. Likewise, fN (λ) = (−λ)n fBA (λ). It follows that (−λ)n fAB (λ) = (−λ)n fBA (λ) and therefore fAB (λ) = fBA (λ), as claimed.

same characteristic polynomial, by Theorem 7.3.6a. Now fM (λ) = det

AB B

0 0

Im 0

A In

Im 0

A In

7.4.57 Modifying the hint in Exercise 56 slightly, we can write = 0 0 AB 0 0 0 . Thus matrix M = is similar to N = . By Theorem 7.3.6a, matrices M B BA B 0 B BA and N have the same characteristic polynomial. AB − λIm 0 Now fM (λ) = det = (−λ)n det(AB − λIm ) = (−λ)n fAB (λ). To understnd the second B −λIn equality, consider Theorem 6.1.5. Likewise, fN (λ) −λIm 0 = det = (−λ)m fBA (λ). B BA − λIn It follows that (−λ)n fAB (λ) = (−λ)m fBA (λ). Thus matrices AB and BA have the same nonzero eigenvalues, with the same algebraic multiplicities. If mult(AB) and mult(BA) are the algebraic multiplicities of 0 as an eigenvalue of AB and BA, respectively, then the equation (−λ)n fAB (λ) = (−λ)m fBA (λ) implies that n + mult(AB) = m + mult(BA). 7.4.58 a. If ~v is in the image of A, then ~v = Aw ~ for some vector w. ~ Now A~v = A2 w ~ = ~0, showing that ~v is in the kernel of A. b. From part (a) we know that dim imA ≤ dim ker A. Also, dim imA > 0 since A is nonzero, and dim ker A + dim imA = 3 by the rank-nullity theorem. This leaves us with only one possibility, namely, dim imA = 1 and dim ker A = 2. c. Consider a relation c1~v1 + c2~v2 + c3~v3 = ~0. Multiplying both sides with A from the left, keeping in mind that ~v1 and ~v3 are in the kernel, we find c2 A~v2 = c2~v1 = ~0, so c2 = 0. Now c1 = c3 = 0 since ~v1 and ~v3 are independent by construction. 0 1 0 d. We have B = 0 0 0 since A~v1 = ~0, A~v2 = ~v1 and A~v3 = ~0. 0 0 0 0 1 0 7.4.59 Yes, A is similar to B, since both A and B are similar to 0 0 0 , by Exercise 58. 0 0 0

7.4.60 We will use the method outlined in Exercise 58. We start

1 Since ~v1 = (first column of A) = A~e1 , we can let ~v2 = ~e1 = 0 0 333

1 with the vector ~v1 = 2 in the image of A. 3 2 . Finally, we let ~v3 = 1 , another vector in 0

Chapter 7

1 the kernel (use Kyle numbers!). Thus S = 2 3

1 2 0 1 will do the job, but there are other answers. 0 0

7.4.61 First we need to verify that the vectors ~v1 , ..., ~vr , w ~ 1 , ..., w ~ r , ~u1 , ..., ~um are linearly independent. Consider a relation a1~v1 + ... + ar ~vr + b1 w ~ 1 + ... + br w ~ r + c1 ~u1 + ... + cm ~um = ~0. Multiplying both sides with A, we find b1~v1 + ... + br ~vr = ~0, so that b1 = ... = br = 0 since the ~vi are independent by construction. Now the ai and the ci must be 0 as well, since the vectors ~v1 , ..., ~vr , ~u1 , ..., ~um are independent by construction. Secondly, we need to show that 2r + m = n, meaning that we have the right number of vectors to form a basis of Rn. Indeed. ~ i gives us a block n = dim ker A + dim imA = (r + m) + r = 2r + m. Each of the r pairs ~vi , w 0 1 J= along the diagonal of B, since A~vi = ~0 and Aw ~ i = ~vi , while all other entries of matrix B are zero. 0 0 7.4.62 A nonzero function f is an eigenfunction of T with eigenvalue λ if T (f ) = f ′′ + af ′ + bf = λf , or, f ′′ + af ′ + (b − λ)f = 0. By Theorem 4.1.7, this differential equation has a two-dimensional solution space. Thus all real numbers are eigenvalues of T , and all the eigenspaces are two-dimensional. 7.4.63 a We need to solve the differential equation f ′′ (x) = f (x). As in Example 18 of Section 4.1, we will look for exponential solutions. The function f (x) = ekx is a solution if k 2 = 1, or k = ±1. Thus the eigenspace E1 is the span of functions ex and e−x . b We need to solve the differential equation f ′′ (x) = 0. Integration gives f ′ (x) = C, a constant. If we integrate again, we find f (x) = Cx + c, where c is another arbitrary constant. Thus E0 = span(1, x). c The solutions of the differential equation f ′′ (x) = −f (x) are the functions f (x) = a cos(x) + b sin(x), so that E−1 = span(cos x, sin x). See the introductory example of Section 4.1 and Exercise 4.1.58. d Modifying part c, we see that the solutions of the differential equation f ′′ (x) = −4f (x) are the functions f (x) = a cos(2x) + b sin(2x), so that E−4 = span(cos(2x), sin(2x)). 1 1 . Arguing as in Exercise 65, we and 7.4.64 The eigenvalues of A are 1 and 3, with associated eigenvectors 1 0 1 0 0 1 find the basis , for V, so that dim(V ) = 2. 0 0 0 1 7.4.65 Let’s write S in terms of its columns, as S = [ ~v w ~ ]. 5 0 , or, [ A~v Aw ~ ] = [ 5~v −w ~ ] , that is, we want ~v to be in the eigenspace We want A [ ~v w ~ ] = [ ~v w ~] 0 −1 1 1 , so that S must be of the form and E−1 = span E5 , and w ~ in E−1 . We find that E5 = span −1 2 0 1 1 0 0 1 1 0 1 1 b a , and dim(V ) = 2. , . Thus, a basis of the space V is +b =a 0 −1 2 0 0 −1 2 0 −1 2 1 0 1 ~ 7.4.66 For A we find the eigenspaces E1 = span 0 , 1 and E2 = span 1 . If we write S = [~u ~v w], 0 −1 0 334

Section 7.4

1 0 then we want A[~u ~v w] ~ = [~u ~v w] ~ 0 1 0 0 w ~ must be in E2 . The matrices S we seek

1 and a basis of V is 0 0 five.

0 0 0 0 0 0, 1 0 0 0 −1 0

0 ~ = [~u ~v 2w], ~ that is, 0 , or [A~u A~v Aw] 2 0 1 are of the form S = a 0 + b 1 −1 0 0 0 1 0 0 0 0 0 0 0,0 0 0,0 1 0,0 0 0 0 0 0 0 −1 0 0 0

~u and ~v must be in E1 , and 1 0 1 c 0 + d 1 e 1 , 0 −1 0 1 1 . The dimension of V is 0

~ 1, w ~ 2 ). As in Exercise 65, we can see that S must be of the 7.4.67 Let Eλ1 = span(~v1 , ~v2 , ~v3 ) and Eλ2 = span(w form [ ~x1 ~x2 ~x3 ~x4 ~x5 ] where ~x1 , ~x2 and ~x3 are in Eλ1 and ~x4 and ~x5 are in Eλ2 . Thus, we can write ~x1 = c1~v1 + c2~v2 + c3~v3 , for example, or ~x5 = d1 w ~ 1 + d2 w ~ 2. Using Summary 4.1.6, we find a basis: [ ~v1 ~0 ~0 ~0 ~0 ] , [ ~v2 ~0 ~0 ~0 ~0 ] , [ ~v3 ~0 ~0 ~0 ~0 ] , [ ~0 ~v1 ~0 ~0 ~0 ] , [ ~0 ~v2 ~0 ~0 ~0 ] , [ ~0 ~v3 ~0 ~0 ~0 ] , [ ~0 ~0 ~v1 ~0 ~0 ] , [ ~0 ~0 ~v2 ~0 ~0 ] , [ ~0 ~0 ~v3 ~0 ~0 ] , [ ~0 ~0 ~0 w ~ 1 ~0 ] , [ ~0 ~0 ~0 w ~ 2 ~0 ] , [ ~0 ~0 ~0 ~0 w ~ 1 ] , [ ~0 ~0 ~0 ~0 w ~2 ] . Thus, the dimension of the space of matrices S is 3 + 3 + 3 + 2 + 2 = 13. 7.4.68 Let ~v1 , . . . , ~vn be an eigenbasis for A, with A~vi = λi~vi . Arguing as in Exercises 64 through 67, we see that the ith column of S must be in Eλi , so that it must be of the form ci~vi for some scalar ci . The matrices S we seek are of the form S = [c1~v1 . . . cn~vn ], involving the n arbitrary constants c1 , . . . , cn , so that the dimension of V is n. 7.4.69 a B is diagonalizable since it has three distinct eigenvalues, so that S −1 BS is diagonal for some invertible S. But S −1 AS = S −1 I3 S = I3 is diagonal as well. Thus A and B are indeed simultaneously diagonalizable. b There is an invertible S such that S −1 AS = D1 and S −1 BS = D2 are both diagonal. Then A = SD1 S −1 and B = SD2 S −1 , so that AB = (SD1 S −1 )(SD2 S −1 ) = SD1 D2 S −1 and BA = (SD2 S −1 )(SD1 S −1 ) = SD2 D1 S −1 . These two results agree, since D1 D2 = D2 D1 for the diagonal matrices D1 and D2 . c Let A be In and B a nondiagonalizable n × n matrix, for example, A =

1 0

0 1

and B =

1 0

1 . 1

d Suppose BD = DB for a diagonal D with distinct diagonal entries. The ij-th entry of the matrix BD = DB is bij djj = dii bij . For i 6= j this implies that bij = 0. Thus B must be diagonal. e Since A has n distinct eigenvalues, A is diagonalizable, that is, there is an invertible S such that S −1 AS = D is a diagonal matrix with n distinct diagonal entries. We claim that S −1 BS is diagonal as well; by part d it suffices to show that S −1 BS commutes with D = S −1 AS. This is easy to verify: (S −1 BS)D = (S −1 BS)(S −1 AS) = S −1 BAS = S −1 ABS = (S −1 AS)(S −1 BS) = D(S −1 BS) .

335

Chapter 7 7.4.70 Consider an n × n matrix A with m distinct eigenvalues λ1 , ..., λm . If ~v is an eigenvector of A with eigenvalue λm , then (A − λm In ) ~v = ~0, so that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) ~v = ~0. Since the factors in the product (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) commute, we can conclude that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) ~v = ~0 for all eigenvectors ~v of A. Now assume that A is diagonalizable. Then there exists a basis consisting of eigenvectors, so that (A − λ1 In ) (A − λ2 In )· · · (A − λm In ) ~v = ~0 for all vectors in Rn . We can conclude that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) = 0. Conversely, assume that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) = 0. Then n = dim ker ((A − λ1 In ) (A − λ2 In ) · · · (A − λm In )) ≤ dim ker (A − λ1 In ) + ... + dim ker (A − λm In ) by Exercise 4.2.83. This means that the geometric multiplicities of the eigenvalues add up to n (we know that this sum cannot exceed n), so that A is diagonalizable, by Theorem 7.3.4b. 7.4.71 The eigenvalues are 1 and 2, and (A − I3 ) (A − 2I3 ) = 0. Thus A is diagonalizable.

0 7.4.72 The eigenvalues of this upper triangular matrix are 0 and 1. Now A (A − I3 ) = 0 0 A is diagonalizable if (and only if) b = −ac.

0 b + ac . Matrix 0 0 0 0

7.4.73 If an n × n matrix A has m distinct eigenvalues λ1 , ..., λm , then we can write its characteristic polynomial as fA (λ) = (λ − λ1 ) (λ − λ2 ) · · · (λ − λm ) g (λ) for some polynomial g(λ) of degree n − m. Now fA (A) = (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) g (A) = 0 for a diagonalizable matrix A since (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) = 0, as we have seen in Exercise 70. (In Exercise 7.3.54 we prove the Cayley-Hamilton theorem for all n × n matrices A, but that proof is a bit longer.) 7.4.74 a For a diagonalizable n × n matrix A with only two distinct eigenvalues, λ1 and λ2 , we have (A − λ1 In )(A − λ2 In ) = 0, by Exercise 70. Thus the column vectors of A − λ2 In are in the kernel of A − λ1 In , that is, they are eigenvectors of A with eigenvalue λ1 (or else they are ~0). Conversely, the column vectors of A − λ1 In are eigenvectors of A with eigenvalue λ2 (or else they are ~0). b If A is a 2 × 2 matrix with distinct eigenvalues λ1 and λ2 , then the nonzero columns of A − λ1 I2 are eigenvectors λ1 0 of A with eigenvalue λ2 , as we observed in part (a). Since the matrices A − and A − λ1 I2 have the 0 λ2 λ1 0 same first column, the first column of A − will be an eigenvector of A with eigenvalue λ2 as well (or 0 λ2 λ 0 it is zero). Likewise, the second column of A − 1 will be an eigenvector of A with eigenvalue λ1 (or it is 0 λ2 zero).

Section 7.5 p √ 7.5.1 z = 3 − 3i so |z| = 32 + (−3)2 = 18 and arg(z) = − π4 , √ so z = 18 cos − π4 + i sin − π4 . 336

Section 7.5 7.5.2 If z = r(cos θ + i sin θ) then z 4 = r4 (cos 4θ + i sin 4θ). z 4 = 1 if r = 1, cos 4θ = 1 and sin 4θ = 0 so 4θ = 2kπ for an integer k, and θ = kπ 2 , + i sin kπ i.e. z = cos kπ 2 2 , k = 0, 1, 2, 3. Thus z = 1, i, −1, −i. See Figure 7.24.

Figure 7.24: for Problem 7.5.2. 7.5.3 If z = r(cos θ + i sin θ), then z n = rn (cos(nθ) + i sin(nθ)). z n = 1 if r = 1, cos(nθ) = 1, sin(nθ) = 0 so nθ = 2kπ for an integer k, and θ = + i sin 2kπ i.e. z = cos 2kπ n n , k = 0, 1, 2, . . . , n − 1. See Figure 7.25.

2kπ n ,

Figure 7.25: for Problem 7.5.3. 7.5.4 Let z = r(cos θ + i sin θ) then w =

√

7.5.5 Let z = r(cos θ + i sin θ) then w =

√ n

r cos r cos

θ+2πk 2

θ+2πk n

+ i sin

θ+2πk 2

+ i sin

θ+2πk n

, k = 0, 1.

, k = 0, 1, 2, . . . , n − 1.

7.5.6 If we have z = r(cos θ + i sin θ) then z1 must have the property that z · z1 = 1 = cos 0 + i sin 0 i.e. |z| · z1 = 1 and arg z · z1 = arg(z) + arg z1 = 0 so z1 = 1r (cos(−θ) + i sin(−θ)) = 1 1 ¯. See Figure 7.26. r (cos θ − i sin θ) (since cosine is even, sine odd). Hence z is a real scalar multiple of z √ 7.5.7 |T (z)| = |z| 2 and arg(T (z)) = arg(1 − i) + arg(z) = − π4 + arg(z) so T is a clockwise rotation by √ by a scaling of 2. 7.5.8 By Theorem 7.5.1, (cos 3θ + i sin 3θ) = (cos θ + i sin θ)3 , i.e. cos 3θ + i sin 3θ = cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 φ 337

π 4

followed

Chapter 7

Figure 7.26: for Problem 7.5.6. = (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ). Equating real and imaginary parts, we get cos 3θ = cos3 θ − 3 cos θ sin2 θ sin 3θ = 3 cos2 θ sin θ − sin3 θ. 7.5.9 |z| =

√

0.82 + 0.72 =

√ 1.15, arg(z) = arctan − 0.7 0.8 ≈ −0.72. See Figure 7.27.

Figure 7.27: for Problem 7.5.9. The trajectory spirals outward, in the clockwise direction. 7.5.10 Let p(x) = ax3 + bx2 + cx + d, where a 6= 0. Since p must have a real root, say λ1 , we can write p(x) = a(x − λ1 )g(x) where g(x) is of the form g(x) = x2 + px + q. On page 350, we see that g(x) = (x − λ2 )(x − λ3 ), so that p(x) = a(x − λ1 )(x − λ2 )(x − λ3 ), as claimed. 7.5.11 Notice that f (1) = 0 so λ = 1 is a root of f (λ). Hence f (λ) = (λ − 1)g(λ), where g(λ) = Setting g(λ) = 0 we get λ = 1 ± 2i so that f (λ) = (λ − 1)(λ − 1 − 2i)(λ − 1 + 2i).

f (λ) λ−1

= λ2 − 2λ + 5.

7.5.12 We will use the facts: i) z + w = z¯ + w ¯ and ii) (z n ) = z¯n which are easy to check. Assume λ0 is a complex root of f (λ) = an λn + · · · + a1 λ + a0 , where the coefficients ai are real. Since λ0 is a root of f , we have an λn0 + an−1 λ0n−1 + · · · + a1 λ0 + a0 = 0. 338

Section 7.5 Taking the conjugate of both sides we get an λn0 + an−1 λ0n−1 + · · · + a1 λ0 + a0 = ¯0 so by fact i), and factoring the real constants we get an λn0 + an−1 λ0n−1 + · · · + a1 λ0 + a0 = 0. Now, by fact ii), an (λ0 )n + an−1 (λ0 )n−1 + · · · + a1 λ0 + a0 = 0, i.e. λ0 is also a root of f , as claimed. 2 2 0 0 ~ ~v = , +i is an eigenvector with eigenvalue 2i, so that we can let S = w 0 0 1 1 | {z } | {z } w ~ ~v 0 −2 −1 with S AS = . 2 0

7.5.13

2i 1

1+i 1

=

1 1 1 1 ~ ~v = 7.5.14 = +i is an eigenvector with eigenvalue i, so that we can let S = w , 1 0 0 1 | {z } | {z } w~ ~v 0 −1 −1 . with S AS = 1 0 1 +i 2 | {z } | ~ v 0 1 , with S −1 AS = 1 2

7.5.15

1 2+i

1 1+i

=

0 ~ ~v = is an eigenvector with eigenvalue 2 + i, so that we can let S = w 1 {z } w ~ 2 −1 . 1 2

0 ~ ~v = is an eigenvector with eigenvalue 4 + i, so that we can let S = w 1 {z } w ~ 4 −1 . 1 4

1 7.5.16 = +i 1 | {z } | ~ v 0 1 −1 , with S AS = 1 1

2 +i −1 | {z } | ~ v 0 2 3 , with S −1 AS = 2 −1 4

7.5.17

2 −1 + 2i

=

0 ~ ~v = is an eigenvector with eigenvalue 3 + 4i, so that we can let S = w 2 {z } w ~ −4 . 3

7.5.18 Let ~v1 , ~v2 be two eigenvectors of A. They define a parallelogram of area S = | det[~v1 ~v2 ]|. Now A~v1 = λ1~v1 and A~v2 = λ2~v2 define a parallelogram of area S1 = | det[λ1~v1 λ2~v2 ]| = |λ1 λ2 det[~v1 ~v2 ]| so SS1 = |λ1 λ2 | = | det(A)|. Hence | det(A)| = |λ1 λ2 |, as claimed. In R3 , a similar argument holds if we replace areas by volumes. See Figure 7.28. 7.5.19 a Since A has eigenvalues 1 and 0 associated with V and V ⊥ respectively and since V is the eigenspace of λ = 1, by Theorem 7.5.5, tr(A) = m, det(A) = 0. b Since B has eigenvalues 1 and −1 associated with V and V ⊥ respectively and since V is the eigenspace associated with λ = 1, tr(A) = m − (n − m) = 2m − n, det B = (−1)n−m . 7.5.20 fA (λ) = (3 − λ)(−3 − λ) + 10 = λ2 + 1 so λ1,2 = ±i. 339

Chapter 7

Figure 7.28: for Problem 7.5.18. 7.5.21 fA (λ) = (11 − λ)(−7 − λ) + 90 = λ2 − 4λ + 13 so λ1,2 = 2 ± 3i. 7.5.22 fA (λ) = (1 − λ)(10 − λ) + 12 = λ2 − 11λ + 22 so λ1,2 =

√ 11± 33 . 2

7.5.23 fA (λ) = −λ3 + 1 = −(λ − 1)(λ2 + λ + 1) so λ1 = 1, λ2,3 =

√ −1± 3i . 2

7.5.24 fA (λ) = −λ3 + 3λ2 − 7λ + 5 so λ1 = 1, λ2,3 = 1 ± 2i. (See Exercise 11.) 7.5.25 fA (λ) = λ4 − 1 = (λ2 − 1)(λ2 + 1) = (λ − 1)(λ + 1)(λ − i)(λ + i) so λ1,2 = ±1 and λ3,4 = ±i 7.5.26 fA (λ) = (λ2 − 2λ + 2)(λ2 − 2λ) = (λ2 − 2λ + 2)(λ − 2)λ = 0, so λ1,2 = 1 ± i, λ3 = 2, λ4 = 0. 7.5.27 By Theorem 7.5.5, tr(A) = λ1 + λ2 + λ3 , det(A) = λ1 λ2 λ3 but λ1 = λ2 6= λ3 by assumption, so tr(A) = 1 = 2λ2 + λ3 and det(A) = 3 = λ22 λ3 . Solving for λ2 , λ3 we get −1, 3 hence λ1 = λ2 = −1 and λ3 = 3. (Note that the eigenvalues must be real; why?) 7.5.28 Suppose the complex eigenvalues are z = a+ib and z¯ = a−ib. By Theorem 7.5.5, we have tr(A) = 2+z + z¯ = 2 + 2a = 8, so that a = 3. Furthermore , det(A) = 2z z¯ = 2(a2 + b2 ) = 2(9 + b2 ) = 50, so that b = 4. Hence the complex eigenvalues are 3 ± 4i. 7.5.29 tr(A) = 0 so λ1 + λ2 + λ3 = 0. Also, we can compute det(A) = bcd > 0 since b, c, d > 0. Therefore, λ1 λ2 λ3 > 0. Hence two of the eigenvalues must be negative, and the largest one (in absolute value) must be positive. 7.5.30 a The ith entry of A~x is

n X

aik xk , so that the sum of all the entries of A~x is

k=1 n n X X

i=1 k=1

aik xk =

n n X X

k=1 i=1

aik xk =

n n X X

k=1

aik

i=1

↑

!

xk =

n X

xk = 1.

k=1

1 340

Section 7.5 b As we do some computer experiments, At appears to approach a matrix with identical columns, with column sum 1. Let ~v1 , ~v2 , . . . , ~vn be an eigenbasis with λ1 = 1 and |λj | < 1 for j = 2, . . . , n. For a fixed i, write ~ei = c1~v1 + c2~v2 + · · · + cn~vn , so that (ith column of At ) = At~ei = c1~v1 + [c2 λt2~v2 + · · · + cn λtn~vn ]. (The term in square brackets goes to zero as t goes to infinity.) Therefore, lim (ith column of At ) = lim (At~ei ) = c1~v1 . t→∞

t→∞

Furthermore, the entries of At~ei add up to 1, for all t, by part a. Therefore, the same is true for the limit (since the limit of a sum is the sum of the limits). It follows that lim (At ) exists and has identical columns, with column sum 1, as claimed. t→∞

1 A is a regular transition matrix, so that lim 7.5.31 No matter how we choose A, 15

t→∞

t

t 1 15 A

is a matrix with identical

columns by Exercise 30. Therefore, the columns of A “become more and more alike” as t approaches infinity, in ij th entry of At = 1 for all i, j, k. the sense that lim ik t→∞ th entry of At 0.6 0.6a(t) + 0.1m(t) + 0.5s(t) a(t) 7.5.32 a ~x(t) = m(t) = 0.2a(t) + 0.7m(t) + 0.1s(t) so A = 0.2 0.2 0.2a(t) + 0.2m(t) + 0.4s(t) s(t)

0.1 0.7 0.2

Note that A is a regular transition matrix.

0.5 0.1 . 0.4

b By Exercise 30, lim (At ) = [~v ~v ~v ], where ~v is the unique eigenvector of A with eigenvalue 1 and column sum 1. t→∞ 0.4 We find that ~v = 0.35 . 0.25 Now lim ~x(t) = lim (At ~x0 ) = lim At ~x0 = [~v ~v ~v ]~x0 = ~v , since the components of ~x0 add up to 1. The market t→∞

t→∞

t→∞

shares approach 40%, 35%, and 25%, respectively, regardless of the initial shares.

7.5.33 a C is obtained from B by dividing each column of B by its first component. Thus, the first row of C will consist of 1’s. b We observe that the columns of C are almost identical, so that the columns of B are “almost parallel” (that is, almost scalar multiples of each other). c Let λ1 , λ2 , . . . , λ5 be the eigenvalues. Assume λ1 real and positive and λ1 > |λj | for 2 ≤ j ≤ 5. Let ~v1 , . . . , ~v5 be corresponding eigenvectors. For a fixed i, write ~ei =

5 X

cj ~vj ; then

j=1

(ith column of At ) = At~ei = c1 λt1~v1 + · · · + c5 λt5~v5 . But in the last expression, for large t, the first term is dominant, so the ith column of At is almost parallel to ~v1 , the eigenvector corresponding to the dominant eigenvalue. 341

Chapter 7 d By part c, the columns of B and C are almost eigenvectors of A associated with the largest eigenvalue, λ1 . Since the first row of C consists of 1’s, the entries in the first row of AC will be close to λ1 . 7.5.34 a The eigenvalues of A − λIn are λ1 − λ, λ2 − λ, . . . , λn − λ, and we were told that |λ1 − λ| < |λi − λ| for i = 2, . . . , n. We may assume that λ1 6= λ (otherwise we are done). The eigenvalues of (A−λIn )−1 are (λ1 −λ)−1 , (λ2 −λ)−1 , . . . , (λn −λ)−1 , and (λ1 −λ)−1 has the largest modulus. The matrices A, A − λIn , and (A − λIn )−1 have the same eigenvectors. For large t, the columns of the tth power of (A − λIn )−1 will be almost eigenvectors of A. If ~v is such a column, compare ~v and A~v to find an approximation of λ1 . b See Figure 7.29.

λ 3 ≈ 17 λ 1 ≈ −1

λ2 ≈ 0 (not to scale)

Figure 7.29: for Problem 7.5.34b. Let λ = −1.

2 A − λI3 = 4 7

9 2 3 6 6 , N = (A − λI3 )−1 = −1 −5 8 11

1 −3 0.5 0 , and B = N 20 . −1 2

Obtain C from B as in Exercise 33: 1 1 1 C ≈ −0.098922005729 −0.098922005729 −0.098922005729 −0.569298722688 −0.569298722688 −0.569298722688 −0.905740179522 −0.905740179522 −0.905740179522 AC ≈ ∗ ∗ ∗ ∗ ∗ ∗

The entries in the first row of AC give us a good approximation for λ1 , and the columns of C give us a good approximation for a corresponding eigenvector.

7.5.35 We have fA (λ) = (λ1 − λ)(λ2 − λ) · · · (λn − λ) = (−λ)n + (λ1 + λ2 + · · · + λn )(−λ)n−1 + · · · + (λ1 λ2 · · · λn ). But, by Theorem 7.2.5, the coefficient of (−λ)n−1 is tr(A). So, tr(A) = λ1 + · · · + λn . 7.5.36 a The entries in the first row are age-specific birth rates and the entries just below the diagonal are age-specific survival rates. For example, the entry 1.6 in the first row tells us that during the next 15 years the people who 342

Section 7.5 are 15–30 years old today will on average have 1.6 children (3.2 per couple) who will survive to the next census. The entry 0.53 tells us that 53% of those in the age group 45–60 today will still be alive in 15 years (they will then be in the age group 60–75). b Using technology, we find the largest eigenvalue λ1 = 1.908 with associated eigenvector 0.574 0.247 0.115 ~v1 ≈ . 0.047 0.014 0.002

The components of ~v1 give the distribution of the population among the age groups in the long run, assuming that current trends continue. λ1 gives the factor by which√the population will grow in the long run in a period of 15 years; this translates to an annual growth factor of 15 1.908 ≈ 1.044, or an annual growth of about 4.4%. 7.5.37 a Use that w + z = w + z and wz = wz. w1 −z 1 w2 −z 2 w1 + w2 −(z1 + z2 ) + = is in H. z1 z2 w1 w2 z1 + z2 w1 + w2

w1 z1

−z 1 w1

w2 z2

−z 2 w2

w1 w2 − z 1 z2 = z1 w2 + w1 z2

−(z1 w2 + w1 z2 ) w1 w2 − z 1 z2

is in H.

b If A in H is nonzero, then det(A) = ww + zz = |w|2 + |z|2 > 0, so that A is invertible. −z , then A−1 = w

w c Yes; if A = z

d For example, if A = 0 0 2 7.5.38 a C4 = 1 0

0 0 0 1

1 0 0 0

i 0

0 −i

1 |w|2 +|z|2

and B =

0 0 1 3 0 , C4 = 0 0 1 0

1 0 0 0

w −z

z w

is in H.

0 −1 0 , then AB = 1 0 −i 0 1 0 0

−i 0

and BA =

0 0 4 4+k , C4 = I4 , then C4 = C4k . 1 0

Figure 7.30 illustrates how C4 acts on the basis vectors ~ei .

Figure 7.30: for Problem 7.5.38a.

343

0 i

i . 0

Chapter 7 λ3k 2 λ b The eigenvalues are λ1 = 1, λ2 = −1, λ3 = i, and λ4 = −i, and for each eigenvalue λk , ~vk = k is an associated λk 1 eigenvector.

c M = aI4 + bC4 + cC42 + dC43 If ~v is an eigenvector of C4 with eigenvalue λ, then M~v = a~v + bλ~v + cλ2~v + dλ3~v = (a + bλ + cλ2 + dλ3 )~v , so that ~v is an eigenvector of M as well, with eigenvalue a + bλ + cλ2 + dλ3 . The eigenbasis for C4 we found in part b is an eigenbasis for all circulant 4 × 4 matrices. 7.5.39 Figure 7.31 illustrates how Cn acts on the standard basis vectors ~e1 , ~e2 , . . . , ~en of Rn .

Figure 7.31: for Problem 7.5.39. 7.5.39 a Based on Figure 7.31, we see that Cnk takes ~ei to ~ei+k “modulo n,” that is, if i + k exceeds n then Cnk takes ~ei to ~ei+k−n (for k = 1, . . . , n − 1). To put it differently: Cnk is the matrix whose ith column is ~ei+k if i + k ≤ n and ~ei+k−n if i + k > n (for k = 1, . . . , n − 1). b The characteristic polynomial is 1 − λn , so that the eigenvalues are the n distinct solutions of the λn = 1 equation 2πk 2πk (the so-called nth roots of unity), equally spaced points along the unit circle, λk = cos n + i sin n , for k = 0, 1, . . . , n − 1 (compare with Exercise 5 and Figure 7.7.). For each eigenvalue λk , n−1 λk . .. ~vk = λ2 is an associated eigenvector. k λk 1 c The eigenbasis ~v0 , ~v1 , . . . , ~vn−1 for Cn we found in part b is in fact an eigenbasis for all circulant n × n matrices. 7.5.40 In Exercise 7.2.50 we derived the formula x =

r 3

for the solution of the equation x3 + px = q. Here r r q q 2 3 3 q q 2 −p 3 + 2q − i − 2q + x= 2 +i − 2 + 3

q 2

+

q 2 2

q

+

−p 3 . 3

344

q 2 2

p 3 3

+

p 3 3

+

r 3

q 2

−

q

q 2 2

+

p 3 3

is negative, and we can write

Section 7.5 Let us write this solution in polar coordinates: q q 3/2 3/2 3 3 (cos α + i sin α) + (cos α − i sin α) x= − p3 − p3 q −p p α+2πk + cos α+2πk + i sin − i sin α+2πk = − p3 cos α+2πk 3 3 3 3 3

p = 2 − p3 cos

α+2πk 3

, k = 0, 1, 2. See Figure 7.32.

Figure 7.32: for Problem 7.5.40. Answer: x1,2,3

q = 2 −p 3 cos

α+2πk 3

, k = 0, 1, 2, where α = arccos q

q

−p 3 ,2

−p 3

Note that x is on the interval q q −p , when k = −1 (Think about it!). − −p 3 3 7.5.41 Substitute ρ = 14 x2

+

12 x3

1 x

(

q 2 p 3/2 −3

)

.

q q −p when k = 0, on −2 3 , − −p when k = 1 and on 3

into 14ρ2 + 12ρ3 − 1 = 0;

−1=0

14x + 12 − x3 = 0 x3 − 14x = 12 Now use the formula derived in Exercise 40 to find x, with p = −14 and q = 12. There is only one positive solution, x ≈ 4.114, so that ρ = x1 ≈ 0.243. 7.5.42 a We will use the fact that for any two complex numbers z and w, z + w = z + w and zw = zw. The ij th entry of AB is

p X

k=1

aik bkj =

p X

k=1

aik bkj =

p X

aik bkj , which is the ij th entry of AB, as claimed.

k=1

b Use part a, where B is the n × 1 matrix ~v + iw. ~ We are told that AB = λB, where λ = p + iq. Then ¯ B, ¯ = AB = λB = λ ¯ or A(~v − iw) AB = A¯ B ~ = (p − iq)(~v − iw). ~ 7.5.43 Note that f (z) is not the zero polynomial, since f (i) = det(S1 + iS2 ) = det(S) 6= 0, as S is invertible. A nonzero polynomial has only finitely many zeros, so that there is a real number x such that f (x) = det(S1 +xS2 ) 6= 345

Chapter 7 0, that is, S1 + xS2 is invertible. Now SB = AS or (S1 + iS2 )B = A(S1 + iS2 ). Considering the real and the imaginary part, we can conclude that S1 B = AS1 and S2 B = AS2 and therefore (S1 + xS2 )B = A(S1 + xS2 ). Since S1 + xS2 is invertible, we have B = (S1 + xS2 )−1 A(S1 + xS2 ), as claimed. 7.5.44 Let A be a complex 2 × 2 matrix. Let λ be a complex eigenvalue of A, and consider an associated eigenvector ~v , so that A~v = λ~v . Now let P be an invertible 2 ×2 matrix ~ (the first column of P is of the form P = [~v w] λ ∗ −1 , so that we have found an upper triangular matrix our eigenvector ~v ). Then P AP will be of the form 0 ∗ similar to A (compare with the proof of Theorem 7.4.1). Yes, any complex square matrix is similar to an upper triangular matrix, although the proof is challenging at this stage of the course. Following the hint, we will assume that the claim holds for n × n matrices, and we will prove it for an (n + 1) × (n + 1) matrix A. As inthe case of a 2 × 2 matrix discussed above, we can find an invertible λ w ~ P such that P −1 AP is of the form for some scalar λ, a row vector w ~ with n components, and an n × n 0 B matrix B (just make the first column of P an eigenvector of A). By the induction hypothesis, B is similar to some upper triangular matrix T , that is, R−1 BR = T for some invertible R. 1 ~0 Now let S = P , an invertible (n + 1) × (n + 1) matrix. Then 0 R ~0 ~0 1 1 0 1 λ w ~ 1 0 λ wR ~ −1 S −1 AS = P AP = = , an upper triangular matrix, 0 R−1 0 R 0 R−1 0 B 0 R 0 T showing that A is indeed similar to an upper triangular matrix. You will see an analogous proof in Section 8.1 (proof of Theorem 8.1.1, Page 371). 7.5.45 If a 6= 0, thenthere are two distinct eigenvalues, 1 ± 1 1 1 1 = fails to be diagonalizable. a 1 0 1

√

a, so that the matrix is diagonalizable. If a = 0, then

7.5.46 6= 0,then there are two distinct eigenvalues, ±ia, so that the matrix is diagonalizable. If a = 0, then If a 0 0 0 −a is diagonalizable as well. Thus the matrix is diagonalizable for all a. = 0 0 a 0 7.5.47 Ifa 6= 0, 0 0 then 1 0 0 1

then there 0 0 a = 1 0 0

√ are three distinct eigenvalues, 0, ± a, so that the matrix is diagonalizable. If a = 0, 0 0 0 0 fails to be diagonalizable. 1 0

7.5.48 The characteristic polynomial is f (λ) = −λ3 + 3λ + a. We need to find the values a such that this polynomial has multiple roots. Now λ is a multiple root if (and only if) f (λ) = f ′ (λ) = 0 (see Exercise 7.2.37). Since f ′ (λ) = −3λ2 + 3 = −3(λ − 1)(λ + 1), the only possible multiple roots are 1 and −1. Now 1 is a multiple root if f (1) = 2 + a = 0, or, a = −2, and −1 is a multiple root if a = 2. Thus, if a is neither 2 nor −2, then the matrix is diagonalizable. Conversely, if a = 2 or a = −2, then the matrix fails to be diagonalizable, since all the eigenspaces will be one-dimensional (verify this!). 7.5.49 The eigenvalues are 0, 1, a − 1. If a is neither 1 nor 2, then there are three distinct eigenvalues, so that the matrix is diagonalizable. Conversely, if a = 1 or a = 2, then the matrix fails to be diagonalizable, since all the eigenspaces will be one-dimensional (verify this!). 346

Section 7.6 0 1 7.5.50 The eigenvalues are 0, 0, 1. Since the kernel is always two-dimensional, with basis 1 , 1 , the matrix is 1 0 diagonalizable for all values of constant a. 7.5.51 Yes, Q is a field. Check the axioms on page 350. 7.5.52 No, Z is not a field since multiplicative inverses do not exist, i.e. division within Z is not possible. (Axiom 8 does not hold). 0 ). 1

7.5.53 Yes, check the axioms on page 350. (additive identity

1 0 0 , multiplicative identity 0 0 0

7.5.54 Yes, check the axioms on page 350. (additive identity

1 0 0 0 ). Also notice , multiplicative identity 0 1 0 0

that rotation-scaling matrices commute when multiplied.

7.5.55 No, since multiplication is not commutative; Axiom 5 does not hold.

Section 7.6 7.6.1 λ1 = 0.9, λ2 = 0.8, so, by Theorem 7.6.2, ~0 is a stable equilibrium. 7.6.2 λ1 = −1.1, λ2 = 0.9, so by Theorem 7.6.2, ~0 is not a stable equilibrium. (|λ1 | > 1) 7.6.3 λ1,2 = 0.8 ± (0.7)i so |λ1 | = |λ2 | =

√

7.6.4 λ1,2 = −0.9 ± (0.4)i so |λ1 | = |λ2 | =

0.64 + 0.49 > 1 so ~0 is not a stable equilibrium.

√

0.81 + 0.16 < 1 so ~0 is a stable equilibrium.

7.6.5 λ1 = 0.8, λ2 = 1.1 so ~0 is not a stable equilibrium. 7.6.6 λ1,2 = 0.8 ± (0.6)i so |λ1 | = |λ2 | = 7.6.7 λ1,2 = 0.9 ± (0.5)i so |λ1 | = |λ2 | =

√

0.64 + 0.36 = 1 and ~0 is not a stable equilibrium.

√

0.81 + 0.25 > 1 and ~0 is not a stable equilibrium.

7.6.8 λ1 = 0.9, λ2 = 0.8 so ~0 is a stable equilibrium. 7.6.9 λ1,2 = 0.8 ± (0.6)i, λ3 = 0.7, so |λ1 | = |λ2 | = 1 and ~0 is not a stable equilibrium. 7.6.10 λ1,2 = 0, λ3 = 0.9 so ~0 is a stable equilibrium. 7.6.11 λ1 = k, λ2 = 0.9 so ~0 is a stable equilibrium if |k| < 1. 7.6.12 λ1,2 = 0.6 ± ik so ~0 is a stable equilibrium if |λ1 | = |λ2 | = 347

√

0.36 + k 2 < 1 i.e. if k 2 < 0.64 or |k| < 0.8.

Chapter 7 7.6.13 Since λ1 = 0.7, λ2 = −0.9, ~0 is a stable equilibrium regardless of the value of k. 7.6.14 λ1 = 0, λ2 = 2k so ~0 is a stable equilibrium if |2k| < 1 or |k| < 12 . 7.6.15 λ1,2 = 1 ±

1 10

√

k

If k ≥ 0 then λ1 = 1 + for any real k.

1 10

√

k ≥ 1. If k < 0 then |λ1 | = |λ2 | > 1. Thus, the zero state isn’t a stable equilibrium

√ √ 1+30k 1 7.6.16 λ1,2 = 2± 10 so |2 ± 1 + 30k| must be less than 10. λ1,2 are real if k ≥ − 30 . In this case it is required √ √ √ 21 that 2 + 1 + 30k < 10 and −10 < 2 − 1 + 30k, which means that 1 + 30k < 8 or k < 10 . 1 97 λ1,2 are complex if k < − 30 . Here it is required that 4 + (−1 − 30k) < 100 or k > − 30 . Overall, ~0 is a stable 21 97 equilibrium if − 30 < k < 10 .

7.6.17 λ1,2 = 0.6 ± (0.8)i = 1(cos θ ± i sin θ), where 0.8 4 θ = arctan 0.8 0.6 = arctan 0.6 = arctan 3 ≈ 0.927. −0.8i −0.8 −1 0 −1 Eλ1 = ker = span so w ~= , ~v = . 0.8 −0.8i i 1 0 0 ~x0 = = 1w ~ + 0~v , so a = 1 and b = 0. Now we use Theorem 7.6.3: 1 0 −1 cos(θt) − sin(θt) 1 0 −1 cos θt − sin θt ~x(t) = = = , where 1 0 sin(θt) cos(θt) 0 1 0 sin θt cos θt θ = arctan 34 ≈ 0.927. The trajectory is the circle shown in Figure 7.33.

Figure 7.33: for Problem 7.6.17.

7.6.18 λ1,2 =

√ −4±2 3i 5

= r(cos θ ± i sin θ), where r ≈ 1.058 and θ = π − arctan 348

2

√ 3 5 4 5

≈ 2.428 (second quadrant).

Section 7.6

Eλ1 = span

√ 3 2

i

√ 3 0 , ~v = 2 . , so w ~= 1 0

~x0 = 1w ~ + 0~v , so a = 1, b = 0. cos(θt) − sin(θt) 0 0.866 1 cos(θt) ~x(t) = rt [ w ~ ~v ] ≈ (1.058)t sin(θt) cos(θt) 1 0 0 sin(θt) 0.866 · sin(2.428t) . See Figure 7.34. ≈ (1.058)t cos(2.428t)

Figure 7.34: for Problem 7.6.18. Spirals slowly outwards (plot the first few points). √ 13, and θ = arctan 32 ≈ 0.98, so √ 0 −1 a 1 ~ ~v ] = , = and ~x(t) λ1 ≈ 13(cos(0.98) + i sin(0.98)), [w 1 0 b 0 √ t − sin(0.98t) . ≈ 13 cos(0.98t)

7.6.19 λ1,2 = 2 ± 3i, r =

The trajectory spirals outwards; see Figure 7.35. 7.6.20 λ1,2 = 4 ± 3i, r = 5, θ = arctan 34 ≈ 0.64, so λ1 ≈ 5(cos(0.64) + i sin(0.64)), [w ~ ~v ] = sin(0.64t) . See Figure 7.36. and ~x(t) ≈ 5t cos(0.64t) Spirals outwards (rotation-dilation). 7.6.21 λ1,2 = 4 ± i, r =

√ 17, θ = arctan

1 4

≈ 0.245 so √ 1 a 0 5 = , ~ ~v ] = λ1 ≈ 17(cos(0.245) + i sin(0.245)), [w 0 b 1 3 √ t 5 sin(0.245t) and ~x(t) ≈ 17 cos(0.245t) + 3 sin(0.245t) 349

0 1 a 1 , = 1 0 b 0

Chapter 7

Figure 7.35: for Problem 7.6.19.

Figure 7.36: for Problem 7.6.20.

Figure 7.37: for Problem 7.6.21.

The trajectory spirals outwards; see Figure 7.37. 7.6.22 λ1,2 = −2 ± 3i, r =

√

13, θ ≈ 2.16 (in second quadrant) 350

Section 7.6

[w ~ ~v ] =

0 1

√ t −5 sin(θt) 1 a −5 , where θ ≈ 2.16. so ~x(t) = 13 = , cos(θt) − 3 sin(θt) 0 b −3

Spirals outwards, as in Figure 7.38.

Figure 7.38: for Problem 7.6.22. 7.6.23 λ1,2 = 0.4 ± 0.3i, r = 12 , θ = arctan 0 5 a 1 [w ~ ~v ] = , = so ~x(t) = 1 3 b 0

0.3 0.4

≈ 0.643 5 sin(θt) 1 t . 2 cos(θt) + 3 sin θ(t)

The trajectory spirals inwards as shown in Figure 7.39.

Figure 7.39: for Problem 7.6.23. 7.6.24 λ1,2 = −0.8 ± 0.6i, r = 1, θ = π − arctan .6 .8 ≈ 2.5 (second quadrant) 0 −5 a 1 −5 sin(θt) [w ~ ~v ] = , = so ~x(t) = , an ellipse, as shown in Figure 7.40. 1 −3 b 0 cos(θt) − 3 sin(θt) 7.6.25 Not stable since if λ is an eigenvalue of A, then

1 λ

is an eigenvalue of A−1 and

351

1 λ

=

1 |λ|

> 1.

Chapter 7

Figure 7.40: for Problem 7.6.24. 7.6.26 Stable since A and AT have the same eigenvalues. 7.6.27 Stable since if λ is an eigenvalue of −A, then −λ is an eigenvalue of −A and | − λ| = |λ|. 7.6.28 Not stable, since if λ is an eigenvalue of A, then (λ − 2) is an eigenvalue of (A − 2In ) and |λ − 2| > 1. 1

2

0

3

2

0

, then A + I2 is 7.6.29 Cannot tell; for example, if A = 1 0 23 1 1 0 2 0 0 −2 A= then A + I2 = 2 1 and the zero state is stable. 0 − 21 0 2

and the zero state is not stable, but if

7.6.30 Consider the dynamical systems ~x(t + 1) = A2 ~x(t) and ~y (t + 1) = A~y (t) with equal initial values, ~x(0) = ~y (0). Then ~x(t) = ~y (2t) for all positive integers t. We know that lim ~y (t) = ~0; thus lim ~x(t) = ~0, proving that the t→∞

zero state is a stable equilibrium of the system ~x(t + 1) = A2 ~x(t).

t→∞

7.6.31 We need to determine for which values of det(A) and tr(A) the modulus of both eigenvalues is less than 1. We will first think about the border line case and examine when one of the moduli is exactly 1: If one of the eigenvalues is 1 and the other is λ, then tr(A) = λ + 1 and det(A) = λ, so that det(A) = tr(A) − 1. If one of the eigenvalues is −1 and the other is λ, then tr(A) = λ − 1 and det(A) = −λ, so that det(A) = −tr(A) − 1. If the eigenvalues are complex conjugates with modulus 1, then det(A) = 1 and |tr(A)| < 2 (think about it!). It is convenient to represent these conditions in the tr-det plane, where each 2 × 2 matrix A is represented by the point (trA, detA), as shown in Figure 7.41. If tr(A) = det(A) = 0, then both eigenvalues of A are zero. We can conclude that throughout the shaded triangle in Figure 7.42 the modulus of both eigenvalues will be less than 1, since the modulus of the eigenvalues changes continuously with tr(A) and det(A) (consider the quadratic formula!). Conversely, we can choose sample points to show that in all the other four regions in Figure 7.42 the modulus of at least one of the eigenvalues exceeds one; consider 0 −2 2 0 −2 0 2 0 . , and , , the matrices 2 0 0 −2 0 0 0 0 ↑ ↑ ↑ ↑ in (I) in (II) in (III) in (IV) 352

Section 7.6

Figure 7.41: for Problem 7.6.31. It follows that throughout these four regions, (I), (II), (III), and (IV), at least one of the eigenvalues will have a modulus exceeding one. The point (trA, detA) is in the shaded triangle if det(A) < 1, det(A) > tr(A) − 1 and det(A) > −tr(A) − 1. This means that |trA| − 1 < det(A) < 1, as claimed. 7.6.32 Take conjugates of both sides of the equation ~x0 = c1 (~v + iw) ~ + c2 (~v − iw): ~ ~x0 = ~x0 = c1 (~v + iw) ~ + c2 (~v − iw) ~ = c¯1 (~v − iw) ~ + c¯2 (~v + iw) ~ = c¯2 (~v + iw) ~ + c¯1 (~v − iw). ~ The claim that c2 = c¯1 now follows from the fact that the representation of ~x0 as a linear combination of the linearly independent vectors ~v + iw ~ and ~v − iw ~ is unique. 7.6.33 Take conjugates of both sides of the equation ~x0 = c1 (~v + iw) ~ + c2 (~v − iw): ~ ~ + c2 (~v − iw) ~ = c¯1 (~v − iw) ~ + c¯2 (~v + iw) ~ = c¯2 (~v + iw) ~ + c¯1 (~v − iw). ~ ~x0 = ~x0 = c1 (~v + iw) The claim that c2 = c¯1 now follows from the fact that the representation of ~x0 as a linear combination of the linearly independent vectors ~v + iw ~ and ~v − iw ~ is unique. 7.6.34 a If | det A| = |λ1 λ2 · · · λn | = |λ1 λ2 | · · · |λn | > 1 then at least one eigenvalue is greater than one in modulus and the zero state fails to be stable. b If | det A| = |λ1 kλ2 | · · · |λn | < 1 we cannot conclude anything about the stability of ~0. |2k0.1| < 1 and |0.2k0.1| < 1 but in the first case we would not have stability, in the second case we would. 7.6.35 a Let ~v1 , . . . , ~vn be an eigenbasis for A. Then ~x(t) =

n X

ci λti~vi and

i=1

k~x(t)k = |

n X i=1

ci λti~vi | ≤

n X i=1

kci λti~vi k =

n X i=1

|λi |t kci~vi k ≤

n X i=1

↑ ≤1 353

kci~vi k.

Chapter 7

The last quantity,

n X i=1

kci~vi k, gives the desired bound M .

t 0 k+1 k 1 1 is = , so that ~x(t) = At = represents a shear parallel to the x-axis, with A 1 1 1 1 0 1 not bounded. This does not contradict part a, since there is no eigenbasis for A.

b A=

7.6.36 If the zero state is stable, then lim (ith column of At ) = lim (At~ei ) = ~0, so that all columns and therefore t→∞

all entries of At approach 0.

t→∞

Conversely, if lim At = 0, then lim (At ~x0 ) = t→∞

t→∞

lim At ~x0 = ~0 for all ~x0 (check the details).

t→∞

7.6.37 a Write Y (t + 1) = Y (t) = Y, C(t + 1) = C(t) = C, I(t + 1) = I(t) = I. Y = C + I + G0 → Y = γY G+0 G0 C = γY Y = 1−γ I=0 Y =

G0 1−γ , C

γG0 1−γ , I

=

b y(t) = Y (t) −

=0

G0 1−γ , c(t)

= C(t) −

γG0 1−γ , i(t)

= I(t)

Substitute to verify the equations. C(t + 1) γ γ c(t) = i(t + 1) αγ − α αγ i(t) c A=

0.2 0.2 −4 1

eigenvalues 0.6 ± 0.8i

not stable d A=

γ γ−1

γ e A= αγ − α

γ , trA = 2γ, detA = γ, stable (use Exercise 31) γ γ trA = γ(1 + α) > 0, detA = αγ αγ

Use Exercise 31; stable if det(A) = αγ < 1 and trA − 1 = αγ + γ − 1 < αγ. The second condition is satisfied since γ < 1. Stable if γ < α1 eigenvalues are real if γ ≥

4α (1+α)2

7.6.38 a T (~v ) = A~v + ~b = ~v if ~v − A~v = ~b or (In − A)~v = ~b. In − A is invertible since 1 is not an eigenvalue of A. Therefore, ~v = (In − A)−1 ~b is the only solution. 354

Section 7.6 b Let ~y (t) = ~x(t) − ~v be the deviation of ~x(t) from the equilibrium ~v . Then ~y (t + 1) = ~x(t + 1) − ~v = A~x(t) + ~b − ~v = A(~y (t) + ~v ) + ~b − ~v = A~y (t) + A~v + ~b − ~v = A~y (t), so that ~y (t) = At ~y (0), or ~x(t) = ~v + At (~x0 − ~v ). lim ~x(t) = ~v for all ~x0 if lim At (~x0 − ~v ) = ~0. This is the case if the modulus of all the eigenvalues of A is less t→∞ than 1. t→∞

7.6.39 Use Exercise 38: ~v = (I2 − A)−1~b =

0.9 −0.4

−0.2 0.7

−1 2 1 . = 4 2

2 is a stable equilibrium since the eigenvalues of A are 0.5 and −0.1. 4 7.6.40 a AT A =

BT −C

b By part a, A−1 =

CT B

B C

−C T BT

1 T p2 +q 2 +r 2 +s2 A

= (p2 + q 2 + r2 + s2 )I4

if A 6= 0.

c (det A)2 = (p2 + q 2 + r2 + s2 )4 , by part a, so that det A = ±(p2 + q 2 + r2 + s2 )2 . Laplace Expansion along the first row produces the term +p4 , so that det(A) = (p2 + q 2 + r2 + s2 )2 . d Consider det(A − λI4 ). Note that the matrix A − λI4 has the same “format” as A, with p replaced by p − λ and q, r, s remaining unchanged. By part c, det(A − λI4 ) = ((p − λ)2 + q 2 + r2 + s2 )2 = 0 when (p − λ)2 = −q 2 − r2 − s2 p p − λ = ±i q 2 + r2 + s2 p λ = p ± i q 2 + r2 + s2

Each of these eigenvalues has algebraic multiplicity 2 (if q = r = s = 0 then λ = p has algebraic multiplicity 4).

p e By part a we can write A = p2 + q 2 + r2 + s2

1 |

p

p2 + q 2 + r2 + s2 {z

!

A , where S is orthogonal.

S

p p Therefore, kA~xk = k p2 + q 2 + r2 + s2 (S~x)k = p2 + q 2 + r2 + s2 k~xk.

3 −3 3 3 f Let A = 4 −5 5 4

}

−4 −5 1 −39 5 −4 2 13 and ~x = ; then A~x = . 3 3 4 18 −3 3 4 13 2

By part e, kA~xk2 = (32 + 32 + 42 + 52 )k~xk , or 392 + 132 + 182 + 132 = (32 + 32 + 42 + 52 )(12 + 22 + 42 + 42 ), as desired. 355

Chapter 7 g Any positive integer m can be written as m = p1 p2 . . . pn . Using part f repeatedly we see that the numbers p1 , p1 p2 , p1 p2 p3 , . . . , p1 p2 p3 · · · pn−1 , and finally m = p1 · · · pn can be expressed as the sums of four squares. 7.6.41 Find the 2 × 2 matrix A

8 6

−3 −8 −3 = 4 −6 4

8 −3 −3 −8 A that transforms into and into : 6 4 4 −6 −1 36 −73 8 −3 −3 −8 1 . = 50 and A = 52 −36 6 4 4 −6

There are many other correct answers. 7.6.42 a x(t + 1) = x(t) − ky(t) y(t + 1) = kx(t) + y(t) = kx(t) + (1 − k 2 )y(t) so

x(t + 1) 1 = y(t + 1) k

−k 1 − k2

x(t) . y(t)

b fA (λ) = λ2 − (2 − k 2 )λ + 1 = 0 The discriminant is (2 − k 2 )2 − 4 = −4k 2 + k 4 = k 2 (k 2 − 4), which is negative if k is a small positive number (k < 2). Therefore, the eigenvalues are complex. By Theorem 7.6.4 the trajectory will be an ellipse, since det(A) = 1.

True or False Ch 7.TF.1 F; If

1 1 , then eigenvalue 1 has geometric multiplicity 1 and algebraic multiplicity. 2. 0 1

Ch 7.TF.2 T, by Theorem 7.4.3 Ch 7.TF.3 T; A = AIn = A[~e1 . . . ~en ] = [λ1~e1 . . . λn~en ] is diagonal. Ch 7.TF.4 T; If A~v = λ~v , then A3~v = λ3~v . Ch 7.TF.5 T; Consider a diagonal 5 × 5 matrix with only two distinct diagonal entries. Ch 7.TF.6 F, by Theorem 7.2.7. Ch 7.TF.7 T, by Summary 7.1.5 Ch 7.TF.8 T, by Theorem 7.2.4 Ch 7.TF.9 T, by Theorem 7.2.2 Ch 7.TF.10 T, by Definition 7.2.3 Ch 7.TF.11 T, by Example 6 of Section 7.5 Ch 7.TF.12 T; The geometric multiplicity of eigenvalue 0 is dim(kerA) = n − rank(A). 356

True or False Ch 7.TF.13 T; If S −1 AS = D, then S T AT (S T )−1 = D.

2 Ch 7.TF.14 F; Let A = 0 0

1 0 0 3 0 and B = 0 0 0 0

0 0 4 0 , for example. 0 0

1 1 Ch 7.TF.15 F; Consider A = . 0 1 Ch 7.TF.16 F; Let A = 8 0 . AB = 0 15

2 0 4 0 , α = 2, B = , β = 5, for example. Then αβ = 10 isn’t an eigenvalue of 0 3 0 5

Ch 7.TF.17 T; If A~v = 3~v , then A2~v = 9~v . Ch 7.TF.18 T; Construct an eigenbasis by concatenating a basis of V with a basis of V ⊥ . Ch 7.TF.19 T, by Theorem 7.5.5

1 1 Ch 7.TF.20 F; Let A = , for example. 0 −1 Ch 7.TF.21 T; If S −1 AS = D, then S −1 A−1 S = D−1 is diagonal. Ch 7.TF.22 F; the equation det(A) = det(AT ) holds for all square matrices, by Theorem 6.2.1. Ch 7.TF.23 T; The sole eigenvalue, 7, must have geometric multiplicity 3. Ch 7.TF.24 F; Let A =

1 1 0 0

and B =

0 1 1 , with A + B = 0 1 0

2 , for example. 1

Ch 7.TF.25 F; Consider the zero matrix. Ch 7.TF.26 T; If A~v = α~v and B~v = β~v , then (A + B)~v = A~v + B~v = α~v + β~v = (α + β)~v .

0 1 0 0 2 Ch 7.TF.27 F; Consider A = , with A = . 0 0 0 0 Ch 7.TF.28 T, by Theorem 7.5.5 1 0 , for example. Ch 7.TF.29 F; Let A = 1 1

1 1 Ch 7.TF.30 F; Let A = 0 0

0 0 1 , with AB = and B = 0 0 1

357

2 , for example. 0

Chapter 7 Ch 7.TF.31 T, Consider the proof of Theorem 7.3.4a. Ch 7.TF.32 T; An eigenbasis for A is an eigenbasis for A + 4I4 as well. Ch 7.TF.33 F; Consider the identity matrix.

1 Ch 7.TF.34 T; Both A and B are similar to 0 0 Ch 7.TF.35 F; Let A =

Ch 7.TF.36 F; Consider

Ch 7.TF.37 F; Let A =

1 1 0 1

1 1 0 1

and ~v =

0 0 2 0 , by Theorem 7.4.1 0 3

1 , for example. 0

2 0 1 0 , ~v = , and w ~= , for example. 0 3 0 1

Ch 7.TF.38 T; A nonzero vector on L and a nonzero vector on L⊥ form an eigenbasis. Ch 7.TF.39 T; The eigenvalues are 3 and −2. Ch 7.TF.40 T, We will us Theorem 7.3.7 throughout: The geometric multiplicity of an eigenvalue is ≤ its algebraic multiplicity. Now let’s show the contrapositive of the given statement: If the geometric multiplicity of some eigenvalue is less than its algebraic multiplicity, then the matrix A fails to be diagonalizable. Indeed, in this case the sum of the geometric multiplicities of all the eigenvalues is less than the sum of their algebraic multiplicities, which in turn is ≤ n (where A is an n × n matrix). Thus the geometric multiplicities do not add up to n, so that A fails to be diagonalizable, by Theorem 7.3.4b. Ch 7.TF.41 F; Consider a rotation through π/2. Ch 7.TF.42 T; Suppose

Ch 7.TF.43 F; Consider

A A ~v A(~v + w) ~ λ~v ~v = = for a nonzero vector . If w ~ is nonzero, then it 0 A w ~ Aw ~ λw ~ w ~ is an eigenvector of A with eigenvalue λ; otherwise ~v is such an eigenvector. 1 0 0 1

and

1 1 . 0 1

Ch 7.TF.44 T; Note that S −1 AS = D, so that D4 = S −1 A4 S = S −1 0S = 0, and therefore D = 0 (since D is diagonal) and A = SDS −1 = 0. Ch 7.TF.45 T; There is an eigenbasis ~v1 , . . . , ~vn , and we can write ~v = c1~v1 + · · · + cn~vn . The vectors ci~vi are either eigenvectors or zero. Ch 7.TF.46 T; If A~v = α~v and B~v = β~v , then AB~v = αβ~v . 358

True or False Ch 7.TF.47 T, by Theorem 7.3.6a

1 0 Ch 7.TF.48 F; Let A = , for example. 0 0 Ch 7.TF.49 T; Recall that the rank is the dimension of the image. If ~v is in the image of A, then A~v is in the image of A as well, so that A~v is parallel to ~v . Ch 7.TF.50 F; Consider

0 1 . 0 0

Ch 7.TF.51 T; If A~v = λ~v for a nonzero ~v , then A4~v = λ4~v = ~0, so that λ4 = 0 and λ = 0.

1 1 Ch 7.TF.52 F; Let A = 0 0

0 1 , for example. and B = 0 1

~ Ch 7.TF.53 T; If the eigenvalue associated with ~v is λ = 0, then A~v = 0, so that ~v is in the kernel of A; otherwise ~v = A λ1 ~v , so that ~v is in the image of A. Ch 7.TF.54 T; either there are two distinct real eigenvalues, or the matrix is of the form kI2 . Ch 7.TF.55 T; Either A~u = 3~u or A~u = 4~u. Ch 7.TF.56 T; Note that (~u~uT )~u = k~uk2 ~u. Ch 7.TF.57 T; Suppose A~vi = αi~vi and B~vi = βi~vi , and let S = [~v1 . . . ~vn ]. Then ABS = BAS = [α1 β1~v1 . . . αn βn~vn ], so that AB = BA. p a b ap + bq Ch 7.TF.58 T; Note that a nonzero vector ~v = is an eigenvector of A = if (and only if) A~v = q c d cp + dq p p ap + bq is parallel to ~v = , that is, if det = 0. Check that this is the case if (and only if) ~v is an q q cp + dq eigenvector of adj(A) (use the same criterion).

359

Chapter 8

Chapter 8 Section 8.1 8.1.1 ~e1 , ~e2 is an orthonormal eigenbasis. 8.1.2

√1 2

1 √1 1 , 2 is an orthonormal eigenbasis. 1 −1

8.1.3

√1 5

2 √1 −1 , 5 is an orthonormal eigenbasis. 1 2

1 1 1 8.1.4 √13 1 , √12 −1 , √16 1 is an orthonormal eigenbasis. −1 0 2

8.1.5 Eigenvalues −1, −1, 2 −1 Choose ~v1 = √12 1 in E−1 and ~v2 = 0 2 8.1.6 31 2 , 1 8.1.7

√1 2

1 , 1

2 1 −1 , 3 −2

√1 2

1 −1

1 √1 1 in E2 and let ~ v3 = ~v1 × ~v2 = 3 1

1 √1 1 . 6 −2

1 1 −2 is an orthonormal eigenbasis. 3 2

is an orthonormal eigenbasis, so S =

√1 2

1 1

1 −1

5 and D = 0

0 . 1

3 −1 1 √ 8.1.8 , 10 is an orthonormal eigenbasis, with λ1 = 4 and λ2 = −6, so S = 3 1 4 0 . D= 0 −6 √1 10

√1 10

1 −1 0 8.1.9 √12 0 , √12 0 , 1 is an orthonormal eigenbasis, with λ1 = 3, λ2 = −3, and λ3 = 2, so 1 1 0 3 0 0 1 −1 √0 S = √12 0 2 and D = 0 −3 0 . 0 0 0 2 1 1 0 8.1.10 λ1 = λ2 = 0 and λ3 = 9. 2 ~v1 = √15 1 is in E0 and ~v2 = 0

1 1 −2 is in E9 . 3 2 360

3 −1 1 3

and

Section 8.1 2 1 Let ~v3 = ~v1 × ~v2 = 3√ −4 ; then ~v1 , ~v2 , ~v3 is an orthonormal eigenbasis. 5 −5

S=

√2 5 √1 5

2 √ 3 5 4 − 3√ 5 √ 5 − 3

1 3

− 32

0

2 3

0 0 and D = 0 9 0 0

0 0 . 0

0 1 −1 8.1.11 √12 0 , √12 0 , 1 is an orthonormal eigenbasis, with λ1 = 2, λ2 = 0, and λ3 = 1, so S = 0 1 1 1 −1 0 2 0 0 √ √1 0 0 2 and D = 0 0 0 . 2 0 0 1 1 1 0 1 8.1.12 a E1 = span 0 and E−1 = (E1 )⊥ . An orthonormal eigenbasis is 2

1 b Use Theorem 7.4.1: B = 0 0 c A = SBS −1

−0.6 = 0 0.8

0 1 √1 0 , 1 , 5 0 2

2 √1 0 . 5 −1

0 0 −1 0 . 0 −1

1 √ 0 0.8 5 0 , where S = −1 0 0 0.6 √2 5

0 1

√2 5

0 .

0 − √15

8.1.13 Yes; if ~v is an eigenvector of A with eigenvalue λ, then ~v = I3~v = A2~v = λ2~v , so that λ2 = 1 and λ = 1 or λ = −1. Since A is symmetric, E1 and E−1 will be orthogonal complements, so that A represents the reflection about E1 . 0 0 0 8.1.14 Let S be as in Example 3. Then S −1 AS = 0 0 0 . 0 0 3 0 0 0 a. This matrix is 2A so that S −1 (2A)S = 0 0 0 . 0 0 6

−3 0 0 b. This is A − 3I3 , so that S −1 (A − 3I3 )S = S −1 AS − 3I3 = 0 −3 0 . 0 0 0 1 0 −2 c. This is 12 (A − I3 ), so that S −1 12 (A − I3 ) S = 21 (S −1 AS − I3 ) = 0 − 12 0

361

0

0 0 . 1

Chapter 8 8.1.15 Yes, if A~v = λ~v , then A−1~v = λ1 ~v , so that an orthonormal eigenbasis for A is also an orthonormal eigenbasis for A−1 (with reciprocal eigenvalues). 8.1.16 a ker(A) is four-dimensional, so that the eigenvalue 0 has multiplicity 4, and the remaining eigenvalue is tr(A) = 5. b B = A + 2I5 , so that the eigenvalues are 2, 2, 2, 2, 7. c det(B) = 24 · 7 = 112 (product of eigenvalues) 8.1.17 If A is the n × n matrix with all 1’s, then the eigenvalues of A are 0 (with multiplicity n − 1) and n. Now B = qA + (p − q)In , so that the eigenvalues of B are p − q (with multiplicity n − 1) and qn + p − q. Thus det(B) = (p − q)n−1 (qn + p − q). p 8.1.18 By Theorem 6.3.6, the volume is |det A| = det(AT A). Now ~vi · ~vj = k~vi kk~vj kcos(θ) = 12 , so that AT A has n−1 1 1 all 1’s on the diagonal and 21 ’s outside. By Exercise 17 with p = 1 and q = 12 , det(AT A) = 21 2n + 2 = p n/2 √ 1 n n + 1. (n + 1), so that the volume is det(AT A) = 21 2

8.1.19 Let L(~x) = A~x. Then AT A is symmetric, since (AT A)T = AT (AT )T = AT A, so that there is an orthonormal eigenbasis ~v1 , . . . , ~vm for AT A. Then the vectors A~v1 , . . . , A~vm are orthogonal, since A~vi · A~vj = (A~vi )T A~vj = ~viT AT A~vj = ~vi · (AT A~vj ) = ~vi · (λj ~vj ) = λj (~vi · ~vj ) = 0 if i 6= j.

8.1.20 By Exercise 19, there is an orthonormal basis ~v1 , . . . , ~vm of Rm such that T (~v1 ), . . . , T (~vm ) are orthogonal. 1 vi ) for Suppose that T (~v1 ), . . . , T (~vr ) are nonzero and T (~vr+1 ), . . . , T (~vm ) are zero. Then let w ~ i = kT (~ vi )k T (~ ⊥ i = 1, . . . , r and choose an orthonormal basis w ~ r+1 , . . . , w ~ n of [span(w ~ 1, . . . , w ~ r )] . Then w ~ 1, . . . , w ~ n does the job. 8.1.21 For each eigenvalue there are two unit eigenvectors: ±~v1 , ±~v2 , and ±~v3 . We have 6 choices for the first column of S, 4 choices remaining for the second column, and 2 for the third. Answer: 6 · 4 · 2 = 48. 8.1.22 a If we let k = 2 then A is symmetric and therefore (orthogonally) diagonalizable. b If we let k = 0 then 0 is the only eigenvalue (but A 6= 0), so that A fails to be diagonalizable. 8.1.23 The eigenvalues are real (by Theorem 8.1.3), so that the only possible eigenvalues are ±1. Since A is symmetric, E1 and E−1 are orthogonal complements. Thus A represents a reflection about E1 . 8.1.24 Note that A is symmetric and orthogonal, so that the eigenvalues are 1 and −1 (see Exercise 23). 1 1 0 0 0 1 0 1 E1 = span , and E−1 = span , , so that 0 0 −1 1 −1 1 0 0 1 0 1 0 0 1 0 1 √1 , √1 , √1 , √12 is an orthonormal eigenbasis. 2 2 2 0 1 0 −1 1 0 −1 0 362

Section 8.1 8.1.25 Note that A is symmetric an orthogonal, so that the eigenvalues of A are 1 and −1. 1 0 0 1 0 0 1 0 0 1 E1 = span 0 , 0 , 1 , E−1 = span 0 , 0 0 1 0 0 −1 1 0 0 −1 0 1 0 0 1 0 0 1 0 1 √0 2 0 0 is one possible choice. The columns of S must form an eigenbasis for A : S = √12 0 0 0 1 0 0 −1 1 0 0 −1 0 8.1.26 Since Jn is both orthogonal and symmetric, the eigenvalues are 1 and −1. If n is even, then both have multiplicity n2 (as in Exercise 24). If n is odd, then the multiplicities are n+1 for 1 and n−1 for −1 (as in 2 2 Exercise 25). One way to see this is to observe that tr(Jn ) is 0 for even n, and 1 for odd n (recall that the trace is the sum of the eigenvalues). 8.1.27 If n is even, then this matrix is Jn + In , for the Jn introduced in Exercise 26, so that the eigenvalues are 0 and 2, with multiplicity n2 each. E2 is the span of all ~ei + ~en+1−i , for i = 1, . . . , n2 , and E0 is spanned by all ~ei −~en+1−i . If n is odd, then E2 is spanned by all ~ei +~en+1−i , for i = 1, . . . , n−1 ei −~en+1−i , 2 ; E0 is spanned by all ~ n+1 . for i = 1, . . . , n−1 , and E is spanned by ~ e 1 2 2

8.1.28 For λ 6= 0 −λ fA (λ) = det 0

1

−λ

1 = λ det

−λ

0

−λ

1

0

.

··· 0

..

.

0 0

..

···

| | | | | | −λ | | | 1 |

| | | | | | −λ | | | 0 |

1 1 .. . 1 1−λ 1 1 .. .

=

1 −λ2 + λ + 12

= −λ11 (λ2 − λ − 12) = −λ11 (λ − 4)(λ + 3)

−λ

1 det λ

Eigenvalues are 0 (with multiplicity 11), 4 and −3. Eigenvalues for 0 are ~e1 − ~ei (i = 2, . . . , 12), 1 1 1 1 . . . . E4 = span . (12 ones), E−3 = span . (12 ones) 1 1 4 −3 363

0

−λ

..

.

0 λ

λ

···

| | | | | | −λ | | | λ |

1 1 .. . 1 λ − λ2

Chapter 8 so 1 1 0 −1 0 −1 0 0 0 0 0 0 S= 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 0 0 0 0 −1 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 −1 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 −1 0 0 0 0 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 −1 0 0

1 1 1 1 1 1 1 1 1 1 1 1 4

1 1 1 1 1 1 1 1 1 1 1 1 −3

diagonalizes A, and D = S −1 AS will have all zeros as entries except d12,

12

= 4 and d13,

13

= −3.

8.1.29 By Theorem 5.4.1 (im A)⊥ = ker(AT ) = ker(A), so that ~v is orthogonal to w. ~ 8.1.30 The columns ~v , ~v2 , . . . , ~vn of R form an orthogonal eigenbasis for A = ~v ~v T , with eigenvalues 1, 0, 0, . . . , 0(n − 1 zeros), since A~v = ~v v T ~v = ~v (~v · ~v ) = ~v , (since ~v · ~v = 1) and A~vi = ~v v T ~vi = ~v (~v · ~vi ) = ~0 (since ~v · ~vi = 0). 1 0 ··· 0 0 0 ··· 0 . Therefore we can let S = R, and D = ... 0

0 ···

0

8.1.31 True; A is diagonalizable, that is, A is similar to a diagonal matrix D; then A2 is similar to D2 . Now rank(D) = rank(D2 ) is the number of nonzero entries on the diagonal of D (and D2 ). Since similar matrices have the same rank (by Theorem 7.3.6b) we can conclude that rank(A) = rank(D) = rank(D2 ) = rank(A2 ). 8.1.32 By Exercise 17, det(A) = (1 − q)n−1 (qn + 1 − q). A is invertible if det(A) 6= 0, that is, if q 6= 1 and q 6= 8.1.33 The angles must add up to 2π, so θ =

2π 3

= 120◦ . (See Figure 8.1.)

v1 θ

v2

θ

unit circle

θ

v3

Figure 8.1: for Problem 8.1.33. 364

1 1−n .

Section 8.1 Algebraically, we can see this as follows: let A = [ ~v1 ~v2 ~v3 ], a 2 × 3 matrix. 1 cos θ cos θ 1 cos θ is a noninvertible 3 × 3 matrix, so that cos θ = Then AT A = cos θ cos θ cos θ 1 ◦ = 120 . and θ = 2π 3 8.1.34 Let ~v1 , ~v2 , ~v3 , ~v4 be such vectors. Form A = [ ~v1

~v2

~v3

1 1−3

= − 21 , by Exercise 32,

~v4 ], a 3 × 4 matrix.

1 cos θ cos θ cos θ 1 cos θ cos θ cos θ Then AT A = is noninvertible, so that cos θ = cos θ cos θ 1 cos θ 1 cos θ cos θ cos θ θ = arccos − 31 ≈ 109.5◦ . See Figure 8.2.

1 1−4

= − 13 , by Exercise 32, and

Figure 8.2: for Problem 8.1.34. The tips of ~v1 , ~v2 , ~v3 , ~v4 form a regular tetrahedron. 8.1.35 Let ~v1 , . . . , ~vn+1 be these vectors. Form A = [ ~v1 · · · ~vn+1 ], an n × (n + 1) matrix. 1 cos θ · · · cos θ cos θ 1 · · · cos θ is a noninvertible (n + 1) × (n + 1) matrix with 1’s on the diagonal and Then AT A = .. ... . cos θ

cos θ outside, so that cos θ =

···

1 1−n ,

1

by Exercise 32, and θ = arccos

1 1−n

.

8.1.36 If ~v is an eigenvector with eigenvalue λ, then λ~v = A~v = A2~v = λ2~v , so that λ = λ2 and therefore λ = 0 or λ = 1. Since A is symmetric, E0 and E1 are orthogonal complements, so that A represents the orthogonal projection onto E1 . 8.1.37 In Example 4 we see that the image of the unit circle is an ellipse with semi-axes 2 and 3. Thus ||A~u|| takes all values in the interval [2, 3]. 8.1.38 The spectral theorem tells us that there exists an orthonormal eigenbasis ~v1 , ~v2 for A, with associated eigenvalues -2 and 3. Consider a unit vector ~u = c1~v1 + c2~v2 in R2 , with c21 + c22 = 1. Then ~u · A~u = (c1~v1 + c2~v2 )·(−2c1~v1 + 3c2~v2 ) = −2c21 +3c22 , which takes all values on the interval [−2, 3] since −2 = −2c21 −2c22 ≤ −2c21 + 3c22 ≤ 3c21 + 3c22 = 3. 8.1.39 The spectral theorem tells us that there exists an orthonormal eigenbasis ~v1 , ~v2 , ~v3 for A, with associated eigenvalues -2, 3 and 4. Consider a unit vector ~u = c1~v1 + c2~v2 + c3~v3 in R3 , with c21 + c22 + c23 = 1. Then 365

Chapter 8 ~u · A~u = (c1~v1 + c2~v2 + c3~v3 ) · (−2c1~v1 + 3c2~v2 + 4c3~v3 ) = −2c21 + 3c22 + 4c23 , which takes all values on the interval [−2, 4] since −2 = −2c21 − 2c22 − 2c23 ≤ −2c21 + 3c22 + 4c23 ≤ 4c21 + 4c22 + 4c23 = 4. 8.1.40 Using the terminology introducedp in Exercise 8.1.39, we have kA~uk = k−2c1~v1 + 3c2~v2 + 4c3~v3 k = 4c21 + 9c22 + 16c22 , which takes all values on the interval [2, 4]. Geometrically, the image of the unit sphere under A is the ellipsoid with semi-axes 2, 3, and 4. 8.1.41 The spectral theorem tells us that there exists an orthogonal matrix S such that S −1 AS = D is diagonal. Let D1 be the diagonal matrix such that D13 = D; the diagonal entries of D1 are the cube roots of those of D. 3 Now B = SD1 S −1 does the job, since B 3 = SD1 S −1 = SD13 S −1 = SDS −1 = A.

8.1.42 We will use the strategy outlined in Exercise 8.1.41. An orthogonal matrix that diagonalizes A = 15 2 0 8 0 1 −2 6 and B = SD1 S −1 = 51 . Now D1 = , with S −1 AS = D = is S = √15 0 1 0 1 2 1 2

12 14 14 33 2 . 9

1 1 1 8.1.43 There is an orthonormal eigenbasis ~v1 = √12 −1 , ~v2 = √16 1 , ~v3 = √13 1 with associated 0 −2 1 eigenvalues -9, -9, 24. We are looking for a nonzero vector ~v = c1~v1 + c2~v2 + c3~v3 such that ~v · A~v = (c1~v1 + c2~v2 + c3~v3 ) · (−9c1~v1 − 9c2~v2 + 24c3~v3 ) = −9c21 −9c22 + 24c23 = 0 or −3c21 − 3c22 + 8c23 = 0. One possible 3 √ √ √ solution is c1 = 8 = 2 2, c2 = 0, c3 = 3, so that ~v = −1 . 1 8.1.44 Use Exercise 8.1.43 as a guide. Consider an orthonormal eigenbasis ~v1 , ...., ~vn for A, with associated eigenvalues λ1 ≤ λ2 ≤ .... ≤ λn , listed in ascending order. If ~v = c1~v1 + ... + cn~vn is any nonzero vector in Rn , then ~v · A~v = (c1~v1 + ... + cn~vn ) · (λ1 c1~v1 + ... + λn cn~vn ) = λ1 c21 + ... + λn c2n . If all the eigenvalues are positive, then ~v · A~v will be positive. Likewise, if all the eigenvalues are negative, then ~v · A~v will be negative. However, if A has positive as well as negative eigenvalues, meaning√that λ1 < √ 0 < λn (as in Example 8.1.43) , then there exist nonzero vectors ~v with ~v · A~v = 0, for example, ~v = λn~v1 + −λ1~vn . 8.1.45 a If S −1 AS is upper triangular then the first column of S is an eigenvector of A. Therefore, any matrix 0 −1 . without real eigenvectors fails to be triangulizable over R, for example, 1 0 b Proof by induction on n: For an n × n matrix A we can choose a complex invertible n × n matrix P whose first λ ~v −1 column is an eigenvector for A. Then P AP = . B is triangulizable, by induction hypothesis, that is, 0 B there is an invertible that Q−1 BQ = triangular. Q such (n − 1) × (n −1) matrix T is upper λ ~ v 1 0 1 0 λ ~ v 1 0 λ ~v Q Now let R = R= . Then R−1 = is upper triangular. 0 B 0 Q−1 0 B 0 Q 0 T 0 Q λ ~v R = R−1 P −1 AP R = S −1 AS, where S = P R, proving our claim. R−1 0 B 8.1.46 a By definition of an upper triangular matrix, ~e1 is in ker U , ~e2 is in ker(U 2 ), . . . , ~en is in ker(U n ), so that all ~x in Cn are in ker(U n ), that is, U n = 0.

366

Section 8.2 b By Exercise 45b, there exists an invertible S such that S −1 AS = U is upper triangular. The diagonal entries of U are all zero, since A and U have the same eigenvalues; therefore U n = 0 by part a. Now A = SU S −1 and An = SU n S −1 = 0, as claimed. 8.1.47 a For all i, j,

"

n X

k=1

#

aik bkj ≤

n X

k=1

|aik bkj | =

n X

k=1

|aik ||bkj |

↑ triangle inequality b By induction on t: |At | = |At−1 A| ≤ |At−1 ||A| ≤ |A|t−1 |A| = |A|t ↑ part a

↑ by induction hypothesis

t t t t 2 n−1 n 8.1.48 If t ≥ n − 1 then (In + U ) = In + U+ U +···+ U , since U = 0. Now ≤ tn 1 2 n−1 k for k = 1, . . . , n − 1, so that (In + U )t ≤ tn (In + U + · · · + U n−1 ), as claimed. Check that the formula holds for t < n − 1 as well. t

8.1.49 Let λ be the largest |rii |; note that λ < 1. Then |R| =

|r11 |

..

∗ .

0 |rnn | λ(In + U ), and |Rt | ≤ |R|t ≤ λt (In + U )t ≤ λt tn (In + U + · · · + U n−1 ).

≤

λ

..

.

0

∗ λ

= λ

1 0

..

.

∗ 1

=

We learn in Calculus that lim (λt tn ) = 0, so that lim (Rt ) = 0. t→∞

t→∞

8.1.50 a From Exercise 45b we know that there is an invertible S and an upper triangular R such that S −1 AS = R, and |rii | < 1 for all i, since the diagonal entries of R are the eigenvalues of A. Now lim Rt = 0 by Exercise 49. Note that A = SRS −1 and At = SRt S −1 , so that lim At = 0, as claimed.

t→∞

t→∞

b See the remark after Definition 7.6.1.

Section 8.2 8.2.1 We have a11 = coefficient of x21 = 6, a22 = coefficient of x22 = 8, a12 = a21 = 6 − 27 So, A = − 72 8 0

1 2

1 2

0

3 8.2.3 A = 0

0 4

3

7 2

5

8.2.2 A =

3

7 2

367

1 2(

coefficient of x1 x2 ) = − 72 .

Chapter 8

8.2.4 A =

6 2 , positive definite 2 3

1 2 , indefinite (since det(A) < 0) 8.2.5 A = 2 1

2 3 , indefinite (since det(A) < 0) 8.2.6 A = 3 4

0 0 2 8.2.7 A = 0 3 0 , indefinite (eigenvalues 2, −2, 3) 2 0 0

8.2.8 If S −1 AS = D is diagonal, then S −1 A2 S = D2 , so that all eigenvalues of A2 are ≥0. So A2 is positive semi-definite; it is positive definite if and only if A is invertible. 8.2.9 a (A2 )T = (AT )2 = (−A)2 = A2 , so that A2 is symmetric. b q(~x) = ~x T A2 ~x = ~x T AA~x = −~x T AT A~x = −(A~x) · (A~x) = −kA~xk2 ≤ 0 for all ~x, so that A2 is negative semi-definite. The eigenvalues of A2 will be ≥ 0. c If ~v is a complex eigenvector of A with eigenvalue λ, then A2~v = λ2~v , and λ2 ≤ 0, by part b. Therefore, λ is imaginary, that is, λ = bi for a real b. Thus, the zero matrix is the only skew-symmetric matrix that is diagonalizable over R. 8.2.10 L(~x) = (~x+~v )T A(~x + ~v ) − ~x T A~x − ~v T A~v =~x T A~x + ~x T A~v + ~v T A~x + ~v T A~v − ~x T A~x − ~v T A~v =~x T A~v + ~v T A~x = ~v T A~x + ~v T A~x = (2~v T A)~x, ↑ note that ~x T A~v is a scalar so that ~x T A~v = (~x T A~v )T = ~v T AT ~x = ~v T A~x if A is symmetric. So L is linear with matrix 2~v T A. 8.2.11 The eigenvalues of A−1 are the reciprocals of those of A, so that A and A−1 have the same definiteness. 8.2.12 det(A) is the product of the two (real) eigenvalues. q is indefinite if an only if those have different signs, that is, their product is negative. 8.2.13 q(~ei ) = ~ei · A~ei = aii > 0 8.2.14 If det(A) is positive then both eigenvalues have the same sign, so that A is positive definite or negative definite. Since ~e1 · A~e1 = a > 0, A is in fact positive definite.

2 ; eigenvalues λ1 = 7 and λ2 = 2 3 2 −1 1 1 √ √ orthonormal eigenbasis ~v1 = 5 , ~v2 = 5 1 2

6 8.2.15 A = 2

368

Section 8.2 λ1 c21 + λ2 c22 = 1 or 7c21 + 2c22 = 1. (See Figure 8.3.)

E7

v2 1 v 2

1 v 1

√2

√7

– 1 v1 √7

v1 1

E2 – 1 v2 √2

Figure 8.3: for Problem 8.2.15.

8.2.16 A =

0 1 2

1 2

0

; eigenvalues λ1 = 12 , and λ2 = − 21

orthonormal eigenbasis ~v1 = 1 2 2 c1

√1 2

1 and ~v2 = 1

√1 2

1 −1

− 21 c22 = 1. (See Figure 8.4.)

2 , eigenvalues λ1 = 4, λ2 = −1 0 2 −1 orthonormal eigenbasis ~v1 = √15 , ~v2 = √15 1 2

8.2.17 A =

3 2

4c21 − c22 = 1 (hyperbola) (See Figure 8.5.) −2 , eigenvalues λ1 = 10, λ2 = 5 6 2 −1 , ~v2 = √15 orthonormal eigenbasis ~v1 = √15 1 2

9 8.2.18 A = −2

10c21 + 5c22 = 1. This is an ellipse, as shown in Figure 8.6. 2 ; eigenvalues λ1 = 5, λ2 = 0 4 1 −2 1 1 √ √ eigenvectors ~v1 = 5 , ~v2 = 5 2 1

1 8.2.19 A = 2

369

Chapter 8

E1 2

√2 v1 v1

1 v2 –√2 v1 E– 1 2

Figure 8.4: for Problem 8.2.16.

E4 v2 1 2 v1

v1

1 – 12 v1

E–1

Figure 8.5: for Problem 8.2.17. 5c21 = 1 (a pair of lines) (See Figure 8.7.) Note that (x21 + 4x1 x2 + 4x22 ) = (x1 + 2x2 )2 = 1, so that x1 + 2x2 = ±1, and the two lines are x2 =

1−x1 2

and x2 =

−1−x1 . 2

370

Section 8.2

E10 v2 1 v √5 2

– 1 v1

1 v √10 1

v1

– 1 v2

1

√5

√10

E5

Figure 8.6: for Problem 8.2.18.

v1

v2 1 v √5 1

– 1 v1 √5

1

E0 E5

Figure 8.7: for Problem 8.2.19.

3 ; eigenvalues λ1 = 6 and λ2 = −4 5 1 −3 , ~v2 = √110 orthonormal eigenbasis ~v1 = √110 3 1

−3 8.2.20 A = 3

6c21 − 4c22 = 1. This is a hyperbola, as shown in Figure 8.8. 8.2.21 a In each case, it is informative to think about the intersections with the three coordinate planes: x1 − x2 , x1 − x3 , and x2 − x3 . 371

Chapter 8

v1 v2

1 v √6 1

– 1 v1 √6

E–4

E6

Figure 8.8: for Problem 8.2.20. •] For the surface x21 + 4x22 + 9x23 = 1, all these intersections are ellipses, and the surface itself is an ellipsoid . 0 1 This surface is connected and bounded; the points closest to the origin are ± 0 , and those farthest ± 0 . 1 0 3 (See Figure 8.9.) x3

x2 x1

Figure 8.9: for Problem 8.2.21a: x21 + 4x22 + 9x23 = 1, an ellipsoid (not to scale). •] In the case of x21 + 4x22 − 9x23 = 1, the intersection with the x1 − x2 plane is an ellipse, and the two other intersections are hyperbolas. The surface is connected and not bounded; the points closest to the origin are 372

Section 8.2 0 ± 12 . (See Figure 8.10.) 0 x3

x2

x1

Figure 8.10: for Problem 8.2.21a: x21 + 4x22 − 9x23 = 1, a hyperboloid of one sheet (not to scale). •] In the case −x21 −4x22 +9x23 = 1, the intersection with the x1 −x2 plane is empty, and the two other intersections 0 are hyperbolas. The surface consists of two pieces and is unbounded. The points closest to the origin are ± 0 . 1 3

(See Figure 8.11.)

x3

x2

x1

Figure 8.11: for Problem 8.2.21a: −x21 − 4x22 + 9x23 = 1, a hyperboloid of two sheets (not to scale). 373

Chapter 8 1

1 2

1

b A = 12 1

2

3 2

3 2

3

is positive definite, with three positive eigenvalues λ1 , λ2 , λ3 .

The surface is given by λ1 c21 + λ2 c22 + λ3 c23 = 1 with respect to the principal axis, an ellipsoid. To find the points closest to and farthest from the origin, use technology to find the eigenvalues and eigenvectors:

eigenvalues: λ1 ≈ 0.56, λ2 ≈ 4.44, λ3 = 1 0.86 0.31 unit eigenvectors: ~v1 ≈ 0.19 , ~v2 ≈ 0.54 , ~v3 = −0.47 0.78 Equation: 0.56c21 + 4.44c22 + c23 = 1

1 √1 −2 6 1

1 Farthest points when c1 = ± √0.56 and c2 = c3 = 0 1 and c1 = c3 = 0 Closest points when c2 = ± √4.44

0.86 1.15 1 0.19 ≈ ± 0.26 Farthest points ≈ ± √0.56 −0.47 −0.63 0.31 0.15 1 0.54 ≈ ± 0.26 Closest points ≈± √4.44 0.78 0.37

−1 0 8.2.22 A = 0 1 5 0

5 0 ; eigenvalues λ1 = 4, λ2 = −6, λ3 = 1 −1

Equation with respect to principal axis: 4c21 − 6c22 + c23 = 1, a hyperboloid of one sheet (see Figure 8.10). Closest to origin when c1 = ± 12 , c2 = c3 = 0.

1 1 0.35 A unit eigenvector for eigenvalue 4 is ~v = √12 0 , so that the desired points are ± 21 √12 0 ≈ ± 0 . 1 1 0.35 8.2.23 Yes; M = 21 (A + AT ) is symmetric, and ~x T M~x = 12 ~x T A~x + 12 ~x T AT ~x = 21 ~x T A~x + 12 ~x T A~x = ~x T A~x Note that ~x T A~x is a 1 × 1 matrix, so that ~x T A~x = (~x T A~x)T = ~x T AT ~x. 8.2.24 q(~e1 ) = ~e1 · A~e1 = ~e1 · (first column of A) = a11 8.2.25 q(~v ) = ~v · A~v = ~v · (λ~v ) = λ(~v · ~v ) = λ ↑ ~v is a unit vector 374

Section 8.2

8.2.26 False; If A =

0 1 1 0

then q

0 1 1 0 1 1 1 = 0. · = · = 1 0 0 1 0 0 0

8.2.27 Let ~v1 , . . . , ~vn be an orthonormal eigenbasis for A with A~vi = λi~vi . We know that q(~vi ) = λi (see Exercise 25), so that q(~v1 ) = λ1 and q(~vn ) = λn are in the image. We claim that all numbers between λn and λ1 are in the image as well. To see this, apply the Intermediate Value π Theorem to the continuous function f (t) = q((cos t)~ v + (sin t)~ v ) on 0, vn ) = λn and n 1 2 (note that f (0) = q(~ f π2 = q(~v1 ) = λ1 ). (See Figure 8.12.) λ1 v1

(cos t)vn + (sin t)v1, a unit vector

f(t) λn 0

π 2

t

vn

Figure 8.12: for Problem 8.2.27. The Intermediate Value Theorem tells us that for any c between λn and λ1 , there is a t0 such that f (t0 ) = q((cos t0 )~vn + (sin t0 )~v1 ) = c. Note that (cos t0 )~vn + (sin t0 )~v1 is a unit vector. Now we will show that, conversely, q(~v ) is on [λn , λ1 ] for all unit vectors ~v . Write ~v = c1~v1 + · · · + cn~vn and note that k~v k2 = c21 + · · · + c2n = 1. Then q(~v ) = λ1 c21 + λ2 c22 + · · · + λn c2n ≤ λ1 c21 + λ1 c22 + · · · + λ1 c2n = λ1 . Likewise, q(~v ) ≥ λn . We have shown that the image of S n−1 under q is the closed interval [λn , λ1 ]. 8.2.28 The hint almost gives it away. Since D is a diagonal matrix with positive diagonal entries, we can write D = D12 , where D1 is diagonal with positive diagonal entries (the square roots of the entries of D). Now A = SDS T = SD1 D1 S T = SD1 (SD1 )T = BB T where B = SD1 . The columns of B are scalar multiples of the corresponding columns of S, so that they are orthogonal. 8.2.29 From Example 1 we have S = 6 2 √1 . 5 −3 4

√1 5

2 1 −1 2

and D =

9 0

3 0 . Let D1 = 0 4

0 2

and B = SD1 =

8.2.30 Define D1 as in Exercise 28. Then A = SDS −1 = SD1 D1 S −1 = (SD1 S −1 )(SD1 S −1 ) = B 2 , where B = SD1 S −1 . B is positive definite, since S −1 BS = D1 is diagonal with positive diagonal entries. 8.2.31 S =

√1 5

2 −1

1 2

and D1 =

3 0 0 2

(see Exercise 29), so that B = SD1 S −1 =

8.2.32 Recall that a = q(~e1 ) > 0 and det A = ac − b2 = λ1 λ2 > 0.

x x 0 a b = 0 y z b c

y z

x2 = xy

xy y2 + z2

x2 = a means that xy = b y2 + z2 = c 375

2.8 −0.4

−0.4 . 2.2

Chapter 8 It is required that x and z be positive. This system has the unique solution √ x= a y=

b x

z=

p

=

√b a

q

c − y2 =

c−

b2 a

=

q

ac−b2 a

8.2.33 Use the formulas for x, y, z derived in Exercise 32. √ √ √ x= a= 8=2 2 y= z=

√b a

q

2 = − √12 = − 2√ 2

ac−b2 a

√ 2 2 L= − √12

=

q

0

√3 2

36 8

=

√3 , 2

so that

.

8.2.34 (i) implies (ii): See the hint at the end of the exercise. (ii) implies (iii): det A(m) is the product of the (positive) eigenvalues. (iii) implies (iv): T A(n−1) ~v B 0 B ~x BB T = = T T ~v k ~x 1 0 t ~x T B T B~x = ~v has the unique solution The system ~x T ~x + t = k A=

B~x ~x T ~x + t

~x = B −1~v t = k − ~x T ~x = k − kB −1~v k2 . (n)

Note that t is positive since 0 < det(A

B ) = det(A) = det ~x T

(iv) ⇒ (i)

T B 0 det 0 1

~x = (det B)2 · t. t

~x T A~x = ~x T LLT ~x = (LT ~x)T (LT ~x) = kLT ~xk2 > 0 if ~x 6= ~0, since L is invertible.

4 8.2.35 Solve the system −4 8

x2 = 4, so x = 2 2y = −4, so y = −2 2z = 8, so z = 4 4 + w2 = 13, so w = 3 −8 + 3t = 1, so t = 3 16 + 9 + s2 = 26, so s = 1

−4 8 x 0 13 1 = y w 1 26 z t 2 0 0 L = −2 3 0 4 3 1

0 x y 00 w s 0 0

z t s

8.2.36 If A = QR, then AT A = (QR)T QR = RT QT QR = RT R = LLT , L = RT . 376

Section 8.2 2

∂q ∂x1

2

2

∂q ∂ q ∂ q ∂ q = 2ax1 + bx2 and ∂x = bx1 + 2cx2 , so that q11 = ∂x = b, and 2 = 2a, q22 = ∂x2 = 2c, and q12 = ∂x ∂x 1 2 1 2 2 2a b q q D = det 11 12 = det = 4ac − b2 > 0. q12 q22 b 2c a 2b The matrix A = b of q is positive definite, since a > 0 and det(A) = 14 D > 0. This means, by definition, c 2 that q has a minimum at ~0, since q(~x) > 0 = q(~0) for all ~x 6= ~0.

8.2.37

8.2.38 The eigenvalues of B are p − q and nq + p − q = p + (n − 1)q, so that B is positive definite if p − q > 0 and p + (n − 1)q > 0. 8.2.39 If ~v1 , . . . , ~vn is such a basis consisting of unit vectors, and we let A = [~v1 · · · ~vn ], then 1 cos θ · · · cos θ .. . cos θ 1 cos θ AT A = . is positive definite, so that, by Exercise 38, 1−cos θ > 0 and 1+(n−1) cos θ > 0 .. .. .. .. . . . cos θ cos θ · · · 1 1 1 or 0 < θ < arccos 1−n . or 1 > cos θ > 1−n

1

cos θ Conversely, if θ is in this range, then the matrix . ..

cos θ 1 .. .

··· .. . .. .

cos θ

cos θ is positive definite, so that it has a .. .

cos θ cos θ · · · 1 Cholesky factorization LLT . The columns of LT give us a basis with the desired property.

8.2.40 Let λ be the smallest eigenvalue of A. If we let k = 1 − λ, then the smallest eigenvalue of the matrix A + kIn will be λ + k = 1, so that all the eigenvalues of A + kIn will be positive. Thus matrix A + kIn will be positive definite, by Theorem 8.2.4. 8.2.41 The functions x21 , x1 x2 , x22 form a basis of Q2 , so that dim(Q2 ) = 3. 8.2.42 The functions xi xj form a basis of Qn , where 1 ≤ i ≤ j ≤ n. A little combinatorics shows that there are 1 + 2 + 3 + · · · + n = n(n + 1)/2 such functions, so that dim(Qn ) = n(n + 1)/2 8.2.43 Note that T (ax21 + bx1 x2 + cx22 ) = ax21 (we let x2 = 0). Thus im(T ) = span(x21 ), rank(T ) = 1, ker(T ) = span(x1 x2 , x22 ), nullity(T ) = 2. 8.2.44 Note that T (ax21 +bx1 x2 +cx22 ) = ax21 +bx1 +c (we let x2 = 1). Thus im(T ) = P2 , rank(T ) = 3, ker(T ) = {0}, nullity(T ) = 0 (T is an isomorphism). 8.2.45 Note that T (ax21 + bx22 + cx23 + dx1 x2 + ex1 x3 + f x2 x3 ) = ax21 + b + c + dx1 + ex1 + f (we let x2 = x3 = 1). Thus im(T ) = P2 and rank(T ) = 3. The kernel of T consists of the quadratic forms with a = 0, d + e = 0, and b + c + f = 0 (consider the coefficients of x21 , x1 , and 1). The general element of the kernel is q(x1 , x2 , x3 ) = (−c − f )x22 + cx23 − ex1 x2 + ex1 x3 + f x2 x3 = c(x23 − x22 ) + e(x1 x3 − x1 x2 ) + f (x2 x3 − x22 ). Thus ker(T ) = span(x23 − x22 , x1 x3 − x1 x2 , x2 x3 − x22 ) and nullity(T ) = 3. 8.2.46 Note that T (ax21 + bx22 + cx23 + dx1 x2 + ex1 x3 + f x2 x3 ) = ax21 + bx22 + cx21 + dx1 x2 + ex21 + f x1 x2 (we let x3 = x1 ). Thus im(T ) = Q2 and rank(T ) = 3. The kernel of T consists of the quadratic forms with 377

Chapter 8 a + c + e = 0, b = 0, and d + f = 0 (consider the coefficients of x21 , x22 , and x1 x2 ). The general element of the kernel is q(x1 , x2 , x3 ) = (−c − e)x21 + cx23 − f x1 x2 + ex1 x3 + f x2 x3 = c(x23 − x21 ) + e(x1 x3 − x21 ) + f (x2 x3 − x1 x2 ). Thus ker(T ) = span(x23 − x21 , x1 x3 − x21 , x2 x3 − x1 x2 ) and nullity(T ) = 3. 8.2.47 T (A + B)(~x) = ~x T (A + B)~x = ~x T A~x + ~x T B~x equals (T (A) + T (B))(~x) = T (A)(~x) + T (B)(~x) = ~x T A~x + ~x T B~x. The verification of the second axiom of linearity is analogous. By definition of a quadratic form, im(T ) = Qn : For every quadratic form q in Qn there is a symmetric n × n matrix A such that q = T (A). Thus, the rank of T is dim(Qn ) = n(n + 1)/2 (see Exercise 42). By the rank nullity theorem, nullity(T ) = dim(Rn×n ) − rank(T ) = n2 −

n(n + 1) n(n − 1) = 2 2

Next, let’s think about the kernel of T . In our solution to Exercise 23 we observed that T (A) = T ( 12 (A + AT )); note that matrix 21 (A + AT ) is symmetric. Now 21 (A + AT ) = 0 if (and only if) AT = −A, that is, if A is skew-symmetric. Thus the skew-symmetric matrices are in the kernel of T . Since the space of skew-symmetric matrices has the same dimension as ker(T ), namely, n(n − 1)/2, we can conclude that ker(T ) consists of all skew-symmetric n × n matrices. 0 0 1 8.2.48 The matrix of T with respect to the basis x21 , x1 x2 , x22 is A = 0 1 0 , with the eigenvalues 1,1, −1 and 1 0 0 0 1 1 corresponding eigenvectors 1 , 0 , 0 . Thus x1 x2 and x21 + x22 are eigenfunctions with eigenvalue 1, and 0 1 −1 x21 − x22 has eigenvalue −1. Yes, T is diagonalizable, since there is an eigenbasis.

1 0 0 8.2.49 The matrix of T with respect to the basis x21 , x1 x2 , x22 is A = 0 2 0 , with the eigenvalues 1,2,4 and 0 0 4 1 0 0 corresponding eigenvectors 0 , 1 , 0 . Thus x21 , x1 x2 , x22 are eigenfunctions with eigenvalues 1, 2, and 4, 0 0 1 respectively. Yes, T is diagonalizable, since there is an eigenbasis.

0 8.2.50 The matrix of T with respect to the basis x21 , x1 x2 , x22 is A = 2 0 1 1 1 corresponding eigenvectors 0 , 2 , −2 . −1 1 1

1 0 0 2 , with the eigenvalues 0,2, −2 and 1 0

Thus x21 − x22 , x21 + 2x1 x2 + x22 , x21 − 2x1 x2 + x22 are eigenfunctions with eigenvalues 0, 2, and −2, respectively. Yes, T is diagonalizable, since there is an eigenbasis.

8.2.51 If B is negative definite, then A = −B is positive definite, so that the determinants of all principal submatrices A(m) are positive. Thus det(B (m) ) = det(−A(m) ) = (−1)m det(A(m) ) is positive for even m and negative for odd m. 378

Section 8.2 8.2.52 Because aij = ~eTj A~ei , we have q(~ ei ) = aii . Further, using linearity q(~ei + ~ej ) = (~ei + ~ej )T A(~ei + ~ej ) = T T T T ~ei A~ei + ~ei A~ej + ~ej A~ei + ~ej A~ej = q(~ei ) + q(~ej ) + 2aij . Solving for aij gives aij = 12 (q(~ei + ~ej ) − q(~ei ) − q(~ej )). T

8.2.53 a. Because p (x, y) = q (x~ei + y~ej ) = (x~ei + y~ej ) A (x~ei + y~ej ) = aii x2 + aij xy + aji yx + ajj y 2 , this is a quadratic form with matrix B. b. If q is positive definite and (x, y) 6= (0, 0) , then p (x, y) = q (x~ei + y~ej ) > 0. c. If q is positive semidefinite, then p (x, y) = q (x~ei + y~ej ) ≥ 0 for all x, y. d. If q (x1 , x2 , x3 ) = x21 + x22 − x23 and we let i = 1, j = 2, then p(x, y) = q(x, y, 0) = x2 + y 2 is positive definite. 8.2.54 The entries a1j = aj1 must all be 0. To see that a1j = 0, consider the function p(x, y) = q(x~e1 + y~ej )defined a11 a1j 0 a1j in Exercise 8.2.53. By Exercise 53a, the symmetric matrix of p will be = . This a a a ajj j1 jj j1 0 a1j matrix is positive semidefinite, by Exercise 53c, implying that det = −a21j ≥ 0. Thus a1j = 0, as aj1 ajj claimed. 8.2.55 As the hint suggests, it suffices to prove that aij < aii or aij < ajj , implying that for every entry off the T diagonal there exists a larger entry on the diagonal. Now q (~ei − ~ej ) = (~ei − ~ej ) A (~ei − ~ej ) = aii −2aij +ajj > 0, or, aii + ajj > 2aij , proving the claim. 8.2.56 Let λ1 ≥ λ2 ≥ ... ≥ λn be the eigenvalues of A. In Exercise 8.2.27 we see that the range of q on unit vectors is the interval [λn , λ1 ]. Since a11 = q (~e1 ) is in that range, we must have a11 ≤ λ1 , as claimed. 8.2.57 Working in coordinates with respect to an orthonormal eigenbasis for A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 + λ3 c23 = 1, where the eigenvalues λ1 , λ2 , λ3 are positive. This level surface is an ellipsoid. 8.2.58 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 = 1, where the eigenvalues λ1 and λ2 are positive. This level surface is a cylinder. 8.2.59 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation√ q (~x) = 1 as λ1 c21 = 1, where the eigenvalue λ1 is positive. This level surface is a pair of parallel planes, c1 = ±1 λ1

8.2.60 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 + λ3 c23 = 1, where the eigenvalue λ1 is positive, while λ2 and λ3 are negative. This level surface is a hyperboloid of two sheets. 8.2.61 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 + λ3 c23 = 1, where the eigenvalues λ1 and λ2 are positive, while λ3 is negative. This level surface is a hyperboloid of one sheet. 8.2.62 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 0 as λ1 c21 + λ2 c22 + λ3 c23 = 0, where the eigenvalues λ1 and λ2 are positive, while λ3 is negative. This level surface is a cone. 8.2.63 Note that w ~i · w ~ i = λ1i . Now q (c1 w ~ 1 + ... + cn w ~ n) = (c1 w ~ 1 + ... + cn w ~ n ) · A (c1 w ~ 1 + ... + cn w ~ n )= (c1 w ~ 1 + ... + cn w ~ n ) · (λ1 c1 w ~ 1 + ... + λn cn w ~ n )= λ1 c21 λ11 + ... + 379

Chapter 8 λn c2n λ1n = c21 + ... + c2n , as claimed.

8 −2 −2 5

, with

8.2.64 We will use the strategy outlined in Exercise 8.2.63. The symmetric matrix of q is A = 2 1 , ~v2 = √15 , with associated eigenvalues λ1 = 9 and λ2 = 4. an orthonormal eigenbasis ~v1 = √15 −1 2 2 1 1 1 1 √ √ Thus the orthogonal basis w ~ 1 = √1λ ~v1 = 3√ , w ~ = has the required property. (See ~ v = 2 λ2 2 5 2 5 1 −1 2 Figure 8.13.)

y 1 √ , √15 2 5

1 2

w2

x

− 21

w1

1 2 2 1 √ , − 3√ 3 5 5

− 21 Figure 8.13: for Problem 8.2.64.

8.2.65 Working in coordinates c1 , c2 with respect to an orthonormal eigenbasis ~v1 , ~v2 for A, we can write q (~x) = λ1 c21 +λ2 c22 , where the eigenvalue λ1 is positive while λ2 is negative. We define the orthogonal vectors w ~ 1 = √1λ ~v1 1 1 1 1 ~2 · w ~ 2 = (−λ2 ) . Now q (c1 w ~ 1 + c2 w ~ 2 ) = (c1 w ~ 1 + c2 w ~ 2) · ~1 · w ~ 1 = λ1 and w and w ~ 2 = √−λ ~v2 . Note that w 2 1 (λ1 c1 w ~ 1 + λ2 c2 w ~ 2 ) = λ1 c21 λ11 + λ2 c22 (−λ = c21 − c22 , as claimed. 2)

3 −5 −5 3

, with

8.2.66 We will use the strategy outlined in Exercise 8.2.65. The symmetric matrix of q is A = 1 1 , ~v2 = √12 an orthonormal eigenbasis ~v1 = √12 , with associated eigenvalues λ1 = 8 and λ2 = −2. −1 1 1 1 1 , w ~ 2 = √−λ has the required property that Thus the orthogonal basis w ~ 1 = √1λ ~v1 = 41 ~v2 = 21 1 2 −1 1 q (c1 w ~ 1 + c2 w ~ 2 ) = c21 − c22 . (See Figure 8.14.) 8.2.67 Consider an orthonormal eigenbasis ~v1 , ..., ~vn for A with associated eigenvalues λ1 , ..., λn , such that the eigenvalues λ1 , ..., λp are positive, λp+1 , ..., λr are negative, and the remaining eigenvalues are 0. Define a 380

Section 8.2

y

1

1 1 2, 2

w2 w1

-1

x

1 1, 1 4 −4

-1

Figure 8.14: for Problem 8.2.66. new orthogonal eigenbasis w ~ 1 , ..., w ~ n by setting w ~i =

.p |λi | ~vi for i = 1, ..., r and w ~ i = ~vi for i = 1

r + 1, ..., n. Note that w ~i · w ~ i = 1/λi for i = 1, ..., p and w ~i · w ~ i = 1/ (−λi ) for i = p + 1, ..., r. Now q (c1 w ~ 1 + ... + cp w ~ p + ... + cr w ~ r + ... + cn w ~ n ) = (c1 w ~ 1 + ... + cp w ~ p + ... + cr w ~ r + ... + cn w ~ n )·(λ1 c1 w ~ 1 + ... + λp cp w ~ p + ... + λ 1 2 2 2 2 = λ1 c21 λ11 + ... + λp c2p λ1p − ... − λr c2r (−λ = c + ... + c − c − ... − c , as claimed. p r 1 p+1 r) T 8.2.68 p (~x) = q (L (~x)) = q (R~x) = (R~x) A (R~x) = ~xT RT AR ~x, proving that p is a quadratic form with symmetric matrix RT AR. T T 8.2.69 If A is positive definite, then ~xT RT AR ~x = n(R~ ox) A (R~x) ≥ 0 for all ~x, meaning that R AR is positive T semidefinite. If A is positive definite and ker R = ~0 , then ~xT RT AR ~x = (R~x) A (R~x) > 0 for all nonzero T T ~x, meaning that RT AR is positive definite. n o Conclusion: R AR is always positive semidefinite, and R AR is positive definite if (and only if) ker R = ~0 , meaning that the rank of R is m.

8.2.70 Since A is indefinite, there exist vectors ~v1 and ~v2 in Rn such that ~v1T A~v1 > 0 and ~v2T A~v2 < 0. Since the n×m matrix R has rank n, we know that the image of R is all of Rn , so that there exist vectors w ~ 1 and w ~ 2 in Rm with T T T T T T ~ 2 = ~v2T A~v2 < 0, ~ 2 R ARw ~ 1 = (Rw ~ 1 ) A (Rw ~ 1 ) = ~v1 A~v1 > 0 and w Rw ~ 1 = ~v1 and Rw ~ 2 = ~v2 . Now w ~ 1 R ARw T proving that matrix R AR is indefinite. 8.2.71 Anything can happen. Consider the example A =

381

1 0 0 −1

1 0 , R1 = I2 , R2 = and R3 = . Then 0 1

Chapter 8 R1T AR1 = A is indefinite, R2T AR2 = [1] is positive definite, and R3T AR3 = [−1] is negative definite.

Section 8.3 8.3.1 σ1 = 2, σ2 = 1 8.3.2 The image of the unit circle is the unit circle, since the transformation defined by A preserves length. Thus σ1 = σ2 = 1 by Theorem 8.3.2. 8.3.3 AT A = In ; the eigenvalues of AT A are all 1, so that the singular values of A are all 1. q √ √ 1 1 , with eigenvalues λ1,2 = 3±2 5 . The singular values of A are σ1 = 3+2 5 = 8.3.4 AT A = 1 2 q √ √ 3− 5 σ2 = = −1+2 5 ≈ 0.62. 2

√ 1+ 5 2

≈ 1.62 and

2 p + q2 0 , with eigenvalues λ1 = λ2 = p2 + q 2 . The singular values of A are σ1 = σ2 = 8.3.5 AT A = 0 p2 + q 2 p p p2 + q 2 . A represents a rotation combined with a scaling, with a scaling factor of p2 + q 2 , so that the image p 2 2 of the unit circle is a circle with radius p + q . 8.3.6 The eigenvalues of AT A are λ1 = 25 and λ2 = 0, so that the singular values of A are σ1 = 5 and σ2 = 0 (these are also the eigenvalues of A; compare with Exercise 24). 1 1 1 works. The image of the unit circle is the line segment connecting the tips E5 = span , so that ~v1 = √5 2 2 of A~v1 = 5~v1 and A(−~v1 ) = −5~v1 . See Figure 8.15. 5 v1 v1

A

– v1 –5 v1

Figure 8.15: for Problem 8.3.6.

1 8.3.7 A A = 0 T

0 4

λ1 = 4, λ2 = 1; σ1 = 2, σ2 = 1 382

Section 8.3 1 0 , ~u1 = , ~v2 = eigenvectors of AT A : ~v1 = 0 1 0 1 2 0 . ,V = Σ= 1 0 0 1

1 v1 ) σ1 (A~

=

0 , ~u2 = −1

1 v2 ) σ2 (A~

8.3.9 AT A =

0 1 1 , , so that U = −1 0 0

p2 + q 2 0

p 0 2 2 p2 + q 2 2 2 ; λ1 = λ2 = p + q ; σ 1 = σ 2 = p +q 0 1 T , ~u1 = σ11 A~v1 = , ~v2 = eigenvectors of A A : ~v1 = 1 0 p √ 21 2 −q , so that U = √ 21 2 p −q , Σ = ( p2 + q 2 )I2 , V = I2 . p +q p +q p q p

8.3.8 AT A =

=

5 10

10 20

√ 21 2 p +q

p , ~u2 q

=

1 v2 σ2 A~

=

(See Exercise 6)

λ1 = 25, λ2 = 0; σ1 = 5, σ2 = 0; eigenvectors of AT A : 1 −2 1 1 1 1 1 √ √ √ , ~v2 = 5 , ~u1 = σ1 A~v1 = 5 , ~u2 = a unit vector orthogonal to ~v1 = 5 2 1 2 −2 1 −2 5 0 ~u1 = √15 so that U = V = √15 ,Σ= . 1 2 1 0 0 8.3.10 In Example 4 we found sides: 6 −7 = 2 6

√1 5

2 1 −1 2 ↑ U

6 −7

10 0 0 5

2 = 6

√1 5

√1 5

↑ Σ

1 2

1 2 −2 1

−2 1

10 0

0 5

√1 5

2 −1 1 2

; now take the transpose of both

.

↑ VT

1 0 8.3.11 A A = ; λ1 = 4, λ2 = 1; σ1 = 2, σ2 = 1 eigenvectors of AT A : 0 4 0 0 1 1 0 , ~u1 = σ11 A~v1 = 1 , ~u2 = σ12 A~v2 = 0 , ~u3 = 0 , , ~v2 = ~v1 = 0 1 1 0 0 0 1 0 2 0 0 1 U = 1 0 0 , Σ = ,V = . 0 1 1 0 0 0 1 T

8.3.12 In Example 5 we see that

0 1 1 1

1 = 0

√1 2

1 1

−1 1

Now take the transpose of both sides. 383

√

3 0 0 1

0 0

√1 6 √1 2 √1 3

√2 6

0 − √13

√1 6 − √12 √1 3

.

Chapter 8

0 1 1

1 √ 6 1 2 √ 1 = 6 0 √1 6

√1 2

0 − √12

√1 3 − √13 √1 3

↑ U

√ 3 0 0

0 1 1 √12 −1 0

↑ Σ

1 1

↑ VT

√ √ 37 16 8.3.13 A A = ; λ1 = 45, λ2 = 5; σ1 = 3 5, σ2 = 5 eigenvectors of AT A : 16 13 2 −1 1 0 , ~v2 = √15 , ~u1 = σ11 A~v1 = ~v1 = √15 , ~v2 = σ12 A~v2 = , so that 1 2 0 1 √ 2 −1 1 0 3 5 √0 1 √ . ,V = 5 U= ,Σ= 1 2 5 0 0 1 T

4 6 ; λ1 = 16, λ2 = 1; σ1 = 4, σ2 = 1 8.3.14 A A = 6 13 1 −2 1 1 T eigenvectors of A A : ~v1 = √5 , ~v2 = √5 , ~u1 = 2 1 −1 , so that = √15 2 2 −1 1 −2 4 0 1 1 √ √ U= 5 ,Σ= . ,V = 5 1 2 2 1 0 1 T

8.3.15 If A~v1 = σ1 ~u1 and A~v2 = σ2 ~u2 , then A−1 ~u1 = are the reciprocals of the singular values of A.

1 v1 σ1 ~

1 v1 σ1 A~

=

√1 5

2 , ~u2 = 1

and A−1 ~u2 =

1 v2 , σ2 ~

1 v2 σ2 A~

so that the singular values of A−1

2 8.3.16 If A = U ΣV T then A−1 = V Σ−1 U T and (A−1 )T A−1 = U Σ−1 U −1 . Thus (A−1 )T A−1 is similar to 2 Σ−1 , so that the eigenvalues of (A−1 )T A−1 are the squares of the reciprocals of the singular values of A. It follows that the singular values of A−1 are the reciprocals of those of A. 8.3.17 We need to check that A But A

~b·~ u1 v1 σ1 ~

+ ··· +

~b·~ um vm σm ~

~b·~ u1 v1 σ1 ~

=

+ ··· +

~b·~ u1 v1 σ1 A~

~b·~ um vm σm ~

+ ··· +

= projimA~b. (see Page 223).

~b·~ um vm σm A~

= (~b · ~u1 )~u1 + · · · + (~b · ~um )~um

= projimA~b, since ~u1 , . . . , ~um is an orthonormal basis of im(A) (see Theorem 5.1.5). 1 1 1 u2 = 2 1 , ~ 1 −0.1 ~b·~ ~b·~ u1 u2 ∗ . ~x = σ1 ~v1 + σ2 ~v2 = −3.2

1 2 8.3.18 ~b = , ~u1 = 3 4

1 1 1 v1 = 2 −1 , ~ −1

1 5

3 , ~v2 = −4

384

1 5

4 , σ1 = 2, σ2 = 1, so that 3

Section 8.3 8.3.19 ~x = c1~v1 + · · · + cm~vm is a least-squares solution if A~x = c1 A~v1 + · · · + cm A~vm = c1 σ1 ~u1 + · · · + cr σr ~ur = projimA~b. But projimA~b = (~b·~u1 )~u1 +· · ·+(~b·~ur )~ur , since ~u1 , . . . , ~ur is an orthonormal basis of im(A). Comparing ~ ui the coefficients of ~ui above we find that it is required that ci σi = ~b · ~ui or ci = b·~ σi , for i = 1, . . . , r, while no u1 v1 + · · · + condition is imposed on cr+1 , . . . , cm . The least-squares solutions are of the form ~x∗ = b·~ σ1 ~ cr+1~vr+1 + · · · cm~vm , where cr+1 , . . . , cm are arbitrary (see Exercise 17 for a special case).

~b·~ ur vr σr ~

+

8.3.20 a A = U ΣV T = U V T V ΣV T = QS, where Q = U V T and S = V ΣV T . Note that Q is orthogonal, being the product of orthogonal matrices; S is symmetric as S T = (V T )T ΣT V T = V ΣV T = S; and S is similar to Σ, so that the eigenvalues of S are the (nonnegative) diagonal entries of Σ. b Yes, write A = U ΣV T = U ΣU T U V T = S1 Q1 where S1 = U ΣU T and Q1 = U V T . 1 1 2 −1 10 0 1 2 √ √ 8.3.21 A = 2 0 5 5 −2 1 5 1 {z } | {z } | {z } | Σ U VT 1 1 2 −1 1 1 2 −1 2 1 10 0 1 2 √ √ √ √ = 2 2 0 5 5 −2 1 5 1 5 −1 2 5 1 {z }| {z }| {z } | {z } | {z } | Σ U V VT VT 1 4 3 9 −2 = −2 6 5 −3 4 | {z }| {z } Q S

8.3.22 a. T1 is the orthogonal projection onto the plane perpendicular to the vector ~v . T2 scales by the length of the vector ~v and T3 is a rotation about the line through the origin spanned by ~v by a rotation angle π/2. Because Q = A3 is orthogonal and S = A2 A1 is symmetric this is a polar decomposition: A = QS. b. Here, A1 represents the orthogonal projection onto the xz plane, A2 represents a scaling by a factor of 2, and A3 represents a rotation about the y axis through an angle of π/2, counterclockwise as viewed from the positive y axis:

0 0 0 0 , A2 0 1 0 Thus Q = A3 = 0 −1 1 A1 = 0 0

2 0 = 0 2 0 0 0 1 1 0 ,S 0 0

0 0 1 0 0 , A3 = 0 1 0 . −1 0 0 2 0 2 0 0 = A2 A1 = 0 0 0 , and A = QS = 0 −2 0 0 2

0 2 0 0 . 0 0

8.3.23 AAT U = U ΣV T V ΣT U T U = U ΣSigmaT , since V T V = Im and U T U = In , so that 2 σi ~ui for i = 1, . . . , r T AA ~ui = ~0 for i = r + 1, . . . , n The nonzero eigenvalues of AT A and AAT are the same. 8.3.24 The eigenvalues of AT A = A2 are the squares of the eigenvalues of A, so that the singular values of A are the absolute values of the eigenvalues of A. 385

Chapter 8 8.3.25 See Figure 8.16.

→

A unit circle

Au

→

σ2

u

σ1

Figure 8.16: for Problem 8.3.25. Algebraically: Write ~u = c1~v1 + c2~v2 and note that k~uk2 = c21 + c22 = 1. Then A~u = c1 σ1 ~u1 + c2 σ2 ~u2 , so that kA~uk2 = c21 σ12 + c22 σ22 ≥ c21 σ22 + c22 σ22 = σ22 and kA~uk ≥ σ2 . Likewise kA~uk ≤ σ1 . 8.3.26 Write ~v = c1~v1 + · · · + cm~vm and note that k~v k2 = c21 + · · · + c2m . Then A~v = c1 σ1 ~u1 + · · · + cr σr ~ur and kA~v k2 = c21 σ12 + c22 σ22 + · · · + c2r σr2 ≤ c21 σ12 + c22 σ12 + · · · + c2r σ12 ≤ σ12 kvk2 so that kA~v k ≤ σ1 k~v k. Likewise, kA~v k ≥ σm k~v k. 8.3.27 Let ~v be a unit eigenvector with eigenvalue λ and use Exercise 26. 8.3.28 If λ1 , . . . , λn are the eigenvalues of AT A, then (det A)2 = det(AT A) = λ1 · · · λn = σ12 · · · σn2 , so that | det A| = σ1 · · · σn . For a 2 × 2 matrix:

A unit circle

σ1 σ2 Ω

A(Ω)

Figure 8.17: for Problem 8.3.28. | det(A)| = expansion factor =

8.3.29 A = U ΣV T

area of ellipse A(Ω) = area of unit circle Ω

= [~u1 · · · ~ur · · ·]

σ1

..

. σr ..

0

.

πσ1 σ2 π

0 T ~v .1 . . T ~vr . .. 0 386

= σ1 σ2 . See Figure 8.17.

σ ~v T 1 1 .. = [~u1 · · · ~ur · · ·] . σr ~vrT 0

True or False = σ1 ~u1 ~v1T + · · · + σr ~ur ~vrT 1 2 2 −1 1 10 0 6 2 1 1 1 √ √ √1 [2 − 1] = 10 8.3.30 = √5 5 1 5 −2 5 −2 1 2 0 5 −7 6 2 4 −2 2 4 √1 [ 1 2 ] = +5 √15 + 5 1 −8 4 1 2

8.3.31 The formula A = σ1 ~u1 ~v1T + · · · + σr ~ur ~vrT gives such a representation. 8.3.32 (SAR)T SAR = RT AT S T SAR = RT AT AR is similar to AT A, so that the matrices AT A and (SAR)T SAR have the same eigenvalues. Thus A and SAR have the same singular values. 8.3.33 Yes; since AT A is diagonalizable and has only 1 as an eigenvalue, we must have AT A = In . 8.3.34 A = U ΣU T means that U T AU = U −1 AU = Σ, i.e., A is orthogonally diagonalizable, with eigenvalues ≥ 0. This is the case if and only if A is symmetric and positive semidefinite. 8.3.35 We will freely use the diagram on Page 393 (with r = m). We have AT A~vi = AT (σi ~ui ) = σi2~vi and therefore (AT A)−1~vi = σ12 ~vi for i = 1, . . . , m. Then (AT A)−1 AT ~ui = (AT A)−1 (σi~vi ) = σ1i ~vi for i = 1, . . . , m i and (AT A)−1 AT ~ui = ~0 for i = m + 1, . . . , n since ~ui is in ker(AT ) in this case. Note that (AT A)−1 AT ~ui is the least-squares solution of the equation A~x = ~ui ; for i = 1, . . . , m this is the exact solution since ~ui is in im(A). 8.3.36 We will freely use the diagram on Page 393. By construction of the ~vi as eigenvectors of AT A we have AT A~vi = λi~vi = σi2~vi , or (AT A)−1~vi = σ12 ~vi . Then A(AT A)−1 AT ~ui = A(AT A)−1 (σi~vi ) = A σ1i ~vi = σ1i A~vi = i T −1 T ~0 for i = m + 1, . . . , n since ~ui is in ker(AT ) in this case. The fact that ~ui for i = 1, . . . , m and A(A A) A ~ u = i ~ u if i = 1, . . . , m i A(AT A)−1 AT ~ui = means that the matrix A(AT A)−1 AT represents the orthogonal ~0 if i = m + 1, . . . , n projection onto im(A) = span(~u1 , . . . , ~um ).

True or False Ch 8.TF.1 F. The orthogonal matrix A =

0 −1 1 0

fails to be diagonalizable (over R).

3 , then the eigenvalue of AT A = [ 3 Ch 8.TF.2 T. If A = 4 √ A is σ = λ = 5.

3 = [25] is λ = 25, so that the singular value of 4] 4

Ch 8.TF.3 F. The last term, 5x2 , does not have the form required in Definition 8.2.1 Ch 8.TF.4 F. The singular values of A are the square roots of the eigenvalues of AT A, by Definition 8.3.1. Ch 8.TF.5 T, by Theorem 8.2.4. Ch 8.TF.6 T, by Definition 8.2.1 387

Chapter 8 2 λ1 . λ1 . 0 Ch 8.TF.7 T. If D = . . . , then DT D = D2 = . . 0 . 0 . λn p p and the singular values of D are λ21 = |λ1 |, . . . , λ2n = |λn |.

Ch 8.TF.8 F, since det

2 5 2

5 2

3

0 . . The eigenvalues of DT D are λ21 , . . . , λ2n , λ2n

= − 14 < 0 (see Theorem 8.2.7).

Ch 8.TF.9 T, by the spectral theorem (Theorem 8.1.1) Ch 8.TF.10 T. Note that [ 1

0]

a b b c

1 = a > 0, by Definition 8.2.3. 0

Ch 8.TF.11 T, by Theorem 8.3.2 Ch 8.TF.12 T. All four eigenvalues are negative, so that their product, the determinant, is positive. Ch 8.TF.13 T, by Theorem 8.1.2 Ch 8.TF.14 F, since the determinant is 0, so that 0 is an eigenvalue. Ch 8.TF.15 F. Consider A =

0 −1 . 1 0

Ch 8.TF.16 T, since AAT is symmetric (use the spectral theorem) Ch 8.TF.17 T, by Theorem 8.2.4: all the eigenvalues are positive. Ch 8.TF.18 T, since the matrix is symmetric. Ch 8.TF.19 F. Consider the shear matrix A = values fail to be 1, 1.

1 0

1 . The unit circle isn’t mapped into itself, so that the singular 1

Ch 8.TF.20 F. In general, (AT A)T = AT A 6= AAT Ch 8.TF.21 F. Let A = I2 , ~v =

1 1 . ,w ~= 1 0

P P Ch 8.TF.22 T. Consider the singular value decomposition A = U V T , or P AV = U , where V is orthogonal (see Theorem 8.3.5). We can let S = V , since the columns of AS = AV = U are orthogonal, by construction.

Ch 8.TF.23 T. By the spectral theorem, A is diagonalizable: S −1 AS = D for some invertible S and a diagonal D. Now Dn = S −1 An S = S −1 0S = 0, so that D = 0 (since D is diagonal). Finally, A = SDS −1 = S0S −1 = 0, as claimed. Ch 8.TF.24 F. If k is negative, then kq(~x) will be negative definite. 388

True or False Ch 8.TF.25 T. The eigenvalues λ1 , . . . , λn of A are nonzero, since A is invertible, so that the eigenvalues λ21 , . . . , λ2n of A2 are positive. Now use Theorem 8.2.4. Ch 8.TF.26 T, by Theorem 8.3.2, since ~v = A~e1 and w ~ = A~e2 are the principal semi-axis of the image of the unit circle. Ch 8.TF.27 F. As a counterexample, consider A = S = 2In . Ch 8.TF.28 T, since ~eiT A~ei = aii < 0. Ch 8.TF.29 T. By Theorem 7.3.6, matrices A and B have the same eigenvalues. Now use Theorem 8.2.4. Ch 8.TF.30 T. The spectral theorem guarantees that there is an orthogonal R such that RT AR is diagonal. Now let S = RT . Ch 8.TF.31 F. Let A =

0 1 0 0

and B =

1 0 . Then 1 is a singular value of BA but not of AB . 0 0

Ch 8.TF.32 T, since A + A−1 = A + AT is symmetric. Ch 8.TF.33 F. For example, (x21 )(x2 x3 ) fails to be a quadratic form. Ch 8.TF.34 T. We can write q(~x) = ~x

−1 0 Ch 8.TF.35 F. Consider A = 0 1 0 0

T

1

5 2

5 2

4

~x.

0 0 , which is indefinite. 1

Ch 8.TF.36 T, by Definition 8.2.3: ~x T (A + B)~x = ~x T A~x + ~x T B~x > 0 for all nonzero ~x. Ch 8.TF.37 T, since ~x · A~x is positive, so that cos θ is positive, where θ is the angle enclosed by ~x and A~x. Ch 8.TF.38 T. Preliminary remark: If σ is the largest singular value of an n×m matrix M , then kM~v k ≤ σk~v k for all ~v in Rm (see Exercise 8.3.26). Now let σ1 , σ2 be the singular values of matrix AB , with σ1 ≥ σ2 , and let ~v1 be a unit vector in R2 such that kAB~v1 k = σ1 (see Theorem 8.3.3). Now σ2 ≤ σ1 = kA(B~v1 )k ≤ 3kB~v1 k ≤ 3 · 5k~v1 k = 15, proving our claim; note that we have used the preliminary remark twice. 0 1 . Ch 8.TF.39 F. Consider A = 0 0

Ch 8.TF.40 T. If λ is the smallest eigenvalue of A, let k = 1 − λ. Then the smallest eigenvalue of A + kIn is λ + k = 1, so that all the eigenvalues of A + kIn are positive. Now use Theorem 8.2.4.

Ch 8.TF.41 T. The quadratic form q(x1 , x2 ) = [ x1

a 0 x2 ] b c 389

b d e

c x1 e 0 = ax21 + 2cx1 x2 + f x22 is positive x2 f

Chapter 8

definite. The matrix of this quadratic form is A = Thus af > c2 , as claimed.

c , and det(A) = af − c2 > 0 since A is positive definite. f

a c

1 −1 Ch 8.TF.42 F. Consider the positive definite matrix A = . −1 2

0 . −1

1 Ch 8.TF.43 F. Consider the indefinite matrix A = 0

cos x Ch 8.TF.44 T. By Theorem 8.3.2., the continuous function f (x) = A sin x

has the global maximum 5 and the cos x . By the global minimum 3. Note that the image of the unit circle consists of all vectors of the form A sin x cos c intermediate value theorem, f (c) = 4 for some c. Let ~u = (draw a sketch!). sin c

Ch 8.TF.45 T, since ~x T A2 ~x = −~x T AT A~x = −(A~x)T A~x = −kA~xk2 ≤ 0 for all ~x. Ch 8.TF.46 T. If λ1 , . . . , λn are the eigenvalues of AT A, then λ1 λ2 . . . λn = det(AT A) = (det A)2 . If √ √ σ1 = λ1 , . . . , σn = λn are the singular values of A, then √ σ1 σ2 . . . σn = λ1 λ2 . . . λn = | det A|, as claimed. Ch 8.TF.47 F. Note that the columns of S must be unit eigenvectors of A. There are two distinct real eigenvalues, λ1 , λ2 , and for each of them there are two unit eigenvectors, ±~v1 (for λ1 ) and ±~v2 (for λ2 ). (Draw a sketch!) Thus there are 8 matrices S, namely S = [ ±~v1 ±~v2 ] and S = [ ±~v2 ±~v1 ] Ch 8.TF.48 T. See the remark following Definition 8.2.1. Ch 8.TF.49 F. Some eigenvalues of A may be negative. Ch 8.TF.50 F. Consider the similar matrices A = and 3, while those of B are 0 and 5.

0 0

0 3

and B =

0 4 . Matrix A has the singular values 0 0 3

Ch 8.TF.51 T. Let ~v1 , ~v2 be an orthonormal eigenbasis, with A~v1 = ~v1 and A~v2 = 2~v2 . Consider a nonzero vector ~x = c1~v1 + c2~v2 ; then A~x = c1~v1 + 2c2~v2 . If c1 = 0, then ~x = c2~v2 and A~x = 2c2~v2 are parallel, and we are all set. Now consider the case when c1 6= 0. Then the angle between ~x and A~x is arctan(2c2 /c1 ) − arctan(c2 /c1 ); to see this, subtract the angle between ~v1 and ~x from the angle between ~v1 and A~x (draw a sketch). Let m = c2 /c1 and use calculus to see that the function f (m) = arctan(2m) − arctan(m) assumes its global maximum at m = √12 . √ √ The maximal angle between ~x and A~x is arctan( 2) − arctan(1/ 2) < 0.34 < π/6. √ a b 1 a Ch 8.TF.52 T. Let A = . By Theorem 8.3.2, A = = a2 + c2 < 5 (since the length of the c d 0 c semi-major axis of the image of the unit circle is less than 5). Thus a < 5 and c < 5. Likewise, b < 5 and d < 5.

390

True or False Ch 8.TF.53 T. We need to show that each entry aij = aji off the diagonal is smaller than some entry on the diagonal. Now (~ei −~ej )T A(~ei −~ej ) = aii + ajj − 2aij > 0, so that aii + ajj > 2aij . Thus the larger of the diagonal entries aii and ajj must exceed aij . Ch 8.TF.54 T. Let λ1 , ..., λm be the Pmdistinct eigenvalues of A, with the associated eigenspaces EA,λi . Since A is diagonalizable, we know that k=1 dim (EA,λi ) = n. By definition of an eigenvector ,EA,λi is a subspace of Pm EA3 ,λ3i , Since k=1 dim EA3 ,λ3i cannot exceed n, we must have EA,λi =EA3 ,λ3i for all eigenvalues. Applying the same reasoning to B and B 3 , we can conclude that EA,λi =EA3 ,λ3i = EB 3 ,λ3i = EB,λi . Since the diagonalizable matrices A and B have the same eigenvectors with the same eigenvalues, they must be equal.

391

Chapter 9

Chapter 9 Section 9.1 9.1.1 x(t) = 7e5t , by Theorem 9.1.1. 9.1.2 x(t) = −e · e−0.71t = −e1−0.71t , by Theorem 9.1.1. 9.1.3 P (t) = 7e0.03t , by Theorem 9.1.1. 2

9.1.4 This is just an antiderivative problem: y(t) = 0.8 t2 + C = 0.4t2 + C, and C = −0.8, so that y(t) = 0.4t2 − 0.8. 9.1.5 y(t) = −0.8e0.8t , by Theorem 9.1.1. 9.1.6 x dx = dt x2 2

= t + C, and

1 2

= 0 + C, so that

x2 2

=t+

1 2

x2 = 2t + 1 √ x(t) = 2t + 1 9.1.7 x−2 dx = dt −x−1 = t + C − x1 = t + C, and −1 = 0 + C, so that − x1 = t − 1 x(t) =

1 1−t ;

note that lim x(t) = ∞. x→1−

9.1.8 x−1/2 dx = dt √ 2x1/2 = t + C, and 2 4 = 0 + C, so that 2x1/2 = t + 4. 2 x(t) = 2t + 2 for t ≥ −4.

9.1.9 x−k dx = dt 1 1−k 1−k x

= t + C, and

1 1−k 1−k x

=t+

1 1−k

= C, so that

1 1−k

x1−k = (1 − k)t + 1 x(t) = ((1 − k)t + 1)1/1−k . 9.1.10 cos x dx = dt sin x = t + C, and C = 0. x(t) = arcsin(t) for |t| < 1. 392

Section 9.1 9.1.11

dx 1+x2

= dt

arctan(x) = t + C and C = 0. x(t) = tan(t) for |t| <

π 2.

9.1.12 We want ekt = 3t or ek = 3 or k = ln(3) :

dx dt

= ln(3)x.

9.1.13 a The debt in millions is 0.45(1.06)212 ≈ 104, 245, or about 100 billion dollars. b The debt in millions is 0.45e0.06·212 ≈ 150, 466, or about 150 billion dollars. t

9.1.14 a x(t) = e− 8270 , by Theorem 9.1.1 T

If T is the half-life, then e− 8270 =

1 2

T or − 8270 = ln

1 2

The half-life is about 5732 years.

or T = −8270 ln

1 2

≈ 5732.

t

t = ln(0.53) or t = −8270 ln(0.53) ≈ 5250. The b We want to find t such that e− 8270 = 1 − 0.47 = 0.53 or − 8270 Iceman died about 5000 years before A.D. 1991, or about 3000 B.C. The Austrian expert was wrong.

9.1.15 If a quantity P grows at an instantaneous rate of k %, then P (t) = P0 ekt/100 , by Theorem 9.1.1. The 2) doubling time T is given by P (T ) = P0 ekT /100 = 2P0 , or ekT /100 = 2 or kT /100 = ln 2 or k = 100(ln ≈ 69 k k since ln 2 ≈ 0.69. 9.1.16 See Figure 9.1. x2

x1

Figure 9.1: for Problem 9.1.16. 9.1.17 See Figure 9.2. 9.1.18 See Figure 9.3. 9.1.19 See Figure 9.4.

0 −1 9.1.20 A~x = 1 0

x1 x2

−x2 = . See Figure 9.5. x1 393

Chapter 9

Figure 9.2: for Problem 9.1.17.

Figure 9.3: for Problem 9.1.18.

Figure 9.4: for Problem 9.1.19. 1 cos(t) we will trace out the unit circle ~x(t) = . 0 sin(t) − sin(t) 0 −1 − sin(t) , as claimed. ~x(t) = equals = cos(t) 1 0 cos(t)

It appears that the trajectories will be circles. If we start at We can verify that

9.1.21 A~x =

0 1 0 0

d~ x dt

x1 x2

=

x2 0

(see Figure 9.6).

p , then the horizontal velocity will be q, so that The trajectories will be horizontal lines. If we start at q x1 (t) p + qt q 0 1 q d~ x ~x(t) = = . We can verify that dt = equals ~x(t) = , as claimed. x2 (t) q 0 0 0 0 394

Section 9.1 x2

(1, 0)

x1

Figure 9.5: for Problem 9.1.20. x2

x1

Figure 9.6: for Problem 9.1.21. x1 9.1.22 We are told that d~ x1 and dt = A~ A(~x1 + ~x2 ) = A~x, as claimed.

9.1.23 We are told that claimed.

d~ x1 dt

d~ x2 dt

= A~x2 . Let ~x(t) = ~x1 (t) + ~x2 (t). Then

= A~x1 . Let ~x(t) = k~x1 (t). Then

d~ x dt

=

d~ c dt

=

d kt x) dt (e ~

x x. Let ~c(t) = ekt ~x(t). Then 9.1.24 We are told that d~ dt = A~ (A + kIn )(ekt ~x) = (A + kIn )~c, as claimed.

d x1 ) dt (k~

=

d~ x dt

=

d~ x1 dt

+

d~ x2 dt

= A~x1 + A~x2 =

x1 = k d~ x1 = A(k~x1 ) = A~x, as dt = kA~

d kt dt e

x kt ~x + ekt d~ x + ekt A~x = dt = ke ~

x 9.1.25 We are told that d~ x. Let ~c(t) = ~x(kt). Using the chain rule we find that dt = A~ kA(~x(kt)) = kA~c(t), as claimed.

d~ c dt

=

d x(kt)) dt (~

x = k d~ dt ]kt =

To get the vector field kA~c we scale the vectors of the field A~x by k. 1 −2 −2t −2 3t 1 9.1.26 λ1 = 3, λ2 = −2; ~v1 = , ~v2 = . −e , c1 = 5, c2 = −1, so that ~x(t) = 5e 1 3 1 3 9.1.27 Use Theorem 9.1.3.

−4 3 2 −3

are λ1 = −6 and λ2 = −1, with associated eigenvectors ~v1 = The eigenvalues of A = 1 1 ~v2 = . The coordinates of ~x(0) = with respect to ~v1 and ~v2 are c1 = − 15 and c2 = 52 . 1 0 395

−3 2

and

Chapter 9 By Theorem 9.1.3 the solution is ~x(t) = − 51 e−6t

−3 1 + 25 e−t . 2 1

−3 1 1 5 1 2t −3 5 10t 1 9.1.28 λ1 = 2, λ2 = 10; ~v1 = , ~v2 = ; c1 = − 8 , c2 = 8 , so that ~x(t) = − 8 e + 8e . 2 2 2 2

−2 1 , ~v2 = ; c1 = −2, c2 = 1, 1 2 1 4 1 −2 . + e5t = + e5t so that ~x(t) = −2 2 −2 2 1

9.1.29 λ1 = 0, λ2 = 5; ~v1 =

−2 1 −2 2 9.1.30 λ1 = 0, λ2 = 5, ~v1 = , ~v2 = ; c1 = −1, c2 = 0, so that ~x(t) = − . = 1 2 −1 1

1 9.1.31 λ1 = 1, λ2 = 6, λ3 = 0; ~v1 = −2 . Since ~x(0) = ~v1 we need not find ~v2 and ~v3 . 1 1 c1 = 1, c2 = c3 = 0, so that ~x(t) = et −2 . 1

In Exercises 32 to 35, find the eigenvalues and eigenspaces. Then determine the direction of the flow along the eigenspaces (outward if λ > 0 and inward if λ < 0). Use Figure 11 of Section 9.1 as a guide to sketch the other trajectories.

9.1.32 See Exercise 26 and Figure 9.7. E3

E–2

Figure 9.7: for Problem 9.1.32. 9.1.33 See Exercise 27 and Figure 9.8. 9.1.34 See Exercise 28 and Figure 9.9. 9.1.35 See Exercise 29 and Figure 9.10. In Exercises 36 to 39, find the eigenvalues and eigenspaces (the eigenvalues will always be positive). Then determine the direction of the flow along the eigenspaces (outward if λ > 1 and inward if 1 > λ > 0). Use Figure 11 of Section 7.1 as a guide to sketch the other trajectories. 396

Section 9.1

E–1

E–6

Figure 9.8: for Problem 9.1.33. E10

E2

Figure 9.9: for Problem 9.1.34. E5

E0

Figure 9.10: for Problem 9.1.35.

9.1.36 See Figure 9.11. 9.1.37 See Figure 9.12. 9.1.38 See Figure 9.13. 9.1.39 See Figure 9.14. 2 3t 3 +e 9.1.40 ~x(t) = e 4 3 2t

397

Chapter 9 E1.3

E0.8

Figure 9.11: for Problem 9.1.36. E1.6

E1.1

Figure 9.12: for Problem 9.1.37.

E0.7

E0.9

Figure 9.13: for Problem 9.1.38.

E1 E1.4

Figure 9.14: for Problem 9.1.39.

398

Section 9.1 2 and We want a 2 × 2 matrix A with eigenvalues λ1 = 2 and λ2 = 3 and associated eigenvectors ~v1 = 3 −1 11 −6 −4 3 4 9 2 3 4 9 4 9 2 3 3 . = = or A = = ; that is A ~v2 = 12 −6 3 −2 6 12 3 4 6 12 6 12 3 4 4 9.1.41 The trajectories are of the form ~x(t) = c1 eλ1 t~v1 + c2 eλ2 t~v2 = c1~v1 + c2 eλ2 t~v2 . See Figure 9.15. span (v2)

span (v1)

Figure 9.15: for Problem 9.1.41. 9.1.42 a The term 0.8x in the second equation indicates that species y is helped by x, while species x is hindered by y (consider the term −1.2y in the first equation). Thus y preys on x. b See Figure 9.16. y E1 = span

1 2

E1 = span

3 1

x

Figure 9.16: for Problem 9.1.42b. y(t) t→∞ x(t)

c If

y(0) x(0)

< 2 then both species will prosper, and lim

If

y(0) x(0)

≥ 2 then both species will die out.

= 31 .

9.1.43 a These two species are competing as each is hindered by the other (consider the terms −y and −2x). b Although only the first quadrant is relevant for our model, it is useful to consider the phase portrait in the other quadrants as well. See Figure 9.17. y(0) x(0) y(t) x(t) =

c If

> 2 then species y wins (x will die out); if

y(0) x(0)

< 2 then x wins. If

2 for all t. 399

y(0) x(0)

= 2 then both will prosper and

Chapter 9 y E3 = span

1 2

x

E6 = span

1 –1

Figure 9.17: for Problem 9.1.43b. 9.1.44 a The two species are in symbiosis: Each is helped by the other (consider the terms 4y and 2x). b See Figure 9.18. y

E3 = span

2 1

x

E–3 = span

1 –1

Figure 9.18: for Problem 9.1.44b. y(t) t→∞ x(t)

c Both populations will prosper and lim

= 21 , regardless of the initial populations.

9.1.45 a Species y has the more vicious fighters, since they kill members of species x at a rate of 4 per time unit, while the fighters of species x only kill at a rate of 1. b See Figure 9.19. c If

y(0) x(0)

<

1 2

then x wins; if

y(0) x(0)

>

1 2

then y wins; if

y(0) x(0)

=

1 2

nobody will survive the battle.

9.1.46 Look at the phase portrait in Figure 9.20. 9.1.47 a The two species are in symbiosis: Each is helped by the other (consider the positive terms kx and ky). b λ1,2 =

√ −5± 9+4k2 2

Both eigenvalues are negative if

√

9 + 4k 2 < 5 or 9 + 4k 2 < 25 or 4k 2 < 16 or k < 2 (recall that k is positive). 400

Section 9.1 y E–2 = span

2 1

x

E2 = span

2 –1

Figure 9.19: for Problem 9.1.45b. y E–√pq = span

√p √q

x

E√pq = span

√p –√q

Figure 9.20: for Problem 9.1.46. If k = 2 then the eigenvalues are −5 and 0. If k > 2 then there is a positive and a negative eigenvalue. c See Figure 9.21.

k=1

k=3

k=2 E0

both species die out

E–5 system approaches an equilibrium state

both species prosper

Figure 9.21: for Problem 9.1.47c.

9.1.48 a Symbiosis 401

Chapter 9 b λ1,2 =

√ −5± 9+4k 2

Both eigenvalues are negative if

√

9 + 4k < 5 or 9 + 4k < 25 or 4k < 16 or k < 4.

If k = 4 then the eigenvalues are −5 and 0. If k > 4 then there is a positive and a negative eigenvalue. c k = 1: See corresponding figure in Exercise 47 and Figure 9.22. k=4

k = 10

E0

E1

E–6 E–5 system approaches an equilibrium state

system approaches an equilibrium state

Figure 9.22: for Problem 9.1.48c. −1 −0.2 , λ1 = −0.4, λ2 = −0.8 9.1.49 A = 0.6 −0.2 −1 1 E−0.4 = span , E−0.8 = span 3 −1 g(0) −1 1 = 15 + 45 , so that c1 = 15, c2 = 45. h(0) 3 −1 1 g(t) −0.8t −0.4t −1 , so that + 45e = 15e −1 3 h(t)

g(t) = −15e−0.4t + 45e−0.8t h(t) = 45e−0.4t − 45e−0.8t . See Figure 9.23. 9.1.50 We want both eigenvalues λ1 and λ2 to be negative, so that tr(A) = λ1 + √λ2 < 0 and det(A) = λ1 λ2 > 0. tr(A)± (trA)2 −4 det(A) are both negative. Conversely, if tr(A) < 0 and det(A) > 0, then the two eigenvalues λ1,2 = 2 Answer: tr(A) < 0 and det(A) > 0. 9.1.51 ith component of

d x) dt (S~

=

d dt (si1 x1 (t)

+ si2 x2 (t) + · · · + sin xn (t))

dx2 dxn 1 = si1 dx dt + si2 dt + · · · + sin dt

402

Section 9.1

30 0

E–0.4

E–0.8

Figure 9.23: for Problem 9.1.49. x = ith component of S d~ dt

0 1 p + qt p = ~x are of the form , where ~x(0) = , by Exercise 21. Since 9.1.52 The solutions of 0 0 q q 0 1 λ 1 λt p + qt , by Exercise 24. , the solutions of the given system are of the form ~x(t) = e = λI2 + q 0 0 0 λ The zero state is a stable equilibrium solution if and only if λ < 0. The case λ = 0 is discussed in Exercise 21. See Figure 9.24. d~ x dt

λ>0

λ<0

Figure 9.24: for Problem 9.1.52.

1 0 −1 cos(t) x 9.1.53 For the initial value , the system d~ = ~ x has the solution ~ x (t) = , by Exercise 20; the dt 0 sin(t) 0 1 cos(qt) 0 −q p −q d~ x d~ x , by Exercise 25; and the system dt = has the solution ~x(t) = ~x has system dt = sin(qt) q 0 q p cos(qt) p −q 0 −q , by Exercise 24 write the solution ~x(t) = ept = pI2 + . See Figure 9.25. sin(qt) q p q 0 9.1.54 A =

0 −2

1 1 1 , λ1 = −1, λ2 = −2; ~v1 = and ~v2 = . See Figure 9.26. −3 −1 −2

In the case of trajectory 3 the door will slam: Initially the door is opened just a little (θ is small) and given a θ(0) strong push to close it (ω is large negative). More generally, the door will slam if the point representing ω(0) 403

Chapter 9

p=0

p>0

p<0

Figure 9.25: for Problem 9.1.53. ω

θ trajectory 1 E–1 trajectory 2

door slams trajectory 3

E–2

Figure 9.26: for Problem 9.1.54.

the initial state is located below the line E−2 = span

1 , that is, if −2

ω(0) θ(0)

< −2.

p 0 1 , λ1,2 = 12 −q ± q 2 − 4p ; note that both eigenvalues are negative. −p −q 1 1 . and Eλ2 = = span λ2 λ1

9.1.55 A = Eλ1

See Figure 9.27. In the case of trajectory 3 the door will slam: Initially the door is opened just a little (θ is small) and given a strong push to close it (ω is large negative). More generally, the door will slam if the point θ(0) 1 representing the initial state is located below the line Eλ2 = span , that is, if ω(0) θ(0) < λ2 . ω(0) λ2

Section 9.2 9.2.1 By Euler’s formula (Theorem 9.2.2), e2πi = cos(2π) + i sin(2π) = 1. 1

9.2.2 By Euler’s formula (Theorem 9.2.2), e 2 πi = cos 9.2.3 r =

p

(−1)2 + 12 =

√ 2

π 2

+ i sin

404

π 2

= i.

Section 9.2 ω

θ trajectory 1 Eλ 1 trajectory 2

door slams

Eλ

trajectory 3

2

Figure 9.27: for Problem 9.1.55. –1 + i

i r

θ

–1

1

Figure 9.28: for Problem 9.2.3. θ=

3π 4 ,

so that z =

√

3

2e 4 πi . See Figure 9.28.

9.2.4 e3it = cos(3t) + i sin(3t) t=

π 6

i t=

π 3

i –i

t = 0,

2π 3

–i t=

π 2

Figure 9.29: for Problem 9.2.4. Period T given by 3T = 2π or T =

2π 3 .

See Figure 9.29.

9.2.5 e−0.1t−2it = e−0.1t e−2it = e−0.1t (cos(2t) − i sin(2t)) spirals inward, in clockwise direction. See Figure 9.30. 9.2.6 λ1,2

3−i = ±i; Ei = ker 5

3−i 3+i −2 . and E−i = span = span 5 5 −3 − i

General solution: 405

Chapter 9

t=

3π 4

–1 t=

π 2

i

t=π 3π t= 2

1 t=0

t= –i

π 4

Figure 9.30: for Problem 9.2.5.

~x(t) = c1 eit

3−i 3+i + c2 e−it 5 5

6 cos(t) − 2 sin(t) 3−i 3+i . = + (cos(t) − i sin(t)) If c1 = c2 = 1 then ~x(t) = (cos(t)+i sin(t)) 10 cos(t) 5 5

9.2.7 det(A) = −2, so that the zero state is not stable, by Theorem 9.2.5. 9.2.8 Recall that all eigenvalues of a symmetric matrix are real. The zero state is stable if (and only if) all eigenvalues are negative (by Theorem 9.2.4); this is the case if (and only if) the matrix is negative definite (by Theorem 8.2.2). 9.2.9 The eigenvalues are conjugate complex, λ1,2 = p ± iq, and tr(A) = 2p < 0, so that p is negative. By Theorem 9.2.4, the zero state is stable. 9.2.10 a The matrix of the system is

2a 2b

2b , which is 2A, so that 2c

d~ x dt

= grad(q) = 2A~x.

b By Theorem 8.2.7, the level curves are ellipses. From multivariable calculus we know that grad (q) is perpendicular to the level curve, pointing inwards, so that all trajectories will approach the origin: The zero state is stable. See Figure 9.31. c By Theorem 8.2.7, the level curves are hyperbolas, as shown in Figure 9.32. λ 1 > 0 > λ2 x d The zero state is a stable equilibrium solution of the system d~ x if (and only if) the eigenvalues dt = grad(q) = 2A~ of 2A (and A) are negative. This means that the quadratic form q(~x) = ~x · A~x is negative definite.

9.2.11 a q(~x) = 2ai1 xi x1 + 2ai2 xi x2 + · · · + aii x2i + · · · + 2ain xi xn + terms not involving xi , so that x x. 2ai2 x2 + · · · + 2aii xi + · · · + 2ain xn and d~ dt = grad(q) = 2A~

∂q ∂xi

= 2ai1 x1 +

The matrix of the system is B = 2A. x x if (and only if) all the eigenvalues b The zero state is a stable equilibrium solution of the system d~ dt = grad(q) = 2A~ of 2A are negative. This means that the quadratic form q(~x) = ~x · A~x is negative definite.

406

Section 9.2

Eλ1

q = –1 q = –4 q = –9

Eλ2

Figure 9.31: for Problem 9.2.10b.

q=4 q = –4

q=1

q = –1

q = –1 q = –4 q=1

Eλ2

q=4

Eλ1

Figure 9.32: for Problem 9.2.10c. 9.2.12 We will show that the real parts of all the eigenvalues are negative, so that the zero state is a stable equilibrium solution. Now the characteristic polynomial of A is fA (λ) = −λ3 − 2λ2 − λ − 1. It is convenient to get rid of all these minus signs: The eigenvalues are the solutions of the equation g(λ) = λ3 + 2λ2 + λ + 1 = 0. Since g(−1) = 1 and g(−2) = −1, there will be an eigenvalue λ1 between -2 and -1. Using calculus (or a graphing calculator), we see that the equation g(λ) = 0 has no other real solutions. Thus there must be two complex 1 conjugate eigenvalues p ± iq. Now the sum of the eigenvalues is λ1 + 2p = tr(A) = −2, and p = −2−λ will be 2 negative , as claimed. The graph of g(λ) is shown in Figure 9.33. 9.2.13 Recall that the zero state is stable if (and only if) the real parts of all eigenvalues are negative. Now the eigen 1 −1 . = pp−iq values of A are the reciprocals of those of A; the real parts have the same sign if λ = p + iq, then λ1 = p+iq 2 +q 2 407

Chapter 9

g(λ) = λ3 + 2λ2 + λ + 1

(–1, 1)

(0, 1) λ1 (–2, –1)

Figure 9.33: for Problem 9.2.12. i 9.2.14 a For i > 1, dx dt = −ki xi + xi−1 . This means that in the absence of quantity xi−1 (t), the quantity xi (t) will decay exponentially, but the presence of xi−1 helps xi to grow.

For i = 1, the beginning of the loop, of x1 .

dx1 dt

= −k1 x1 − bxn , so that the presence of xn contributes to the decrease

−k1 b If n = 2 then the matrix of the system is A = 1 so that the zero state is stable, by Theorem 9.2.5.

−b −k2

with tr(A) = −k1 −k2 < 0 and det(A) = k1 k2 +b > 0,

−1 0 −b c No; consider the case k1 = k2 = k3 = 1 for simplicity. Then the matrix of the system is A = 1 −1 0 0 1 −1 √ 3 3 and fA (λ) = − (λ + 1) − b. Matrix A has the real eigenvalue λ1 = − b − 1 and the √complex . eigenvalues √ λ2,3 = 3 b e±πi/3 − 1. (Compare with Exercise 7.5.5.) The real part of λ2 and λ3 is 3 b 2 − 1, which is positive if b exceeds 8. The zero state fails to be stable in that case. 9.2.15 The eigenvalues are λ1 = tr(A) > 0 and λ2 = 0. See Figure 9.34.

E0

Eλ1

Figure 9.34: for Problem 9.2.15.

0 1 a b both negative.

9.2.16 If A =

then tr(A) = b and det(A) = −a. By Theorem 9.2.5, the zero state is stable if a and b are

408

Section 9.2 −1 k then tr(A) = −2 and det(A) = 1 − k 2 . By Theorem 9.2.5, the zero state is stable if k −1 det(A) = 1 − k 2 > 0, that is, if |k| < 1.

9.2.17 If A =

9.2.18 If λ1 , λ2 , λ3 are real and negative, then tr(A) = λ1 + λ2 + λ3 < 0 and det(A) = λ1 λ2 λ3 < 0. If λ1 is real and negative and λ2,3 = p ± iq, where p is negative, then tr(A) = λ1 + 2p < 0 and det(A) = λ1 (p2 + q 2 ) < 0. Either way, both trace and determinant are negative.

1 0 0 9.2.19 False, consider A = 0 2 0 . 0 0 −4 9.2.20 Use Theorem 9.2.6, with p = 0, q = π; a = 1, b = 0. 1 cos(πt) − sin(πt) = (cos(πt))w ~ + (sin(πt))~v . See Figure 9.35. ~x(t) = [ w ~ ~v ] 0 sin(πt) cos(πt)

( )

x 12 = v x(0) = x(2) = w

x(1) = –w –v

Figure 9.35: for Problem 9.2.20.

9.2.21 a

db

dt ds dt

= 0.05b+ =

s 0.07s

b λ1 = 0.07, λ2 = 0.05; ~v1 =

and

b(0) 1, 000 = s(0) 1, 000

1 50 ; ~x(0) = 1, 000~v1 − 49, 000~v2 ; so that , ~v2 = 0 1

b(t) = 50, 000e0.07t − 49, 000e0.05t and s(t) = 1, 000e0.07t . 0 1 , E0.5 = span 9.2.22 λ1 = 3, λ2 = 0.5; E3 = span 1 −1

System is discrete so choose VII. 9.2.23 λ1,2 = − 12 ± i, r > 1, so that trajectory spirals outwards. Choose II. 9.2.24 λ1 = 3, λ2 = 0.5, E3 =

0 1 . , E0.5 = 1 −1

System is continuous, so choose I. 1 9.2.25 negative so that trajectories spiral inwards in the counterclockwise direction λ1,2 = − 2 ± i; realpart is −1.5 1 x . Choose IV. then d~ if ~x = dt = 2 0

409

Chapter 9

9.2.26 λ1 = 1, λ2 = −2; E1 = span

1 0 . , E−2 = span −1 1

System is continuous so choose V. 9.2.27 λ1,2 = ±3i, E3i = span

1 0 0 1 +i , so that p = 0, q = 3, w ~= , ~v = . 0 −1 −1 0

Now use Theorem 9.2.6: 0 1 cos(3t) ~x(t) = e0t −1 0 sin(3t)

− sin(3t) cos(3t)

a sin(3t) = b − cos(3t)

2 0 9.2.28 λ1,2 = ±6i, E6i = span +i , so that 0 3 0 2 cos(6t) − sin(6t) a 2 sin(6t) ~x(t) = = 3 0 sin(6t) cos(6t) b 3 cos(6t)

cos(3t) sin(3t)

2 cos(6t) −3 sin(6t)

1 0 +i , so that 0 1 sin(4t) a − sin(4t) = e2t cos(4t) b cos(4t)

a b

a . b

9.2.29 λ1,2 = 2 ± 4i, E2+4i = span ~x(t) = e2t

0 1 1 0

cos(4t) sin(4t)

cos(4t) − sin(4t)

a . b

0 5 , so that +i 1 3 5 sin(3t) a − sin(3t) = e−2t cos(3t) + 3 sin(3t) b cos(3t)

9.2.30 λ1,2 = −2 ± 3i, E−2+3i = span ~x(t) = e−2t

0 5 1 3

cos(3t) sin(3t)

9.2.31 λ1,2 = −1 ± 2i, E−1+2i

5 cos(3t) − sin(3t) + 3 cos(3t)

1 0 0 = span +i , so that p = −1, q = 2, w ~= , 0 −1 −1

1 1 ~v = . Now = ~x(0) = w ~ + ~v , so that a = 1 and b = 1. 0 −1 0 1 cos(2t) − sin(2t) sin(2t) + cos(2t) 1 Then ~x(t) = e−t . = e−t −1 0 sin(2t) cos(2t) sin(2t) − cos(2t) 1 See Figure 9.36. 1 0 0 1 9.2.32 λ1,2 = ±2i, E2i = span +i , ~x(0) = 0 +1 , so that a = 0 and b = 1. 0 2 2 0 0 1 cos(2t) − sin(2t) 0 cos(2t) ~x(t) = = . See Figure 9.37. 2 0 sin(2t) cos(2t) 1 −2 sin(2t) 1 0 9.2.33 λ1,2 = ±i, Ei = span +i 1 1 0 1 cos(t) a = 1, b = 0, so that ~x(t) = 1 1 sin(t)

− cos(t) cos(t)

1 sin(t) = 0 sin(t) + cos(t)

410

a . b

Section 9.2

1

1

Figure 9.36: for Problem 9.2.31.

1 x (0) = x(π) = 1

1 0

–1 x π2 = 0

( )

0 x π4 = –2

( )

Figure 9.37: for Problem 9.2.32.

= cos(t)

0 1 + sin(t) . See Figure 9.38. 1 1

3 1 +i −2 0 1 3 cos(2t) − sin(2t) cos(2t) + 3 sin(2t) 1 a = 1, b = 0, so that ~x(t) = et . See Figure 9.39. = et 0 −2 sin(2t) cos(2t) −2 sin(2t) 0

9.2.34 λ1,2 = 1 ± 2i, E1+2i = span

9.2.35 If z = f + ig and w = p + iq then zw = (f p − gq) + i(f q + gp), so that (zw)′ = (f ′ p + f p′ − g ′ q − gq ′ ) + i(f ′ q + f q ′ + g ′ p + gp′ ). Also z ′ w = (f ′ + ig ′ )(p + iq) = (f ′ p − g ′ q) + i(f ′ q + g ′ p) and zw′ = (f + ig)(p′ + iq ′ ) = (f p′ − gq ′ ) + i(gp′ + f q ′ ). We can see that (zw)′ = z ′ w + zw′ , as claimed. 9.2.36 A =

0 −b

1 −c

and fA (λ) = λ2 + cλ + b, with eigenvalues λ1,2 = 411

√ −c± c2 −4b . 2

Chapter 9

x (0) = x(2π) =

0 1

1 x π2 = 1

( )

x(π) =

0 –1

Figure 9.38: for Problem 9.2.33.

Figure 9.39: for Problem 9.2.34. √ 9.2.36 a If c = 0 then λ1,2 = ±i b. The trajectories are ellipses. See Figure 9.40. v

x

Figure 9.40: for Problem 9.2.36a. The block oscillates harmonically, with period

2π √ . b

The zero state fails to be asymptotically stable. 412

Section 9.2 b λ1,2 =

√ −c±i 4b−c2 2

The trajectories spiral inwards, since Re(λ1 ) = Re(λ2 ) = − 2c < 0. This is the case of a damped oscillation. The zero state is asymptotically stable. See Figure 9.41.

Figure 9.41: for Problem 9.2.36b. c This case is discussed in Exercise 9.1.55. The zero state is stable here. is differentiable when z(t) 6= 0, since both the real and the imaginary parts are differentiable if z = p + iq then ′ ′ ′ ′ To find z1 , apply the product rule to the equation z z1 = 1: z ′ z1 + z z1 = 0, so that z1 = − zz2 .

9.2.37 a

b

z ′ w

9.2.38 a

1 z(t)

′

= z ′ w1 + z

=

z ′ 1 z2 −z1 z ′ 2 z22

= z w1

z1 z2

′

1 ′ w

=

=

z′ w

−

zw′ w2

λz1 z2 −λz1 z2 z22

=

z ′ w−zw′ w2

= 0, so that

z1 (t) z2 (t)

= k, a constant.

Now z1 (t) = kz2 (t); substituting t = 0 gives 1 = z1 (0) = kz2 (0) = k, so that z1 (t) = z2 (t), as claimed. b Let z2 (t) = ept (cos(qt) + i sin(qt)) be the solution constructed in the text (see Page 413). Since z2 (t) 6= 0 for all t, this is the only solution, by Part a.

λ 1 9.2.39 Let A = 0 λ 0 0 3 c3 (t), dc = 0. dt

0 1 . We first solve the system λ

c3 (t) = k3 , a constant, so that

dc2 dt

d~ c dt

0 = (A − λI3 )~c = 0 0

= k3 and c2 (t) = k3 t + k2 . Likewise c1 (t) =

1 0 0 1 ~c, or 0 0 k3 2 2 t

dc1 dt

2 = c2 (t), dc dt =

+ k2 t + k1 .

Applying Exercise 9.1.24, with k = −λ, we find that ~c(t) = e−λt ~x(t) or ~x(t) = eλt~c(t) k1 + k2 t + k23 t2 where k1 , k2 , k3 are arbitrary constants. The zero state is stable if (and only if) the real k2 + k3 t = eλt k3 part of λ is negative. 9.2.40 a B(t) = 1000(1 + 0.05i)t = 1000(r(cos θ + i sin θ))t = 1000rt (cos(θt) + i sin(θt)), where √ r = 1 + 0.052 > 1 and θ = arctan(0.05) ≈ 0.05. See Figure 9.42. 413

1 z

=

Chapter 9 b B(t) = 1000e0.05i = 1000(cos(0.05t) + i sin(0.05t)). See Figure 9.42.

1000 1000

1000

trajectory to part a slowly spiral outwards

trajectory to part b circle 2π ≅ period = 126 (years) 0.05

Figure 9.42: for Problem 9.2.40. c We would choose an account with annual compounding, since the modulus of the balance grows in this case. In the case of continuous compounding the modulus of the balance remains unchanged.

Section 9.3 9.3.1 The characteristic polynomial of this differential equation is λ − 5, so that λ1 = 5. By Theorem 9.3.8 the general solution is f (t) = Ce5t , where C is an arbitrary constant. −3t , where C is an arbitrary constant, and the differential 9.3.2 The solutions of dx dt +3x = 0 are of the form x(t) = Ce dx equation dt + 3x = 7 has the particular solution xp (t) = 73 , so that the general solution is x(t) = Ce−3t + 37 (where C is a constant). Alternatively, we could use Theorem 9.3.13.

9.3.3 Use Theorem 9.3.13, where a = −2 and g(t) = e3t : Z Z 1 5t 1 f (t) = e−2t e2t e3t dt = e−2t e5t dt = e−2t e + C = e3t + Ce−2t , where C is a constant. 5 5 9.3.4 We can look for a sinusoidal solution xp (t) = P cos(3t) + Q sin(3t), as in Example 7. P and Q needto be −2P + 3Q = 1 with chosen in such a way that −3P sin(3t) + 3Q cos(3t) − 2P cos(3t) − 2Q sin(3t) = cos(3t) or −3P − 2Q = 0 3 dx 2 2t solution P = − 13 and Q = 13 . Since the general solution of dt − 2x = 0 is x(t) = Ce , the general solution of dx 2 3 2t dt − 2x = cos(3t) is x(t) = Ce − 13 cos(3t) + 13 sin(3t), where C is an arbitrary constant. t

9.3.5 Using Theorem 9.3.13, f (t) = e stant.

Z

2t

9.3.6 Using Theorem 9.3.13, f (t) = e

Z

e−t t dt = et (−te−t − e−t + C) = Cet − t − 1, where C is an arbitrary con-

−2t 2t

e

2t

e dt = e

Z

414

dt = e2t (t + C), where C is an arbitrary constant.

Section 9.3 9.3.7 By Definition 9.3.6, pT (λ) = λ2 + λ − 12 = (λ + 4)(λ − 3). Since pT (λ) has distinct roots λ1 = −4 and λ2 = 3, the solutions of the differential equation are of the form f (t) = c1 e−4t + c2 e3t , where c1 and c2 are arbitrary constants (by Theorem 9.3.8). 9.3.8 pT (λ) = λ2 + 3λ − 10 = (λ + 5)(λ − 2) = 0 x(t) = c1 e−5t + c2 e2t , where c1 , c2 are arbitrary constants. 9.3.9 pT (λ) = λ2 − 9 = (λ − 3)(λ + 3) = 0 f (t) = c1 e3t + c2 e−3t , where c1 , c2 are arbitrary constants. 9.3.10 pT (λ) = λ2 + 1 = 0 has roots λ1,2 = ±i. By Theorem 9.3.9, f (t) = c1 cos(t) + c2 sin(t), where c1 , c2 are arbitrary constants. 9.3.11 pT (λ) = λ2 − 2λ + 2 = 0 has roots λ1,2 = 1 ± i. By Theorem 9.3.9, x(t) = et (c1 cos(t) + c2 sin(t)), where c1 , c2 are arbitrary constants. 9.3.12 pT (λ) = λ2 − 4λ + 13 = 0 has roots λ1,2 = 2 ± 3i. By Theorem 9.3.9, f (t) = e2t (c1 cos(3t) + c2 sin(3t)), where c1 , c2 are arbitrary constants. 9.3.13 pT (λ) = λ2 + 2λ + 1 = (λ + 1)2 = 0 has the double root λ = −1. Following Summary 9.3.15 in the text, we find f (t) = e−t (c1 t + c2 ), where c1 , c2 are arbitrary constants. 9.3.14 pT (λ) = λ2 + 3λ = λ(λ + 3) = 0 has roots λ1 = 0, λ2 = −3. 9.3.15 By integrating twice we find f (t) = c1 + c2 t, where c1 , c2 are arbitrary constants. 9.3.16 By Theorem 9.3.10, the differential equation has a particular solution of the form fp (t) = P cos(t) + Q sin(t). Plugging fp into the equation we find (−P cos(t) − Q sin(t)) + 4(−P sin(t) + Q cos(t)) + 13(P cos(t) + Q sin(t)) = cos(t) or 12P + 4Q = 1 , so −4P + 12Q = 0 P =

3 40

Q=

1 40 .

Therefore, fp (t) =

3 40

cos(t) +

1 40

sin(t).

Next we find a basis of the solution space of f ′′ (t) + 4f ′ (t) + 13f (t) = 0. pT (λ) = λ2 + 4λ + 13 = 0 has roots −2 ± 3i. By Theorem 9.3.9, f1 (t) = e−2t cos(3t) and f2 (t) = e−2t sin(3t) is a basis of the solution space. By Theorem 9.3.4, the solutions of the original differential equation are of the form f (t) = c1 f1 (t) + c2 f2 (t) + 1 3 cos(t) + 40 sin(t), where c1 , c2 are arbitrary constants. fp (t) = c1 e−2t cos(3t) + c2 e−2t sin(3t) + 40 415

Chapter 9 9.3.17 By Theorem 9.3.10, the differential equation has a particular solution of the form fp (t) = P cos(t) + Q sin(t). Plugging fp into we find (−P cos(t) − Q sin(t)) + 2(−P sin(t) + Q cos(t)) + P cos(t) + Q sin(t) = sin(t) the equation P = − 12 2Q = 0 . or , so −2P = 1 Q=0 Therefore, fp (t) = − 21 cos(t). Next we find a basis of the solution space of f ′′ (t) + 2f ′ (t) + f (t) = 0. In Exercise 13 we see that f1 (t) = e−t , f2 (t) = te−t is such a basis. By Theorem 9.3.4, the solutions of the original differential equation are of the form f (t) = c1 f1 (t) + c2 f2 (t) + fp (t) = c1 e−t + c2 te−t − 12 cos(t), where c1 , c2 are arbitrary constants. 9.3.18 We follow the approach outlined in Exercises 16 and 17. • Particular solution fp =

1 10

cos(t) +

3 10

sin(t)

• Solutions of f ′′ (t) + 3f ′ (t) + 2f (t) = 0 are f1 (t) = e−t and f2 (t) = e−2t . • The solutions of the original differential equation are of the form f (t) = c1 e−t + c2 e−2t + where c1 and c2 are arbitrary constants.

1 10

cos(t) +

3 10

sin(t),

9.3.19 We follow the approach outlined in Exercise 17. • Particular solution xp (t) = cos(t) • Solutions of

d2 x dt2

√ √ + 2x = 0 are x1 (t) = cos( 2t) and x2 (t) = sin( 2t).

√ √ • The solutions of the original differential equation are of the form x(t) = c1 cos( 2t) + c2 sin( 2t) + cos(t), where c1 and c2 are arbitrary constants. 9.3.20 pT (λ) = λ3 − 3λ2 + 2λ = λ(λ − 1)(λ − 2) = 0 has roots λ1 = 0, λ2 = 1, λ3 = 2. By Theorem 9.3.8, the general solution is f (t) = c1 + c2 et + c3 e2t , where c1 , c2 , c3 are arbitrary constants. 9.3.21 pT (λ) = λ3 + 2λ2 − λ − 2 = (λ − 1)(λ + 1)(λ + 2) = 0 has roots λ1 = 1, λ2 = −1, λ3 = −2. By Theorem 9.3.8, the general solution is f (t) = c1 et + c2 e−t + c3 e−2t , where c1 , c2 , c3 are arbitrary constants. 9.3.22 pT (λ) = λ3 − λ2 − 4λ + 4 = (λ − 1)(λ − 2)(λ + 2) = 0 has roots λ1 = 1, λ2 = 2, λ3 = −2. By Theorem 9.3.8, the general solution is f (t) = c1 et + c2 e2t + c3 e−2t , where c1 , c2 , c3 are arbitrary constants. 9.3.23 General solution f (t) = Ce5t Plug in: 3 = f (0) = Ce0 = C, so that f (t) = 3e5t . 9.3.24 General solution x(t) = Ce−3t +

7 3

(see Exercise 2).

Plug in: 0 = x(0) = C + 37 , so that C = − 37 and x(t) = − 73 e−3t + 37 . 9.3.25 General solution f (t) = Ce−2t Plug in: 1 = f (1) = Ce−2 , so that C = e2 and f (t) = e2 e−2t = e2−2t . 416

Section 9.3 9.3.26 General solution f (t) = c1 e3t + c2 e−3t (see Exercise 9), with f ′ (t) = 3c1 e3t − 3c2 e−3t Plug in: 0 = f (0) = c1 + c2 and 1 = f ′ (0) = 3c1 − 3c2 , so that c1 = 61 , c2 = − 16 , and f (t) = 16 e3t − 61 e−3t . 9.3.27 General solution f (t) = c1 cos(3t) + c2 sin(3t) (Theorem 9.3.9) Plug in: 0 = f (0) = c1 and 1 = f π2 = −c2 , so that c1 = 0, c2 = −1, and f (t) = − sin(3t). 9.3.28 General solution f (t) = c1 e−4t + c2 e3t , with f ′ (t) = −4c1 e−4t + 3c2 e3t

Plug in: 0 = f (0) = c1 + c2 and 0 = f ′ (0) = −4c1 + 3c2 , so that c1 = c2 = 0 and f (t) = 0. 9.3.29 General solution f (t) = c1 cos(2t) + c2 sin(2t) + (use the approach outlined in Exercise 17)

1 3

sin(t), so that f ′ (t) = −2c1 sin(2t) + 2c2 cos(2t) +

Plug in: 0 = f (0) = c1 and 0 = f ′ (0) = 2c2 + 13 , so that c1 = 0, c2 = − 61 , and f (t) = − 61 sin(2t) +

1 3

1 3

cos(t)

sin(t).

9.3.30 a k is a positive constant that depends on the rate of cooling of the coffee (it varies with the material of the cup, for example). A is the room temperature. b T ′ (t) + kT (t) = kA Constant particular solution: Tp (t) = A General solution of T ′ (t) + kT (t) = 0 is T (t) = Ce−kt . General solution of the original differential equation: T (t) = Ce−kt + A Plug in: T0 = T (0) = C + A, so that C = T0 − A and T (t) = (T0 − A)e−kt + A. 9.3.31

dv dt

+

k mv

=g

constant particular solution: vp = General solution of

dv dt

+

k mv

mg k k

= 0 is v(t) = Ce− m t . k

General solution of the original differential equation: v(t) = Ce− m t + mg k k mg mg t −m Plug in: 0 = v(0) = C + mg 1 − e , so that C = − and v(t) = k k k lim v(t) =

t→∞

mg k

(the “terminal velocity”). See Figure 9.43.

Figure 9.43: for Problem 9.3.31.

417

Chapter 9 dB dt

9.3.32

= kB − r or

dB dt

− kB = −r

↑ ↑ interest withdrawals constant particular solution Bp = General solution of

dB dt

r k

− kB = 0 is B(t) = Cekt

General solution of the original differential equation: B(t) = Cekt + Plug in: B0 = B(0) = C + kr , so that C = B0 −

r k

and B(t) = B0 −

r k r k

ekt +

r k

if B0 >

r k

then interest will exceed withdrawals and balance will grow.

if B0 <

r k

then withdrawals will exceed interest and account will eventually be depleted.

if B0 =

r k

then the balance will remain the same.

The graph in Figure 9.44 shows the three possible scenarios.

Figure 9.44: for Problem 9.3.32. 9.3.33 By Theorem 9.3.9, x(t) = c1 cos

L

√

2 = P = 2π √Lg or L =

√ pg pg 2π √ √L . It is required that g = 2π L t + c2 sin L t , with period P = g

g π2

≈ 0.994 (meters).

9.3.34 a We will take downward forces as positive. Let g = acceleration due to gravity, ρ = density of block a = length of edge of block Then (weight of block) = (mass of block) · g = (density of block)(volume of block) g buoyancy = (weight of displaced water) = (mass of displaced water) · g = (density of water) (volume of displaced water) g = 1a2 x(t)g = a2 gx(t). b Newton’s Second Law of Motion tells us that 2

m ddt2x = F = weight − buoyancy = ρa3 g − a2 gx(t), where m = ρa3 is the mass of the block. 2

ρa3 ddt2x = ρa3 g − a2 gx(t) 418

=

ρa3 g

Section 9.3 d2 x dt2

=g−

d2 x dt2

+

g ρa x

g ρa x(t)

=g

constant solution xp = ρa general solution (use Theorem 9.3.9): x(t) = c1 cos Now c2 = 0 since block is at rest at t = 0.

q

g ρa t

+ c2 sin

q

g ρa t

+ ρa

Plug in: a = x(0) = c1 + ρa, so that c1 = a − ρa and q g x(t) = (a − ρa) cos ρa t + ρa ≈ 2 cos(11t) + 8 (measured in centimeters) c The period is P = √2πg = ρa

√ 2π ρa √ g .

Thus the period increases as ρ or a increases (denser wood or larger block), or

as g decreases (on the moon). The period is independent of the initial state. 9.3.35 a pT (λ) = λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0 with roots λ1 = −1 and λ2 = −2, so x(t) = c1 e−t + c2 e−2t . b x′ (t) = −c1 e−t − 2c2 e−2t Plug in: 1 = x(0) = c1 + c2 and 0 = x′ (0) = −c1 − 2c2 , so that c1 = 2, c2 = −1 and x(t) = 2e−t − e−2t . See Figure 9.45.

Figure 9.45: for Problem 9.3.35b. c Plug in: 1 = x(0) = c1 + c2 and −3 = x′ (0) = −c1 − 2c2 , so that c1 = −1, c2 = 2, and x(t) = −e−t + 2e−2t . See Figure 9.46.

Figure 9.46: for Problem 9.3.35c. d The oscillator in part (b) never reaches the equilibrium, while the oscillator in part (c) goes through the equilibrium once, at t = ln(2). Take another look at Figures 9.45 and 9.46. 419

Chapter 9 9.3.36 fT (λ) = λ2 + 2λ + 101 = 0 has roots λ1,2 = −1 ± 20i. By Theorem 9.3.9, x(t) = e−t (c1 cos(20t) + c2 sin(20t)). Any nonzero solution goes through the equilibrium infinitely many times. See Figure 9.47.

Figure 9.47: for Problem 9.3.36. 9.3.37 fT (λ) = λ2 + 6λ + 9 = (λ + 3)2 has roots λ1,2 = −3. Following the method of Example 10, we find the general solution x(t) = e−3t (c1 + c2 t) with x′ (t) = e−3t (c2 − 3c1 − 3c2 t). Plug in: 0 = x(0) = c1 , and 1 = x′ (0) = c2 − 3c1 , so that c1 = 0, c2 = 1, and x(t) = te−3t . See Figure 9.48.

Figure 9.48: for Problem 9.3.37. The oscillator does not go through the equilibrium at t > 0. 9.3.38 a (D − λ)(p(t)eλt ) = [p(t)eλt ]′ − λp(t)eλt = p′ (t)eλt + λp(t)eλt − λp(t)eλt = p′ (t)eλt , as claimed. b Applying the result from part (a) m times we find (D − λ)m (p(t)eλt ) = p(m) (t)eλt = 0, since p(m) (t) = 0 for a polynomial of degree less than m. c By Theorem 9.3.3, we are looking for m linearly independent functions. By part (b), the functions eλt , teλt , t2 eλt , . . . , tm−1 eλt do the job. d Note that the kernel of (D − λi )mi is contained in the kernel of (D − λ1 )m1 · · · (D − λr )mr , for any 1 ≤ i ≤ r. Therefore, we have the following basis: eλ1 t , teλ1 t , . . . , tm1 −1 eλ1 t , eλ2 t , teλ2 t , . . . , tm2 −1 eλ2 t , . . . eλr t , teλr t , . . . , tmr −1 eλr t . 420

Section 9.3 9.3.39 fT (λ) = λ3 + 3λ2 + 3λ + 1 = (λ + 1)3 = 0 has roots λ1,2,3 = −1. In other words, we can write the differential equation as (D + 1)3 f = 0. By Exercise 38, part (c), the general solution is f (t) = e−t (c1 + c2 t + c3 t2 ). 9.3.40 fT (λ) = λ3 + λ2 − λ − 1 = (λ + 1)2 (λ − 1) = 0 has roots λ1,2 = −1, λ3 = 1. In other words, we can write the differential equation as (D + 1)2 (D − 1) = 0. By Exercise 38, part (d), the general solution is x(t) = e−t (c1 + c2 t) + c3 et . 9.3.41 We are looking for functions x such that T (x) = λx, or T (x) − λx = 0. Now T (x) − λx is an nth-order linear differential operator, so that its kernel is n-dimensional, by Theorem 9.3.3. Thus λ is indeed an eigenvalue of T , with an n-dimensional eigenspace. 2

9.3.42 a We need to solve the second-order differential equation T x = D2 x = ddt2x = λx. This differential equation has a two-dimensional solution space Eλ for any λ, so that all λ are eigenvalues of T . √ √ if λ > 0 then Eλ = span e λt , e− λt if λ = 0 then Eλ = span(1, t)

if λ < 0 then Eλ = span sin

√ √ −λt , cos −λt

b Among the eigenfunctions of T we found in part (a), we seek those of period 1. In the case λ < 0 the shortest . Now 1 is a period if P = √2π = k1 for some positive integer k, or, λ = −4π 2 k 2 . Then period is P = √2π −λ −λ Eλ = span(cos(2πkt), (sin(2πkt)). In the case λ > 0 there are no periodic solutions. In the case λ = 0 we have the constant solutions, so that λ = 0 is an eigenvalue with E0 = span(1). Summary: λ = −4π 2 k 2 is an eigenvalue, for k = 1, 2, 3, . . ., with Eλ = span(cos(2πkt), (sin(2πkt)). λ = 0 is an eigenvalue, with E0 = span(1). 9.3.43 a Using the approach of Exercise 17, we find x(t) = c1 e−2t + c2 e−3t + b For large t, x(t) ≈

1 10

cos t +

1 10

1 10

cos t +

1 10

sin t.

sin t.

9.3.44 a Using the approach of Exercises 16 and 17 we find x(t) = e−2t (c1 cos t + c2 sin t) − 1 b For large t, x(t) ≈ − 40 cos(3t) +

9.3.45 We can write the system as

3 40

sin(3t).

dx1 dt

= x1 + 2x2 dx2 dt = x2

with x1 (0) = 1, x2 (0) = −1.

The solution of the second equation, with the given initial value, is x2 (t) = −et . Now the first equation takes the form

dx1 dt

− x1 = −2et .

Using Example 9 (with a = 1 and c = −2) we find x1 (t) = et (−2t + C). 421

1 40

cos(3t) +

3 40

sin(3t).

Chapter 9

plug in: 1 = x1 (0) = C, so that x1 (t) = et (1 − 2t) and ~x(t) = et dx1

9.3.46 We can write the system as

dt dx2 dt dx3 dt

= = =

We solve for x2 and x3 as in Exercise 45:

1 − 2t . −1

2x1 + 3x2 + x3 x2 + 2x3 with x1 (0) x3

=

2, x2 (0)

=

1, x3 (0)

= −1 .

x2 (t) = et (1 − 2t) x3 (t) = −et Now the first equation takes the form

dx1 dt

− 2x1 = 3et (1 − 2t) − et = et (2 − 6t), x1 (0) = 2.

We use Theorem 9.3.13 to solve this differential equation: Z Z x1 (t) = e2t e−2t et (2 − 6t) dt = e2t (2e−t − 6te−t ) dt = e2t [−2e−t + 6te−t + 6e−t + C] plug in: 2 = x1 (0) = (−2 + 6 + c), so that c = −2 and x1 (t) = e2t (4e−t + 6te−t − 2) = 4et + 6tet − 2e2t . t 4e + 6tet − 2e2t et − 2tet ~x(t) = t −e 9.3.47 a We start with a preliminary remark that will be useful below: If f (t) = p(t)eλt , where p(t) is a polynomial, then f (t) has an antiderivative of the form q(t)eλt , where q(t) is another polynomial. We leave this remark as a calculus exercise. The function xn (t) satisfies the differential equation sired form.

dxn dt

= ann xn , so that xn = Ceann t , which is of the de-

Now we will show that xk is of the desired form, assuming that xk+1 , . . . , xn have this form. xk satisfies the dxk k differential equation dx dt = akk xk + ak,k+1 xk+1 + · · · + akn xn or dt − akk xk = ak,k+1 xk+1 + · · · + akn xn . Note that, by assumption, the function on the right-hand side has the form p1 (t)eλ1 t + · · · + pm (t)eλm t . If we set λ1 t k akk = a for simplicity, we can write dx + · · · + pm (t)eλm t . dt − axk = p1 (t)e By Theorem 9.3.13, the solution is Z xk (t) = eat e−at (p1 (t)eλ1 t + · · · + pm (t)eλm t ) dt = eat

Z

(p1 (t)e(λ1 −a)t + · · · + pm (t)e(λm −a)t ) dt

= eat (q1 (t)e(λ1 −a)t + · · · + qm (t)e(λm −a)t + C) = q1 (t)eλ1 t + · · · + qm (t)eλm t + Ceat as claimed (note that a is one of the λi ). The constant C is determined by xk (0). Note that we used the preliminary remark in the second to last step. b It is shown in introductory calculus classes that lim (tm eλt ) = 0 if and only if λ is negative (here m is a fixed t→∞

positive integer). In light of part (a), this proves the claim. 422

Section 9.3 9.3.48 a By Exercise 47, the system solution ~x(t) = S~u(t).

d~ u dt

= B~u has a unique solution ~u(t). Then the system

b It suffices to note that lim ~x(t) = ~0 if and only if lim ~u(t) = ~0, where ~u = S −1 ~x. t→∞

t→∞

423

d~ x dt

= A~x has the unique

1/12/09

8:40 AM

Page 1

INSTRUCTOR’S SOLUTIONS MANUAL KYLE BURKE Boston University with art by

GEORGE WELCH

LINEAR ALGEBRA WITH APPLICATIONS FOURTH EDITION

Otto Bretscher Colby College

Bretscher_600928X_ISM_TTL.qxd:au_123456_ttl

1/12/09

8:40 AM

Page 2

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Prentice Hall from electronic files supplied by the author. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Prentice Hall, Upper Saddle River, NJ 07458. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-600928-3 ISBN-10: 0-13-600928-X 1 2 3 4 5 6 OPM 10 09 08

Table of Contents Chapter 1: Linear Equations ..……………………………………………………… 1 Chapter 2: Linear Transformations...………………………………………………… 48 Chapter 3: Subspaces of Rn and Their Dimensions….…………………………….... 120 Chapter 4: Linear Spaces…………………………………………………...…………171 Chapter 5: Orthogonality and Least Squares...……………………………………..... 217 Chapter 6: Determinants …………………………………………………………….. 266 Chapter 7: Eigenvalues and Eigenvectors…………………………………………… 297 Chapter 8: Symmetric Matrices and Quadratic Forms.……………..……………..... 360 Chapter 9: Linear Differential Equations.…………………………………………… 392

Section 1.1

Chapter 1 Section 1.1 x + 2y = 1 x + 2y = 1 → → 2x + 3y = 1 −2 × 1st equation −y = −1 ÷(−1) x + 2y = 1 −2 × 2nd equation x = −1 → , so that (x, y) = (−1, 1). y=1 y =1

1.1.1

1.1.2

x + 34 y = 21 = 2 ÷4 → → =3 7x + 5y = 3 −7 × 1st equation = 21 x + 43 y = 12 x = −1 → , → 3 y = 2 y = 2 − 4 × 2nd equation = − 12 ×(−4)

4x + 3y 7x + 5y

x + 34 y − 14 y

so that (x, y) = (−1, 2). 1.1.3

2x + 4y 3x + 6y

=3 =2

÷2

→

x + 2y 3x + 6y

= 23 =2

→

x + 2y 3x + 6y

=1 =3

−3 × 1st equation

→

x + 2y 0

= 32 . = − 52

→

x + 2y 0

=1 =0

So there is no solution. 1.1.4

2x + 4y 3x + 6y

=2 =3

÷2

−3 × 1st equation

This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation x + 2y = 1 gives us x = 1 − 2y = 1 − 2t. Therefore the general solution is (x, y) = (1 − 2t, t), where t is an arbitrary real number. 1.1.5

x + 23 y = 0 = 0 ÷2 → → =0 4x + 5y = 0 −4 × 1st equation =0 x =0 x + 23 y = 0 − 23 × 2nd equation , → → y =0 y =0 = 0 ÷(−1)

2x + 3y 4x + 5y

x + 23 y −y

so that (x, y) = (0, 0). x + 2y + 3z x + 2y + 3z = 8 y 1.1.6 x + 3y + 3z = 10 −I → z x + 2y + 4z = 9 −I x + 3z = 4 −3(III) x=1 y = 2 → y = 2 , so that z =1 z=1

x + 2y + 3z 1.1.7 x + 3y + 4z x + 4y + 5z

x + 2y + 3z =1 y+z = 3 −I → 2y + 2z = 4 −I

This system has no solution.

=8 −2(II) → = 2 =1 (x, y, z) = (1, 2, 1).

x+z = 1 −2(II) → y +z = 2 0 = 3 −2(II) 1

= −3 = 2 = −1

Chapter 1 x + 2y + 3z = 0 x + 2y + 3z =0 = 0 ÷(−3) → 1.1.8 4x + 5y + 6z = 0 −4(I) → −3y − 6z −6y − 11z = 0 7x + 8y + 10z = 0 −7(I) x + 2y + 3z = 0 −2(II) x−z =0 +III x =0 y + 2z = 0 → y + 2z = 0 −2(III) → y = 0 , −6y − 11z = 0 +6(II) z =0 z =0

so that (x, y, z) = (0, 0, 0).

x + 2y + 3z = 1 x + 2y + 3z = 1 1.1.9 3x + 2y + z = 1 −3(I) → −4y − 8z = −2 ÷(−4) → 7x + 2y − 3z = 1 −7(I) −12y − 24z = −6 −2(II) x−z =0 x + 2y + 3z = 1 y + 2z → y + 2z = 12 = 12 −12y − 24z = −6 +12(II) 0 =0

This system has infinitely many solutions: if we choose z = t, an arbitrary real number, then we get x = z = t and y = 21 − 2z = 12 − 2t. Therefore, the general solution is (x, y, z) = t, 21 − 2t, t , where t is an arbitrary real number. x + 2y + 3z =1 z = 2 −2(I) → y + 2z = 8 −3(I)

x + 2y + 3z 1.1.10 2x + 4y + 7z 3x + 7y + 11z

x + 2y + 3z y + 2z z

x−z = 1 −2(II) = 5 → y + 2z =0 z

so that (x, y, z) = (−9, 5, 0). 1.1.11

x − 2y 3x + 5y

=2 = 17

−3(I)

→

x − 2y 11y

= −9

=1 Swap : = 0 → II ↔ III =5

+III x = −9 = 5 −2(III) → y = 5 , z =0 =0

=2 = 11

÷11

→

x − 2y y

=2 =1

so that (x, y) = (4, 1). See Figure 1.1.

Figure 1.1: for Problem 1.1.11.

1.1.12

x − 2y 2x − 4y

=3 =6

−2(I)

x − 2y → 0

=3 =0

2

+2(II)

→

x =4 , y =1

Section 1.1 This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation x − 2y = 3 gives us x = 3 + 2y = 3 + 2t. Therefore the general solution is (x, y) = (3 + 2t, t), where t is an arbitrary real number. (See Figure 1.2.)

Figure 1.2: for Problem 1.1.12.

1.1.13

x − 2y 2x − 4y

=3 =8

−2(I)

→

x − 2y 0

=3 , which has no solutions. (See Figure 1.3.) =2

Figure 1.3: for Problem 1.1.13.

x + 5z 1.1.14 The system reduces to y − z 0 planes.

=0 = 0 , so that there is no solution; no point in space belongs to all three =1

Compare with Figure 2b. x =0 1.1.15 The system reduces to y = 0 so the unique solution is (x, y, z) = (0, 0, 0). The three planes intersect at z =0 the origin.

x + 5z 1.1.16 The system reduces to y − z 0 arbitrary number. The three planes 1.1.17

x + 2y 3x + 5y

=a =b

=0 = 0 , so the solutions are of the form (x, y, z) = (−5t, t, t), where t is an =0 intersect in a line; compare with Figure 2a.

x + 2y → −3(I) −y

=a = −3a + b

x + 2y → ÷(−1) y

3

=a = 3a − b

−2(II)

Chapter 1

x = −5a + 2b , so that (x, y) = (−5a + 2b, 3a − b). y = 3a − b

x + 2y + 3z = a x + 2y + 3z =a −2(II) 1.1.18 x + 3y + 8z = b −I → y + 5z = −a + b → x + 2y + 2z = c −I −z = −a + c x = 10a − 2b − 7c x − 7z = 3a − 2b +7(III) x − 7z = 3a − 2b y + 5z = −a + b → y + 5z = −a + b −5(III) → y = −6a + b + 5c , z =a−c z =a−c −z = −a + c ÷(−1) so that (x, y, z) = (10a − 2b − 7c, −6a + b + 5c, a − c).

1.1.19 a Note that the demand D1 for product 1 increases with the increase of price P2 ; likewise the demand D2 for product 2 increases with the increase of price P1 . This indicates that the two products are competing; some people will switch if one of the products gets more expensive.

70 − 2P1 + P2 105 + P1 − P2 which yields the unique solution P1 = 26 and P2 = 46.

b Setting D1 = S1 and D2 = S2 we obtain the system

−5P1 + P2 = −14 + 3P1 , or P1 − 3P2 = −7 + 2P2

= −84 , = 112

1.1.20 The total demand for the product of Industry A is 1000 (the consumer demand) plus 0.1b (the demand from Industry B). The output a must meet this demand: a = 1000 + 0.1b. a = 1000 + 0.1b a − 0.1b = 1000 Setting up a similar equation for Industry B we obtain the system or , b = 780 + 0.2a −0.2a + b = 780 which yields the unique solution a = 1100 and b = 1000. 1.1.21 The total demand for the products of Industry A is 310 (the consumer demand) plus 0.3b (the demand from Industry B). The output a must meet this demand: a = 310 + 0.3b. a = 310 + 0.3b a − 0.3b = 310 Setting up a similar equation for Industry B we obtain the system or , b = 100 + 0.5a −0.5a + b = 100 which yields the solution a = 400 and b = 300. 1.1.22 Since x(t) = a sin(t) + b cos(t) we can compute

dx dt

= a cos(t) − b sin(t) and 2

d2 x dt2

= −a sin(t) − b cos(t). Substituting these expressions into the equation ddt2x − dx dt − x = cos(t) and simplifying gives (b − 2a) sin(t) + (−a − 2b) cos(t) = cos(t). Comparing the coefficients of sin(t) and cos(t) on both sides of −2a + b = 0 , so that a = − 51 and b = − 25 . See Figure 1.4. the equation then yields the system −a − 2b = 1

Figure 1.4: for Problem 1.1.22.

4

Section 1.1 1.1.23 a Substituting λ = 5 yields the system 2x − y 2x − y =0 7x − y = 5x or or 0 −6x + 3y = 0 −6x + 8y = 5y

=0 . =0 There are infinitely many solutions, of the form (x, y) = 2t , t , where t is an arbitrary real number.

b Proceeding as in part (a), we find (x, y) = − 13 t, t .

c Proceedings as in part (a), we find only the solution (0, 0).

1.1.24 Let v be the speed of the boat relative to the water, and s be the speed of the stream; then the speed of the boat relative to the land is v + s downstream and v − s upstream. Using the fact that (distance) = (speed)(time), we obtain the system # " 8 = (v + s) 31 ← downstream ← upstream 8 = (v − s) 2 3

The solution is v = 18 and s = 6. x+z 1.1.25 The system reduces to y − 2z 0

=1 = −3 . =k−7

a. The system has solutions if k − 7 = 0, or k = 7. b. If k = 7 then the system has infinitely many solutions. c. If k = 7 then we can choose z = t freely and obtain the solutions (x, y, z) = (1 − t, −3 + 2t, t). x − 3z y + 2z 1.1.26 The system reduces to (k 2 − 4)z

= 1 = 1 = k−2

This system has a unique solution if k 2 − 4 6= 0, that is, if k 6= ±2.

If k = 2, then the last equation is 0 = 0, and there will be infinitely many solutions. If k = −2, then the last equation is 0 = −4, and there will be no solutions. 1.1.27 Let x = the number of male children and y = the number of female children. Then the statement “Emile has twice as many sisters as brothers” translates into y = 2(x − 1) and “Gertrude has as many brothers as sisters” translates into x = y − 1. Solving the system

−2x + y x−y

= −2 = −1

gives x = 3 and y = 4.

There are seven children in this family. 5

Chapter 1 1.1.28 The thermal equilibrium condition requires −4T1 + T2 We can rewrite this system as T1 − 4T2 + T3 T2 − 4T3 The solution is (T1 , T2 , T3 ) = (75, 100, 125).

, T2 = that T1 = T2 +200+0+0 4 = −200 = −200 = −400

T1 +T3 +200+0 , 4

and T3 =

T2 +400+0+0 . 4

1.1.29 To assure that the graph goes through the point (1, −1), we substitute t = 1 and f (t) = −1 into the equation f (t) = a + bt + ct2 to give −1 = a + b + c. a+b+c = −1 Proceeding likewise for the two other points, we obtain the system a + 2b + 4c = 3 . a + 3b + 9c = 13 The solution is a = 1, b = −5, and c = 3, and the polynomial is f (t) = 1 − 5t + 3t2 . (See Figure 1.5.)

Figure 1.5: for Problem 1.1.29. a + b + c = p 1.1.30 Proceeding as in the previous exercise, we obtain the system a + 2b + 4c = q . a + 3b + 9c = r a = 3p − 3q + r The unique solution is b = −2.5p + 4q − 1.5r . c = 0.5p − q + 0.5r

Only one polynomial of degree 2 goes through the three given points, namely, f (t) = 3p − 3q + r + (−2.5p + 4q − 1.5r)t + (0.5p − q + 0.5r)t2 .

1.1.31 f (t) is of the form at2 + bt + c. So f (1) = a(12 ) + b(1) + c = 3, and f (2) = a(22 ) + b(2) + c = 6. Also, f ′ (t) = 2at + b, meaning that f ′ (1) = 2a + b = 1. a+b+c=3 So we have a system of equations: 4a + 2b + c = 6 2a + b = 1

a=2

which reduces to b = −3 . c=4 6

Section 1.1 Thus, f (t) = 2t2 − 3t + 4 is the only solution. 1.1.32 f (t) is of the form at2 + bt + c. So, f (1) = a(12 ) + b(1) + c = 1 and f (2) = 4a + 2b + c = 0. Also, R2 R2 f (t)dt = 1 (at2 + bt + c)dt 1 = a3 t3 + 2b t2 + ct|21

= 83 a + 2b + 2c − ( a3 +

b 2

+ c)

= 73 a + 23 b + c = −1.

a+b+c=1 So we have a system of equations: 4a + 2b + c = 0 7 3 3 a + 2 b + c = −1 a=9 which reduces to b = −28 . c = 20

Thus, f (t) = 9t2 − 28t + 20 is the only solution.

1.1.33 f (t) is of the form at2 + bt + c. f (1) = a + b + c = 1, f (3) = 9a + 3b + c = 3, and f ′ (t) = 2at + b, so f ′ (2) = 4a + b = 1. a+b+c=1 Now we set up our system to be 9a + 3b + c = 3 . 4a + b = 1 a − 3c = 0 This reduces to b + 43 c = 1 . 0=0 We write everything in terms of a, revealing c = 3a and b = 1 − 4a. So, f (t) = at2 + (1 − 4a)t + 3a for an arbitrary a. 1.1.34 f (t) = at2 + bt + c, so f (1) = a + b + c = 1, f (3) = 9a + 3b + c = 3. Also, f ′ (2) = 3, so 2(2)a + b = 4a + b = 3. a+b+c=1 Thus, our system is 9a + 3b + c = 3 . 4a + b = 3 When we reduce this, however, our last equation becomes 0 = 2, meaning that this system is inconsistent.

1.1.35 f (t) = ae3t + be2t , so f (0) = a + b = 1 and f ′ (t) = 3ae3t + 2be2t , so f ′ (0) = 3a + 2b = 4. a+b=1 , Thus we obtain the system 3a + 2b = 4 a=2 . which reveals b = −1 So f (t) = 2e3t − e2t . 7

Chapter 1 1.1.36 f (t) = a cos(2t) + b sin(2t) and 3f (t) + 2f ′ (t) + f ′′ (t) = 17 cos(2t). f ′ (t) = 2b cos(2t) − 2a sin(2t) and f ′′ (t) = −4b sin(2t) − 4a cos(2t). So, 17 cos(2t) = 3(a cos(2t) + b sin(2t)) + 2(2b cos(2t) − 2a sin(2t)) + (−4b sin(2t) − 4a cos(2t)) = (−4a + 4b + 3a) cos(2t) + (−4b − 4a + 3b) sin(2t) = (−a + 4b) cos(2t) + (−4a − b) sin(2t). −a + 4b = 17 . So, our system is: −4a − b = 0 a = −1 This reduces to: . b=4 So our function is f (t) = − cos(2t) + 4 sin(2t). 1.1.37 Plugging the three points (x, y) into the equation a + bx + cy + x2 + y 2 = 0, leads to a system of linear equations for the three unknowns (a, b, c). a + 5b + 5c + 25 + 25 a + 4b + 6c + 16 + 36

= =

a + 6b + 2c + 36 + 4 =

0 0 0.

The solution is a = −20, b = −2, c = −4. −20 − 2x − 4y + x2 + y 2 = 0 is a circle of radius 5 centered at (1, 2). 1.1.38 Plug the three points into the equation ax2 + bxy + cy 2 = 1. We obtain a system of linear equations a + 2b + 4c 4a + 4b + 4c

= =

9a + 3b + c =

1 1 1.

The solution is a = 3/20, b = −9/40, c = 13/40. This is the ellipse (3/20)x2 − (9/40)xy + (13/40)y 2 = 1.

x−z

1.1.39 The given system reduces to y + 2z 0

−5a+2b 3 = 4a−b 3

=

= a − 2b + c

.

This system has solutions (in fact infinitely many) if a − 2b + c = 0. The points (a, b, c) with this property form a plane through the origin.

1.1.40 a x1 = −3 x2 = 14 + 3x1 = 14 + 3(−3) = 5 x3 = 9 − x1 − 2x2 = 9 + 3 − 10 = 2 x4 = 33 + x1 − 8x2 + 5x3 − x4 = 33 − 3 − 40 + 10 = 0, so that (x1 , x2 , x3 , x4 ) = (−3, 5, 2, 0). 8

Section 1.1 b x4 = 0 x3 = 2 − 2x4 = 2 x2 = 5 − 3x3 − 7x4 = 5 − 6 = −1 x1 = −3 − 2x2 + x3 − 4x4 = −3 + 2 + 2 = 1, so that (x1 , x2 , x3 , x4 ) = (1, −1, 2, 0).

Figure 1.6: for Problem 1.1.41a.

1.1.41 a The two lines intersect unless t = 2 (in which case both lines have slope −1). To draw a rough sketch of x(t), note that limt→∞ x(t) = limt→−∞ x(t) = −1 the line x + 2t y = t becomes almost horizontal and limt→2− x(t) = ∞, limt→2+ x(t) = −∞. Also note that x(t) is positive if t is between 0 and 2, and negative otherwise. Apply similar reasoning to y(t). (See Figures 1.6 and 1.7.)

Figure 1.7: for Problem 1.1.41a. 9

Chapter 1 b x(t) =

−t t−2 ,

and y(t) =

2t−2 t−2 .

1.1.42 We can think of the line through the points (1, 1, 1) and (3, 5, 0) as the intersection of any two planes through these two points; each of these planes will be defined by an equation of the form ax + by + cz = d. It is required that 1a + 1b + 1c = d and 3a + 5b + 0c = d. a +b +c −d = 0 Now the system reduces to 3a +5b −d = 0 a + 25 c −2d = 0 . b − 32 c +d = 0 We can choose arbitrary real numbers for c and d; then a = − 25 c + 2d and b = 23 c − d. For example, if we choose c = 2 and d = 0, then a = −5 and b = 3, leading to the equation −5x + 3y + 2z = 0. If we choose c = 0 and d = 1, then a = 2 and b = −1, giving the equation 2x − y = 1. −5x +3y +2z = 0 . We have found one possible answer: 2x −y = 1 1.1.43 To eliminate the arbitrary constant t, we can solve the last equation for t to give t = z − 2, and substitute x = 6 + 5(z − 2) x − 5z = −4 z − 2 for t in the first two equations, obtaining or . y = 4 + 3(z − 2) y − 3z = −2 This system does the job. 1.1.44 Let b = Boris’ money, m = Marina’s money, and c = cost of a chocolate bar. b + 21 m = c We are told that 1 , with solution (b, m) = (0, 2c). 2 b + m = 2c Boris has no money. 1.1.45 Let us start by reducing the system: x + 2y + 3z = 39 x + 2y + 3z x + 3y + 2z = 34 −I → y−z 3x + 2y + z = 26 −3(I) −4y − 8z

= 39 = −5 = −91

Note that the last two equations are exactly those we get when we substitute y−z = −5 x = 39 − 2y − 3z: either way, we end up with the system . −4y − 8z = −91

1.1.46 a We set up two equations here, with our variables: x1 = servings of rice, x2 = servings of yogurt. 3x1 +12x2 = 60 So our system is: . 30x1 +20x2 = 300 Solving this system reveals that x1 = 8, x2 = 3. b Again, we set up our equations:

3x1 30x1

+12x2 +20x2

P + and reduce them to find that x1 = − 15

C 25 ,

=P , =C

while x2 = 10

P 10

−

C 100 .

Section 1.2 1.1.47 Let x1 = number of one-dollarbills, x2 = the number offive-dollar bills, and x3 = the number of ten-dollar x1 + x2 + x3 = 32 , bills. Then our system looks like: x1 + 5x2 + 10x3 = 100 which reduces to give us solutions that fit: x1 = 15 + 45 x3 , x2 = 17 − 49 x3 , where x3 can be chosen freely. Now let’s keep in mind that x1 , x2 , and x3 must be positive integers and see what conditions this imposes on the variable x3 . We see that since x1 and x2 must be integers, x3 must be a multiple of 4. Furthermore, x3 must be positive, and x2 = 17 − 94 x3 must be positive as well, meaning that x3 < 68 9 . These constraints leave us with only one possibility, x3 = 4, and we can compute the corresponding values x1 = 15 + 45 x3 = 20 and x2 = 17 − 94 x3 = 8. Thus, we have 20 one-dollar bills, 8 five-dollar bills, and 4 ten-dollar bills. 1.1.48 Let x1 , x2 , x3 be the number of 20 cent, 50 cent, and 2 Euro coins, respectively. Then we need solutions to x1 +x2 +x3 = 1000 the system: .2x1 +.5x2 +2x3 = 1000 x1 −5x3 = − 5000 3 . this system reduces to: x2 +6x3 = 8000 3 5x3 − 5000 x1 3 . Unfortunately for the meter maids, there are no Our solutions are then of the form x2 = −6x3 + 8000 3 x3 x3 integer solutions to this problem. If x3 is an integer, then neither x1 nor x2 will be an integer, and no one will ever claim the Ferrari.

Section 1.2

. 1 −2.. . 3 4..

. 4 −1.. . 8 −2..

. −2.. . 8..

. 0 −10.. . 1 8..

1 5 1 1 5 −II 1 1.2.1 → → 2 2 −2(I) 0 1 −8 0 x − 10z = 13 x = 13 + 10z −→ y + 8z = −8 y = −8 − 8z x 13 + 10t y = −8 − 8t , where t is an arbitrary real number. z t 3 1.2.2 6

8 ÷3 3

→

1 6

4 3

8

.

− 13 .. . −2..

8 3

3

−6(I)

→

1

4 3

0

0

.

− 13 .. . 0..

13 −8

8 3

−13

This system has no solutions, since the last row represents the equation 0 = −13.

1.2.3 x = 4 − 2y − 3z y and z are free variables; let y = s and z = t. 4 − 2s − 3t x , where s and t are arbitrary real numbers. y = s t z 11

Chapter 1

1 1.2.4 2 3

1 0

0

. 0.. . 1.. . 0..

0

0 1.2.5 1

. 1.. . −1.. . 4..

. 1.. 1 1 . 5 −2(I) → 0 −3.. . 0 1.. 2 −3(I)

2 x = 2 −1 , so that y = −1 . 0 0 1 1 1 1 0

. 1.. . 0.. . 0.. . 1..

0

.. 1 1 1.. 3 ÷(−3) → 0 1.. . 0 1.. −1

1

0 swap : → 0 I ↔ III 0 0

1 0 1 1 0 1

1 0 0 .. +III 1 0 −1 0 . 0 1 0 0 .. 0 1 −III → 0 1 0 1 0. 0 . .. 0 0 0 1 1 0 0 1 .. 0 0 1 1. 0 −III 0 0 0 x1 x1 + x4 = 0 x2 − x4 = 0 −→ x2 x3 x3 + x4 = 0 1

0 0

0

. 0.. . 0.. . 1.. . 1..

. 1.. . −1.. . 1.. . 0..

0

0 0

0 −I

1 −II → −1 −II

−1

1

0 → 0

1 1 0

0 −1

. 0 0.. . 1 0.. . 1 1.. . 0 1..

0 0 0 0

= −x4 = x4 = −x4

x1 −t x2 t = , where t is an arbitrary real number. x3 −t x4 t

1.2.6 The system is in rref already. x1 = 3 + 7x2 − x5 x3 = 2 + 2x5 x4 = 1 − x5 Let x2 = t and x5 = r. 3 + 7t − r x1 t x2 x3 = 2 + 2r 1−r x4 r x5

1 0 1.2.7 0 0

2 0 0 1 0 1 0 0

. 3.. . 3 2.. . 4 −1.. . 0 1.. 2

0 1 0 0 → 0 0 −II 0 0

2 0 0 1 0 0 0 0

. 3.. . 3 2.. . 1 −3.. . 0 1.. 2

12

0 −2(III) 0 −3(III) → 0 0

0

0 0

0

−II → +II

Section 1.2 . 9.. 0 −9(IV ) 1 2 0 0 .. 0 11. 0 −11(IV ) 0 0 1 0 → .. +3(IV ) 0 0 0 1 1 −3. 0 .. 0 0 0 0 0 1. 0 =0 x1 = −2x2 = 0 =0 x3 −→ =0 x4 =0 =0 x5 =0 −2t x1 x2 t Let x2 = t. x3 = 0 , where t is an arbitrary 0 x4 0 x5

1 2 0 0 0 1 0 0 0 0 0 0 x1 + 2x2 x3 x4 x5

1.2.8

"

0 0

. 3.. . 8..

1 0 2 0 0 4

x2 − x5 x4 + 2x5

. 0.. . 0.. . 0.. . 1..

0

0

#

0

÷4

=0 x2 −→ x4 =0

→

"

0 1 0 0 0 0

. 2 3.. . 1 2..

0

0 0 0

real number.

0 0

#

= x5 = −2x5

−2(II) 0 1 0 0

0 0 0 1

.. −1. 0 . 2.. 0

Let x1 = r, x3 = s, x5 = t. r x1 x2 t x3 = s , where r, t and s are arbitrary real numbers. −2t x4 t x5 . . 1 2 0 0 1 −1.. 0 −1.. 2 swap : . . → 1.2.9 1 2 0 0 1 −1.. 0 I ↔ II → 0 0 0 1 2 −1.. 2 .. .. 1 2 2 0 −2 1. 2 −I 1 2 2 0 −1 1. 2 . . 1 2 0 0 1 −1.. 0 1 2 0 0 1 −1.. 0 swap : . . 2.. 2 ÷2 → 0 0 0 1 2 −1.. 2 II ↔ III → 0 0 2 0 −2 .. .. 0 0 0 1 2 −1. 2 0 0 2 0 −2 2. 2 . 1 2 0 0 1 −1.. 0 . 1.. 1 0 0 1 0 −1 .. 0 0 0 1 2 −1. 2 x1 + 2x2 + x5 − x6 = 0 x1 = −2x2 − x5 + x6 = 1 + x5 − x6 x3 − x5 + x6 = 1 −→ x3 x4 = 2 − 2x5 + x6 x4 + 2x5 − x6 =2

0

0 0

1

2

Let x2 = r, x5 = s, and x6 = t.

13

Chapter 1 −2r − s + t x1 r x2 x3 1 + s − t , where r, s and t are arbitrary real numbers. = x4 2 − 2s + t s x5 t x6

x1

1.2.10 The system reduces to

x2 x3

+ x4 − 3x4 + 2x4

x1 = 1 = 2 −→ x2 x3 = −3

= 1 − x4 = 2 + 3x4 = −3 − 2x4

x1 = 0 = 4 −→ x2 x4 = −2

= −2x3 = 4 + 3x3 . = −2

Let x4 = t. 1−t x1 x2 2 + 3t , where t is an arbitrary real number. = −3 − 2t x3 t x4

x1

x1

1.2.11 The system reduces to

+ 2x3 − 3x3

x2

x4

Let x3 = t. −2t x1 x2 4 + 3t = t x3 −2 x4

x2

1.2.12 The system reduces to x1 x2 x3 x4

+ +

3.5x5 x5

+

3x5

x3 x4

= −3.5x5 − x6 = −x5 . = 53 x6 = −3x5 − x6

Let x5 = r and x6 = t. −3.5r − t x1 −r x2 5 t x3 3 = x4 −3r − t x5 r x6 t

1.2.13 The system reduces to There are no solutions.

x y

− +

z 2z 0

= 0 = 0 . = 1 14

+

x6

− +

5 3 x6 x6

= 0 = 0 −→ = 0 = 0

Section 1.2

1.2.14 The system reduces to

x = −2 − 2y = −2 . −→ z =2 = 2

x +

x

x1 + 2x2 + 3x3

2y z

Let y = t. −2 − 2t x y = t 2 z 1.2.15 The system reduces to 1.2.16 The system reduces to

x1 x4

y z

4 2 . 1

= = =

x4

+5x5 +2x5

= =

6 −→ 7

= 6 − 2x2 − 3x3 − 5x5 . = 7 − 2x5

Let x2 = r, x3 = s, and x5 = t. 6 − 2r − 3s − 5t x1 r x2 s x3 = 7 − 2t x4 t x5

8221 = − 4340

x1

1.2.17 The system reduces to

x2

= x3

= x4

= x5

=

4695 . 434 459 − 434 8591 8680

699 434

1.2.18 a No, since the third column contains two leading ones. b Yes c No, since the third row contains a leading one, but the second row does not. d Yes 0 1 0 0 1.2.19 and 0 0 0 0 1.2.20 Four, namely

1 0 0 , 0 0 0

1 0 0 1 k (k is an arbitrary constant.) , , 0 1 0 0 0 15

Chapter 1

0 1.2.21 Four, namely 0 0 1.2.22 Seven, namely

1 0 0 1 1 k 0 0 , 0 0 , 0 0 , 0 1 (k is an arbitrary constant.) 0 0 0 0 0 0 0

0 0 0 0

0 1 1 a b 0 , , 0 0 0 0 0 0

0 c , 0 0

1 0 1 , 0 0 0

1 0 d , 0 1 e

f 0

0 0 , 0 1

1 0 . 0 1

Here, a, b, . . . , f are arbitrary constants. 1.2.23 The conditions a, b, and c for the reduced row-echelon form correspond to the properties P1, P2, and P3 given on Page 13. The Gauss-Jordan algorithm, summarized on Page 15, guarantees that those properties are satisfied. 1.2.24 Yes; each elementary row operation is reversible, that is, it can be “undone.” For example, the operation of row swapping can be undone by swapping the same rows again. The operation of dividing a row by a scalar can be reversed by multiplying the same row by the same scalar. 1.2.25 Yes; if A is transformed into B by a sequence of elementary row operations, then we can recover A from B by applying the inverse operations in the reversed order (compare with Exercise 24). 1.2.26 Yes, by Exercise 25, since rref(A) is obtained from A by a sequence of elementary row operations.

1 1.2.27 No; whatever elementary row operations you apply to 4 7 to zero.

2 3 5 6 , you cannot make the last column equal 8 9

a11 x1 + a12 x2 + · · · + a1n xn = b1 1.2.28 Suppose (c1 , c2 , . . . , cn ) is a solution of the system a21 x1 + a22 x2 + · · · + a2n xn = b2 . .........

To keep the notation simple, suppose we add k times the first equation to the second; then the second equation of the new system will be (a21 + ka11 )x1 + · · · + (a2n + ka1n )xn = b2 + kb1 . We have to verify that (c1 , c2 , . . . , cn ) is a solution of this new equation. Indeed, (a21 + ka11 )c1 + · · · + (a2n + ka1n )cn = a21 c1 + · · · + a2n cn + k(a11 c1 + · · · + a1n cn ) = b2 + kb1 .

We have shown that any solution of the “old” system is also a solution of the “new.” To see that, conversely, any solution of the new system is also a solution of the old system, note that elementary row operations are reversible (compare with Exercise 24); we can obtain the old system by subtracting k times the first equation from the second equation of the new system. 1.2.29 Since the number of oxygen atoms remains constant, we must have 2a + b = 2c + 3d. 2a + b = 2c + Considering hydrogen and nitrogen as well, we obtain the system 2b = c + a = c + a − 2d = 0 2a + b − 2c − 3d = 0 b − d = 0 . 2b − c − d = 0 , which reduces to c − d = 0 a − c − d = 0 16

3d d or d

Section 1.2 2t a b t The solutions are = . t c t d

To get the smallest positive integers, we set t = 1: 2N O2 + H2 O −→ HN O2 + HN O3 1.2.30 a a a a

Plugging the points into f (t), we obtain the system = 1 + b + c + d = 0 − b + c − d = 0 + 2b + 4c + 8d = −15

with unique solution a = 1, b = 2, c = −1, and d = −2, so that f (t) = 1 + 2t − t2 − 2t3 . (See Figure 1.8.)

Figure 1.8: for Problem 1.2.30. 1.2.31 a a a a a

Let f (t) = a + bt + ct2 + dt3 + et4 . Substituting the points in, we get + b + c + d + e = 1 + 2b + 4c + 8d + 16e = −1 + 3b + 9c + 27d + 81e = −59 − b + c − d + e = 5 − 2b + 4c − 8d + 16e = −29

This system has the unique solution a = 1, b = −5, c = 4, d = 3, and e = −2, so that f (t) = 1−5t+4t2 +3t3 −2t4 . (See Figure 1.9.)

Figure 1.9: for Problem 1.2.31.

17

Chapter 1 ′ ′′ 1.2.32 The requirement fi′ (ai ) = fi+1 (ai ) and fi′′ (ai ) = fi+1 (ai ) ensure that at each junction two different cubics fit “into” one another in a “smooth” way, since they must have the same slope and be equally curved. The requirement that f1′ (a0 ) = fn′ (an ) = 0 ensures that the track is horizontal at the beginning and at the end. How many unknowns are there? There are n pieces to be fit, and each one is a cubic of the form f (t) = p+qt+rt2 +st3 , with p, q, r, and s to be determined; therefore, there are 4n unknowns. How many equations are there?

fi (ai ) = bi fi (ai−1 ) = bi−1 ′ (ai ) fi′ (ai ) = fi+1 ′′ fi′′ (ai ) = fi+1 (ai ) f1′ (a0 ) = 0, fn′ (an ) = 0

for for for for

i = 1, 2, . . . , n i = 1, 2, . . . , n i = 1, 2, . . . , n − 1 i = 1, 2, . . . , n − 1

gives gives gives gives gives

n equations n equations n − 1 equations n − 1 equations 2 equations

Altogether, we have 4n equations; convince yourself that all these equations are linear. 1.2.33 Let f (t) = a + bt + ct2 + dt3 , so that f ′ (t) = b + 2ct + 3dt2 . Substituting a + b a + 2b b b

the + + + +

given points into f (t) and f ′ (t) we obtain the system c + d = 1 4c + 8d = 5 2c + 3d = 2 4c + 12d = 9

This system has the unique solution a = −5, b = 13, c = −10, and d = 3, so that f (t) = −5 + 13t − 10t2 + 3t3 . (See Figure 1.10.)

Figure 1.10: for Problem 1.2.33. x x 1 1.2.34 We want all vectors y in R3 such that y · 3 = x + 3y − z = 0. The endpoints of these vectors z z −1 form a plane. x −3r + t , where r and t are arbitrary real numbers. These vectors are of the form y = r z t

x1 + x2 + x3 + x4 1.2.35 We need to solve the system x1 + 2x2 + 3x3 + 4x4 x1 + 9x2 + 9x3 +7x4 x1 + 0.25x4 = 0 which reduces to x2 − 1.5x4 = 0 . x3 + 2.25x4 = 0 18

=0 = 0 , =0

Section 1.2 −0.25t x1 x 1.5t The solutions are of the form 2 = , where t is an arbitrary real number. −2.25t x3 t x4

1.2.36 Writing the equation ~b = x1~v1 + x2~v2 + x3~v3 in terms of its components, we obtain the system x1 + 2x2 + 4x3 = −8 4x1 + 5x2 + 6x3 = −1 7x1 + 8x2 + 9x3 = 9 5x1 + 3x2 + x3 = 15 The system has the unique solution x1 = 2, x2 = 3, and x3 = −4.

1.2.37 Compare with the solution of Exercise 1.1.21. x1 − 0.2x2 − 0.3x3 x1 = 0.2x2 + 0.3x3 + 320 The diagram tells us that x2 = 0.1x1 + 0.4x3 + 90 or −0.1x1 + x2 − 0.4x3 −0.2x1 − 0.5x2 + x3 x3 = 0.2x1 + 0.5x2 + 150 This system has the unique solution x1 = 500, x2 = 300, and x3 = 400.

= 320 = 90 . = 150

320 0.3 0.2 0 1.2.38 a ~v1 = 0.1 , ~v2 = 0 , ~v3 = 0.4 , ~b = 90 150 0 0.5 0.2

b Recall that xj is the output of industry Ij , and the ith component aij of ~vj is the demand of Industry Ij on industry Ij for each dollar of output of industry Ij . Therefore, the product xj aij (that is, the ith component of xj ~vj ), represents the total demand of industry Ij on Industry Ii (in dollars). c x1~v1 + · · · + xn~vn + ~b is the vector whose ith component represents the total demand on industry Ii (consumer demand and interindustry demand combined). d The ith component of the equation x1~v1 + · · · + xn~vn + ~b = ~x expresses the requirement that the output xi of industry Ii equal the total demand on that industry. 1.2.39 a These components are zero because neither manufacturing not the energy sector directly require agricultural products. b We have to solve the system x1~v1 + x2~v2 + x3~v3 + ~b = ~x or 0.707x1 = 13.2 −0.014x1 + 0.793x2 − 0.017x3 = 17.6 −0.044x1 + 0.01x2 + 0.784x3 = 1.8

The unique solution is approximately x1 = 18.67, x2 = 22.60, and x3 = 3.63.

1.2.40 We want to find m1 , m2 , m3 such that m1 + m2 + m3 = 1 and 19

Chapter 1 2 4 2 1 , that is, we have to solve the system = + m3 + m2 2 1 3 2 m1 + m2 + m3 =1 m1 + 2m2 + 4m3 = 2 . 2m1 + 3m2 + m3 = 2 1 1

m1

The unique solution is m1 = 12 , m2 = 14 , and m3 = 14 . 1 We will put 12 kg at the point and 41 kg at each of the two other vertices. 2 1.2.41 We know that −3m1 + 2m2 or −6m1 + 4m2 −3m1 + 2m2

m1~v1+ m2~v2 = m1 w ~ 1 + m2 w ~ 2 or m1 (~v1 − w ~ 1 ) + m2 (~v2 − w ~ 2 ) = ~0 =0 = 0 . =0

We can conclude that m1 = 23 m2 .

1.2.42 Let x1 , x2 , x3 , and x4 be the traffic volume at the four locations indicated in Figure 1.11.

Figure 1.11: for Problem 1.2.42. We are told that x1 + 300 x2 + 300 x3 + x4 + 100 150 + 120

the number of cars coming into each intersection x1 − x2 = 320 + x2 x2 − x3 = 400 + x3 or = 250 x3 + x4 = x1 + x4 x1 + x4 270 − t x1 x2 250 − t The solutions are of the form = . 150 − t x3 t x4 20

is the same as the number of cars coming out: = 20 = 100 = 150 = 270

Section 1.2 Since the xi must be positive integers (or zero), t must be an integer with 0 ≤ t ≤ 150. The lowest possible values are x1 = 120, x2 = 100, x3 = 0, and x4 = 0, while the highest possible values are x1 = 270, x2 = 250, x3 = 150, and x4 = 150. 1.2.43 Plugging the data into the function S(t) we obtain the system + c sin 2π·47 = 11.5 a + b cos 2π·47 365 365 a + b cos 2π·74 + c sin 2π·74 = 12 365 365 2π·273 a + b cos 2π·273 + c sin = 12 365 365

The unique solution is approximately a = 12.17, b = −1.15, and c = 0.18, so that 2πt 2πt + 0.18 sin 365 . S(t) = 12.17 − 1.15 cos 365

The longest day is about 13.3 hours. (See Figure 1.12.)

Figure 1.12: for Problem 1.2.43. x1 +x2 +x3 = 24 . 3x1 +2x2 + 21 x3 = 24 x1 −1.5x3 = −24 This system reduces to . x2 +2.5x3 = 48 1.5x3 − 24 x1 Thus, our solutions will be of the form x2 = −2.5x3 + 48 . Since all of our values must be non-negative x3 x3 lilies 0 3 integers (and x3 must be even), we find the following solutions for roses : 8 and 3 . Since Olivia daisies 16 18 loves lilies, Kyle spends his 24 dollars on 3 lilies, 3 roses and 18 daisies.

1.2.44 Kyle first must solve the following system:

1.2.45 a When k 6= 1 and k 6= 2, we can see that this will continue to reduce to a consistent system with a unique solution. b When k = 1, our bottom row reveals the inconsistency 0 = 2. c When k = 2, the second row and third row both represent the equation z = 2, meaning that the third row will be replaced with the equation 0 = 0 during further reduction. This reveals that we will have an infinite number of solutions. 21

Chapter 1

0 1.2.46 a We reduce our matrix in the following steps: 1 k

1 0

2

6

1

2k

k

0

2

1 0

0

0

1

.. . 2 1 .. → . 0 0 .. −k(I) 0 . 1 6 − 4k 2k

0 2 − 6k + 4k

2

.. . .. . .. .

2 0 1 − 2k

2

6

1

2k

−2k

2 − 6k

1 0 → 0 1

0 0

.. . .. . .. .

.. . 0 .. swap : . 2 I ↔ II → .. . 1

1 2k 2

6

0

2

2

0 1 − 2k

−2(II)

→

+2k(II)

6 − 4k 2k

2(2k − 1)(k − 1)

.. . 2 .. . . 0 .. . −(2k − 1)

We see that there will be a unique solution when the 2(2k − 1)(k − 1) term is not equal to zero, when 2k − 1 6= 0 and k − 1 6= 0, or k 6= 21 and k 6= 1. b We will have no solutions when the term 2(2k − 1)(k − 1) is equal to zero, but the term −(2k − 1) is not. This occurs only when k = 1. c We will have infinitely many solutions when the last row represents the equation 0 = 0. This occurs when 2k − 1 = 0, or k = 21 . 1.2.47 a So − 21 x1 + x2 − 21 x3 = 0 and − 21 x2 + x3 − 21 x4 = 0. After reduction of the system, we find that our solutions are all of the form −2 3 x1 −1 2 x2 = s + t . x3 0 1 x4 1 0

b Yes, from our solution in part (a), if we plug in 1 for x1 and 13 for x4 , we obtain 3t − 2s = 1 and s = 13, which leads to t = 9, and x2 = 5, x3 = 9. So we have the solution: x1 = 1, x2 = 5, x3 = 9 and x4 = 13, which is an arithmetic progression. 1.2.48 It is required that xk = 12 (xk−1 + xk+1 ), or 2xk = xk−1 + xk+1 , or xk − xk−1 = xk+1 − xk . This means that the difference of any two consecutive terms must be the same; we are looking at the finite arithmetic sequences. Thus the solutions are of the form (x1 , x2 , x3 , . . . , xn ) = (t, t + r, t + 2r, . . . , t + (n − 1)r), where t and r are arbitrary constants.

2 1 1.2.49 We begin by solving the system. Our augmented matrix begins as: 0 3 1 0

22

.. 0. C . 1.. C . 4.. C

Section 1.2 . 9 C 0 0.. 25 .. 7 . In order for x, y and z to be integers, C must be a multiple of 25. We 1 0. 25 C .. 4 0 0 1. 25 C want the smallest positive choice, so C = 25.

1 and is reduced to 0

1.2.50 f (t) = a + bt + ct2 + dt3 and we learn that f (0) = a = 3, f (1) = a + b + c + d = 2, f (2) = a + 2b + 4c + 8d = 0. Also, Z 2 1 1 8 1 f (t)dt = at + bt2 + ct3 + dt4 |20 = 2a + 2b + c + 4d = 4. 2 3 4 3 0

1

0

0

0

1 Now, we set up our matrix, 1

1

1

1

2

4

8

8 3

2 2 4 meaning that the system is inconsistent.

.. . .. . .. . .. .

3

2 . However, when we reduce this, the last line becomes 0 = 1, 0 4

In introductory calculus you may have seen the approximation formula: Z

b

a

f (t)dt ≈

b−a a+b (f (a) + 4f ( ) + f (b)), 6 2

the simplest form of Simpson’s Rule. For polynomials f (t) of degree ≤ 3, Simpson’s Rule gives the exact value of the integral. Thus, for the f (t) in our problem, Z

2

f (t)dt =

0

2 1 11 (f (0) + 4f (1) + f (2)) = (3 + 8 + 0) = . 6 3 3

Thus it is impossible to find such a cubic with Z

2

f (t)dt = 4,

0

as required. 1.2.51 The system of linear equations is c1 c1 + c2 + c4

= =

0 0

c1 + 2c2 + 4c4 c1 + c3 + c6 c1 + 2c3 + 4c6

= = =

0 0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, 0, 0, c5 , 0). This is the conic xy = 0. 23

Chapter 1 y 3

2

1

x -1

1

2

3

-1

1.2.52 The system of linear equations is c1 c1 + 2c2 + 4c4

= =

0 0

c1 + 2c3 + 4c6 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 c1 + c2 + 3c3 + c4 + 3c5 + 9c6

= = =

0 0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, −6c6 , −2c6 , 3c6 , 0, c6 ). This√is the conic −6x − 2y + 3x2 + y 2 = 0 or 3(x − 1)2 + (y − 1)2 = 4, the ellipse centered at (1, 1) with semiaxis 2/ 3 and 2. y 3

2

1

x -1

1

2

3

-1

1.2.53 The system of linear equations is c1

=

0

c1 + c2 + c4 c1 + 2c2 + 4c4 c1 + 3c2 + 9c4

= = =

0 0 0

c1 + c2 + c3 + c4 + c5 + c6

=

0

This system has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, −c5 − c6 , 0, c5 , c6 ). Setting c5 = a and c6 = b, we get the family of conics a(xy − y) + b(y 2 − y) = ay(x − 1) + by(y − 1) = 0 where a 6= 0 or b 6= 0. Each such conic is the union of the x axis with some line through the point (1, 1). Two sample solutions are shown in the accompanying figures. 24

Section 1.2 y

y 3

3

2

2

1

1

x

x -1

1

2

-1

3

1

2

3

-1

-1

1.2.54 The system of linear equations is c1

=

0

c1 + c2 + c3 + c4 + c5 + c6 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6

= =

0 0

c1 + 3c2 + 3c3 + 9c4 + 9c5 + 9c6 c1 + c2 + c4

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, c5 + c6 , −c5 − c6 , −c5 − c6 , c5 , c6 ). Setting c5 = a and c6 = b, we get the family of conics a(x − y + xy − x2 ) + b(x − y − x2 + y 2 ) = a(x − y)(1 − x) + b(x − y)(1 − x − y) = 0 where a 6= 0 or b 6= 0. Each such conic is the union of the line y = x with some line through the point (1, 0). Two sample solutions are shown in the accompanying figures. y

y

3

3

2

2

1

1

x -1

1

2

3

x -1

-1

1

2

3

-1

1.2.55 The system of linear equations is c1

=

0

c1 + c2 + c4 c1 + c3 + c6

= =

0 0

c1 + c2 + c3 + c4 + c5 + c6

=

0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, −c4 , −c6 , c4 , 0, c6 ). Setting c4 = a and c6 = b, we get the family of conics a(x2 − x) + b(y 2 − y) = ax(x − 1) + by(y − 1) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. 25

Chapter 1 y

y

1

1

x

x

1

1

1.2.56 The system of linear equations is c1 c1 + c2 + c4

= =

0 0

c1 + c3 + c6 c1 + c2 − c3 + c4 − c5 + c6

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, −c4 , −c6 , c4 , 2c6 , c6 ). Setting c4 = a and c6 = b, we get the family of conics a(x2 − x) + b(y 2 + 2xy − y) = ax(x − 1) + by(y + 2x − 1) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. y

y

2

2

1

1

x 1

-1

x

3

2

1

-1

-1

-1

-2

-2

2

3

1.2.57 The system of linear equations is c1 + 5c2 + 25c4 c1 + c2 + 2c3 + c4 + 2c5 + 4c6

= =

0 0

c1 + 2c2 + c3 + 4c4 + 2c5 + c6 c1 + 8c2 + c3 + 64c4 + 8c5 + c6

= =

0 0

c1 + 2c2 + 9c3 + 4c4 + 18c5 + 81c6

=

0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (25c6 , −10c6 , −10c6 , c6 , 0, c6 ). This is the conic 25−10x−10y +x2 +y 2 = 0, a circle of radius 5 centered at (5, 5). y 12

8 (5, 5)

4

x -4

4 -4

26

8

12

Section 1.2 1.2.58 The system of linear equations is c1 + c2 + c4

=

0

c1 + 2c2 + 4c4 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6

= =

0 0

c1 + 5c2 + 2c3 + 25c4 + 10c5 + 4c6 c1 + 5c2 + 6c3 + 25c4 + 30c5 + 36c6

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (2c6 , −3c6 , 2c6 , c6 , −2c6 , c6 ). This is the conic 2−3x+2y+x2 −2xy+y 2 = 0. y 6

4

2

x -2

2

4

6

-2

1.2.59 The system of linear equations is c1 c1 + c2 + c4

= =

0 0

c1 + 2c2 + 4c4 c1 + c3 + c6

= =

0 0

c1 + 2c3 + 4c6 c1 + c2 + c3 + c4 + c5 + c6

= =

0 0

It has the only solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, 0, 0, 0, 0). There is no conic which goes through these points. Alternatively, note that the only conic through the first five points is xy = 0, according to Exercise 51. But that conic fails to run through the point (1, 1), so that there is no conic through all six points. y 3

2

1

x -1

1

2

3

-1

1.2.60 The system of linear equations is c1 27

=

0

Chapter 1 c1 + 2c2 + 4c4

=

0

c1 + 2c3 + 4c6 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6

= =

0 0

c1 + c2 + 3c3 + c4 + 3c5 + 9c6 c1 + 4c2 + c3 + 16c4 + 4c5 + c6

= =

0 0

It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (0, 0, 0, 0, 0, 0). There is no conic which goes through these points. y 4 3 2 1

x -1

1

2

3

4

-1

1.2.61 Let x1 be the cost of the environmental statistics book, x2 be the cost of the set theory textand x3 be the cost .. 1 1 0.. 178 of the educational psychology book. Then, from the problem, we deduce the augmented matrix 2 1 1.. 319 . . 0 1 1.. 147 .. 1 0 0.. 86 We can reduce this matrix to 0 1 0.. 92 , revealing that x1 = 86, x2 = 92 and x3 = 55. Thus, the . 0 0 1.. 55 environmental statistics book costs $ 86, the set theory book costs $ 92 and the educational psychology book is only priced at $ 55. grammar x1 1.2.62 Let our vectors x2 represent the numbers of the books W erther . Then we can set up the matrix LinearAlg. x3 .. 43 1 1 0 .. 64 . This system yields one solution, 21 , meaning that the grammar book costs 43 Euro, the . 1 0 1 . 98 . 55 0 1 1 .. 76 novel costs 21 Euro, and the linear algebra text costs 55 Euro.

1.2.63 The difficult part of this problem lies in setting up a system from which we can derive our matrix. We will define x1 to be the number of “liberal” students at the beginning of the class, and x2 to be the number of “conservative” students at the beginning. Thus, since there are 260 students in total, x1 + x2 = 260. We need one more equation involving x1 and x2 in order to set up a useful system. Since we know that the number of conservative students at the end of the semester is equal to the number of liberal student initially, we obtain the 6 7 6 3 x1 + 10 x2 = x1 , or − 10 x1 + 10 x2 = 0. equation 10 " # " # .. .. 1 1 . 260 to obtain 1 0. 120 . We then use . . 7 6 . − . 0 0 1.. 140 10

10

28

Section 1.2 Thus, there are initially 120 liberal students, and 140 conservative students. Since the number of liberal students initially is the same as the number of conservative students in the end, the class ends with 120 conservative students and 140 liberal students. 1.2.64 Let x1 and x2 be the initial number of students in Sections A and B, respectively. Then, since there are 55 students total, x1 + x2 = 55. Also, interpreting the change of students from the perspective of Section B, we gain 0.2x1 , lose 0.3x lose 4 students. Thus, 0.2x1 − 0.3x2 = −4.0. Our matrix becomes # 2 , and in the process, " " # .. .. . 55 , which reduces to 1 0 . 25 . This reveals that there are initially 25 students in Section 1 1 . . .2 −.3 .. −4 0 1 .. 30 A and 30 students in Section B. 1.2.65 We are told that five cows and two sheep cost ten liang, and two cows and five sheep silver. cost eight liang of 5C +2S = 10 . So, we let C be the cost of a cow, and S be the cost of a sheep. From this we derive 2C +5S = 8 C = 34 34 21 which gives the prices: 21 liang silver for a cow, and 20 This reduces to 21 liang silver for a sheep. S = 20 21 1.2.66 Letting x 1 , x2 , and x3 be the prize, incoins, of cows, sheep and pigs, respectively, we can represent the system .. .. 2 5 −13 . 1000 1 0 0 .. 1200 .. . We reduce this matrix to in a matrix: 3 −9 . 0 3 0 1 0 .. 500 . The prize of a cow, . .. 0 0 1 .. 300 . −600 −5 6 8 a sheep, and a pig is 1200, 500 and 300 coins, respectively. 1.2.67 The second measurement in the problem tells us that 4 sparrows and 1 swallow weigh as much as 1 sparrow and 5 swallows. We will immediately interpret this as 3 sparrows weighing the same as 4 swallows. The other measurement we use is that all the birds together weigh 16 liang. Setting x1 to be the weight of a sparrow, and . 3 −4.. 0 representing these two equations. x2 to be the weight of a swallow, we find the augmented matrix . 5 6.. 16 .. 32 . 1 0 19 , meaning that each sparrow weighs 32 liang, and each swallow weighs 24 liang. We reduce this to 19 19 .. 24 0 1. 19 1.2.68 This problem gives us three different combinations of horses that can pull exactly 40 dan up a hill. We condense the statements to fit our needs, saying that, One military horse and one ordinary horse can pull 40 dan, two ordinary and one weak horse can pull 40 dan and one military and three weak horses can also pull 40 dan. .. 40 .. 7 1 0 0 . 1 1 0 .. 40 With this information, we set up our matrix: 0 2 1 .. 40 , which reduces to 0 1 0 ... 120 . 7 . . 1 0 3 .. 40 0 0 1 .. 40 7

Thus, the military horses can pull dan each.

40 7

dan, the ordinary horses can pull

1.2.69 Here, let W be the depth of the well. 29

120 7

dan and the weak horses can pull

40 7

Chapter 1

2A

Then our system becomes

+B 3B

+C 4C

+D 5D

A

+E +6E

−W −W −W −W −W

= 0 = 0 = 0 . = 0 = 0

We 1 0 0 0

transform this system into an augmented matrix, then perform a prolonged reduction to reveal . 265 . . 0 0 0 0 0 − 721 . 191 . 1 0 0 0 − 721 . 0 . 191 129 76 265 148 . W , B = 721 W , C = 148 . Thus, A = 721 0 1 0 0 − 721 . 0 721 W , D = 721 W and E = 721 W . . 129 . . 0 0 0 1 0 − 721 . 76 . 0 0 0 0 1 − 721 . 0

If we choose 721 to be the depth of the well, then A = 265, B = 191, C = 148, D = 129 and E = 76. 1.2.70 We let x1 , x2 and x3 be the numbers of roosters, hens and chicks respectively. Then, since we buy a total of a hundred birds, and spend a hundred coins on them, we find the equations x1 + x2 + x3 = 100 and 5x1 + 3x2 + 13 x3 = 100. We fit these into our matrix,

which reduces to

4 1 0 −3

0 1

7 3

"

1 1

1

5 3

1 3

.. . .. .

−100

# .. . 100 , .. . 100 .

200

x1 So, x1 − 43 x3 = −100, and x2 + 37 x3 = 200. Now, we can write our solution vectors in terms of x3 : x2 = x3 4 x − 100 3 3 − 7 x3 + 200 . Since all of our values must be non-negative, x1 must be greater than or equal to zero, or 3 x3 4 x − 100 ≥ 0, which means that x3 ≥ 75. 3 3

Also, x3 must be greater than or equal to zero, meaning that − 37 x3 + 200 ≥ 0 or x3 ≤ 600 7 . Since x3 must be an integer, this forces x3 ≤ 85. Thus, we are looking for solutions where 75 ≤ x3 ≤ 85. We notice, however, that x1 and x2 are only integers when x3 is a multiple of 3. Thus, the possible values for x3 are 75, 78, 81 and 84. roosters 0 4 8 12 Now the possible solutions for hens are 25 , 18 , 11 , and 4 . chicks 75 78 81 84 1.2.71 We let x1 , x2 , x3 and x4 be the numbers of pigeons, sarasabirds, swans and peacocks respectively. We first determine the cost of each bird. Each pigeon costs 35 panas, each sarasabird costs 57 panas, the swans cost 79 panas apiece and each peacock costs 3 panas. We use these numbers to set up our system, but we must remember to make sure we are buying the proper amount of each to qualify for these deals when we find our solutions (for example, the number of sarasabirds we buy must be a multiple of 7). 30

Section 1.2

Our matrix then is

which reduces to

"

1

1

1

3 5

5 7

7 9

1 0 0 1

− 95 14 9

# . 1.. 100 . 3.. 100 . −20.. . 21..

−250 . 350

Thus, x1 = 95 x3 + 20x4 − 250 and x2 = − 14 9 x3 − 21x4 + 350. Then our solutions are of the form

5 9 x3 + 20x4 − 250 14 − 9 x3 − 21x4 + 350 . x3

x4

We determine the possible solutions by choosing combinations of x3 and x4 of the correct multiples (9 for x3 , 3 for x4 ) that give us non-negative integer solutions for x1 and x2 . Thus it is required that x1 = 59 x3 + 20x4 − 250 ≥ 0 and x2 = − 14 9 x3 − 21x4 + 350 ≥ 0. Solving for x3 we find that 225 −

27 2 x4

≥ x3 ≥ 450 − 36x4 .

To find all the solutions, we can begin by letting x4 = 0, and finding all corresponding values of x3 . Then we can increase x4 in increments of 3, and find the corresponding x3 values in each case, until we are through. For x4 = 0 we have the inequality 225 ≥ x3 ≥ 450, so that there aren’t any solutions for x3 . Likewise, there are no feasible x3 values for x4 = 3, 6 and 9, since 450 − 36x4 exceeds 100 in these cases. In the case of x4 = 12 our inequality becomes 63 ≥ x3 ≥ 18, so that x3 could be 18, 27, 36, 45, 54 or 63. In the next case, x4 = 15, we have

45 2

≥ x3 ≥ −90, so that the non-negative solutions are 0, 9 and 18.

If x4 is 18 or more, then the term 225 − 27 2 x4 becomes negative, so that there are only negative solutions for x3 . (Recall that it is required that 225 − 27 2 x4 ≥ x3 .) We have found nine solutions. If we compute the corresponding values of number of pigeons number of sarasabirds x1 = 95 x3 +20x4 −250 and x2 = − 14 9 x3 −21x4 +350, we end up with the following vectors for: number of swans number of peacocks 0 5 10 15 20 25 50 55 60 70 56 42 28 14 0 35 21 7 to be: , , , , , , , , . 18 27 36 45 54 63 0 9 18 12 12 12 12 12 12 15 15 15

"

1.2.72 We follow the outline of Exercise 70 to find the matrix 1 4 400−4x3 19

1

1

1 5

1

# .. . 100 , which reduces to 1 0 .. . 100 0 1

4 19 15 19

.. . .. .

3 . We find that our solutions are bound by 0 ≤ x3 ≤ 100. However, Thus, our solutions are of the form 1500−15x 19 x3 3 3 3 3 3 since both 400−4x = 4 100−x and 1500−15x = 15 100−x must be non-negative integers, the quantity 100−x must 19 19 19 19 19 be a non-negative integer, k, so that x3 = 100 − 19k. The condition x3 ≥ 0 now leaves us with the possibilities k = 0, 1, 2, 3, 4, 5.

31

400 19 1500 19

.

Chapter 1 20 16 12 8 4 0 ducks Thus, we find our solutions for sparrows : 0 . 15 , 30 , 45 , 60 and 75 . 5 24 43 62 81 100 roosters

1.2.73 We let x1 be the number of sheep, x2 be the number of goats, and x3 be the number of hogs. We can then 1 4 7 use the two equations 2 x1 + 3 x2 + 2 x3 = 100 and x1 + x2 + x3 = 100 to generate the following augmented matrix: . 4 7. 1 2 3 2 . 100 . 1 1 1.. 100 then reduce it to

1 0

.. 0 − 13 5 . . 18 . 1 5 .

40 . 60

40 + 13 5 s With this, we see that our solutions will be of the form 60 − 18 5 s . Now all three components of this vector s must be non-negative integers, meaning that s must be a non-negative multiple of 5 (that is, s = 0, 5, 10, . . .) such 50 that 60 − 18 5 s ≥ 0, or, s ≤ 3 . This leaves the possible solutions x3 = s = 0, 5, 10 and 15, and we can compute 18 the corresponding values of x1 = 40 + 13 5 s and x2 = 60 − 5 s in each case. 40 53 66 79 So we find the following solutions: 60 , 42 , 24 and 6 . 0 5 10 15

1.2.74 This problem is similar in nature to Exercise that example, revealing the ma 70, and we will follow # " .. .. 3 1 1 1 . 100 . We reduce this to 1 0 − 2 . −100 , which yields solutions of the form trix: . .. 5 3 2 21 .. 100 0 1 . 200 2 3 x − 100 3 2 − 5 x3 + 200 . Since all the values must be positive (there are at least one man, one woman and one child), 2 x3 we see 80, and beeven. From this, we use x3 to find our solutions: x3 must x3< < 66 that 17 14 11 8 5 2 30 , 25 , 20 , 15 , 10 and 5 . 78 76 74 72 70 68 1.2.75 Rather than setting up a huge system, here we will reason this out logically. Since there are 30 barrels, each son will get 10 of them. If we use the content of a full barrel as our unit for wine, we see that each brother will get 15 3 = 5 barrel-fulls of wine. Thus, the ten barrels received by each son will, on average, be half full, meaning that for every full barrel a son receives, he also receives an empty one. Now let x1 , x2 , and x3 be the numbers of half-full barrels received by each of the three sons. The first son, receiving x1 half-full barrels will also gain 10 − x1 other barrels, half of which must be full and half of which 1 must be empty, each equal to the quantity 10−x . Thus, x1 must be even. The same works for x2 and x3 . Since 2 x1 + x2 + x3 = 10, we have boiled down our problem to simply finding lists of three non-negative even numbers that add up to 10. We find our solutions by inspection: 4 4 4 4 6 6 6 8 8 10 0 ,2,0,4,2,0,6,4,2,0, 6 4 2 0 4 2 0 2 0 0 32

Section 1.2 0 0 0 0 0 0 2 2 2 2 2 8 , 6 , 4 , 2 , 0 , 10 , 8 , 6 , 4 , 2 and 0 . 10 8 6 4 2 0 8 6 4 2 0

As we stated before, the number of full and empty barrels is dependent on the number of half-full barrels. Thus, each solutionhere translates into exactly one solution for the overall problem. Here we list those solutions, for first son second son , using triples of the form (full barrels, half-full barrels, empty barrels) as our entries: third son (2, 6, 2) (2, 6, 2) (2, 6, 2) (1, 8, 1) (1, 8, 1) (0, 10, 0) (5, 0, 5) , (4, 2, 4) , (5, 0, 5) , (3, 4, 3) , (4, 2, 4) , (5, 0, 5) , (3, 4, 3) (4, 2, 4) (5, 0, 5) (4, 2, 4) (5, 0, 5) (5, 0, 5) (3, 4, 3) (3, 4, 3) (3, 4, 3) (3, 4, 3) (4, 2, 4) (4, 2, 4) (2, 6, 2) , (3, 4, 3) , (4, 2, 4) , (5, 0, 5) , (1, 8, 1) , (2, 6, 2) , (5, 0, 5) (4, 2, 4) (3, 4, 3) (2, 6, 2) (5, 0, 5) (4, 2, 4) (5, 0, 5) (5, 0, 5) (5, 0, 5) (4, 2, 4) (4, 2, 4) (4, 2, 4) (3, 4, 3) , (4, 2, 4) , (5, 0, 5) , (0, 10, 0) , (1, 8, 1) , (2, 6, 2) , (3, 4, 3) (4, 2, 4) (5, 0, 5) (1, 8, 1) (2, 6, 2) (3, 4, 3) (5, 0, 5) (5, 0, 5) (5, 0, 5) (3, 4, 3) , (4, 2, 4) and (5, 0, 5) . (2, 6, 2) (1, 8, 1) (0, 10, 0) 1.2.76 We let x1 be the amount of gold in the crown, x2 be the amount of bronze, x3 be the amount of tin and x4 be the amount of iron. Then, for example, since the first requirement in the problem is: “Let the gold and bronze together form two-thirds,” we will interpret this as x1 + x2 = 32 (60). We do this for all three requirements, and use the fact. that allcombined will be the total weight of the crown as our fourth. So we find the matrix 1 1 0 0 .. 23 (60) 30.5 gold . 1 0 1 0 .. 3 (60) bronze 9.5 4 , which has the solution . = . 14.5 tin 1 0 0 1 .. 3 (60) 5 5.5 iron .. 1 1 1 1 . 60 1.2.77 Let xi be the number of coins the ith merchant has. We interpret the statement of the first merchant, “If I keep the purse, I shall have twice as much money as the two of you together” as x1 + 60 = 2(x2 + x3 ), or −x1 + 2x2 + 2x3 = 60. We interpret the other statements in a similar fashion, translating this into the augmented .. −1 2 2. 60 . matrix, 3.. 60 . 3 −1 . 5 5 −1.. 60 . 0 0.. 4 . 1 0.. 12 . Thus we deduce that the first merchant has 4 . 0 0 1.. 20 coins, the second has 12, and the third is the richest, with 20 coins.

1 The reduced row echelon form of this matrix is 0

33

Chapter 1 1.2.78 For each of the three statements, we set up an equation of the form (initial amount of grass) + (grass growth) = (grass consumed by cows), or (#of f ields)x + (#of f ields)(#of days)y = (#of cows)(#of days)z. For the first statement, this produces the equation x + 2y = 6z, or x + 2y − 6z = 0. Similarly, we obtain the equations 4x + 16y − 28z = 0 and 2x + 10y − 15z = 0 for the other two statements. From this information, we . . 1 0 −5 .. 0 1 2 −6 .. 0 .. . 1 write the matrix 4 16 −28 .. 0 , which reduces to 0 1 − 2 . 0 . Thus our solutions are of the .. . 2 10 −15 . 0 0 0 0 .. 0 x 5t form y = 21 t , where t is an arbitrary positive real number. t z

Section 1.3 1.3.1 a No solution, since the last row indicates 0 = 1. b The unique solution is x = 5, y = 6. c Infinitely many solutions; the first variable can be chosen freely. 1.3.2 The rank is 3 since each row contains a leading one.

1 1 1 1.3.3 This matrix has rank 1 since its rref is 0 0 0 . 0 0 0

1 0 −1 1.3.4 This matrix has rank 2 since its rref is 0 1 2 0 0 0 1 2 7 1.3.5 a x +y = 3 1 11 b The solution of the system in part (a) is x = 3, y = 2. (See Figure 1.13.) 1.3.6 No solution, since any linear combination x~v1 + y~v2 of ~v1 and ~v2 will be parallel to ~v1 and ~v2 . 1.3.7 A unique solution, since there is only one parallelogram with sides along ~v1 and ~v2 and one vertex at the tip of ~v3 . 1.3.8 Infinitely many solution. There are at least two obvious solutions. Write ~v4 as a linear combination of ~v1 and ~v2 alone or as a linear combination of ~v3 and ~v2 alone. Therefore, this linear system has infinitely many solutions, by Theorem 1.3.1. 34

Section 1.3

Figure 1.13: for Problem 1.3.5.

1 1.3.9 4 7

2 3 1 x 5 6y = 4 8 9 9 z

1 1 1.3.10 2 · −2 = 1 · 1 + 2 · (−2) + 3 · 1 = 0 3 1 1.3.11 Undefined since the two vectors do not have the same number of components. 5 6 1.3.12 [1 2 3 4] · = 1 · 5 + 2 · 6 + 3 · 7 + 4 · 8 = 70 7 8 1.3.13

1 3

2 4

7 1 2 29 1 2 7 1 · 7 + 2 · 11 29 =7 + 11 = or = = 11 3 4 65 3 4 11 3 · 7 + 4 · 11 65

−1 2 3 1 2 3 6 1 1.3.14 2 = −1 +2 +1 = or 3 4 2 3 4 8 2 1 1 · (−1) + 2 · 2 + 3 · 1 6 = 2 · (−1) + 3 · 2 + 4 · 1 8

1 2

5 6 1.3.15 [1 2 3 4] = 5 · 1 + 6 · 2 + 7 · 3 + 4 · 8 = 70 either way. 7 8 1.3.16

0 3

1 2

−3 0 · 2 + 1 · (−3) 2 = = 0 3 · 2 + 2 · (−3) −3 35

−1 2 3 2 = 3 4 1

Chapter 1 1.3.17 Undefined, since the matrix has three columns, but the vector has only two components. 5 2 1 1 2 1 = 1 3 + 2 4 = 11 1.3.18 3 4 2 17 6 5 5 6

0 −1 1 1 1 −1 1 2 = 1 −5 + 2 1 + 3 1 = 0 0 3 −5 1 3 3

1 1 1 1.3.19 −5 1 −5

9 1.3.20 a 7 6 b

8 6 6

9 −9 18 27 36 45

158 70 1.3.21 81 123

1 0 1.3.22 By Theorem 1.3.4, the rref is 0 1 0 0

0 0 1

1 0 1.3.23 All variables are leading, that is, there is a leading one in each column of the rref: 0 0

1 0 1.3.24 By Theorem 1.3.4, rref (A) = 0 0

0 1 0 0

0 0 1 0

0 1 0 0

0 0 . 1 0

0 0 . 0 1

1.3.25 In this case, rref(A) has a row of zeros, so that rank(A) < 4; there will be a free variable. The system A~x = ~c could have infinitely many solutions (for example, when ~c = ~0) or no solutions (for example, when ~c = ~b), but it cannot have a unique solution, by Theorem 1.3.4. 1 0 1.3.26 From Example 3d we know that rank(A) = 3, so that rref(A) = 0 0

0 1 0 0

0 0 . 1 0

Since all variables are leading, the system A~x = ~c cannot have infinitely many solutions, but it could have a unique solution (for example, if ~c = ~b) or no solutions at all (compare with Example 3c). 36

Section 1.3 1 0 1.3.27 By Theorem 1.3.4, rref (A) = 0 0

0 1 0 0

0 0 1 0

0 0 . 0 1

1 0 1.3.28 There must be a leading one in each column: rref (A) = 0 0 0

1.3.29 A is

a and 0 0

a 0 0 of the form 0 b 0 0 0 c 0 0 5 5a 2 b 0 3 = 3b = 0 . 0 c −9 −9c 1 2

0 1 0 0 0

0 0 1 . 0 0

5

So a = 25 , b = 0 and c = − 19 , and A = 0 0

0 0 0 0 0 − 19

b c 5 2 e f 3 = 0 . Thus, 5a + 3b − 9c = 2, 5d + 3e − 9f = 0, and h i −9 1 matrix to have rank 1 is to make all the entries 2 in the second and third 0 0 5 columns zero, meaning that a = 52 , d = 0, and g = 15 . Thus, one possible matrix is 0 0 0 . 1 0 0 5

a 1.3.30 We must satisfy the equation d g 5g + 3h − 9i = 1. One way to force our

1.3.31 A is

a and 0 0

a b c of the form 0 d e 0 0 f 5 5a + 3b − 9c 2 b c d e 3 = 3d − 9e = 0 . −9 −9f 1 0 f

Clearly, f must equal − 91 . Then, since 3d = 9e, we can choose any non-zero value for the free variable e, and d will be 3e. So, if we choose 1 for e, then d = 3e = 3. Lastly, we must resolve 5a + 3b − 9c = 2. Here, b and c are . If we let b = c = 1. Then, a = 2−3(1)+9(1) = 58 . the free variables, and a = 2−3b+9c 5 5 8 1 1 5 So, in our example, A = 0 3 1 0 0 − 91 1.3.32 For this problem, we set up the same three equations as in Exercise 30. However, here, we must enforce that −2 −2 −2 our matrix, A, contains no zero entries. One possible solution to this problem is the matrix 3 1 2 . −1 −1 −1 37

Chapter 1 x1 x2 ... 1.3.33 The ith component of A~x is [0 0 . . . 1 . . . 0] = xi . (The 1 is in the ith position.) xi ... xn

Therefore, A~x = ~x.

c b a 1.3.34 a A~e1 = d , A~e2 = e , and A~e3 = f . k h g 1 b B~e1 = [~v1 ~v2 ~v3 ] 0 = 1~v1 + 0~v2 + 0~v3 = ~v1 . 0 Likewise, B~e2 = ~v2 and B~e3 = ~v3 .

1.3.35 Write A = [~v1 ~v2 . . . ~vi . . . ~vm ], then 0 0 ... A~ei = [~v1 ~v2 . . . ~vi . . . ~vm ] = 0~v1 + 0~v2 + · · · + 1~vi + · · · + 0~vm = ~vi = ith column of A. 1 ... 0

1 1.3.36 By Exercise 35, the ith column of A is A~ei , for i = 1, 2, 3. Therefore, A = 2 3

4 7 5 8 . 6 9

2 x1 = 2 − 2x2 1.3.37 We have to solve the system or . x3 1 x3 =1 2 − 2t x1 Let x2 = t. Then the solutions are of the form x2 = t , where t is an arbitrary real number. x3 1 x1

+

2x2

= =

1.3.38 We will illustrate our reasoning with an example. We generate the “random” 3 × 3 matrix 0.141 0.592 0.653 A = 0.589 0.793 0.238 . 0.462 0.643 0.383

Since the entries of this matrix are chosen from a large pool of numbers (in our case 1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and even less likely that the whole first column will consist of zeros). means that we will usually be able to to turn the first apply the Gauss-Jordan column This algorithm 1 4.199 4.631 0.141 0.592 0.653 1 into 0 ; this is indeed possible in our example: 0.589 0.793 0.238 −→ 0 −1.680 −2.490 . 0 −1.297 −1.757 0.462 0.643 0.383 0 38

Section 1.3 Again, it is unlikely thatany entries in the second column of the new matrix will be zero. Therefore, we can turn 0 the second column into 1 . 0 1 0 0 Likewise, we will be able to clear up the third column, so that rref(A) = 0 1 0 . 0 0 1

We summarize:

As we apply Gauss-Jordan elimination to a random matrix A (of any size), it is unlikely that we will ever encounter a zero on the diagonal. Therefore, rref(A) is likely to have all ones along the diagonal.

0 a 0 b , where a, b, and c are arbitrary. 1 c

1 0 1.3.39 We will usually get rref(A) = 0 1 0 0 1 0 1.3.40 We will usually have rref(A) = 0 0

0 1 0 0

0 0 . 1 0

(Compare with the summary to Exercise 38.)

1 1.3.41 If A~x = ~b is a “random” system, then rref(A) will usually be 0 0 solution.

0 0 1 0 , so that we will have a unique 0 1

1.3.42 If A~x = ~b is a “random” system of three equations with four unknowns, then rref(A) will usually be 1 0 0 a 0 1 0 b (by Exercise 39), so that the system will have infinitely many solutions (x4 is a free variable). 0 0 1 c . 1.3.43 If A~x = ~b is a “random” system of equations with three unknowns, then rref[A..~b] will usually be . 1 0 0 ..0 0 1 0 ...0 . , so that the system is inconsistent. 0 0 1 ..0 . 0 0 0 ..1 1.3.44 Let E = rref(A), and note that all the entries in the last row of E must be zero, by the definition of rref. If ~c is any vector in Rn whose last component isn’t zero, then the system E~x = ~c will be inconsistent. Now consider the elementary row operations that in reversed i i transform A into E, and apply the opposite h operations, h order, to the augmented matrix E ... ~c . You end up with an augmented matrix A ... ~b that represents an inconsistent system A~x = ~b, as required. 39

Chapter 1

x1 kx1 1.3.45 Write A = [~v1 ~v2 . . . ~vm ] and ~x = . . . . Then A(k~x) = [~v1 . . . ~vm ] . . . = kx1~v1 + · · · + kxm~vm and xm kxm k(A~x) = k(x1~v1 + · · · + xm~vm ) = kx1~v1 + · · · + kxm~vm . The two results agree, as claimed. 1.3.46 Since a, d, and f are all nonzero, we can divide the first row by a, the second row by d, and the third row by f to obtain 1 ab ac 0 1 e . d 0 0 1 It follows that the rank of the matrix is 3.

1.3.47 a ~x = ~0 is a solution. b This holds by part (a) and Theorem 1.3.3. c If ~x1 and ~x2 are solutions, then A~x1 = ~0 and A~x2 = ~0. Therefore, A(~x1 + ~x2 ) = A~x1 + A~x2 = ~0 + ~0 = ~0, so that ~x1 + ~x2 is a solution as well. Note that we have used Theorem 1.3.10a. d A(k~x) = k(A~x) = k~0 = ~0 We have used Theorem 1.3.10b. 1.3.48 The fact that ~x1 is a solution of A~x = ~b means that A~x1 = ~b. a. A(~x1 + ~xh ) = A~x1 + A~xh = ~b + ~0 = ~b b. A(~x2 − ~x1 ) = A~x2 − A~x1 = ~b − ~b = ~0 c. Parts (a) and (b) show that the solutions of A~x = ~b are exactly the vectors of the form ~x1 + ~xh , where ~xh is a solution of A~x = ~0; indeed if ~x2 is a solution of A~x = ~b, we can write ~x2 = ~x1 + (~x2 − ~x1 ), and ~x2 − ~x1 will be a solution of A~x = ~0, by part (b). Geometrically, the vectors of the form ~x1 + ~xh are those whose tips are on the line L in Figure 1.14; the line L runs through the tip of ~x1 and is parallel to the given line consisting of the solutions of A~x = ~0.

Figure 1.14: for Problem 1.3.48c.

40

Section 1.3 . 1.3.49 a This system has either infinitely many solutions (if the right-most column of rref[A..b] does not contain a leading one), or no solutions (if the right-most column does contain a leading one). . . b This system has either a unique solution (if rank[A..~b] = 3), or no solution (if rank[A..~b] = 4). . c The right-most column of rref[A..~b] must contain a leading one, so that the system has no solutions. d This system has infinitely many solutions, since there is one free variable. . 1.3.50 The right-most column of rref[A..~b] must contain a leading one, so that the system has no solutions. 1.3.51 For B~x to be defined, the number of columns of B, which is s, must equal the number of components of ~x, which is p, so that we must have s = p. Then B~x will be a vector in Rr ; for A(B~x) to be defined we must have m = r. Summary: We must have s = p and m = r.

0 −1 1.3.52 A(B~x) = A 1 0 0 −1 so that C = . 2 −1

x1 x2

1 0 = 1 2

−x2 x1

−x2 = 2x1 − x2

0 −1 = 2 −1

x1 , x2

x1 1.3.53 Yes; write A = [~v1 . . . ~vm ], B = [w ~1 . . . w ~ m ], and ~x = . . . . xm x1 ~ 1 ) + · · · + xm (~vm + w ~ m ) and Then (A + B)~x = [~v1 + w ~ 1 . . . ~vm + w ~ m ] . . . = x1 (~v1 + w xm x1 x1 ~ 1 + · · · + xm w ~ m. ~1 . . . w ~ m ] . . . = x1~v1 + · · · + xm~vm + x1 w A~x + B~x = [~v1 . . . ~vm ] . . . + [w xm xm The two results agree, as claimed.

1.3.54 The vectors of the form c1~v1 + c2~v2 form a plane through the origin containing ~v1 and ~v2 ; in Figure 1.15 we draw a typical vector in this plane. 7 4 1 1.3.55 We are looking for constants a and b such that a 2 + b 5 = 8 . 9 6 3 7 a + 4b = 7 The resulting system 2a + 5b = 8 has the unique solution a = −1, b = 2, so that 8 is indeed a linear 9 3a + 6b = 9 1 4 combination of the vector 2 and 5 . 3 6 41

Chapter 1

Figure 1.15: for Problem 1.3.54. −2 9 5 1 30 −5 2 6 7 −1 1.3.56 We can use technology to determine that the system 38 = x1 1 + x2 3 + x3 3 + x4 4 is 7 5 2 9 56 9 2 8 4 62 30 −1 inconsistent; therefore, the vector 38 fails to be a linear combination of the other four vectors. 56 62

2 1 on y = x2 and on y = 3x. 1 3 7 1 2 1 2 7 . = +b : a and as a linear combination of Then write 11 3 1 3 1 11 3 4 7 . + = The unique solution is a = 2, b = 3, so that the desired representation is 9 2 11 4 3 x is on the line y = 2 ; is on line y = 3x. 2 9

1.3.57 Pick a vector on each line, say

−1 2 1 3 1.3.58 We want b = k1 3 + k2 6 + k3 −3 , for some k1 , k2 and k3 . −2 4 2 c 1 1 1 Note that we can rewrite this right-hand side as k1 3 + 2k2 3 − k3 3 2 2 2 1 = (k1 + 2k2 − k3 ) 3 . It follows that k1 + 2k2 − k3 = 3, so that b = 9 and c = 6. 2 42

Section 1.3 a+b 1 1 5 2 a + 2b 1 7 1.3.59 = a + b = . a + 3b 3 1 c a + 4b 4 1 d a +b = 5 , which we quickly solve to find a = 3 and b = 2. Then, c = So we have a small system: a +2b = 7 a + 3b = 3 + 6 = 9 and d = a + 4b = 3 + 8 = 11. a 0 1 2 k2 + 2k3 0 b 0 0 0 1.3.60 We need = k1 + k2 + k3 = . From this we see that a, c and d can c 3 4 5 3k1 + 4k2 + 5k3 d 0 0 6 6k3 be any value, while b must equal zero. 1.3.61 We need to solve the system 1 1 1 c = x 2 + y 3 9 4 c2 with augmented matrix

1 2 4

The matrix reduces to

1 0 0

. 1.. . 1.. . 0..

. 1.. . 3.. . 9..

1 . c

c2 1

c−2 2

c − 5c + 6

.

This system is consistent if and only if c = 2 or c = 3. Thus the vector is a linear combination if c = 2 or c = 3. 1.3.62 We need to solve the system with augmented matrix

1 1 1 c = x a + y b c2 a2 b2

1 a

a2

The matrix reduces to

. 1.. 1 . 0 b − a.. . 0 0..

. 1.. . b.. . b2 ..

1 . c

c2

1 c−a (c − a)(c − b)

.

This system is consistent if and only if c = a or c = b. Thus the vector is a linear combination if c = a or c = b. 43

Chapter 1 1.3.63 This is the line parallel to w ~ which goes through the end point of the vector ~v . 1.3.64 This is the line segment connecting the head of the vector ~v to the head of the vector ~v + w. ~ 1.3.65 This is the full parallelogram spanned by the two vectors ~v and w. ~ 1.3.66 Write b = 1 − a and a~v + bw ~ = a~v + (1 − a)w ~ =w ~ + a(~v − w) ~ to see that this is the line segment connecting the head of the vector ~v to the head of the vector w. ~ 1.3.67 This is the full triangle with its vertices at the origin and at the heads of the vectors ~v and w. ~ 1.3.68 Writing ~u · ~v = ~u · w ~ as ~u · (~v − w) ~ = 0, we see that this is the line perpendicular to the vector ~v − w. ~

0 1 1.3.69 We write out the augmented matrix: 1 0 1 1

So x =

−a+b+c , 2

y=

a−b+c 2

and z =

a+b−c . 2

. 1.. . 1.. . 0..

.. a 1 0 0. .. b and reduce it to 0 1 0. . c 0 0 1..

−a+b+c 2 a−b+c 2 a+b−c 2

.

1.3.70 We find it useful to let s = x1 + x2 + · · · + xn . Adding up all n equations of the system, and realizing that n . Now the ith the term xi is missing from the ith equation, we see that (n − 1)s = b1 + · · · + bn , or, s = b1 +···+b n−1 n − bi . equation of the system can be written as s − xi = bi , so that xi = s − bi = b1 +···+b n−1

True or False Ch 1.TF.1 F, by Example 3a of Section 1.3 Ch 1.TF.2 T, by Definition 1.3.7 Ch 1.TF.3 T, by Theorem 1.3.4 Ch 1.TF.4 F, by Theorem 1.3.1 Ch 1.TF.5 F, by Theorem 1.3.4 Ch 1.TF.6 F; As a counter-example, consider the zero matrix. Ch 1.TF.7 T, by Theorem 1.3.8 Ch 1.TF.8 T, by Definition 1.3.9 Ch 1.TF.9 T, by Definition. Ch 1.TF.10 F; Consider the equation x + y + z = 0, repeated four times. 44

True or False Ch 1.TF.11 T; Find rref. Ch 1.TF.12 T; Find rref Ch 1.TF.13 F; Consider the 4 × 3 matrix A that contains all zeroes, except for a 1 in the lower left corner. 2 1 Ch 1.TF.14 F; Note that A = 2A for all 2 × 2 matrices A. 2 1 Ch 1.TF.15 F; The rank is 1. Ch 1.TF.16 F; The product on the left-hand side has two components.

−3 Ch 1.TF.17 T; Let A = −5 −7

0 0 , for example. 0

7 4 1 Ch 1.TF.18 T; We have 2 = 2 5 − 8 . 9 6 3 Ch 1.TF.19 T; The last component of the left-hand side is zero for all vectors ~x.

3 Ch 1.TF.20 T; A = 4

0 , for example. 0

0 Ch 1.TF.21 F; Let A = 0 0 to A all we want, we will

1 1 0 0 1 and B = 0 0 0 0 always end up with a

0 0 1 0 , for example. We can apply elementary row operations 0 0 matrix that has all zeros in the first column.

Ch 1.TF.22 T; If ~u = a~v + bw ~ and ~v = c~ p + d~q + e~r, then ~u = ac~ p + ad~q + ae~r + bw. ~ Ch 1.TF.23 F; The system x = 2, y = 3, x + y = 5 has a unique solution.

0 1 Ch 1.TF.24 F; Let A = , for example. 0 0

1 Ch 1.TF.25 F; Let A = 0 1

2 0 1 and ~b = 3 , for example. 5 1

Ch 1.TF.26 T, by Exercise 1.3.44. Ch 1.TF.27 F; Find rref to see that the rank is always 2. Ch 1.TF.28 T; Note that ~v = 1~v + 0w. ~ 45

Chapter 1

Ch 1.TF.29 F; Let ~u =

0 2 1 , for example. ,w ~= , ~v = 1 0 0

Ch 1.TF.30 T; Note that ~0 = 0~v + 0w ~

Ch 1.TF.31 1 0 0 0

0 1 0 0

1 1 . . F; If A 2 = ~0, then ~x = 2 is a solution to A..~0 . However, since rank(A) = 3, rref A..~0 = 3 3 0 0 0 .. 0 . , meaning that only ~0 is a solution to A~x = ~0. 1 0 0 0

Ch 1.TF.32 F; If ~b = ~0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. Ch 1.TF.33 T; By Example 3c of Section 1.3, the equation A~x = ~0 has the unique solution ~x = ~0. Now note that A(~v − w) ~ = ~0, so that ~v − w ~ = ~0 and ~v = w. ~ Ch 1.TF.34 T; Note that rank(A) = 4, by Theorem 1.3.4 Ch 1.TF.35 F; Let ~u =

2 1 0 , ~v = , w ~= , for example.. 0 0 1

−2t Ch 1.TF.36 T; We use rref to solve the system A~x = ~0 and find ~x = −3t , where t is an arbitrary constant. t −2 ~ = ~0, so that w ~ = 2~u + 3~v . Letting t = 1, we find [~u ~v w] ~ −3 = −2~u − 3~v + w 1

Ch 1.TF.37 F; Let A = B =

1 0 , for example. 0 1

1 0 ... 0 1 ... Ch 1.TF.38 T; Matrices A and B can both be transformed into I = ... ... ... 0 0 0 operations backwards, we can transform I into B. Thus we can first transform A

0 0 . Running the elementary 0 1 into I and then I into B.

Ch 1.TF.39 T; If ~v = a~u + bw, ~ then A~v = A(a~u + bw) ~ = A(a~u) + A(bw) ~ = aA~v + bAw. ~ Ch 1.TF.40 T; check that the three defining properties of a matrix in rref still hold. F; If ~b = ~0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. . ~ Ch 1.TF.41 T; A~x = b is inconsistent if and only if rank A..~b = rank(A)+1, since there will be an extra leading one in the last column of the augmented matrix: (See Figure 1.16.) 46

True or False

Figure 1.16: for Problem T/F 41. Ch 1.TF.42 T; The system A~x = ~b is consistent, by Example 3b, and there are, in fact, infinitely many solutions, by Theorem 1.3.3. Note that A~x = ~b is a system of three equations with four unknowns. .. .. .. ~ ~ ~ ~ Ch 1.TF.43 T; Recall that we use rref A.0 to solve the system A~x = 0. Now, rref A.0 = rref(A).0 = .. .. .. .. ~ ~ ~ ~ rref(B).0 = rref B .0 . Then, since rref(A).0 = rref(B).0 , they must have the same solutions. Ch 1.TF.44 F; Consider

1 2 . If we remove the first column, then the remaining matrix fails to be in rref. 0 0

Ch 1.TF.45 T; First we list all possible matrices rref(M ), where M is a 2 × 2 matrix, and show the corresponding solutions for M~x = ~0: rref(M solutions of M~x = ~0 ) 1 0 {~0} 0 1 −at 1 a , for an arbitrary t 0 0 t 0 1 t , for an arbitrary t 0 0 0 0 0 R2 0 0 Now, we see that if rref(A) 6= rref(B), then the systems A~x = ~0 and B~x = ~0 must have different solutions. Thus, it must be that if the two systems have the same solutions, then rref(A) = rref(B).

47

Chapter 2

Chapter 2 Section 2.1 2.1.1 Not a linear transformation, since y2 = x2 + 2 is not linear in our sense.

0 2.1.2 Linear, with matrix 0 1

2 0 0 3 0 0

2.1.3 Not linear, since y2 = x1 x3 is nonlinear. 9 3 −3 1 2 −9 2.1.4 A = 4 −9 −2 5 1 5

2.1.5 By Theorem 2.1.2, the three columns of the 2 × 3 matrix A are T (~e1 ), T (~e2 ), and T (~e3 ), so that 7 6 −13 . A= 11 9 17 1 4 1 2.1.6 Note that x1 2 + x2 5 = 2 3 6 3 2.1.7 Note that x1~v1 + · · · + xm~vm

2.1.8 Reducing the system

4 1 4 x 5 1 , so that T is indeed linear, with matrix 2 5 . x2 6 3 6

x1 = [~v1 . . . ~vm ] · · · , so that T is indeed linear, with matrix [~v1 ~v2 · · · ~vm ]. xm

x1 + 7x2 3x1 + 20x2

x1 = y1 , we obtain = y2

x2

= −20y1 = 3y1

+ −

7y2 . y2

y1 2 3 x1 2.1.9 We have to attempt to solve the equation = for x1 and x2 . Reducing the system y2 x2 6 9 x1 + 1.5x2 = 0.5y1 2x1 + 3x2 = y1 . we obtain 0 = −3y1 + y2 6x1 + 9x2 = y2 No unique solution (x1 , x2 ) can be found for a given (y1 , y2 ); the matrix is noninvertible. 1 2 y1 x1 = for x1 and x2 . Reducing the system 2.1.10 We have to attempt to solve the equation y2 4 9 x2 x1 + 2x2 = y1 x1 = 9y1 + 2y2 x1 9 −2 y1 we find that or = . 4x1 + 9x2 = y2 x2 = −4y1 + y2 x2 y2 −4 1 9 −2 . The inverse matrix is −4 1 48

Section 2.1 y1 x1 1 2 2.1.11 We have to attempt to solve the equation = for x1 and x2 . Reducing the system y2 3 9 x2 3 − 32 x1 = 3y1 − 23 y2 x1 + 2x2 = y1 . The inverse matrix is . we find that 1 3x1 + 9x2 = y2 −1 x2 = −y1 + 31 y2 3 2.1.12 Reducing the system 1 −k . 0 1

x1 + kx2 x2

= y1 = y2

we find that

x1 x2

2.1.13 a First suppose that a 6= 0. We have to attempt to solve the equation

ax1 cx1

x1

x1

+ bx2 + dx2 + (d +

= y1 =

b a x2 − bc a )x2

b a x2 ( ad−bc a )x2

y2

÷a

1 = a y1 = − ac y1

= =

1 a y1 − ac y1

x1 + → cx1 + → + y2

b a x2 dx2

= =

1 a y1

y2

− ky2 . The inverse matrix is y2

= y1 =

−c(I)

y1 y2

=

a b c d

x1 x2

for x1 and x2 .

→

+ y2

We can solve this system for x1 and x2 if (and only if) ad − bc 6= 0, as claimed. If a = 0, then we have to consider the system bx2 = y1 cx1 swap : I ↔ II cx1 + dx2 = y2

+ dx2 bx2

= = y1

y2

We can solve for x1 and x2 provided that both b and c are nonzero, that is if bc 6= 0. Since a = 0, this means that ad − bc 6= 0, as claimed. b First suppose that ad − bc 6= 0 and a 6= 0. Let D = ad − bc for simplicity. We continue our work in part (a): 1 x1 + ab x2 = a y1 a → D c ·D a x2 = − a y1 + y2 b 1 − a (II) x1 + ab x2 = a y1 → a x2 = − Dc y1 + D y2 bc x1 = ( a1 + aD )y1 − Db y2 a x2 = − Dc y1 + D y2 d − Db y2 x1 = D y1 a x2 = − Dc y1 + D y2 bc ad d Note that a1 + aD = D+bc aD = aD = D . It follows that system

a b c d

−1

=

1 ad−bc

d −c

−b , as claimed. If ad − bc 6= 0 and a = 0, then we have to solve the a 49

Chapter 2

cx1 + dx2 bx2

= y2 = y1

÷c ÷b x1 + dc x2 = 1c y2 − dc (II) x2 = 1b y1 d x1 = − bc y1 + 1c y2 1 x2 = b y1 −1 d − bc a b It follows that = 1 c d b 2.1.14 a By Exercise 13a,

b By Exercise 13b,

2 3 5 k

2 3 5 k −1

=

1 c

0

=

1 ad−bc

d −c

−b a

1 2n .

Since

k 2k−15

(recall that a = 0), as claimed.

is invertible if (and only if) 2k − 15 6= 0, or k 6= 7.5. 1 2k−15

−3 . 2

k −5

2 3 − 2k−15 2k−15 + 21 is an integer

If all entries of this inverse are integers, then or k = 7.5 +

= kn = 7.5n

=

1 2k−15

is a (nonzero) integer n, so that 2k −15 =

1 n

as well, n must be odd.

1 , where n is an odd integer. The We have shown: If all entries of the inverse are integers, then k = 7.5 + 2n −1 2 3 converse is true as well: If k is chosen in this way, then the entries of will be integers. 5 k

a b

−b a

is invertible if (and only if) a2 + b2 6= 0, which is the case unless 2.1.15 By Exercise 13a, the matrix a −b a b 1 is invertible, then its inverse is a2 +b a = b = 0. If , by Exercise 13b. 2 b a −b a 3 0 , then A~x = 3~x for all ~x in R2 , so that A represents a scaling by a factor of 3. Its inverse is a 0 3 1 0 3 1 −1 . (See Figure 2.1.) scaling by a factor of 3 : A = 0 13

2.1.16 If A =

2.1.17 If A =

−1 0

0 , then A~x = −~x for all ~x in R2 , so that A represents a reflection about the origin. −1

This transformation is its own inverse: A−1 = A. (See Figure 2.2.) 2.1.18 Compare with Exercise 16: This matrix represents a scaling by the factor of 12 ; the inverse is a scaling by 2. (See Figure 2.3.) x1 x 1 0 , so that A represents the orthogonal projection onto the ~e1 axis. (See , then A 1 = 2.1.19 If A = 0 0 0 x2 1 Figure 2.1.) This transformation is not invertible, since the equation A~x = has infinitely many solutions ~x. 0 (See Figure 2.4.)

50

Section 2.1

· ¸ 0 6

· ¸ 0 2

·

3 0 0 3

¸

· ¸ 1 0

· ¸ 3 0 Figure 2.1: for Problem 2.1.16.

· ¸ 0 2

·

−1 0 0 −1

¸

·

−1 0

¸

· ¸ 1 0

·

0 −2

¸

Figure 2.2: for Problem 2.1.17.

· ¸ 0 2

·

0.5 0 0 0.5

· ¸ 0 1

¸

·

· ¸ 1 0

0.5 0

¸

Figure 2.3: for Problem 2.1.18.

2.1.20 If A =

0 1 1 0

, then A

x1 x2

=

x2 , so that A represents the reflection about the line x2 = x1 . This x1

transformation is its own inverse: A−1 = A. (See Figure 2.5.)

2.1.21 Compare with Example 5. x 0 1 x2 , then A 1 = If A = . Note that the vectors ~x and A~x are perpendicular and have the same −1 0 x2 −x1 51

Chapter 2

· ¸ 0 2

·

1 0 0 0

¸

· ¸ 0 0 · ¸ 1 0

· ¸ 1 0

Figure 2.4: for Problem 2.1.19.

· ¸ 0 2

·

0 1 1 0

¸

· ¸ 0 1

· ¸ 1 0

· ¸ 2 0 Figure 2.5: for Problem 2.1.20.

length. If ~x is in the first quadrant, then A~x is in the fourth. Therefore, A represents the rotation through an 0 −1 angle of 90◦ in the clockwise direction. (See Figure 2.6.) The inverse A−1 = represents the rotation 1 0 ◦ through 90 in the counterclockwise direction.

· ¸ 0 2

·

0 1 −1 0

· ¸ 2 0

¸ ·

· ¸ 1 0

0 −1

¸

Figure 2.6: for Problem 2.1.21.

2.1.22 If A =

1 0 0 −1

x , then A 1 x2

=

x1 , so that A represents the reflection about the ~e1 axis. This −x2

transformation is its own inverse: A−1 = A. (See Figure 2.7.) 2.1.23 Compare with Exercise 21.

0 1 Note that A = 2 , so that A represents a rotation through an angle of 90◦ in the clockwise direction, −1 0 followed by a scaling by the factor of 2. 52

Section 2.1

· ¸ 0 2

·

1 0 0 −1

· ¸ 1 0

¸

· ¸ 1 0

·

0 −2

¸

Figure 2.7: for Problem 2.1.22.

−1

The inverse A

=

0 1 2

− 12 0

represents a rotation through an angle of 90◦ in the counterclockwise direction,

followed by a scaling by the factor of 12 . (See Figure 2.8.)

· ¸ 0 2

· · ¸ 1 0

0 2 −2 0

· ¸ 4 0

¸

·

0 −2

¸

Figure 2.8: for Problem 2.1.23. 2.1.24 Compare with Example 5. (See Figure 2.9.)

Figure 2.9: for Problem 2.1.24. 2.1.25 The matrix represents a scaling by the factor of 2. (See Figure 2.10.) 2.1.26 This matrix represents a reflection about the line x2 = x1 . (See Figure 2.11.) 2.1.27 This matrix represents a reflection about the ~e1 axis. (See Figure 2.12.) 53

Chapter 2

Figure 2.10: for Problem 2.1.25.

Figure 2.11: for Problem 2.1.26.

Figure 2.12: for Problem 2.1.27.

1 0 x x1 , then A 1 = , so that the x2 component is multiplied by 2, while the x1 component x2 0 2 2x2 remains unchanged. (See Figure 2.13.)

2.1.28 If A =

Figure 2.13: for Problem 2.1.28.

54

Section 2.1 2.1.29 This matrix represents a reflection about the origin. Compare with Exercise 17. (See Figure 2.14.)

Figure 2.14: for Problem 2.1.29.

2.1.30 If A = 2.15.)

0 0

0 x1 0 , then A = , so that A represents the projection onto the ~e2 axis. (See Figure x2 1 x2

Figure 2.15: for Problem 2.1.30.

2.1.31 The image must be reflected about the ~e2 axis, that is

x1 x2

must be transformed into −1 0 be accomplished by means of the linear transformation T (~x) = ~x. 0 1

3 0 2.1.32 Using Theorem 2.1.2, we find A = ...

0

else.

0 3 .. .

· · .. .

0 ···

−x1 : This can x2

0 0 . This matrix has 3’s on the diagonal and 0’s everywhere .. . 3

1 0 2.1.33 By Theorem 2.1.2, A = T T . (See Figure 2.16.) 0 1 √1 − √12 2 . Therefore, A = 1 1 √

√

2

2

2.1.34 As in Exercise 2.1.33, we find T (~e1 ) and T (~e2 ); then by Theorem 2.1.2, A = [T (~e1 ) T (~e2 )]. (See Figure 2.17.) 55

Chapter 2

Figure 2.16: for Problem 2.1.33.

Figure 2.17: for Problem 2.1.34.

cos θ Therefore, A = sin θ

− sin θ . cos θ

2.1.35 We want to find a matrix A = 5a + 42b 6a + 41b solving the system

a c

5c + 42d 6c + 41d

b d

such that A

= 89 = 88 . = 52 = 53

88 6 89 5 . This amounts to = and A = 53 41 52 42

(Here we really have two systems with two unknowns each.) The unique solution is a = 1, b = 2, c = 2, and d = 1, so that A =

1 2 . 2 1

2.1.36 First we draw w ~ in terms of ~v1 and ~v2 so that w ~ = c1~v1 + c2~v2 for some c1 and c2 . Then, we scale the ~v2 -component by 3, so our new vector equals c1~v1 + 3c2~v2 . 2.1.37 Since ~x = ~v + k(w ~ − ~v ), we have T (~x) = T (~v + k(w ~ − ~v )) = T (~v ) + k(T (w) ~ − T (~v )), by Theorem 2.1.3 Since k is between 0 and 1, the tip of this vector T (~x) is on the line segment connecting the tips of T (~v ) and T (w). ~ (See Figure 2.18.) 56

Section 2.1

Figure 2.18: for Problem 2.1.37.

2.1.38 T

2 = [~v1 −1

2 ~v2 ] = 2~v1 − ~v2 = 2~v1 + (−~v2 ). (See Figure 2.19.) −1

Figure 2.19: for Problem 2.1.38.

x1 x1 T (~ e ) . . . T (~ e ) 1 m 2.1.39 By Theorem 2.1.2, we have T . . . = . . . = x1 T (~e1 ) + · · · + xm T (~em ). xm xm

2.1.40 These linear transformations are of the form [y] = [a][x], or y = ax. The graph of such a function is a line through the origin. x1 , or y = ax1 + bx2 . The graph of such a function 2.1.41 These linear transformations are of the form [y] = [a b] x2 is a plane through the origin.

2.1.42 a See Figure 2.20. 1 0 b The image of the point 21 is the origin, . 0 1 2

57

Chapter 2

Figure 2.20: for Problem 2.1.42. # " 1 x1 − 2 x1 + x2 =0 0 x2 = . (See Figure 2.16.) c Solve the equation , or 0 − 21 0 1 x3 = 0 − 21 x1 + x3 2t x1 The solutions are of the form x2 = t , where t is an arbitrary real number. For example, for t = 12 , we t x3 1 find the point 12 considered in part b.These points are on the line through the origin and the observer’s eye.

− 21

1

0

1 2

x1 2 x1 2.1.43 a T (~x) = 3 · x2 = 2x1 + 3x2 + 4x3 = [2 3 4] x2 x3 4 x3 The transformation is indeed linear, with matrix [2 3 4].

v1 b If ~v = v2 , then T is linear with matrix [v1 v2 v3 ], as in part (a). v3

a a x1 x1 x1 c Let [a b c] be the matrix of T . Then T x2 = [a b c] x2 = ax1 + bx2 + cx3 = b · x2 , so that ~v = b c c x3 x3 x3 does the job.

0 v2 x3 − v3 x2 x1 v1 x1 2.1.44 T x2 = v2 × x2 = v3 x1 − v1 x3 = v3 −v2 v1 x2 − v2 x1 x3 v3 x3 0 −v3 v2 v3 0 −v1 . −v2 v1 0

58

−v3 0 v1

x1 v2 −v1 x2 , so that T is linear, with matrix x3 0

Section 2.1 2.1.45 Yes, ~z = L(T (~x)) is also linear, which we will verify using Theorem 2.1.3. Part a holds, since L(T (~v + w)) ~ = L(T (~v ) + T (w)) ~ = L(T (~v )) + L(T (w)), ~ and part b also works, because L(T (k~v )) = L(kT (~v )) = kL(T (~v )). pa + qc a 1 1 = =B =B A ra + sc c 0 0 0 0 b pb + qd T =B A =B = 1 1 d rb + sd 0 1 x1 b pb + qd So, T = + x2 T = x1 T = x2 1 0 d rb + sd

2.1.46 T

2.1.47 Write w ~ as a linear combination of ~v1 and ~v2 : w ~ = c1~v1 + c2~v2 . (See Figure 2.21.)

Figure 2.21: for Problem 2.1.47. Measurements show that we have roughly w ~ = 1.5~v1 + ~v2 . Therefore, by linearity, T (w) ~ = T (1.5~v1 + ~v2 ) = 1.5T (~v1 ) + T (~v2 ). (See Figure 2.22.)

Figure 2.22: for Problem 2.1.47. 2.1.48 Let ~x be some vector in R2 . Since ~v1 and ~v2 are not parallel, we can write ~x in terms of components of ~v1 and ~v2 . So, let c1 and c2 be scalars such that ~x = c1~v1 + c2~v2 . Then, by Theorem 2.1.3, T (~x) = T (c1~v1 + c2~v2 ) = T (c1~v1 ) + T (c2~v2 ) = c1 T (~v1 ) + c2 T (~v2 ) = c1 L(~v1 ) + c2 L(~v2 ) = L(c1~v1 + c2~v2 ) = L(~x). So T (~x) = L(~x) for all ~x in R2 .

2x1 2.1.49 a Let x1 be the number of 2 Franc coins, and x2 be the number of 5 Franc coins. Then x1 59

+5x2 +x2

= =

144 . 51

Chapter 2

From this we easily find our solution vector to be

b

total value of coins 2x1 = x1 total number of coins 2 5 So, A = . 1 1

+5x2 +x2

=

2 5 1 1

37 . 14

x1 . x2

c By Exercise 13, matrix A is invertible (since ad − bc = −3 6= 0), and A−1 = Then

− 13

part a.

1 −5 −1 2

144 51

=

− 13

144 −144

−5(51) +2(51)

=

− 31

−111 −42

=

1 ad−bc

d −b 1 = − 31 −c a −1

−5 . 2

37 , which was the vector we found in 14

p mass of the platinum alloy 2.1.50 a Let = . Using the definition density = mass/volume, or volume = s mass of the silver alloy mass/density, we can set up the system: p +s = 5, 000 , with the solution p = 2, 600 and s = 2, 400. We see that the platinum alloy makes up p s + 10 = 370 20 only 52 percent of the crown; this gold smith is a crook! p total mass 1 p+s b We seek the matrix A such that A = = p 1 s . Thus A = s total volume + 20 10 20

1 1 10

.

2 −20 p c Yes. By Exercise 13, A−1 = . Applied to the case considered in part a, we find that = −1 20 s 2, 600 5, 000 2 −20 total mass , confirming our answer in part a. = = A−1 2, 400 370 −1 20 total volume 5 5 C (F − 32) F− 9 2.1.51 a = = 9 1 1 1 5 − 160 9 . So A = 9 0 1

160 9

=

5 9

0

− 160 9 1

F . 1

5 b Using Exercise 13, we find 95 (1) − (− 160 9 )0 = 9 6= 0, so A is invertible. 9 1 160 32 9 5 = A−1 = 95 . So, F = 59 C + 32. 5 0 1 0 9

2.1.52 a A~x =

300 , meaning that the total value of our money is C$300, or, equivalently, ZAR2400. 2, 400

b From Exercise 13, we test the value ad − bc and find it to be zero. Thus A is not invertible. To determine when .. A is consistent, we begin to compute rref A.~b : 60

Section 2.2 .. .. 1 1 . b1 . b 1 8 . → .. .. −8I 8 1 . b2 0 0 . b2 − 8b1 Thus, the system is consistent only when b2 = 8b1 . This makes sense, since b2 is the total value of our money in terms of Rand, while b1 is the value in terms of Canadian dollars. Consider the example in part a. If the system A~x = ~b is consistent, then there will be infinitely many solutions ~x, representing various compositions of our portfolio in terms of Rand and Canadian dollars, all representing the same total value.

1

1 8

2.1.53 All four entries along the diagonal must be 1: they represent the process of converting a currency to itself. We also know that aij = 1/aji for all i, j because converting from one currency i to currencty jis the inverse 1 5/8 1/170 ∗ 8/5 1 ∗ 2 . Next, let’s to converting currency j to currency i. This gives us 3 more entries: 170 ∗ 1 ∗ ∗ 1/2 ∗ 1 find the entry a41 , giving the value of one Euro expressed in Pounds. Now E1 = $(8/5) = $1.60 and $1 = £(1/2) = £(0.50) so that E1 = £(1/2)(8/5) = £(4/5) = £0.80. We have found that a41 = a42 a21 = 4/5 and 1 5/8 1/170 5/4 8/5 1 ∗ 2 . Similarly, we have aij = aik akj for all indices i, j, k = 1, 2, 3, 4. This the matrix is 170 ∗ 1 ∗ 4/5 1/2 ∗ 1 gives a32 = a31 a12 = 170 ∗ 5/8 = 425/4 and a43 = a41 a13 = (4/5)(1/170) = 2/425. Using the fact that aij = a−1 ji , 1 5/8 1/170 5/4 8/5 1 4/425 2 . we can complete the matrix: 170 425/4 1 425/2 4/5 1/2 2/425 1 2.1.54 a 1: this represents converting a currency to itself. b aij is the reciprocal of aji , meaning that aij aji = 1. This represents converting on currency to another, then converting it back. c Note that aik is the conversion factor from currency k to currency i meaning that (1 unit of currency k) = (aik units of currency i) Likewise, (1 unit of currency j) = (akj units of currency k). It follows that (1 unit of currency j) = (akj aik units of currency i) = (aij units of currency i), so that aik akj = aij . d The rank of A is only 1, because every row is simply a scalar multiple of the top row. More precisely, since aij = ai1 a1j , by part c, the ith row is ai1 times the top row. When we compute the rref, every row but the top will be removed in the first step. Thus, rref(A) is a matrix with the top row of A and zeroes for all other entries.

Section 2.2 1 3 1 1 3 2.2.1 The standard L is transformed into a distorted L whose foot is the vector T = = . 0 1 2 0 1 61

Chapter 2

Meanwhile, the back becomes the vector T

2.2.2 By Theorem 2.2.3, this matrix is

2 0 3 1 0 . = = 4 2 1 2 2 ◦

cos(60 ) sin(60◦ )

◦

− sin(60 ) = cos(60◦ )

√ 1 2 √ 3 2

−

3 2 1 2

.

2.2.3 If ~x is in the unit square in R2 , then ~x = x1~e1 + x2~e2 with 0 ≤ x1 , x2 ≤ 1, so that T (~x) = T (x1~e1 + x2~e2 ) = x1 T (~e1 ) + x2 T (~e2 ). The image of the unit square is a parallelogram in R3 ; two of its sides are T (~e1 ) and T (~e2 ), and the origin is one of its vertices. (See Figure 2.23.)

Figure 2.23: for Problem 2.2.3.

2.2.4 By Theorem 2.2.4, this is a rotation√combined with a scaling. The transformation rotates 45 degrees counterclockwise, and has a scaling factor of 2. 2.2.5 Note that cos(θ) = −0.8, so that θ = arccos(−0.8) ≈ 2.498. 2 1 1 2.2.6 By Theorem 2.2.1, projL 1 = ~u · 1 ~u, where ~u is a unit vector on L. To get ~u, we normalize 1 : 2 1 1 2 1 ~u = 31 1 , so that projL 1 = 2 1

5 3

10 2 9 · 31 1 = 59 . 10 2 9

1 1 1 2.2.7 According to the discussion in the text, refL 1 = 2 ~u · 1 ~u − 1 , where ~u is a unit vector on L. To 1 1 1 11 1 1 2 2 2 9 get ~u, we normalize 1 : ~u = 31 1 , so that refL 1 = 2( 53 ) 31 1 − 1 = 19 . 11 1 1 2 2 2 9

62

Section 2.2 2.2.8 From Definition 2.2.2, we can see that this is a reflection about the line x1 = −x2 . 2.2.9 By Theorem 2.2.5, this is a vertical shear. 0.8 4 . Then = 2.2.10 By Theorem 2.2.1, projL ~x = (~u ·~x)~u, where ~u is a unit vector on L. We can choose ~u = 0.6 3 x1 0.64 0.48 0.64x1 + 0.48x2 0.8 0.8 x 0.8 x . = = = (0.8x1 + 0.6x2 ) · 1 = projL 1 0.48 0.36 x2 0.6 0.48x1 + 0.36x2 0.6 0.6 x2 x2 0.64 0.48 The matrix is A = . 0.48 0.36 1 5

2.2.11 In Exercise 10 we found the matrix A =

0.64 0.48

0.48 0.36

of the projection onto the line L. By Theorem 2.2.2, 0.28 0.96 refL ~x = 2(projL ~x) − ~x = 2A~x − ~x = (2A − I2 )~x, so that the matrix of the reflection is 2A − I2 = . 0.96 −0.28

p 2.2.12 Let ~u = (1/||w||) ~ w ~ be the unit vector in the direction of w. ~ It has the components u1 = w1 / w12 + w22 and 2 u2 = √ww2 +w 2 . On Pages 57/58, we see that the matrix representing the projection is 1

2

This can be written as

u21 u1 u2

1 w12 + w22

u1 u2 u22

w12 w1 w2

.

w1 w2 w22

,

as claimed. 2.2.13 By Theorem 2.2.2, x u1 x u1 x1 − 1 · 1 =2 refL x2 u2 x2 u2 x2 . 2 u x (2u1 − 1)x1 + 2u1 u2 x2 = 2(u1 x1 + u2 x2 ) 1 − 1 = . u2 x2 2u1 u2 x1 + (2u22 − 1)x2 2 a b 2u1 − 1 2u1 u2 The matrix is A = . Note that the sum of the diagonal entries is a + d = = 2u1 u2 2u22 − 1 c d a b 2 2 2(u1 + u2 ) − 2 = 0, since ~u is a unit vector. It follows that d = −a. Since c = b, A is of the form . Also, b −a 2 2 2 4 2 2 2 2 2 2 a + b = (2u1 − 1) + 4u1 u2 = 4u1 − 4u1 + 1 + 4u1 (1 − u1 ) = 1, as claimed.

2.2.14 a Proceeding as on Page 57/58 in the text, we find that A is the matrix whose ijth entry is ui uj : 2 u1 u2 u1 u3 u1 u22 u2 u3 A = u2 u1 un u1 un u2 u23 b The sum of the diagonal entries is u21 + u22 + u23 = 1, since ~u is a unit vector. 63

Chapter 2 2.2.15 According to the discussion on Page 60 in the text, refL (~x) = 2(~x · ~u)~u − ~x x1 u1 = 2(x1 u1 + x2 u2 + x3 u3 ) u2 − x2 x3 u3 (2u21 − 1)x1 +2u2 u1 x2 +2u1 u3 x3 2x1 u21 +2x2 u2 u1 +2x3 u3 u1 −x1 +(2u22 − 1)x2 +2u2 u3 x3 . +2x2 u22 +2x3 u3 u2 −x2 = 2u1 u2 x1 = 2x1 u1 u2 2u1 u3 x1 +2u2 u3 x2 +(2u23 − 1)x3 2x1 u1 u3 +2x2 u2 u3 +2x3 u23 −x3 (2u21 − 1) 2u2 u1 2u1 u3 (2u22 − 1) 2u2 u3 . So A = 2u1 u2 2u1 u3 2u2 u3 (2u23 − 1) 2.2.16 a See Figure 2.24.

Figure 2.24: for Problem 2.2.16a. b By Theorem 2.1.2, the matrix of T is [T (~e1 )

T (~e2 )].

Figure 2.25: for Problem 2.2.16b. 64

Section 2.2

T (~e2 ) is the unit vector in the fourth quadrant perpendicular to T (~e1 ) =

cos(2θ) , so that sin(2θ)

sin(2θ) cos(2θ) sin(2θ) . The matrix of T is therefore . − cos(2θ) sin(2θ) − cos(2θ) u1 cos θ Alternatively, we can use the result of Exercise 13, with = to find the matrix u2 sin θ 2 cos2 θ − 1 2 cos θ sin θ . 2 cos θ sin θ 2 sin2 θ − 1

T (~e2 ) =

You can use trigonometric identities to show that the two results agree. (See Figure 2.25.) 2.2.17 We want,

a b b −a

v1 v2

av1 = bv1

+bv2 −av2

v1 . = v2

Now, (a − 1)v1 + bv2 = 0 and bv1 − (a + 1)v2 , which is a system with solutions of the form is an arbitrary constant.

bt , where t (1 − a)t

b . Let’s choose t = 1, making ~v = 1−a

a−1 Similarly, we want Aw ~ = −w. ~ We perform a computation as above to reveal w ~ = b A quick check of ~v · w ~ = 0 reveals that they are indeed perpendicular.

as a possible choice.

Now, any vector ~x in R can be written in terms of components with respect to L = span(~v ) as ~x = ~x|| + ~x⊥ = c~v + dw. ~ Then, T (~x) = A~x = A(c~v + dw) ~ = A(c~v ) + A(dw) ~ = cA~v + dAw ~ = c~v − dw ~ = ~x|| − ~x⊥ = refL (~x), by Definition 2.2.2. (The vectors ~v and w ~ constructed above are both zero in the special case that a = 1 and b = 0. In that case, we can let ~v = ~e1 and w ~ = ~e2 instead.)

b 0.8 2 2.2.18 From Exercise 17, we know that the reflection is about the line parallel to ~v = = = 0.4 . 1 − a 0.4 1 x 2 So, every point on this line can be described as =k . So, y = k = 21 x, and y = 12 x is the line we are y 1 looking for.

1 0 2.2.19 T (~e1 ) = ~e1 , T (~e2 ) = ~e2 , and T (~e3 ) = ~0, so that the matrix is 0 1 0 0

0 0 −1 0 . 0 1

−1 0 0 0 . (See Figure 2.26.) 0 1

1 2.2.20 T (~e1 ) = ~e1 , T (~e2 ) = −~e2 , and T (~e3 ) = ~e3 , so that the matrix is 0 0 0 2.2.21 T (~e1 ) = ~e2 , T (~e2 ) = −~e1 , and T (~e3 ) = ~e3 , so that the matrix is 1 0 65

0 0 . 0

Chapter 2

Figure 2.26: for Problem 2.2.21. 2.2.22 Sketch the ~e1 − ~e3 plane, as viewed from the positive ~e2 axis.

Figure 2.27: for Problem 2.2.22.

cos θ Since T (~e2 ) = ~e2 , the matrix is 0 − sin θ

0 sin θ 1 0 . (See Figure 2.27.) 0 cos θ

0 2.2.23 T (~e1 ) = ~e3 , T (~e2 ) = ~e2 , and T (~e3 ) = ~e1 , so that the matrix is 0 1

0 1 1 0 . (See Figure 2.28.) 0 0

Figure 2.28: for Problem 2.2.23. 0 1 = w. ~ Since A preserves length, both ~v and w ~ must be unit = ~v and A 2.2.24 a A = [ ~v w ~ ] , so A 1 0 1 0 vectors. Furthermore, since A preserves angles and and are clearly perpendicular, ~v and w ~ must also 0 1 66

Section 2.2 be perpendicular. b Since w ~ is a unit vector perpendicular to ~v , it can be obtained by rotating ~v through 90 degrees, either in the counterclockwise or in the clockwise direction. Using the corresponding rotation matrices, we see that 0 −1 −b 0 1 b w ~= ~v = or w ~= ~v = . 1 0 a −1 0 −a c Following part b, A is either of the form reflection.

a b

−b a b , representing a rotation, or A = , representing a a b −a

1 k 1 −k represents a horizontal shear, and its inverse A−1 = represents such a 0 1 0 1 shear as well, but “the other way.”

2.2.25 The matrix A =

2.2.26 a

k 0

0 k

4 0 8 2k 2 . . So k = 4 and A = = = 0 4 −4 −k −1

b This is the orthogonal projection onto the horizontal axis, with matrix B = 4 3 −5b 0 a −b 4 3 5 . So a = 5 , b = − 5 , and C = = = c 4 5a − 35 5 b a a rotation matrix.

3 5 4 5

1 0

0 . 0

. Note that a2 + b2 = 1, as required for

d Since the x1 term is being modified, this must be a horizontal shear. 1 2 7 1 + 3k 1 1 k . . So k = 2 and D = = = Then 0 1 3 3 3 0 1 4 −5 7a + b 7 a b − 4 3 . So a = − 5 , b = 5 , and E = 35 = = e 5 7b − a 1 b −a 5 for a reflection matrix.

3 5 4 5

. Note that a2 + b2 = 1, as required

2.2.27 Matrix B clearly represents a scaling. Matrix C represents a projection, by Definition 2.2.1, with u1 = 0.6 and u2 = 0.8. Matrix E represents a shear, by Theorem 2.2.5. Matrix A represents a reflection, by Definition 2.2.2. Matrix D represents a rotation, by Definition 2.2.3. 2.2.28 a D is a scaling, being of the form

k 0

0 . k

b E is the shear, since it is the only matrix which has the proper form (Theorem 2.2.5). c C is the rotation, since it fits Theorem 2.2.3. 67

Chapter 2 d A is the projection, following the form given in Definition 2.2.1. e F is the reflection, using Definition 2.2.2. 2.2.29 To check that L is linear, we verify the two parts of Theorem 2.1.3: a) Use the hint and apply L to both sides of the equation ~x + ~y = T (L(~x) + L(~y )): L(~x + ~y ) = L(T (L(~x) + L(~y ))) = L(~x) + L(~y ) as claimed. b)L (k~x) = L (kT (L (~x))) = L (T (kL (~x))) = kL (~x) , as claimed ↑ ↑ ~x = T (L (~x)) T is linear

x1 2.2.30 Write A = [ ~v1 ~v2 ]; then A~x = [~v1 ~v2 ] = x1~v1 + x2~v2 . We must choose ~v1 and ~v2 in such a way that x2 1 , for all x1 and x2 . This is the case if (and only if) both ~v1 and x1~v1 + x2~v2 is a scalar multiple of the vector 2 1 ~v2 are scalar multiples of . 2 1 0 0 1 . , so that A = and ~v2 = For example, choose ~v1 = 2 0 0 2 x1 2.2.31 Write A = [~v1 ~v2 ~v3 ]; then A~x = [~v1 ~v2 ~v3 ] x2 = x1~v1 + x2~v2 + x3~v3 . x3

1 We must choose ~v1 , ~v2 , and ~v3 in such a way that x1~v1 + x2~v2 + x3~v3 is perpendicular to w ~ = 2 for all 3 x1 , x2 , and x3 . This is the case if (and only if) all the vectors ~v1 , ~v2 , and ~v3 are perpendicular to w, ~ that is, if ~v1 · w ~ = ~v2 · w ~ = ~v3 · w ~ = 0. −2 −2 0 0 For example, we can choose ~v1 = 1 and ~v2 = ~v3 = ~0, so that A = 1 0 0 . 0 0 0 0

2.2.32 a See Figure 2.29.

cos α b Compute D~v = sin α

− sin α cos α

cos β sin β

cos α cos β − sin α sin β = . sin α cos β + cos α sin β

Comparing this result with our finding in part (a), we get the addition theorems cos(α + β) = cos α cos β − sin α sin β sin(α + β) = sin α cos β − cos α sin β 68

Section 2.2

Figure 2.29: for Problem 2.2.32a.

Figure 2.30: for Problem 2.2.33. 2.2.33 Geometrically, we can find the representation ~v = ~v1 + ~v2 by means of a parallelogram, as shown in Figure 2.30. To show the existence and uniqueness of this representation algebraically, ~ 1 in L1 and choose a nonzero vector w x 1 a nonzero w ~ 2 in L2 . Then the system x1 w ~ 1 + x2 w ~ 2 = ~0 or [w ~1 w ~ 2] = ~0 has only the solution x1 = x2 = 0 x2 (if x1 w ~ 1 + x2 w ~ 2 = ~0 then x1 w ~ 1 = −x2 w ~ 2 is both in L1 and in L2 , so that it must be the zero vector). x Therefore, the system x1 w ~ 1 + x2 w ~ 2 = ~v or [w ~1 w ~ 2 ] 1 = ~v has a unique solution x1 , x2 for all ~v in R2 (by x2 Theorem 1.3.4). Now set ~v1 = x1 w ~ 1 and ~v2 = x2 w ~ 2 to obtain the desired representation ~v = ~v1 + ~v2 . (Compare with Exercise 1.3.57.) To show that the transformation T (~v ) = ~v1 is linear, we will verify the two parts of Theorem 2.1.3. Let ~v = ~v1 + ~v2 , w ~ =w ~1 + w ~ 2 , so that ~v + w ~ = (~v1 + w ~ 1 ) + (~v2 + w ~ 2 ) and k~v = k~v1 + k~v2 . ↑ ↑ ↑ ↑ in L1 in L2 in L1 in L2

↑ in L1

a. T (~v + w) ~ = ~v1 + w ~ 1 = T (~v ) + T (w), ~ and b. T (k~v ) = k~v1 = kT (~v ), as claimed. 69

↑ in L2

↑ ↑ in L1 in L2

Chapter 2 2.2.34 Keep in mind that the columns of the matrix of a linear transformation T from R3 to R3 are T (~e1 ), T (~e2 ), and T (~e3 ). If T is the orthogonal projection onto a line L, then T (~x) will be on L for all ~x in R3 ; in particular, the three columns of the matrix of T will be on L, and therefore pairwise parallel. This is the case only for matrix B: B represents an orthogonal projection onto a line. A reflection transforms orthogonal vectors into orthogonal vectors; therefore, the three columns of its matrix must be pairwise orthogonal. This is the case only for matrix E: E represents the reflection about a line. 2.2.35 If the vectors ~v1 and ~v2 are defined as shown in Figure 2.27, then the parallelogram P consists of all vectors of the form ~v = c1~v1 + c2~v2 , where 0 ≤ c1 , c2 ≤ 1. The image of P consists of all vectors of the form T (~v ) = T (c1~v1 + c2~v2 ) = c1 T (~v1 ) + c2 T (~v2 ). These vectors form the parallelogram shown in Figure 2.31 on the right.

Figure 2.31: for Problem 2.2.35. 2.2.36 If the vectors ~v0 , ~v1 , and ~v2 are defined as shown in Figure 2.28, then the parallelogram P consists of all vectors ~v of the form ~v = ~v0 + c1~v1 + c2~v2 , where 0 ≤ c1 , c2 ≤ 1. The image of P consists of all vectors of the form T (~v ) = T (~v0 + c1~v1 + c2~v2 ) = T (~v0 ) + c1 T (~v1 ) + c2 T (~v2 ). These vectors form the parallelogram shown in Figure 2.32 on the right.

Figure 2.32: for Problem 2.2.36.

2.2.37 a By Definition 2.2.1, a projection has a matrix of the form 70

u21 u1 u2

u1 u1 u2 , where is a unit vector. u2 u22

Section 2.2 So the trace is u21 + u22 = 1. b By Definition 2.2.2, reflection matrices look like

a b , so the trace is a − a = 0. b −a

cos θ c According to Theorem 2.2.3, a rotation matrix has the form sin θ for some θ. Thus, the trace is in the interval [−2, 2].

− sin θ , so the trace is cos θ +cos θ = 2 cos θ cos θ

1 k 1 0 , depending on whether it represents or 0 1 k 1 a vertical or horizontal shear. In both cases, however, the trace is 1 + 1 = 2.

d By Theorem 2.2.5, the matrix of a shear appears as either

2.2.38 a A =

b A=

u21 u1 u2

u1 u2 , so det(A) = u21 u22 − u1 u2 u1 u2 = 0. u22

a b , so det(A) = −a2 − b2 = −(a2 + b2 ) = −1. b −a

a −b c A= , so det(A) = a2 − (−b2 ) = a2 + b2 = 1. b a

1 k d A= 0 1

or

1 k

0 , both of which have determinant equal to 12 − 0 = 1. 1

1 1 1 1 1 1 2 . The matrix 2 2 represents an orthogonal projection (Definition 2.2.39 a Note that = 2 2 1 1 1 1 1 1 2 2 2 " √2 #2 1 1 u1 represents a projection combined with a scaling by a factor of 2. = √22 . So, 2.2.1), with ~u = 1 1 u2

2

b This lookssimilar to a shear, with the one zero off the diagonal. Since the two diagonal entries are identical, we 1 0 3 0 =3 can write , showing that this matrix represents a vertical shear combined with a scaling − 13 1 −1 3 by a factor of 3. c We are asked to write " 3 4 # k

4 k 3 2 (k)

k

3 4 4 −3

=k

"

3 k

4 k

4 k

− k3

#

, with our scaling factor k yet to be determined. This matrix,

a b

b −a

has the form of a reflection matrix . This form further requires that 1 = a2 + b2 = − k3 + ( k4 )2 , or k = 5. Thus, the matrix represents a reflection combined with a scaling by a factor of 5.

2.2.40 ~x = projP ~x + projQ ~x, as illustrated in Figure 2.33. 2.2.41 refQ ~x = −refP ~x since refQ ~x, refP ~x, and ~x all have the same length, and refQ ~x and refP ~x enclose an angle of 2α + 2β = 2(α + β) = π. (See Figure 2.34.) 71

Chapter 2

Figure 2.33: for Problem 2.2.40.

Figure 2.34: for Problem 2.2.41. 2.2.42 T (~x) = T (T (~x)) since T (~x) is on L hence the projection of T (~x) onto L is T (~x) itself. 2.2.43 Since ~y = A~x is obtained from ~x by a rotation through θ in the counterclockwise direction, ~x is obtained from ~y by a rotation through θ in the clockwise direction, that is, a rotation through −θ. (See Figure 2.35.)

Figure 2.35: for Problem 2.2.43. Therefore, the matrix of the inverse transformation is A−1 = use the formula in Exercise 2.1.13b to check this result. 2.2.44 By Exercise 1.1.13b, A−1 =

a b

−b a

−1

=

1 a2 +b2

cos(−θ) − sin(−θ) cos θ sin θ = . You can sin(−θ) cos(−θ) − sin θ cos θ

a b . −b a

If A represents a rotation through θ followed by a scaling by r, then A−1 represents a rotation through −θ followed by a scaling by 1r . (See Figure 2.36.) −1

2.2.45 By Exercise 2.1.13, A

=

1 −a2 −b2

−a −b = −b a

1 −(a2 +b2 )

a b −a −b −a −b . = = −1 b −a −b a −b a

So A−1 = A, which makes sense. Reflecting a vector twice about the same line will return it to its original state. 72

Section 2.2

Figure 2.36: for Problem 2.2.44.

2.2.46 We want to write A = k

a k b k

b k − ka

, where the matrix B =

a k b k

b k − ka

represents a reflection. It is required √ a b 1 b 2 a 2 −1 2 2 2 2 2 = k12 A = k1 B, that ( k ) + ( k ) = 1, meaning that a + b = k , or, k = a + b . Now A = a2 +b2 b −a for the reflection matrix B and the scaling factor k introduced above. In summary: If A represents a reflection combined with a scaling by k, then A−1 represents the same reflection combined with a scaling by k1 .

a b ax1 + bx2 x1 . = cx1 + dx2 c d x2 −a sin t + b cos t a cos t + b sin t − sin t cos t · = · T a. f (t) = T −c sin t + d cos t c cos t + d sin t cos t sin t

2.2.47 Write T

x1 x2

=

= (a cos t + b sin t)(−a sin t + b cos t) + (c cos t + d sin t)(−c sin t + d cos t) This function f (t) is continuous, since cos(t), sin(t), and constant functions are continuous, and sums and products of continuous functions are continuous. 0 −1 0 1 π b. f 2 = T ·T =− T ·T , since T is linear. 1 0 1 0 1 0 0 1 f (0) = T ·T =T ·T . The claim follows. 0 1 1 0 c. By part (b), the numbers f (0) and f π2 have different signs (one is positive and the other negative), or they are both zero. Since f (t) is continuous, by part (a), we can apply the intermediate value theorem. (See Figure 2.37.)

Figure 2.37: for Problem 2.2.47c. d. Note that

cos(t) sin(t)

and

− sin(t) cos(t)

are perpendicular unit vectors, for any t. If we set 73

Chapter 2 − sin(c) cos(c) , with the number c we found in part (c), then f (c) = T (~v1 ) · T (~v2 ) = 0, so , ~v2 = cos(c) sin(c) that T (~v1 ) and T (~v2 ) are perpendicular, as claimed. Note that T (~v1 ) or T (~v2 ) may be zero.

~v1 =

2.2.48 We find 0 4 − sin(t) cos(t) 0 4 · f (t) = cos(t) 5 −3 sin(t) 5 −3 =

4 sin(t) 4 cos(t) · 5 cos(t) − 3 sin(t) −5 sin(t) − 3 cos(t)

= 15(sin2 t − cos2 t) = 15(2 sin2 t − 1). See Figure 2.38.

Figure 2.38: for Problem 2.2.48. The only zero of f (t) between 0 and π2 is at c = π4 . √ # √2 " π 2 − 2 − sin( ) cos( π4 ) 4 = √ work. Note that T (~v1 ) = Therefore, ~v1 = = √2 and ~v2 = π π 2 2 sin( 4 ) cos( 4 ) 2

and T (~v2 ) =

√1 2

4 −8

2

are indeed perpendicular. See Figure 2.39.

Figure 2.39: for Problem 2.2.48.

cos(t) 2.2.49 If ~x = sin(t)

5 0 then T (~x) = 0 2

cos(t) 5 cos(t) 5 0 = = cos(t) + sin(t) . sin(t) 2 sin(t) 0 2 74

√1 2

4 2

Section 2.2 Thesevectors form an consider the characterization of an ellipse given in the footnote on Page 69, with ellipse; 0 5 . (See Figure 2.40.) and w ~2 = w ~1 = 2 0

Figure 2.40: for Problem 2.2.49. 2.2.50 Use the hint: Since the vectors on the unit circle are of the form ~v = cos(t)~v1 + sin(t)~v2 , the image of the unit circle consists of the vectors of the form T (~v ) = T (cos(t)~v1 + sin(t)~v2 ) = cos(t)T (~v1 ) + sin(t)T (~v2 ).

75

Chapter 2

Figure 2.41: for Problem 2.2.50. These vectors form an ellipse: Consider the characterization of an ellipse given in the footnote, with w ~ 1 = T (~v1 ) and w ~ 2 = T (~v2 ). The key point is that T (~v1 ) and T (~v2 ) are perpendicular. See Figure 2.41. 2.2.51 Consider the linear transformation T with matrix A = [w ~1 x x x1 ~ 1 + x2 w ~ 2. ~1 w ~ 2 ] 1 = x1 w = A 1 = [w T x2 x2 x2

w ~ 2 ], that is,

cos(t) is on the unit circle, sin(t) then T (~v ) = cos(t)w ~ 1 + sin(t)w ~ 2 is on the curve C. Therefore, C is an ellipse, by Exercise 50. (See Figure 2.42.) The curve C is the image of the unit circle under the transformation T : if ~v =

Figure 2.42: for Problem 2.2.51. 2.2.52 By definition, the vectors ~v on an ellipse E are of the form ~v = cos(t)~v1 + sin(t)~v2 , for some perpendicular vectors ~v1 and ~v2 . Then the vectors on the image C of E are of the form T (~v ) = cos(t)T (~v1 ) + sin(t)T (~v2 ). These vectors form an ellipse, by Exercise 51 (with w ~ 1 = T (~v1 ) and w ~ 2 = T (~v2 )). See Figure 2.43.

76

Section 2.3

Figure 2.43: for Problem 2.2.52.

Section 2.3

2.3.1

4 6 3 4

2.3.2

4 4 −8 −8

2.3.3 Undefined

2 2.3.4 2 7

2 0 4

a b 2.3.5 c d 0 0 2.3.6

ad − bc 0 0 ad − bc

−1 2.3.7 5 −6

1 0 3 4 −2 −4

2.3.8

0 0 0 0

2.3.9

0 0 0 0

2.3.10 [0 1] 2.3.11 [10] 77

Chapter 2 3 6 9

1 2 2.3.12 2 4 3 6 2.3.13 [h]

−2 2 2 2.3.14 A2 = , BC = [14 8 2], BD = [6], C 2 = 4 2 2 10 5 DE = 5 , EB = [5 10 15], E 2 = [25] 5

3 3 , 3

1 2 0 −2 −2 1 −2 , CD = 3 , DB = 1 2 1 2 6 4 −2

"

" # " # #" # " # # " # " # " 1 0 0 0 0 1 1 0 1 0 + + 1 0 [ 4 ] [ 3 ] 0 1 0 0 2 2 0 = " # 0 1" #0 2.3.15 = 2 0 1 0 +[ 4 ][ 3 ] [ 1 3 ] +[ 4 ][ 4 ] [1 3] 19 16 [ 19 ] [ 16 ] 2 0 "

1 0 2.3.16 " 0 0

0 1# 0 0

1 3 " 1 3

# "

2 1 + 4# " 0 2 1 + 4 0 # "

0 0 1 #" 0 0 0 1 0 #"

# " 0 1 0 # " 0 0 0 0 0

0 2 1 #" 4 0 2 0 4 #"

2.3.17 We must find all S such that SA = AS, or

3 1 + 5# " 0 3 1 + 5 0 # "

a b c d

0 1 1 #" 3 0 1 1 3 #"

" # 2 1 4# = " 3 2 0 0 4

1 1 0 = 0 0 2

0 2

2 4# 0 0

3 7 " 1 3

# "

# 1 5 3 9# = 2 0 4 0

2 4 0 0

3 7 1 3

5 9 2 4

a b . c d

a 2b a = c 2d 2c

b , meaning that b = 2b and c = 2c, so b and c must be zero. 2d 1 0 a 0 . ) commute with We see that all diagonal matrices (those of the form 0 2 0 d So

a b 1 2 1 2 a b a b . = . Now we want 2.3.18 As in Exercise 2.3.17, we let A = c d 0 1 0 1 c d c d a + 2c b + 2d a 2a + b , revealing that c = 0 (since a + 2c = a) and a = d (since b + 2d = 2a + b). = So, c d c 2c + d a b . Thus B is any matrix of the form 0 a

a b 0 −2 0 −2 a b a b . = . We want c d 2 0 2 0 c d c d 2b −2a −2c −2d Thus, = , meaning that c = −b and d = a. 2d −2c 2a 2b 0 −2 a b . commute with We see that all matrices of the form 2 0 −b a

2.3.19 Again, let A =

78

Section 2.3

2.3.20 Following the form of Exercise 17, we let A =

Now we want

a b c d

2 3 2 = −3 2 −3

3 2

a b . c d

a b . c d

2a − 3b 3a + 2b 2a + 3c 2b + 3d So, = , revealing that a = d (since 3a + 2b = 2b + 3d) and −b = 2c − 3d 3c + 2d −3a + 2c −3b + 2d c (since 2a + 3c = 2a − 3b). a b Thus B is any matrix of the form . −b a 2.3.21 Now we want

a b c d

2 1 2 a b = . −1 2 −1 c d

1 2

a + 2c b + 2d a + 2b 2a − b . So a + 2b = a + 2c, or c = b, and 2a − b = b + 2d, revealing = Thus, 2a − c 2b − d c + 2d 2c − d d = a − b. (The other two equations are redundant.) 1 2 a b . commute with All matrices of the form 2 −1 b a−b

a b a b 1 1 1 1 a b 2.3.22 As in Exercise 17, we let A = . Now we want = . c d c d 1 1 1 1 c d So,

a+c a+b a+b = a+c c+d c+d

b+d , revealing that a = d (since a + b = b + d) and b = c (since a + c = a + b). b+d

Thus B is any matrix of the form

2.3.23 We want

a c

b d

a b . b a

a b 1 3 1 3 . = c d 2 6 2 6

a + 2b 3a + 6b a + 3c b + 3d Then, = . So a + 2b = a + 3c, or c = 23 b, and 3a + 6b = b + 3d, revealing c + 2d 3c + 6d 2a + 6c 2b + 6d d = a + 35 b. The other two equations are redundant. Thus all matrices of the form

a b 5 2 b a + 3 3b

commute with

a b 2.3.24 Following the form of Exercise 2.3.17, we let A = d e g h

a b Now we want d e g h

c 2 0 f 0 3 i 0 0

2 0 0 0 = 0 3 0 0 4

0 a b 0d e 4 g h 79

1 2

3 . 6

c f . i

c f . i

Chapter 2 2a 2a 3b 4c So, 2d 3e 4f = 3d 4g 2g 3h 4i chosen freely.

2b 3e 4h

2c 3f , which forces b, c, d, f, g and h to be zero. a, e and i, however, can be 4i

a Thus B is any matrix of the form 0 0

0 0 e 0. 0 i

0 2 0 = 0 2 0

a 2.3.25 Now we want d g 2a 3b 2c or, 2d 3e 2f = 2g 3h 2i must all be zero.

b c 2 0 e f 0 3 h i 0 0 2a 2b 2c 3d 3e 3f . 2g 2h 2i

a 0 Thus all matrices of the form 0 e g 0

0 0 a b 3 0d e 0 2 g h

c f , i

So, 3b = 2b, 2d = 3d, 3f = 2f and 3h = 2h, meaning that b, d, f and h

2 0 0 c 0 commute with 0 3 0 . 0 0 2 i

a 2.3.26 Following the form of Exercise 2.3.17, we let A = d g a 2 0 0 2 0 0 a b c Then we want d e f 0 2 0 = 0 2 0 d g 0 0 3 0 0 3 g h i 2a 2b 2c 2a 2b 3c So, 2d 2e 3f = 2d 2e 2f . Thus c, f, g and h 3g 3h 3i 2g 2h 3i a b 0 d e 0. 0 0 i

c f . i b c e f . h i

b e h

must be zero, leaving B to be any matrix of the form

2.3.27 We will prove that A(C + D) = AC + AD, repeatedly using Theorem 1.3.10a: A(~x + ~y ) = A~x + A~y . Write B = [~v1 . . . ~vm ] and C = [w ~1 . . . w ~ m ]. Then A(C + D) = A[~v1 + w ~ 1 · · · ~vm + w ~ m ] = [A~v1 + Aw ~ 1 · · · A~vm + Aw ~ m ], and AC + AD = A[~v1 · · · ~vm ] + A[w ~1 · · · w ~ m ] = [A~v1 + Aw ~ 1 · · · A~vm + Aw ~ m ]. The results agree. 2.3.28 The ijth entries of the three matrices are p X

h=1

.

(kaih )bhj ,

p X

aih (kbhj ), and k

h=1

p X

h=1

80

aih bhj

!

Section 2.3 The three results agree. 2.3.29 a Dα Dβ and Dβ Dα are the same transformation, namely, a rotation through α + β.

cos α − sin α b Dα Dβ = sin α cos α

cos β − sin β sin β cos β

=

cos α cos β − sin α sin β sin α cos β + cos α sin β

=

cos(α + β) sin(α + β)

− cos α sin β − sin α cos β − sin α sin β + cos α cos β

− sin(α + β) cos(α + β)

Dβ Dα yields the same answer. 2.3.30 a See Figure 2.44.

Figure 2.44: for Problem 2.4.30. The vectors ~x and T (~x) have the same length (since reflections leave the length unchanged), and they enclose an angle of 2(α + β) = 2 · 30◦ = 60◦ b Based on the answer in part (a), we conclude that T is a rotation through 60◦ .

c The matrix of T is

cos(60◦ ) sin(60◦ )

1 − sin(60◦ ) = √2 ◦ 3 cos(60 )

√

−

2

3 2 1 2

.

w ~1 ~ w 2.3.31 Write A in terms of its rows: A = 2 (suppose A is n × m). ··· w ~n

81

Chapter 2 We can think of this as a partition into n w ~ 1B w ~1 ~ B ~ w w 1 × m matrices. Now AB = 2 B = 2 (a product of partitioned matrices). ··· ··· w ~ nB w ~n

We see that the ith row of AB is the product of the ith row of A and the matrix B.

a b 1 0 1 0 a b 1 0 1 0 a b 2.3.32 Let X = . Then we want X = X, or = , or cd 0 0 0 0 c d 0 0 0 0 c d a 0 a b 0 1 0 1 a 0 0 1 = , meaning that b = c = 0. Also, we want X = X, or = c 0 0 0 0 0 0 0 0 d 0 0 0 1 a 0 0 a 0 d a 0 , or = so a = d. Thus, X = = aI2 must be a multiple of the identity 0 0 0 d 0 0 0 0 0 a matrix. (X will then commute with any 2 × 2 matrix M , since XM = aM = M X.) 2.3.33 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between A = −I2 and I2 . The matrix A describes a reflection about the origin. Alternatively one can say A represents a rotation by 180◦ = π. Since A2 is the −1 0 identity, A1000 is the identity and A1001 = A = . 0 −1 2.3.34 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between A and I2 . The matrix a reflection A describes 1 0 2 1000 1001 . about the x axis. Because A is the identity, A is the identity and A =A= 0 −1 2.3.35 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between A and I2 . The matrix A describes a reflection 0 1 2 1000 1001 about the diagonal x = y. Because A is the identity, A is the identity and A =A= . 1 0

1 0

1 0 −2 1

1 3 ,A = 0

2.3.36 A = and A = . The power An represents a horizontal shear along the 1 1001 x-axis. The shear strength increases linearly in n. We have A1001 = . 0 1 2

2 1

3 1

4

1 0

4 1

1 0 1 0 , A3 = and A4 = . The power An represents a vertical shear along −3 1 −4 1 1 0 the y axis. The shear magnitude increases linearly in n. We have A1001 = . −1001 1

2.3.37 A2 =

−1 0 , A3 = −A, A4 = I2 . The matrix A represents the rotation through π/2 in the 0 −1 counterclockwise Since A4 is the identity matrix, we know that A1000 is the identity matrix and direction. 0 −1 A1001 = A = . 1 0

2.3.38 A2 =

−1 1 0 1 , A4 = −I2 . The matrix A describes a rotation by π/4 in the , A3 = √12 −1 1 −1 0 clockwise direction. Because A8 is the identity matrix, we know that A1000 is the identity matrix and A1001 =

2.3.39 A2 =

82

Section 2.3 √ A = (1/ 2)

1 −1

1 1

.

√ −1 3 √ , A3 = I2 , A4 = A. The matrix A describes a rotation by 120◦ = 2π/3 in the − 3 −1 3 999 counterclockwise direction. Because is the identity matrix and √ A is the identity matrix, we know that A 3 −1 √ . A1001 = A2 = A−1 = 21 − 3 −1

2.3.40 A2 =

1 2

2.3.41 A2 = I2 , A3 = A, A4 = I2 . The power An alternates between I2 for even n and A for odd n. Therefore A1001 = A. The matrix represents a reflection about a line. 2.3.42 An = A. The matrix A represents a projection on the line x = y spanned by the vector 1 1 A1001 = A = (1/2) . 1 1 2.3.43 An example is A =

1 0

0 −1

−1 √ 3

0 −1 1 0

√ − 3 , the rotation through 2π/3. See Problem 2.3.40. −1

2.3.46 For example, A =

1 1

1 1

1 , the orthogonal projection onto the line spanned by . 1

2.3.47 For example, A =

1 2

1 1

1 1

1 . , the orthogonal projection onto the line spanned by 1

. We have

.

1 2

2.3.48 For example, the shear A =

1 1

, representing the reflection about the horizontal axis.

2.3.44 A rotation by π/2 given by the matrix A =

2.3.45 For example, A = (1/2)

1 1/10 0 1

.

1 0 −1 0 represents the reflection about the x-axis, while F A = represents the reflection 0 −1 0 1 about the y-axis. (See Figure 2.45.)

2.3.49 AF =

0 0 1 represents a reflection about the line x = y, while GC = −1 1 0 about the line x = −y. (See Figure 2.46.)

2.3.50 CG =

2.3.51 F J = JF = Figure 2.47.)

−1 −1 1 −1

−1 0

represents a reflection

both represent a rotation through 3π/4 combined with a scaling by

83

√ 2. (See

Chapter 2

A

F

AF

A

F

FA

Figure 2.45: for Problem 2.3.49. 0.2 −1.4 . Since H represents a rotation and J represents a rotation through π/4 combined 2.3.52 JH = HJ = 1.4 0.2 √ with a scaling √ by 2, the products in either order will be the same, representing a rotation combined with a scaling by 2. (See Figure 2.48.)

0 −1 0 1 2.3.53 CD = represents the rotation through π/2, while DC = represents the rotation 1 0 −1 0 through −π/2. (See Figure 2.49.)

−0.6 −0.8 2.3.54 BE = represents the rotation through the angle θ = arccos(−0.6) ≈ 2.21, while EB = 0.8 −0.6 −0.6 0.8 represents the rotation through −θ. (See Figure 2.50.) −0.8 −0.6 2.3.55 We need to solve the matrix equation

1 2

2 4

a b c d 84

=

0 0 0 0

,

Section 2.3

C

G

CG

C

G

GC

Figure 2.46: for Problem 2.3.50. which amounts to solving the system a + 2c = 0, 2a + 4c = 0, b + 2d = 0 and 2b + 4d = 0. The solutions are of −2c −2d , where c, d are arbitrary constants. the form a = −2c and b = −2d. Thus X = c d 2.3.56 Proceeding as in Exercise 55, we find X = 2.3.57 We need to solve the matrix equation

1 3

2 5

, where b and d are arbitrary.

−2b b −2d d

a b c d

=

1 0 0 1

,

which amounts to solving the system a + 2c = 1, 3a + 5c = 0, b + 2d = 0 and 3b + 5d = 1. The solution is −5 2 X= . 3 −1 2.3.58 Proceeding as in Exercise 57, we find X =

−5 2 3 −1 85

.

Chapter 2

F

J

and FJ

and F

J

JF

Figure 2.47: for Problem 2.3.51. 2.3.59 The matrix equation

a c

b d

2 1 4 2

=

1 0 0 1

has no solutions, since we have the inconsistent equations 2a + 4b = 1 and a + 2b = 0. 2.3.60 Proceeding as in Exercise 59, we find that this equation has no solutions. 2.3.61 We need to solve the matrix equation

1 2 0 1

3 2

a b c d = 1 0 e f

0 1

,

which amounts to solving the system a+ 2c + 3e = 0, c + 2e = 0, b + 2d + 3f = 0 and d + 2f = 1. The solutions e+1 f −2 are of the form X = −2e 1 − 2f , where e, f are arbitrary constants. e f 86

Section 2.3

37 ◦

H

J

and HJ

and H

J

37 ◦

JH

Figure 2.48: for Problem 2.3.52.

e − 5/3 2.3.62 Proceeding as in Exercise 61, we find X = −2e + 4/3 e

f + 2/3 −2f − 1/3 , where e, f are arbitrary constants. f

2.3.63 The matrix equation

1 4 2 5 a b d e 3 6

c f

1 0 = 0 1 0 0

0 0 1

has no solutions, since we have the inconsistent equations a + 4d = 1, 2a + 5d = 0, and 3a + 6d = 0. 2.3.64 The matrix equation

1 0 2 1 a b d e 3 2

c f

1 0 = 0 1 0 0

0 0 1

has no solutions, since we have the inconsistent equations a = 1, 2a + d = 0 and 3a + 2d = 0. 87

Chapter 2

C

D

CD

C

D

DC

Figure 2.49: for Problem 2.3.53.

2 2.3.65 With X = , we have to solve X = 0 can be arbitrary. The general solution is X = 0 a b 0 c

a2 ab + bc 0 c2 b . 0

=

0 0 0 0

. This means a = 0, c = 0 and b

0 0 c 0 then the diagonal entries of X 3 will be a3 , c3 , and f 3 . Since we want X 3 = 0, we must e f 0 0 0 have a = c = f = 0. If X = b 0 0 , then a direct computation shows that X 3 = 0. Thus the solutions d e 0 0 0 0 are of the form X = b 0 0 , where b, d, e are arbitrary. d e 0

a 2.3.66 If X = b d

2.3.67 For a horizontal shear, A =

1 k 0 1

0 2 , we have (A − I2 ) = 0 88

k 0

2

=

0 0 0 0

. Note that A~x − ~x =

Section 2.4

B

E

BE

B

E

EB

Figure 2.50: for Problem 2.3.54. A2 ~x − A~x for all vectors ~x, as illustrated in the accompanying figure. This equation means that A2 ~x − 2A~x + ~x = (A − I2 )2 ~x = ~0. Analogous results hold for vertical shears. x

A2 x

2.3.68 Let ~v1 , . . . , ~vn be the columns of the matrix X. Solving the matrix equation AX = In amounts to solving the linear systems A~vi = ~ei for i = 1, . . . , n. Since A is a n × m matrix of rank n, all these systems are consistent, so that the matrix equation AX = In does have at least one solution. If n < m, then each of the systems A~vi = ~ei has infinitely many solutions, so that the matrix equation AX = In has infinitely many solutions as well. See the examples in Exercices 2.3.57,2.3.61 and 2.3.62. 2.3.69 Let ~v1 , . . . , ~vn be the columns of the matrix X. Solving the matrix equation AX = In amounts to solving the linear systems A~vi = ~ei for i = 1, . . . , n. Since A is an n × n matrix of rank n, all these systems have a unique solution, by Theorem 1.3.4, so that the matrix equation AX = In has a unique solution as well.

89

Chapter 2

Section 2.4 .. 2.4.1 rref 2 3.. 5 8.. "

1

0

0

1

.. 2.4.2 rref 1 1.. 1 1..

1

0

0

1

.. . 0 2 2.4.3 rref . 1 1..

1

0

0

1

"

"

#

. 1 0.. = . 0 1..

#

. 1 1.. = . 0 0..

#

. 1 0.. = . 0 1..

8

1 2.4.5 rref 1 1

2 2 1 3 1 = 0 1 3 0

−3 , so that 2

−5

0

1 , so that −1

1 − 21

1

1 2

0

2.4.4 Use Theorem 2.4.5; the inverse is

1 2 3 0 2 = 0 0 0 3

0 1

0 0 2.4.8 Use Theorem 2.4.5; the inverse is 0 1 1 0 1 2.4.9 rref 1 1

1 1 1 1 1 = 0 1 1 0

−1

fails to be invertible.

−1

=

=

8 −3 . −5 2

− 21 1 2

1 0

.

.

1 −2 . 1

2 0 0 1 , so that the matrix fails to be invertible, by Theorem 2.4.3. 0 0

0 2 1 1

1 2 1 −2 1 2

−1

1 −2 1 2.4.6 Use Theorem 2.4.5; the inverse is 0 0 0 1 2.4.7 rref 0 0

1 1

3 8

0 4 1 −1 , so that the matrix fails to be invertible, by Theorem 2.4.3. 0 0

1 1

, so that

3 2 1 2 − 32

2 5

1 0 . 0

1 1 0 0 , so that the matrix fails to be invertible, by Theorem 2.4.3. 0 0

3 2.4.10 Use Theorem 2.4.5; the inverse is −3 1

1 2.4.11 Use Theorem 2.4.5; the inverse is 0 0

−3 1 5 −2 . −2 1 0 −1 1 0 . 0 1 90

Section 2.4 5 0 2.4.12 Use Theorem 2.4.5; the inverse is −2 0

−20 −1 6 3

1 −2 2.4.13 Use Theorem 2.4.5; the inverse is 1 0

0 0 1 0 −2 1 1 −2

0 0 . 0 1

3 −1 2.4.14 Use Theorem 2.4.5; the inverse is 0 0

−5 0 2 0 0 5 0 −2

0 0 . −2 1

−6 9 2.4.15 Use Theorem 2.4.5; the inverse is −5 1

9 −5 −1 −5 −5 9 2 −3

1 2 −3 1

−2 −7 0 0 1 2 0 1

2.4.16 Solving for x1 and x2 in terms of y1 and y2 we find that x1 x2

= −8y1 + 5y2 = 5y1 − 3y2

2.4.17 We make an attempt to solve for x1 and x2 in terms of y1 and y2 : x1 + 2x2 = y1 x1 + 2x2 = y1 −−−→ . 0 = −4y1 + y2 4x1 + 8x2 = y2 −4(I) This system has no solutions (x1 , x2 ) for some (y1 , y2 ), and infinitely many solutions for others; the transformation fails to be invertible. 2.4.18 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 we find that x1 x2 x3

= y3 = y1 = y2

2.4.19 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 , we find that x1 x2 x3

= 3y1 − 52 y2 + 21 y3

= −3y1 + 4y2 − y3 = y1 − 32 y2 + 21 y3

2.4.20 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 we find that x1 x2 x3

= −8y1 − 15y2 + 12y3 = 4y1 + 6y2 − 5y3 = −y1 − y2 + y3

2.4.21 f (x) = x2 fails to be invertible, since the equation f (x) = x2 = 1 has two solutions, x = ±1. 91

Chapter 2 2.4.22 f (x) = 2x fails to be invertible, since the equation f (x) = 2x = 0 has no solution x. 2.4.23 Note that f ′ (x) = 3x2 + 1 is always positive; this implies that the function f (x) = x3 + x is increasing throughout. Therefore, the equation f (x) = b has at most one solution x for all b. (See Figure 2.51.) Now observe that limx→∞ f (x) = ∞ and limx→−∞ f (x) = −∞; this implies that the equation f (x) = b has at least one solution x for a given b (for a careful proof, use the intermediate value theorem; compare with Exercise 2.2.47c).

Figure 2.51: for Problem 2.3.23. 2.4.24 We can write f (x) = x3 − x = x(x2 − 1) = x(x − 1)(x + 1). The equation f (x) = 0 has three solutions, x = 0, 1, −1, so that f (x) fails to be invertible. 2.4.25 Invertible, with inverse

x1 x2

=

√ 3 y 1 y2

2.4.26 Invertible, with inverse

x1 x2

=

√ 3 y2 − y1 y1

2.4.27 This transformation fails to be invertible, since the equation

x1 + x2 x1 x2

22 13 −16 −3 2.4.28 We are asked to find the inverse of the matrix A = 8 9 5 4 1 −2 9 − 25 5 −22 60 −2 −1 We find that A = . 4 −9 41 −112 −9 17 80 222

1 1 −II → 0 k−1 0 k 2 − 1 −3(II)

0 2−k 1 k−1 0 k 2 − 3k + 2 92

0 has no solution. = 1

8 3 −2 −2 . 7 2 3 1

T −1 is the transformation from R4 to R4 with matrix A−1 .

2.4.29 Use Theorem 2.4.3: 1 1 1 1 1 1 2 k −I → 0 1 1 4 k 2 −I 0 3

Section 2.4 The matrix is invertible if (and only if) k 2 − 3k + 2 = (k − 2)(k − 1) 6= 0, in which case we can further reduce it to I3 . Therefore, the matrix is invertible if k 6= 1 and k 6= 2. 2.4.30 Use Theorem 2.4.3: −1 0 1 b −−−−→ −1 0 c I ↔ II 0 −b −b −c 0

0 1 −c

−−−−→ 1 c ÷(−1) 0 b −b 0

1 0 −c −→ 0 1 b 0 0 0 +b(I) + c(II)

0 1 −c

This matrix fails to be invertible, regarless of the values of b and c.

−c b 0

2.4.31 Use Theorem 2.4.3; first assume that a 6= 0.

0 −a −b

a 0 −c

0

− ac

1

0

a b ÷a → 0 −c − bc 0 a

1

1

0 0

−a b swap : c → 0 I ↔ II −b 0

−c

1 0 c ÷(−a) → 0 a b −c 0 −b − ac

b a − bc a

Now consider the case when a = 0: −b −c 0 0 b swap : 0 → 0 0 0 c I ↔ III 0 0 −b −c 0

0 It follows that the matrix −a −b

+c(II)

1

→ 0 0

− ac

0

→ b +b(I) 0

a −c

0 − ac b a

1 0

0

0 c : The second entry on the diagonal of rref will be 0. b

a b 0 c fails to be invertible, regardless of the values of a, b, and c. −c 0

2.4.32 Use Theorem 2.4.9. a b If A = is a matrix such that ad − bc = 1 and A−1 = A, then c d A−1 =

1 ad−bc

d −c

−b d = a −c

−b a b = , so that b = 0, c = 0, and a = d. a c d

The condition ad − bc = a2 = 1 now implies that a = d = 1 or a = d = −1. This leaves only two matrices A, namely, I2 and −I2 . Check that these two matrices do indeed satisfy the given requirements. 2.4.33 Use Theorem 2.4.9. 1 The requirement A−1 = A means that − a2 +b 2

a2 + b2 = 1.

−a −b 93

−b a

=

a b

b . This is the case if (and only if) −a

Chapter 2 2.4.34 a By Theorem 2.4.3, A is invertible if (and only if) a, b, and c are all nonzero. 1 0 0 a 0 1b 0 . 0

0

In this case, A−1 =

1 c

b In general, a diagonal matrix is invertible if (and only if) all of its diagonal entries are nonzero. 2.4.35 a A is invertible if (and only if) all its diagonal entries, a, d, and f , are nonzero. b As in part (a): if all the diagonal entries are nonzero. . c Yes, A−1 will be upper triangular as well; as you construct rref[A..In ], you will perform only the following row operations: • divide rows by scalars • subtract a multiple of the jth row from the ith row, where j > i. Applying these operations to In , you end up with an upper triangular matrix. d As in part (b): if all diagonal entries are nonzero. 2.4.36 If a matrix A can be transformed into B by elementary row operations, then A is invertible if (and only if) B is invertible. The claim now follows from Exercise 35, where we show that a triangular matrix is invertible if (and only if) its diagonal entries are nonzero. 2.4.37 Make an attempt to solve the linear equation ~y = (cA)~x = c(A~x) for ~x: A~x = 1c ~y , so that ~x = A−1 1c ~y = 1c A−1 ~y . This shows that cA is indeed invertible, with (cA)−1 = 1c A−1 .

2.4.38 Use Theorem 2.4.9; A−1 =

1 −1

−1 −k 1 = 0 1 0

k (= A). −1

2.4.39 Suppose the ijth entry of M is k, and all other entries are as in the identity matrix. Then we can find . rref[M ..In ] by subtracting k times the jth row from the ith row. Therefore, M is indeed invertible, and M −1 differs from the identity matrix only at the ijth entry; that entry is −k. (See Figure 2.52.) 2.4.40 If you apply an elementary row operation to a matrix with two equal columns, then the resulting matrix will also have two equal columns. Therefore, rref(A) has two equal columns, so that rref(A) 6= In . Now use Theorem 2.4.3. 2.4.41 a Invertible: the transformation is its own inverse. b Not invertible: the equation T (~x) = ~b has infinitely many solutions if ~b is on the plane, and none otherwise. 94

Section 2.4

Figure 2.52: for Problem 2.3.39. c Invertible: The inverse is a scaling by

1 5

(that is, a contraction by 5). If ~y = 5~x, then ~x = 51 ~y .

d Invertible: The inverse is a rotation about the same axis through the same angle in the opposite direction. 2.4.42 Permutation matrices are invertible since they row reduce to In in an obvious way, just by row swaps. The . . inverse of a permutation matrix A is also a permutation matrix since rref[A..In ] = [In ..A−1 ] is obtained from . [A..In ] by a sequence of row swaps. 2.4.43 We make an attempt to solve the equation ~y = A(B~x) for ~x: B~x = A−1 ~y , so that ~x = B −1 (A−1 ~y ). 1 0 2.4.44 a rref(M4 ) = 0 0

0 −1 −2 1 2 3 , so that rank(M4 ) = 2. 0 0 0 0 0 0

b To simplify the notation, we introduce the row vectors ~v = [1 1 . . . 1] and w ~ = [0 n 2n . . . (n − 1)n] with n components. ~v + w ~ ~ −2(I) 2~v + w Then we can write Mn in terms of its rows as Mn = . ··· ... n~v + w ~ −n(I) ~v + w ~ −w ~ Applying the Gauss-Jordan algorithm to the first column we get −2w ~ . ... −(n − 1)w ~ All the rows below the second are scalar multiples of the second; therefore, rank(Mn ) = 2. c By part (b), the matrix Mn is invertible only if n = 1 or n = 2. 2.4.45 a Each of the three row divisions requires three multiplicative operations, and each of the six row subtractions requires three multiplicative operations as well; altogether, we have 3 · 3 + 6 · 3 = 9 · 3 = 33 = 27 operations. 95

Chapter 2 . b Suppose we have already taken care of the first m columns: [A..In ] has been reduced the matrix in Figure 2.53.

Figure 2.53: for Problem 2.3.45b. Here, the stars represent arbitrary entries. Suppose the (m+1)th entry on the diagonal is k. Dividing the (m+1)th row by k requires n operations: n−m−1 to the left of the dotted line not counting the computation kk = 1 , and m + 1 to the right of the dotted line including k1 . Now the matrix has the form shown in Figure 2.54.

Figure 2.54: for Problem 2.4.45b. Eliminating each of the other n − 1 components of the (m + 1)th column now requires n multiplicative operations (n − m − 1 to the left of the dotted line, and m + 1 to the right). Altogether, it requires n + (n − 1)n = n2 operations to process the mth column. To process all n columns requires n · n2 = n3 operations. c The inversion of a 12 × 12 matrix requires 123 = 43 33 = 64 · 33 operations, that is, 64 times as much as the inversion of a 3 × 3 matrix. If the inversion of a 3 × 3 matrix takes one second, then the inversion of a 12 × 12 matrix takes 64 seconds. 2.4.46 Computing A−1~b requires n3 + n2 operations: First, we need n3 operations to find A−1 (see Exercise 45b) and then n2 operations to compute A−1~b (n multiplications for each component). . How many operations are required to perform Gauss-Jordan eliminations on [A..~b]? Let us count these operations “column by column.” If m columns of the coefficient matrix are left, then processing the next column requires nm operations (compare with Exercise 45b). To process all the columns requires = n · n + n(n − 1) + · · · + n · 2 + n · 1 = n(n + n − 1 + · · · + 2 + 1) = n n(n+1) 2 96

n3 +n2 2

operations.

Section 2.4 only half of what was required to compute A−1~b. We mention in passing that one can reduce the number of operations further (by about 50% for large matrices) by performing the steps of the row reduction in a different order. 2.4.47 Let f (x) = x2 ; the equation f (x) = 0 has the unique solution x = 0. 2.4.48 Consider the linear system A~x = ~0. The equation A~x = ~0 implies that BA~x = ~0, so ~x = ~0 since BA = Im . Thus the system A~x = ~0 has the unique solution ~x = ~0. This implies m ≤ n, by Theorem 1.3,3. Likewise the linear system B~y = ~0 has the unique solution ~y = ~0, implying that n ≤ m. It follows that n = m, as claimed.

0.293 0 0 0.707 2.4.49 a A = 0.014 0.207 0.017 , I3 − A = −0.014 0.044 0.01 0.216 −0.044 1.41 0 0 0.0274 (I3 − A)−1 = 0.0267 1.26 0.0797 0.0161 1.28

0 −0.017 0.784

0 0.793 −0.01

1.41 1 b We have ~b = 0 , so that ~x = (I3 − A)−1~e1 = first column of (I3 − A)−1 ≈ 0.0267 . 0.0797 0 c As illustrated in part (b), the ith column of (I3 − A)−1 gives the output vector required to satisfy a consumer demand of 1 unit on industry i, in the absence of any other consumer demands. In particular, the ith diagonal entry of (I3 − A)−1 gives the output of industry i required to satisfy this demand. Since industry i has to satisfy the consumer demand of 1 as well as the interindustry demand, its total output will be at least 1. d Suppose the consumer demand increases from ~b to ~b + ~e2 (that is, the demand on manufacturing increases by one unit). Then the output must change from (I3 − A)−1~b to (I3 − A)−1 (~v + ~e2 ) = (I3 − A)−1~b + (I3 − A)−1~e2 = (I3 − A)−1~b+ (second column of (I3 − A)−1 ). The components of the second column of (I3 −A)−1 tells us by how much each industry has to increase its output. e The ijth entry of (In − A)−1 gives the required increase of the output xi of industry i to satisfy an increase of the consumer demand bj on industry j by one unit. In the language of multivariable calculus, this quantity is ∂xi ∂bj . 2.4.50 Recall that 1 + k + k 2 + · · · =

1 1−k .

The top left entry of I3 − A is I − k, and the top left entry of (I3 − A)−1 will therefore be

1 − k ∗ ∗

0 0 ∗ ∗ ∗ ∗

.. . 1 .. . 0 .. . 0

0 0 ÷(1 − k) −→ 1 0 0 1

1 0

∗ ∗ ∗ ∗

. 0 .. . ∗ .. . ∗ ..

1 1−k

0 0

→ . . . (first row will remain unchanged). 97

0 0

1 0 0 1

1 1−k ,

as claimed:

Chapter 2 In terms of economics, we can explain this fact as follows: The top left entry of (I3 − A)−1 is the output of industry 1 (Agriculture) required to satisfy a consumer demand of 1 unit on industry 1. Producting this one unit to satisfy the consumer demand will generate an extra demand of k = 0.293 units on industry 1. Producting these k units in turn will generate an extra demand of k · k = k 2 units, and so forth. We are faced with an infinite series of (ever smaller) demands, 1 + k + k 2 + · · · . 2.4.51 a Since rank(A)< n, the matrix E =rref(A) will not have a leading one in the last row, and all entries in the last row of E will be zero. 0 0 . . Let ~c = . . Then the last equation of the system E~x = ~c reads 0 = 1, so this system is inconsistent. 0 1 .. ~ Now, we can “rebuild” b from ~c by performing the reverse row-operations in the opposite order on E .~c until . we reach A..~b . Since E~x = ~c is inconsistent, A~x = ~b is inconsistent as well. b Since rank(A)≤ min(n, m), and m < n, rank(A) < n also. Thus, by part a, there is a ~b such that A~x = ~b is inconsistent. 0 0 . 0 0 2.4.52 Let ~b = . Then A..~b = 1 0 0 1 an inconsistency in the third row. 2.4.53 a A − λI2 =

.. . .. . .. . .. .

1 2 2 4 3 6 4 8

0

1 0

. 0 . We find that rref A..~b = 0 1 0 0 1 0 0

0

. 0 .. 0 . 2 .. 0 , which has . 0 .. 1 . 0 .. 0

1 . 5−λ

3−λ 3

This fails to be invertible when (3 − λ)(5 − λ) − 3 = 0, or 15 − 8λ + λ2 − 3 = 0, or 12 − 8λ + λ2 = 0 or (6 − λ)(2 − λ) = 0. So λ = 6 or λ = 2. b For λ = 6, A − λI2 =

−3 1 . 3 −1

The system (A − 6I2 )~x = ~0 has the solutions

1 t , for example. , where t is an arbitrary constant. Pick ~x = 3 3t

1 1 For λ = 2, A − λI2 = . 3 3 98

Section 2.4 The system (A − 2I2 )~x = ~0 has the solutions example. c For λ = 6, A~x =

3 3

For λ = 2, A~x =

3 3

1 t , for , where t is an arbitrary constant. Pick ~x = −1 −t

1 6 1 . =6 = 3 18 3 1 1 2 1 = =2 . 5 −1 −2 −1 1 5

1−λ 10 2.4.54 A − λI2 = . This fails to be invertible when det(A − λI2 ) = 0, −3 12 − λ so 0 = (1 − λ)(12 − λ) + 30 = 12 − 13λ + λ2 + 30 = λ2 − 13λ + 42 = (λ − 6)(λ − 7). In order for this to be zero, λ must be 6 or 7. −5 10 . We solve the system (A − 6I2 ) ~x = ~0 and find that the solutions are of the If λ = 6, then A − 6I2 = −3 6 2t 2 form ~x = . For example, when t = 1, we find ~x = . t 1 −6 10 If λ = 7, then A − 7I2 = . Here we solve the system (A − 7I2 ) ~x = ~0, this time finding that our −3 5 5 5t . . For example, for t = 1, we find ~x = solutions are of the form ~x = 3 3t 1/2 0 . The linear transformation defined by A is a 0 1/2 −1 scaling by a factor 2 and A defines a scaling by 1/2. The determinant of A is the area of the square spanned 2 0 by ~v = and w ~= . The angle θ from ~v to w ~ is π/2. (See Figure 2.55.) 0 2

2.4.55 The determinant of A is equal to 4 and A−1 =

w=

0 2 θ= π 2

v=

2 0

Figure 2.55: for Problem 2.4.55.

−3−1 0 2.4.56 The determinant of A is 9. The matrix is invertible with inverse A = . The linear 0 −3−1 transformation defined by A is a reflection about the origin combined with a scaling by a factor 3. The inverse −1

99

Chapter 2 defines a reflection about combined with the origin a scaling by a factor 1/3. The determinant is the area of the −3 0 square spanned by ~v = and w ~= . The angle θ from ~v to w ~ is π/2. (See Figure 2.56.) 0 −3

v=

−3 0 θ= π 2

w=

0 −3

Figure 2.56: for Problem 2.4.56. 2.4.57 The determinant of A is −1. Matrix A is invertible, with A−1 = A. Matrices A and A−1 define reflection cos(α/2) . The absolute value of the determinant of A is the area of the about the line spanned by the ~v = sin(α/2) cos(α) sin(α) unit square spanned by ~v = and w ~ = . The angle θ from ~v to w ~ is −π/2. (See Figure sin(α) − cos(α) 2.57.)

v=

cos α sin α θ=− π 2

w=

sin α − cos α

Figure 2.57: for Problem 2.4.57.

cos(α) sin(α) 2.4.58 The determinant of A is 1. The matrix is invertible with inverse A = . The linear − sin(α) cos(α) transformation defined by A is a rotation by angle α in the counterclockwise direction. The inverse represents a rotation by the angle α in the clockwise direction. The determinant of A is the area of the unit square spanned −1

100

Section 2.4

by ~v =

cos(α) sin(α)

and ~v =

θ= π 2

w=

v=

− sin(α) cos(α)

. The angle θ from ~v to w ~ is π/2. (See Figure 2.58.)

cos α sin α

α

− sin α cos α

Figure 2.58: for Problem 2.4.58.

0.6 0.8 . The matrix A −0.8 0.6 represents the rotation through the angle α = arccos(0.6). Its inverse represents a rotation by thesame angle 0.6 in the clockwise direction. The determinant of A is the area of the unit square spanned by ~v = and 0.8 −0.8 . The angle θ from ~v to w ~ is π/2. (See Figure 2.59.) w ~= 0.6

2.4.59 The determinant of A is 1. The matrix A is invertible with inverse A−1 =

θ= π 2

w=

v=

0.6 0.8

−0.8 0.6

Figure 2.59: for Problem 2.4.59. −1 −1 2.4.60 The determinant of A is −1. The matrix A is invertible with inverse A = A. Matrices A and A define cos(α/2) , where α = arccos(−0.8). The absolute value of the the reflection about the line spanned by ~v = sin(α/2) −0.8 0.6 determinant of A is the area of the unit square spanned by ~v = and w ~= . The angle θ from v 0.6 0.8

101

Chapter 2 to w is −π/2. (See Figure 2.60.)

θ=− π 2

v=

w=

0.6 0.8

−0.8 0.6

Figure 2.60: for Problem 2.4.60.

1 −1 2.4.61 The determinant of A is 2 and A = . The matrix A represents a rotation through the angle 1 1 √ √ −π/4 combined with scaling by 2. describes through a rotation π/4 and scaling by 1/ 2. The determinant of √ 1 1 A is the area of the square spanned by ~v = and w ~ = with side length 2. The angle θ from ~v to −1 1 w ~ is π/2. (See Figure 2.61.) −1

w=

1 2

1 1 θ= π 2

v=

1 −1

Figure 2.61: for Problem 2.4.61.

2.4.62 The determinant of A is 25. The matrix A is a rotation dilation matrix with scaling factor 5 and rotation by 3 −4 is a rotation dilation too an angle arccos(0.6) in the clockwise direction. The inverse A−1 = (1/25) 4 3 with a scaling factor 1/5 angle arccos(0.6). The determinant of A is the area of the parallelogram and rotation 4 3 with side length 5. The angle from ~v to w ~ is π/2. (See Figure 2.62.) and w ~= spanned by ~v = 3 −4 102

Section 2.4

w=

4 3

θ= π 2

v=

3 −4

Figure 2.62: for Problem 2.4.62.

−3 4 1 A. The matrix A represents a reflection 2.4.63 The determinant of A is −25 and A = (1/25) = 25 4 3 −1 about a line combined with a scaling by 5 whilc A represents a reflection about the same line combined with −3 a scaling by 1/5. The absolute value of the determinant of A is the area of the square spanned by ~v = 4 4 and w ~= with side length 5. The angle from ~v to w ~ is −π/2. (See Figure 2.63.) 3 −1

v=

−3 4

θ=− π 2

w=

4 3

Figure 2.63: for Problem 2.4.63.

−1

1 1 0 1

. Both A and A−1 represent horizontal shears. The −1 1 . The angle from and w ~ = determinant of A is the area of the parallelogram spanned by ~v = 1 0 ~v to w ~ is 3π/4. (See Figure 2.64.)

2.4.64 The determinant of A is 1 and A

−1

2.4.65 The determinant of A is 1 and A

=

=

1 −1

0 1

. Both A and A−1 represent vertical shears. The determinant 103

Chapter 2

w=

−1 1

θ= 3π 4

v=

1 0

Figure 2.64: for Problem 2.4.64.

of A is the area of the parallelogram spanned by ~v = Figure 2.65.)

w=

0 1

θ= π 4

v=

1 1

and w ~=

0 1

. The angle from ~v to w ~ is π/4. (See

1 1

Figure 2.65: for Problem 2.4.65. 2.4.66 We can write AB(AB)−1 = A(B(AB)−1 ) = In and (AB)−1 AB = ((AB)−1 A)B = In . By Theorem 2.4.8, A and B are invertible. 2.4.67 Not necessarily true; (A + B)2 = (A + B)(A + B) = A2 + AB + BA + B 2 6= A2 + 2AB + B 2 if AB 6= BA. 2.4.68 True; apply Theorem 2.4.7 to B = A. 2.4.69 Not necessarily true; consider the case A = In and B = −In . 104

Section 2.4 2.4.70 Not necessarily true; (A − B)(A + B) = A2 + AB − BA − B 2 6= A2 − B 2 if AB 6= BA. 2.4.71 True; ABB −1 A−1 = AIn A−1 = AA−1 = In . 2.4.72 Not necessarily true; the equation ABA−1 = B is equivalent to AB = BA (multiply by A from the right), which is not true in general. 2.4.73 True; (ABA−1 )3 = ABA−1 ABA−1 ABA−1 = AB 3 A−1 . 2.4.74 True; (In + A)(In + A−1 ) = In2 + A + A−1 + AA−1 = 2In + A + A−1 . 2.4.75 True; (A−1 B)−1 = B −1 (A−1 )−1 = B −1 A (use Theorem 2.4.7).

1 2 2 1 2 2.4.76 We want A such that A = , so that A = 2 5 1 3 1

1 3

1 2

2 5

−1

8 −3 = . −1 1

2.4.77 We want A such that A~vi = w ~ i , for i = 1, 2, . . . , m, or A[~v1 ~v2 . . . ~vm ] = [w ~1 w ~2 . . . w ~ m ], or AS = B. Multiplying by S −1 from the right we find the unique solution A = BS −1 .

1 2 2 5

3 1 2.4.79 Use the result of Exercise 2.4.77, with S = 1 2 9 3 . A = BS −1 = 15 −2 16

2.4.78 Use the result of Exercise 2.4.77, with S =

A = BS −1

−13 − 8 −3

33 = 21 9

T

T

T

7 and B = 5 3

1 2 ; 3

6 3 and B = ; 2 6

T

2.4.80 P0 −→ P1 , P1 −→ P3 , P2 −→ P2 , P3 −→ P0 L

L

L

L

P0 −→ P0 , P1 −→ P2 , P2 −→ P1 , P3 −→ P3 a. T −1 is the rotation about the axis through 0 and P2 that transforms P3 into P1 . b. L−1 = L c. T 2 = T −1 (See part (a).) d. P0 −→ P1

T ◦L

P0 −→ P2

P1 −→ P2 P2 −→ P3 P3 −→ P0

P1 −→ P3 P2 −→ P1 P3 −→ P0

e.

L◦T

The transformations T ◦ L and L ◦ T are not the same.

105

Chapter 2 P0 P1 P2 P3

L◦T ◦L

−→ P2 −→ P1 −→ P3 −→ P0

This is the rotation about the axis through 0 and P1 that sends P0 to P2 . 2.4.81 Let A be the matrix of T and C the matrix of L. We want that AP0 = P1 , AP1 = P3 , and AP2 = P2 . We 1 −1 −1 1 1 −1 1 . 1 and B = −1 −1 can use the result of Exercise 77, with S = 1 −1 −1 1 −1 1 −1 −1 0 0 1 0 0 . Then A = BS −1 = −1 0 −1 0 0 1 0 Using an analogous approach, we find that C = 1 0 0 . 0 0 1 a b c 2.4.82 a EA = d − 3a e − 3b f − 3c g h k

The matrix EA is obtained from A by an elementary row operation: subtract three times the first row from the second.

a 1 b EA = 4 d g

b 1 4e h

c 1 4f k

The matrix EA is obtained from A by dividing the second row of A by 4 (an elementary row operation).

1 0 c If we set E = 0 0 0 1

1 0 0 1 then 0 0 0 1 0

a b a b c 0 1d e f = g h d e g h k 0

c k , as desired. f

d An elementary n × n matrix E has the same form as In except that either • eij = k(6= 0) for some i 6= j [as in part (a)], or • eii = k(6= 0, 1) for some i [as in part (b)], or • eij = eji = 1, eii = ejj = 0 for some i 6= j [as in part (c)].

2.4.83 Let E be an elementary n × n matrix (obtained from In by a certain elementary row operation), and let F be the elementary matrix obtained from In by the reversed row operation. Our work in Exercise 2.4.82 [parts (a) through (c)] shows that EF = In , so that E is indeed invertible, and E −1 = F is an elementary matrix as well. 2.4.84 a The matrix rref(A) is obtained from A by performing a sequence of p elementary row operations. By Exercise 2.4.82 [parts (a) through (c)] each of these operations can be represented by the left multiplication with an elementary matrix, so that rref(A) = E1 E2 . . . Ep A. 106

Section 2.4

b

A=

0 1

2 3

3 2

3 1

0 1

swap rows 1 and 2, represented by

0 1

1 0

↓

1 0

÷2

, represented by

1 0

0 1 2

↓

1 0

−3(II)

, represented by

1 −3 0 1

1 0

0

↓ rref(A) =

1 0

1 0 1 Therefore, rref(A) = = 0 1 0

−3 1

1 2

0 1

1 0

0 1

2 = E1 E2 E3 A. 3

2.4.85 a Let S = E1 E2 . . . Ep in Exercise 2.4.84a. By Exercise 2.4.83, the elementary matrices Ei are invertible: now use Theorem 2.4.7 repeatedly to see that S is invertible. b A=

1 2 4 8

2 4 4 8

÷2

, represented by

, represented by

−4(I) 1 2 rref(A) = 0 0

Therefore, rref(A) =

S=

1 −4

0 1

1 2

0

1

0 1

1 0 −4 1

2

0

1 2 1 = 0 0 −4

1 0 2 = 1 −2

0 1

1

2

0

0 1

2 4

4 = E1 E2 A = SA, where 8

0 . 1

(There are other correct answers.) 2.4.86 a By Exercise 2.4.84a, In = rref(A) = E1 E2 . . . Ep A, for some elementary matrices E1 , . . . , Ep . By Exercise 2.4.83, the Ei are invertible and their inverses are elementary as well. Therefore, A = (E1 E2 . . . Ep )−1 = Ep−1 . . . E2−1 E1−1 expresses A as a product of elementary matrices. b We can use out work in Exercise 2.4.84 b: −1 −1 −1 1 3 1 0 1 0 0 1 0 1 0 1 1 −3 1 0 1 −3 0 2 = = = 0 1 0 2 0 21 1 0 1 0 1 0 0 1 0 12 0 1 1 3 107

Chapter 2

2.4.87

k 0

1 0

0 1

1 0 1 k represents a vertical shear, represents a horizontal shear, k 1 0 1 0 represents a “scaling in ~e1 direction” (leaving the ~e2 component unchanged), 1 0 represents a “scaling in ~e2 direction” (leaving the ~e1 component unchanged), and k 1 1 . represents the reflection about the line spanned by 1 0

2.4.88 Performing a sequence of p elementary row operations on a matrix A amounts to multiplying A with E1 E2 . . . Ep from the left, where the Ei are elementary matrices. If In = E1 E2 . . . Ep A, then E1 E2 . . . Ep = A−1 , so that a. E1 E2 . . . Ep AB = B, and b. E1 E2 . . . Ep In = A−1 . 2.4.89 Let A and B be two lower triangular n×n matrices. We need to show that the ijth entry of AB is 0 whenever i < j. This entry is the dot product of the ith row of A and the jth column of B, 0 .. . 0 [ai1 ai2 . . . aii 0 . . . 0] · b , which is indeed 0 if i < j. jj .. . bnj

1 2.4.90 a 2 2 ↓

2 3 1 0 6 7 −2I , represented by 0 1 2 4 −2I −2 0

1 0 0

2 3 1 2 1 represented by 0 −2 −2 +II 0

2 3 2 1 , so that 0 −1 1 1 0 0 2 3 2 1 = 0 1 0 0 −2 0 1 1 0 −1

1 0 0 1 0 0

↓

↑ U

↑ E3

0 0 1 0 1 1

1 0 0 1 0 −2 0 0 1

↑ E2

0 1 0 0 0 −2 1 0 1 0 0 1

1 2 0 0 1 02 6 2 2 0 1

↑ A

↑ E1

108

3 7 4

Section 2.4 1 b A = (E3 E2 E1 )−1 U = E1−1 E2−1 E3−1 U = 2 0

↑ M1

c Let L = M1 M2 M3 in part (b);

1 2 Then 2 6 2 2

↑ A

1 0 3 1 7 = 2 2 −1 4

↑ L

1 0 0 0 1 1 00 0 −1 0 1

1 0 0 1 00 2 0 1

↑ M2

↑ M3

1 2 3 0 1 00 2 0 0 −1 1 ↑ U

0 0 1 0 . −1 1

1 we compute L = 2 2 1 2 3 0 1 00 2 0 0 −1 1 ↑ U

1 d We can use the matrix L we found in part (c), but U needs to be modified. Let D = 0 0 (Take the 1 Then 2 2

diagonal entries of the matrix U in part (c)). 2 3 1 0 0 1 0 0 1 2 6 7 = 2 1 00 2 00 1 0 0 2 4 2 −1 1 0 0 −1 ↑ A

↑ L

↑ D

3 1 2

1

↑ U

.

2.4.91 a Write the system L~y = ~b in components: y1 = −3 = 14 −3y1 + y2 , so that y1 = −3, y2 = 14 + 3y1 = 5, y1 + 2y2 + y3 = 9 −y1 + 8y2 − 5y3 + y4 = 33 y3 = 9 − y1 − 2y2 = 2, and y4 = 33 + y1 − 8y2 + 5y3 = 0: −3 5 ~y = . 2 0

1 −1 b Proceeding as in part (a) we find that ~x = . 2 0

a 0 2.4.92 We try to find matrices L = b c

d and U = 0 109

e f

such that

0 0 2 0 . 0 −1

Chapter 2

d a 0 0 1 = 0 b c 1 0

e f

=

ad ae . bd be + cf

Note that the equations ad = 0, ae = 1, and bd = 1 cannot be solved simultaneously: If ad = 0 then a or d is 0 so that ae or bd is zero. 0 1 Therefore, the matrix does not have an LU factorization. 1 0 (m) L(m) 0 U U2 and U = . L3 L4 0 U4 (m) (m) L U L(m) U2 Then A = LU = , so that A(m) = L(m) U (m) , as claimed. L3 U (m) L3 U2 + L4 U4

2.4.93 a Write L =

b By Exercise 2.4.66, the matrices L and U are both invertible. By Exercise 2.4.35, the diagonal entries of L and U are all nonzero. For any m, the matrices L(m) and U (m) are triangular, with nonzero diagonal entries, so that they are invertible. By Theorem 2.4.7, the matrix A(m) = L(m) U (m) is invertible as well.

A(n−1) c Using the hint, we write A = w ~

~v k

L′ = ~x

0 t

U′ 0

~y . s

We are looking for a column vector ~y , a row vector ~x, and scalars t and s satisfying these equations. The following equations need to be satisfied: ~v = L′ ~y , w ~ = ~xU ′ , and k = ~x~y + ts. We find that ~y = (L′ )−1~v , ~x = w(U ~ ′ )−1 , and ts = k − w(U ~ ′ )−1 (L′ )−1~v . We can choose, for example, s = 1 and t = k − w(U ~ ′ )−1 (L′ )−1~v , proving that A does indeed have an LU factorization. Alternatively, one can show that if all principal submatrices are invertible then no row swaps are required in the Gauss-Jordan Algorithm. In this case, we can find an LU -factorization as outlined in Exercise 2.4.90. 2.4.94 a If A = LU is an LU factorization, then the diagonal entries of L and U are nonzero (compare with Exercise 2.4.93). Let D1 and D2 be the diagonal matrices whose diagonal entries are the same as those of L and U , respectively. Then A = (LD1−1 )(D1 D2 )(D2−1 U ) is the desired factorization ↑ new L

↑ D

↑ new U

(verify that LD1−1 and D2−1 U are of the required form). b If A = L1 D1 U1 = L2 D2 U2 and A is invertible, then L1 , D1 , U1 , L2 , D2 , U2 are all invertible, so that we can −1 multiply the above equation by D2−1 L−1 from the right: 2 from the left and by U1 D2−1 L2−1 L1 D1 = U2 U1−1 . Since products and inverses of upper triangular matrices are upper triangular (and likewise for lower triangular −1 is both upper and lower triangular, that is, it is diagonal. Since matrices), the matrix D2−1 L−1 2 L1 D1 = U2 U1 the diagonal entries of U2 and U1 are all 1, so are the diagonal entries of U2 U1−1 , that is U2 U1−1 = In , and thus U2 = U1 . 110

Section 2.4 −1 Now L1 D1 = L2 D2 , so that L−1 is diagonal. As above, we have in fact L−1 2 L1 = D2 D1 2 L1 = In and therefore L2 = L1 .

2.4.95 Suppose A11 is a p×p matrix and A22 is a q ×q matrix. For B to be the inverse of A we must have AB = Ip+q . Let us partition B the same way as A: B11 B12 B= , where B11 is p × p and B22 is q × q. B21 B22 I 0 A11 B11 A11 B12 B11 B12 A11 0 means that = p = Then AB = 0 Iq A22 B21 A22 B22 B21 B22 0 A22 A11 B11 = Ip , A22 B22 = Iq , A11 B12 = 0, A22 B21 = 0. −1 This implies that A11 and A22 are invertible, and B11 = A−1 11 , B22 = A22 .

This in turn implies that B12 = 0 and B21 = 0. We summarize: A is invertible if (and only if) both A11 and A22 are invertible; in this case −1 0 A11 −1 A = . 0 A−1 22 2.4.96 This exercise is very similar to Example 7 in the text. We outline the solution: I 0 A11 0 B11 B12 = p means that B21 B22 0 Iq A21 A22 A11 B11 = Iq , A11 B12 = 0, A21 B11 + A22 B21 = 0, A21 B12 + A22 B22 = Iq . −1 This implies that A11 is invertible, and B11 = A−1 11 . Multiplying the second equation with A11 , we conclude that −1 B12 = 0. Then the last equation simplifies to A22 B22 = Iq , so that B22 = A22 . −1 −1 Finally, B21 = −A−1 22 A21 B11 = −A22 A21 A11 .

We summarize: A is invertible if (and only if) both A11 and A22 are invertible. In this case, 0 A−1 −1 11 . A = −1 A−1 −A−1 22 22 A21 A11 2.4.97 Suppose A11 is a p × p matrix. Since A11

I is invertible, rref(A) = p 0

A12 0

∗ , so that rref(A23 )

rank(A) = p + rank(A23 ) = rank(A11 ) + rank(A23 ). 2.4.98 Try to find a matrix B =

I AB = n w ~

~v 1

X ~y

X ~y

~x t

~x X + ~v~y = t wX ~ + ~y

(where X is n × n) such that ~x + t~v I = n w~ ~x + t 0

0 . 1

We want X + ~v~y = In , ~x + t~v = ~0, wX ~ + ~y = ~0, and w~ ~ x + t = 1. Substituting ~x = −t~v into the last equation we find −tw~ ~ v + t = 1 or t(1 − w~ ~ v ) = 1. 111

Chapter 2 v~y into the This equation can be solved only if w~ ~ v 6= 1, in which case t = 1−1w~ ~ v . Now substituting X = In − ~ 1 ~ third equation, we find w ~ − w~ ~ v~y + ~y = 0 or ~y = − 1−w~ ~ = −tw. ~ ~v w In + t~v w ~ −t~v −1 We summarize: A is invertible if (and only if) w~ ~ v 6= 1. In this case, A = , where t = 1−1w~ ~v . −tw ~ t . The same result can be found (perhaps more easily) by working with rref[A..In+1 ], rather than partitioned matrices. 2.4.99 Multiplying both sides with A−1 we find that A = In : The identity matrix is the only invertible matrix with this property. 2.4.100 Suppose the entries of A are all a, where a 6= 0. Then the entries of A2 are all na2 . The equation na2 = a 1 1 · · · n1 n n 1 1 ··· 1 n n n 1 . is satisfied if a = n . Thus the solution is A = .. . 1 1 1 ··· n n n 2.4.101 The ijth entry of AB is n X

aik bkj .

k=1

Then

n X

k=1

aik bkj ≤

n X

sbkj = s

k=1

n X

bkj

k=1

!

≤ sr.

↑ ↑ since aik ≤ s this is ≤ r, as it is the j th column sum of B. 2.4.102 a We proceed by induction on m. Since the column sums of A are ≤ r, the entries of A1 = A are also ≤ r1 = r, so that the claim holds for m = 1. Suppose the claim holds for some fixed m. Now write Am+1 = Am A; since the entries of Am are ≤ rm and the column sums of A are ≤ r, we can conclude that the entries of Am+1 are ≤ rm r = rm+1 , by Exercise 101. b For a fixed i and j, let bm be the ijth entry of Am . In part (a) we have seen that 0 ≤ bm ≤ rm . Note that limm→∞ rm = 0 (since r < 1), so that limm→∞ bm = 0 as well (this follows from what some calculus texts call the “squeeze theorem”). c For a fixed i and j, let cm be the ijth entry of the matrix In + A + A2 + · · · + Am . By part (a), cm ≤ 1 + r + r2 + · · · + rm <

1 1−r .

Since the cm form an increasing bounded sequence, limm→∞ cm exists (this is a fundamental fact of calculus). 112

Section 2.4 d (In − A)(In + A + A2 + · · · + Am ) = In + A + A2 + · · · Am − A − A2 − · · · − Am − Am+1 = In − Am+1 Now let m go to infinity; use parts (b) and (c). (In − A)(In + A + A2 + · · · + Am + · · ·) = In , so that (In − A)−1 = In + A + A2 + · · · + Am + · · ·. 2.4.103 a The components of the jth column of the technology matrix A give the demands industry Jj makes on the other industries, per unit output of Jj . The fact that the jth column sum is less than 1 means that industry Jj adds value to the products it produces. b A productive economy can satisfy any consumer demand ~b, since the equation (In − A)~x = ~b can be solved for the output vector ~x : ~x = (In − A)−1~b (compare with Exercise 2.4.49). c The output ~x required to satisfy a consumer demand ~b is ~x = (In − A)−1~b = (In + A + A2 + · · · + Am + · · ·) ~b = ~b + A~b + A2~b + · · · + Am~b + · · ·. To interpret the terms in this series, keep in mind that whatever output ~v the industries produce generates an interindustry demand of A~v . The industries first need to satisfy the consumer demand, ~b. Producing the output ~b will generate an interindustry demand, A~b. Producing A~b in turn generates an extra interindustry demand, A(A~b) = A2~b, and so forth. For a simple example, see Exercise 2.4.50; also read the discussion of “chains of interindustry demands” in the footnote to Exercise 2.4.49. 2.4.104 a We write our three equations below: I L S

1 = 13 R + 31 G + 13 B 3 =R−G , so that the matrix is P = 1 = − 12 R − 12 G + B − 21

R R 1 b G is transformed into G , with matrix A = 0 B 0 0

c This matrix is P A =

1 3

1 3

1

−1

− 21

− 12

1 3

−1 − 12

1 3

0 . 1

0 0 1 0 . 0 0

0 0 (we apply first A, then P .) 0

d See Figure 2.66. A “diagram chase” shows that M = P AP −1 = 2.4.105 a A−1

0 = 1 0

1 0 0 1 0 0 and B −1 = 0 0 0 1 1 0

0 1 . 0 113

2 3

0 −1

0 − 29

1 0

0 . 1 3

Chapter 2

Figure 2.66: for Problem 2.4.104d. Matrix A−1 transforms a wife’s clan into her husband’s clan, and B −1 transforms a child’s clan into the mother’s clan. b B 2 transforms a women’s clan into the clan of a child of her daughter. c AB transforms a woman’s clan into the clan of her daughter-in-law (her son’s wife), while BA transforms a man’s clan into the clan of his children. The two transformations are different. (See Figure 2.67.)

Figure 2.67: for Problem 2.4.105c. d The matrices for the four given diagrams 0 0 1 0 BAB −1 = 1 0 0 , B(BA)−1 = 0 0 1 0 1 e Yes; since BAB −1 = A−1 clan.

0 0 = 1 0 0 1

(in the same order) are BB −1 = I3 , 1 0 0 1 , BA(BA)−1 = I3 . 0 0

1 0 , in the second case in part (d) the cousin belongs to Bueya’s husband’s 0

2.4.106 a We need 8 multiplications: 2 to compute each of the four entries of the product. b We need n multiplications to compute each of the mp entries of the product, mnp multiplications altogether. 2.4.107 g(f (x)) = x, for all x, so that g ◦ f is the identity, but f (g(x)) = 114

x x+1

if x is even . if x is odd

True or False

2.4.108 a The formula

1 − Rk y = −k n

L + R − kLR 1 − kL

x m

is given, which implies that

y = (1 − Rk)x + (L + R − kLR)m. In order for y to be independent of x it is required that 1 − Rk = 0, or k =

1 R

= 40 (diopters).

1 k

then equals R, which is the distance between the plane of the lens and the plane on which parallel incoming rays focus at a point; thus the term “focal length” for k1 . b Now we want y to be independent of the slope m (it must depend on x alone). In view of the formula above, 1 1 10 L+R = + = 40 + ≈ 43.3 (diopters). this is the case if L + R − kLR = 0, or k = LR R L 3 c Here the transformation is y 1 0 1 D 1 = n −k1 1 0 1 −k1

0 1

x 1 − k1 D = m k1 k2 D − k1 − k2

D 1 − k2 D

x . m

We want the slope n of the outgoing rays to depend on the slope m of the incoming rays alone, and not on x; 1 1 2 this forces k1 k2 D − k1 − k2 = 0, or, D = kk11+k k2 = k1 + k2 , the sum of the focal lengths of the two lenses. See Figure 2.68.

Figure 2.68: for Problem 2.4.108c.

True or False Ch 2.TF.1 T, by Theorem 2.4.3. Ch 2.TF.2 T; Let A = B in Theorem 2.4.7. Ch 2.TF.3 F, by Theorem 2.3.3. Ch 2.TF.4 T, by Theorem 2.4.8. Ch 2.TF.5 F; Matrix AB will be 3 × 5, by Definition 2.3.1b. 0 0 . A linear transformation transforms ~0 into ~0. = Ch 2.TF.6 F; Note that T 1 0 115

Chapter 2 Ch 2.TF.7 T, by Theorem 2.2.4. Ch 2.TF.8 T, by Theorem 2.4.6. Ch 2.TF.9 T; The matrix is

1 −1 . −1 1

Ch 2.TF.10 F; The columns of a rotation matrix are unit vectors; see Theorem 2.2.3. Ch 2.TF.11 F; Note that det(A) = (k − 2)2 + 9 is always positive, so that A is invertible for all values of k. Ch 2.TF.12 T; Note that the columns are unit vectors, since (−0.6)2 + (±0.8)2 = 1. The matrix has the form presented in Theorem 2.2.3. Ch 2.TF.13 F; Consider A = I2 (or any other invertible 2 × 2 matrix).

1 2 Ch 2.TF.14 T; Note that A = 3 4

−1

1 1

1 1

5 6 7 8

−1

is the unique solution.

Ch 2.TF.15 F, by Theorem 2.4.9. Note that the determinant is 0. Ch 2.TF.16 T, by Theorem 2.4.3.

1 Ch 2.TF.17 T; The shear matrix A = 0 Ch 2.TF.18 T; Simplify to see that T

1 2

1

works.

x 4y 0 4 x = = . y −12x −12 0 y

Ch 2.TF.19 T; The equation det(A) = k 2 − 6k + 10 = 0 has no real solution. Ch 2.TF.20 T; The matrix fails to be invertible for k = 5 and k = −1, since the determinant det A = k 2 − 4k − 5 = (k − 5)(k + 1) is 0 for these values of k. Ch 2.TF.21 T; The product is det(A)I2 . Ch 2.TF.22 T; Writing an upper triangular matrix A = 0 b , where b is any nonzero constant. that A = 0 0 Ch 2.TF.23 T; Note that the matrix of 4) works.

0 −1 1 0

a b 0 c

and solving the equation A2 =

0 0

0 0

we find

represents a rotation through π/2. Thus n = 4 (or any multiple

−1

Ch 2.TF.24 F; If a matrix A is invertible, then so is A

. But

116

1 1 1 1

fails to be invertible.

True or False Ch 2.TF.25 F; If matrix A has two identical rows, then so does AB, for any matrix B. Thus AB cannot be In , so that A fails to be invertible. Ch 2.TF.26 T, by Theorem 2.4.8. Note that A−1 = A in this case.

1 1 Ch 2.TF.27 F; For any 2 × 2 matrix A, the two columns of A 1 1 Ch 2.TF.28 T; One solution is A =

1 0

will be identical.

1 . 0

Ch 2.TF.29 F; A reflection matrix is of the form

a b , where a2 + b2 = 1. Here, a2 + b2 = 1 + 1 = 2. b −a

Ch 2.TF.30 T; Just multiply it out.

0 Ch 2.TF.31 F; Consider matrix 0 1

0 1 1 0 , for example. 0 0

Ch 2.TF.32 T; Apply Theorem 2.4.8 to the equation (A2 )−1 AA = In , with B = (A2 )−1 A. Ch 2.TF.33 F; Consider the matrix A that represents a rotation through the angle 2π/17. Ch 2.TF.34 F; Consider the reflection matrix A =

1 0 . 0 −1

Ch 2.TF.35 T; We have (5A)−1 = 15 A−1 . Ch 2.TF.36 T; The equation A~ei = B~ei means that the ith columns of A and B are identical. This observation applies to all the columns. Ch 2.TF.37 T; Note that A2 B = AAB = ABA = BAA = BA2 . Ch 2.TF.38 T; Multiply both sides of the equation A2 = A with A−1 . Ch 2.TF.39 F; Consider A = I2 and B = −I2 . Ch 2.TF.40 T; Since A~x is on the line onto which we project, the vector A~x remains unchanged when we project again: A(A~x) = A~x, or A2 ~x = A~x, for all ~x. Thus A2 = A. Ch 2.TF.41 T; If you reflect twice in a row (about the same line), you will get the original vector back: A(A~x) = ~x, or, A2 ~x = ~x = I2 ~x. Thus A2 = I2 and A−1 = A. 0 1 1 1 , for example. ,w ~= , ~v = Ch 2.TF.42 F; Let A = 1 0 0 1

117

Chapter 2 1 0 1 0 0 , B = 0 1 , for example. Ch 2.TF.43 T; Let A = 0 1 0 0 0

Ch 2.TF.44 F; By Theorem 1.3.3, there is a nonzero vector ~x such that B~x = ~0, so that AB~x = ~0 as well. But I3 ~x = ~x 6= ~0, so that AB 6= I3 . Ch 2.TF.45 T; We can rewrite the given equation as A2 + 3A = −4I3 and − 41 (A + 3I3 )A = I3 . By Theorem 2.4.8, the matrix A is invertible, with A−1 = − 14 (A + 3I3 ). Ch 2.TF.46 T; Note that (In + A)(In − A) = In2 − A2 = In , so that (In + A)−1 = In − A. Ch 2.TF.47 F; A and C can be two matrices which fail to commute, and B could be In , which commutes with anything. Ch 2.TF.48 F; Consider T (~x) = 2~x, ~v = ~e1 , and w ~ = ~e2 . Ch 2.TF.49 F; Since there are only eight entries that are not 1, there will be at least two rows that contain only ones. Having two identical rows, the matrix fails to be invertible. Ch 2.TF.50 F; Let A = B =

0 0 , for example. 0 1

0 1 a b S fails to be diagonal, for an arbitrary invertible matrix S = Ch 2.TF.51 F; We will show that S −1 . 0 0 c d 0 1 c d d −b cd d2 1 1 S = ad−bc Now, S −1 = ad−bc . Since c and d cannot both be zero (as S 2 0 0 0 0 −c a −c −cd must be invertible), at least one of the off-diagonal entries (−c2 and d2 ) is nonzero, proving the claim. Ch 2.TF.52 T; Consider an ~x such that A2 ~x = ~b, and let ~x0 = A~x. Then A~x0 = A(A~x) = A2 ~x = ~b, as required.

a b d −b −a −b 1 . Now we want A−1 = −A, or ad−bc = . This holds if c d −c a −c −d ad − bc = 1 and d = −a. These equations have many solutions: for example, a = d = 0, b = 1, c = −1. More 2 generally, we can choose an arbitrary a and an arbitrary nonzero b. Then, d = −a and c = − 1+a b .

Ch 2.TF.53 T; Let A =

2 a b a + bc Ch 2.TF.54 F; Consider a 2×2 matrix A = . We make an attempt to solve the equation A2 = c d ac + cd 2 a + bc b(a + d) 1 0 = . Now the equation b(a + d) = 0 implies that b = 0 or d = −a. c(a + d) d2 + bc 0 −1

ab + bd = cb + d2

If b = 0, then the equation d2 + bc = −1 cannot be solved. If d = −a, then the two diagonal entries of A2 , a2 + bc and d2 + bc, will be equal, so that the equations a2 + bc = 1 and d2 + bc = −1 cannot be solved simultaneously. 1 0 2 cannot be solved. In summary, the equation A = 0 −1 118

True or False u21 u1 u2 u1 , where is a u1 u2 u22 u2 unit vector. Thus, a2 + b2 + c2 + d2 = u41 + (u1 u2 )2 + (u1 u2 )2 + u42 = u41 + 2(u1 u2 )2 + u42 = (u21 + u22 )2 = 12 = 1.

Ch 2.TF.55 T; Recall from Definition 2.2.1 that a projection matrix has the form

Ch 2.TF.56 T; We observe that the systems AB~x = 0 and B~x = 0 have the same solutions (multiply with A−1 and A, respectively, to obtain one system from the other). Then, by True or False Exercise 45 in Chapter 1, rref(AB) =rref(B).

119

Chapter 3

Chapter 3 Section 3.1 3.1.1 Find all ~x such that A~x = ~0: " # " # . . 1 2.. 0 −→ 1 0.. 0 , so that x = x = 0. 1 2 . . 3 4.. 0 0 1.. 0 ker(A) = {~0}. 3.1.2 Find all ~x such that A~x = ~0: " # . .. 3. 3t . 0 1 2 3. 0 −→ 2 , so that x1 = − 2 . . x2 t 6 9.. 0 0 0.. 0 Setting t = 2 we find ker(A) = span

−3 . 2

3.1.3 Find all ~x such that A~x = ~0; note that all ~x in R2 satisfy the equation, so that ker(A) = R2 = span(~e1 , ~e2 ). 3.1.4 Find all ~x such that A~x = ~0, or x1 + 2x2 + 3x2 = 0. −2t − 3r −2 −3 x1 = t 1 + r 0 , so that t The solutions are of the form x2 = r 0 1 x2 −2 −3 ker(A) = span 1 , 0 . 0 1 3.1.5 Find all ~x such that A~x = ~0. . .. 1 0 −1.. 1 1 1 . 0 . . 2.. 1 2 3.. 0 −→ 0 1 . . 1 3 5.. 0 0 0 0..

0

x1 0 ; x2 0

1 ker(A) = span −2 . 1

3.1.6 Find all ~x such that A~x = ~0. .. .. 1 1 1.. 1 1 1.. 0 1 1 1.. 0 −→ 0 0 0.. . . 0 0 0.. 1 1 1.. 0

t x = x3 1 ; x2 = −2t = −2x3 t x3

0 0 ; x1 + x2 + x3 = 0 0

120

Section 3.1 −1 −1 −r − t x1 x2 = r = r 1 + t 0 1 0 t x3

−1 −1 ker(A) = span 1 , 0 . 1 0

3.1.7 Find all ~x such that A~x = ~0. Since rref(A) = I3 we have ker(A) = {~0}.

1 3.1.8 Find all ~x such that A~x = ~0. Solving this system yields ker(A) = span −2 . 1 3.1.9 Find all ~x such that A~x = ~0. Solving this system yields ker(A) = {~0}. 1 −2 3.1.10 Solving the system A~x = ~0 we find that ker(A) = span . 1 0

−2 3 3.1.11 Solving the system A~x = ~0 we find that ker(A) = span . 1 0

1 −2 1 0 3.1.12 Solving the system A~x = ~0 we find that ker(A) = span 0 , −1 . 0 1 0 0 0 −3 −2 1 0 0 0 −2 0 3.1.13 Solving the system A~x = ~0 we find that ker(A) = span , . , 0 −1 0 0 1 0 1 0 0

3.1.14 By Theorem 3.1.3, the image of A is the span of the column vectors of A: 1 1 1 2 im(A) = span , . 1 3 1 4 3.1.15 By Theorem 3.1.3, the image of A is the span of the columns of A: 121

Chapter 3

im(A) = span

1 1 1 1 . , , , 4 3 2 1

Since any two of these vectors span all of R2 already, we can write 1 1 . , im(A) = span 2 1 3.1.16 By Theorem 3.1.3, the image of A is the span of the column vectors of A: 3 2 1 im(A) = span 1 , 2 , 3 . 3 2 1

Since these three vectors are parallel, we need only one of them to span the image: 1 im(A) = span 1 . 1

2 1 = R2 (the whole plane). , 3.1.17 By Theorem 3.1.3, im(A) = span 4 3 1 4 1 (a line in R2 ). = span , 3.1.18 By Theorem 3.1.3, im(A) = span 3 12 3 3.1.19 Since the four column vectors of A are parallel, we have im(A) = span

1 , a line in R2 . −2

1 3.1.20 Since the three column vectors of A are parallel, we have im(A) = span 1 , a line in R3 . 1 3 7 4 3.1.21 By Theorem 3.1.3, im(A) = span 1 , 9 , 2 . 8 6 5 We We 4 1

must simply find out can detemine this by . .. 1 0.. 7. 3 . . 9.. 2 −→ 0 1.. . . 0 0.. 5 6.. 8 R3 .

how many of the column vectors are not redundant to determine a basis of the image. taking the rref of the matrix: 0 0 , which shows us that all three column vectors are independent: the span is all of 1

3.1.22 Compare with the solution to Exercise 21. . .. 2 1 0.. 2 1 . 3 . . 3 4.. 2 −→ 0 1.. −1 .. . 6 5. 7 0 0.. 0 122

Section 3.1 This computation shows that the third column vector of A, ~v3 , is a linear combination of the first two, Thus, only the first two vectors are independent, and the image is a plane in R3 . 3.1.23 im(T ) = R2 and ker(T ) = {~0}, since T is invertible (see Summary 3.1.8). 3.1.24 im(T ) is the plane x + 2y + 3z = 0, and ker(T ) is the line perpendicular to this plane, spanned by the vector 1 2 (compare with Examples 5 and 9). 3 3.1.25 im(T ) = R2 and ker(T ) = {~0}, since T is invertible (see Summary 3.1.8). 3.1.26 Since limt→∞ f (t) = ∞ and limt→−∞ f (t) = −∞, we have im(f ) = R. A careful proof involves the intermediate value theorem (see Exercise 2.2.47c),

Figure 3.1: for Problem 3.1.26. Any horizontal line intersects this graph at least once (compare with Example 3 and see Figure 3.1). 3.1.27 Let f (x) = x3 − x = x(x2 − 1) = x(x − 1)(x + 1). Then im(f ) = R, since lim f (x) = ∞ and

x→∞

lim f (x) = −∞

x→−∞

but the function fails to be invertible since the equation f (x) = 0 has three solutions, x = 0, 1, and −1. 3.1.28 This ellipse can be obtained from the unit circle by means of the linear transformation with matrix as shown in Figure 3.2 (compare with Exercise 2.2.49). 1 0 cos(t) cos(t) We obtain the parametrization = for the ellipse. 0 2 sin(t) 2 sin(t) We can check that x2 +

y2 4

= cos2 (t) +

4 sin2 (t) 4

= 1.

sin(φ) cos(θ) φ = sin(φ) sin(θ) 3.1.29 Use spherical coordinates (see any good text on multivariable calculus): f θ cos(φ) 123

1 0 , 0 2

Chapter 3

Figure 3.2: for Problem 3.1.28. 1 does the job. There are many other possible answers: any nonzero 2 × n matrix 5 1 A whose column vectors are scalar multiples of vector . 5

3.1.30 By Theorem 3.1.3, A =

−2 −3 3.1.31 The plane x + 3y + 2z = 0 is spanned by the two vectors 0 and 1 , for example. Therefore, 1 0 −2 −3 A= 0 1 does the job. There are many other correct answers. 1 0 7 3.1.32 By Theorem 3.1.3, A = 6 does the job. There are many other correct answers: any nonzero 3 × n matrix 5 7 A whose column vectors are scalar multiples of 6 . 5 x 3.1.33 The plane is the kernel of the linear transformation T y = x + 2y + 3z from R3 to R. z 3.1.34 To describe a subset of R3 as a kernel means to describe it as an intersection of planes (think about it). By inspection, the given line is the intersection of the planes x+y 2x + z

= =

0 0.

and

x x+y from R3 to R2 . This means that the line is the kernel of the linear transformation T y = 2x + z z 3.1.35 ker(T ) = {~x : T (~x) = ~v · ~x = 0} = the plane with normal vector ~v . im(T ) = R, since for every real number k there is a vector ~x such that T (~x) = k, for example, ~x = 3.1.36 ker(T ) = {~x : T (~x) = ~v × ~x = ~0} = the line spanned by ~v 124

k v. ~ v ·~ v~

Section 3.1 (see Theorem A.10d in the Appendix) im(T ) = the plane with normal vector ~v By Definition A.9, T (~x) = ~v × ~x is in this plane, for all ~x in R3 . Conversely, for every vector w ~ in this plane ~ (verify this!). there is an ~x in R3 such that T (~x) = w, ~ namely ~x = − ~v1·~v T (w)

0 1 3.1.37 A = 0 0 0 0

0 0 0 1 0 0 1 , A2 = 0 0 0 , A3 = 0 0 0 0 0 0 0 0

0 0 , so that 0

ker(A) = span(~e1 ), ker(A2 ) = span(~e1 , ~e2 ), ker(A3 ) = R3 , and im(A) = span(~e1 , ~e2 ), im(A2 ) = span(~e1 ), im(A3 ) = {~0}.

3.1.38 a If a vector ~x is in ker(Ak ), that is, Ak ~x = ~0, then ~x is also in ker(Ak+1 ), since Ak+1 ~x = AAk ~x = A~0 = ~0. Therefore, ker(A) ⊆ ker(A2 ) ⊆ ker(A3 ) ⊆ . . . Exercise 37 shows that these kernels need not be equal. b If a vector ~y is in im(Ak+1 ), that is, ~y = Ak+1 ~x for some ~x, then ~y is also in im(Ak ), since we can write ~y = Ak (A~x). Therefore, im(A) ⊇ im(A2 ) ⊇ im(A3 ) ⊇ . . .. Exercise 37 shows that these images need not be equal. 3.1.39 a If a vector ~x is in ker(B), that is, B~x = ~0, then ~x is also in ker(AB), since AB(~x) = A(B~x) = A~0 = ~0: ker(B) ⊆ ker(AB). Exercise 37 (with A = B) illustrates that these kernels need not be equal. b If a vector ~y is in im(AB), that is, ~y = AB~x for some ~x, then ~y is also in im(A), since we can write ~y = A(B~x): im(AB) ⊆ im(A). Exercise 37 (with A = B) illustrates that these images need not be equal. 3.1.40 For any ~x in Rm , the vector B~x is in im(B) = ker(A), so that AB~x = ~0. If we apply this fact to ~x = ~e1 , ~e2 , . . . , ~em , we find that all the columns of the matrix AB are zero, so that AB = 0. 4 −4 3.1.41 a rref(A) = 1 3 , so that ker(A) = span . 3 0 0 0.36 3 im(A) = span = span . 0.48 4 Note that im(A) and ker(A) are perpendicular lines. b A2 = A If ~v is in im(A), with ~v = A~x, then A~v = A2 ~x = A~x = ~v . 125

Chapter 3

Figure 3.3: for Problem 3.1.41c. c Any vector ~v in R2 can be written uniquely as ~v = ~v1 + ~v2 , where ~v1 is in im(A) and ~v2 is in ker(A). (See Figure 3.3.) Then A~v = A~v1 + A~v2 = ~v1 (A~v1 =~v1 by part b, A~v2 = ~0 since ~v2 is in ker(A)), so that A represents the 3 orthogonal projection onto im(A) = span . 4 y1 y 3.1.42 Using the hint, we see that the vector ~y = 2 is in the image of A if y3 y4

y1

y2

−3y3 −2y3

+2y4 +y4

=0 = 0.

and

This means that im(A) is the kernel of the matrix

1 0

0 −3 2 . 1 −2 1

3.1.43 Using our work in Exercise 42 as a guide, we come up with the following procedure to express the image of an n × m matrix A as the kernel of a matrix B: If rank(A) = n, let B be the n × n zero matrix. If r = rank(A) < n, let B be the (n − r) × n matrix obtained by omitting the first r rows and the first m columns . of rref[A..I ]. n

3.1.44 a Yes; by construction of the echelon form, the systems A~x = ~0 and B~x = ~0 have the same solutions (it is the whole point of Gaussian elimination not to change the solutions of a system). b No; as a counterexample, consider A = im(B) = span(~e1 ).

0 1

1 0 0 , with , with im(A) = span(~e2 ), but B = rref(A) = 0 0 0

3.1.45 As we solve the system A~x = ~0, we obtain r leading variables and m − r free variables. The “general vector” in ker(A) can be written as a linear combination of m − r vectors, with the free variables as coefficients. (See Example 11, where m − r = 5 − 3 = 2.) 126

Section 3.1 3.1.46 If rank(A) = r, then im(A) = span(~e1 , . . . , ~er ). See Figure 3.4.

Figure 3.4: for Problem 3.1.46. 3.1.47 im(T ) = L2 and ker(T ) = L1 . 3.1.48 a w ~ = A~x, for some ~x, so that Aw ~ = A2 ~x = A~x = w. ~ b If rank(A) = 2, then A is invertible, and the equation A2 = A implies that A = I2 (multiply by A−1 ). 0 0 If rank(A) = 0 then A = . 0 0 c First note that im(A) and ker(A) are lines (there is one nonleading variable).

Figure 3.5: for Problem 3.1.48c. By definition of a projection, we need to verify that ~x − A~x is in ker(A). This is indeed the case, since A(~x − A~x) = A~x − A2 ~x = A~x − A~x = ~0 (we are told that A2 = A). See Figure 3.5. 3.1.49 If ~v and w ~ are in ker(T ), then T (~v + w) ~ = T (~v ) + T (w) ~ = ~0 + ~0 = ~0, so that ~v + w ~ is in ker(T ) as well. If ~v is in ker(T ) and k is an arbitrary scalar, then T (k~v ) = kT (~v ) = k~0 = ~0, so that k~v is in ker(T ) as well. 3.1.50 From Exercise 38 we know that ker(A3 ) ⊆ ker(A4 ). Conversely, if ~x is in ker(A4 ), then A4 ~x = A3 (A~x) = ~0, so that A~x is in ker(A3 ) = ker(A2 ), which implies that A2 (A~x) = A3 ~x = ~0, that is, ~x is in ker(A3 ). We have shown that ker(A3 ) = ker(A4 ). 3.1.51 We need to find all ~x such that AB~x = ~0. If AB~x = ~0, then B~x is in ker(A) = {~0}, so that B~x = ~0. Since ker(B) = {~0}, we can conclude that ~x = ~0. It follows that ker(AB) = {~0}. 127

Chapter 3 A~x A , we can conclude that C~x = ~0 if (and only if) both A~x = ~0 and B~x = ~0. It ~x = B~x B follows that ker(C) is the intersection of ker(A) and ker(B): ker(C) = ker(A) ∩ ker(B).

3.1.52 Since C~x =

3.1.53 a Using the equation 1 + 1 = 0 (or −1 = 1), we can write the general vector ~x in ker(H) as x1 p+r+s x2 p + q + s x3 p + q + r ~x = x4 = p q x5 x6 r x7 s 1 0 1 1 1 1 0 1 1 1 1 0 = p1 + q0 + r0 + s0 0 1 0 0 0 0 1 0 0 0 0 1 ↑ ~v1

↑ ~v2

↑ ~v3

↑ ~v4

b ker(H) = span(~v1 , ~v2 , ~v3 , ~v4 ) by part (a), and im(M ) = span(~v1 , ~v2 , ~v3 , ~v4 ) by Theorem 3.1.3, so that im(M ) = ker(H). M~x is in im(M ) = ker(H), so that H(M~x) = ~0. 3.1.54 a If no error occurred, then w ~ = ~v = M~u, and H w ~ = H(M~u) = ~0, by Exercise 53b. If an error occurred in the ith component, then w ~ = ~v + ~ei = M~u + ~ei , so that Hw ~ = H(M~u) + H~ei = ith column of H. Since the columns of H are all different, this method allows us to find out where an error occurred. 1 b Hw ~ = 1 = seventh column of H: an error occurred in the seventh component of ~v . 0 1 0 0 1 1 Therefore ~v = w ~ + ~e7 = 0 and ~u = . 0 1 1 0 1

Section 3.2 3.2.1 Not a subspace, since W does not contain the zero vector. 128

Section 3.2 −1 1 3.2.2 Not a subspace, since W contains the vector ~v = 2 but not the vector (−1)~v = −2 . −3 3

1 2 3.2.3 W = im 4 5 7 8

3 6 is a subspace of R3 , by Theorem 3.2.2. 9

3.2.4 span(~v1 , . . . , ~vm ) = im[~v1 . . . ~vm ] is a subspace of Rn , by Theorem 3.2.2. 3.2.5 We have subspaces {~0}, R3 , and all lines and planes (through the origin). To prove this, mimic the reasoning in Example 2. 3.2.6 a Yes! • The zero vector is in V ∩ W , since ~0 is in both V and W . • If ~x and ~y are in V ∩ W , then both ~x and ~y are in V , so that ~x + ~y is in V as well, since V is a subspace of Rn . Likewise, ~x + ~y is in W , so that ~x + ~y is in V ∩ W . • If ~x is in V ∩ W and k is an arbitrary scalar, then k~x is in both V and W , since they are subspaces of Rn . Therefore, k~x is in V ∩ W . b No; as a counterexample consider V = span(~e1 ) and W = span(~e2 ) in R2 . 3.2.7 Yes; we need to show that W contains the zero vector. We are told that W is nonempty, so that it contains some vector ~v . Since W is closed under scalar multiplication, it will contain the vector 0~v = ~0, as claimed. 1 2 3 1 2 3.2.8 We need to solve the system c1 + c2 + c3 = 2 3 4 2 3 t c1 The general solution is c2 = −2t . t c3

c1 3 0 c2 = . 4 0 c3

1 2 3 0 Picking t = 1 we find the nontrivial relation 1 −2 +1 = . 2 3 4 0

3.2.9 These vectors are linearly dependent, since ~vm = 0~v1 + 0~v2 + · · · + 0~vm−1 . 3.2.10 Linearly dependent, since

6 2 6 is redundant. . Thus, the vector =3 3 1 3

3.2.11 Linearly independent, since the two vectors are not parallel, and therefore not redundant. 7 0 . Thus, the vector ~0 is redundant. =0 3.2.12 Linearly dependent, since 11 0 129

Chapter 3

3.2.13 Linearly dependent, since the second vector is redundant

1 1 3.2.14 Linearly independent, since ref 0 2 0 0

1 1 . =1 2 2

1 2 = I3 (use Theorem 3.2.6). 3

3.2.15 Linearly dependent. By Theorem 3.2.8, since we have three vectors in R2 , at least one must be redundant. We can perform a straightforward computation to reveal that ~v3 = −~v1 + 2~v2 . 6 3 1 6 1 3 3.2.16 Certainly 2 is not a multiple of 1 , so it is not redundant. However, 5 = 3 1 + 1 2 , so 5 4 1 1 4 1 1 is redundant. Thus, these vectors are linearly dependent.

1 1 1 3.2.17 Linearly independent. The first two vectors are clearly not redundant, and since rref 1 2 3 = I3 , the 1 3 6 last vector is also not redundant. Thus, the three vectors turn out to be linearly independent. 1 1 3.2.18 Linearly dependent, since rref 1 1 be redundant.

1 2 3 4

1 1 4 0 = 0 7 0 10

1 0 −2 1 3 4 . So, we find that the vector turns out to 7 0 0 10 0 0

1 0 3.2.19 Linearly dependent. First we see that is not redundant, because it is first, and non-zero. However, 0 0 2 1 0 0 = 2 , so it is redundant. 0 0 0 0 0 0 3 1 0 0 1 0 4 0 1 0 and are clearly not redundant, but = 3 + 4 + 5 , so it is redundant. 0 1 5 0 0 1 0 0 0 0 0 0 0 3.2.20 0 is redundant, simply because it is the zero vector. 0 1 0 is our first non-zero vector, and thus, is not redundant. 0 1 3 0 = 3 0 and is redundant. 0 0 130

Section 3.2 1 0 1 is not a multiple of 0 and is not redundant. 0 0 0 1 4 5 = 4 0 + 5 1 and is redundant. 0 0 0 0 1 6 Similarly, 7 = 6 0 + 7 1 and is also redundant. 0 0 0 0 1 0 However, by inspection, 0 is not a linear combination of 0 and 1 , meaning that this last vector is not 0 0 1 redundant. Thus, the seven vectors are linearly dependent. 3.2.21 Certainly, sincethe second vector equals the first, the second is redundant. So ~v1 = ~v2 , 1~v1 − 1~v2 = ~0, 1 is in ker(A). revealing that −1 3.2.22

1 3 3 1 1 3 3 is in the kernel of = ~0. Thus, −1 . So, 3 =3 is redundant, because 2 −1 6 2 2 6 6

3.2.23 The first column is ~0, so it is redundant. 1~v1 = ~0, so

3 . 6

1 is in ker(A). 0

6 6 1 3 1 3 6 3 3.2.24 5 is redundant, because 5 = 3 1 + 1 2 . Thus, 3 1 + 1 2 − 1 5 = ~0 and 1 is in 4 4 1 1 1 1 4 −1 1 3 6 the kernel of 1 2 5 . 1 1 4

1 3.2.25 The third column equals the first, so it is redundant and ~v1 = ~v3 , or 1~v1 + 0~v2 − 1~v3 = ~0. Thus, 0 is in −1 ker(A). 2 1 0 3.2.26 3 = 2 0 + 3 1 , 0 0 0 2 1 3 is in the kernel of 0 −1 0 0

2 1 0 2 0 so 3 is redundant. Now, 2 0 + 3 1 − 1 3 + 0 0 = ~0, revealing that 0 0 0 0 1 0 2 0 1 3 0. 0 0 1

1 1 3.2.27 A basis of im(A) is 1 , 2 , by Theorem 3.2.4. 3 1 131

Chapter 3 1 1 1 3.2.28 The three column vectors are linearly independent, since rref 1 2 5 = I3 . 1 3 7

Therefore, the three columns form a basis of im(A)(= R3 ): 1 1 1 1 , 2 , 5 . 1 3 7 Another sensible choice for a basis of im(A) is ~e1 , ~e2 , ~e3 .

3.2.29 The three column vectors of A span all of R2 , so that im(A) = R2 . We can choose any two of the columns of A to form a basis of im(A); another sensible choice is ~e1 , ~e2 . 3.2.30 im(A) = span(~e1 , ~e2 ) We can choose ~e1 , ~e2 as a basis of im(A). 3.2.31 The two column vectors of the given matrix A are linearly independent (they are not parallel), so that they form a basis of im(A). 3.2.32 By inspection, the first, third and sixth columns are redundant. Thus, a basis of the image consists of the 1 0 0 0 1 0 remaining column vectors: , , . 0 0 1 0 0 0 3.2.33 im(A) = span(~e1 , ~e2 , ~e3 ), so that ~e1 , ~e2 , ~e3 is a basis of im(A). 1 2 3.2.34 The fact that is in ker(A) means that 3 4 1 1 2 2 A = [~v1 ~v2 ~v3 ~v4 ] = ~v1 + 2~v2 + 3~v3 + 4~v4 = ~0, so that ~v4 = − 14 ~v1 − 12 ~v2 − 43 ~v3 . 3 3 4 4 3.2.35 If ~vi is a linear combination of the other vectors in the list, ~vi = c1~v1 + · · · + ci−1~vi−1 + ci+1~vi+1 + · · · + cn~vn , then we can subtract ~vi from both sides to generate a nontrivial relation (the coefficient of ~vi will be -1). Conversely, if there is a nontrivial relation c1~v1 + · · · + ci~vi + · · · + cn~vn = ~0, with ci 6= 0, then we can solve for vector ~vi and thus express ~vi as a linear combination of the other vectors in the list. 3.2.36 Yes; we know that there is a nontrivial relation c1~v1 + c2~v2 + · · · + cm~vm = ~0. Now apply the transformation T to the vectors on both sides, and use linearity: T (c1~v1 + c2~v2 + · · · + cm~vm ) = T (~0), so that c1 T (~v1 ) + c2 T (~v2 ) + · · · + cm T (~vm ) = ~0. 132

Section 3.2 This is a nontrivial relation among the vectors T (~v1 ), . . . , T (~vm ), so that these vectors are linearly dependent, as claimed. 3.2.37 No; as a counterexample, consider the extreme case when T is the zero transformation, that is, T (~x) = ~0 for all ~x. Then the vectors T (~v1 ), . . . , T (~vm ) will all be zero, so that they are linearly dependent. 3.2.38 a Using the terminology introduced in the exercise, we need to show that any vector ~v in V is a linear combination of ~v1 , . . . , ~vm . Choose a specific vector ~v in V . Since we can find no more than m linearly independent vectors in V , the m + 1 vectors ~v1 , . . . , ~vm , ~v will be linearly dependent. Since the vectors ~v1 , . . . , ~vm are independent, ~v must be redundant, meaning that ~v is a linear combination of ~v1 , . . . , ~vm , as claimed. b With the terminology introduced in part a, we can let V = im [ ~v1

· · · ~vm ] .

3.2.39 Yes; the vectors are linearly independent. The vectors in the list ~v1 , . . . , ~vm are linearly independent (and therefore non-redundant), and ~v is non-redundant since it fails to be in the span of ~v1 , . . . , ~vm . 3.2.40 Yes; by Theorem 3.2.8, ker(A) = {~0} and ker(B) = {~0}. Then ker(AB) = {~0} by Exercise 3.1.51, so that the columns of AB are linearly independent, by Theorem 3.2.8. 3.2.41 To show that the columns of B are linearly independent, we show that ker(B) = {~0}. Indeed, if B~x = ~0, then AB~x = A~0 = ~0, so that ~x = ~0 (since AB = Im ). By Theorem 3.2.8, rank(B) = # columns = m, so that m ≤ n and in fact m < n (we are told that m 6= n). This implies that the rank of the m × n matrix A is less than n, so that the columns of A are linearly dependent (by Theorem 3.2.8). 3.2.42 We can use the hint and form the dot product of ~vi and both sides of the relation c1~v1 + · · · + ci~vi + · · · + cm~vm = ~0: (c1~v1 + · · · + ci~vi + · · · + cm~vm ) · ~vi = ~0 · ~vi , so that c1 (~v1 · ~vi ) + · · · + ci (~vi · ~vi ) + · · · + cm (~vm · ~vi ) = 0. Since ~vi is perpendicular to all the other ~vj , we will have ~vi · ~vj = 0 whenever j 6= i; since ~vi is a unit vector, we will have ~vi · ~vi = 1. Therefore, the equation above simplifies to ci = 0. Since this reasoning applies to all i = 1, . . . , m, we have only the trivial relation among the vectors ~v1 , ~v2 , . . . , ~vm , so that these vectors are linearly independent, as claimed. 3.2.43 Consider a linear relation c1~v1 + c2 (~v1 +~v2 ) + c3 (~v1 +~v2 +~v3 ) = ~0, or, (c1 + c2 + c3 )~v1 + (c2 + c3 )~v2 + c3~v3 = ~0. Since there is only the trivial relation among the vectors ~v1 , ~v2 , ~v3 , we must have c1 + c2 + c3 = c2 + c3 = c3 = 0, so that c3 = 0 and then c2 = 0 and then c1 = 0, as claimed. 3.2.44 Yes; this is a special case of Exercise 40 (recall that ker(A) = {~0}, by Theorem 3.1.7b). 3.2.45 Yes; if A is invertible, then ker(A) = {~0}, so that the columns of A are linearly independent, by Theorem 3.2.8. 3.2.46 Solve the system

x1

+

2x2 x3

+ +

3x4 4x4

+ +

The solutions are of the form 133

5x5 6x5

=0 . =0

Chapter 3 −5 −3 −2 −2s − 3t − 5r x1 s 0 0 1 x2 x3 = −4t − 6r = s 0 + t −4 + r −6 . 0 1 0 t x4 1 0 0 r x5 −2 −3 −5 1 0 0 The vectors 0 , −4 , −6 span the kernel, by construction, and they are linearly independent, by 0 1 0 0 0 1 Theorem 3.2.5. Therefore, the three vectors form a basis of the kernel.

1 0 3.2.47 By Theorem 3.2.8, the rank of A is 3. Thus, rref(A) = 0 0

0 1 0 0

0 0 . 1 0

x1 3.2.48 We can write 3x1 + 4x2 + 5x3 = [3 4 5] x2 = 0, so that V = ker[3 4 5]. x3 0 4 To express V as an image, choose a basis of V , for example, −3 , 5 . −4 0 4 0 Then, V = im −3 5 . 0 −4

There are other solutions.

1 3.2.49 L = im 1 1

To write L as a kernel, think of L as the intersection of the planes x = y and y = z, that is, as the solution set x − y =0 of the system . y − z =0 1 −1 0 Therefore, L = ker . 0 1 −1

There are other solutions. 3.2.50 The verification of the three properties listed in Definition 3.2.1 is straightforward. Alternatively, we can choose a basis ~v1 , . . . , ~vp of V and a basis w ~ 1, . . . , w ~ q of W (see Exercise 38a) and show that V + W = span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ) (compare with Exercise 4). Indeed, if ~v + w ~ is in V +W , then ~v is a linear combination of ~v1 , . . . , ~vp and w ~ is a linear combination of w ~ 1, . . . , w ~q, so that ~v + w ~ is a linear combination of ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q . Conversely, if ~x is in span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ), then ~x = (c1~v1 + · · · + cp~vp ) + (d1 w ~ 1 + · · · + dq w ~ q ), so that ~x is in V + W . 134

Section 3.2 If V and W are distinct lines in R3 (spanned by ~v and w, ~ respectively), then V + W is the plane spanned by ~v and w. ~ 3.2.51 a Consider a relation c1~v1 + · · · + cp~vp + d1 w ~ 1 + · · · + dq w ~ q = ~0. Then the vector c1~v1 + · · · + cp~vp = −d1 w ~ 1 − · · · − dq w ~ q is both in V and in W , so that this vector is ~0 : ~ ~ c1~v1 + · · · + cp~vp = 0 and d1 w ~ 1 + · · · + dq w ~ q = 0. Now the ci are all zero (since the ~vi are linearly independent) and the dj are zero (since the w ~ j are linearly independent). Since there is only the trivial relation among the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q , they are linearly independent. b In Exercise 50 we show that V + W = span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ), and in part (a) we show that these vectors are linearly independent. a b d 1 0 0 0 c e 0 1 0 3.2.52 If a, c and f are nonzero, then rref = , and the three vectors are linearly independent, 0 0 f 0 0 1 0 0 0 0 0 0 by Theorem 3.2.6. If at least one of the constants a, c or f is zero, then at least one column of rref will not contain a leading one, so that the three vectors are linearly dependent.

3.2.53 The zero vector is in V ⊥ , since ~0 · ~v = 0 for all ~v in V . If w ~ 1 and w ~ 2 are both in V ⊥ , then (w ~1 + w ~ 2 ) · ~v = w ~ 1 · ~v + w ~ 2 · ~v = 0 + 0 = 0 for all ~v in V , so that w ~1 + w ~ 2 is in ⊥ V as well. If w ~ is in V ⊥ and k is an arbitrary constant, then (k w) ~ · ~v = k(w ~ · ~v ) = k0 = 0 for all ~v in V , so that k w ~ is in V ⊥ as well. 1 x x 3.2.54 We need to find all vectors y in R3 such that y · 2 = x + 2y + 3z = 0. 3 z z −3 −2 −2s − 3t x = s 1 + t 0 . s These vectors have the form y = 1 0 t z −3 −2 Therefore, 1 , 0 is a basis of L⊥ . 1 0 1 x1 x2 2 3.2.55 We need to find all vectors ~x in R5 such that x3 · 3 = x1 + 2x2 + 3x3 + 4x4 + 5x5 = 0. 4 x4 5 x5

These vectors are of the form 135

Chapter 3 −5 −4 −3 −2 −2a − 3b − 4c − 5d x1 a 0 1 0 0 x2 b = a 0 + b 1 + c 0 + d 0 . x3 = 0 1 0 0 c x4 1 0 0 0 d x5

The four vectors to the right form a basis of L⊥ ; they span L⊥ , by construction, and they are linearly independent, by Theorem 3.2.5. 3.2.56 Consider a linear relation c1~v1 + c2~v2 + c3~v3 + c4~v4 = ~0 among the four given vectors. The last component of the vector on the left hand side is c3 , so that c3 = 0. Now the fifth component on the left is c1 , so that c1 = 0. The third component is now c4 , so c4 = 0. It follows that c2 = 0 as well. We have shown that there is only the trivial relation among the given vectors, so that they are linearly independent, regardless of the values of the constants a, b . . . , m. 3.2.57 We will begin to go through the possibilties for j until we see a pattern: 1 0 0 j = 1: Yes, because 0 is in ker(A) (the first column is ~0). 0 0 0 j = 2: No, this would just be a multiple of the second column, and only ~0 if the jth component is zero. 0 2 −1 j = 3: Yes, since 0 is in ker(A). 0 0 0

At this point, we realize that we are choosing the redundant columns. Thus, j can also be 6 and 7, because 0 0 0 3 0 0 4 , and 0 are in ker(A). 0 5 0 −1 0 1 3.2.58 This occurs for each column, j, that is redundant. If ~x is in the kernel, and the j th component of ~x is the last non-zero component, then x1~v1 + · · · + xj ~vj + xj+1~vj+1 + · · · + xm~vm = ~0, but xj+1 = · · · = xm = 0, so x1~v1 + · · · + xj ~vj = ~0. 136

Section 3.3 Thus, since xj 6= 0, ~vj = −

x1 ~ v1 +···+xj−1 ~ vj−1 xj

and ~vj is redundant. Conversely, if ~vj is redundant, with ~vj = c1 .. . cj−1 c1~v1 + · · · + cj−1~vj−1 , then the vector ~x = −1 is in the kernel of A. The last non-zero component of ~x is the 0 . .. 0 j th , as required.

Section 3.3 3.3.1 Clearly the second column is just three time the first, and thus is redundant. Applying the notion of Kyle Numbers, we see: 3 1 2

−1 3 3 3 , so is in the ker(A). No other vectors belong in our list, so a basis of the kernel is , −1 −1 6 1 and a basis of the image is . 2 3.3.2 Using Kyle Numbers, we see that the second column is redundant: 4 1 2

−1 4 4 , 4 , so is in the ker(A). No other vectors belong in our list, so a basis of the kernel is −1 −1 8 1 . and a basis of the image is 2 3.3.3 The two columns here are independent, so there redundant vectors. Thus, ∅ is a basis of the kernel, are no 1 2 and the two columns form a basis of the image: , . 3 4 3.3.4 The first column is redundant. We use the following Kyle Numbers: 1 0 0

0 1 1 1 , so is in the ker(A). No other vectors belong in our list, so a basis of the kernel is , and 0 0 2 1 a basis of the image is . 2 3.3.5 The first two vectors are non-redundant, but the third is a multiple of the first. We see: 3 0 −1 3 1 −2 3 , so a basis of the kernel is 0 , and a basis of the image consists of the non-redundant 2 4 6 −1 −2 1 . , columns, or 4 2 137

Chapter 3 3.3.6 The first two vectors are non-redundant, but the third is a combination of the first two: 1 2 −1 1 1 1 1 1 3 , so a basis of the kernel is , and a basis of the image is 2 , . 2 1 2 1 4 −1 3.3.7 We immediately see fitting Kyle numbers for one relation: 2 −1 1 2 1 2 zero above

0 3 . Now, since the second column is redundant, we remove it from further inspection and keep a 4 it:

0 2 3 . However, in this case, there are no more redundant vectors. Thus, a basis of the kernel is 2 4 2 1 3 −1 , and a basis of the image is , . 1 4 0

1 1

3.3.8 Here the second column is redundant, with Kyle Numbers as: 3 1 2 3

1 1 −3 3 . This reveals a basis of our kernel as and a basis of the image to be 2 . −6 1 3 −9

3.3.9 The second column is redundant, and we can choose Kyle numbers as follows: 2 1 1 1

−1 2 2 2

0 2 1 , but the third column is non-redundant. Thus, a basis of the kernel is −1 , while a basis 2 0 3 1 1 of the image is 1 , 2 . 3 1 1 3.3.10 The first column is redundant, and 0 is in the kernel: 0 1 0 0 0

0 1 1 1

0 1 1 0 , while . No other columns are redundant, however, meaning that a basis of the kernel is 2 0 3 1 1 a basis of the image is 1 , 2 . 1 3

3.3.11 Here itis clear third the that only column is redundant, since it is equal to the first. Thus, a basis of the 1 1 0 kernel is 0 , and 0 , 1 is a basis of the image. −1 0 1 138

Section 3.3 3.3.12 The third vector is the only redundant vector here, shown by: 0 1 1 1

1 0 0 1

−1 0 0 . No other columns are redundant, however, meaning that a basis of the kernel is 1 , while 0 −1 1 0 1 a basis of the image is 1 , 0 . 1 1 3.3.13 Here we first see

2 [1

−1 0 3 , then 2 3] [1

0 −1 , 2 3]

3 2 so both the second and third columns are redundant, and a basis of the kernel is −1 , 0 . This leaves −1 0 ([ 1 ]) to be a basis of the image.

3.3.14 The first and the third columns are redundant, as the Kyle Numbers show us: 0 1 0 2 −1 1 0 0 , so that a basis of the kernel is 0 , 2 . This leaves ([ 1 ]) to be a basis , then 2] [0 1 [0 1 2] −1 0 of the image. 3.3.15 We quickly find that the third column is redundant, with the Kyle numbers 2 1 0 1 0

2 0 1 0 1

−1 2 2 2 2

0 1 0 1 0

0 0 1 0 1

0 2 2 2 2

0 0 0 , then see that the fourth column is also redundant, 0 0 1 0 0. 0 0

0 1 0 2 0 1 2 0 Thus, a basis of our kernel is , , while , is a basis of our image. 0 1 0 −1 1 0 1 0

3.3.16 The third column is redundant, as we find with 3 1 0 0 0

2 1 1 1 1

−1 5 2 2 2

0 1 2 . The fourth column, however, fails to be redundant. 3 4 139

Chapter 3 1 1 1 3 0 1 2 2 Thus, a basis of our kernel is , while , , is a basis of our image. 3 1 0 −1 4 1 0 0

3.3.17 For this problem, we again successively use Kyle Numbers to find our kernel, investigating the columns from left to right. We initially see that the first column is redunant: 0 0 2 −1 3 , then the third column: 0 1 2 4 0 0 0 0 0 1 0 2 3 Thus, 0 , −1 , 0 is a basis of the kernel, and 4 0 0 −1 0 0 1 0 0

0 1 0

0 2 0

0 0 1

0 0 1

0 0 3 3 , followed by the fifth column: 0 1 0 0 4

0 2 0

1 0 , is a basis of the image. 0 1

3.3.18 This matrix is already in rref, and we see that there are two columns without leading ones. These will be our redundant columns. Thus we see 2 1 0 0

1 −2 0 0

0 0 1 0

0 −1 5 0

0 −1 0 1 0 , and 0 1 0

0 −2 0 0

5 0 1 0

−1 −1 5 0

0 0 . 0 1

−1 2 0 0 1 1 0 Then 0 , 5 is a basis of the kernel, and 0 , 1 , 0 is a basis of the image. 1 0 0 −1 0 0 0 3.3.19 We see that the third column is redundant, and choose Kyle numbers as follows: 5 1 0 0 0

3 1 0 0 0

4 0 1 0 0 2 0 1 0 0

−1 5 4 0 0

0 3 2 0 0

0 0 0 , then we see that the fourth column is also redundant, 1 0

0 −1 0 5 3 0 4 2 0. 0 0 1 0 0 0 3 5 1 0 0 4 2 0 1 0 Thus, −1 , 0 is a basis of the kernel, and , , is a basis of the image. 0 0 1 −1 0 0 0 0 0 0 3.3.20 Although this matrix is not quite in rref, we can still quickly see that columns 2, 3, and 5 are the redundant columns: 140

4 0 1

−1 3 . 4

Section 3.3 0 5 −3 1 3 , 0 0 0 0 0 −12 5 0 1 0 0 So, 0 , −1 , 0 is 3 0 0 −1 0 0 0 1 0 0 0

1 0 0 0 0

0 5 0 0 0

0 3 1 0 0

0 −1 0 5 0 0 0 0 0 0

0 3 1 0 0

0 −3 3 , 0 0

−12 1 0 0 0

0 0 0 0 0

0 5 0 0 0

3 3 1 0 0

−1 −3 3 . 0 0

1 3 0 1 a basis of the kernel, and , is a basis of the image. 0 0 0 0

−3 1 0 −3 1 3 9 3.3.21 rref 4 5 8 = 0 1 4 , which we can use to “spot” a vector in the kernel: 4 . Since the third −1 0 0 0 7 6 3 column is the only redundant one, thisforms abasis of the kernel, and implies that the third column of A is also 1 3 redundant. Thus, a basis of im(A) is 4 , 5 . 7 6

2 4 8 1 0 −6 3.3.22 rref 4 5 1 = 0 1 5 . It is clear that the third vector is redundant, and we quickly see that the 0 0 0 7 9 3 −6 −6 vector 5 is in the kernel. Since this is the only redundant column, 5 is a basis of the kernel. Thus, −1 −1 2 4 a basis of im(A) is 4 , 5 . 7 9 1 0 3.3.23 rref 3 0

0 2 4 1 1 −3 −1 0 = 4 −6 8 0 −1 3 1 0

0 1 0 0

2 4 −3 −1 . 0 0 0 0

Using the method of Exercises 17 and 19, we find the kernel: 4 −1 0 −1 0 4 4 1 0 2 −1 , then 0 1 −3 −1 . 0 0 0 0 0 0 0 0 0 0 4 2 −3 −1 So a basis of ker(A) is , . The non-redundant column vectors of A form a basis of im(A): −1 0 0 −1 0 1 0 1 . , 4 3 −1 0 2 1 0 0 0

−3 0 1 0 0

−1 2 −3 0 0

141

Chapter 3 4 3 3.3.24 rref 2 1 2 −1 0 , 0 0

1 3 3.3.25 rref 1 2

8 6 4 2

1 1 1 3

1 1 6 2 5 0 = 0 9 10 0 2 0

2 0 0 0

0 1 0 0

0 0 1 0

0 0 . Here our kernel is the span of only one vector: 0 1

4 1 1 6 3 1 2 5 while a basis of the image of A is , , , . 2 1 9 10 1 3 2 0 2 6 2 4

3 9 4 9

2 6 1 1

1 1 3 0 = 2 0 2 0

2 0 0 0

0 1 0 0

5 −1 0 0

5 0 0 1 0 , then 0 1 0 0 0 5 2 −1 0 So a basis of ker(A) is 0 , −1 and −1 0 0 0

2 1 kernel: 0 0 0

−1 2 0 0 0

0 0 1 0 0

0 5 −1 0 0

0 0 . We will emulate Exercise 23 to find the 1 0 0 −1 2 0 0 1 0 0 0 0

−1 5 −1 0 0

0 0 0. 1 0

1 3 1 3 9 3 a basis of im(A) is , , . 1 4 2 2 9 2

3.3.26 a We notice that each of the six matrices has two identical In matrices C and L, the second column columns. 0 is identical to the third, so that ker(C) = ker(L) = span 1 . In matrices H, T, X and Y, the first column is −1 1 identical to the third, so that ker(H) = ker(T ) = ker(X) = ker(Y ) = span 0 . Thus, only L has the same −1 kernel as C. b We observe that each of the six matrices in the list has two identical rows. For example, the first and the y1 last row of matrix C are identical, so that any vector y2 in im(C) will satisfy the equation y1 = y3 . We y3 y1 y1 can conclude that im(C) = im(H) = im(X) = y2 : y1 = y3 , im(L) = y2 : y1 = y2 , and im(T ) = y3 y3 y1 im(Y ) = y2 : y2 = y3 . y3 c Our discussion in part b shows that the answer is matrix L.

3.3.27 Form a 4 × 4 matrix A with the given vectors as its columns. We find that rref(A) = I4 , so that the vectors 142

Section 3.3 do indeed form a basis of R4 , by Summary 3.3.10. 3.3.28 Form a 4 × 4 matrix A with the given vectors as its columns. The matrix A reduces to 1 0 0 0

0 1 0 0

0 0 1 0

2 3 . 4 k − 29

This matrix can be reduced further to I4 if (and only if) k − 29 6= 0, that is, if k 6= 29. By Summary 3.3.10, the four given vectors form a basis of R4 unless k = 29. 3.3.29 x1 = − 23 x2 − 12 x3 ; let x2 = s and x3 = t. Then the solutions are of the form 3 1 3 − 2 s − 21 t x1 −2 −2 x2 = = s 1 + t 0 . s x3 0 1 t

−1 −3 Multiplying the two vectors by 2 to simplify, we obtain the basis 2 , 0 . 2 0

1 −1 −2 2 0 0 3.3.30 Proceeding as in Exercise 29, we find the basis , , . 0 1 0 0 0 1 1 −2 −4 1 0 0 3.3.31 Proceeding as in Exercise 29, we can find the following basis of V : , , . 0 1 0 0 0 1 Now let A be the 4 × 3 matrix with these three vectors as its columns. Then im(A) = V by Theorem 3.1.3, and ker(A) = {~0} by Theorem 3.2.8, so that A does the job. 1 −2 −4 0 0 1 A= . 0 1 0 0 0 1

0 1 x1 x1 x2 1 x2 0 4 3.3.32 We need to find all vectors ~x in R such that · = 0 and · = 0. 2 −1 x3 x3 3 x4 1 x4 x1 − x3 + x4 = 0 , which in turn amounts to finding the This amounts to solving the system x + 2x3 + 3x4 = 0 2 1 0 −1 1 kernel of . 0 1 2 3

143

Chapter 3 −1 1 −2 −3 Using Kyle Numbers, we find the basis . , 0 1 1 0

3.3.33 We can write V = ker(A), where A is the 1 × n matrix A = [c1 c2 · · · cn ]. Since at least one of the ci is nonzero, the rank of A is 1, so that dim(V ) = dim(ker(A)) = n − rank(A) = n − 1, by Theorem 3.3.7. A “hyperplane” in R2 is a line, and a “hyperplane” in R3 is just a plane. 3.3.34 We can write V = ker(A), where A is the n × m matrix with entries aij . Note that rank(A) ≤ n. Therefore, dim(V ) = dim(ker(A)) = m − rank(A) ≥ m − n, by Theorem 3.3.7. 3.3.35 We need to find all vectors ~x in Rn such that ~v · ~x = 0, or v1 x1 + v2 x2 + · · · + vn xn = 0, where the vi are the components of the vector ~v . These vectors form a hyperplane in Rn (see Exercise 33), so that the dimension of the space is n − 1. 3.3.36 No; if im(A) = ker(A) for an n × n matrix A, then n = dim(ker(A)) + dim(im(A)) = 2 dim(im(A)), so that n is an even number. 3.3.37 Since 1 0 A= 0 0

dim(ker(A)) = 5 − rank(A), any 4 × 5 matrix with rank 2 will do; for example, 0 0 0 0 1 0 0 0 . 0 0 0 0 0 0 0 0

3.3.38 a The rank of a 3 × 5 matrix A is 0,1,2, or 3, so that dim(ker(A)) = 5 − rank(A) is 2,3,4, or 5. b The rank of a 7 × 4 matrix A is at most 4, so that dim(im(A)) = rank(A) is 0,1,2,3, or 4. 3.3.39 Note that ker(C) 6= {~0}, by Theorem 3.1.7a, and ker(C) ⊆ ker(A). Therefore, ker(A) 6= {~0}, so that A is not invertible. 3.3.40 Substituting a point (x, y) = (xi , yi ) into the equation c1 + c2 x + c3 y + c4 x2 + c5 xy + c6 y 2 = 0 produces a linear equation in the six unknows c1 , ..., c6 . Fitting a conic through m points Pi (xi , yi ), for i = 1, ..., m, amounts to solving a system of m homogenous linear equations in six unknowns. This system can be written in matrix form as A~x = ~0, where A is an m × 6 matrix. The ith row of A is 1 xi yi x2i xi yi yi2 .

3.3.41 The kernel of a 4 × 6 matrix is at least two-dimensional. Since every one-dimensional subspace of this kernel defines a conic through the four given points, there will be infinitely many such conics. 3.3.42 The kernel of a 5 × 6 matrix is at least one-dimensional, so that there is at least one conic passing through five given points. Exercise 1.2.52 provides an example with exactly one solution, while there are infinitely many solutions in Exercise 1.2.54. 3.3.43 Here, “anything can happen”: 144

Section 3.3 n o 1. If ker A = ~0 , then there is no solution, as in Exercise 1.2.59

2. If dim ker A = 1, then there is a unique solution. For example, the only conic through the six points (0, 0) , (1, 0) , (2, 0) , (3, 0) , (0, 1) , (0, 2) is xy = 0. 3. If dim ker A > 1, then there are infinitely many solutions. For example, any conic consisting of the x axis and some other line runs through the six points (0, 0) , (1, 0) , (2, 0) , (3, 0) , (4, 0) , (5, 0).

3.3.44 To run through the points (0, 0) , (1, 0) , (2, 0), and (3, 0), the cubic must satisfy the equations c1 = 0, c1 + c2 + c4 + c7 = 0, c1 + 2c2 + 4c4 + 8c7 = 0, and c1 + 3c2 + 9c4 + 27c7 = 0. This means that c1 = c2 = c4 = c7 = 0. Likewise, the cubic runs through the points (0, 0) , (0, 1) , (0, 2) and (0, 3) if (and only if) c1 = c3 = c6 = c10 = 0. Therefore, to run through the first seven given points, the equation must be of the form c5 xy + c8 x2 y + c9 xy 2 = 0. The last point, (1, 1) imposes the condition c5 + c8 + c9 = 0, or c5 = −c8 − c9 . Setting c8 = a and c9 = b, we find the family of cubics (−a − b) xy + ax2 y + bxy 2 = xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Such a cubic is the union of the x axis, the y axis, and some line through the point (1, 1). Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x 1

-1

2

1

-1

4

3

2

3

4

-1

-1

3.3.45 To run through the points (0, 0) , (1, 0) , (2, 0), and (3, 0), the cubic must satisfy the equations c1 = 0, c1 + c2 + c4 + c7 = 0, c1 + 2c2 + 4c4 + 8c7 = 0, and c1 + 3c2 + 9c4 + 27c7 = 0. This means that c1 = c2 = c4 = c7 = 0. Likewise, the cubic runs through the points (0, 0) , (0, 1) , (0, 2) and (0, 3) if (and only if) c1 = c3 = c6 = c10 = 0. Therefore, to run through the first seven given points, the equation must be of the form c5 xy + c8 x2 y + c9 xy 2 = 0 . The last two point, (1, 1) and (2, 2), impose the conditions c5 + c8 + c9 = 0 and 4c5 + 8c8 + 8c9 = 0 , implying that c5 = 0 and c8 = −c9 . Letting c9 = 1, we find that there is one such cubic, −x2 y + xy 2 = xy (y − x) = 0, the union of the x axis, the y axis, and the line y = x. y 4

3

2

1

x 1

-1

2

3

4

-1

3.3.46 Here we add the point (4, 0) to the list of points in Exercise 44 . In Exercise 44 we found the family of cubics xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Since all these cubics run through the point (4, 0), we find the same solutions here: xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. 145

Chapter 3 y

y 4

4

3

3

2

2

1

1

x

x 1

-1

2

1

-1

4

3

2

3

4

-1

-1

3.3.47 Here we add the point (2, 1) to the list of points in Exercise 45 . In Exercise 45 we found the cubic xy (y − x) = 0. Since this cubic fails to run through the point (2, 1), there are no solutions here. y 4

3

2

1

x 1

-1

2

4

3

-1

3.3.48 Here we add the point (3, 3) to the list of points in Exercise 45 . In Exercise 45 we found the cubic xy (y − x) = 0. Since this cubic runs though the point (3, 3), we get the same solution here, xy (y − x) = 0. y 4

3

2

1

x 1

-1

2

4

3

-1

3.3.49 Here we add the point (0, 4) to the list of points in Exercise 46 . In Exercise 46 we found the family of cubics xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Since all these cubics run through the point (0, 4), we find the same solutions here: xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. Each such cubic is the union of the x axis, the y axis, and some line through the point (1,1). Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x 1

-1

2

3

1

-1

4

2

3

4

-1

-1

3.3.50 To run through the points (0, 0) , (1, 0) , and (2, 0), the cubic must satisfy the equations c1 = 0, c1 +c2 +c4 + c7 = 0, and c1 + 2c2 + 4c4 + 8c7 = 0. This means that c1 = 0, c2 = 2c7 and c4 = −3c7 . Likewise, the cubic runs 146

Section 3.3 through the points (0, 0) , (0, 1) , and (0, 2) if (and only if) c1 = 0, c3 = 2c10 and c6 = −3c10 . Therefore, to run through the points (0, 0), (1, 0), (2,0), (0, 1), (0, 2) the cubic must be of the form c5 xy + c7 x3 − 3x2 + 2x + c8 x2 y + c9 xy 2 + c10 y 3 − 3y 2 + 2y = 0. The three additional points, (1, 1) , (2, 1) , and (1, 2) impose the conditions c5 + c8 + c9 = 0, 2c5 + 4c8 + 2c9 = 0 and 2c5 + 2c8 + 4c 9 = 0, implying that c5 = c8 = c9 = 0. Letting c7 = a and c10 = b, we find the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x -1

1

2

3

1

-1

4

2

3

4

-1

-1

3.3.51 Here we add the point (3, 2) to the list of points in Exercise 50 . In Exercise 50 we found the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. The point (3, 2) now imposed the condition a = 0, so that the only solution here is y 3 − 3y 2 + 2y = 0, the union of the lines y = 0, y = 1, and y = 2. y 4

3

2

1

x 1

-1

2

3

4

-1

3.3.52 Here we add the point (2, 2) to the list of points in Exercise 50 . In Exercise 50 we found the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0or b 6= 0. Since all these cubics run through the point (2, 2), we find the same solutions here: a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. Two sample solutions are shown in the accompanying figures. y

y 4

4

3

3

2

2

1

1

x

x -1

1

2

3

1

-1

4

2

3

4

-1

-1

3.3.53 Here we add the point (3, 3) to the list of points in Exercise 52 . In Exercise 52 we found the family of cubics a x3 − 3x2 + 2x + b y 3 − 3y 2 + 2y = 0 , where a 6= 0 or b 6= 0. The point (3, 3) now imposes the condition a = −b, so that the only cubic here (with b = 1) is −2x + 2y + 3x2 − 3y 2 − x3 + y 3 = 0, or x (x − 1) (x − 2) = y (y − 1) (y − 2). 147

Chapter 3 y 4

3

2

1

x 1

-1

2

3

4

-1

3.3.54 Substituting a point (x, y) = (xi , yi ) into the equation c1 + c2 x + ... + c9 xy 2 + c10 y 3 = 0 produces a linear equation in the ten unknows c1 , ..., c10 . Fitting a cubic through m points Pi (xi , yi ), for i = 1, ..., m, amounts to solving a system of m homogenous linear equations in ten unknowns. This system can be written in matrix form as A~x = ~0, where A is an m × 10 matrix. The ith row of A is 1 xi yi x2i xi yi yi2 x3i x2i yi xi yi2 yi3 . 3.3.55 The kernel of a 8 × 10 matrix is at least two-dimensional. Since every one-dimensional subspace of this kernel defines a cubic through the eight given points, there will be infinitely many such cubics. 3.3.56 The kernel of a 9 × 10 matrix is at least one-dimensional, so that there is at least one cubic passing through nine given points. Exercise 51 provides an example with exactly one solution, while there are infinitely many solutions in Exercise 52. 3.3.57 Here, “anything can happen”: n o If ker A = ~0 , then there is no solution, as in Exercise 47

If dim ker A = 1, then there is a unique solution, as in Exercise 48 If dim ker A > 1, then there are infinitely many solutions, as in Exercise 49

3.3.58 It is not true that “Any nine distinct points determine a unique cubic,” as illustrated in Exercises 46 and 52, although “most” sets of nine distinct points do determine a unique cubic. 3.3.59 Here we add the point P to the list of points in Exercise 44 . In Exercise 44 we found the family of cubics xy (ax + by − a − b) = 0, where a 6= 0 or b 6= 0. As long as we choose a point P on the x or on the y axis (with y = 0 or x = 0), all these cubics will run through P, so that there are infinitely many cubics through the nine given points. (Another, “cheap” solution is P = (1, 1)). However, if P is neither (1, 1) nor located on one of the axes, then the only cubic through the nine points is the union of the two axes and the line through (1, 1) and P.

148

Section 3.3 3.3.60 We can choose a basis ~v1 , . . . , ~vp in V , where p = dim(V ). Then ~v1 , . . . , ~vp are linearly independent vectors in W , so that dim(V ) = p ≤ dim(W ), by Theorem 3.3.4a, as claimed. 3.3.61 We can choose a basis ~v1 , . . . , ~vp of V , where p = dim(V ) = dim(W ). Then ~v1 , . . . , ~vp is a basis of W as well, by Theorem 3.3.4c, so that V = W = span(~v1 , . . . , ~vp ), as claimed. 3.3.62 Consider a basis ~v1 , . . . , ~vn of V . Since the ~vi are n linearly independent vectors in Rn , they form a basis of Rn (by parts (vii) and (ix) of Summary 3.3.10), so that V = span(~v1 , . . . , ~vn ) = Rn , as claimed. (Note that Exercise 3.3.62 is a special case of Exercise 3.3.61.) 3.3.63 dim(V + W ) = dim(V ) + dim(W ), by Exercise 3.2.51b. 3.3.64 Suppose that V ∩ W = {~0} and dim(V ) + dim(W ) = n. Choose a basis ~v1 , . . . , ~vp of V and a basis w ~ 1, . . . , w ~ q in W ; note that p + q = n. By Exercise 3.2.51b, the n vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q in Rn are linearly independent, so that they form a basis of Rn (by parts (vii) and (ix) of Summary 3.3.10). By Theorem 3.2.10, any vector ~x can be written uniquely as ~x = (c1~v1 + · · · + cp~vp ) + (d1 w ~ 1 + · · · + dq w ~ q ), with ~v = c1~v1 + · · · + cp~vp in V and w ~ = d1 w ~ 1 + · · · + dq w ~ q in W , which gives the desired representation. Conversely, suppose V and W are complements. Let us first show that V ∩ W = {~0} in this case. Indeed, if ~x is in V ∩ W , then we can write ~x = ~x + ~0 = ~0 + ~x ↑ ↑ in in V W

↑ ↑ in in V W

Since this representation is unique (by definition of complements), we must have ~x = ~0, so that V ∩ W = {~0}. By definition of complements, we have Rn = V + W , so that n = dim(V + W ) = dim(V ) + dim(W ), by Exercise 63. 3.3.65 Note that im(A) = span(~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q ) = V , since the w ~ j alone span V . To find a basis of V = im(A), we omit the redundant vectors from the list ~v1 , . . . , ~vp , w ~ 1, . . . w ~ q , by Theorem 3.2.4. Since the vectors ~v1 , . . . , ~vp are linearly independent, none of them are redundant, so that our basis of V contains all vectors ~v1 , . . . , ~vp and some of the vectors from the list w ~ 1, . . . , w ~q.

3.3.66 Use Exercise 65

1 2 Now rref 3 4

1 4 6 8

1 0 0 0

1 1 4 2 with ~v1 = , ~v2 = , and 6 3 8 4 1 0 2 0 0 0 0 0 0 1 −1 0 0 1 0 0 = 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1

w ~ i = ~ei for i = 1, 2, 3, 4.

− 14

1 4 − 12 − 34

.

0 0 1 1 2 4 1 0 Picking the non-redundant columns gives the basis , , , . 1 0 6 3 0 0 8 4 149

Chapter 3 3.3.67 Using the terminology suggested in the hint, we need to show that ~u1 , . . . , ~um , ~v1 , . . . , ~vp , w ~ 1, . . . , w ~q is a basis of V + W . Then dim(V + W ) + dim(V ∩ W ) = (m + p + q) + m = (m + p) + (m + q) = dim(V ) + dim(W ), as claimed. Any vector ~x in V + W can be written as ~x = ~v + w, ~ where ~v is in V and w ~ is in W . Since ~v is a linear combination of the ~ui and the ~vj , and w ~ is a linear combination of the ~ui and w ~ j , ~x will be a linear combination of the ~ui , ~vj , and w ~ k ; this shows that the vectors ~u1 , . . . , ~um , ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q span V + W . To show linear independence, consider the relation a1 ~u1 + · · · + am ~um + b1~v1 + · · · + bp~vp + c1 w ~ 1 + · · · + cq w ~ q = ~0. Then the vector a1 ~u1 + · · · + am ~um + b1~v1 + · · · + bp~vp = −c1 w ~ 1 − · · · − cq w ~ q is in V ∩ W , so that it can be expressed uniquely as a linear combination of ~u1 , . . . , ~um alone; this implies that the bi are all zero. Now our relation simplifies to a1 ~u1 + · · · + am ~um + c1 w ~ 1 + · · · + cq w ~ q = ~0, which implies that the ai and the cj are zero as well (since the vectors ~u1 , . . . , ~um , w ~ 1, . . . , w ~ q are linearly independent). 3.3.68 By Exercise 3.3.67, dim(V ∩ W ) = dim(V ) + dim(W ) − dim(V + W ) = 13 − dim(V + W ). The dimension of V + W is at least 7 (since W ⊆ V + W ) and at most 10 (since V + W ⊆ R10 ); therefore the dimension of V ∩ W is at least 3 and at most 6. 3.3.69 The nonzero rows of E span the row space, and they are linearly independent (consider the leading ones), so that they form a basis of the row space: [0 1 0 2 0], [0 0 1 3 0], [0 0 0 0 1]. 3.3.70 As in Exercise 3.3.69, we observe that the nonzero rows of E form a basis of the row space, so that dim(row space of E) = rank(E). 3.3.71 a All elementary row operations leave the row space unchanged, so that A and rref(A) have the same row space. b By part (a) and Exercise 3.3.70, dim(row space of A) = dim(row space of rref(A)) = rank(rref(A)) = rank(A). 1 0 3.3.72 rref(A) = 0 0

0 −1 1 2 0 0 0 0

−2 3 0 0

By Exercises 3.3.70 and 3.3.71a, [1 0 − 1 − 2], [0 1 2 3] is a basis of the row space of A.

3.3.73 Using the terminology suggested in the hint, we observe that the vectors ~v , A~v , . . . , An~v are linearly dependent (by Theorem 3.2.8), so that there is a nontrivial relation c0~v + c1 A~v + · · · + cn An~v = ~0. We can rewrite this relation in the form (c0 In + c1 A + · · · + cn An )~v = ~0.

The nonzero vector ~v is in the kernel of the matrix c0 In + c1 A + · · · + cn An , so that this matrix fails to be invertible. 1 3.3.74 We can use the approach outlined in Exercise 3.3.73, with ~v = , say. 0 −3 1 −3 −4 1 1 2 . = , and A ~v = , A~v = Then ~v = 4 0 4 −3 2 0 We find the relation 5~v − 2A~v + A2~v = ~0, so that the matrix 5I2 − 2A + A2 does the job. 150

Section 3.3 3.3.75 If rank(A) = n, then the n non-redundant columns of A form a basis of im(A) = Rn , so that the matrix formed by the non-redundant columns is invertible (by Summary 3.3.10). Conversely, if A has an invertible n × n submatrix B, then the columns of B form a basis of Rn (again by Summary 3.3.10), so that im(A) = Rn and therefore rank(A) = dim(im(A)) = n. 3.3.76 Using the terminology suggested in the Exercise, we multiply the relation c0~v + c1 A~v + · · · + cm−1 Am−1~v = ~0 with Am−1 and obtain c0 Am−1~v = ~0 (all other terms vanish since Am = 0). Since the vector Am−1~v is nonzero (by construction), the scalar c0 must be zero, and our relation simplifies to c1 A~v + c2 A2~v + · · · + cm−1 Am−1~v = ~0. Now we multiply both sides with Am−2 and obtain c1 Am−1~v = ~0, so that c1 = 0 as above. Continuing like this we conclude that all the ci must be zero, as claimed. 3.3.77 As in Exercise 76, let m be the smallest positive integer such that Am = 0. In Exercise 76 we construct m linearly independent vectors ~v , A~v , . . . , Am−1~v in Rn ; now m ≤ n by Theorem 3.2.8. Therefore An = Am An−m = 0An−m = 0, as claimed. 3.3.78 If the vectors w ~ 1, . . . , w ~ q span an m-dimensional space V (with basis ~v1 , . . . , ~vm ), then m ≤ q by Theorem 3.3.1 (since the vectors ~vi are linearly independent). 3.3.79 Prove Theorem 3.3.4d: If m vectors ~v1 , . . . , ~vm span an m-dimensional space V , then they form a basis of V . We need to show that the vectors ~vi are linearly independent. We will argue indirectly, assuming that the vectors are linearly dependent; this means that at least one of the vectors ~vi is redundant, say ~vp . But then V = span(~v1 , . . . , ~vp , . . . , ~vm ) = span(~v1 , . . . , ~vp−1 , ~vp+1 , . . . , ~vm ), contradicting Theorem 3.3.4b. 3.3.80 im(A) is the plane onto which we project, so that rank(A) = dim(im(A)) = 2. 3.3.81 a Note that rank(B) ≤ 2, so that dim(ker(B)) = 5 − rank(B) ≥ 3 and dim(ker(AB)) ≥ 3 since ker(B) ⊆ ker(AB). Since ker(AB) is an subspace of R5 , dim(ker(AB)) could be 3,4, or 5. It is easy to give an example 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 1 , then AB = for each case; for example, if A = and and B = 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 dim(ker(AB)) = 3. b Since dim(im(AB)) = 5 − dim(ker(AB)), the possible values of dim(im(AB)) are 0,1, and 2, by part a. 3.3.82 Write A = [~v1 . . . ~vm ] and B = [w ~1 . . . w ~ m ], so that A + B = [~v1 + w ~ 1 · · · ~vm + w ~ m ]. Any linear combination of the columns of A + B, ~y = c1 (~v1 + w ~ 1 ) + · · · + cm (~vm + w ~ m ), can be written as ~ 1 + · · · + cm w ~ m) ~y = (c1~v + · · · + cm~vm ) + (c1 w {z } | {z } | in im(A)

in im(B)

so that im(A + B) ⊆ im(A) + im(B) (see Exercise 3.2.50). Since dim(V + W ) ≤ dim(V ) + dim(W ), by Exercise 3.3.67, we can conclude that rank(A+B) = dim(im(A+B)) ≤ dim(im(A))+dim(im(B)) = rank(A)+rank(B). Summary: rank(A + B) ≤ rank(A) + rank(B). 151

Chapter 3 3.3.83 a By Exercise 3.1.39b, im(AB) ⊆ im(A), and therefore rank(AB) ≤ rank(A). b Write B = [~v1 · · · ~vm ] and AB = [A~v1 · · · A~vm ]. If r = rank(B), then the r non-redundant columns of B will span im(B), and the corresponding r columns of AB will span im(AB), by linearity of A. By Theorem 3.3.4b, rank(AB) = dim(im(AB)) ≤ r = rank(B). Summary: rank(AB) ≤ rank(A), and rank(AB) ≤ rank(B). 3.3.84 Same answer as Exercise 3.3.85. 3.3.85 Let ~v1 , . . . , ~v6 be the columns of matrix A. Following the hint, we observe that ~v5 = 4~v1 + 5~v2 + 6~v4 , which gives the relation 4~v1 + 5~v2 + 6~v4 − ~v5 = ~0. Thus the vector 4 5 0 ~x = 6 −1 0 is in the kernel of matrix A. Since ~x fails to be in the kernel of matrix B, the two kernels are different, as claimed.

3.3.86 We will freely use the terminology introduced in the hint. First we need to show that at ileast one of the h ~ ~ column vectors ~ak and bk fails to contain a leading 1. If rank[~a1 · · · ~ak−1 ] = rank b1 · · · ~bk−1 = r, and if ~ak contains a leading 1, then ~ak is the standard vector ~er+1 ; likewise for ~bk . Since ~ak and ~bk are different vectors, they cannot both contain a leading 1. Without loss of generality, we can assume that ~ak fails to contain a leading 1, so that ~ak is redundant: We can write ~ak = c1~a1 + · · · + ck−1~ak−1 . Then the vector c1 .. . ck−1 ~x = −1 is in the kernel of A. We will show that ~x fails to be in the kernel of matrix B, so that ker(A) 6= 0 . .. 0 ker(B), as claimed. Indeed, B~x = c1~b1 + · · · + ck−1~bk−1 − ~bk = c1~a1 + · · · + ck−1~ak−1 − ~bk = ~ak − ~bk 6= ~0. We have used the fact that the first k − 1 columns of B are identical to those of A, while ~bk 6= ~ak . 3.3.87 Exercise 86 shows that if two matrices A and B of the same size are both in rref and have the same kernel, then A = B. Apply this fact to A and B = rref(M ).

Section 3.4 2 0 1 2 . , so [~x]B = +3 =2 3.4.1 3 1 0 3 3.4.2

0 3 1 −4 = −4 +3 . , so [~x]B = −4 1 3 0 152

Section 3.4

3.4.3

0 31 23 31 . , so [~x]B = +1 =0 1 37 29 37

3.4.4

23 = 29

3.4.5

−4 5 2 7 . , so [~x]B = +3 = −4 3 12 5 16

1 2

1 61 46 , so [~x]B = 2 . +0 0 67 58

This may not be as obvious as Exercises 1 and 3, but we can find our coefficients simply by reducing the matrix 2 5 .. 7 . . 5 12 16 3.4.6

−4 1 5 11 = 11 −3 , so [~x]B = . 4 2 6 −3

We arrive at this solution by reducing the matrix

1 2

5 .. −4 . . 6 4

0 1 3 3.4.7 We need to find the scalars c1 and c2 such that 1 = c1 −1 + c2 1 . Solving a linear system gives −1 0 −4 c1 3 c1 = 3, c2 = 4. Thus [~x]B = = . c2 4

2 1 2 3.4.8 We need to find the scalars c1 and c2 such that 3 = c1 1 + c2 0 . Attempting to solve the linear 1 0 4 system reveals an inconsistency; ~x is not in the span of ~v1 and ~v2 . 3.4.9 We can solve this by inspection: Note that our first coefficient must be 3 because of the first terms of the vectors. Also, the second coefficient must be 2 due to the last terms. 3 0 3 However, 3~v1 + 2~v2 = 3 + −2 = 1 . Thus, we reason that ~x is not in the span of ~v1 and ~v2 . 4 4 0 1 0 .3 We can also see this by attempting to solve 1 −1 .. 3 , which turns out to be inconsistent. Thus, ~x is not in 0 2 4 V. 3 3.4.10 Proceeding as in Example 1, we find [~x]B = . 1 3.4.11 Proceeding as in Example 1, we find [~x]B =

1 2 1 2

.

−3 . 3.4.12 Proceeding as in Example 1, we find [~x]B = 5

153

Chapter 3 3.4.13 Here, we quickly see that since x1 = 1 = 1c1 + 0c2 + 0c3 , c1 must equal 1. We find c2 = −1 similarly, since x2 = 1 = 2(1) + 1c2 + 0c3 . Finally, now that x3 = 3(1) + 2(−1) + 1c3 , c3 must be zero. 1 0 0 1 1 So 1 = 1 2 − 1 1 + 0 0 , and [~x]B = −1 . 0 1 2 3 1 3.4.14 We proceed by inspection here, noting that we need c1 = 3, then see that c2 must be 4. Finally, c3 must be 6. 3 0 0 1 3 Thus, 7 = 3 1 + 4 1 + 6 0 , and [~x]B = 4 . 6 1 1 1 13 1 1.1 4 .. 0 to 0 0 8 0

1 1 3.4.15 This may be a bit too difficult to do by inspection. Instead we reduce 2 3 1 4

0 0. 8 1 0 .. −12 , 0 1 5

8 revealing that ~x = 8~v1 − 12~v2 + 5~v3 , and [~x]B = −12 . 5

1 1 3.4.16 We reduce 1 2 1 3

1 0 1.7 3 .. 1 to 0 1 0 0 6 3

0 . 21 0 .. −22 , 1 8

21 revealing that ~x = 21~v1 − 22~v2 + 8~v3 , and [~x]B = −22 . 8 3.4.17 By inspection, we see that in order for ~x to be in V , ~x = 1~v1 +1~v2 −1~v3 (by paying attention to the first, second, and fourth terms). Now we need to verify that the third terms “work out”. So, 1(2)+1(3)−1(4) = 5−4 = 1 = x3 .

1 Thus ~x is in V , and [~x]B = 1 . −1 3.4.18 Here, ~x is not in V, as we find an inconsistency while attempting to solve the system.

1 3.4.19 a S = 1 Then B = S

−1

1 , and we find the inverse S −1 to be equal to −1 AS =

1 2

1 1

1 −1

0 1

1 0

1 1 = 1 −1

1 2

b Our commutative diagram: 154

1 1 −1 1

1 2

1 1 . 1 −1

1 1 = 1 −1

1 2

2 0

0 1 = −2 0

0 . −1

Section 3.4 1 1 + c2 ~x = c1 −1 1 y

c [~x]B = 1 c2 So,

a b c d

c1 c2

1 1 T (~x) = A~x = c1 A + c2 A −1 1 1 −1 1 1 = c1 + c2 = c1 − c2 1 1 1 −1 y c1 −−−→ B [T (~x)]B = −c2 −−→ T

c B = [[T (~v1 )]B [T (~v2 )]B ] =

3.4.20 a S =

0 1

1 0

1 1 B

0 1 1 0

1 −1

−1

AS =

1 2

1 1

1 −1

1 1

1 1

1 1 = 1 −1

1 2

=

B

1 , and we find the inverse S −1 to be equal to −1

1 1

Then B = S

0 . −1

c1 1 = , and we quickly find B = −c2 0

2 2 0 0

1 2

1 −1 1 0 = . 1 B 1 B 0 −1

1 1 . 1 −1

1 1 = 1 −1

1 2

0 2 0 = . 0 0 0

4 0

b Our commutative diagram: 1 1 + c2 A T (~x) = A~x = c1 A −1 1 2 0 1 1 = c1 + c2 = 2c +0 2 0 1 1 −1 y 2c1 −−−→ B [T (~x)]B = 0

1 1 + c2 ~x = c1 −1 1 y

[~x]B =

So,

a b c d

c1 c2

c1 c2

−−→ T

2c1 2 0 . = , and we quickly find B = 0 0 0

c B = [[T (~v1 )]B [T (~v2 )]B ] =

3.4.21 a S =

1 3

Then B = S

−1

1 1

1 1

1 1 B

1 1 1 1

1 −1

B

−2 , and we find the inverse S −1 to be equal to 1 AS =

1 7

1 −3

2 1

1 3

2 6

1 −2 = 3 1

1 7

b Our commutative diagram: 155

7 14 0 0

1 7

2 0 2 0 = = . 2 B 0 B 0 0

1 2 . −3 1

1 −2 = 3 1

1 7

49 0

0 7 0 = . 0 0 0

Chapter 3 −2 1 + c2 ~x = c1 1 3

1 −2 T (~x) = A~x = c1 A + c2 A 3 1 7 0 1 = c1 + c2 = 7c1 21 0 3 y y 7c1 c1 −−−→ B [T (~x)]B = [~x]B = 0 c2 c1 a b 7c1 7 0 . So, = , and we quickly find B = c d c2 0 0 0 −−→ T

c B = [[T (~v1 )]B [T (~v2 )]B ] = =

7 21

B

1 2 3 6 0 . 0

0 7 = 0 B 0

1 3 B

1 2 3 6

−2 1

B

1 2 −2 1 −1 . , and we find the inverse S to be equal to 5 −2 1 1 1 2 5 10 −3 4 1 −2 1 −2 Then B = S −1 AS = 15 = = 15 −2 1 10 −5 4 3 2 1 2 1

1 3.4.22 a S = 2

1 5

25 0

b Our commutative diagram: −2 1 −−→ T + c2 ~x = c1 1 2

−2 1 + c2 A T (~x) = A~x = c1 A 1 2 5 10 1 −2 = c1 + c2 = 5c1 − 5c2 10 −5 2 1 y y 5c1 c1 −−−→ B [T (~x)]B = [~x]B = −5c2 c2 a b c1 5c1 5 0 So, = , and we quickly find B = . c2 c d −5c2 0 −5

c B = [[T (~v1 )]B [T (~v2 )]B ] = =

5 10

B

10 −5

B

=

5 0

−3 4

4 3

1 2 B

−3 4 4 3

−2 1

B

0 . −5

2 −1 1 −1 . , and we find the inverse S to be equal to −1 1 2 2 0 1 1 4 −2 1 1 5 −3 2 −1 −1 . = = Then B = S AS = 0 −1 1 2 1 −1 1 2 6 −4 −1 1

1 3.4.23 a S = 1

156

0 5 0 = . −25 0 −5

Section 3.4 b Our commutative diagram: 1 1 −−→ + c2 ~x = c1 T 2 1 y

c [~x]B = 1 c2 So,

a b c d

c1 c2

1 1 T (~x) = A~x = c1 A + c2 A 1 2 1 1 −1 2 − 1c2 = 2c1 + c2 = c1 2 1 −2 2 y 2c1 −−−→ B [T (~x)]B = −c2

2c1 2 = , and we quickly find B = −c2 0

c B = [[T (~v1 )]B [T (~v2 )]B ] = 2 = 2 B 3.4.24 a S =

−1 −2

5 6

−3 −4

1 1 B

5 −3 6 −4

1 2 B

2 0 = . 0 −1

B

3 −5 5 . , and we find the inverse S −1 to be equal to −1 2 3

2 1

Then B = S

0 . −1

−1

3 −5 AS = −1 2

13 6

−20 −9

2 5 9 −15 2 = 1 3 −1 2 1

5 3 = 3 0

0 . 1

b Our commutative diagram: 5 2 −−→ T + c2 ~x = c1 3 1 y

[~x]B =

So,

a b c d

c1 c2

c1 c2

5 2 + c2 A T (~x) = A~x = c1 A 3 1 6 5 2 5 = c1 + c2 = 3c + 1c2 3 3 1 1 3 y 3c1 −−−→ B [T (~x)]B = c2

=

3c1 3 0 . , and we quickly find B = c2 0 1

c B = [[T (~v1 )]B [T (~v2 )]B ] = 6 = 3 B

13 6

−20 −9

2 1 B

13 6

−20 −9

5 3 B

5 3 0 . = 3 B 0 1

3.4.25 We will use the commutative diagram method here (though any method suffices). 157

Chapter 3 1 1 + c2 ~x = c1 2 1

1 2 1 2 1 1 T (~x) = A~x = c1 +c 3 4 1 2 3 4 2 3 5 = c1 + c2 7 11 1 1 1 1 +6 + c2 −1 +4 = c1 −1 1 2 1 2 1 1 + (4c1 + 6c2 ) = (−c1 − c2 ) 2 1 y y c1 −c1 − c2 −−−→ [~x]B = B [T (~x)]B = c2 4c1 + 6c2 c −c1 − c2 −1 −1 B 1 = , so B = . c2 4c1 + 6c2 4 6 −−→ T

3.4.26 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B ] 1 0 1 1 0 1 = 1 B 2 3 2 B 2 3 1 2 6 4 = . = 5 B 8 B −4 −3 3.4.27 We use a commutative diagram: 2 0 1 −−→ ~x = c1 1 + c2 2 + c3 0 T −2 1 1

y

c1 [~x]B = c2 c3 9 9c1 c1 B c2 = 0 , so B = 0 0 0 c3

T (~x) = A~x

1 0 2 = c1 A 1 + c2 A 2 + c3 A 0 1 1 −2 2 18 = c1 9 + ~0 + ~0 = 9c1 1 −2 −18 y 9c1 −−−→ B [T (~x)]B = 0 0 0 0 0 0 . 0 0

3.4.28 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] 158

Section 3.4

5 −4 −2 5 2 −4 = −4 5 −2 2 −2 −2 8 −2 1 B 0 9 0 0 −9 9 = 0 = 0 0 B 0 B −18 B 0

−4 −2 1 5 −2 −1 −2 8 0 B 0 0 9 0 . 0 9

5 −4 −2 0 −4 5 −2 1 −2 −2 8 −2 B

3.4.29 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] −1 1 0 1 −1 0 = 0 −2 2 1 3 −9 6 1 3 B 0 1 2 0 2 6 = 0 = 0 0 B 3 B 12 B 0

1 0 1 −2 2 2 −9 6 3 B 0 0 1 0 . 0 2

0 1 2 3 6 6 B

1 −1 0 2 4 1 B

−1 1 0 −2 3 −9

3.4.30 Let’s build B “column-by-column”: B = [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] 0 2 1 0 2 −1 2 −1 = 2 −1 0 1 4 −4 1 4 −4 1 B 0 0 1 1 0 0 = 0 −1 −1 = 1 0 B −2 B 1 B 0 0

0 −1 0 1 2 1 B 0 0 . 0

0 2 2 −1 4 −4

3.4.31 We can use a commutative diagram to see how this works: ~x = c1~v1 + c2~v2 + c3~v3 y

c1 [~x]B = c2 c3 c3 c1 B c2 = 0 , −c1 c3

−−→ T

T (~x) = ~v2 × ~x = c1 (~v2 × ~v1 ) + c2 (~v2 × ~v2 ) + c3 (~v2 × ~v3 ) = c1 (−~v3 ) + c2 (~0) +c3 (~v1 ) = c3~v1 − c1~v3 y c3 −−−→ B [T (~x)]B = 0 −c1 0 0 1 so B = 0 0 0 . −1 0 0

3.4.32 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

= [ [~v1 × ~v3 ]B

[~v2 × ~v3 ]B

[T (~v3 )]B ] [~v3 × ~v3 ]B ] = [ [−~v2 ]B

[~v1 ]B ~0 ], since all three are perpendicular unit vectors. 159

Chapter 3

0 So, B = −1 0

1 0 0 0 . 0 0

3.4.33 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

[T (~v3 )]B ]

= [ [(~v2 · ~v1 )~v2 ]B [(~v2 · ~v2 )~v2 ]B 0 0 0 So, B = 0 1 0 . 0 0 0

[(~v2 · ~v3 )~v2 ]B ] = [ ~0

[1~v2 ]B ~0 ], since all three are perpendicular unit vectors.

3.4.34 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

[T (~v3 )]B ]

= [ [~v1 − 2(~v3 · ~v1 )~v3 ]B [~v2 − 2(~v3 · ~v2 )~v3 ]B [~v3 − 2(~v3 · ~v3 )~v3 ]B ] = [ [~v1 ]B [~v2 ]B [−~v3 ]B ]. 1 0 0 So, B = 0 1 0 . This is the reflection about the plane spanned by ~v1 and ~v2 . 0 0 −1 3.4.35 Using another commutative diagram: −−→ T ~x = c1~v1 + c2~v2 + c3~v3 T (~x) = c1 T (~v1 ) + c2 T (~v2 ) + c3 T (~v3 ) = c1 (~v1 − 2(~v1 · ~v1 )~v2 ) + c2 (~v2 − 2(~v1 · ~v2 )~v2 )+ c3 (~v3 − 2(~v1 · ~v3 )~v2 ) = c1 (~v1 − 2~v2 ) + c2 (~v2 − ~0) + c3 (~v3 − ~0) = c1~v1 + (−2c1+ c2 )~v2 + c3~v3 y y c1 c1 − − − → B [T (~x)]B = −2c1 + c2 [~x]B = c2 c3 c3 1 0 0 So B = −2 1 0 . This is a shear along the second term. 0 0 1 3.4.36 Here we will build B column-by-column: B = [ [T (~v1 )]B

[T (~v2 )]B

= [ [~v1 × ~v1 + (~v1 · ~v1 )~v1 ]B = [ [~v1 ]B

[T (~v3 )]B ] [~v1 × ~v2 + (~v1 · ~v2 )~v1 ]B

[~v1 × ~v3 + (~v1 · ~v3 )~v1 ]B ]

[~v3 ]B

[−~v2 ]B ]. 1 0 0 So, B = 0 0 −1 . This is a 90-degree rotation about the line spanned by ~v1 . The rotation is counterclockwise 0 1 0 when looking from the positive ~v1 direction. 160

Section 3.4 3.4.37 We want a basis B = (~v1 , ~v2 ) such that T (~v1 ) = a~ v1 a the B-matrix of T will be B = [ [T (~v1 )]B [T (~v2 )]B ] = 0 that T (~v ) = ~v = 1~v for vectors parallel to the line L onto perpendicular L. Thus, we can pick a basis where ~v1 is to 1 −2 B= , . 2 1

and T (~v2 ) = b~v2 for some scalars a and b. Then 0 , which is a diagonal matrix as required. Note b which we project, and T (w) ~ = ~0 = 0w ~ for vectors parallel to L and ~v2 is perpendicular, for example,

3.4.38 We want a basis B = (~v1 , ~v2 ) such that T (~v1 ) = T (~v2 ) = b~v2 for some scalars a and b. Then the a~v1 and a 0 , which is a diagonal matrix as required. Note that B-matrix of T will be B = [ [T (~v1 )]B [T (~v2 )]B ] = 0 b T (~v ) = ~v = 1~v for vectors parallel to the line L about which we reflect, and T (w) ~ = −w ~ = (−1)w ~ for vectors perpendicular to L. Thus, we can pick a basis where ~ v is parallel to L and ~ v is perpendicular, for example, 1 2 −3 2 . , B= 2 3 3.4.39 Using the same approach as in Exercise 37, we want a basis, ~v1 , ~v2 , ~v3 such that T (~v1 ) = a~v1 , T (~v2 ) = b~v2 and 1 T (~v3 ) = c~v3 . First we see that if ~v1 = 2 , then T (~v1 ) = ~v1 . Next we notice that if ~v2 and ~v3 are perpendicular 3 −3 −2 to ~v1 , then T (~v2 ) = −~v2 and T (~v3 ) = −~v3 . So we can pick ~v2 = 1 and ~v3 = 0 , for example. 1 0 3.4.40 From Exercise 37, we see that we want one of our basis vectors to parallel be to the line, while the others 1 1 1 must be perpendicular the line. We can easily find such a basis: B = 1 , −1 , 0 . 1 0 −1 3.4.41 We will use the same approach as in Exercises 37 and 39. Any basis with 2vectors and one in the plane 0 −1 perpendicular to it will work nicely here! So, let ~v1 , ~v2 be in the plane. ~v1 can be 3 , and ~v2 = −2 (note 0 1 3 that these must be independent). Then ~v3 should be perpendicular to the plane. We will use ~v3 = 1 —the 2 3 coefficient vector. This is perpendicular to the plane because all vectors perpendicular to 1 lie in the plane. 2

−1 0 3 So, our basis is: 3 , −2 , 1 . 0 1 2 3.4.42 From Exercise 38, we deduce that one of our vectors should be perpendicular to this plane, while two should 1 fall inside it. Finding the perpendicular is not difficult: we simply take the coefficient vector: −2 . Then 2 161

Chapter 3 0 2 we add two linearly independent vectors on the plane, 1 , 1 , for instance. These three vectors form one 1 0 possible basis. 4 −2 −1 3.4.43 By definition of coordinates (Definition 3.4.1), ~x = 2 0 + (−3) 1 = −3 . 2 0 1

11 5 8 3.4.44 By definition of coordinates, ~x = 2 4 + (−1) 2 = 6 . −1 −1 −1

3.4.45 If ~v1 , ~v2 is a basis with the desired property, then ~x = 2~v1 + 3~v2 , or ~v2 = 13 ~x − 32 ~v1 . Thus we can make ~v 1 any 3 vector in the plane that is not parallel to ~x, and then let ~v2 = 31 ~x − 32 ~v1 . For example, if we choose ~v1 = 2 , 0 −4 then ~v2 = 13 −4 . −1 3.4.46 As in Exercise 3.4.45, we canmake~v1 any vectorin the plane that is not parallel to ~x, and then let ~v2 = 2~v1 −~x. 1 1 For example, if we choose ~v1 = 0 , then ~v2 = 1 . −3 −1 3.4.47 By Theorem 3.4.4, we have A = SBS

3.4.48 [~x]B =

−1 2

−1

0 1 = 1 0

a b c d

0 1 1 0

−1

d = b

means that ~x = −~v + 2w. ~ See Figure 3.6

Figure 3.6: for Problem 3.4.48. −1 . 3.4.49 ~u + ~v = −w, ~ so that w ~ = −~u − ~v , i.e., [w] ~B= −1

1 2 −−→ −−→ −−→ −−→ . , OQ = ~v + 2w, ~ so that [OQ]B = 3.4.50 a OP = w ~ + 2~v , so that [OP ]B = 2 1 −−→ b OR = 3~v + 2w. ~ See Figure 3.7. 162

c . a

Section 3.4

Figure 3.7: for Problem 3.4.50. c If the tip of ~u is a vertex, then so is the tip of ~u + 3~v and also the tip of ~u + 3w ~ (draw a sketch!). We know that −→ the tip P of 2~v + w ~ is a vertex (see part a.). Therefore, the tip S of OS = 17~v + 13w ~ = (2~v + w) ~ + 5(3~v ) + 4(3w) ~ is a vertex as well. 3.4.51 Let B = (~v1 , ~v2 , · · · , ~vm ). Then, let ~x = a1~v1 + a2~v2 + · · · + am~vm and ~y = b1~v1 + b2~v2 + · · · + bm~vm . Then [~x + ~y ]B = [a1~v1 + a2~v2 + · · · + am~vm + b1~v1 + b2~v2 + · · · + bm~vm ]B = [(a1 + b1 )~v1 + (a2 + b2 )~v2 + · · · + (am + bm )~vm ]B b1 a1 a1 + b1 a2 + b2 a2 b2 = . + . = [~x]B + [~y ]B = .. .. .. . am

am + bm

bm

3.4.52 Yes; T (~x) = [~x]B = S −1 ~x, so that T is “given by a matrix.” (See Definition 2.1.1.)

1 3 3.4.53 By Definition 3.4.1, we have ~x = S[~x]B = 2 4

7 40 = . 11 58

3.4.54 Let Q be the matrix whose columns are the vectors of the basis T . Then [[~v1 ]T . . . [~vn ]T ] = [Q−1~v1 . . . Q−1~vn ] = Q−1 [~v1 . . . ~vn ] is an invertible matrix, so that the vectors [~v1 ]T . . . [~vn ]T form a basis of Rn . 1 1 1 1 3 1 1 [~x]B = [~x]R , so that [~x]B and ~x = 3.4.55 By Definition 3.4.1, we have ~x = 2 1 2 2 4 1" 2 # −1 1 − 1 1 3 1 1 2 [~x]R = [~x]B , i.e., P = . 1 2 4 1 2 0 2 | {z }

3 [~x]R and 4

P

1 3 3 2 3.4.56 Let S = [~v1~v2 ] where ~v1 , ~v2 is the desired basis. Then by Theorem 3.4.1, =S and =S , 2 5 4 3 −1 3 2 1 3 1 3 3 2 12 −7 12 −7 i.e. S = . Hence S = = . The desired basis is , . 5 3 2 4 2 4 5 3 14 −8 14 −8 3.4.57 If we can find a basis B = (~v1 , ~v2 , ~v3 ) such that the B-matrix of A is 1 0 0 1 0 0 B = 0 1 0 , then A must be similar to 0 1 0 . Because of the entries in the matrix B, it is required 0 0 −1 0 0 −1 that A~v1 = ~v1 , A~v2 = ~v2 and A~v3 = −~v3 . So, all we need for our basis is to pick independent ~v1 , ~v2 in the plane, and ~v3 perpendicular to the plane. 3.4.58 a Consider a linear relation c1 A2~v + c2 A~v + c3~v = ~0. 163

Chapter 3 Multiplying A2 with the vectors on both sides and using that A3~v = ~0 and A4~v = ~0, we find that c3 A2~v = ~0 and therefore c3 = 0, since A2~v 6= ~0. Therefore, our relation simplifies to c1 A2~v + c2 A~v = ~0. Multiplying A with the vectors on both sides we find that c2 A2~v = ~0 and therefore c2 = 0. Then c1 = 0 as well. We have shown that there is only the trivial relation among the vectors A2~v , A~v , and ~v , so that these three vectors from a basis of R3 , as claimed. 0 b T (A2~v ) = A3~v = ~0 so [T (A2~v )]B = 0 . 0 1 T (A~v ) = A2~v so [T (A~v )]B = 0 . 0 0 T (~v ) = A~v so [T (~v )]B = 1 . 0

0 Hence, by Theorem 3.4.3, the desired matrix is 0 0

1 0 0 1 . 0 0

x y x y 2 1 = , or z t z t 0 3 −y y 2x x + 3y 2x 2y , where y and t are arbitrary constants. . The solutions are of the form S = = 0 t 2z z + 3t 3z 3t 2 0 2 1 Since there are invertible solutions S (for example, let y = t = 1), the matrices and are indeed 0 3 0 3 similar.

3.4.59 First we find the matrices S =

x y z t

such that

2 0

0 3

1 0 x y x y 0 1 x y 3.4.60 First we find the matrices S = such that = , or, = z t z t 1 0 −z −t 0 −1 y x y y . The solutions are of the form S = , where y and t are arbitrary constants. Since there are t z −t t 0 1 1 0 are indeed similar. and invertible solutions S (for example, let y = t = 1), the matrices 1 0 0 −1 x z

y t

x y x y 3.4.61 We seek a basis ~v1 = , ~v2 = such that the matrix S = [~v1 ~v2 ] = satisfies the equation z t z t 3z z 3t −2 4− 2 1 1 x y x y −5 −9 . . Solving the ensuing linear system gives S = = 0 1 z t z t 4 7 z t −9 0 We need to choose z and t so that S will be invertible. For example, if we let z = 6 and t = 1, then S = , 6 1 0 −9 . , ~v2 = so that ~v1 = 1 6 164

Section 3.4 x y y x satisfies the equation such that the matrix S = [~v1 ~v2 ] = , ~v2 = 3.4.62 We seek a basis ~v1 = t zz t z −t 5 0 x y x y 1 2 . We need to choose . Solving the ensuing linear system gives S = 2 = z t 0 −1 z t z t 4 3 1 −1 both z and t nonzero to make S invertible. For example, if we let z = 2 and t = 1, then S = , so that 2 1 1 −1 ~v1 = , ~v2 = . 2 1 px − qz py − qt p q x y x y p −q x y = , or, = such that 3.4.63 First we find the matrices S = qx + pz qy + pt −q p z t z t q p z t px − qy qx + py −t z . If q 6= 0, then the solutions are of the form S = , where z and t are arbitrary conpz − qt qz + pt z t p q p −q and stants. Since there are invertible solutions S (for example, let z = t = 1), the matrices −q p q p are indeed similar. (If q = 0, then the two matrices are equal.)

3.4.64 If b and c are both zero, then the given matrices are equal, so that they are similar, by Theorem 3.4.6.a. Let’s now assume that at least one of the scalars b and c is nonzero; reversing the roles of b and c if necessary, we can assume that c 6= 0. x y a b x y x y a c Let’s find the matrices S = such that = , or z t c d z t z t b d ax + by cx + dy ax + bz ay + bt . The solutions are of the form = az + bt cz + dt cx + bz cy + dt " # (a−d)z+b z c S= , where z and t are arbitrary constants. Since there are invertible solutions S (for example, z t a b a c let z = 1, t = 0), the matrices and are indeed similar. c d b d 3.4.65 a If S = In , then S −1 AS = A. b If S −1 AS = B, then SBS −1 = A. If we let R = S −1 , then R−1 BR = A, showing that B is similar to A. 3.4.66 We build B “column-by-column”: 2 a + b2 − a ab + b − ba a−1 b = T B= T b2 + a2 − a B ba − b − ab B b 1−a B B b 1−a 1 0 = . = 1−a B −b B 0 −1 b b Thus, this matrix represents the reflection about the line spanned by . Note that the two vectors 1−a 1−a a−1 are perpendicular. and b 165

Chapter 3

3.4.67 The matrix we seek is

2 a a + bc 1 a 0 T T = = c B ac + cd B 0 B c B 1

3.4.68 Using Exercise 67 as a guide, consider the basis

bc − ad . a+d

1 a 1 1 1 , = , and let S = . 0 c 3 0 3

1 2 3 −2 3.4.69 The matrix of the transformation T (~x) = A~x with respect to the basis , is D = = 2 1 6 B −1 B 3 0 1 2 . Thus S −1 AS = D for S = . 0 −1 2 1

a b , then 0 c

3.4.70 Suppose such a basis ~v1 , ~v2 exists. If B = [[T (~v1 )]B [T (~v2 )]B ] is upper triangular, of the form a , so that T (~v1 ) = a~v1 , that is, T (~v1 ) is parallel to ~v1 . But this is impossible, since T is a rotation [T (~v1 )]B = 0 π through 2 . 3.4.71 a Note that AS = SB. If ~x is in ker(B), then A(S~x) = SB~x = S~0 = ~0, so that S~x is in ker(A), as claimed.

b We use the hint and observe that nullity (B) = dim(ker B) = p ≤ dim(ker A) = nullity(A), since S~v1 , . . . , S~vp are p linearly independent vectors in ker(A). Reversing the roles of A and B shows that, conversely, nullity(A) ≤ nullity(B), so that the equation nullity(A) = nullity(B) holds, as claimed. 3.4.72 If A and B are similar n × n matrices, then rank(A) = n − nullity(A) = n − nullity(B) = rank(B), by Exercise 71 and the rank nullity theorem (Theorem 3.3.7). 0 0.6 3.4.73 a By inspection, we can find an orthonormal basis ~v1 = ~v , ~v2 , ~v3 of R3 , ~v1 = ~v = 0.8 , ~v2 = 0 , 1 0 0.8 ~v3 = −0.6 . 0

Figure 3.8: for Problem 3.4.73b.

b Now T (~v1 ) = ~v1 , T (~v2 )= 1 basis ~v1 , ~v2 , ~v3 is B = 0 0

~v3 and T B of T with respect to the 3.8), so that the matrix (~v3 ) = −~v2 (see Figure 0 0 0.36 0.48 0.8 0 −1 .Then A = SBS −1 = 0.48 0.64 −0.6 . 1 0 −0.8 0.6 0 166

Section 3.4 0 −1 −1 1 1 3.4.74 a ~v0 + ~v1 + ~v2 + ~v3 = 1 + −1 + 1 + −1 = 0 . 0 1 −1 −1 1 −1 b If B is the basis ~v1 , ~v2 , ~v3 , then ~v0 + ~v1 + ~v2 + ~v3 = ~0 (by part a) so ~v0 = −~v1 − ~v2 − ~v3 , i.e. [~v0 ]B = −1 . −1

c T (~v2 ) = T (−~v0 − ~v1 − ~v3 ) = −T (~v0 ) − T (~v1 ) − T (~v3 ) = −~v3 − ~v0 − ~v1 = ~v2 Hence, T is a rotation through 120◦ about the line spanned by ~v2 . Its matrix, B, is given by [[T (~v1 )]B [T (~v2 )]B [T (~v3 )]B ] where −1 T (~v1 ) = ~v0 = −~v1 − ~v2 − ~v3 so [T (~v1 )]B = −1 −1 0 T (~v2 ) = ~v2 so [T (~v2 )]B = 1 0 1 T (~v3 ) = ~v1 so [T (~v3 )]B = 0 0 −1 0 1 and B = −1 1 0 . 1 0 0

B 3 = I3 since if the tetrahedron rotates through 120◦ three times, it returns to the original position.

3.4.75 B = S

−1

3.4.76 B = S

−1

0 AS, where S = A = 1

0 −1 −1 . Thus B = A = 1 0 0

cos(t) AS, where S = A = sin(t)

− sin(t) cos(t) . Thus B = A = cos(t) sin(t)

− sin(t) . cos(t)

3.4.77 Let S be the n × n matrix whose columns are ~en , ~en−1 , . . . , ~e1 . Note that S has all 1’s on “the other diagonal” and 0’s elsewhere: 1 if i + j = n + 1 sij = 0 otherwise Also, S −1 = S. Now, B = S −1 AS = SAS; the entries of B are bij = si,n+1−i an+1−i,n+1−j sn+1−j,j = an+1−i,n+1−j . Answer: bij = an+1−i,n+1−j B is obtained from A by reversing the order of the rows and of the columns. 167

Chapter 3 3.4.78 Note first that the diagonal entry sij of S gives the unit price of good i. If aij tells us how many dollars’ worth of good i are required to produce one dollar’s worth of good j, then aij sjj tells us how many dollars’ worth of good i are required to produce one unit of good j, and s−1 ii aij sjj is the number of units of good i required to produce one unit of good j. Thus bij = s−1 a s , and B = S −1 AS. ij jj ii

True or False Ch 3.TF.1 T, by Summary 3.3.10. Ch 3.TF.2 F, by Theorem 3.3.7. Ch 3.TF.3 T, by Summary 3.3.10. Ch 3.TF.4 F; The identity matrix is similar only to itself. Ch 3.TF.5 T; We have the nontrivial relation 3~u + 3~v + 3w ~ = ~0. Ch 3.TF.6 F; The columns could be ~e1 , ~e2 , ~e3 , ~e4 in R5 , for example. Ch 3.TF.7 T, by Theorem 3.3.2. Ch 3.TF.8 F; The nullity is 6 − 4 = 2, by Theorem 3.3.7. Ch 3.TF.9 F; It’s a subspace of R3 . Ch 3.TF.10 T; by Definition 3.1.2. Ch 3.TF.11 T, by Definition 3.2.3. Ch 3.TF.12 T, by Definition 3.2.1. Ch 3.TF.13 T; Check that

1 0

0 −1

1 −1

1 1 1 0 = 1 −1 1 1

1 . 0

Ch 3.TF.14 T, by Theorem 3.3.9. Ch 3.TF.15 T, by Theorem 3.2.8. Ch 3.TF.16 T, by Summary 3.3.10. Ch 3.TF.17 F; The number n may exceed 4. Ch 3.TF.18 T, by Definition 3.2.1 (V is closed under linear combinations) Ch 3.TF.19 T, by Theorem 3.4.6, parts b and c. 168

True or False

Ch 3.TF.20 F; Let V = span

1 in R2 , for example. 1

Ch 3.TF.21 T, since AB~v = A~0 = ~0. Ch 3.TF.22 T, by Definition 3.2.3. Ch 3.TF.23 F; Suppose ~v2 = 2~v1 . Then T (~v2 ) = 2T (~v1 ) = 2~e1 cannot be ~e2 . Ch 3.TF.24 F; Consider ~u = ~e1 , ~v = 2~e1 , and w ~ = ~e2 . Ch 3.TF.25 T, since A−1 (AB)A = BA. Ch 3.TF.26 T, since both kernels consist of the zero vector alone. Ch 3.TF.27 F; There is no invertible matrix S as required in the definition of similarity. Ch 3.TF.28 F; Five vectors in R4 must be dependent, by Theorem 3.2.8. Ch 3.TF.29 T, by Definition 3.2.1 (all vectors in R3 are linear combinations of ~e1 , ~e2 , ~e3 ). Ch 3.TF.30 T; Use a basis with one vector on the line and the other perpendicular to it. Ch 3.TF.31 F; Note that R2 isn’t even a subset of R3 . A vector in R2 , with two components, does not belong to R3 . Ch 3.TF.32 T; If B = S −1 AS, then B + 7In = S −1 (A + 7In )S. Ch 3.TF.33 T; for any ~v1 , ~v2 , ~v3 of V also k~v1 , ~v2 , ~v3 is a basis too, for any nonzero scalar k. Ch 3.TF.34 F; The identity matrix is similar only to itself. Ch 3.TF.35 F; Consider

0 1 0 1

1 0 0 1 1 , but = 0 0 0 0 0

1 0

0 0

0 1 = 0 1

2 . 0

Ch 3.TF.36 F; Let A = I2 , B = −I2 and ~v = ~e1 , for example. Ch 3.TF.37 F; Let V = span(~e1 ) and W = span(~e2 ) in R2 , for example. Ch 3.TF.38 T; If A~v = Aw, ~ then A(~v − w) ~ = ~0, so that ~v − w ~ = ~0 and ~v = w. ~ Ch 3.TF.39 T; Consider the linear transformation with matrix [w ~1 . . . w ~ n ][~v1 . . . ~vn ]−1 . Ch 3.TF.40 F; Suppose A were similar to B. Then A4 = I2 were similar to B 4 = −I2 , by Example 7 of Section 3.4. But this isn’t the case: I2 is similar only to itself. 169

Chapter 3

Ch 3.TF.41 T; Let A =

0 1 , for example, with ker(A) = im(A) = span(~e1 ). 0 0

Ch 3.TF.42 F; Consider In and 2In , for example. Ch 3.TF.43 T; Matrix B = S −1 AS is invertible, being the product of invertible matrices. Ch 3.TF.44 T; Note that im(A) is a subspace of ker(A), so that dim(im A) = rank(A) ≤ dim(ker A) = 10 − rank(A). Ch 3.TF.45 T; Pick three vectors ~v1 , ~v2 , ~v3 that span V . Then V = im[~v1 ~v2 ~v3 ]. Ch 3.TF.46 T; Check that

0 0

1 0

is similar to

0 0

2 . 0

Ch 3.TF.47 T; Pick a vector ~v that is neither on the line nor perpendicular to it. Then the matrix of the linear 0 1 transformation T (~x) = R~x with respect to the basis ~v , R~v is , since R(R~v ) = ~v . 1 0 Ch 3.TF.48 F; If B = S −1 AS, then B = (2S)−1 A(2S) as well. Ch 3.TF.49 T; Note that A(B − C) = 0, so that all the columns of matrix B − C are in the kernel of A. Thus B − C = 0 and B = C, as claimed. Ch 3.TF.50 T; Suppose ~v is in both ker(A) and im(A), so that ~v = Aw ~ for some vector w. ~ Then ~0 = A~v = A2 w ~= Aw ~ = ~v , as claimed. Ch 3.TF.51 F; Suppose such a matrix A exists. Then there is a vector ~v in R2 such that A2~v 6= ~0 but A3~v = ~0. As in Exercise 3.4.58a we can show that vectors ~v , A~v , A2~v are linearly independent, a contradiction (we are looking at three vectors in R2 ). Ch 3.TF.52 T; The ith column ~ai of A, being in the image of A, is also in the image of B, so that ~ai = B~ci for some ~ci in Rm . If we let C = [~c1 · · · ~cm ] , then BC = [B~c1 · · · B~cm ] = [~a1 · · · ~am ] = A, as required. Ch 3.TF.53 F; Think about this problem in terms of “building” such an invertible matrix column by column. If we wish the matrix to be invertible, then the first column can be any column other than ~0 (7 choices). Then the second column can be any column other than ~0 or the first column (6 choices). For the third column, we have at most 5 choices (not ~0 or the first or second columns, as well as possibly some other columns). For some choices of the first two columns there will be other columns we have to exclude (the sum or difference of the first two), but not for others. Thus, in total, fewer than 7 × 6 × 5 = 210 matrices are invertible, out of a total 29 = 512 matrices. Thus, most are not invertible.

170

Section 4.1

Chapter 4 Section 4.1 4.1.1 Not a subspace since it does not contain the neutral element, that is, the function f (t) = 0, for all t. 4.1.2 This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V . • If f and g are in V (so that f (2) = g(2) = 0), then (f + g)(2) = f (2) + g(2) = 0 + 0 = 0, so that f + g is in V . • If f is in V (so that f (2) = 0), and k is any constant, then (kf )(2) = kf (2) = 0, so that kf is in V . A polynomial f (t) = a + bt + ct2 is in V if f (2) = a + 2b + 4c = 0, or a = −2b − 4c. The general element of V is of the form f (t) = (−2b − 4c) + bt + ct2 = b(t − 2) + c(t2 − 4), so that t − 2, t2 − 4 is a basis of V . 4.1.3 This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V since f ′ (1) = f (2) = 0. • If f and g are in V (so that f ′ (1) = f (2) and g ′ (1) = g(2)), then (f + g)′ (1) = (f ′ + g ′ )(1) = f ′ (1) + g ′ (1) = f (2) + g(2) = (f + g)(2), so that f + g is in V . • If f is in V (so that f ′ (1) = f (2)) and k is any constant, then (kf )′ (1) = (kf ′ )(1) = kf ′ (1) = kf (2) = (kf )(2), so that kf is in V . If f (t) = a + bt + ct2 then f ′ (t) = b + 2ct, and f is in V if f ′ (1) = b + 2c = a + 2b + 4c = f (2), or a + b + 2c = 0. The general element of V is of the form f (t) = (−b − 2c) + bt + ct2 = b(t − 1) + c(t2 − 2), so that t − 1, t2 − 2 is a basis of V . 4.1.4 This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V since • If f and g are in V • If f is in V

so that

so that

Z

1

f=

0

Z

0

1

Z

0

1

Z

1

0 dt = 0.

0

Z Z 1 (f + g) = g = 0 then 0

Z

0

1

f+

1

kf = k

Z

1

g = 0, so that f + g is in V .

0

0

0

Z f = 0 and k is any constant, then

If f (t) = a + bt + ct2 then

1

Z

1

f = 0, so that kf is in V .

0

1 b c b c b c f (t)dt = at + t2 + t3 = a + + = 0 if a = − − . 2 3 0 2 3 2 3

b 1 1 1 c 1 2 2 The general element of V is f (t) = − − + bt + ct = b t − +c t − , so that t − , t2 − is a 2 3 2 3 2 3 basis of V . 171

Chapter 4 4.1.5 If p(t) = a + bt + ct2 then p(−t) = a − bt + ct2 and −p(−t) = −a + bt − ct2 . Comparing coefficients we see that p(t) = −p(−t) for all t if (and only if) a = c = 0. The general element of the subset is of the form p(t) = bt. These polynomials form a subspace of P2 , with basis t. 4.1.6 Not a subspace, since I3 and −I3 are invertible, but their sum is not. 4.1.7 a The set V of diagonal 3 × 3 matrices is a subspace of R3×3 :

0 0 0 b The zero matrix 0 0 0 is in V , 0 0 0 p 0 0 a 0 0 c If A = 0 b 0 and B = 0 q 0 are in V , then so is their sum 0 0 r 0 0 c a+p 0 0 b+q 0 . A+B = 0 0 0 c+r

a 0 0 ka 0 d If A = 0 b 0 is in V , then so is kA = 0 kb 0 0 c 0 0

0 0 , for all constants k. kc

4.1.8 This is a subspace; the justification is analogous to Exercise 7. 4.1.9 Not a subspace; consider multiplication with a negative scalar. I3 belongs to the set, but −I3 doesn’t. 1 4.1.10 a Let ~v = 2 . Let V be the set of all 3 × 3 matrices A such that A~v = ~0. Then V is a subspace of R3×3 : 3 b The zero matrix 0 is in V , since 0~v = ~0. c If A and B are in V , then so is A + B, since (A + B)~v = A~v + B~v = ~0 + ~0 = ~0. d If A is in V , then so is kA for all scalars k, since (kA)~v = k(A~v ) = k~0 = ~0. 4.1.11 Not a subspace: I3 is in rref, but the scalar multiple 2I3 isn’t. 4.1.12 Yes, the set W of all arithmetic sequences is a subspace of V . Use the fact that a sequence (x0 , x1 , x2 , . . .) is arithmetic if xn = x0 + kn for some constant k. • The sequence (0, 0, 0, . . .) is an arithmetic sequence, with k = 0. 172

Section 4.1 • If (xn ) and (yn ) are arithmetic sequences (with xn = x0 + pn and yn = y0 + qn), then xn + yn = x0 + y0 + (p + q)n, so that (xn + yn ) is an arithmetic sequence as well. • If (xn ) is an arithmetic sequence (with xn = x0 + pn) and k is an arbitrary constant, then kxn = kx0 + (kp)n, so that (kxn ) is an arithmetic sequence as well. 4.1.13 Not a subspace: (1, 2, 4, 8, . . .) and (1, 1, 1, 1, . . .) are both geometric sequences, but their sum (2, 3, 5, 9, . . .) is not, since the ratios of consecutive terms fail to be equal, for example, 32 6= 35 . 4.1.14 Yes • (0, 0, 0, . . . , 0, . . .) converges to 0. • If limn→∞ xn = 0 and limn→∞ yn = 0, then limn→∞ (xn + yn ) = limn→∞ xn + limn→∞ yn = 0. • If limn→∞ xn = 0 and k is any constant, then limn→∞ (kxn ) = k limn→∞ xn = 0. 4.1.15 The set W of all square-summable sequences is a subspace of V : • The sequence (0, 0, 0, . . .) is in W . • Suppose (xn ) and (yn ) are in W . Note that the inequality (xn +P yn )2 ≤ 2x2n + 2yn2 holds n, since P∞ for all P ∞ ∞ 2 2 2 2 2 2 2xn + 2yn − (xn + yn ) = xn + yn − 2xn yn = (xn − yn ) ≥ 0. Thus n=1 (xn + yn )2 ≤ 2 n=1 x2n + 2 n=1 yn2 converges, so that the sequence (xn + yn ) is in W as well. P∞ • If (xn ) is in W so that n=1 x2n converges , then (kxn ) is in W as well, for any constant k, since ∞ X

(kxn )2 = k 2

n=1

∞ X

x2n

n=1

will converge. 0 0 1 1 0 a b 4.1.16 c d = a 0 0 + b 0 0 + c 1 0 0 0 0 0 e f 0 0 0 0 1 1 0 The matrices 0 0 , 0 0 , 1 0 , 0 0 0 0 0 0 0 0 6.

0 0 0 0 0 0 0 0 + d0 1 + e0 0 + f 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 , 0 0 , 0 0 form a basis of R3×2 , so that dim(R3×2 ) = 0 1 1 0 0

4.1.17 Let Eij be the n × m matrix with a 1 as its ijth entry, and zeros everywhere else. Any A in Rn×m can be written as the sum of all aij Eij , and the Eij are linearly independent, so that they form a basis of Rn×m . Thus dim(Rn×m ) = nm. 4.1.18 Any f in Pn can be written as a linear combination of 1, t, t2 , . . . , tn , by definition of Pn . Also, 1, t, . . . , tn are linearly independent; to see this consider a relation c0 +c1 t+· · ·+cn tn = 0; since the polynomial c0 +c1 t+· · ·+cn tn has more than n zeros, we must have c0 = c1 = · · · = cn = 0, as claimed. Thus, dim(Pn ) = n + 1. 173

Chapter 4 0 0 i 1 a + bi +d +c +b =a i 1 0 0 c + di 0 0 i 1 form a basis of C2 as a real linear space, so that dim(C2 ) = 4. , , , The vectors i 1 0 0

4.1.19

4.1.20 We use Summary 4.1.6. We have a = −d, so that the general element of the subspace is 0 1 0 0 −1 0 b +c +d . 0 0 1 0 0 1 0 1 0 0 −1 0 Thus , , is a basis of the subspace; the dimensions is 3. 0 0 1 0 0 1

−d c

b d

=

4.1.21 Use Summary 4.1.6. The general element of the subspace is 0 0 1 0 0 0 1 0 a 0 is a basis of the subspace; the dimension is 2. , . Thus +d =a 0 1 0 0 0 1 0 0 0 d 4.1.22 Using Exercise 21 as a guide, we find the basis E11 , E22 , . . . , Enn , where Eii is the n × n matrix with all 0 entries, except for a 1 at the ith place on the diagonal. The dimension of this space is n. 4.1.23 Proceeding as in Exercise 21, we find the basis

0 ; the dimension is 3. 1

0 0 , 0 0

0 1 0 , 1 0 0

4.1.24 Proceeding as in Exercise 21, we find the basis E11 , E12 , E13 , E22 , E23 , E33 . Here Eij is the 3 × 3 matrix with all 0 entries, except for a 1 in the ith component of the jth column; the dimension is 6. 4.1.25 A polynomial f (t) = a + bt + ct2 is in this subspace if f (1) = a + b + c = 0, or a = −b − c. The polynomials in the subspace are of the form f (t) = (−b − c) + bt + ct2 = b(t − 1) + c(t2 − 1), so that t − 1, t2 − 1 is a basis of the subspace, whose dimension is 2. 4.1.26 Denote the subspace by W . A polynomial f (t) = a + bt + ct2 + dt3 is in W if f (1) = a + b + c + d = 0 and 1 Z 1 b c d 2 f (t)dt = at + t2 + t3 + t4 = 2a + c = 0. 2 3 4 −1 3 −1 − 31 c " # " # 2 a+b+c+d =0 − c − d =0 a + 13 c 3 . The system reduces to , with general solution 2a + 23 c b + 23 c + d = 0 =0 c d The polynomials in W are of the form f (t) = − 31 c − 32 c + d t + ct2 + dt3 = c t2 − 23 t − 31 + d(t3 − t), so that t2 − 23 t − 13 , t3 − t is a basis of W , and dim(W ) = 2.

4.1.27

a b c d

is in the subspace if

a b c d

1 0 0 2

= 174

a 2b c 2d

equals

1 0 0 2

a b c d

=

a b , which is 2c 2d

Section 4.1

the case if b = c = 0. The matrices in the subspace are of the form 0 0 1 0 is a basis, and the dimension is 2. , 0 1 0 0

a 0 0 d

=a

0 1 0 +d 0 0 0

0 , so that 1

a b a b 1 1 a a+b 1 1 a b a+c b+d 4.1.28 is in the subspace if = equals = c d c d 0 1 c c+d 0 1 c d c d which is the case if c = 0 and a = d. 0 1 1 0 0 1 1 0 a b is a , , so that +b = a The matrices in the subspace are of the form 0 0 0 1 0 0 0 1 0 a basis, and the dimension is 2.

a b c d

a b c d

1 4.1.29 We are looking for the matrices such that 1 −b is required that a = −b and c = −d. The general element is −d −1 1 0 0 , is a basis, and the dimension is 2. 0 0 −1 1

1 a+b a+b 0 0 = = . It 1 c+d c+d 0 0 b −1 1 0 0 = b +d . Thus d 0 0 −1 1

0 0 a + 2c b + 2d a b . It = = 0 0 3a + 6c 3b + 6d c d −2c −2d −2 0 0 −2 is required that a = −2c and b = −2d. Thus the general element is =c +d . c d 1 0 0 1

4.1.30 We are looking for the matrices

Thus

a b c d

such that

1 2 3 6

0 −2 −2 0 is a basis, and the dimension is 2. , 0 1 1 0

a b 0 1 a 4.1.31 We are looking for the matrices such that c d 1 0 c a −b c d . It is required that a = c and b = −d. = or, c −d a b The general element is is 2.

1 c −d =c 1 c d

0 0 +d 0 0

1 −1 . Thus 1 1

b a b 1 = d c d 0

0 −1 0 is a basis, and the dimension , 0 1 0

2 a b a b 1 1 a b = such that 4.1.32 We are looking for the matrices 0 c d c d 1 1 c d a+c b+d 2a 0 or, = . It is required that a = c and b = −d. a+c b+d 2c 0

0 , −1

0 , 0

a b 1 0 0 1 1 0 0 1 The general element is =a +b . Thus , is a basis, and the dimension a −b 1 0 0 −1 1 0 0 −1 is 2.

a b 1 1 a b a b 4.1.33 Let S = . Then = , meaning c d 1 1 c d c d 175

Chapter 4 b . So a + c = a, b + d = b, a + c = c and b + d = d. These imply, respectively, that d b = 0. 0 0 , and the basis is ∅. Thus, S can only equal 0 0 a a+c b+d = c a+c b+d c = 0, d = 0, a = 0 and

a b 3 2 a b a b . We want = , meaning c d 4 5 c d c d 3a + 2c 3b + 2d a b = . So 3a + 2c = a, 3b + 2d = b, 4a + 5c = c and 4b + 5d = d. These imply that 4a + 5c 4b + 5d c d a = −c and b = −d. 0 1 1 0 0 1 1 0 a b is a basis, and the , . Thus +b =a So the general element is 0 −1 −1 0 0 −1 −1 0 −a −b dimension is 2. a b c 4.1.35 Let A = d e f . We want AB = BA, or g h i a b c 2 0 0 2 0 0 a b c d e f 0 3 0 = 0 3 0 d e f , or g h i 0 0 4 0 0 4 g h i 2a 2b 2c 2a 3b 4c 2d 3e 4f = 3d 3e 3f . 4g 4h 4i 2g 3h 4i 4.1.34 Let S =

We note that b, c, d, f, g, h must be zero, but a, e, and i are chosen freely. So, our space, V , consists of all matrices a 0 0 of the form 0 e 0 0 0 i 0 0 0 0 0 0 1 0 0 = a 0 0 0 + e 0 1 0 + i 0 0 0 . 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 Thus, 0 0 0 , 0 1 0 , 0 0 0 is a basis of V , and dim(V ) = 3. 0 0 1 0 0 0 0 0 0

a b c 4.1.36 Let A = d e f . We want g h i a b c 2 0 0 2 0 d e f 0 3 0 = 0 3 g h i 0 0 3 0 0 2a 2b 2c 2a 3b 3c 2d 3e 3f = 3d 3e 3f . 3g 3h 3i 2g 3h 3i

AB = BA, or 0 a b 0d e 3 g h

c f , or i

176

Section 4.1 We note that b, c, d, g must be zero, but a, e, f, h and i are chosen freely. So, our space, V , consists of all matrices a 0 0 of the form 0 e f 0 h i

1 = a0 0

0 0 0 0 0 0 + e0 1 0 0 0 0

1 Thus, 0 0

0 0 0 0 0 0,0 1 0 0 0 0

0 0 0 + f 0 0 0

0 0 0 0,0 0 0 0 0

0 0 0 0 0 1 + h0 0 0 0 0 1

0 0 0 + i0 0 0

0 0 0 0 is a basis of V , and dim(V ) = 5. 0 1

0 0 0 0 0,0 0 1 0

0 0 1,0 0 0

0 0 0 0 . 0 1

4.1.37 If all diagonal entries of B are different, then dim(V ) = 3, as in Exercise 35. If two of the diagonal entries are equal, then dim(V ) = 5, as in Example 36. If all three entries are equal, then B is a scalar multiple of I3 , and will commute with all 3x3 matrices, so that dim(V ) = dim(R3×3 ) = 9. So, in summary, dim(V ) could be 3, 5 or 9. 4.1.38 We look at similar cases mentioned in Exercise 37, and see that the different possibilities occur when all four entries are different (dim(V ) = 4), when exactly two are the same, but the other two are different (dim(V ) = 6), when exactly three are the same (dim(V ) = 10), when all four are the same (dim(V ) = 16) and when two of the terms of B are equal, and the other two diagonal terms of B are also equal, but different from the first pair (dim(V ) = 8). a11 0 4.1.39 An upper-triangular matrix has the form: ...

0

1 0 0 0 = a11 ... ... 0 0

0 0 0 1 + a22 ... ... 0 0

··· ··· .. . ··· ··· ··· .. . ···

0 1 0 0 0 0 + a12 .. ... ... . 0 0 0

··· ··· .. . ···

0 0 0 0 + · · · + a2n .. ... .

0

0

a12 a22 .. .

··· ··· .. .

0

···

a1n a2n .. .

ann

0 0 0 0 + · · · + a1n .. ... .

0

0

0 ··· 0 ··· .. . . . . 0 ···

0 ··· 0 ··· .. . . . . 0 ···

1 0 .. . 0

0 0 0 0 0 1 + · · · + ann . . .. .. .. . 0 0 0

So the dimension of this space is n + (n − 1) + · · · + 1 =

··· ··· .. . ···

0 0 . .. .

1

n(n+1) . 2

4.1.40 Let V be the space of all n × n matrices A such that A~v = ~0. We look at some possibilities for ~v . If ~v = ~0, then any matrix A will work, and dim(V ) = n2 . Now assume that ~v 6= 0, and suppose that the ith component vi is nonzero. If we denote the columns of A by w ~ 1, . . . , w ~ n , then the condition A~v = ~0 means that v1 w ~ 1 + · · · + vi w ~ i + · · · + vn w ~ n = ~0. We can solve this relation for w ~ i and express w ~ i in terms of the other n − 1 column vectors of A. This means that we can choose n − 1 columns (or n(n − 1) entries) of A freely; the column w ~ i is then determined by these choices. Thus dim(V ) = n(n − 1) in this case. In summary, we have dim(V ) = n2 if ~v = ~0, and dim(V ) = n(n − 1) otherwise. 177

Chapter 4 4.1.41 Let V be the space of all matrices A such that BA = 0. Note that A is in V if (and only if) all the columns of A are in the kernel of B. Since the columns of A can be chosen independently, it is plausible that dim(V ) = 3dim(ker(B)). We show this more clearly by investigating the different possibilities for ker(B). In the case where ker(B) 6= {~0}, let ~v1 , . . . , ~vn be a basis of the kernel of B, and note that n = nullity(B) can be either 1, 2 or 3. Then, the general element of V is of the form: [ a1~v1 + . . . + an~vn b1~v1 + . . . + bn~vn c1~v1 . . . + cn~vn ], which has the 3n arbitrary constants a1 , . . . , an , b1 , . . . , bn , c1 , . . . , cn . Using Summary 4.1.6, we can construct a basis of V with 3n elements, proving our claim that dim(V ) = 3dim(ker(B)). Thus, dim(V ) can be 0, 3, 6 or 9. 4.1.42 Let B be a matrix such that dim(ker(B))= k. Then, it is required that the columns of A contain only vectors in the kernel of B. Thus, each column of A can be written as: c1~v1 + c2~v2 + · · · + ck~vk , where the vectors ~vi form as basis of the kernel of B. Thus, each of the n columns in A involves k arbitrary constants, and matrix A involves nk arbitrary constants overall. The space of matrices A has dimension nk, where k is an integer in the range [0, n]. 1 0 . Using the terminology introduced in the 4.1.43 Let V be the space of all matrices S such that AS = S 0 −1 1 0 hint, we want A [ ~v w ~ ] = [ ~v w ~] or [ A~v Aw ~ ] = [ ~v −w ~ ]. Thus, ~v must be parallel to L, and w ~ 0 −1 must be perpendicular to L. If ~v1 is a nonzero vector parallel to L, and ~v2 a nonzero vector perpendicular to L, then the general element of V is of the form S = [ ~v w ~ ] = [ a~v1 b~v2 ] = a [ ~v1 0 ] + b [ 0 ~v2 ] , where a and b are arbitrary constants. We see that [ ~v1 0 ] , [ 0 ~v2 ] is a basis of V and dim(V ) = 2.

1 0 0 4.1.44 Let V be the space of all matrices S such that AS = S 0 1 0 . Let’s denote the column vectors of S 0 0 0 1 0 0 by ~u, ~v and w. ~ The condition AS = S 0 1 0 means that A~u = ~u, A~v = ~v and Aw ~ = ~0. This in turn means 0 0 0 that the vectors ~u and ~v have to be on the plane V, while w ~ is perpendicular to V . If we choose a basis ~v1 , ~v2 of V and a nonzero vector ~v3 perpendicular to V, then we can write ~u = a~v1 + b~v2 , ~v = c~v1 + d~v2 , w ~ = e~v3 , and

S = [~u ~v w] ~ = [a~v1 + b~v2 c~v1 + d~v2 e~v3 ] = a[~v1 ~0 ~0] + b[~v2 ~0 ~0] + c[~0 ~v1 ~0] + d[~0 ~v2 ~0] + e[~0 ~0~v3 ]. Thus dim(V ) = 5; the five matrices in the linear combination above form a basis of V .

4.1.45 Let

a d g 0 0 0

b e h a d g

a b A = d e g h 0 1 c f 0 0 0 0 i b d e = g h 0

c f . We want i 0 1 0 1 = 0 0 0 0 0 e f h i . 0 0

AB = BA, or a b 0 1d e g h 0

c f , or i

1 a b c So d, g, h = 0, a = e = i and f = b. Our space V consists of all matrices of the form 0 a b = a 0 0 0 0 a

178

0 0 1 0 + 0 1

Section 4.1 0 0 1 0 1 0 b 0 0 1 + c 0 0 0 . 0 0 0 0 0 0 0 0 0 1 0 1 0 0 Thus, 0 1 0 , 0 0 1 , 0 0 0 0 0 0 0 0 0 1

1 0 is a basis of V and dim(V ) = 3. 0

4.1.46 The arithmetic sequences are of the form (a, a + k, a + 2k, a + 3k, . . .) = a(1, 1, 1, 1, . . .) + k(0, 1, 2, 3, . . .), so that the sequences (1, 1, 1, . . .) (all 1’s) and (0, 1, 2, 3, . . .) (the nth entry is n) form a basis of this space, whose dimension is 2. 4.1.47 We show that the set of all even functions is a subspace of F (R, R): • If f (t) = 0 for all t, then f (−t) = f (t) = 0 for all t. • If f and g are even (that is, f (−t) = f (t) and g(−t) = g(t) for all t), then (f + g)(−t) = f (−t) + g(−t) = f (t) + g(t) = (f + g)(t), so that f + g is even as well. • If f is even and k is any constant, then (kf )(−t) = kf (−t) = kf (t) = (kf )(t), so that kf is even as well. An analogous proof shows that the odd functions form a subspace of F (R, R). 4.1.48 a f is even if f (−t) = f (t) for all t. Comparing coefficients we find that b = d = 0, so that f (t) is of the form f (t) = a + ct2 + et4 , with basis 1, t2 , t4 . The dimension is 3. b f is odd if f (−t) = −f (t), which is the case if a = c = e = 0. The odd polynomials are of the form f (t) = bt+dt3 , with basis t, t3 and dimension 2. 4.1.49 We show that L(Rm , Rn ) is a subspace of F (Rm , Rn ): • The zero transformation T (~x) = ~0 (for all ~x) is linear, represented by the zero matrix. • If T and S are linear transformations from Rm to Rn (with T (~x) = A~x and S(~x) = B~x for some n × m matrices A and B), then (T + S)(~x) = T (~x) + S(~x) = A~x + B~x = (A + B)~x, so that T + S is linear as well, given by the matrix A + B. • If T is a linear transformation from Rm to Rn (with T (~x) = A~x) and k is any constant, then (kT )(~x) = kT (~x) = kA~x = (kA)~x, so that kT is linear as well, with matrix kA. 4.1.50 Using Example 18 as a guide, we first look for solutions of the form f (x) = ekx . It is required that f ′′ (x) + 8f ′ (x) − 20f (x) = k 2 ekx + 8kekx − 20ekx = 0 for all x, or k 2 + 8k − 20 = (k − 2)(k + 10) = 0. Thus k = 2 or k = −10. By Theorem 4.1.5, the solutions of the differential equation are of the form f (x) = c1 e2x + c2 e−10x , where c1 and c2 are arbitrary constants. 4.1.51 Using Example 18 as a guide, we first look for solutions of the form f (x) = ekx . It is required that f ′′ (x) − 7f ′ (x) + 12f (x) = k 2 ekx − 7kekx + 12ekx = 0 for all x, or k 2 − 7k + 12 = (k − 3)(k − 4) = 0. Thus k = 3 or k = 4. By Theorem 4.1.7, the solutions of the differential equation are of the form f (x) = c1 e3x + c2 e4x , where c1 and c2 are arbitrary constants. 179

Chapter 4 4.1.52 We have to find constants a and b such that the functions e−x and e−5x are solutions of the differential equation f ′′ (x) + af ′ (x) + bf (x) = 0. Thus it is required that e−x − ae−x + be−x = 0, or 1 − a + b = 0, and also that 25 − 5a + b = 0. The solution of this system of two equations in two unknowns is a = 6, b = 5, so that the desired differential equation is f ′′ (x) + 6f ′ (x) + 5f (x) = 0. 4.1.53 Let B = (f1 , . . . , fn ) be a basis of V and suppose that the elements g1 , . . . , gm in V are linearly independent. In the proof of Theorem 4.1.5, we show that the coordinate vectors [g1 ]B , . . . , [gm ]B in Rn are linearly independent, so that m ≤ n by Theorem 3.2.8. 4.1.54 We can adapt the answer to Exercise 3.2.38a. Let m be the largest number of linearly independent elements we can find in W ; note that m ≤ n, by Exercise 53. Choose linearly independent elements f1 , . . . , fm in W . We claim that the elements f1 , . . . , fm span W . Indeed, if f is any element of W, then the m + 1 elements f1 , . . . , fm , f are linearly dependent, so that f is redundant: f is a linear combination of f1 , . . . , fm . It follows that f1 , . . . , fm is a basis of W, so that dim(W ) = m ≤ n =dim(V ), as claimed. 4.1.55 We will argue indirectly, assuming that F (R, R) is n-dimensional for some n. Now, the n + 1 polynomials 1, x, x2 , . . . , xn in F (R, R) are linearly independent, contradicting the fact that we can find at most n linearly independent elements in an n-dimensional space (see Exercise 53). We conclude that F (R, R) is infinite dimensional, as claimed. 4.1.56 Argue indirectly and assume that the space V of infinite sequences is finite-dimensional, with dim(V ) = n. According to the solution to Exercise 57, there can be at most n linearly independent elements in V . But here is our contradiction: It is easy to give n + 1 linearly independent infinite sequences, namely, (1, 0, 0, 0, . . .), (0, 1, 0, 0, . . .), (0, 0, 1, 0, . . .), . . . , (0, 0, 0, . . . , 0, 1, 0, . . .); in the last sequence the 1 is in the (n + 1)th place. 4.1.57 We can construct a basis of V by omitting the redundant elements from the list g1 , . . . , gm . It follows that V is finite-dimensional, and, in fact, dim(V ) ≤ m, since our basis is a “sub-list” of the original list g1 , . . . , gm . 4.1.58 a Let g(x) be a function in V . Thus, g ′′ (x) = −g(x). Now, if f (x) = g(x)2 +g ′ (x)2 , then f ′ (x) = 2(g(x))g ′ (x)+ 2(g ′ (x))g ′′ (x) = 2(g(x))g ′ (x) − 2(g(x))g ′ (x) = 0. So, f (x) = g(x)2 + g ′ (x)2 is a constant function. b Let g(x) be a function in V such that g(0) = g ′ (0) = 0. From part a we know that g(x)2 + g ′ (x)2 = k, a constant. Now g(0)2 + g ′ (0)2 = 02 + 02 = 0, so that k = 0. The equation g(x)2 + g ′ (x)2 = 0 means that g(x) = g ′ (x) = 0 for all x, as claimed. c First we note that g(x) = f (x) − f (0) cos(x) − f ′ (0) sin(x) is in V, since the functions f (x), cos(x) and sin(x) are all in V, and V is a subspace of F (R, R). Note that g(0) = f (0) − f (0) cos(0) − f ′ (0) sin(0) = f (0) − f (0) = 0. Also, g ′ (x) = f ′ (x) + f (0) sin(x) − f ′ (0) cos(x), so that g ′ (0) = f ′ (0) − f ′ (0) = 0. By part b, we can conclude that g(x) = 0 for all x, so that f (x) = f (0) cos(x) + f ′ (0) sin(x), as claimed.

Section 4.2 4.2.1 Fails to be linear, since T (A + B) = A + B + I2 doesn’t equal T (A) + T (B) = A + I2 + B + I2 = A + B + 2I2 . 180

Section 4.2 4.2.2 Linear, since T (A + B) = 7(A + B) = 7A + 7B equals T (A) + T (B) = 7A + 7B, and T (kA) = 7kA equals kT (A) = k(7A) = 7kA. Yes, T is an isomorphism, with T −1 (A) = 71 A.

a b p q a+p b+q + =T =a+p+d+s c d r s c+r d+s a b p q a b ka kb equals T +T = a + d + p + s, and T k =T = ka + kd c d r s c d kc kd a b equals kT = k(a + d) = ka + kd. c d

4.2.3 Linear, since T

No, T fails to be an isomorphism, since 4 = dim(R2×2 ) 6= dim(R) = 1; see Theorem 4.2.4b. 4.2.4 Fails to be linear, since T (2I2 ) = det(2I2 ) = 4 does not equal 2T (I2 ) = 2 det(I2 ) = 2. 4.2.5 Fails to be linear, since T (2I2 ) = 4I2 does not equal 2T (I2 ) = 2I2 . 4.2.6 Let P =

1 3

2 . Transformation T is linear, since 6

T (A + B) = (A + B)P = AP + BP equals T (A) + T (B) = AP + BP , and T (kA) = (kA)P = kAP equals kT (A) = kAP . No, T isn’t an isomorphism, since ker(T ) 6= {0}; the matrix A = rem 4.2.4a).

3 −1 0 0

is in the kernel of T (see Theo-

1 2 1 2 1 2 (M + N ) = M+ N = T (M ) + T (N ). Also, T (kM ) = 3 4 3 4 3 4

4.2.7 Linear, since T (M + N ) = 1 2 1 2 (kM ) = k M = kT (M ). 3 4 3 4

−1 1 2 1 2 M for M to find the inverse M = N= This is also an isomorphism. Solve the equation N = 3 4 3 4 2 1 −4 N. 2 3 −1

1 1 2 2 (kM ) = k M 4.2.8 Not linear, since T (kM ) = (kM ) 3 3 4

2 M = k 2 T (M ) 6= kT (M ), in general. 4

4.2.9 Linear, since T (A + B) = S −1 (A + B)S = S −1 AS + S −1 BS equals T (A) + T (B) = S −1 AS + S −1 BS, and T (kA) = S −1 (kA)S = kS −1 AS equals kT (A) = kS −1 AS. 181

Chapter 4 Yes, T is an isomorphism. Solve the equation B = S −1 AS for A to find the inverse A = SBS −1 . 4.2.10 Linear, since T (A + B) = P (A + B)P −1 = P AP −1 + P BP −1 equals T (A) + T (B) = P AP −1 + P BP −1 , and T (kA) = P (kA)P −1 = kP AP −1 equals kT (A) = kP AP −1 . Yes, T is an isomorphism. Solve the equation B = P AP −1 for A to find the inverse A = P −1 BP . 4.2.11 Linear, since T (M + N ) = P (M + N )Q = (P M + P N )Q = P M Q + P N Q = T (M ) + T (N ). Also, T (kM ) = P (kM )Q = kP M Q = kT (M ). This is also an isomorphism. Solve the equation N = P M Q for M to find the inverse M = P −1 N Q−1 .

2 4.2.12 Linear. Let A = 4

3 . Then T (c + d) = (c + d)A = cA + dA equals T (c) + T (d) = cA + dA, and 5

T (kc) = kcA equals kT (c) = kcA. No, T isn’t an isomorphism, since domain and codomain have different dimensions.

1 4.2.13 Let Q = 0

2 . Transformation T is linear, since 1

T (M + N ) = (M + N )Q − Q(M + N ) = M Q + N Q − QM − QN equals T (M ) + T (N ) = M Q − QM + N Q − QN , and T (kM ) = (kM )Q − Q(kM ) = kM Q − kQM equals kT (M ) = k(M Q − QM ) = kM Q − kQM . No, T isn’t an isomorphism, since A = I2 is in ker(T ); see Theorem 4.2.4a.

2 4.2.14 Let Q = 5

3 . Transformation T is linear, since 7

T (M + N ) = Q(M + N ) − (M + N )Q = QM + QN − M Q − N Q equals T (M ) + T (N ) = QM − M Q + QN − N Q, and T (kM ) = Q(kM ) − (kM )Q = kQM − kM Q equals kT (M ) = k(QM − M Q) = kQM − kM Q. No, T isn’t an isomorphism, since A = I2 is in ker(T ); see Theorem 4.2.4a. 2 0 4 0 2 0 2 4.2.15 Linear, since T (M + N ) = (M + N ) − (M + N ) = M+ 0 3 0 5 0 3 0 4 0 2 0 4 0 2 0 4 0 = T (M ) + T (N ). N −N + M −M = N 0 5 0 3 0 5 0 3 0 5 2 0 4 0 2 0 4 0 Also, T (kM ) = (kM ) − (kM ) =k M − kM = kT (M ). 0 3 0 5 0 3 0 5 182

0 4 0 N −M − 3 0 5

Section 4.2 This is also an isomorphism. We first of V and W are show that the kernel then equal, the dimensions see that 4a 5b 2a 2b 4 0 a b a b 2 0 a b = − = − = contains only zero. Let T (M ) = T 4c 5d 3c 3d 0 5 c d c d 0 3 c d −2a −3b . Clearly the only time this matrix will equal zero is when a, b, c, d = 0. Thus, ker(T ) = { 0}. −c −2d 3 0 2 0 2 0 3 0 2 0 M− − +N (M + N ) = M − 4.2.16 Linear, since T (M + N ) = (M + N ) 0 4 0 3 0 3 0 3 0 4 3 0 2 0 3 0 2 0 3 0 N =M − M +N − N = T (M ) + T (N ). 0 4 0 3 0 4 0 3 0 4 2 0 3 0 2 0 3 0 Also, T (kM ) = (kM ) − (kM ) = k M − M = kT (M ). 0 3 0 4 0 3 0 4 2 0 a b a b − = To determine whether T is an isomorphism, we find the kernel of T . Now, T (M ) = T 0 3 c d c d 3 0 a b 2a 3b 3a 3b −a 0 a b = − = . We see that M = is in the kernel of T if 0 4 c d 2c 3d 4c 4d −2c −d c d 0 b a = c = d = 0. Thus, the kernel consists of all matrices , where b is arbitrary. Since the kernel is nonzero, 0 0 T fails to be an isomorphism.

4.2.17 T is linear: T ((a + ib) + (c + id)) = T (a + c + i(b + d)) = a + c = T (a + ib) + T (c + id), and T (k(x + iy)) = T (kx + iky) = kx = kT (x + iy). However, T (5i) = 0, so T fails to be an isomorphism. 4.2.18 This transformation fails to be linear, since 2T (3) = 2(9) = 18 6= T (6) = 36. 4.2.19 T is linear: if w and z are complex numbers, then T (w + z) = i(w + z) = iw + iz = T (w) + T (z) and T (kz) = i(kz) = k(iz) = kT (z). T is also an isomorphism. Solve the equation w = iz for z to find that z = 1i w = −iw. We have used the fact that i2 = −1, so that 1i = −i. 4.2.20 Linear, since T ((x + iy) + (z + it)) = T (x + z + i(y + t)) = x + z − i(y + t) equals T (x + iy) + T (z + it) = x − iy + z − it = x + z − i(y + t) and T (k(x + iy)) = T (kx + iky) = kx − iky equals kT (x + iy) = k(x − iy) = kx − iky. Yes, T is an isomorphism; it’s its own inverse, since T (T (x + iy)) = T (x − iy) = x + iy. 4.2.21 Linear, since T ((x + iy) + (z + it)) = T (x + z + i(y + t)) = y + t + i(x + z) equals T (x + iy) + T (z + it) = y + ix + t + iz = y + t + i(x + z), and T (k(x + iy)) = T (kx + iky) = ky + ikx equals kT (x + iy) = k(y + ix) = ky + ikx. Yes, T is an isomorphism; it’s its own inverse, since T (T (x + iy)) = T (y + ix) = x + iy. 183

Chapter 4

4.2.22 Linear, since T (f + g) =

Z

3

(f + g) =

Z

3

−2

f+

Z

3

Z

3

f+

−2

−2

T (f ) + T (g) =

Z

3

g, and T (kf ) =

Z

3

kf = k

Z

3

f equals kT (f ) = k

−2

−2

−2

g equals

−2

Z

3

f.

−2

No, T isn’t an isomorphism, since domain and codomain have different dimensions. 4.2.23 T is linear, because T (f (t) + g(t)) = f (7) + g(7) = T (f (t)) + T (g(t)), and T (kf (t)) = kf (7) = kT (f (t)). However, T cannot be an isomorphism, because the dimensions of the domain and codomain fail to be equal. 4.2.24 T is not linear. 2T (t2 ) = 2(2(t2 )) = 4t2 6= T (2t2 ) = 4(2t2 ) = 8t2 . 4.2.25 Linear, since T (f + g) = (f + g)′′ + 4(f + g)′ = f ′′ + g ′′ + 4f ′ + 4g ′ equals T (f ) + T (g) = f ′′ + 4f ′ + g ′′ + 4g ′ , and T (kf ) = (kf )′′ + 4(kf )′ = kf ′′ + 4kf ′ equals kT (f ) = k(f ′′ + 4f ′ ) = kf ′′ + 4kf ′ . No, it isn’t an isomorphism, since the constant function f (x) = 1 is in ker(T ). 4.2.26 Linear, since T (f (t) + g(t)) = f (−t) + g(−t) equals T (f (t)) + T (g(t)) = f (−t) + g(−t), and T (kf (t)) = kf (−t) equals kT (f (t)) = kf (−t). Yes, T is an isomorphism; it’s its own inverse, since T (T (f (t))) = T (f (−t)) = f (t). 4.2.27 Linear, since T (f (t) + g(t)) = f (2t) + g(2t) equals T (f (t)) + T (g(t)) = f (2t) + g(2t), and T (kf (t)) = kf (2t) equals kT (f (t)) = kf (2t). Yes, T is an isomorphism; the inverse is T −1 (g(t)) = g 2t .

4.2.28 T is linear, since T (f (t) + g(t)) = f (2t) + g(2t) − f (t) − g(t) = f (2t) − f (t) + g(2t) − g(t) = T (f (t)) + T (g(t)), and T (kf (t)) = kf (2t) − kf (t) = k(f (2t) − f (t)) = kT (f (t)). T is not an isomorphism, however, since T (3) = 3 − 3 = 0. 4.2.29 Linear, because T (f (t) + g(t)) = f ′ (t) + g ′ (t) = T (f (t)) + T (g(t)) and T (kf (t)) = kf ′ (t) = kT (f (t)). However, since T (5) = 0, the kernel is nonzero, and T fails to be an isomorphism. 4.2.30 Linear, since T (f (t) + g(t)) = t(f ′ (t) + g ′ (t)) = t(f ′ (t)) + t(g ′ (t)) equals T (f (t)) + T (g(t)) = t(f ′ (t)) + t(g ′ (t)), and T (kf (t)) = t(kf ′ (t)) = kt(f ′ (t)) equals kT (f (t)) = kt(f ′ (t)). No, T isn’t an isomorphism, since the constant function f (t) = 1 is in ker(T ).

f (0) + g(0) f (1) + g(1) f (0) f (1) g(0) 4.2.31 T is linear: T (f (t) + g(t)) = = + f (2) + g(2) f (3) + g(3) f (2) f (3) g(2) kf (0) kf (1) f (0) f (1) = T (f (t)) + T (g(t)), and T (kf (t)) = =k = kT (f (t)). kf (2) kf (3) f (2) f (3) 184

g(1) g(3)

Section 4.2 However, the dimensions here are different, so that T fails to be an isomorphism. 4.2.32 T is not linear: T (2) = 0 + t2 6= 2T (1) = 2(0 + t2 ). 4.2.33 Linear, since T ((x0 , x1 , x2 , . . .) + (y0 , y1 , y2 , . . .)) = T (x0 + y0 , x1 + y1 , x2 + y2 , . . .) = (x0 + y0 , x2 + y2 , . . .) equals T (x0 , x1 , x2 , . . .) + T (y0 , y1 , y2 , . . .) = (x0 , x2 , . . .) + (y0 , y2 , . . .) = (x0 + y0 , x2 + y2 , . . .), and T (k(x0 , x1 , x2 , . . .)) = T (kx0 , kx1 , kx2 , . . .) = (kx0 , kx2 , . . .) equals kT (x0 , x1 , x2 , . . .) = k(x0 , x2 , . . .) = (kx0 , kx2 , . . .). No, T isn’t an isomorphism, since (0, 1, 0, 0, 0, . . .) is in ker(T ). 4.2.34 Linear, since T ((x0 , x1 , x2 , . . .) + (y0 , y1 , y2 , . . .)) = T (x0 + y0 , x1 + y1 , x2 + y2 , . . .) = (0, x0 + y0 , x1 + y1 , x2 + y2 , . . .) equals T (x0 , x1 , x2 , . . .) + T (y0 , y1 , y2 , . . .) = (0, x0 , x1 , x2 , . . .) + (0, y0 , y1 , y2 , . . .) = (0, x0 + y0 , x1 + y1 , x2 + y2 , . . .), and T (k(x0 , x1 , x2 , . . .)) = T (kx0 , kx1 , kx2 , . . .) = (0, kx0 , kx1 , kx2 , . . .) equals kT (x0 , x1 , x2 , . . .) = k(0, x0 , x1 , x2 , . . .) = (0, kx0 , kx1 , kx2 , . . .). No, T isn’t an isomorphism, since (1, 0, 0, 0, . . .) isn’t in im(T ). 4.2.35 Linear, since T (f (t) + g(t)) = (f (0) + g(0), f ′ (0) + g ′ (0), · · ·) = (f (0), f ′ (0), · · ·) + (g(0), g ′ (0), · · ·) = T (f (t)) + T (g(t)) and T (kf (t)) = (kf (0), kf ′ (0), · · ·) = k(f (0), f ′ (0), · · ·) = kT (f (t)). T fails to be an isomorphism. Note that the sequences in the image of T have only finitely many nonzero entries, so that a sequence like (1, 1, 1, 1, . . .), with all 1’s, fails to be in the image of T . Now use Theorem 4.2.4a. 4.2.36 T is linear: T (f (t)+g(t)) = (f (0)+g(0), f (1)+g(1), ...) = (f (0), f (1), ...)+(g(0), g(1), ...) = T (f (t))+T (g(t)), and T (kf (t)) = (kf (0), kf (1), ...) = k(f (0), f (1), ...) = kT (f (t)). We will show that the image of T isn’t all of V , so that T fails to be an isomorphism. More specifically, we claim that the sequence (1, 0, 0, 0, . . .), a 1 followed by all 0’s, fails to be in the image. We make an attempt to find a polynomial f (t) such that T (f (t)) = (f (0), f (1), f (2), . . .) = (1, 0, 0, . . .). This polynomial f (t) is required to have infinitely many zeros, at t = 1, 2, 3, . . . , so that f (t) must be the zero polynomial, and the equation f (0) = 1 isn’t satisfied. Thus there is no polynomial f (t) such that T (f (t)) = (1, 0, 0, 0, . . .). 4.2.37 Linear, since T (f +g) = f +g+f ′ +g ′ = f +f ′ +g+g ′ = T (f )+T (g) and T (kf ) = kf +kf ′ = k(f +f ′ ) = kT (f ). However, T (e−x ) = e−x − e−x = 0, so T fails to be an isomorphism. 4.2.38 Linear, just as in Exercise 37. T is not an isomorphism, since T (sin(x)) = sin(x) − sin(x) = 0. 185

Chapter 4 4.2.39 Linear; the proof is analogous to Exercise 25. No, T isn’t an isomorphism, since the kernel of T is two-dimensional, by Theorem 4.1.7. 4.2.40 Same answer as in Exercise 39. 4.2.41 Not linear, because T (f (t) + g(t)) = f (t) + g(t) + f ′′ (t) + g ′′ (t) + sin(t) does not equal T (f (t)) + T (g(t)) = f (t) + f ′′ (t) + sin(t) + g(t) + g ′′ (t) + sin(t).

f (7) + g(7) 4.2.42 Linear, since T (f (t) + g(t)) = equals T (f (t)) + T (g(t)) = f (11) + g(11) kf (7) f (7) + g(7) g(7) f (7) equals , and T (kf (t)) = = + kf (11) f (11) + g(11) g(11) f (11) f (7) kf (7) kT (f (t)) = k = . f (11) kf (11) Not an isomorphism, since domain and codomain have different dimensions. g(5) f (5) f (5) + g(5) 4.2.43 Linear, since T (f (t) + g(t)) = f (7) + g(7) = f (7) + g(7) = T (f (t)) + T (g(t)), and T (kf (t)) = g(11) f (11) f (11) + g(11) f (5) kf (5) kf (7) = k f (7) = kT (f (t)). f (11) kf (11)

T is an isomorphism; the proof is analogous to Example 6b.

g(1) f (1) f (1) + g(1) 4.2.44 Linear, since T (f (t) + g(t)) = f ′ (2) + g ′ (2) = f ′ (2) + g ′ (2) = T (f (t)) + T (g(t)), and g(3) f (3) f (3) + g(3) f (1) kf (1) T (kf (t)) = kf ′ (2) = k f ′ (2) = kT (f (t)). f (3) kf (3)

Not an isomorphism, since T ((t − 1)(t − 3)) = T (t2 − 4t + 3) = ~0.

4.2.45 Linear, since T (f (t) + g(t)) = t(f (t) + g(t)) = tf (t) + tg(t) = T (f (t)) + T (g(t)) and T (kf (t)) = t(kf (t)) = ktf (t) = kT (f (t)). This is not an isomorphism, since the constant function f (t) = 1 isn’t in the image. 4.2.46 Linear, since T (f (t) + g(t)) = (t − 1)(f (t) + g(t)) = (t − 1)f (t) + (t − 1)g(t) = T (f (t)) + T (g(t)) and T (kf (t)) = (t − 1)(kf (t)) = k(t − 1)f (t) = kT (f (t)). This is not an isomorphism, since the constant function f (t) = 1 isn’t in the image. Rt Rt Rt 4.2.47 T is linear, because T (f (t) + g(t)) = 0 (f (x) + g(x))dx = 0 f (x)dx + 0 g(x)dx = T (f (t)) + T (g(t)). Also, Rt Rt T (kf (t)) = 0 kf (x)dx = k 0 f (x)dx = kT (f (t)). 186

Section 4.2 However, there is no polynomial f (t) such that T (f (t)) =

Rt 0

f (x) = 6. Thus, T is not an isomorphism.

4.2.48 Linear, since T (f (t)+g(t)) = (f +g)′ (t) = f ′ (t)+g ′ (t) = T (f (t))+T (g(t)) and T (kf (t)) = kf ′ (t) = kT (f (t)). This is not an isomorphism, however, since T (5) = 0. 4.2.49 Linear, since T (f (t) + g(t)) = f (t2 ) + g(t2 ) = T (f (t)) + T (g(t)) and T (kf (t)) = kf (t2 ) = kT (f (t)). However, there is no f (t) in P such that T (f (t)) = f (t2 ) = t. Thus, the image of T fails to be all of P , and T fails to be an isomorphism. (t)−g(t) = 4.2.50 Linear. T (f (t)+g(t)) = f (t+2)+g(t+2)−f 2 kf (t+2)−kf (t) f (t+2)−f (t) =k = kT (f (t)). 2 2

This is not an isomorphism, however, since T (5) =

f (t+2)−f (t) + g(t+2)−g(t) 2 2

5−5 2

= T (f (t))+T (g(t)), and T (kf (t) =

= 0.

x y 1 2 that commute with , that is, z t 0 1 x + 2z y + 2t x 2x + y x y 1 2 1 2 x y . = , or, = z t z 2z + t z t 0 1 0 1 z t

4.2.51 We need to find the matrices M =

It follows that z = 0 and x = t, so that the kernel of T consists of all matrices of the form (i.e., the dimension of the kernel of T ) is 2. 4.2.52 We need to find the matrices A =

x z

y t

t y . The nullity 0 t

such that

0 0 = . It is required that x = −3y and z = −3t, so that the kernel 0 0 −3y y of T consists of all matrices of the form . −3t t x y z t

1 2 3 6

=

x + 3y z + 3t

2x + 6y 2z + 6t

The nullity (i.e., the dimension of the kernel of T ) is 2. 4.2.53 Note that T (a + bt + ct2 ) = 2c + 4b + 8ct. Thus the kernel consists of all constant polynomials f (t) = a(when b = c = 0), and the nullity is 1. The image consists of all linear polynomials f (t) = p + qt, and the rank is 2. 4.2.54 Use calculus to see that T (a + bt + ct2 ) = 5a + 52 b +

35 3 c.

The image is all of

R, so that the rank is 1. The kernel consists of all polynomials of the form f (t) = − 21 b − 37 c + bt + ct2 , and thus the nullity is 2. 4.2.55 The kernel consists of all infinite sequences (x0 , x1 , x2 , x3 , . . .) such that T (x0 , x1 , x2 , x3 , x4 , . . .) = (x0 , x2 , x4 , . . .) = (0, 0, 0, . . .), that is, all terms xk with even k must be 0. Thus the kernel consists of all sequences of the form (0, x1 , 0, x3 , 0, . . .). The image consists of all infinite sequences (y0 , y1 , y2 , . . .), since (y0 , y1 , y2 , . . .) = 187

Chapter 4 T (y0 , 0, y1 , 0, y2 , 0, . . .) for example. 4.2.56 Note that T (a + bt + ct2 ) = bt + 2ct2 . Thus the kernel consists of all constant polynomials f (t) = a (when b = c = 0), and the nullity is 1. The image consists of all polynomials of the form f (t) = pt + qt2 , and the rank is 2. 4.2.57 The kernel consists of the solutions of the differential equation f ′′ (t) − 5f ′ (t) + 6f (t) = 0. Using the approach outlined in Example 18 of Section 4.1 (involving a trial solution f (t) = ekt ), we find the general solution f (t) = c1 e2t + c2 e3t ; thus the nullity is 2. 4.2.58 The kernel consists of all infinite sequences such that T (x0 , x1 , x2 , . . .) = (0, x0 , x1 , x2 , . . .) = (0, 0, 0, 0, . . .), that is, all terms xk must be 0. Thus the kernel consists of the zero sequence (0, 0, 0, . . .) alone. The image consists of all infinite sequences of the form (0, x0 , x1 , x2 , . . .). 4.2.59 To find the kernel, we solve the equation T (f (t)) = T (a + bt + ct2 ) = a + 7b + 49c = 0. It follows that a = −7b − 49c, and the general element of the kernel is (−7b − 49c) + bt + ct2 = b(−7 + t) + c(−49 + t2 ). Then a basis of the kernel is −7 + t, −49 + t2 , and the nullity of T is 2. Now the rank of T must be 1, and the image is all of R. a + 7b + 49c a + 7b + 49c = . To find the kernel, solve the linear system 4.2.60 Note that T (a+bt+ct ) = a + 11b + 121c a + 11b + 121c 0 . The solution is a = 77c, b = −18c, so that the kernel consists of all polynomials of the form f (t) = c(77 − 0 18t + t2 ) = c(t − 11)(t − 7). You can also see directly that the quadratic polynomials f (t) with f (7) = f (11) = 0 are of this form. The nullity is 1. The image consists of all of R2 , so that the rank is 2. 2

4.2.61 The kernel consists of all polynomials f (t) such that t(f (t)) = 0 for all t, that is, the zero polynomial f (t) = 0 alone. The image consists of all polynomials g(t) that can be written as g(t) = t(f (t)), meaning that we can factor out a t. These are the polynomials with constant term 0, of the form g(t) = a1 t + a2 t2 + · · · + an tn . 4.2.62 The image of this transformation consists of all polynomials, since any polynomial is the derivative of another. The kernel of this transformation consists of all constant functions, or the span of the function f (t) = 1. 4.2.63 This is impossible, since dim(P3 ) = 4 and dim(R3 ) = 3. See Theorem 4.2.4b.

a b 4.2.64 Consider T (a + bt + ct + dt ) = , for example. c d 2

3

4.2.65 a First, we need to show that T (A + B) = T (A) + T (B) for all n × m matrices A and B, that is (T (A + B))(~v ) = (T (A) + T (B))(~v ) for all ~v in Rm . Indeed (T (A + B))(~v ) = (A + B)~v = A~v + B~v equals (T (A) + T (B))(~v ) = T (A)(~v ) + T (B)(~v ) = A~v + B~v . Also (T (kA))(~v ) = (kA)~v = kA~v equals (kT (A))(~v ) = k(T (A))(~v ) = kA~v . 188

Section 4.2 b The kernel of T consists of all n × m matrices A such that T (A) = 0, that is (T (A))(~v ) = A~v = ~0 for all ~v in Rm . This holds for the zero matrix only. Thus ker(T ) = {0}. c This is true by definition of a linear transformation (Definition 2.1.1). d Note that T gives an isomorphism from Rn×m to L(Rm , Rn ), by parts a, b, and c. Since dim(Rn×m ) = nm, by Exercise 17 of Seciton 4.1, we have dim(L(Rm , Rn )) = nm, by Theorem 4.2.4b. 4.2.66 The kernel of T consists of all smooth functions f (t) such that T (f (t)) = f (t) − f ′ (t) = 0, or f ′ (t) = f (t). As you may recall from a discussion of exponential functions in calculus, those are the functions of the form f (t) = Cet , where C is a constant. Thus the nullity of T is 1. 4.2.67 To show that T is linear, proceed as in Exercise 15. Now let M =

a b . c d

3 0 a b a b 2 3 − Then T (M ) = 0 k c d c d 0 4 2a + 3c 2b + 3d 3a kb −a + 3c (2 − k)b + 3d = − = . 4c 4d 3c kd c (4 − k)d a b The matrix is in the kernel of T if a = c = 0, (2 − k)b + 3d = 0 and (4 − k)d = 0. c d

If k is neither 2 nor 4, then the equation (4 − k)d = 0 implies that d = 0, and (2 − k)b + 3d = 0 then implies that b = 0. Thus the kernel of T is 0, and T is an isomorphism. If k is 2, then bis arbitrary (while a = c = d = 0), and T fails to be an isomorphism. A nonzero matrix in 0 1 kernel is in this case. 0 0 0 2 If k is 4, then b is arbitrary, and d = 3 b, so that, again, the kernel contains nonzero matrices for example, 0 and T fails to be an isomorphism.

the

3 2

,

In summary, T is an isomorphism if (and only if) k is neither 2 nor 4.

a b a b 5 0 4.2.68 We find the constants k which make the kernel of T nonzero. Let M = . Then T (M ) = − c d c d 0 1 0 0 0 0 3a −b 2a 2b 5a b a b 2 0 . = . We see that if k = 1, T = − = 0 0 0 1 (5 − k)c (1 − k)d kc kd 5c d c d 0 k 0 0 0 0 Similarly, if k = 5, T = . Thus, for k = 1, 5, the kernel of T is non-zero, and T is not an 1 0 0 0 isomorphism. For all other values of k, however, T is an isomorphism. 189

Chapter 4 4.2.69 No, since A is similar to B, there exists an invertible S such that AS = SB. Now T (S) = AS − SB = 0, so that the nonzero matrix S is in the kernel of T . 4.2.70 Since the dimensions of the domain and codomain are equal, it suffices to examine when the kernel is zero. Now f (t) is in the kernel of T if f (c0 ) = f (c1 ) = · · · = f (cn ) = 0. Recall that a nonzero polynomial of degree ≤ n has at most n zeros. If the n + 1 numbers c0 , c1 , . . . , cn are all different, then the only polynomial in Pn with f (c0 ) = f (c1 ) = . . . . = f (cn ) = 0 is the zero polynomial, so that the kernel of T is zero, and T is an isomorphism. However, if some of the numbers are equal, for example, cp = cq , then there will be nonzero polynomials in the kernel (for example, the product of all (t − ci ) where i 6= q), so that T fails to be an isomorphism. 4.2.71 Exercise 70 tells us that T is an isomorphism so long as c0 , c1 , · · · , cn are all different. This condition is met f (2) f (3) here, so T (f (t)) = f (5) is an isomorphism from P4 to R5 . Thus, there is exactly one polynomial f (t) with f (7) f (11) the required properties. 4.2.72 We satisfy all the requirements of Definition 4.1.2. Clearly 0 is an element of Zn . Let h = f + g, where f and g are elements of Zn . Then h(0) = f (0) + g(0) = 0 + 0 = 0. Also, if f is in Zn , then kf (0) = k(0) = 0. We notice that the space Zn has the basis, t, t2 , . . . , tn . Thus, the dimension of Zn is n. 4.2.73 Yes. T is an isomorphism; the inverse transformation is D(f (t)) = f ′ (t) = df dt , the derivative. We will check that the composite of T with D is the identity, in either order. Indeed Z d d t D(T (f (t))) = (T (f (t)) = f (x)dx = f (t) dt dt 0 by what is sometimes called the first fundamental theorem. And Z t f ′ (x)dx = f (t) − f (0) = f (t) for all f (t) in Zn T (D(f (t))) = T (f ′ (t)) = 0

by the ”second” fundamental theorem. 4.2.74 Let T (f (t)) = f ′ (t). This is an isomorphism, since the spaces have the same dimension, and the kernel consists only of the zero polynomial. 4.2.75 To see that 0 + 0 = 0, apply the formula f + 0 = f to f = 0 (see Definition 4.1.1). Then 0 = k0 − k0 = k(0 + 0) − k0 = k0 + k0 − k0 = k0. 4.2.76 0W = T (0V ) − T (0V ) = T (0V + 0V ) − T (0V ) = T (0V ) + T (0V ) − T (0V ) = T (0V ) We have used the equation 0V + 0V = 0V derived in Exercise 41. 4.2.77 If T and L are linear, then (L ◦ T )(f + g) = L(T (f + g)) = L(T (f ) + T (g)) = L(T (f )) + L(T (g)) 190

Section 4.2 = (L ◦ T )(f ) + (L ◦ T )(g), and (L ◦ T )(kf ) = L(T (kf )) = L(kT (f )) = kL(T (f )) = k(L ◦ T )(f ), so that L ◦ T is linear as well. If T and L are isomorphism, then L ◦ T is an isomorphism as well, since the composite of invertible functions is invertible. (See Figure 4.1.)

Figure 4.1: for Problem 4.2.77.

4.2.78 a Check all the conditions in Definition 4.1.1. A basis is 2. b T (x ⊕ y) = T (xy) = ln(xy) = ln(x) + ln(y) = T (x) + T (y) and T (k ⊙ x) = T (xk ) = ln(xk ) = k ln(x) = kT (x). The inverse of T is L(y) = ey , so that T is indeed an isomorphism. 4.2.79 Yes; let T be an invertible function from R2 to R. On R2 we can define the “exotic” operations ~v ⊕ w ~ = T −1 (T (~v ) + T (w)) ~ and k ⊙ ~v = T −1 (kT (~v )) (check that the conditions of Definition 4.1.1 hold). Then T is a linear transformation from R2 (with these exotic operations) to R (with the usual operations): T (~v ⊕ w) ~ = T (T −1 (T (~v ) + T (w))) ~ = T (~v ) + T (w), ~ and T (k ⊙ v) = kT (~v ). Now T is an isomorphism, so that the dimension of R2 (with our exotic operations) equals the dimension of the “usual” R, which is 1. 4.2.80 If a real linear space V has more than one element, then it is infinite; to see this, note that the scalar multiples kf of a nonzero f in V are all distinct (think about it!). Since there is more than one student in your class, but the number of students is finite, we cannot make the set X into a real linear space. 4.2.81 a ker(T ) is a subspace of V , so by Exercise 4.1.54, ker(T ) must also be finite-dimensional. Also, im(T ) is finite-dimensional, because it is finitely-generated by some elements T (f1 ), T (f2 ), · · · , T (fk ), where f1 , f2 , · · · , fk is a basis of V . b Following the hint: T (c1 u1 + · · · + cr ur + d1 v1 + · · · + dn vn ) = T (0), so c1 T (u1 ) + · · · + cr T (ur ) + d1 T (v1 ) + · · · + dn T (vn ) = c1 w1 + · · · + cr wr + 0 + · · · + 0 = 0. So, since w1 , · · · , wr is a basis and must be linearly independent, c1 , · · · cr must all be zero. Now, d1 v1 + · · · + dn vn = 0, but v1 , · · · , vn is also a basis, so d1 , · · · dn also must all equal zero. Thus, the elements u1 , · · · , ur , v1 , · · · , vn are linearly independent. c The hint guides us right along here. Let v be in V . Then T (v) is in im(T ), so T (v) = d1 w1 + · · · + dr wr . Now, T (v − d1 u1 − · · · − dr ur ) = T (v) − d1 T (u1 ) − · · · − dr T (ur ) = d1 w1 + · · · + dr wr − d1 w1 − · · · − dr wr = 0. Thus, v − d1 u1 − · · · − dr ur is in the kernel of T , and there are some c1 , · · · , cn such that v − d1 u1 − · · · − dr ur = c1 v1 + · · · + cn vn . Thus, v = d1 u1 + · · · + dr ur + c1 v1 + · · · + cn vn . Thus the elements u1 , · · · , ur , v1 , · · · , vn span V. 191

Chapter 4 4.2.82 Consider a basis v1 , ..., vn of ker T and a basis w1 , ..., wr of imT . Consider elements u1 , ..., ur in V such that T (ui ) = wi for i = 1, ..., r. In Exercise 81, parts b and c, we prove that the elements v1 , ..., vn , u1 , ..., ur form a basis of V, proving our claim. 4.2.83 The transformation T induces a transformation T˜ from ker (L ◦ T ) to ker L, with ker T˜ = ker T . Since ker L is assumed to be finite-dimensional, its subspace imT˜ will be finite-dimensional as well, with dim imT˜ ≤ dim ker L. Thus we can apply the rank-nullity-theorem as presented in Exercise 82 to T˜, finding that ker (L ◦ T ) is finitedimensional, with dim ker(L ◦ T ) = dim ker T˜ + dim imT˜ ≤ dim ker T + dim ker L, as claimed. 4.2.84 Using the terminology and the results introduced in Exercise 83, we observe that imT˜ = ker L. Indeed, if w is in ker L, then w = T (v) for some v in V, since imT = W . But this v will be in ker (L ◦ T ) since L (T (v)) = L(w) = 0. Thus w is in imT˜. Therefore dim ker(L◦T ) = dim ker T˜ +dim imT˜ = dim ker T +dim ker L, as claimed.

Section 4.3 2 4.3.1 Let B bethe of the given polynomials with respect to B t, t . Then the coordinates of P2 : 1, standard basis 7 9 3 3 9 7 are [f ]B = 3 , [g]B = 9 , [h]B = 2 . Finding rref 3 9 2 = I3 , we conclude that [f ]B , [g]B , [h]B are 1 4 1 1 4 1 linearly independent, hence so are f, g, h, since the coordinate transformation is an isomorphism.

0 0 0 0 0 1 1 0 of R2×2 . Then the coordinates of the given matrices , , , 0 1 1 0 0 0 0 0 1 1 2 1 1 1 1 1 2 2 2 3 3 1 4 4 = , = , = , = . Finding with respect to B are 1 3 5 1 1 B 6 3 4 B 5 7 B 6 8 B 1 4 7 8 1 1 2 1 1 0 0 −1 1 1 2 1 1 2 3 4 0 1 0 4 1 2 3 4 rref = 6= I4 , we conclude that the four vectors , , , are 1 3 5 6 0 0 1 −1 1 3 5 6 1 4 7 8 0 0 0 0 1 4 7 8 1 4 1 1 1 2 2 3 linearly dependent, and so are the four given matrices. In fact =− +4 − . 6 8 1 1 3 4 5 7

4.3.2 Let B be the basis

1 2 4.3.3 We proceed as in Exercise 1. Since rref 9 1

1 7 0 7

1 8 1 5

1 8 = I4 , the four given polynomials do form a basis of P3 . 4 8

4.3.4 Consider the coordinate vectors of the 3 given polynomials with respect to the standard basis of P2 : 1, t, t2 . 2k 0 1 1,1,2 + k 1 1 0 1 0 2k 1 0 2k Since matrix 1 1 2 + k reduces to 0 1 2 − k , these three vectors form a basis of R3 unless k = 1. 0 0 k−1 0 1 1 192

Section 4.3 Therefore, the three polynomials f (t), tf (t), g(t) form a basis of P2 unless k = 1. 4.3.5 Use a diagram: a b + 2c a b a b 1 2 −−→ = T 0 3c 0 c 0 c 0 3 y y a a − − → b b + 2c A c 3c 1 a a The matrix A that transforms b into b + 2c is A = 0 0 3c c A is invertible, so T is an isomorphism.

0 0 1 2 . 0 3

4.3.6 Use Theorem 4.3.2. to construct matrix B column by column: 1 0 0 1 0 1 B= T T T 0 0 B 0 0 B 0 1 B =

1 0

0 0

B

0 0

1 0

B

0 0

3 3

B

1 0 = 0 1 0 0

0 0 . 3

This matrix is invertible, so T is an isomorphism. 4.3.7 Use Theorem 4.3.2 to construct B column by column: 1 0 0 1 1 0 0 0 0 0 0 4 B= T T T = 0 1 B 0 0 B 0 −1 B 0 0 B 0 0 B 0 0 B 0 0 1 0 0 0 = 0 0 4 . 0 , 1 is a basis of the kernel of B, and 1 is a basis of the image of B. This implies that 0 0 0 0 0 0 0 1 0 1 1 0 is a basis of the image of T . Thus, T fails to be an is a basis of the kernel of T , and , 0 0 0 0 0 1 isomorphism.

4.3.8 Use a diagram: 0 2a − 2c a b −−→ T 0 0 c 0 y y 0 a − − → 2a − 2c b A 0 c

193

Chapter 4 0 0 0 0 a The matrix that transforms b into 2a − 2c is A = 2 0 −2 . 0 0 0 0 c 0 1 0 We see that a basis of the kernel of A is 1 , 0 and a basis of the image of A is 2 . Thus, a basis of 0 1 0 0 1 0 2 the kernel of T is , I2 while a basis of the image is . Thus, the rank of T is 1 and T is not an 0 0 0 0 isomorphism. 4.3.9 Use a diagram: a b a 2b −−→ T 0 c 0 c y y a a −−→ b 2b A c c

a a 1 The matrix A that transforms b into 2b is A = 0 c c 0 A is invertible, so T is an isomorphism.

4.3.10 Use a diagram: a b −−→ 1 3 T 3 0 0 c y a −−→ b A c 1 The matrix is A = 2 0

−2 1

0 3 0

0 −2 . 1

0 0 2 0 . 0 1

1 2 a 2a + 3b − 2c = 0 3 0 c y a 2a + 3b − 2c c

a b 0 c

Since A is invertible, T is an isomorphism.

4.3.11 Use Theorem 4.3.2. to construct matrix B column by column: 1 −1 0 1 0 1 1 −1 0 1 0 3 B= T T T = 0 0 B 0 1 B 0 0 B 0 0 B 0 1 B 0 0 B 1 0 0 = 0 1 0 . 0 0 3 A is invertible, so T is an isomorphism.

194

Section 4.3 4.3.12 Use a diagram as in Definition 4.3.1: a b 2a 3b −−→ T cd 2c 3d y y a 2a −−→ b 3b A c 2c d 3d 2 0 0 0 0 3 0 0 The matrix is A = . 0 0 2 0 0 0 0 3

Since A is invertible, T is an isomorphism.

4.3.13 Use a diagram: a b a+c b+d −−→ T cd 2a + 2c 2b + 2d y y a a+c −−→ b b+d A c 2a + 2c d 2b + 2d 1 0 1 0 0 1 0 1 The matrix is A = . 2 0 2 0 0 2 0 2 0 1 0 1 0 1 0 1 We find to be a basis of the kernel of A, and , to be a basis of the image of A. Thus, a , 0 2 0 −1 2 0 −1 0 1 0 0 1 1 0 0 1 basis of the kernel of T is , , a basis of the image of T is , , and T fails to be −1 0 0 −1 2 0 0 2 an isomorphism. 4.3.14 Use Theorem 4.3.2 to construct matrix B column by column: 0 0 0 0 0 0 0 0 3 0 0 3 0 0 0 0 B= = 0 0 B 0 0 B 6 0 B 0 6 B 0 0 3 0 0 0 0 3

1 0 0 We find bases of the kernel and image of B, and use them to find that a basis of the kernel of T is , −1 0 0 0 1 1 0 . Thus the rank of T is 2, and T is not an isomorphism. , and a basis of the image is 0 2 2 0 195

1 , −1

Chapter 4 4.3.15 We use a diagram again: x+ iy y x y

−−→ T

x− iy y x −−→ A −y 1 0 . Thus A = 0 −1 A is invertible, so T is an isomorphism. 4.3.16 B = [[1 − i]B [1 + i]B ] =

0 1 . Since B is invertible, T must be an isomorphism. 1 0

4.3.17 Another diagram: −−→ x+ T −y + ix iy y y −y x −−→ A x y 0 −1 A= . 1 0 A is invertible, so T is an isomorphism. −−→ x+ T 2x − 3y +i(3x + 2y) iy y 4.3.18 y . x 2x − 3y −−→ A y 3x + 2y 2 −3 . Since A is invertible, T must be an isomorphism. A= 3 2 4.3.19 We use a diagram to show our work: −−→ x+ T px − qy +i(qx + py) iy y y x px − qy −−→ A y qx + py p −q A = . If p = q = 0, then T (z) = 0 for all z, so that the kernel is all of C, while the image is {0}. q p Otherwise, T is an isomorphism. 196

Section 4.3 −−→ T a + bt+ ct2 b +2ct y y 4.3.20 a b −−→ b A 2c c 0 0 1 0 1 1 0 Thus A = 0 0 2 . We see that a basis of the kernel of A is 0 and a basis of the image of A is 0 , 2 . 0 0 0 0 0 0 Thus, a basis of the kernel of T is 1, while a basis of the image of T is 1, 2t. Thus, the rank of T is 2, and T is not an isomorphism. 4.3.21 We use a diagram to show our work: −−→ a + bt+ ct2 T b − 3a + (2c− 3b)t − 3ct2 y y a b − 3a −−→ b 2c − 3b A c −3c −3 1 0 2 . Thus A = 0 −3 0 0 −3 A is invertible, so T is an isomorphism.

−−→ a + bt+ ct2 T 4b + 2c + 8ct y y 4.3.22 a 4b + 2c −−→ b A 8c c 0 0 4 2 1 4 2 Thus A = 0 0 8 . We find a basis of the kernel of A to be 0 , a basis of the image of A to be 0 , 8 . 0 0 0 0 0 0 From this we see that a basis of kernel of T is 1, while a basis of the image of T is 4, 2 + 8t. Thus, T is not an isomorphism, since the rank of T is only 2. 4.3.23 A diagram shows our work: −−→ a + bt+ ct2 T a + 3b + 9c y y a + 3b + 9c a − − → b 0 A 0 c 1 3 9 Thus A = 0 0 0 . 0 0 0 197

Chapter 4 1 9 3 We find −1 , 0 to be a basis of the kernel of A, and 0 to be a basis of the image of A. Thus, a basis 0 −1 0 of the kernel of T is 3 − t, 9 − t2 , a basis of the image of T is 1 (the image is R), and T fails to be an isomorphism.

4.3.24 B = [[1]B [0]B respectively.

1 0 [0]B ] = 0 0 0 0

1 0 0 0 0 . A basis of the kernel and image of B are obvious, 1 , 0 and 0 0 1 0 0

So we find a basis of the kernel of T to be t − 3, (t − 3)2 , and a basis of the image of T to be 1. Thus T has a rank of 1, and is not an isomorphism. 4.3.25 We use the following diagram: −−→ a + bt+ ct2 T a − bt+ ct2 y y a a −−→ −b b A c c 1 0 0 Thus A = 0 −1 0 . 0 0 1

A is invertible, so T is an isomorphism.

−−→ a + bt+ ct2 T a + 2bt+ 4ct2 y y 4.3.26 a a −−→ 2b b A 4c c 1 0 0 Thus A = 0 2 0 . Since A is invertible, T must be an isomorphism. 0 0 4 4.3.27 We use a diagram: a + b(2t − 1) + c(2t − 1)2 −−→ a + bt + ct2 T = a − b + c +(2b − 4c)t + 4ct2 y y a a−b+c −−→ b 2b − 4c A c 4c 1 −1 1 Thus A = 0 2 −4 . 0 0 4 198

Section 4.3 A is invertible, so T is an isomorphism. 1 0 0 4.3.28 B = [1]B [2t − 2]B 4(t − 1)2 B = 0 2 0 . Since B is invertible, T will be an isomorphism. 0 0 4 4.3.29 This diagram shows our work: a + bt+ ct2 y a b c

−−→ T

2a + 2b+ 8c/3 y

2a + 2b + 8c/3 −−→ 0 A 0

2 2 Thus A = 0 0 0 0

8/3 0 . 0

4 1 1 3 We find −1 , 0 to be a basis of the kernel of A, and 0 to be a basis of the image of A. Thus, a basis 0 0 −1 of the kernel of T is 1 − t, 34 − t2 , a basis of the image of T is 1 (the image is R), and T fails to be an isomorphism.

4.3.30

a + bt + ct y a b c

2

−−→ T

a+b(t+h)+c(t+h)2 −a−bt−ct2 h

= b +ch + 2ct y b + ch 2c 0

−−→ A

0 1 h 1 1 h Thus A = 0 0 2 . A basis of the kernel of A is 0 , while a basis of the image of A is 0 , 2 . Thus, 0 0 0 0 0 0 a basis of the kernel of T is 1 while a basis of the image of T is 1, h + 2t. So the rank of T is 2, and it is not an isomorphism.

Recall that in calculus

f (t+h)−f (t) h

is called the difference quotient of function f (t).

Geometrically, it represents the slope of the secant through points (t, f (t)) and (t + h, f (t + h)). The limit of the difference quotient, as h approaches 0, is the derivative f ′ (t). Compare with Exercise 20. 199

Chapter 4 a+b(t+h)+c(t+h)2 −a−b(t−h)−c(t−h)2

−−→ 2h a + bt + ct2 T = b+ 2ct y y 4.3.31 a b −−→ b 2c A c 0 0 1 0 Thus A = 0 0 2 . 0 0 0 0 1 1 We find 0 to be a basis of the kernel of A, and 0 , 2 to be a basis of the image of A. Thus, a basis of 0 0 0 the kernel of T is 1, a basis of the image of T is 1, 2t, and T fails to be an isomorphism. (t−h) Geometrically, f (t+h)−f is the slope of the secant through points (t − h, f (t − h)) and (t + h, f (t + h)). This 2h exercise shows that for a quadratic polynomial f (t) this secant is parallel to the tangent at the midpoint (t, f (t)), a fact that was known to Archimedes. Draw a sketch!

Thus T (f (t)) =

f (t+h)−f (t−h) 2h

= f ′ (t) in this case. Compare with Exercise 20.

a + b + c + (b + 2c)(t − 1) −−→ a + bt + ct2 T = a − c+ (b + 2c)t y y 4.3.32 a−c a − − → b + 2c b A 0 c −1 1 0 −1 2 . We find a basis of the kernel of A to be 2 , while a basis of the image of A is Thus A = 0 1 −1 0 0 0 0 1 0 , 1 . Thus, a basis of the kernel of T is −1 + 2t − t2 and a basis of the image of T is 1, t. The rank of T 0 0 is only 2, and T is not an isomorphism. Note that T (f (t)) = f (1) + f ′ (1)(t − 1) is the equation of the tangent to the graph of f (t) at the point (1, f (1)).

4.3.33 B = [[1]B [t − 1]B

1 [0]B ] = 0 0

0 0 1 0 . 0 0

0 1 0 We find 0 to be a basis of the kernel of B, and 0 , 1 to be a basis of the image of B. Thus, a basis of 0 0 1 the kernel of T is (t − 1)2 , a basis of the image of T is 1, t − 1, and T fails to be an isomorphism. 200

Section 4.3

4.3.34 A =

0 0 1 1 0 T T T 1 0 0 A 0 0 A

0 0

A

0 T 0

0 1

A

0 0 0 0 0 −3 0 0 = = . 0 0 3 0 A A A A 0 0 0 0 0 0 0 1 0 0 −3 0 We find , to be a basis of the kernel of A, and , to be a basis of the image of A. Thus, a 0 0 3 0 0 0 1 0 1 0 0 0 0 −3 0 0 basis of the kernel of T is , , a basis of the image of T is , , meaning that T has 0 0 0 1 0 0 3 0 a rank of 2 and T fails to be an isomorphism.

0 0

0 0

0 −3 0 0

0 0 3 0

0 0 0 0

4.3.35 Again, we use a diagram to show our work: a b c d−a 1 0 0 1 0 0 −−→ M= T T (M ) = =c + (d − a) −c 0 −c 0 0 0 0 0 1 c d y y a c b −−→ d − a [M ]A = A [T (M )]A = c 0 d −c 0 0 1 0 −1 0 0 1 Thus, A = . 0 0 0 0 0 0 −1 0 0 1 1 0 By inspection, a basis of the kernel and image of this matrix are , and 0 0 0 1 0 1 1 0 0 1 −1 0 , , , , respectively. Thus, we see that a basis of ker(T ) is 0 1 0 0 0 0 0 −1 0 −1 1 0 and a basis of im(T ) is , . Thus, T fails to be an isomorphism. 0 0 0 −1 4.3.36 We will build A column-by-column: 1 0 0 1 0 0 0 0 A= T T T T 0 0 A 0 0 A 1 0 A 0 1 A =

0 1

−1 0

A

−1 0 0 1

A

1 0 0 −1

A

0 1 −1 0

A

0 −1 1 0 −1 0 = 1 0 0 0 1 −1

201

0 1 . −1 0

Chapter 4 −1 0 0 1 −1 0 0 1 We find to be a basis of the image of A. , to be a basis of the kernel of A, and , 0 1 −1 0 1 0 0 −1 1 0 0 1 0 −1 −1 0 Thus, a basis of the kernel of T is , , a basis of the image of T is , , the 0 −1 −1 0 1 0 0 1 rank of T is 2 and T fails to be an isomorphism.

4.3.37 We will construct our matrix B column-by-column: −2 0 0 0 2 2 2 −2 0 2 0 0 B = 0− −0 [0 − 0]B [0 − 0]B = . −2 −2 B 0 0 0 0 2 −2 B 0 0 0 0 0 −2 0 0 0 2 0 0 We find , to be a basis of the kernel of B, and , to be a basis of the image of B. Thus, a 0 0 0 1 0 0 1 0 2 −2 −2 −2 0 1 , and T fails to be , , and a basis of the image of T is basis of the kernel of T is I2 , 2 −2 2 2 1 0 an isomorphism.

4.3.38 We will construct our matrix B column-by-column: 0 0 0 0 −2 0 0 2 0 −2 0 0 B = [0]B [0]B = . 2 0 B 0 2 B 0 0 2 0 0 0 0 0 1 0 0 0 0 0 −2 0 We find , to be a basis of the kernel of B, and , to be a basis of the image of B. From this 0 0 0 2 0 1 0 0 1 0 0 1 −2 0 0 2 we conclude that a basis of ker(T ) is , , a basis of im(T ) is , and the rank of 1 0 0 −1 2 0 0 2 T is 2. So, T fails to be an isomorphism. 4.3.39 We use a diagram to show our work: a b c−a d−b −−→ M= T T (M ) = a− c b − d c d y y Let a c−a b −−→ d−b [M ]A = A [T (M )]A = c a−c d b−d −1 0 1 0 1 0 −1 0 Then our A-matrix of T is: . 1 0 −1 0 0 1 0 −1 202

Section 4.3 0 −1 0 1 0 −1 0 1 Thus, , is a basis of the kernel, and is a basis of the image. From this we conclude , 0 1 0 1 1 0 1 0 1 0 0 1 −1 0 0 −1 that a basis of ker(T ) is , and a basis of im(T ) is , . So, T fails to be an 1 0 0 1 1 0 0 1 isomorphism. 4.3.40 We will construct our matrix A column-by-column: −4 0 2 0 −4 0 0 2 2 0 0 2 0 2 0 2 A= = . 4 0 A 0 4 A −2 0 A 0 4 A 4 0 −2 0 0 4 0 4 1 0 −4 0 2 0 1 0 2 We find , to be a basis of the kernel of A, and , to be a basis of the image of A. 1 0 4 0 0 −1 0 4 1 0 2 −4 0 0 1 0 2 . , and a basis of im(T ) is , From this we conclude that a basis of ker(T ) is 0 4 4 0 0 −1 1 0 So, the rank of T is 2 and T fails to be an isomorphism.

4.3.41 a In Exercise 5 we consider the standard basis A basis B consisting of 1 0 0 1 0 1 , , . 0 0 0 0 0 1 1 1 0 0 1 0 1 Now SB→A = = 0 0 0 A 0 0 A 0 1 A 0

1 b Check that AS = SB = 0 0 −1 c SA→B = SB→A

1 = 0 0

of U 2×2 , and in Exercise 6 we work with the alternative

0 1 0

0 1 . 1

0 0 1 3 . 0 3

0 0 1 −1 . 0 1

4.3.42 a If A is the standard basis considered in Exercise 8 and B is the basis in Exercise 7, then 1 0 1 1 0 0 1 1 0 0 . = 0 1 S= 0 1 A 0 0 A 0 −1 A 1 0 −1

0 b Check that AS = SB = 0 0

0 0 0 4 . 0 0 203

Chapter 4 1

2

−1 c SA→B = SB→A =0 1 2

0 1 0

1 2

0 . − 21

4.3.43 a If A is the standard basis considered in Exercise 10 and B is the basis in Exercise 11, then 1 0 0 1 −1 0 1 0 1 SB→A = = −1 1 1 . 0 0 A 0 1 A 0 0 A 0 1 0 0 0 1 3 . 1 0

1 b Check that AS = SB = −1 0

1 = 0 1

−1 c SA→B = SB→A

0 0 0 1 . 1 −1

4.3.44 a If A is the standard basis considered in Exercise 13 and B is the basis in Exercise 14, then

1 0 2 0

0 0 b Check that AS = SB = 0 0

0 0 0 0

3 0 6 0

S=

1 −1

0 0

A

0 1 0 −1

A

A

0 1 0 2

A

1 0 1 0 = −1 0 0 −1

1 0 2 0

0 1 . 0 2

0 3 . 0 6

4.3.45 a If A is the standard basis considered in Exercise 15 and B is the basis in Exercise 16, then 1 1 . SB→A = [[1 + i]A [1 − i]A ] = 1 −1 b In Exercises 16 and 15 we found B = 1 1 . −1 1 −1 c SA→B = SB→A =

1 2

0 1 1 0

and A =

1 0

0 , respectively. Now check that AS = SB = −1

1 1 . 1 −1

4.3.46 a If A is the standard basis considered in Exercise 23 and B is the basis in Exercise 24, then 1 −3 9 S = [1]A [−3 + t]A 9 − 6t + t2 A = 0 1 −6 . 0 0 1 204

Section 4.3

1 b Check that AS = SB = 0 0 c SA→B =

−1 SB→A

1 = 0 0

0 0 0 0 . 0 0

3 −9 1 6. 0 1

4.3.47 a If A is the standard basis considered in Exercise 27 and B is the basis in Exercise 28, then 1 −1 1 SB→A = [1]A [−1 + t]A 1 − 2t + t2 A = 0 1 −2 . 0 0 1 b In Exercises 28

1 AS = SB = 0 0 −1 c SA→B = SB→A

1 0 and 27 we found B = 0 2 0 0 −2 4 2 −8 . 0 4

1 = 0 0

1 0 0 and A = 0 0 4

−1 1 2 −4 , respectively. Now check that 0 4

1 1 1 2 . 0 1

4.3.48 Use a diagram: −−→ a cos(t) + b sin(t) T b cos(t) − a sin(t) y y a b −−−→ B b −a 0 1 Thus B = . −1 0 4.3.49 B = [[2 cos(t) − 2 sin(t)]B [2 cos(t) + 2 sin(t)]B ] = invertible. 4.3.50 B = [[(b − 1) cos(t) − a sin(t)]B

2 −2

2 . Yes, T is an isomorphism, since matrix B is 2

b−1 [a cos(t) + (b − 1) sin(t)]B ] = −a

a . b−1

4.3.51 Note that cos(t − π/2) = sin(t) and sin(t − π/2) = − cos(t). Thus 0 −1 . Yes, T is an isomorphism. B = [[cos(t − π/2)]B [sin(t − π/2)]B ] = [[sin(t)]B [− cos(t)]B ] = 1 0 4.3.52 Recall that cos(t − δ) = cos(δ) cos(t) + sin(δ) sin(t) and 205

Chapter 4 sin(t − δ) = cos(δ) sin(t) − sin(δ) cos(t). Also, cos(π/4) = sin(π/4) =

√

2/2.

Thus B = [[cos(t − π/4)]B [sin(t − π/4)]B ] =

hh √

2 2

√

cos(t) +

2 2

i sin(t)

B

h

√

−

2 2

√

cos(t) +

2 2

i i sin(t) = B

√ 2 2

1 1

−1 1

4.3.53 Recall that cos(t − θ) = cos(θ) cos(t) + sin(θ) sin(t) and sin(t − θ) = cos(θ) sin(t) − sin(θ) cos(t). Thus B = [[cos(t − θ)]B [sin(t − θ)]B ] cos(θ) − sin(θ) . Yes, T is an isomor= [[cos(θ) cos(t) + sin(θ) sin(t)]B [− sin(θ) cos(t) + cos(θ) sin(t)]B ] = sin(θ) cos(θ) phism. Note that B is a rotation matrix. 5 1 4.3.54 Note that the two basis vectors 1 and −4 are perpendicular. 1 −1 0 1 0 5 1 0 . T −4 = 0 . Now B = 1 0 = 0 0 −1 B 0 B 0 1

1 1 Thus T 1 = 1 and −1 −1

1 1 4.3.55 Let ~u = √16 −2 be the unit vector in the direction of −2 . 1 1 1 1 1 1 1 Now T 1 = 16 −2 · 1 −2 = − 31 −2 and −1 1 −1 1 1 1 5 1 1 5 T −4 = 16 −2 · −4 −2 = 73 −2 , by Theorem 2.2.1. 1 1 1 1 1 1 1 5 1 2 1 1 −2 Also note that −2 = − 3 1 + 3 −4 , so that −2 . =3 1 1 −1 1 1 B 1 1 2 −14 7 1 1 =9 − 3 −2 −2 Now B = . 3 −1 7 1 1 B B

1 1 1 −5 5 1 5 14 4.3.56 T 1 = 2 × 1 = 4 and T −4 = 2 × −4 = 14 . Thus −1 3 −1 −1 1 3 1 −14 14 −5 0 14 B= = . 14 4 −1 0 −1 B −14 B 206

Section 4.3 1 3 5 5 1 −6 1 4.3.57 T 1 = 3 = − 1 − −4 and T −4 = 3 = 3 1 . −1 −3 1 1 −1 0 −1 −6 3 −1 3 Thus B = 3 3 = . −1 0 0 B −3 B

1 3 1 5 0 3 0 3 0 4.3.58 T 1 = 3 = 3 1 and T −4 = 0 . Thus B = 3 0 = . 0 0 −1 −3 −1 1 0 −3 B 0 B 4.3.59 T (f ) = t · f is linear and ker(T ) = {0}, but T is not an isomorphism since the constant function 1 is not in the image of T .

4.3.60 a We will use Theorem 4.3.3. SB→A

−1 b Here, we know SA→B = SB→A =

3 0 = 1 0 3 A

1 = 2 2 A

1 . 1

1 −1 . 0 1

c Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A 4.3.61 a We will use Theorem 4.3.3. SB→A = reflection combined with a scaling. b Here, we know SA→B =

−1 SB→A

=

1 −9−16

3 −4

5 −10

A

10 5

A

−4 3 1 = − 25 −3 −4

c Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A

=

−3 4

−4 = −3

1 25

4 3

−3 4

= 5

"

− 35

4 5

4 5

3 5

#

, and is thus a

4 . 3

4.3.62 a Finding this basis is equivalent to finding of[ 1 −22 ] that does not contain any of the kernel a basis 4 2 4 2 zeroes. We can quickly spot the vectors 2 and 1 , so B = 2 , 1 , for example. −1 1 −1 1 b From Theorem 4.3.3, SB→A

c SA→B =

−1 SB→A

=

1 −3

4 1 = 2 1 −1 A

2 = 2 1 A

−1 −1 = −1 2

1 3

1 1

1 . −2

d Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A

207

1 . −1

Chapter 4 4.3.63 a Finding this basis to finding is equivalent a basis of the kernel of [ 1 2 0 2 0 vectors 0 and 2 , so B = 0 , 2 , for example. 1 3 1 3 b From Theorem 4.3.3, SB→A

−1 c SA→B = SB→A =

1 −3+1

2 = 0 1 A

0 2 = −1 1 3 A

3 1 3 = − 12 −1 −1 −1

1 = −1

1 2

−1 . 3

−3 −1 . 1 1

d Theorem 4.3.4 reveals that ~b1 ~b2 = [ ~a1 ~a2 ] SB→A

4.3.64 a We find A column-by-column: A = [I2 ]B [P ]B P

3 −2 ]. We can quickly spot the

2 B

1 1 = 0 2 1 3

1 8. 9

1 1 −3 −3 b We find a basis of the kernel of A to be 4 , and a basis of the image of A to be 0 , 2 . Thus, 0 1 3 −1 is a basis of the kernel of T , and I2 , P is a basis of the image of T.

2 2 a b a + bc 4.3.65 a P 2 = = c d ac + cd bc − ad . [P 2 ]B = a+d

ab + bd bc + d2

=

a2 + bc (a + d)b (a + d)c bc + d2

= (a + d)

4 −1

a b 1 0 + (bc − ad) . So c d 0 1

b We will do this column-by-column: B = [[T (I2 )]B [T (P )]B ] = [P ]B P 2 B

0 = 1

bc − ad , by part a. T is an isomorphism if bc − ad 6= 0, that is, if P is invertible. a+d

0 0 a+d c Assume that bc − ad = 0. Then the B-matrix of T is B = . So, ker(B) = span and −1 d 1 a+ d −b 0 , a basis of im(T ) is P , and the rank of . Thus a basis of ker(T ) is (a + d)I2 − P = im(B) =span −c a 1 T is 1. 0 −1 0 4.3.66 a B = T (x21 ) B [T (x1 x2 )]B T (x22 ) B = 2 0 1

1 b Note that rref(B) = 0 0

0 −2 . 0

−1 0 0 −1 1 0 . Now im(B) = span 2 , 0 and 1 0 0 0 208

Section 4.3 1 ker(B) = span 0 . Thus 2x1 x2 , −x21 + x22 is a basis of im(T ) and x21 + x22 is a basis of ker(T ). 1 4.3.67 a

[T (cos t)]B [T (sin t)]B

= [0]B = ~0 = [0]B = ~0

0 −2 [T (t cos(t))]B = [−2 sin(t)]B = 0 0

2 0 [T (t sin(t))]B = [2 cos(t)]B = 0 0 0 0 so M = 0 0

0 0 0 0

0 −2 0 0

2 0 . 0 0

1 p 0 q b The equation T (f ) = cos(t) corresponds to M~x = , with solutions ~x = , where p and q are arbitrary. 0 0 1 0 2 1 Thus f (t) = p cos(t) + q sin(t) + 2 t sin(t). In Figure 4.2 we graph f (t) for p = q = 0; note that p cos(t) + q sin(t) is just a sinusoidal function 2π

t 2 1 t sin (t) 2

π –π

π

2π

3π

t

–π –2π

– 2t

Figure 4.2: for Problem 4.3.67b.

4.3.68 a The sequence (0, 0, 0, . . .) is in W . • If the sequences (xn ) and (yn ) are in W (that is, xn+2 = xn+1 + 6xn and yn+2 = yn+1 + 6yn for all n), then xn+2 + yn+2 = xn+1 + yn+1 + 6(xn + yn ) so that the sequence (xn + yn ) is in W as well. • If the sequence (xn ) is in W and k is any constant, then kxn+2 = kxn+1 + 6kxn , so that the sequence (kxn ) is in W as well. b A sequence in W is determined by its first two components (a and b, say), which we can choose freely. All the later components can then be expressed in terms of a and b, since xn+2 = xn+1 + 6xn : (a, b, b + 6a, 6a + 7b, 42a + 13b, . . .) = a(1, 0, 6, 6, 42, . . .) + b(0, 1, 1, 7, 13, . . .). The two sequences (1, 0, 6, 6, 42, . . .) and (0, 1, 1, 7, 13, . . .) form a basis of W , so that dim(W ) = 2. 209

Chapter 4 c It is required that cn+2 = cn+1 + 6cn for all n ≥ 0 or c2 = c + 6 or c2 − c − 6 = 0 or (c − 3)(c + 2) = 0 The solution are c1 = 3 and c2 = −2. The geometric sequences (1, 3, 9, 27, 81, . . .) and (1, −2, 4, −8, 16, . . .) are in W . d Yes, the two sequences we found in part c do the job, since dim(W ) = 2. e (x0 , x1 , x2 , x3 , x4 , . . .) = (0, 1, 1, 7, 13, . . .). We are looking for constants p and q such that (x0 , x1 , x2 , x3 , . . .) = (0, 1, 1, 7, . . .) = p(1, 3, 9, 27, . . .) + q(1, −2, 4, −8, . . .). 0=p+q , so that p = To find p and q, it suffices to consider the first two components: 1 = 3p − 2q Thus xn = 51 3n − 15 (−2)n . f (a ) 1 1 f1 (a2 ) 4.3.69 As the hint suggests, we find the kernel of M = .. . f (a ) c1 1 1 c2 f1 (a2 ) If M ... = .. .

cn

f2 (a1 ) f2 (a2 ) .. .

··· ···

f1 (an ) f2 (an ) · · ·

f2 (a1 ) f2 (a2 ) .. .

··· ···

1 5

and q = − 15 .

fn (a1 ) fn (a2 ) .. . .

f1 (an ) f2 (an ) · · · fn (an ) c f (a ) + · · · + c f (a ) fn (a1 ) c1 0 1 1 1 n n 1 fn (a2 ) c2 c1 f1 (a2 ) + · · · + cn fn (a2 ) 0 = . , .. .. .. ... = . . 0 cn c1 f1 (an ) + · · · + cn fn (an ) fn (an )

then the polynomial c1 f1 + · · · + cn fn in Pn−1 has at least n zeros, namely, a1 , a2 , · · · , an . It follows that c1 f1 + c2 f2 + · · · + cn fn = 0 and therefore c1 = c2 = · · · = cn = 0 since the fi are linearly independent. We have shown that ker(M ) = {~0}, so that M is invertible.

4.3.70 Yes; suppose dim(V ) = dim(W ) = n, and let B and C be bases of V and W , respectively. Then the coordinate transformation TB and TC define isomorphisms from V to Rn and from W to Rn , respectively. TC−1 ◦ TB is an isomorphism from V to W (see Exercise 4.2.77). 4.3.71 We need to show that there are constants w1 , . . . , wn such that Z 1 f1 w1 f1 (a1 ) + w2 f1 (a2 ) + · · · + wn f1 (an ) = −1 1

w1 f2 (a1 ) + w2 f2 (a2 ) + · · · + wn f2 (an ) = .. .

.. .

w1 fn (a1 ) + w2 fn (a2 ) + · · · + wn fn (an ) =

Z

f2

−1

Z

1

fn .

−1

210

True or False In order to be able to use the a somewhat unusual way, as f1 (a1 ) f1 (a2 ) [ w1 w2 . . . wn ] ··· f1 (an ) |

matrix M introduced in Exercise 69, as the hint suggests, we write this system in f2 (a1 ) f2 (a2 ) ··· f2 (an ) {z M

· · · fn (a1 ) Z 1 · · · fn (a2 ) f1 = ··· ··· −1 · · · fn (an ) }

Z

1

f2

−1

···

Z

1

−1

fn

Since M is invertible (Exercise 69), we have the unique solution Z 1 Z 1 Z 1 [ w1 w2 . . . wn ] = fn M −1 . f1 f2 · · · −1

−1

−1

Now if f is any polynomial in Pn−1 and f =

n X

cj fj , then

j=1

Z

1

f=

−1

n X j=1

cj

Z

1

fj =

−1

n n n n n X X X X X cj wi cj fj (ai ) = wi fj (ai ) = wi f (ai ), j=1

i=1

i=1

i=1

j=1

as claimed. 4.3.72 If we work with the basis f1 (t) = 1, f2 (t) = t, and f3 (t) = t2 of P2 , then we have to solve the system w1 + w2 + w3 =2 −w1 + w2 = 0 (see Exercise 71). The solution is w1 = w3 = 13 and w2 = 34 , so that w1 + w2 = 23 1

4 1 1 f (−1) + f (0) + f (1) for f in P2 . This is what you get when you apply Simpson’s Rule (with two 3 3 3 −1 subintervals) to f ; note that Simpson’s Rule gives the exact value of the integral in this case. Z

f =

True or False Ch 4.TF.1 F; A basis of R2×3 is

0 0 0 0

1 0

0 0 0 , 0 0 0

1 0 0 0 , 0 0 0 0

1 0 0 , 0 1 0

0 0 , 0 0

0 0 , 1 0

0 , so it has a dimension of 6. 1

Ch 4.TF.2 T; check with Definition 4.1.3c. Ch 4.TF.3 T; The linear transformation T (ax + b) = a + ib is an isomorphism from P1 to C, with the inverse T −1 (a + ib) = ax + b. Ch 4.TF.4 T; by Theorem 4.2.4c. Ch 4.TF.5 T; This fits all properties of Definition 4.1.2. 211

Chapter 4

Ch 4.TF.6 F; The transformation T could be: T (f ) = dimension of the kernel would be 7.

0 0 , in which case the kernel would be all of P6 and the 0 0

Ch 4.TF.7 T; We are looking at P6 , with a basis 1, t, t2 , t3 , t4 , t5 , t6 , which has seven elements. Ch 4.TF.8 True, we can check both requirements of Definition 4.2.1. Ch 4.TF.9 T; check the three properties listed in Definition 4.1.2. Ch 4.TF.10 T; by Definition 4.2.1. Ch 4.TF.11 F; T (sin(x)) = sin(x) − sin(x) = 0. Ch 4.TF.12 F; T (f ) =

0 0 0 0

is not an isomorphism.

1 1 −1 . . Now im(A) = ker(A) = span Ch 4.TF.13 F; Let V =R , A = 1 1 −1 2

Ch 4.TF.14 T; the dimensions of both spaces are the same: 10. Ch 4.TF.15 F; dim(P3 )= 4, so the three given polynomials cannot span P3 . Ch 4.TF.16 T; We can construct a basis of V by omitting the redundant elements from a list of ten elements that span V . Thus dim(V ) ≤ 10. 0 1 0 + det Ch 4.TF.17 F; det 0 0 0

1 0 = 0 6= det 0 1

0 = 1. 1

Ch 4.TF.18 F; For any matrix A, the space of matrices commuting with A is at least two-dimensional. Indeed, if A is a scalar multiple of I2 , then A commutes with all 2 × 2 matrices, and if A fails to be a scalar multiple of I2 , then A commutes with the linearly independent matrices A and I2 . Ch 4.TF.19 F; t3 , t3 + t2 , t3 + t, t3 + 1 is a basis of P3 . Ch 4.TF.20 T; If T is linear and invertible, then T −1 will be linear and invertible as well. Ch 4.TF.21 F; T (f (t)) = f ′ (t) is not an isomorphism. Ch 4.TF.22 T; We need only show that either the new list contains no redundant elements, or spans the whole space. The latter is slightly easier to show. Since f1 , f2 , f3 form a basis of V , it suffices to show that these three elements are in the span of f1 , f1 + f2 , f1 + f2 + f3 . This is simple to demonstrate: f2 = (f1 + f2 ) − f1 , and f3 = (f1 + f2 + f3 ) − (f1 + f2 ). Ch 4.TF.23 T; We show that none of the polynomials is redundant; let’s call them f (x), g(x) and h(x). Now g(x) isn’t a multiple of f (x) since f (b) = 0, but g(b) 6= 0. Likewise, h(x) isn’t a linear combination of f (x) and g(x) since f (c) = g(c) = 0, but h(c) 6= 0. 212

True or False Ch 4.TF.24 T; Make the substitution 4t − 3 = s to see that the inverse is T −1 (g(s)) = g( s+3 4 ). Ch 4.TF.25 F; T

1 = (a + 2c) 3

a b 1 = c d 3 0 0 + (b + 2d) 0 0

a b a + 2c b + 2d = c d 3a + 6c 3b + 6d 0 1 0 1 and . So the image is the span of 0 3 0 3

2 6

1 , and rank(T )= 2. 3

Ch 4.TF.26 T; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . Thus f = f1 does the job. Ch 4.TF.27 T; If T (f (t)) = f (t2 ) = 0, f (t) must also be zero. Ch 4.TF.28 T; The inverse is T −1 (N ) = S −1 N S −1 .

1 1 1 1 a b a b 1 1 Ch 4.TF.29 T; Let A = . Then we want = , or 0 0 0 0 c d c d 0 0 0 1 0 a a a+c b+d and . Thus, c = 0 and a = b + d. So our space is the span of = 1 0 1 c c 0 0 Ch 4.TF.30 T; Let our basis be clearly none are redundant.

1 0

0 . −1

0 1 0 1 1 0 0 . Each matrix here is invertible, and also , , , −1 0 1 0 0 −1 1

Ch 4.TF.31 F; P2 is a subspace of P , and P is infinite dimensional.

a b Ch 4.TF.32 T; Let T d e g h are exactly as required.

c b f = 0 i

Ch 4.TF.33 F; The space spanned by

1 0

0 0

c . We can easily see that the kernel and image of this transformation f

and

0 1 0 0

contains no invertible matrices.

Ch 4.TF.34 F; This is the change of basis matrix from B to A. The change of basis matrix we are looking for is: 1 −1 . 0 1

1 2 1 Ch 4.TF.35 F; Let B = (f, g) and C = (g, f ). The fact that is the B- matrix of T implies that [T (f )]B = , 3 4 3 3 3 . This , meaning that the second column of the C-matrix of T is or T (f ) = f + 3g. But then [T (f )]C = 1 1 2 1 shows that the matrix fails to be the C-matrix of T . 4 3 Ch 4.TF.36 T; The image of T is Pn−1 , so that rank(T ) = dim (imT ) = dim(Pn−1 ) = n. 213

Chapter 4 Ch 4.TF.37 T; because the matrix is invertible. Ch 4.TF.38 T; The dimension of P9 is 10, and the dimension of R3×4 is 12. Thus, any 10-dimensional subspace of R3×4 will be acceptable. For example, we can consider the space of all 3 × 4 matrices A with a11 = a12 = 0. Ch 4.TF.39 T; let W1 be {~0}. Then any other subspace W2 unioned with W1 will simply be W2 again, which we know is a subspace. Ch 4.TF.40 T; Let T (a0 +a1 t+a2 t2 +· · ·+a5 t5 +· · ·) = a0 +a1 t+a2 t2 +· · ·+a5 t5 . The image of this transformation is clearly all of P5 , and T satisfies the requirements of Definition 4.2.1. Ch 4.TF.41 T; there will be no redundant elements in this list. Ch 4.TF.42 F; The kernel of T consists of all constant functions. Ch 4.TF.43 T; We apply the rank-nullity theorem: dim(W ) = dim(im(T )) = dim(P4 ) − dim(ker(T )) = 5 − dim(ker(T )) ≤ 5. Ch 4.TF.44 F; We can construct as many linearly independent elements in ker(T ) as we want, for example, the 1 , for all positive integers n. polynomials f (t) = tn − n+1 Ch 4.TF.45 T; 0 is in our set, and if f and g are in our set, then T (f + g) = T (f ) + T (g) = f + g so that f + g is in V as well. Also, if f is in V and k is an arbitrary scalar, then T (kf ) = kT (f ) = kf , so kf is in V as well. Ch 4.TF.46 F; Consider the transformation that drops the constant term: T (c0 + c1 t + · · · + c6 t6 ) = c1 t + · · · + c6 t6 . Ch 4.TF.47 T; Let P = I2 , Q = −I2 . Then T (M ) = I2 M − M (−I2 ) = 2M , which is an isomorphism. Ch 4.TF.48 F; We use dimension arithmetic here to show that this cannot happen. Any transformation T from P6 to C must have a kernel of at least 5 dimensions, since P6 is 7-dimensional and C is only a 2-dimensional space. Thus, any such kernel cannot be isomorphic to R2×2 , which is a 4-dimensional space. Ch 4.TF.49 F; If f = −f1 , then 0 is a member of the list! Ch 4.TF.50 T; Consider the space of all matrices of the form

a −b , for example. b a

Ch 4.TF.51 T; note that dim(P11 ) = 12 = dim(R3×4 ). The linear spaces P11 and R3×4 are both isomorphic to R12 , via the coordinate transformation, and thus they are isomorphic to each other. Ch 4.TF.52 F; Consider the linear transformation T (f (t)) = f (t) from P2 to P , for example. Ch 4.TF.53 T; We use the rank-nullity theorem: dim(V ) =dim(im(T ))+dim(ker(T ))=dim(im(T ))≤ dim(R2×2 ) = 4. Ch 4.TF.54 T; Using the fundamental theorem of calculus, we can write g(t) = T (f (t)) = 3f (3t + 4). Make the substitution 3t + 4 = s to see that the inverse is T −1 (g(s)) = g((s − 4)/3)/3. 214

True or False Ch 4.TF.55 T; Using a coordinate transformation, it suffices to show this for R4 . For every real number k, we define the three dimensional subspace Vk of R4 consisting of all vectors ~x such that x4 = kx3 . If c is different from k, 0 0 4 then Vc and Vk will be different subspaces of R , since Vk contains the vector , but Vc does not. Thus we 1 k have generated infinitely many distinct three-dimensional subspaces Vk of R4 , one for every real number k. Ch 4.TF.56 T; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . We are looking for a nonzero f = af1 + bf2 such that T (f ) = 4f . Now T (f ) = aT (f1 ) + bT (f2 ) = 3af1 + 5bf1 + 4bf2 = (3a + 5b)f1 + 4bf2 must be equal to 4f = 4af1 + 4bf2 . Thus it is required that 3a + 5b = 4a, or a = 5b. For example, f = 5f1 + f2 does the job. Ch 4.TF.57 T; This is logically equivalent to the following statement: If the domain of T is finite dimensional, then so is the image of T . Compare with Exercises 4.2.81a and 4.1.57. Ch 4.TF.58 F; If A is a scalar multiple of I2 , then all2 × 2 matrices commute with A, so that the space of a b commuting matrices is 4 - dimensional. If A = fails to be a scalar multiple of I2 , consider the equation c d a b x y x y a b = , which amounts to the system cy − bz = 0, bx + (d − a)y − bt = 0, cx + c d z t z t c d (d − a)z − ct = 0. If b 6= 0, then the first two equations are independent; if c 6= 0, then the first and the third equation are independent; and if a 6= d, then the second and the third equation are independent. Thus the rank of the system is at least two and the solution space is at most two-dimensional. (The solution space is in fact two -dimensional, since A and I2 are independent solutions.) Ch 4.TF.59 T; If A = 0, then we are done. If rank(A) = 1, then the image h of itheh linear i transformation T (M ) = AM 2×2 2×2 ~ ~ from R to R is two dimensional (if ~v is a basis of im(A), then ~v 0 , 0 ~v is a basis of im(T )). Since the three matrices AB = T (B), AC = T (C) , and AD = T (D) are all in im(T ), they must be linearly dependent.

Ch 4.TF.60 F; Consider two distinct three-dimensional subspaces W1 and W2 of P4 . Since the spaces W1 and W2 are distinct, neither of them is a subspace of the other, so that we can find a polynomial f1 that is in W1 but not in W2 as well as an f2 that is in W2 but not in W1 . Then f1 and f2 are both in the union of W1 and W2 , but f1 + f2 isn’t. Ch 4.TF.61 T; Pick the first redundant element fk in the list. Since the elements f1 , . . . , fk−1 are linearly independent, the representation of fk as a linear combination of the preceding elements will be unique. Ch 4.TF.62 F; T (I3 ) = P − P = 0, and T can never be an isomorphism. Ch 4.TF.63 T; Let W = span(f1 , f2 , f3 , f4 , f5 ) = span(f2 , f4 , f5 , f1 , f3 ). If we omit the two redundant elements from the first list, f1 , f2 , f3 , f4 , f5 , we end up with a basis of W with three elements, so that dim(W ) = 3. If we omit the redundant elements from the second list, f2 , f4 , f5 , f1 , f3 , we end up with a (possibly different) basis of W , but that basis must consist of 3 elements as well. Thus there must be two redundant elements in the second list. Ch 4.TF.64 F; The dimensions of the kernel and image would have to be equal, and both add up to the dimension of P6 , which is the odd number 7. 215

Chapter 4 Ch 4.TF.65 T; Consider the proof of the rank nullity theorem outlined in Exercise 4.2.81. In the proof, we use bases of ker(T ) and im(T ) to construct a basis of the domain. Ch 4.TF.66 F; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . We are looking for a nonzero f = af1 + bf2 such that T (f ) = 5f . Now T (f ) = aT (f1 ) + bT (f2 ) = 3af1 + 5bf1 + 4bf2 = (3a + 5b)f1 + 4bf2 must be equal to 5f = 5af1 + 5bf2 . Thus it is required that 3a + 5b = 5a and 4b = 5b, implying that a = b = 0. We are unable to find a nonzero f with the desired property. x y in W can be described by a z t single linear equation ax + by + cz + dt = 0 , where at least one of the coefficients is nonzero. Suppose x is the leading variable (meaning that a 6= 0), and y, z and t are the free variables. We can choose y = z = 1 and t = 0, ∗ 1 in W will be invertible. We represent x by a star, since its value does not affect and the resulting matrix 1 0 the invertibility. If y is the leading variable and the other three are the free variables, then we canconstruct the 1 0 1 ∗ . Finally, for in W . If z is the leading variable, we have the invertible matrix invertible matrix ∗ 1 0 1 0 1 the leading variable t we have . 1 ∗

Ch 4.TF.67 T; Consider a 3-dimensional subspace W of R2×2 . The matrices

216

Section 5.1

Chapter 5 Section 5.1 5.1.1 k~v k = 5.1.2 k~v k = 5.1.3 k~v k =

√

72 + 112 =

√

22 + 32 + 42 =

√

22 + 32 + 42 + 52 =

√

49 + 121 =

√ 170 ≈ 13.04

√ √ 4 + 9 + 16 = 29 ≈ 5.39 √ √ 4 + 9 + 16 + 25 = 54 ≈ 7.35

·~ v √ √7+11 √18 5.1.4 θ = arccos k~u~ukk~ v k = arccos 2 170 = arccos 340 ≈ 0.219 (radians) 2+6+12 ·~ v √ √ 5.1.5 θ = arccos k~u~ukk~ v k = arccos 14 29 ≈ 0.122 (radians) ·~ v 2−3+8−10 √ √ 5.1.6 θ = arccos k~u~ukk~ ≈ 1.700 (radians) v k = arccos 10 54

5.1.7 Use the fact that ~u · ~v = k~ukk~v k cos θ, so that the angle is acute if ~u · ~v > 0, and obtuse if ~u · ~v < 0. Since ~u · ~v = 10 − 12 = −2, the angle is obtuse. 5.1.8 Since ~u · ~v = 4 − 24 + 20 = 0, the two vectors enclose a right angle. 5.1.9 Since ~u · ~v = 3 − 4 + 5 − 3 = 1, the angle is acute (see Exercise 7). 5.1.10 ~u · ~v = 2 + 3k + 4 = 6 + 3k. The two vectors enclose a right angle if ~u · ~v = 6 + 3k = 0, that is, if k = −2. ·~ v √1 5.1.11 a θn = arccos k~u~ukk~ v k = arccos n

θ2 = arccos √12 =

π 4 (=

45◦ )

θ3 = arccos √13 ≈ 0.955 (radians) θ4 = arccos 12 =

π 3 (=

60◦ )

b Since y = arccos(x) is a continuous function, lim θn = arccos lim √1n = arccos(0) = π2 (= 90◦ ) n→∞

n→∞

5.1.12 k~v + wk ~ 2 = (~v + w) ~ · (~v + w) ~ (by hint) = k~v k2 + kwk ~ 2 + 2(~v · w) ~ (by definition of length) ≤ k~v k2 + kwk ~ 2 + 2k~v kkwk ~ (by Cauchy-Schwarz) = (k~v k + kwk) ~ 2 , so that k~v + wk ~ 2 ≤ (k~v k + kwk) ~ 2 Taking square roots of both sides, we find that k~v + wk ~ ≤ k~v k + kwk, ~ as claimed. 217

Chapter 5 kF~2 k = 20 cos θ2 . It is required that kF~2 + F~3 k = 16, so that 20 cos θ2 = 16, or θ = 2 arccos(0.8) ≈ 74◦ .

5.1.13 Figure 5.1 shows that kF~2 + F~3 k = 2 cos

θ 2

Figure 5.1: for Problem 5.1.13. 5.1.14 The horizontal components of F~1 and F~2 are −kF~1 k sin β and kF~2 k sin α, respectively (the horizontal component of F~3 is zero). Since the system is at rest, the horizontal components must add up to 0, so that −kF~1 k sin β + kF~2 k sin α = 0 or ~ kF~1 k sin β = kF~2 k sin α or kF1 k = sin α . ~2 k kF

To find

EA , EB

sin β

note that EA = ED tan α and EB = ED tan β so that

α and β are two distinct acute angles, it follows that

EA EB

6=

~1 k kF ~2 k , kF

EA EB

=

tan α tan β

=

sin α sin β

·

cos β cos α

=

~1 k cos β kF ~2 k cos α . kF

Since

so that Leonardo was mistaken.

5.1.15 The subspace consists of all vectors ~x in R4 such that 1 x1 x2 2 ~x · ~v = · = x1 + 2x2 + 3x3 + 4x4 = 0. 3 x3 4 x4 −2r −3s −4t −2 −3 −4 r 1 0 0 These are vectors of the form = r + s + t . s 0 1 0 t 0 0 1 The three vectors to the right form a basis.

5.1.16 You may be able to find the solutions by educated guessing. Here is the systematic approach: we first find all vectors ~x that are orthogonal to ~v1 , ~v2 , and ~v3 , then we identify the unit vectors among them. Finding the vectors ~x with ~x · ~v1 = ~x · ~v2 = ~x · ~v3 = 0 amounts to solving the system x1 + x2 + x3 + x4 = 0 x1 + x2 − x3 − x4 = 0 x1 − x2 + x3 − x4 = 0 we can omit all the coefficients 12 . t x1 x2 −t The solutions are of the form ~x = = . −t x3 t x4

218

Section 5.1 1 2

Since k~xk = 2|t|, we have a unit vector if t = 1 2 − 12 −1 2 1 2

and

− 21

or t = − 21 . Thus there are two possible choices for ~v4 :

1 2 1 2 − 21

.

5.1.17 The orthogonal complement W ⊥ of W consists of the vectors ~x in R4 such that 5 1 x1 x1 2 x x 2 6 2 · = 0 and · = 0. 7 3 x3 x3 8 x4 4 x4 x1 + 2x2 + 3x3 + 4x4 = 0 . Finding these vectors amounts to solving the system 5x1 + 6x2 + 7x3 + 8x4 = 0 The solutions are of the form s + 2t 1 2 x1 −2 −3 x2 −2s − 3t = = s + t . x3 s 1 0 x4 t 0 1 The two vectors to the right form a basis of W ⊥ . 5.1.18 a k~xk2 = 1 + k~xk =

√2 3

1 4

+

1 16

+

1 64

≈ 1.155.

+ ··· =

1 1− 41

=

4 3

use the formula for a geometric series, with a =

1 4

, so that

b If we let ~u = (1, 0, 0, . . .) and ~v = 1, 12 , 41 , · · · , then √

·~ v θ = arccos k~uu~kk~ v k = arccos

1

2 √ 3

= arccos

3 2

=

π 6 (=

30◦ ).

c ~x = 1, √12 , √13 , · · · , √1n , · · · does the job, since the harmonic series 1 + introductory calculus classes).

1 2

+

1 3

+ · · · diverges (a fact discussed in

√

d If we let ~v = (1, 0, 0, . . .), ~x = 1, 21 , 41 , · · · and ~u = projL~v = (~u · ~v )~u =

5.1.19 See Figure 5.2.

3 4

~ x k~ xk

=

1, 21 , 41 , · · · .

3 2

1, 12 , 41 , · · · then

5.1.20 On the line L spanned by ~x we want to find the vector m~x closest to ~y (that is, we want km~x − ~y k to be minimal). We want m~x − ~y to be perpendicular to L (that is, to ~x), which means that ~x · (m~x − ~y ) = 0 or ·~ y 4182.9 m(~x · ~x) − ~x · ~y = 0 or m = ~x~x·~ x ≈ 198.532 ≈ 0.106. Recall that the correlation coefficient r is r =

~ x·~ y k~ xkk~ yk ,

so that m =

219

k~ yk k~ xk r.

See Figure 5.3.

Chapter 5

Figure 5.2: for Problem 5.1.19.

Figure 5.3: for Problem 5.1.20. 5.1.21 Call the three given vectors ~v1 , ~v2 , and ~v3 . Since ~v2 is required to be a unit vector, we must have b = g = 0. Now ~v1 · ~v2 = d must be zero, so that d = 0. Likewise, ~v2 · ~v3 = e must be zero, so that e = 0.

√

Since ~v3 must be a unit vector, we have k~v3 k2 = c2 +

1 4

= 1, so that c = ±

3 2 .

√

Since we are asked to find just one solution, let us pick c = √

The condition ~v1 · ~v3 = 0 now implies that

3 2 a

3 2 .

√ + 12 f = 0, or f = − 3a.

Finally, it is required that k~v1 k2 = a2 + f 2 = a2 + 3a2 = 4a2 = 1, so that a = ± 12 . √

Let us pick a = 12 , so that f = −

3 2 .

220

Section 5.1 Summary:

√ 3 0 2 ~v1 = √0 , ~v2 = 1 , ~v3 = 0 1 0 − 3 1 2

2

2

There are other solutions; some components will have different signs.

5.1.22 Let W = {~x in Rn : ~x · ~vi = 0 for all i = 1, . . . , m}. We are asked to show that V ⊥ = W , that is, any ~x in V ⊥ is in W , and vice versa. If ~x is in V ⊥ , then ~x · ~v = 0 for all ~v in V ; in particular, x · ~vi = 0 for all i (since the ~vi are in V ), so that ~x is in W. Conversely, consider a vector ~x in W . To show that ~x is in V ⊥ , we have to verify that ~x · ~v = 0 for all ~v in V . Pick a particular ~v in V . Since the ~vi span V , we can write ~v = c1~v1 + · · · + cm~vm , for some scalars ci . Then ~x · ~v = c1 (~x · ~v1 ) + · · · + cm (~x · ~vm ) = 0, as claimed. 5.1.23 We will follow the hint. Let ~v be a vector in V . Then ~v · ~x = 0 for all ~x in V ⊥ . Since (V ⊥ )⊥ contains all vectors ~y such that ~y · ~x = 0, ~v is in (V ⊥ )⊥ . So V is a subspace of (V ⊥ )⊥ . Then, by Theorem 5.1.8c, dim (V ) + dim(V ⊥ ) = n and dim(V ⊥ ) + dim((V ⊥ )⊥ ) = n, so dim (V ) + dim(V ⊥ ) = dim(V ⊥ ) + dim((V ⊥ )⊥ ) and dim (V ) = dim((V ⊥ )⊥ ). Since V is a subspace of (V ⊥ )⊥ , it follows that V = (V ⊥ )⊥ , by Exercise 3.3.61. 5.1.24 Write T (~x) = projV (~x) for simplicity. To prove the linearity of T we will use the definition of a projection: T (~x) is in V , and ~x − T (~x) is in V ⊥ . To show that T (~x + ~y ) = T (~x) + T (~y ), note that T (~x) + T (~y ) is in V (since V is a subspace), and ~x + ~y − (T (~x) + T (~y )) = (~x − T (~x)) + (~y − T (~y )) is in V ⊥ (since V ⊥ is a subspace, by Theorem 5.1.8a). To show that T (k~x) = kT (~x), note that kT (~x) is in V (since V is a subspace), and k~x − kT (~x) = k(~x − T (~x)) is in V ⊥ (since V ⊥ is a subspace). 5.1.25 a kk~v k2 = (k~v ) · (k~v ) = k 2 (~v · ~v ) = k 2 k~v k2 Now take square roots of both sides; note that is negative). kk~v k = |k|k~v k, as claimed.

b k~uk = k~v1k ~v =

1 vk k~ v k k~

√ k 2 = |k|, the absolute value of k (think about the case when k

= 1, as claimed.

↑

by part a 5.1.26 The two given vectors spanning the subspace are orthogonal, but they are not unit vectors: both have length 7. To obtain an orthonormal basis ~u1 , ~u2 of the subspace, we divide by 7: 2 3 ~u1 = 17 3 , ~u2 = 17 −6 . 6 2 221

Chapter 5 49 Now we can use Theorem 5.1.5, with ~x = 49 : 49 19 3 2 projV ~x = (~u1 · ~x)~u1 + (~u2 · ~x)~u2 = 11 3 − −6 = 39 . 64 2 6

5.1.27 Since the two given vectors in the subspace are orthogonal, we have the orthonormal basis 2 −2 2 2 ~u1 = 31 , ~u2 = 13 . 1 0 0 1 Now we can use Theorem 5.1.5, with ~x = 9~e1 : projV ~x = (~u1 · ~x)~u1 + (~u2 · ~x)~u2 2 −2 8 2 2 0 = 2 − 2 = . 1 0 2 0 1 −2 5.1.28 Since the three given vectors in the subspace are orthogonal, we have the orthonormal basis 1 1 1 1 1 −1 ~u1 = 12 , ~u2 = 12 , ~u3 = 21 . 1 −1 −1 1 −1 1 3 1 Now we can use Theorem 5.1.5, with ~x = ~e1 : projV ~x = (~u1 · ~x)~u1 + (~u2 · ~x)~u2 + (~u3 · ~x)~u3 = 14 . −1 1

5.1.29 By the Pythagorean theorem (Theorem 5.1.9), k~xk2

= k7~u1 − 3~u2 + 2~u3 + ~u4 − ~u5 k2 = k7~u1 k2 + k3~u2 k2 + k2~u3 k2 + k~u4 k2 + k~u5 k2 . = 49 + 9 + 4 + 1 + 1 = 64, so that k~xk = 8.

5.1.30 Since ~y = projV ~x, the vector ~x − ~y is orthogonal to ~y , by definition of an orthogonal projection (see Theorem 5.1.4): (~x − ~y ) · ~y = 0 or ~x · ~y − k~y k2 = 0 or ~x · ~y = k~y k2 . See Figure 5.4.

Figure 5.4: for Problem 5.1.30.

222

Section 5.1 5.1.31 If V = span(~u1 , . . . , ~um ), then projV ~x = (~u1 · ~x)~u1 + · · · + (~um · ~x)~um , by Theorem 5.1.5, and kprojV ~xk2 = (~u1 ·~x)2 +· · ·+(~um ·~x)2 = p, by the Pythagorean theorem (Theorem 5.1.9). Therefore p ≤ k~xk2 , by Theorem 5.1.10. The two quantities are equal if (and only if) ~x is in V . 5.1.32 By Theorem 2.4.9a, the matrix G is invertible if (and only if) (~v1 · ~v1 )(~v2 · ~v2 ) − (~v1 · ~v2 )2 = k~v1 k2 k~v2 k2 − (~v1 · ~v2 )2 6= 0. The Cauchy-Schwarz inequality (Theorem 5.1.11) tells us that k~v1 k2 k~v2 k2 − (~v1 · ~v2 )2 ≥ 0; equality holds if (and only if) ~v1 and ~v2 are parallel (that is, linearly dependent). 1 x1 5.1.33 Let ~x = · · · be a vector in Rn whose components add up to 1, that is, x1 + · · · + xn = 1. Let ~y = · · · 1 xn (all n components are √ 1). The Cauchy-Schwarz inequality (Theorem 5.1.11) tells us that |~x · ~y | ≤ k~xkk~y k, or, |x1 + · · · + xn | ≤ k~xk n, or k~xk ≥ √1n . By Theorem 5.1.11, the equation k~xk = √1n holds if (and only if) the vectors ~x and ~y are parallel, that is, x1 = x2 = · · · = xn = n1 . Thus the vector of minimal length is 1 n ~x = · · · all components are n1 .

1 n

Figure 5.5 illustrates the case n = 2. x2

1 → = x X

1 2 1 2

1

x1

x1 + x2 = 1

Figure 5.5: for Problem 5.1.33.

1 5.1.34 Let ~x be a unit vector in Rn , that is, k~xk = 1. Let ~y = . . . (all n components are 1). The Cauchy-Schwarz 1 √ √ inequality (Theorem 5.1.11) tells us that |~x · ~y | ≤ k~xkk~y k, or, |x1 + . . . + xn | ≤ k~xk n = n. By Theorem 5.1.11, k √ the equation x1 +. . .+xn = n holds if ~x = k~y for positive k. Thus ~x must be a unit vector of the form ~x = . . . k 1 √ n for some positive k. It is required that nk 2 = 1, or, k = √1n . Thus ~x = . . . all components are √1n .

√1 n

Figure 5.6 illustrates the case n = 2. 223

Chapter 5 x2 1 √2 1 √2

→ x = X

x1

1

x1 + x2 =√ 2

Figure 5.6: for Problem 5.1.34. 1 x 5.1.35 Applying the Cauchy-Schwarz inequality to ~u = y and ~v = 2 gives |~u · ~v | ≤ k~ukk~v k, or |x + 2y + 3z| ≤ 3 z √ √ 14. The minimal value x + 2y + 3z = − 14 is attained when ~u = k~v for negative k. Thus ~u must be a unit − √114 k √2 vector of the form ~u = 2k , for negative k. It is required that 14k 2 = 1, or, k = − √114 . Thus ~u = − 14 . 3k − √3 14

a 0.2 5.1.36 Let ~x = b and ~y = 0.3 . It is required that ~x · ~y = 0.2a + 0.3b + 0.5c = 76. Our goal is to minimize c 0.5 2 2 2 quantity ~x · ~x = a + b + c . The Cauchy-Schwarz inequality (squared) tells us that (~x · ~y )2 ≤ k~xk2 k~y k2 , or 762 2 2 2 762 ≤ (a2 + b2 + c2 )(0.22 + 0.32 + 0.52 ) or a2+ b2 + c2 ≥ 0.38 . The quantity a + b + c is minimal when 0.2k a 762 a2 + b2 + c2 = 0.38 . This is the case when ~x = b = 0.3k for some positive constant k. It is required that 0.5k c 0.2a + 0.3b + 0.5c = (0.2)2 k + (0.3)2 k + (0.5)2 k = 0.38k = 76, so that k = 200. Thus a = 40, b = 60, c = 100: The student must study 40 hours for the first exam, 60 hours for the second, and 100 hours for the third. 5.1.37 Using Definition 2.2.2 as a guide, we find that ref V ~x = 2(projV~x) − ~x = 2(~u1 · ~x)~u1 + 2(~u2 · ~x)~u2 − ~x. 5.1.38 Since ~v1 and ~v2 are unit vectors, the condition ~v1 · ~v2 = k~v1 kk~v2 k cos(α) = cos(α) = 12 implies that ~v1 and ~v2 enclose an angle of 60◦ = π3 . The vectors ~v1 and ~v3 enclose an angle of 60◦ as well. In the case n = 2 there are two possible scenarios: either ~v2 = ~v3 , or ~v2 and ~v3 enclose an angle of 120◦ . Therefore, either ~v2 · ~v3 = 1 or ~v2 · ~v3 = cos(120◦ ) = − 21 . In the case n = 3, the vectors ~v2 and ~v3 could enclose any angle between 0◦ (if ~v2 = ~v3 ) and 120◦ , as illustrated in Figure 5.7. We have − 12 ≤ ~v2 · ~v3 ≤ 1. √ 3 cos θ 2 0 0 √ √ 3 For example, consider ~v1 = 0 , ~v2 = 23 , ~v3 = sin θ 2 1 1 1 2

2

Note that ~v2 · ~v3 = claimed.

3 4

sin θ +

1 4

could be anything between − 12 (when sin θ = −1) and 1 (when sin θ = 1), as 224

Section 5.1

Figure 5.7: for Problem 5.1.38. If n exceeds three, we can consider the orthogonal projection w ~ of ~v3 onto the plane E spanned by ~v1 and ~v2 . ~ = (~v1 · w)~ ~ v1 = 21 ~v1 , and since kwk Since proj~v1 w ~ ≤ k~v3 k = 1, (by Theorem 5.1.10), the tip of w ~ will be on the line segment in Figure 5.8. Note that the angle φ enclosed by the vectors ~v2 and w ~ is between 0◦ and 120◦ , so that cos φ is between − 12 and 1. Therefore, ~v2 · ~v3 = ~v2 · w ~ = kwk ~ cos φ is between − 12 and 1. This implies that ∠(~v2 , ~v3 ) is between 0◦ and 120◦ as well. To see that all these values are attained, add (n − 3) zeros to the three vectors ~v1 , ~v2 , ~v3 in R3 given above. v1

w v2

φ

Figure 5.8: for Problem 5.1.38.

5.1.39 No! By definition of a projection, the vector ~x − projL ~x is perpendicular to projL ~x, so that (~x − projL ~x) · (projL ~x) = ~x · projL ~x − kprojL ~xk2 = 0 and ~x · projL ~x = kprojL ~xk2 ≥ 0. (See Figure 5.9.)

Figure 5.9: for Problem 5.1.39.

5.1.40 ||~v2 || =

√ √ ~v2 · ~v2 = a22 = 3.

·~ v3 20 √ a23 √ 5.1.41 θ =arccos( ||~v~v22||||~ v3 || ) =arccos( a22 a33 ) =arccos( 21 ) ≈ 0.31 radians.

5.1.42 ||~v1 + ~v2 || =

p √ √ (~v1 + ~v2 ) · (~v1 + ~v2 ) = a11 + 2a12 + a22 = 22. 225

Chapter 5 ~ v2 u is an orthonormal basis 3 . Then, ~ ~ v2 ~ v2 1 ( 3 · ~v1 ) 3 = 3 (~v2 · ~v1 ) ~v32 = 31 (a12 ) ~v32 = 95 ~v2 .

5.1.43 Let ~u = (~u · ~v1 )~u =

~ v2 ||~ v2 ||

=

for span(~v2 ). Using Theorem 5.1.5, proj~v2 (~v1 ) =

5.1.44 One method to solve this is to take ~v = ~v2 − proj~v3 ~v2 = ~v2 −

20 v3 . 49 ~

5.1.45 Write the projection as a linear combination of ~v2 and ~v3 , c2~v2 + c3~v3 . Now you want ~v1 − c2~v2 − c3~v3 to be perpendicular to V , that is, perpendicular to both ~v2 and ~v3 . Using dot products, this boils down to two linear 25 1 equation in two unknowns, 9c2 + 20c3 = 5, and 20c2 + 49c3 = 11, with the solution c2 = 41 and c3 = − 41 . Thus 1 25 the answer is 41 ~v2 − 41 ~v3 . 5.1.46 Write the projection as a linear combination of ~v1 and ~v2 : c1~v1 + c2~v2 . Now we want ~v3 − c1~v1 + c2~v2 to be perpendicular to V , that is, perpendicular to both ~v1 and ~v2 . Using dot products, this boils down to two linear equations in two unknowns, 11 = 3c1 + 5c2 and 20 = 5c1 + 9c2 , with the solution c1 = − 12 , c2 = 52 . Thus, the answer is − 21 ~v1 + 52 ~v2 .

Section 5.2 In Exercises 1–14, we will refer to the given vectors as ~v1 , . . . , ~vm , where m = 1, 2, or 3.

2 5.2.1 ~u1 = k~v11 k ~v1 = 31 1 −2 6 5.2.2 ~u1 = k~v11 k ~v1 = 71 3 2 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

Note that ~u1 · ~v2 = 0. 4 5.2.3 ~u1 = k~v11 k ~v1 = 51 0 3 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

4 5.2.4 ~u1 = 51 0 and ~u2 = 3

2 = 17 −6 3

3 = 15 0 −4

3 1 0 as in Exercise 3. 5 −4

0 Since ~v3 is orthogonal to ~u1 and ~u2 , ~u3 = k~v13 k ~v3 = −1 . 0 226

Section 5.2 2 5.2.5 ~u1 = k~v11 k ~v1 = 31 2 1 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 = √118 −1 = 4

−1 1 √ −1 3 2 4

1 5.2.6 ~u1 = k~v11 k ~v1 = 0 = ~e1 0 ~u2 =

~ v2⊥ k~ v2⊥ k

~u3 =

~ v3⊥ k~ v3⊥ k

0 u1 ·~ v2 )~ u1 1 = ~e2 = = k~~vv22 −(~ −(~ u1 ·~ v2 )~ u1 k 0 0 u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 0 = ~e3 = k~~vv33 −(~ = −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k 1

2 5.2.7 Note that ~v1 and ~v2 are orthogonal, so that ~u1 = k~v11 k ~v1 = 13 2 and ~u2 = 1 1 2 ~ v⊥ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 √1 −4 = 1 −2 . ~u3 = k~v3⊥ k = k~~vv33 −(~ −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k = 3 36 3 4 2 5 1 4 1 5.2.8 ~u1 = k~v1 k ~v1 = 7 2 2 −2 2 ~u2 = k~v12 k ~v2 = 71 5 −4 1 1 1 1 5.2.9 ~u1 = k~v1 k ~v1 = 2 1 1 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 1 7 = 10 −7 1

1 1 1 1 5.2.10 ~u1 = k~v1 k ~v1 = 2 1 1 227

−2 1 v2 = 31 1 . Then k~ v2 k ~ 2

Chapter 5

~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

1 −1 = 12 1 −1

4 0 5.2.11 ~u1 = k~v11 k ~v1 = 51 0 3 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−3 1 2 = √225 = 14 4

−3 1 2 15 14 4

2 1 3 5.2.12 ~u1 = 7 0 6 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

0 −2 = 13 2 1

1 1 1 1 5.2.13 ~u1 = k~v1 k ~v1 = 2 1 1 ~u2 =

~u3 =

~ v2⊥ k~ v2⊥ k

~ v3⊥ k~ v3⊥ k

=

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

=

1 2 −1 2 1 −2 1 2

~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k

=

1 2 1 2 −1 2 − 12

1 7 1 5.2.14 ~u1 = k~v11 k ~v1 = 10 1 7 ~u2 =

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 0 = √12 1 0

228

Section 5.2

~u3 =

~ v3⊥ k~ v3⊥ k

=

0 1 = √12 0 −1

~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k~ v3 −(~ u1 ·~ v3 )~ u1 −(~ u2 ·~ v3 )~ u2 k

In Exercises 15–28, we will use the results of Exercises 1–14 (note that Exercise k, where k = 1, . . . , 14, gives the QR factorization of the matrix in Exercise (k + 14)). We can set Q = [~u1 . . . ~um ]; the entries of R are r11 r22 r33 rij

= k~v1 k = k~v2⊥ k = k~v2 − (~u1 · ~v2 )~u1 k = k~v3⊥ k = k~v3 − (~u1 · ~v3 )~u1 − (~u2 · ~v3 )~u2 k = ~ui · ~vj , where i < j.

2 5.2.15 Q = 13 1 , R = [3] −2

2 7 0 −6 , R = 0 7 3

3 5 5 0,R = 0 35 −4

5 5 0 3 0 0 −5 , R = 0 35 0 0 0 2 −4 0

6 5.2.16 Q = 17 5 2 4 5.2.17 Q = 15 0 5 4 5.2.18 Q = 15 0 5

5.2.19 Q =

1 3

2

2 1

− √12

1 − √12 , R = 3 0

√1 2

√4 2

5.2.20 Q = I3 , R = [ ~v1

~v2

2 3 ~v3 ] = 0 4 0 0

2 5.2.21 Q = 31 2 1

−2 1 3 0 1 −2 , R = 0 3 2 2 0 0

5 4 5.2.22 Q = 71 2 2

−2 2 7 7 , R = 5 0 7 −4

5 6 7 12 −12 6

229

Chapter 5 0.5 0.5 5.2.23 Q = 0.5 0.5

−0.1 2 0.7 , R = 0 −0.7 0.1

1 1 1 5.2.24 Q = 2 1 1

1 −1 2 10 , R = 1 0 2 −1

4 10

12 −3 2 5 1 0 5.2.25 Q = 15 , R = 0 14 0 9 4

2 7 3 7

5.2.26 Q = 0 6 7

0

− 23 7 ,R= 2 0 3

1 1 5.2.27 Q = 21 1 1 5.2.28 Q = 5.2.29 ~u1 =

1 10 7 10 1 10 7 10

10 15

14 3

1 3

1 1 2 1 −1 1 , R = 0 1 −1 −1 0 0 1 −1

− √12 0

1 v1 k~ v1 k ~

0

10 , R = 0 0 0

√1 2

√1 2

− √12

0

=

1 5

−3 4

~u2 =

~ v2⊥ k~ v2⊥ k

=

1 −2 1 10 10 √ 2 √0 0 2 ~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

=

1 5

4 . (See Figure 5.10.) 3

Figure 5.10: for Problem 5.2.29.

230

Section 5.2 5.2.30 See Figure 5.11.

Figure 5.11: for Problem 5.2.30. 1 5.2.31 ~u1 = k~v11 k ~v1 = 0 = ~e1 0

0 b b ~v2⊥ = ~v2 − projV1 ~v2 = c − 0 = c , so that ~u2 = 0 0 0

Here V1 = span(~e1 ) = x axis. d d 0 ~v3⊥ = ~v3 − projV2 ~v3 = e − e = 0 , so that ~u3 = f 0 f Here V2 = span(~e1 , ~e2 ) = x-y plane. (See Figure 5.12.)

~ v2⊥ k~ v2⊥ k

~ v3⊥ k~ v3⊥ k

0 = 1 = ~e2 0 0 = 0 = ~e3 . 1

Figure 5.12: for Problem 5.2.31. −1 −1 5.2.32 A basis of the plane is ~v1 = 1 , ~v2 = 0 . 1 0

231

Chapter 5 Now apply the Gram-Schmidt process. ~u1

~u2

1 v1 k~ v1 k ~

=

=

~ v2⊥ k~ v2⊥ k

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

−1 = √12 1 0 −1 = √16 −1 2

Your solution may be different if you start with a different basis ~v1 , ~v2 of the plane.

1 0 0 5.2.33 rref(A) = 0 1 1

1 0

−1 0 0 −1 A basis of ker(A) is ~v1 = , ~v2 = . 0 1 1 0

−1 0 Since ~v1 and ~v2 are orthogonal already, we obtain ~u1 = √12 , ~u2 = 0 1

5.2.34 rref(A) =

1 0 −1 0 1 2

−2 3

0 −1 √1 . 2 1 0

1 2 −2 −3 A basis of ker(A) is ~v1 = , ~v2 = . 1 0 0 1

We apply the Gram-Schmidt process and obtain

~u1

~u2

1 5.2.35 rref(A) = 0 0

0 1 0

1 3 1 3

0

=

=

1 v1 k~ v1 k ~

~ v2⊥ k~ v2⊥ k

1 −2 = √16 1 0 2 −1 = √130 −4 3

=

~ v2 −(~ u1 ·~ v2 )~ u1 k~ v2 −(~ u1 ·~ v2 )~ u1 k

The non-redundant columns of A give us a basis of im(A): 232

Section 5.2 2 1 ~v1 = 2 , ~v2 = 1 −2 2

1 Since ~v1 and ~v2 are orthogonal already, we obtain ~u1 = 13 2 , ~u2 = 2

1 1 1 1 −1 5.2.36 Write M = 2 1 −1 1 1

↑ Q0

1 2 3 −1 0 −4 1 0 0 −1

2 1 1 . 3 −2

5 6 7

↑ R0

This is almost the QR factorization of M : the matrix Q0 has orthonormal columns and R0 is upper triangular; the only problem is the entry −4 on the diagonal of R0 . Keeping in mind how matrices are multiplied, we can change all the signs in the second column of Q0 and in the second row of R0 to fix this problem: 1 −1 1 2 3 5 1 1 −1 M = 12 0 4 −6 1 1 −1 0 0 7 1 −1 1 ↑ Q

↑ R

1 1 1 −1 5.2.37 Write M = 21 1 −1 1 1

↑ Q0

1 1 3 −1 1 0 1 −1 0 −1 −1 0

↑ R0

4 5 0 0

Note that the last two columns of Q0 and the last two rows of R0 have no effect on the product Q0 R0 ; if we drop them, we have the QR factorization of M : 1 1 3 4 1 −1 M = 12 1 −1 0 5 1 1 ↑ Q

↑ R

5.2.38 Since ~v1 = 2~e3 , ~v2 = −3~e1 and ~v3 0 −1 ~v1 0 0 Q = k~v1 k k~~vv22 k k~~vv33 k = 1 0 0 0

= 4~e4 are orthogonal, we have 0 k~v1 k 0 0 2 0 0 k~v2 k 0 = 0 3 and R = 0 0 0 0 k~v3 k 0 0 1 233

0 0. 4

Chapter 5 1 5.2.39 ~u1 = √114 2 , ~u2 = 3

1 √1 1 , ~ u3 = ~u1 × ~u2 = 3 −1

5.2.40 If ~v1 , . . . , ~vn are the columns of A, then Q =

−5 √1 4 42 −1

~ v1 k~ v1 k

~ vn k~ vn k

···

(See Exercise 38 as an example.)

and R =

k~v1 k 0

0 ..

. k~vn k

.

5.2.41 If all diagonal entries of A are positive, then we have Q = In and R = A. A small modification is necessary if A has negative entries on the diagonal: if aii < 0 we let rij = −aij for all j, and we let qii = −1; if aii > 0 we let rij = aij and qii = 1. Furthermore, qij = 0 if i 6= j (that is, Q is diagonal). −1 2 3 −1 0 0 1 −2 −3 For example, 0 4 5 = 0 1 0 0 4 5 0 0 −6 0 0 −1 0 0 6 ↑ A

↑ Q

↑ R

5.2.42 We have r11 = k~v1 k and r22 = k~v2⊥ k = k~v2 − projL~v2 k, so that r11 r22 is the area of the parallelogram defined by ~v1 and ~v2 . See Figure 5.13.

Figure 5.13: for Problem 5.2.42. 5.2.43 Partition the matrices Q and R in the QR factorization of A as follows: R1 R2 [ A1 A2 ] = A = QR = [ Q1 Q2 ] = [ Q1 R1 Q1 R2 + Q2 R3 ] , 0 R3 where Q1 is n × m1 , Q2 is n × m2 , R1 is m1 × m1 , and R3 is m2 × m2 . Then, A1 = Q1 R1 is the QR factorization of A1 : note that the columns of A1 are orthonormal, and R1 is upper triangular with positive diagonal entries. 5.2.44 No! If m exceeds n, then there is no n × m matrix Q with orthonormal columns (if the columns of a matrix are orthonormal, then they are linearly independent). 5.2.45 Yes. Let A = [ ~v1 with ~um = k~v1m k ~vm .

· · · ~vm ]. The idea is to perform the Gram-Schmidt process in reversed order, starting

Then we can express ~vj as a linear combination of ~uj , . . . , ~um , so that [ ~v1 for some lower triangular matrix L, with 234

· · · ~vj

· · · ~vm ] = [ ~u1

· · · ~uj

· · · ~um ] L

Section 5.3

~vj = [ ~u1

· · · ~uj

l1j ··· · · · ~um ] ljj = ljj ~uj + · · · + lmj ~um . ··· lmj

Section 5.3 5.3.1 Not orthogonal, the column vectors fail to be perpendicular to each other. 5.3.2 This matrix is orthogonal. Check that the column vectors are unit vectors, and that they are perpendicular to each other. 5.3.3 This matrix is orthogonal. Check that the column vectors are unit vectors, and that they are perpendicular to each other. 5.3.4 Not orthogonal, the first and third column vectors fail to be perpendicular to each other. 5.3.5 3A will not be orthogonal, because the length of the column vectors will be 3 instead of 1, and they will fail to be unit vectors. 5.3.6 −B will certainly be orthogonal, since the columns will be perpendicular unit vectors. 5.3.7 AB is orthogonal by Theorem 5.3.4a. 5.3.8 A + B will not necessarily be orthogonal, because the columns may not be unit vectors. For example, if A = B = In , then A + B = 2In , which is not orthogonal. 5.3.9 B −1 is orthogonal by Theorem 5.3.4b. 5.3.10 This matrix will be orthogonal, by Theorem 5.3.4. 5.3.11 AT is orthogonal. AT = A−1 , by Theorem 5.3.7, and A−1 is orthogonal by Theorem 5.3.4b. 5.3.13 3A is symmetric, since (3A)T = 3AT = 3A. 5.3.14 −B is symmetric, since (−B)T = −B T = −B. 5.3.15 AB is not necessarily symmetric, since (AB)T = B T AT = BA, which is not necessarily the same as AB. (Here we used Theorem 5.3.9a.) 5.3.16 A + B is symmetric, since (A + B)T = AT + B T = A + B. 5.3.17 B −1 is symmetric, because (B −1 )T = (B T )−1 = B −1 . In the first step we have used 5.3.9b. 5.3.18 A10 is symmetric, since (A10 )T = (AT )10 = A10 . 235

Chapter 5 5.3.19 This matrix is symmetric. First note that (A2 )T = (AT )2 = A2 for a symmetric matrix A. Now we can use the linearity of the transpose, (2In + 3A − 4A2 )T = 2InT + 3AT − (4A2 )T = 2In + 3A − 4(AT )2 = 2In + 3A − 4A2 . 5.3.20 AB 2 A is symmetric, since (AB 2 A)T = (ABBA)T = (BA)T (AB)T = AT B T B T AT = AB 2 A. 5.3.21 Symmetric. (AT A)T = AT (AT )T = AT A. 5.3.22 BB T is symmetric: (BB T )T = (B T )T B T = BB T . 5.3.23 Not necessarily symmetric. (A − AT )T = AT − A = −(A − AT ). 5.3.24 Not necessarily symmetric. (AT BA)T = AT (AT B)T = AT B T A. 5.3.25 Symmetric, because (AT B T BA)T = AT B T (B T )T (AT )T = AT B T BA. 5.3.26 Symmetric, since (B(A + AT )B T )T = ((A + AT )B T )T B T = B(A + AT )T B T = B(AT + A)T B T = B((AT )T + AT )B T = B(A + AT )B T . 5.3.27 Using Theorems 5.3.6 and 5.3.9a, we find that (A~v ) · w ~ = (A~v )T w ~ = ~v T AT w ~ = ~v · (AT w), ~ as claimed. 5.3.28 Write L(~x) = A~x; by Definition 5.3.1, A is an orthogonal n × n matrix, so that AT A = In , by Theorem 5.3.7. Now L(~v ) · L(w) ~ = (A~v ) · (Aw) ~ = (A~v )T Aw ~ = ~v T AT Aw ~ = ~v T In w ~ = ~v T w ~ = ~v · w, ~ as claimed. Note that we have used Theorems 5.3.6 and 5.3.9a. 5.3.29 We will use the fact that L preserves length (by Definition 5.3.1) and the dot product (by Exercise 28): L(~ v )·L(w) ~ ·w ~ = arccos k~v~vkk v , w). ~ ∠(L(~v ), L(w)) ~ = arccos kL(~ v )kkL(w)k ~ wk ~ = ∠(~

5.3.30 If L(~x) = ~0, then kL(~x)k = k~xk = 0, so that ~x = ~0. Therefore, ker(L) = {~0}. By Theorem 3.3.7, dim(im(L)) = m − dim(ker(L)) = m. Since Rn has an m-dimensional subspace (namely, im(L)), the inequality m ≤ n holds. The transformation L preserves right angles (the proof of Theorem 5.3.2 applies), so that the columns of A are orthonormal (since they are L(~e1 ), . . . , L(~em )). Therefore, we have AT A = Im (the proof of Theorem 5.3.7 applies). Since the vectors ~v1 , . . . , ~vm form an orthonormal basis projection onto im(A), by Theorem 5.3.10. 1 A simple example of such a transformation is L(~x) = 0 0

of im(A), the matrix AAT represents the orthogonal 0 x1 x 1 ~x, that is, L 1 = x2 . x2 0 0

5.3.31 Yes! If A is orthogonal, then so is AT , by Exercise 11. Since the columns of AT are orthogonal, so are the rows of A. 236

Section 5.3 1 0 5.3.32 a No! As a counterexample, consider A = 0 1 (see Exercise 30). 0 0

b Yes! More generally, if A and B are n × n matrices such that BA = In , then AB = In , by Theorem 2.4.8c. cos(φ) , for some φ. Then ~v2 will be one 5.3.33 Write A = [ ~v1 ~v2 ]. The unit vector ~v1 can be expressed as ~v = sin(φ) − sin(φ) sin(φ) of the two unit vectors orthogonal to ~v1 : ~v2 = or ~v2 = . (See Figure 5.7.) cos(φ) − cos(φ)

cos(φ) − sin(φ) cos(φ) sin(φ) Therefore, an orthogonal 2 × 2 matrix is either of the form A = or A = , sin(φ) cos(φ) sin(φ) − cos(φ) representing a rotation or a reflection. Compare with Exercise 2.2.24. See Figure 5.14.

Figure 5.14: for Problem 5.3.33.

a b 5.3.34 Since the first two columns are orthogonal to the third, we have c = d = 0. Then is an e f cos(φ) − sin(φ) 0 gonal 2 × 2 matrix; By Exercise 33, the 3 × 3 matrix A is either of the form A = 0 sin(φ) cos(φ) cos(φ) sin(φ) 0 0 1 . A= 0 sin(φ) − cos(φ) 0 5.3.35 Let us first think about the inverse L = T −1 of T . 2 3

Write L(~x) = A~x = [ ~v1

~v2

~v3 ] ~x. It is required that L(~e3 ) = ~v3 = 23 . 1 3

237

ortho 0 1 or 0

Chapter 5

Furthermore, the vectors ~v1 , ~v2 , ~v3 must form an orthonormal basis of R3 . By inspection, we find ~v1 =

− 13

− 23 1 3 2 3

.

−1 2 2 2 ~x. −2 1

−2 Then ~v2 = ~v1 × ~v3 = 23 does the job. In summary, we have L(~x) = 13 1 2 − 23

Since the matrix of L is orthogonal, the matrix of T = L−1 is the transpose of the matrix of L: −2 1 2 T (~x) = 31 −1 2 −2 ~x. 2 2 1 There are many other answers (since there are many choices for the vector ~v1 above). 2

5.3.36 Let the third column be the cross product of the first two: A =

3 2 3 1 3

√1 2 − √12

0

√1 18 √1 18 − √418

There is another solution, with the signs in the last column reversed.

.

2 3 −3 2 5.3.37 No, since the vectors 3 and 2 are orthogonal, whereas 0 and −3 are not (see Theorem 5.3.2). 0 2 0 0

0 5.3.38 a The general form of a skew-symmetric 3 × 3 matrix is A = −a −b 2 −a − b2 −bc ac −a2 − c2 −ab , a symmetric matrix. A2 = −bc ac −ab −b2 − c2

a b 0 c , with −c 0

b By Theorem 5.3.9.a, (A2 )T = (AT )2 = (−A)2 = A2 , so that A2 is symmetric. 5.3.39 By Theorem 5.3.10, the matrix of the projection is ~u~uT ; the ij th entry of this matrix is ui uj . 0.5 −0.1 0.5 0.7 5.3.40 An orthonormal basis of W is ~u1 = , ~u2 = (see Exercise 5.2.9). 0.5 −0.7 0.5 0.1

By Theorem 5.3.10, the matrix of the projection onto W is QQT , where Q = [ ~u1 26 18 1 QQT = 100 32 24

18 74 −24 32

32 −24 74 18

24 32 18 26 238

~u2 ].

Section 5.3 1 . 1 5.3.41 A unit vector on the line is ~u = √n .. . 1

The matrix of the orthogonal projection is ~u~uT , the n × n matrix whose entries are all cise 39).

1 n

(compare with Exer-

5.3.42 a Suppose we are projecting onto a subspace W of Rn . Since A~x is in W already, the orthogonal projection of A~x onto W is just A~x itself: A(A~x) = A~x, or A2 ~x = A~x. Since this equation holds for all ~x, we have A2 = A. b A = QQT , for some matrix Q with orthonormal columns ~u1 , . . . , ~um . Note that QT Q = Im , since the ij th entry of QT Q is ~ui · ~uj . Then A2 = QQT QQT = Q(QT Q)QT = QIm QT = QQT = A. 5.3.43 Examine how A acts on ~u, and on a vector ~v orthogonal to ~u: A~u = (2~u~uT − I3 )~u = 2~u~uT ~u − ~u = ~u, since ~uT ~u = ~u · ~u = k~uk2 = 1. A~v = (2~u~uT − I3 )~v = 2~u~uT ~v − ~v = −~v , since ~uT ~v = ~u · ~v = 0. Since A leaves the vectors in L = span(~u) unchanged and reverses the vectors in V = L⊥ , it represents the reflection about L. Note that B = −A, so that B reverses the vectors in L and leaves the vectors in V unchanged; that is, B represents the reflection about V . 5.3.44 Note that AT is an m × n matrix. By Theorems 3.3.7 and 5.3.9c we have dim(ker(AT )) = n − rank(AT ) = n − rank(A). By Theorem 3.3.6, dim(im(A)) = rank(A), so that dim(im(A)) + dim(ker(AT )) = n. 5.3.45 Note that AT is an m × n matrix. By Theorems 3.3.7 and 5.3.9c, we have dim(ker(A)) = m − rank(A) and dim(ker(AT )) = n − rank(AT ) = n − rank(A), so that dim(ker(A)) = dim(ker(AT )) if (and only if) A is a square matrix. 5.3.46 By Theorem 5.2.2, the columns ~u1 , . . . , ~um of Q are orthonormal. Therefore, QT Q = Im , since the ij th entry of QT Q is ~ui · ~uj . If we multiply the equation M = QR by QT from the left then QT M = QT QR = R, as claimed. 5.3.47 By Theorem 5.2.2, the columns ~u1 , . . . , ~um of Q are orthonormal. Therefore, QT Q = Im , since the ij th entry of QT Q is ~ui · ~uj . By Theorem 5.3.9a, we now have AT A = (QR)T QR = RT QT QR = RT R. 5.3.48 As suggested, we consider the QR factorization AT = P R 239

Chapter 5 of AT , where P is orthogonal and R is upper triangular with positive diagonal entries. By Theorem 5.3.9a, A = (P R)T = RT P T . Note that L = RT is lower triangular and Q = P T is orthogonal. 5.3.49 Yes! By Exercise 5.2.45, we can write AT = P L, where P is orthogonal and L is lower triangular. By Theorem 5.3.9a, A = (P L)T = LT P T . Note that R = LT is upper triangular, and Q = P T is orthogonal (by Exercise 11). 5.3.50 a If an n×n matrix A is orthogonal and upper triangular, then A−1 is both lower triangular (since A−1 = AT ) and upper triangular (being the inverse of an upper triangular matrix; compare with Exercise 2.4.35c). Therefore, A−1 = AT is a diagonal matrix, and so is A itself. Since A is orthogonal with positive diagonal entries, all the diagonal entries must be 1, so that A = In . −1 b Using the terminology suggested in the hint, we observe that Q−1 2 Q1 is orthogonal (by Theorem 5.3.4) and R2 R1 −1 −1 is upper triangular with positive diagonal entries. By part a, the matrix Q2 Q1 = R2 R1 is In , so that Q1 = Q2 and R1 = R2 , as claimed.

5.3.51 a Using the terminology suggested in the hint, we observe that Im = QT1 Q1 = (Q2 S)T Q2 S = S T QT2 Q2 S = S T S, so that S is orthogonal, by Theorem 5.3.7. b Using the terminology suggested in the hint, we observe that R2 R1−1 is both orthogonal (let S = R2 R1−1 in part a) and upper triangular, with positive diagonal entries. By Exercise 50a, we have R2 R1−1 = Im , so that R1 = R2 . Then Q1 = Q2 R2 R1−1 = Q2 , as claimed.

5.3.52 Applying the strategy outlined in Summary 4.1.6 to the

1 0 0

0 0 0 0 0,1 0 0 0

1 0 0 0 0 0,0 0 0 0 1 0

5.3.53 Applying the strategy

0 1 0 0 basis −1 0 0 , 0 0 0 0 −1

1 0 0 0,0 1 0 0 0

0 0 0 · 0 0 0

0 0 1

a b c general element b d e of V , we find the basis c e f 0 0 0 0 1 , 0 0 0 , so that dim(V ) = 6. 0 0 0 1

0 outlined in Summary 4.1.6 to the general element −b −c 0 1 0 0 0 0 0,0 0 1 , so that dim(V ) = 3. 0 0 0 −1 0

b c 0 e of V , we find the −e 0

5.3.54 To write the general form of a skew-symmetric n × n matrix A, we can place arbitrary constants above the diagonal, the opposite entries below the diagonal (aij = −aji ), and zeros on the diagonal (since aii = −aii ). See Exercise 53 for the case n = 3. Thus the dimension of the space equals the number of entries above the diagonal of an n × n matrix. In Exercise 55 we will see that there are (n2 − n)/2 such entries. Thus dim(V ) = (n2 − n)/2. 5.3.55 To write the general form of a symmetric n × n matrix A, we can place arbitrary constants on and above the diagonal, and then write the corresponding entries below the diagonal (aij = aji ). See Exercise 52 for the 240

Section 5.3 case n = 3. Thus the dimension of the space equals the number of entries on and above the diagonal of an n × n matrix. Now there are n2 entries in the matrix, n2 − n off the diagonal, and half of them, (n2 − n)/2, above the diagonal. Since there are n entries on the diagonal, we have dim(V ) = (n2 − n)/2 + n = (n2 + n)/2. 5.3.56 Yes and yes (see Exercise 57). 5.3.57 Yes, L is linear, since L(A + B) = (A + B)T = AT + B T = L(A) + L(B) and L(kA) = (kA)T = kAT = kL(A). Yes, L is an isomorphism; the inverse is the transformation R(A) = AT from Rn×m to Rm×n . 5.3.58 Adapting the solution of Exercise 59, we see that the kernel consists of all skew-symmetric matrices, and the image consists of all symmetric matrices. 5.3.59 The kernel consists of all matrixes A such that L(A) = symmetric matrices.

1 2 (A

− AT ) = 0, that is, AT = A; those are the

Following the hint, let’s apply L to a skew-symmetric matrix A, with AT = −A. Then L(A) = (1/2)(A − AT ) = (1/2)2A = A, so that A is in the image of L. Conversely, if A is any 2 × 2 matrix, then L(A) will be skewsymmetric, since (L(A))T = (1/2)(A − AT )T = (1/2)(AT − A) = −L(A). In conclusion: The kernel of L consists of all symmetric matrices, and the image consists of all skew-symmetric matrices. 1 0 5.3.60 Using Theorem 4.3.2, we find the matrix 0 0

0 1 0 0

0 0 1 0

0 0 . 0 −1

5.3.61 Note that the first three matrices of the given basis B are symmetric, so that L(A) = A − AT = 0, and the coordinate vector [L(A)]B is ~0 for all three of them. The last matrix of the basis is skew-symmetric, so that 0 0 0 0 0 0 0 0 L(A) = 2A, and [L(A)]B = 2~e4 . Using Theorem 4.3.2, we find that the B-matrix of L is . 0 0 0 0 0 0 0 2

1 0 5.3.62 By Theorem 5.3.9a, AT = 1 1 −1 2

0 1 0 0 0 −1 1 0 0

0 1 3 2 0 0 1 0 . 2 0 0 1

5.3.63 By Exercise 2.4.94b, the given LDU factorization of A is unique. By Theorem 5.3.9a, A = AT = (LDU )T = U T DT LT = U T DLT is another way to write the LDU factorization of A (since U T is lower triangular and LT is upper triangular). By the uniqueness of the LDU factorization, we have U = LT (and L = U T ), as claimed. A −B T C −DT A + C −B T − DT 5.3.64 a + = AT D CT B+D AT + C T B T A + C −(B + D) = is of the required form. B+D (A + C)T

A b k B

−B T AT

kA = kB

−kB T kAT

kA = kB

−(kB)T (kA)T

is of the required form. 241

Chapter 5 p −q −r −s p s −r q c The general element of H is M = , with four arbitrary constants, r,s,p, and q. Thus r −s p q s r −q p dim(H) = 4; use the strategy outlined in Summary 4.1.6 to construct a basis.

d

A −B T C −DT B AT D CT is of the required form.

=

AC − B T D BC + AT D

−ADT − B T C T −BDT + AT C T

=

AC − B T D BC + AT D

−(BC + AT D)T (AC − B T D)T

Note that A, B, C, D, and their transposes are rotation-dilation matrices, so that they all commute. e

A B

−B T AT

T

=

AT −B

BT (AT )T

is of the required form.

p f Note that the columns ~v1 , ~v2 , ~v3 , ~v4 or M are orthogonal, and they all have length p2 + q 2 + r2 + s2 . Now M T M is the 4 × 4 matrix whose ij th entry is ~vi · ~vj , so that M T M = (p2 + q 2 + r2 + s2 )I4 . g If M 6= 0, then k = p2 + q 2 + r2 + s2 > 0, and M −1 =

1 T p2 +q 2 +r 2 +s2 M .

1 T kM

M = I4 , so that M is invertible, with

By parts b and e, M −1 is in H as well. 0 −1 0 1 h No! A = 0 0 0 0

0 0 0 −1

0 0 0 0 0 0 and B = 1 1 0 0 0 1

−1 0 0 −1 do not commute (AB = −BA). 0 0 0 0

a b a b 10 must be an orthogonal ; it is required that a, b, c and d be integers. Now A = 10 c d c d 10 10 a 2 c 2 matrix, implying that ( 10 ) + ( 10 ) = 1, or a2 + c2 = 100. Checking the squares of all integers from 1 to 9, we see that there are only two ways to write 100 as a sum of two positive perfect squares: 100 = 36 + 64 = 64 + 36. Since a and c are required to be positive, we have either a = 6 and c = 8 or a = 8 and c = 6. In each case we have two options for the second column of A, namely, the two unit vectors perpendicular to the first column vector. Thus we end up with four solutions: .6 −.8 .6 .8 .8 −.6 .8 .6 A= , , or . .8 .6 .8 −.6 .6 .8 .6 −.8

5.3.65 Write 10A =

5.3.66 One approach is to take one of the solutions from Exercise 65, say, the rotation matrix B = 0.28 −0.96 and then let A = B 2 = . Matrix A is orthogonal by Theorems 5.3.4a. 0.96 0.28

0.8 0.6

−0.6 , 0.8

5.3.67 a We need to show that AT A~c = AT ~x, or, equivalently, that AT (~x − A~c) = ~0. But AT (~x − A~c) = AT (~x − c1~v1 − · · · − cm~vm ) is the vector whose ith component is (~vi )T (~x − c1~v1 − · · · − cm~vm ) = ~vi · (~x − c1~v1 − · · · − cm~vm ), which we know to be zero. 242

Section 5.4 b The system AT A~c = AT ~x has a unique solution ~c for a given ~x, since ~c is the coordinate vector of projV ~x with respect to the basis ~v1 , . . . , ~vm . Thus the coefficient matrix AT A must be invertible, so that we can solve for ~c and write ~c = (AT A)−1 AT ~x. Then projV ~x = c1~v1 + · · · + cm~vm = A~c = A(AT A)−1 AT ~x. 5.3.68 If A = QR, then A(AT A)−1 AT = QR(RT QT QR)−1 RT QT = QR(RT R)−1 RT QT = QRR−1 (RT )−1 RT QT = QQT , as in Theorem 5.3.10. The equation QT Q = Im holds since the columns of Q are orthonormal.

Section 5.4 T

5.4.1 A basis of ker(A ) is

−3 . (See Figure 5.15.) 2

Figure 5.15: for Problem 5.4.1.

1 5.4.2 A basis of ker(AT ) is −2 . im(A) is the plane perpendicular to this line. 1 5.4.3 We will first show that the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q span Rn . Any vector ~v in Rn can be written as ⊥ ~v = ~vk + ~v⊥ , where ~vk is in V and ~v⊥ is in V (by definition of orthogonal projection, Theorem 5.1.4). Now ~vk is a linear combination of ~v1 , . . . , ~vp , and ~v⊥ is a linear combination of w ~ 1, . . . , w ~ q , showing that the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q span Rn . Note that p + q = n, by Theorem 5.1.8c; therefore, the vectors ~v1 , . . . , ~vp , w ~ 1, . . . , w ~ q form a basis of Rn , by Theorem 3.3.4d. 5.4.4 By Theorem 5.4.1, the equation (im B)⊥ = ker(B T ) holds for any matrix B. Now let B = AT . Then (im(AT ))⊥ = ker(A). Taking transposes of both sides and using Theorem 5.1.8d we obtain im(AT ) = (kerA)⊥ , as claimed. 243

Chapter 5

5.4.5 V = ker(A), where A =

1 1

1 1 1 . 2 5 4

Then V ⊥ = (kerA)⊥ = im(AT ), by Exercise 4. The two columns of AT form a basis of V ⊥ : 1 1 1 2 , 5 1 4 1 5.4.6 Yes! For any matrix A, im(A)

= (ker(AT ))⊥ ↑

Theorem 5.4.1

= ↑

(ker(AAT ))⊥

=

(ker(AAT )T )⊥

Theorem 5.4.2a

= ↑

im(AAT ).

Theorems 5.4.1 and 5.1.8d.

5.4.7 im(A) and ker(A) are orthogonal complements by Theorem 5.4.1: (imA)⊥ = ker(AT ) = ker(A) 5.4.8 a By Theorem 5.4.6, L+ (~y ) = (AT A)−1 AT ~y . The transformation L+ is linear since it is “given by a matrix,” by Definition 2.1.1. b If L (and therefore A) is invertible, then L+ (~y ) = A−1 (AT )−1 AT ~y = A−1 ~y = L−1 ~y , so that L+ = L−1 . c L+ (L(~x)) =

the unique least-squares solution ~u of L(~u) = L(~x) = ~x.

d L(L+ (~y )) = A(AT A)−1 AT ~y = projV ~y , where V = im(A), by Theorem 5.4.7.

1 e Here A = 0 0

0 1 0 1 . Then L+ (~y ) = (AT A)−1 AT ~y = 0 1 0

0 ~y . 0

5.4.9 ~x0 is the shortest of all the vectors in S. (See Figure 5.16.) 5.4.10 a If ~x is an arbitrary solution of the system A~x = ~b, let ~xh = projV ~x, where V = ker(A), and ~x0 = ~x − projV ~x. Note that ~b = A~x = A(~xh + ~x0 ) = A~xh + A~x0 = A~x0 , since ~xh is in ker(A). b If ~x0 and ~x1 are two solutions of the system A~x = ~b, both from (kerA)⊥ , then ~x1 − ~x0 is in the subspace (kerA)⊥ as well. Also, A(~x1 − ~x0 ) = A~x1 − A~x0 = ~b − ~b = ~0, so that ~x1 − ~x0 is in ker(A). By Theorem 5.1.8b, it follows that ~x1 − ~x0 = ~0, or ~x1 = ~x0 , as claimed. c Write ~x1 = ~xh + ~x0 as in part a; note that ~xh is orthogonal to ~x0 . The claim now follows from the Pythagorean Theorem (Theorem 5.1.9). 5.4.11 a Note that L+ (~y ) = AT (AAT )−1 ~y ; indeed, this vector is in im(AT ) = (kerA)⊥ , and it is a solution of L(~x) = A~x = ~y . 244

Section 5.4

Figure 5.16: for Problem 5.4.9. b L(L+ (~y )) = ~y , by definition of L+ . c L+ (L(~x)) = AT (AAT )−1 A~x = projV ~x, where V = im(AT ) = (kerA)⊥ , by Theorem 5.4.7. d im(L+ ) = im(AT ), by part c, and ker(L+ ) = {~0} (if ~y is in ker(L+ ), then ~y = L(L+ (~y )) = L(~0) = ~0, by part b). e Let A =

1 0

1 0 0 0 ; then the matrix of L+ is AT (AAT )−1 = AT = 0 1 . 1 0 0 0

5.4.12 By Theorem 5.4.5, the least-squares solutions of the linear system A~x = ~b are the exact solutions of the (consistent) system AT A~x = AT ~b. The minimal solution of this normal equation (in the sense of Exercise 10) is called the minimal least-squares solution of the system A~x = ~b. Equivalently, the minimal least-squares solution of A~x = ~b can be defined as the minimal solution of the consistent system A~x = projV ~b, where V = im(A). 5.4.13 a Suppose that L+ (~y1 ) = ~x1 and L+ (~y2 ) = ~x2 ; this means that ~x1 and ~x2 are both in (kerA)⊥ = im(AT ), AT A~x1 = AT ~y1 , and AT A~x2 = AT ~y2 . Then ~x1 + ~x2 is in im(AT ) as well, and AT A(~x1 + ~x2 ) = AT (~y1 + ~y2 ), so that L+ (~y1 + ~y2 ) = ~x1 + ~x2 . The verification of the property L+ (k~y ) = kL+ (~y ) is analogous. b L+ (L(~x)) is the orthogonal projection of ~x onto (kerA)⊥ = im(AT ). c L(L+ (~y )) is the orthogonal projection of ~y onto im(A) = (kerAT ))⊥ . d im(L+ ) = im(AT ) and ker(L+ ) = ker(AT ), by parts b and c. 245

Chapter 5

e L+

y1 y2

y1

1

2

2

= 0 , so that the matrix of L+ is 0 0 0

0 0 . 0

Figure 5.17: for Problem 5.4.14. 5.4.14 L+ (w ~ 1 ) is the minimal solution of the system L(~x) = w ~ 1 . The line S in Figure 5.17 shows all solutions of the system L(~x) = w ~ 1 (compare with Exercise 9). The minimal solution, L+ (w ~ 1 ), is perpendicular to ker(L). L+ (w ~ 2 ) = L+ (projim(L) w ~ 2 ) = L+ (~0) = ~0 L+ (w ~ 3 ) = L+ (projim(L) w ~ 3 ) ≈ L+ (0.55w ~ 1 ) = 0.55L+ (w ~ 1) 5.4.15 Note that (AT A)−1 AT A = In ; let B = (AT A)−1 AT . 5.4.16 If A is an m × n matrix, then dim(imA)⊥

= m − dim(imA) ↑

Theorem 5.1.8c and dim(ker(AT ))

= m − rank(A) ↑ Theorem 3.3.6

= m − rank(AT ). ↑

Theorem 3.3.7 It follows that rank(A) = rank(AT ), as claimed. 5.4.17 Yes! By Theorem 5.4.2, ker(A) = ker(AT A). Taking dimensions of both sides and using Theorem 3.3.7, we find that n − rank(A) = n − rank(AT A); the claim follows. 5.4.18 Yes! By Exercise 17, rank(A) = rank(AT A). Substituting AT for A in Exercise 17 and using Theorem 5.3.9c, we find that rank(A) = rank(AT ) = rank(AAT ). The claim follows. ∗

T

−1

5.4.19 ~x = (A A)

1 , by Theorem 5.4.6. A b= 1 T~

246

Section 5.4 −1 2 and ~b − A~x∗ = 1 . 5.4.20 Using Theorem 5.4.6, we find ~x∗ = 2 1 Note that ~b − A~x∗ is perpendicular to the two columns of A.

5.4.21 Using Theorem 5.4.6, we find ~x∗ =

∗

5.4.22 Using Theorem 5.4.6, we find ~x = exact solution; the error k~b − A~x∗ k is 0.

−1 2

−12 and ~b − A~x∗ = 36 , so that k~b − A~x∗ k = 42. −18

and ~b − A~x∗ = ~0. This system is in fact consistent and ~x∗ is the

3 −2

5.4.23 Using Theorem 5.4.6, we find ~x∗ = ~0; here ~b is perpendicular to im(A). 5.4.24 Using Theorem 5.4.6, we find ~x∗ = [2]. 5 15 x1 5 5.4.25 In this case, the normal equation AT A~x = AT ~b is = , which simplifies to x1 + 3x2 = 1, x 15 45 15 2 1 − 3t , where t is an arbitrary constant. or x1 = 1 − 3x2 . The solutions are of the form ~x∗ = t

66 5.4.26 Here, the normal equation AT A~x = AT ~b is 78 90 where t is an arbitrary constant.

78 93 108

t − 67 1 90 x1 108 x2 = 2 , with solutions ~x∗ = 1 − 2t , 3 126 x3 t

5.4.27 The least-squares solutions of the system SA~x = S~b are the exact solutions of the normal equation (SA)T SA~x = (SA)T S~b. Note that S T S = In , since S is orthogonal; therefore, the normal equation simplifies as follows: (SA)T SA~x = AT S T SA~x = AT A~x and (SA)T S~b = AT S T S~b = AT ~b, so that the normal equation is AT A~x = AT ~b, the same as the normal equation of the A~x = ~b. Therefore, the systems A~x = ~b and SA~x = S~b have the same system 7 . least-squares solution, ~x∗ = 11 5.4.28 The least-squares solutions of the system A~x = ~un are the exact solutions of A~x = projim(A) ~un . Note that ~un is orthogonal to im(A), so that projim(A) ~un = ~0, and the unique least-squares solution is ~x∗ = ~0. 5.4.29 By Theorem 5.4.6, ~x∗ = (AT A)−1 AT ~b =

1+ε 1

1 1+ε

−1

1+ε = 1+ε

1 2+ε

1 1+ε ≈ 21 , where ε = 10−20 . 1+ε 2

If we use a hand-held calculator, due to roundoff errors we find the normal equation infinitely many solutions. 247

1 1 1 1

x1 x2

=

1 , with 1

Chapter 5 5.4.30 We attempt to solve the system c0 + 0c1 = 0 1 0 0 c c0 + 0c1 = 1 , or 1 0 0 = 1 . c1 c0 + 1c1 = 1 1 1 1 This system cannot be solved exactly; the least-squares solution is points best is f ∗ (t) =

1 2

+ 12 t.

c∗0 c∗1

1 = 21 . The line that fits the data 2

Figure 5.18: for Problem 5.4.30. The line goes through the point (1, 1) and “splits the difference” between (0, 0) and (0, 1). See Figure 5.18. 5.4.31 We want

c0 c1

such that 3 1 0 3 = c0 + 0c1 c 3 = c0 + 1c1 or 1 1 1 = 3 . c2 6 6 = c0 + 1c1 1 1

−1 ∗ 1 0 1 0 1 1 1 c 1 Since ker 1 1 = {~0}, 0 = 1 1 c1 0 1 1 0 1 1 1 1 −1 3 3 2 12 = = 3 so f ∗ (t) = 3 + 23 t. (See Figure 5.19.) 2 2 9 2

3 1 1 3 1 1 6

Figure 5.19: for Problem 5.4.31. c0 5.4.32 We want c1 of f (t) = c0 + c1 t + c2 t2 such that c2

248

Section 5.4 1 27 = c0 + 0c1 + 0c2 0 = c0 + 1c1 + 1c2 1 or 1 0 = c0 + 2c1 + 4c2 1 0 = c0 + 3c1 + 9c2

0 1 2 3

27 0 c0 1 0 c1 = 0 4 c2 0 9

If we call the coefficient matrix A, we notice that ker(A) = {~0} so ∗ 27 c0 25.65 0 c1 = (AT A)−1 AT = −28.35 so f ∗ (t) = 25.65 − 28.35t + 6.75t2 . 0 c2 6.75 0

c0 5.4.33 We want c1 such that c2

0 1 c cos(1) 1 1 c2 = . 2 cos(2) c3 3 cos(3) ∗ c1 Since the coefficient matrix has kernel {~0}, we compute c2 using Theorem 5.4.6, obtaining c3 1 0 = c0 + sin(0)c1 + cos(0)c2 1 = c0 + sin(1)c1 + cos(1)c2 1 or 2 = c0 + sin(2)c1 + cos(2)c2 1 1 3 = c0 + sin(3)c1 + cos(3)c2

0 sin(1) sin(2) sin(3)

∗ 1.5 c0 c1 ≈ 0.1 so f ∗ (t) ≈ 1.5 + 0.1 sin t − 1.41 cos t. −1.41 c2

c0 c1 5.4.34 We want c2 such that c3 c4

1 1 1 1 1 1 1

0 sin(0.5) sin(1) sin(1.5) sin(2) sin(2.5) sin(3)

1 cos(0.5) cos(1) cos(1.5) cos(2) cos(2.5) cos(3)

0 sin(1) sin(2) sin(3) sin(4) sin(5) sin(6)

0 1 cos(1) c0 0.5 cos(2) c1 1 cos(3) c2 = 1.5 cos(4) c3 2 2.5 c4 cos(5) 3 cos(6)

Since the columns of the coefficient matrix are linearly independent, its kernel is {~0}. We can use Theorem 5.4.6 1.5 c0 c1 0.109 to compute c2 ≈ −1.537 so f ∗ (t) ≈ 1.5 + 0.109 sin(t) − 1.537 cos(t) + 0.303 sin(2t) + 0.043 cos(2t). 0.303 c3 0.043 c4

5.4.35 a The ij th entry of ATn An is the dot product of the ith row of ATn and the jth column of An , i.e. 249

Chapter 5 n X

n n X sin ai ATn An = i=1 X n cos ai

i=1 n X

n X

sin ai

i=1

sin2 ai sin ai cos ai

i=1

2π Z 2π T b lim 2π A A = sin t dt n n n n→∞ 0 Z 2π cos t dt 0

Z

i=1

i=1

Z

2π

sin t dt 2π

sin2 t dt

0

Z

2π

Z

cos t dt

2π 2π 0 = sin t cos t dt 0 0 Z 2π cos2 t dt 0

0

Z

g(ai )

i=1 n X g(ai ) sin ai sin ai cos ai and ATn~b = . i=1 i=1 n n X X 2 g(ai ) cos ai cos ai

i=1 n X

i=1

n X

X n

cos ai

2π

sin t cos t dt

Z

0 π 0

0 0 π

0

0

2π

g(t) dt 0 Z 2π T~ and lim 2π b = A g(t) sin t dt n n n→∞ 0 Z 2π g(t) cos t dt 0

( Here

2π n

=

2π n→∞ n

∆t so lim

n X

cos(ti )

=

i=1

lim

n→∞

All other limits are obtained similarly. )

2π cn c c p = lim pn = 0 n→∞ 0 qn q

0 π 0

where c, p, q are given above.

Z

n X

cos(ti )∆t

=

Z

2π

cos t dt for instance.

0

i=1

Z 2π 1 g(t) dt g(t) dt 2π 0 −1 0 0 1 Z 2π Z 2π 0 g(t) sin t dt = g(t) sin t dt and f (t) = c+p sin t+q cos t, π 0 0 π Z 2π Z 2π 1 g(t) cos t dt g(t) cos t dt π 0 0 2π

a 5.4.36 We want b such that c

2π 2π a + b sin 32 + c cos 32 366 366 2π 2π a + b sin 77 + c cos 77 366 366 2π 2π 121 + c cos 121 a + b sin 366 366 250

=

10

=

12

=

14

Section 5.4

a + b sin

1

sin

1 Using A = 1

sin

sin

1

sin

2π 366 32 2π 366 77 2π 366 121 2π 366 152

cos cos cos cos

2π 2π 152 + c cos 152 366 366

2π 366 32 2π 366 77 2π 366 121 2π 366 152

∗ 10 a 12 and ~b = , we compute b 14 c 15

c0 c1

∗

= (AT A)−1 AT ~b ≈

2π 366 t

− 2.899 cos

2π 366 t

.

35 log 35 46 c0 log 46 = 59 c1 log 77 69 log 133

↑

↑

~b

A so

15

12.26 = (AT A)−1 AT ~b ≈ 0.431 and f ∗ (t) ≈ 12.26 + 0.431 sin −2.899 5.4.37 a We want c0 , c1 such that c0 + c1 (35) = log 35 1 c0 + c1 (46) = log 46 1 or c0 + c1 (59) = log 77 1 c0 + c1 (69) = log 133 1

=

0.915 0.017

so log(d) ≈ 0.915 + 0.017t.

b d ≈ 100.915 · 100.017t ≈ 8.22 · 100.017t c If t = 88 then d ≈ 258. Since the Airbus has only 93 displays, new technologies must have rendered the old trends obsolete. c0 5.4.38 We want c1 such that c2

110 2 1 12 0 c0 180 5 1 c1 = 120 . 160 c2 11 1 160 6 0 ∗ 125 c0 5 , so that w∗ = 125 + 5h − 25g. The least-squares solution is c1 = −25 c2 1 110 = c0 + 2c1 + c2 180 = c0 + 12c1 + 0c2 1 120 = c0 + 5c1 + c2 or 1 1 160 = c0 + 11c1 + c2 160 = c0 + 6c1 + 0c2 1

For a general population, we expect c0 and c1 to be positive, since c0 gives the weight of a 5′ male, and increased height should contribute positively to the weight. We expect c2 to be negative, since females tend to be lighter than males of equal height. 5.4.39 a We want

c0 c1

such that 251

Chapter 5 log(250) = c0 + c1 log(600, 000) log(60) = c0 + c1 log(200, 000) log(25) = c0 + c1 log(60, 000) log(12) = c0 + c1 log(10, 000) log(5) = c0 + c1 log(2500) The least-squares solution to the above system is

c0 c1

∗

≈

−1.616 0.664

so log z ≈ −1.616 + 0.664 log g.

b Exponentiating both sides of the answer to a, we get z ≈ 10−1.616 · g 0.664 ≈ 0.0242 · g 0.664 . c This model is close since 5.4.40 First we look for

c0 c1

√

g = g 0.5 .

such that log D = c0 + c1 log a.

Proceeding as in Exercise 39, we get

c0 c1

∗

0 ≈ , i.e. log D ≈ 1.5 log a, hence D ≈ 101.5 log a = a1.5 . 1.5

Note that the formula D = a1.5 is Kepler’s third law of planetary motion. 5.4.41 a We want log 533 log 1, 823 log 4, 974 log 7, 933

c0 c1

such that log D = c0 + c1 t (t in years since 1975), i.e.

= c0 + c1 (0) = c0 + c1 (10) = c0 + c1 (20) = c0 + c1 (30)

The least-squares solution to the system is

c0 c1

∗

≈

2.8 , i.e. log D ≈ 2.8 + 0.04 or D ≈ 102.8 · 100.040t . 0.040

b In the year 2015, we have t = 40 and D ≈ 24, 200. The formula predicts a debt of about US$24 trillion. 5.4.42 Clearly, L is a linear transformation. We will use Theorem 4.2.4a and show that ker(L) = {~0} and im(L) = im(A). Now ker(L) = ker(A)∩im(AT ) = {~0}, by Theorems 5.4.1 and 5.1.8b. Also, im(L) = {A~v : ~v in im(AT )} = im(AAT ) = im(A), by Exercise 6.

Section 5.5 5.5.1 Since f is nonzero, there is a c on [a, b] such that f (c) = d = 6 0. By continuity, there is an ε > 0 such that for all x on the open interval (c − ε, c + ε) where f is defined (see any good Calculus text). |f (x)| > |d| 2 252

Section 5.5

Figure 5.20: for Problem 5.5.1.

We assume that c − ε ≥ a and c + ε ≤ b and leave the other cases to the reader. Then hf, f i = Z c+ε 2 d2 ε d dt = > 0, as claimed. (See Figure 5.20.) 2 2 c−ε

Z

a

b

(f (t))2 dt ≥

5.5.2 We perform the following operations: hf, g + hi = hg + h, f i ↑ Definition 5.5.1a

= hg, f i + hh, f i ↑

Definition 5.5.1b

= hf, gi + hf, hi ↑

Definition 5.5.1a

5.5.3 a Note that h~x, ~y i = (S~x)T S~y = S~x · S~y . We will check the four parts of Definition 5.5.1 α. h~x, ~y i = S~x · S~y = S~y · S~x = h~y , ~xi β. h~x + ~y , ~zi = S(~x + ~y ) · S~z = (S~x + S~y ) · S~z = (S~x · S~z) + (S~y · S~z) = h~x, ~zi + h~y , ~zi γ.

hc~x, ~y i = S(c~x) · S~y = c(S~x) · S~y = ch~x, ~y i

δ. If ~x 6= 0, then h~x, ~xi = S~x · S~x = kS~xk2 is positive if S~x 6= ~0, that is, if ~x is not in the kernel of S. It is required that S~x 6= ~0 whenever ~x 6= ~0, that is, ker(S) = {~0}. Answer: S must be invertible. b It is required that h~x, ~y i = (S~x)T S~y = ~xT S T S~y equal ~x · ~y = ~xT ~y for all ~x and ~y . This is the case if and only if S T S = In , that is, S is orthogonal. 5.5.4 a For column vectors ~v , w, ~ we have h~v , wi ~ = trace(~v T w) ~ = trace(~v · w) ~ = ~v · w, ~ the dot product. b For row vectors ~v , w, ~ the ij th entry of ~v T w ~ is vi wj , so that h~v , wi ~ = trace(~v T w) ~ = product.

m X i=1

vi wi = ~v · w, ~ again the dot

5.5.5 a hhA, Bii = tr(AB T ) = tr((AB T )T ) = tr(BAT ) = hhB, Aii In the second step we have used the fact that tr(M ) = tr(M T ), for any square matrix M . 253

Chapter 5 b hhA + B, Cii = tr((A + B)C T ) = tr(AC T + BC T ) = tr(AC T ) + tr(BC T ) = hhA, Cii + hhB, Cii c hhcA, Bii = tr(cAB T ) = ctr(AB T ) = chhA, Bii d hhA, Aii = tr(AAT ) =

n X i=1

k~vi k2 > 0 if A 6= 0, where ~vi is the ith row of A.

We have shown that hh·, ·ii does indeed define an inner product. 5.5.6 a The ii th entry of PQ is

m X

pik qki , so that tr(P Q) =

n m X X

pik qki .

i=1 k=1

k=1

Likewise, tr(QP ) =

m n X X

qik pki .

i=1 k=1

Reversing the roles of i and k and the order of summation we see that tr(P Q) = tr(QP ), as claimed. b Using part a and the fact that tr(M ) = tr(M T ) for any square matrix M , we find that hA, Bi = tr(AT B) = tr(BAT ) = tr((BAT )T ) = tr(AB T ) = hhA, Bii 5.5.7 Axioms a, b, and c hold for any choice of k (check this!). Also, it is required that hhv, vii = khv, vi be positive for nonzero v. Since hv, vi is positive, this is the case if (and only if) k is positive. 5.5.8 By parts b and c of Definition 5.5.1, we have T (u + v) = hu + v, wi = hu, wi + hv, wi = T (u) + T (v) and T (cv) = hcv, wi = chv, wi = cT (v), so that T is linear. If w = 0, then im(T ) = {0} and ker(T ) = V . If w 6= 0, then im(T ) = R and ker(T ) consists of all v perpendicular to w. 5.5.9 If f is even and g is odd, then fg is odd, so that hf, gi =

Z

1

f g = 0.

−1

5.5.10 A function g(t) = a + bt + ct2 is orthogonal to f (t) = t if Z 1 2 1 (at + bt2 + ct3 ) dt = [ a2 t2 + 3b t3 + 4c t4 ]−1 = b = 0, that is, if b = 0. hf, gi = 3 −1 Thus, the functions 1 and t2 form a basis of the space of all functions in P2 orthogonal to f (t) = t. To find an Z 1 orthonormal basis g1 (t), g2 (t), we apply Gram-Schmidt. Now k1k = 12 1 dt = 1, so that we can let g1 (t) = 1. Then g2 (t) =

t2 −h1,t2 i1 kt2 −h1,t2 i1k

=

t2 − 13 kt2 − 13 k

−1

√

=

5 2 2 (3t

− 1)

√

Answer : g1 (t) = 1, g2 (t) =

5 2 2 (3t

− 1)

5.5.11 hf, gi = hcos(t), cos(t + δ)i = hcos(t), cos(t) cos(δ) − sin(t) sin(δ)i = cos(δ)hcos(t), cos(t)i − sin(δ) hcos(t), sin(t)i = cos(δ), by Theorem 5.5.4. Also, hg, gi = 1 (left to reader). 254

Section 5.5 Thus, ∠(f, g) = arccos

5.5.12 By Theorem 5.5.5Z E D a0 = |t|, √12 = √12π

hf,gi kf kkgk

= arccos(cos δ) = δ.

π

π 1 |t| dt = √ , bk = h|t|, sin(kt)i = π 2 −π

function.

Z

π

−π

|t| sin(kt) dt = 0, since the integrand is an odd

π

Z 2 π |t| cos(kt) dt = ck = h|t|, cos(kt)i = t cos(kt) dt π 0 −π R π π = π2 k1 t sin(kt) 0 − 0 k1 sin(kt) dt 0 if k is even = π2 k12 [cos(kt)]π0 = πk2 2 (cos(kπ) − 1) = − k24π if k is odd In summary: 1 π

a0 bk ck

=0 =0 0 = − k24π

Z

if k is even if k is odd

5.5.13 The sequence (a0 , b1 , c1 , b2 , c2 , . . .) is “square-summable” by Theorem 5.5.6, so that it is in ℓ2 . Also, k(a0 , b1 , c1 , b2 , c2 , . . .)k a20 + b21 + c21 + b22 + c22 + · · · = kf k2 , by Theorem 5.5.6, so that the two norms are equal. 5.5.14 a This is not an inner product since there are nonzero polynomials f (t) in P2 with f (1) = f (2) = 0, so that hf, f i = (f (1))2 + (f (2))2 = 0. (For example, let f (t) = (t − 1)(t − 2).) b This is an inner product. We leave it to the reader to check axioms a to c. As for d: A nonzero polynomial f in P2 has at most two zeros, so that f (1) 6= 0 or f (2) 6= 0 or f (3) 6= 0, and hf, f i = (f (1))2 + (f (2))2 + (f (3))2 > 0. 1 0 0 1 , = , = c, by part a of Definition 5.5.1, so that b = c. Check 0 1 1 0 that if b = c then h~v , wi ~ = hw, ~ ~v i for all ~v , w ~ in R2 . Check that parts (b) and (c) of Definition 5.5.1 are satisfied for of constants c,d. Next, we need to worry about part allvalues (d) of that definition. It is required that x1 x1 x1 2 2 , = x1 + 2bx1 x2 + dx2 be positive for all nonzero . We can complete the square and write x2 x2 x2 x1 if (and only if) x21 + 2bx1 x2 + dx22 = (x1 + bx2 )2 + (d − b2 )x22 ; this quantity is positive for all nonzero x2 d − b2 > 0, or, d > b2 . In summary, the function is an inner product if (and only if) b = c and d > b2 .

5.5.15 First note that b =

5.5.16 a We start with the standard basis 1, t and use the Gram-Schmidt process to construct an orthonormal basis g1 (t), g2 (t). s Z 1 √ t− 1 t−h1,ti1 dt = 1, so that we can let g1 (t) = 1. Then g2 (t) = kt−h1,ti1k = kt− 21 k = 3(2t − 1). k1k = 2

0

Summary: g1 (t) = 1 and g2 (t) =

√ 3(2t − 1) is an orthonormal basis.

b We are looking for projP1 (t2 ) = hg1 (t), t2 ig1 (t) + hg2 (t), t2 ig2 (t), by Theorem 5.5.3. 255

Chapter 5

We find that hg1 (t), t2 i =

Z

1

t2 dt =

0

1) = t − 16 . See Figure 5.21.

√ 1 and hg2 (t), t2 i = 3 3

Z

0

1

(2t3 − t2 ) dt =

√

3 , so that projP1 t2 = 6

1 3

+ 21 (2t −

Figure 5.21: for Problem 5.5.16b. 5.5.17 We leave it to the reader to check that the first three axioms are satisfied for any such T . As for axiom d: It is required that hv, vi = T (v) · T (v) = kT (v)k2 be positive for any nonzero v, that is, T (v) 6= ~0. This means that the kernel of T must be {0}. c1 c2 5.5.18 Let the orthonormal basis be f1 , . . . , fn and f = c1 f1 + · · · + cn fn ; then [f ]B = ... . The Pythagorean

Theorem tells us that kf k2 = c21 + · · · + c2n = k[f ]B k2 , so that kf k = k[f ]B k.

cn

x1 y p q x1 y ,w ~ = 1 , and A = , then , 1 x2 y2 x2 r s y2 p q y1 = [x1 x2 ] = px1 y1 + qx1 y2 + rx2 y1 + sx2 y2 . Note that in Exercise 15 we considered the special case y2 r s 1 1 p = 1. First it is required that p = , be positive. 0 0 i h x1 y Now we can write , 1 = p x1 y1 + pq x1 y2 + pr x2 y1 + ps x2 y2 and use our work in Exercise 15 with b = pq , c = pr , d x2 y2 2 to see that the conditions q = r and q < ps must hold. In summary, the function is an inner product if (and p q only if) the entries of matrix A = satisfy the conditions p > 0, q = r and det(A) = ps − q 2 > 0. r s

5.5.19 If we write ~v =

5.5.20 a −2 . 1

x1 1 1 = [x1 x2 ] , 2 x2 0

2 8

1 = x1 + 2x2 = 0 when x1 = −2x2 . This is the line spanned by vector 0

1 −2 b Since vectors and are orthogonal, we merely have to multiply each of them with the recipro0 1 256

Section 5.5

2

1 1 2 1 1

cal of its norm. Now = 1, so that is a unit vector, and

= [1 0] 2 8 0 0 0

−2 −1 1 −2 1 2

is an orthonormal basis. = 4, so that [ −2 1 ] 1

1 = 2. Thus 0 , 1 2 8 2

−2 2

1 =

5.5.21 Note that h~v , wi ~ = (~v + w) ~ · (~v + w) ~ − ~v · ~v − w ~ ·w ~ = 2(~v · w), ~ double the dot product. Thus it’s an inner product, by Exercise 7. 5.5.22 Apply the Cauchy-Schwarz inequality to f (t) and g(t) = 1; note that kgk = 1: |hf, gi| ≤ kf kkgk = kf k or hf, gi2 ≤ kf k2 or

Z

1

f (t) dt

0

2

≤

R1 0

(f (t))2 dt.

5.5.23 We start with the standard basis 1, t of P1 and use the Gram-Schmidt process to construct and orthonormal basis g1 (t), g2 (t). q t− 1 t−h1,ti1 = kt− 21 k = 2t − 1. k1k = 12 (1 · 1) + 1 · 1 = 1, so that we can let g1 (t) = 1. Then g2 (t) = kt−(1,t)1k 2

Summary: g1 (t) = 1 and g2 (t) = 2t − 1 is an orthonormal basis.

5.5.24 a hf, g + hi = hf, gi + hf, hi = 0 + 8 = 8 b kg + hk =

p

hg + h, g + hi =

p

hg, gi + 2hg, hi + hh, hi =

√ √ 1 + 6 + 50 = 57

c Since hf, gi = 0, kgk = 1, and kf k = 2, we know that f2 , g is an orthonormal basis of span (f, g). E D Now projE h = f2 , h f2 + hg, hig = 14 hf, hif + hg, hig = 2f + 3g. d From part c we know that 12 f, g are orthonormal, so we apply Theorem 5.2.1 to obtain the third polynomial in an orthonormal basis of span(f, g, h): h−projE h kh−projE hk

=

h−2f −3g kh−2f −3gk

Orthonormal basis:

=

h−2f −3g 5

2 1 2 f, g, − 5 f

= − 25 f − 53 g + 15 h

− 53 g + 15 h

5.5.25 Using the inner product defined in Example 2, we find that q q p 2 k~xk = h~x, ~xi = 1 + 41 + 91 + · · · + n12 + · · · = π6 = √π6 (see the text right after Theorem 5.5.6). 5.5.26 a0 = bk =

1 π

Z

√1 2π

Z

π

f (t) dt = 0

−π

π

f (t) sin(kt) dt =

−π

=

2 − kπ [cos(kt)]π0

=

0 4 πk

1 π

Z −

0

sin(kt) dt +

−π

Z

π

sin(kt) dt

0

if k is even if k is odd 257

=

2 π

Z

0

π

sin(kt) dt

Chapter 5

Figure 5.22: for Problem 5.5.26.

ck =

1 π

Z

π

f (t) cos(kt) dt = 0, since the integrand is odd.

−π

f1 (t) = f2 (t) =

4 π

sin(t). See Figure 5.22.

Figure 5.23: for Problem 5.5.26.

f3 (t) = f4 (t) =

4 π

sin(t) +

4 3π

sin(3t). See Figure 5.23.

f5 (t) = f6 (t) =

4 π

sin(t) +

4 3π

sin(3t) +

4 5π

sin(5t). See Figure 5.24. 258

Section 5.5

Figure 5.24: for Problem 5.5.26.

bk =

1 π

ck =

1 π

π

1 f (t) dt = √ . 2 −π Z π Z 0 1 π 1 π f (t) sin(kt) dt = sin(kt) dt = − [cos(kt)]0 = 2 π 0 kπ −π kπ Z π Z 1 π 1 f (t) cos(kt) dt = [sin(kt)]π0 = 0 cos(kt) dt = π kπ −π 0

5.5.27 a0 =

√1 2π

Z

+

2 π

sin(t), f3 (t) = f4 (t) =

5.5.28 kf k = hf, f i =

1 π

Z

f1 (t) = f2 (t) =

1 2

1 2

+

2 π

sin(t) +

2 3π

if k is even if k is odd

sin(3t)

.. . 2

π

1 (f (t)) dt = π −π 2

Now Theorem 5.5.6 tells us that

5.5.29 kf k2 = hf, f i =

1 π

Z

1 dt = 2

−π

16 16 16 π 2 + 9π 2 + 25π 2 +· · ·

π

(f (t))2 dt =

−π

Now Theorem 5.5.6 tells us that

1 2

+

X 1 1 1 1 =1+ + + + ··· = k2 9 25 49

k odd

π

Z

1 π

4 π2

1 2 4 π2

Z

π

+

4 9π 2

=

16 π2

X 1 k2

k odd

!

= 2, or

X 1 1 1 1 π2 = 1+ + + +· · · = . k2 9 25 49 8

k odd

1 dt = 1

0

=

+

4 25π 2

+ · · · = 1, or

π2 . 8

1 2

+

4 π2

X 1 k2

k odd

!

= 1 or

5.5.30 There is an invertible linear transformation T (~x) = A~x from R2 to R2 that transforms E into the unit circle. −1 (If E is parametrized by cos(t)w ~ 1 + sin(t)w ~ 2 , let A = [ w ~1 w ~ 2 ] . Compare with Exercises 2.2.50 and 2.2.51.) 259

Chapter 5 Now ~x is on E if kT (~x)k = 1, or T (~x) · T (~x) = 1. This means that the inner product h~x, ~y i = T (~x) · T (~y ) does the job (see Exercise 17). (There are other, very different approaches to this problem.) √1 , f1 (t) 2

5.5.31 An orthonormal basis of P2 of the desired form is f0 (t) = with Exercise 10), and the zeros of f2 (t) are a1,2 = Next we find the weights w1 , w2 such that

Z

± √13 .

1

f (t) dt =

−1

2 X

=

q

3 2 t, f2 (t)

=

1 2

q

5 2 2 (3t

− 1) (compare

wi f (ai ) for all f in P1 . We need to make sure that

i=1

the equation holds for 1 and t. # " 2 = w1 + w2 , with solution w1 = w2 = 1. 0 = √13 w1 − √13 w2 1

1 1 √ + f −√ holds for all polynomials f in 3 3 −1 P1 . We can check that it holds for t2 and t3 as well, that is, it holds in fact for all cubic polynomials. Z 1 Z 1 2 2 3 3 2 t2 dt = equals √13 + − √13 = 32 , and t3 dt = 0 equals √13 + − √13 = 0. 3 −1 −1

It follows that the equation

5.5.32 a.htn , tm i = b. ktn k =

p

1 2

R1

Z

f (t) dt = f (a1 ) + f (a2 ) = f

tn+m dt = −1

htn , tn i =

q

1 2n+1

1 2

h

tn+m+1 n+m+1

i1

=

−1

1 n+m+1

if n + m is even if n + m is odd

0

by part a.

c. It helps to observe that 1 and t2 are orthogonal to t and t3 . Accordingly, many terms in the Gram-Schmidt formulas vanish.

√ √ t2 − t2 , 1 1 5 1 t g0 (t) = = 1, g1 (t) = = 3 t, g2 (t) = 2 = 3t2 − 1 2 k1k ktk kt − ht , 1i 1k 2

√ √ √ t3 − t3 , 3 t 3 t t3 − 3t/5 7

√ √

= 3 g3 (t) = = 5t3 − 3t

t3 − t3 , 3 t 3 t kt − 3t/5k 2

d. The first few Legendre Polynomials are

g0 (t) g1 (1)

= 1,

g1 (t) g1 (1)

= t,

g2 (t) g2 (1)

= 3t2 − 1 /2,

g3 (t) g3 (1)

= 5t3 − 3t /2.

1 e. Since f (t) = 1+t 2 is an even function, we need to consider g0 (t) and g2 (t) only. The solution is hf (t), g0 (t)i g0 (t)+

15 3π 2 hf (t), g2 (t)i g2 (t) = hf (t), 1i 1 + 45 f (t), 3t2 − 1 3t2 − 1 = π4 + 85 (6 − 2π) 3t2 − 1 = 15 4 (3 − π) t − 4 + 2 . 1

f (t) = g(t) =

15 4 (3

− π)t2 −

-1

15 4

+ 32 π 1

260

1 1 + t2

t

Section 5.5 5.5.33 a The first property of an inner product, hf, gi = hg, f i, follows from the fact that f (t)g(t) = g(t)f (t). The property hf + h, gi = hf, gi + hh, gi follows from the sum rule for integrals, and the property hcf, gi = c hf, gi follows from the constant multiple rule for integrals. Proceed as in Exercise 1 to show that hf, f i > 0 for all nonzero f. b If f (t) = 1 then kf k =

qR qR p b b 2 dt = hf, f i = w(t) · 1 w(t)dt = 1. a a

√ R1 5.5.34 a. Note that the graph of 1 − t2 is the upper half of the unit circle centered at the origin. Thus −1 w(t) dt = R √ 2 1 1 − t2 dt = π2 (half the area of the unit circle) = 1 π −1 p b. If f (t) = 1 then kf k = hf, f i = 1 by part a and Exercise 33b. R1 √ c. If n + m is odd then htn , tm i = π2 −1 1 − t2 tn+m dt = 0 since the integrand is an odd function.

d. It follows from the definition of this inner product that ht, ti = 1, t2 = 1/4 and t2 , t2 = t, t3 = 1/8 so p

2 p √ that ktk = 1/4 = 1/2 and t = 1/8 = 2/4.

e. It helps to observe that 1 and t2 are orthogonal to t and t3 . Accordingly, many terms in the Gram-Schmidt formulas vanish. The first few Chebyshev Polynomials of the Second Kind are

t2 − t2 , 1 1 1 t t2 − 1/4 p g0 (t) = = 1, g1 (t) = = 2 t, g2 (t) = 2 = = 4t2 − 1 k1k ktk kt − ht2 , 1i 1k 1/16 g3 (t) =

t3 − t3 , 2 t 2 t t3 − t/2 = p = 8t3 − 4t. 3 3 kt − ht , 2 ti 2 tk 1/64

f. Since f (t) = t4 is an even we need to consider g0 (t) and g2 (t) only. The solution is hf (t), g0 (t)i g0 (t) + function,

5 1 − 18 4t2 − 1 = 43 t2 − 16 . hf (t), g2 (t)i g2 (t) = t4 , 1 1 + t4 , 4t2 − 1 4t2 − 1 = 81 + 16 f (t) = t4

1

g(t) = 34 t2 −

-1

1

1 16

t

q R p p p 1 5.5.35 a. ktk32 = 12 −1 t2 dt = 1/3 and ktk34 = ht, ti34 = h1, t2 i34 = 1/2, so that ktk32 > ktk34 . Note that

2 1, t 34 = 1/4 is given in Exercise 34. 1

w32 (t) =

w34 (t) =

1 2

2 π

√ 1 − t2

-1

t

1

1

w32 (t) t2 = 12 t2 w34 (t) t2 =

-1

2 2 πt

√ 1 − t2

1

261

t

Chapter 5 q R q √ p √ √ 1 1 2 ) dt = = (1 − t b. For f (t) = 1 − t2 we have kf k32 = 2/3 and kf k 1 − t2 , 1 − t2 34 = 34 2 −1 p p p h1, 1 − t2 i34 = 1 − 1/4 = 3/4. 1

w32 (t)

w34 (t)

√ √ 1 − t2 = 21 1 − t2

√ 1 − t2 =

2 π (1

-1

− t2 ) 1

t

True or False Ch 5.TF.1 T, since (A + B)T = AT + B T = A + B Ch 5.TF.2 T, by Theorem 5.3.4 Ch 5.TF.3 F. Consider

1 1

1 . 1

Ch 5.TF.4 T. First note that AT = A−1 , by Theorem 2.4.8. Thus A is orthogonal, by Theorem 5.3.7. Ch 5.TF.5 F. The correct formula is projL (~x) = (~x · ~u)~u, by Definition 2.2.1. Ch 5.TF.6 T, since (7A)T = 7AT = 7A. Ch 5.TF.7 F. Consider T (~x) =

1 ~x. 0

1 0

Ch 5.TF.8 T, by Theorem 5.3.9.b Ch 5.TF.9 T, by Theorem 5.3.4a Ch 5.TF.10 F. We have (AB)T = B T AT , by Theorem 5.3.9a. Ch 5.TF.11 T. If A is orthogonal, then AT = A−1 , and A−1 is orthogonal by Theorem 5.3.4b. Ch 5.TF.12 F. Consider

0 1 1 0

0 −1 1 0 . = 1 0 0 −1

0 Ch 5.TF.13 F. Consider A = B = 0

1 1 0 0 T T . Then AB = isn’t equal to B A = 0 0 0 0

0 . 1

Ch 5.TF.14 that the columns of A be orthonormal (Theorem 5.3.10). As a counterexample, consider F. It is required 4 0 2 . , with AAT = A= 0 0 0 262

True or False Ch 5.TF.15 T, since (ABBA)T = AT B T B T AT = ABBA, by Theorem 5.3.9a Ch 5.TF.16 T, since AT B T = (BA)T = (AB)T = B T AT , by Theorem 5.3.9a Ch 5.TF.17 F. dim(V ) + dim(V ⊥ ) = 5, by Theorem 5.1.8c. Thus one of the dimensions is even and the other odd. Ch 5.TF.18 T. Consider the QR factorization (Theorem 5.2.2) Ch 5.TF.19 F. The Pythagorean Theorem holds for orthogonal vectors ~x, ~y only (Theorem 5.1.9) a c a b Ch 5.TF.20 T. det = ad − bc = det . b d c d Ch 5.TF.21 F. As a counterexample, consider

1 0

0 −1

and

0 1

−1 . 0

Ch 5.TF.22 T, by Theorem 5.4.1. Ch 5.TF.23 T, by Theorem 5.4.2a. Ch 5.TF.24 F. Consider A =

2 0 , or any other symmetric matrix that fails to be orthogonal. 0 2

−1 0 −1 0 Ch 5.TF.25 F. det = −1 − 0 = −1, yet is orthogonal. 0 1 0 1 Ch 5.TF.26 T.

1

2 (A

T − AT ) = 12 (A − AT )T = 12 (AT − A) = − 21 (A − AT ) .

Ch 5.TF.27 T, since the columns are unit vectors.

Ch 5.TF.28 T. Use the Gram-Schmidt process to construct such a basis (Theorem 5.2.1) Ch 5.TF.29 F. The columns fail to be unit vectors (use Theorem 5.3.3b) Ch 5.TF.30 T, by definition of an orthogonal projection (Theorem 5.1.4). 1 0 cos θ − sin θ 1 + cos θ − sin θ Ch 5.TF.31 T. Try A = and B = , so that A + B = . It is required sin θ cos θ sin θ 1 + cos θ 0 1 1 + cos θ − sin θ that and be unit vectors, meaning that 1 + 2 cos θ + cos2 θ + sin2 θ = 2 + 2 cos θ = 1, or sin θ 1 + cos θ √ 3 1 √ − − 1 0 2 is a solution. and B = √2 cos θ = − 12 , and sin θ = ± 23 . Thus A = 3 0 1 −1 2

Ch 5.TF.32 F. Consider A =

0 1 2

− 21

0

2

, for example, representing a rotation combined with a scaling. 263

Chapter 5

Ch 5.TF.33 F. Consider A =

−1 1 . 0 1

Ch 5.TF.34 T. By Definition 5.1.12, quantity cos(θ) =

~ v ·w ~ k~ v kkwk ~

is positive, so that θ is an acute angle.

Ch 5.TF.35 T. In Theorem 5.4.1, let A = B T to see that (im(B T ))⊥ = ker(B). Now take the orthogonal complements of both sides and use Theorem 5.1.8d. Ch 5.TF.36 T, since (AT A)T = AT (AT )T = AT A, by Theorem 5.3.9a. Ch 5.TF.37 F. Verify that matrices A =

Ch 5.TF.38 F. Consider B = 5.4.2.

1 0

0 −1

and B =

1 0

1 −1

are similar.

0 1 . The correct formula im(B) = im(BB T ) follows from Theorems 5.4.1 and 0 0

Ch 5.TF.39 T. We know that AT = A and S −1 = S T . Now (S −1 AS)T = S T AT (S −1 )T = S −1 AS, by Theorem 5.3.9a. Ch 5.TF.40 T. By Theorem 5.4.2, we have ker(A) = ker(AT A). Replacing A by AT in this formula, we find that ker(AT ) = ker(AAT ). Now ker(A) = ker(AT A) = ker(AAT ) = ker(AT ). Ch 5.TF.41 T. We attempt to write A = S + Q, where S is symmetric and Q is skew-symmetric. Then AT = S T + QT = S − Q. Adding the equations A = S + Q and AT = S − Q together gives 2S = A + AT and S = 12 (A + AT ). Similarly we find Q = 21 (A − AT ). Check that the decomposition A = S + Q = ( 12 (A + AT )) + ( 21 (A − AT )) does the job.

x1 1 Ch 5.TF.42 T. Apply the Cauchy-Schwarz inequality (squared), (~x · ~y )2 ≤ k~xk2 k~y k2 , to ~x = . . . and ~y = . . . xn 1 (all n entries are 1). 2 2 x y x + y 2 xz + yt x + yz xy + yt T 2 Ch 5.TF.43 T. Let A = . We know that AA = A , or = . We z t xz + yt z 2 + t2 zx + tz yz + t2 2 2 2 need to show that y = z. If y 6= 0, this follows from the equation x + y = x + yz; if z 6= 0, it follows from z 2 + t2 = yz + t2 ; if both y and z are zero, we are all set.

Ch 5.TF.44 T, since ~x · (projV ~x) = (projV ~x + (~x − projV ~x)) · projV ~x = kprojV ~xk2 ≥ 0. Note that ~x − projV ~x is orthogonal to projV ~x, by the definition of a projection.

~ Ch 5.TF.45 T. Note that 1 = A k~x1k ~x = k~x1k Ax

= Definition 5.3.1.

1 xk k~ xk kA~

for all nonzero ~x, so that kA~xk = k~xk. See

a b a−x b is a symmetric matrix, then A − xI2 = . This matrix fails to be b c b c−x invertible if (and only if) det(A − xI2 ) = (a − x)(c − x) − b2 = 0. We use the quadratic formula to find the (real)

Ch 5.TF.46 T. If A =

264

True or False

solutions x = or zero.

a+c±

√

(a+c)2 −4ac+4b2 2

Ch 5.TF.47 T; one basis is:

1 0

=

1 0 , 0 1

a+c±

√

(a−c)2 +4b2 . 2

Note that the discriminant (a − c)2 + 4b2 is positive

0 −1 0 1 0 . , , 1 0 1 0 −1

Ch 5.TF.48 F; A direct computation or a geometrical argument shows that Q = reflection, not a rotation.

√1 5

1 2

2 , representing a −1

Ch 5.TF.49 F; dim(R3×3 )= 9, dim(R2×2 )= 4, so dim(ker(L))≥ 5, but the space of all 3×3 skew-symmetric matrices has dimension of 3. 0 0 0 0 0 −1 0 −1 0 A basis is 1 0 0 , 0 0 0 , 0 0 −1 . 0 1 0 1 0 0 0 0 0 Ch 5.TF.50 T; Consider an orthonormal basis ~v1 , ~v2 of V, and a unit vector ~v3 perpendicular to V, and form the 1 0 0 i h orthogonal matrix S = [~v1 ~v2 ~v3 ]. Now AS = ~v1 ~v2 ~0 = S 0 1 0 . Since S is orthogonal, we have 0 0 0 1 0 0 S T AS = S −1 AS = 0 1 0 , a diagonal matrix. 0 0 0

265

Chapter 6

Chapter 6 Section 6.1 1 2 = 6 − 6 = 0. 6.1.1 Fails to be invertible; since det 3 6 2 3 6.1.2 Invertible; since det = 10 − 12 = −2. 4 5 3 5 6.1.3 Invertible; since det = 33 − 35 = −2. 7 11 1 4 6.1.4 Fails to be invertible; since det = 8 − 8 = 0. 2 8

2 5 6.1.5 Invertible; since det 0 11 0 0

6 0 6.1.6 Invertible; since det 5 4 3 2

7 7 = 2 · 11 · 5 + 0 + 0 − 0 − 0 − 0 = 110. 5 0 0 = 6 · 4 · 1 + 0 + 0 − 0 − 0 − 0 = 24. 1

6.1.7 This matrix is clearly not invertible, so the determinant must be zero. 6.1.8 This matrix fails to be invertible, since the det(A) = 0.

0 1 6.1.9 Invertible; since det 7 8 6 5

1 6.1.10 Invertible; since det 1 1

2 3 = 0 + 3 · 6 + 2 · 7 · 5 − 7 · 4 − 2 · 8 · 6 = −36. 4 1 1 2 3 = 1 · 2 · 6 + 1 · 3 · 1 + 1 · 1 · 3 − 3 · 3 · 1 − 2 · 1 · 1 − 6 · 1 · 1 = 1. 3 6

k 6.1.11 det 3

2 6 0 when 4k 6= 6, or k 6= 32 . = 4

1 6.1.12 det k

k 6 0 when k 2 6= 4, or k 6= 2, −2. = 4

k 6.1.13 det 0 0

3 2 0

5 6 = 8k, so k 6= 0 will ensure that this matrix is invertible. 4 266

Section 6.1

0 k 1

0 0 = 0, so the matrix will never be invertible, no matter which k is chosen. 0

k 3 6

1 4 = 6k − 3. This matrix is invertible when k 6= 12 . 7

2 k 7

3 5 = 60 + 84 + 8k − 18k − 35 − 64 = 45 − 10k. So this matrix is invertible when k 6= 4.5. 8

1 k k2

1 −1 = 2k 2 − 2 = 2(k 2 − 1) = 2(k − 1)(k + 1). So k cannot be 1 or -1. 1

1 2k 7

k 5 = 30 + 21k − 18k 2 = −3(k − 2)(6k + 5). So k cannot be 2 or − 56 . 5

1 k k

4 6.1.14 det 3 2 0 6.1.15 det 2 5 1 6.1.16 det 4 6 1 6.1.17 det 1 1 0 6.1.18 det 3 9 1 6.1.19 det 1 k

k k = −k 3 + 2k 2 − k = −k(k − 1)2 . So k cannot be 0 or 1. k

1 k 1 6.1.20 det 1 k + 1 k + 2 = (k + 1)(2k + 4) + k(k + 2) + (k + 2) − (k + 1) − k(2k + 4) − (k + 2)(k + 2) = 1 k + 2 2k + 4 (k + 1)(3k + 6) − (3k 2 + 9k + 5) = 1. Thus, A will always be invertible, no matter the value of k, meaning that k can have any value.

k 6.1.21 det 1 1

1 k 1

cos k 6.1.22 det 0 sin k

1 1 = k 3 − 3k + 2 = (k − 1)2 (k + 2). So k cannot be -2 or 1. k 1 − sin k 2 0 = 2 cos2 k + 2 sin2 k = 2. So k can have any value. 0 cos k

6.1.23 det(A − λI2 ) = det

1−λ 0

2 = (1 − λ)(4 − λ) = 0 if λ is 1 or 4. 4−λ

6.1.24 det(A − λI2 ) = det

2−λ 1

0 = (2 − λ)(−λ) = 0 if λ is 2 or 0. 0−λ

4−λ 6.1.25 det(A − λI2 ) = det 4

2 = (4 − λ)(6 − λ) − 8 = (λ − 8)(λ − 2) = 0 if λ is 2 or 8. 6−λ 267

Chapter 6

6.1.26 det(A − λI2 ) = det

4−λ 2

2 = (4 − λ)(7 − λ) − 4 = (λ − 8)(λ − 3) = 0 if λ is 3 or 8. 7−λ

6.1.27 A−λI3 is a lower triangular matrix with the diagonal entries (2−λ), (3−λ) and (4−λ). Now, det(A−λI3 ) = (2 − λ)(3 − λ)(4 − λ) = 0 if λ is 2, 3 or 4. 6.1.28 A−λI3 is an upper triangular matrix with the diagonal entries (2−λ), (3−λ) and (5−λ). Now, det(A−λI3 ) = (2 − λ)(3 − λ)(5 − λ) = 0 if λ is 2, 3 or 5. 3−λ 5 6 4−λ 2 = (3 − λ)(λ − 8)(λ − 3) = 0 if λ is 3 or 8. 6.1.29 det(A − λI3 ) = det 0 0 2 7−λ

4−λ 2 0 6−λ 0 = (4 − λ)(6 − λ)(3 − λ) − 8(3 − λ) 6.1.30 det(A − λI3 ) = det 4 5 2 3−λ

= (3 − λ)(8 − λ)(2 − λ) = 0 if λ is 3, 8 or 2.

6.1.31 This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 24. 6.1.32 This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 210. 6.1.33 The determinant of this block matrix is det 6.1.5. 6.1.34 The determinant of this block matrix is det 6.1.5.

1 8

4 3

2 7

5 6

det

det

2 3 7 5

1 4 2 3

= (7 − 16) (10 − 21) = 99,by Theorem

= (24 − 15) (3 − 8) = −45, by Theorem

6.1.35 There are two patterns with a nonzero product, (a12 , a23 , a31 , a44 ) = (3, 2, 6, 4), with two inversions, and (a12 , a23 , a34 , a41 ) = (3, 2, 3, 7), with 3 inversions. Thus det A = 3 · 2 · 6 · 4 − 3 · 2 · 3 · 7 = 18. 6.1.36 There is one pattern with a nonzero product, containing all the 1’s, with six inversions. Thus det A = 1. 6.1.37 The determinant block matrix is of this 5 6 7 5 4 det det 0 1 2 = (35 − 24) (5 · 1 · 1) = 55, by Theorem 6.1.5. 6 7 0 0 1 6.1.38 The 1 det 3 2

determinant of this block matrix is 2 3 6 5 0 4 det = (2 · 4 · 2 + 3 · 3 · 1 − 1 · 4 · 1 − 2 · 3 · 2) (6 · 6 − 5 · 5) = 99, by Theorem 6.1.5. 5 6 1 2

6.1.39 There is only one pattern with a nonzero product, containing all the nonzero entries of the matrix, with eight inversions. Thus det A = 1 · 2 · 4 · 3 · 5 = 120. 268

Section 6.1 6.1.40 There is only one pattern with a nonzero product, containing all the nonzero entries of the matrix, with seven inversions. Thus det A = −3 · 2 · 4 · 1 · 5 = −120. 6.1.41 There are two patterns with a nonzero product, (a15 , a24 , a32 , a41 , a53 ) = (2, 2, 3, 2, 3), with eight inversions, and (a13 , a24 , a32 , a41 , a55 ) = (1, 2, 3, 2, 4), with five inversions. Thus det A = 2 · 2 · 3 · 2 · 3 − 1 · 2 · 3 · 2 · 4 = 24. 6.1.42 There is only one pattern with a nonzero product, (a13 , a24 , a32 , a45 , a51 ) = (2, 2, 9, 5, 3), with six inversions. Thus det A = 2 · 2 · 9 · 5 · 3 = 540. 6.1.43 For each pattern P in A, consider the corresponding pattern Popp in −A, with all the n entries being opposites. Then prod (Popp ) = (−1)n prod(P ) and sgn (Popp ) = sgn (P ), so that det(−A) = (−1)n det A. 6.1.44 For each pattern P in A, consider the corresponding pattern Pm in kA, with all the n entries being multiplied by the scalar k. Then prod (Pm ) = k n prod(P ) and sgn (Pm ) = sgn (P ), so that det(kA) = k n det A. 6.1.45 If A =

6.1.46 Let A =

a c a b = ad − cb = det(A). It turns out that det(AT ) = det(A). , then det(AT ) = det b d c d

a1 a3

a2 . If a1 a4 − a2 a3 6= 0, then A−1 = a4

By Exercise 44, det(A−1 ) =

1 det(A)

2

(a1 a4 − a2 a3 ) =

1 det(A)

1 det(A)

a4 −a3

2

−a2 . a1

· det(A) so det(A−1 ) =

1 det(A) .

6.1.47 We have det(A) = (ah − cf )k + bef + cdg − aeg − bdh. Thus matrix A is invertible for all k if (and only if) the coefficient (ah − cf ) of k is 0, while the sum bef + cdg − aeg − bdh is nonzero. A numerical example is a = c = d = f = h = g = 1 and b = e = 2, but there are infinitely many other solutions as well. 0 0 0 1 0 0 1 0 so det(A) = det(B) = det(C) = det(D) = 0 ,D = ,C = ,B = 0 0 1 0 0 0 0 1 A B hence det(A) det(D) − det(B) det(C) = 0 but det = −1. C D

6.1.48 Consider A =

6.1.49 The kernel of T consists of all vectors ~x such that the matrix [~x ~v w] ~ fails to be invertible. This is the case if ~x is a linear combination of ~v and w ~ as discussed on Pages 249 and 250. Thus ker(T ) = span(~v , w). ~ The image of T isn’t {0}, since T (~v × w) ~ 6= 0, for example. Being a subspace of R, the image must be all of R. 6.1.50 Theorem 6.1.1 tells us that det[~u ~v w] ~ = ~u · (~v × w) ~ = k~uk cos(θ)k~v × wk ~ = k~uk cos(θ)k~v k sin(α)kwk ~ = cos(θ) sin(α), where θ is the angle enclosed by vectors ~u and ~v × w, ~ and α is the angle between ~v and w. ~ Thus det[~u ~v w] ~ can be any number on the closed interval [−1, 1]. 6.1.51 Let aii be the first entry on the diagonal that fails to belong to the pattern. The pattern must contain an entry in the ith row to the right of aii , above the diagonal, and also an entry in the ith column below aii , below the diagonal.

~ ~v 6.1.52 By Definition 6.1.1, we have det ~v × w

2 w ~ = (~v × w) ~ · (~v × w) ~ = k~v × wk ~ .

269

Chapter 6 6.1.53 There is one pattern with a nonzero product, containing all the 1’s. We have n2 inversions, since each of the 1’s in the lower left block forms an inversion with each of the 1’s in the upper right block. Thus det A = n2 n (−1) = (−1) . 5

6.1.54 The pattern containing all the 1000’s has 4 inversions so it contributes (1000) = 1015 to the determinant. There are 5! − 1 = 119 other patterns with at most 3 entries being 1000, the others being ≤ 9. Thus the product 3 2 associated with each of those patterns is less than(1000) (10) = 1011 . Now det A > 1015 − 119 · 1011 > 0. 6.1.55 By Exercise 2.4.93, a square matrix admits an LU factorization if (and only if) all its principal submatrices are invertible. Now 7 4 2 7 4 A(1) = [7], A(2) = , A(3) = A = 5 3 1 , 5 3 3 1 4 with det(A(1) ) = 7, det(A(2) ) = 1, det(A(3) ) = 1. Since all principal submatrices turn out to be invertible, the matrix A does indeed admit an LU factorization. 6.1.56 There is only one pattern with a nonzero product, containing all the 1’s. The number of inversions is Pn−1 . This number is even if either n or n − 1 is divisible by 4, that (n − 1) + (n − 2) + ... + 2 + 1 = k=1 k = (n−1)n 2 is, for n = 4, 5, 8, 9, 12, 13, .... a. det M4 = det M5 = 1, b. det Mn = (−1)

n(n−1)/2

det M2 = det M3 = det M6 = det M7 = −1.

6.1.57 In a permutation matrix P, there is only one pattern with a nonzero product, containing all the 1’s. Depending on the number of inversions in that pattern, we have det P = 1 or det P = −1. 6.1.58 a If a, b, c, d are prime numbers, then ad 6= bc, since the prime factorization of a positive integer is distinct a b 6= 0: No matrix of the required form exists. unique. Thus det c d b We are looking for a noninvertible matrix A = [~u ~v w] ~ whose entries are nine distinct prime numbers. The last column vector, w, ~ must be redundant; to keep things simple, we will make w ~ = ~u + 2~v . Now we have to pick six distinct prime entries for the first two columns, ~u and ~v , suchthat the entries ~ = ~u + 2~v are prime as well. of w 7 2 11 This can be done in many different ways; one solution is A = 17 3 23 . 19 5 29 6.1.59 F

a b c d

=

a c

b · = ab + cd d

a. Yes, F is linear in both columns. To prove linearity in the second column, observe that ab + cd is a linear combination of the variables b and d, with the constant coefficients a and c. An analogous argument proves linearity in the first column. b. No, since ab + cd fails to be a linear combination of a and b. c. No, F

w ~ ~v

=F

~v

w ~

= ~v · w: ~ swapping the columns leaves F unchanged. 270

Section 6.1 6.1.60 The functions in parts a, b, and d are linear in both columns; the functions in parts a, c, and d are linear in both rows; and F (A) = − det A in part d is alternating on the columns. For example, the function F (A) = cd in part b is linear in the first column since cd is a linear combination of the entries a and c in the first column. However, F (A) = cd fails to be linear in the cd fails second row since to be a linear combination of the entries c a b b a and d in the second row. Furthermore, F =F = cd, showing that F fails to be alternating c d d c on the columns. 6.1.61 The function F (A) = bf g is linear in all three columns and in all three rows since the product bf g contains exactly one factor from each row and from each column; it is the product associated with a pattern. For example, F is linear in the second row since bf g is a scalar multiple of f and combination d, e, f of the entries thus a linear 1 0 0 0 1 0 in the second row. F fails to be alternating; for example, F 0 0 1 = 1 but F 0 0 1 = 0 after 0 1 0 1 0 0 swapping the first two columns. 6.1.62 If A =

a a c c

a a , then D(A) = D c c

−D = |{z} swap columns

a a c c

= −D(A), so that D(A) = 0.

a b a b a 0 1 1 1 0 = D + D = ab D + ad D = ad. In Step 1, we 0 d |{z} 0 0 0 d |{z} 0 0 0 1 |{z} Step 2 Step 3 Step1 b b 0 a 1 write = + and use linearity in the second column. In Step 2, we write =a etc. d 0 d 0 0 and use linearity in both columns. In Step 3, we use Exercise 62 for the first summand and the given property D (I2 ) = 1 for the second summand.

6.1.63 D

a a 0 6.1.64 Writing = + and repeatedly using linearity in the columns, we find 0 c c a b a b 0 b 0 b 0 0 D =D +D = ad + D +D c d 0 d c d |{z} c 0 c d Step 2 0 1 0 0 1 0 = ad + bc D + cd D = ad − bc D = ad − bc = det A 1 0 1 1 |{z} 0 1 Step 4 a b = ad from Exercise See the analogous computations in Exercise 63. In Step 2, we are using the result D 0 d 63. In Step 4, we swap the columns of the matrix in the second summand and we apply Exercise 62 to the third summand. 6.1.65 Freely using the linearity and alternating properties in the columns, and omitting terms with two equal columns (see Exercise 62), we find 0 b c 0 b c 1 b c a b c D d e f = a D 0 e f +d D 1 e f +g D 0 e f 0 h i 0 h g h i 1 h i i 0 0 c 0 1 c 1 0 c 1 0 c = ae D 0 1 f + ah D 0 0 f + db D 1 0 f + dh D 1 0 f 0 1 i 0 0 i 0 1 i 0 0 i 271

Chapter 6

0 +gb D 0 1

1 0 0 0 1 0 + ahf D 0 0 0 1 0 1

1 0 0 c 1 c 0 f + ge D 0 1 f = aei D 0 0 1 0 i 0 i

0 1 0 dbi D 1 0 0 + dhc D 0 0 1 = aei − ahf − dbi + dhc + gbf

0 1 0

0 1 1 0 0 0

0 0 1 0 0 0 1 1 0 0 + gbf D 0 0 1 + gec D 0 1 1 0 0 0 1 0 − gec = det A

In the penultimate step, we perform up to two column swaps on each of the six matrices to make them I3 , recalling that D (I3 ) = 1.

a c

b d

1 1 0 0

1 0 0 1

+ + ad F = ab F 6.1.66 a. Proceeding as in Exercises 63 and 64, we can see that F 0 1 0 0 cb F + cd F for all functions F in V. Thus the functions ab, ad, cb, and cd span V . Apply 1 0 1 1 1 0 1 1 , , these functions to the matrices 0 1 0 0 0 1 0 0 , and to see that the four functions are linearly independent. Thus the functions ab, ad, cb, 1 0 1 1 and cd form a basis of V , and dim V = 4. a b 1 0 1 0 b. In Exercise 64 we see that the formula F =F (ad − bc) = F (det A) holds for all c d 0 1 0 1 functions F in W. Thus W is one-dimensional, with det A as a basis.

Section 6.2

1 1 1 1 6.2.1 A = 1 3 3 −I → B = 0 2 2 5 −2I 0

1 1 2 2 . Now det(A) = det(B) = 6, by Algorithm 6.2.5b. 0 3

1 2 3 1 6.2.2 A = 1 6 8 −I → B = 0 −2 −4 0 +2I 0 1 1 6.2.3 A = 1 2

3 6 3 6

2 4 0 4

4 1 8 −I 0 →B= 0 −I 0 12 −2I 0

1 −1 2 1 −1 2 6.2.4 A = 2 1 14 −2 6 10

1 0 0 0

−1 2 1 3 3 10 4 14

2 3 4 5 . Now det(A) = det(B) = 24, by Algorithm 6.2.5b. 0 6 3 2 4 3 2 4 . Now det(A) = det(B) = −24, by Algorithm 6.2.5b. 0 −2 −4 0 0 4

−2 6 +I → 10 −2I 33 +2I

−2 4 → 14 −3II 29 −4II 272

Section 6.2 1 0 0 0

−1 1 0 0

2 3 1 2

1 −2 4 0 →B= 0 2 0 13 −2III

−1 1 0 0

2 3 1 0

1 0 6.2.5 After three row swaps, we end up with B = 0 0 −24.

1 1 1 1 1 1 4 4 −I 6.2.6 A = → 1 −1 2 −2 −I 1 −1 8 −8 −I 1 1 1 1 0 0 3 3 swap: → 0 −2 1 −3 II ↔ III 0 −2 7 −9 1 1 1 1 0 −2 1 −3 ÷ − 2 → 0 0 3 3 0 −2 7 −9 1 1 1 1 1 3 0 1 −2 2 → 0 0 3 3 0 −2 7 −9 +2II 1 1 1 1 1 1 1 3 0 1 0 1 −2 2 → B= 0 0 3 3 0 0 0 0 6 −6 −2III 0 0

−2 4 . Now det(A) = det(B) = 9, by Algorithm 6.2.5b. 2 9 2 2 0 0

3 3 3 0

4 4 . Now, by Algorithm 6.2.5b, det(A) = (−1)3 det(B) = 4 4

1 − 21 3 0

1 3 2

3 −12

det(A) = − 21 (−1) det(B) = −72, by Algorithm 6.2.5b.

6.2.7 After two row swaps, we end up with an upper triangular matrix B with all 1’s along the diagonal. Now det(A) = (−1)2 det(B) = 1, by Algorithm 6.2.5b. 6.2.8 After four row swaps, we end up with an upper triangular matrix B with all 1’s along the diagonal, except for a 2 in the bottom right corner. Now det(A) = (−1)4 det(B) = 2, by Algorithm 6.2.5b. 6.2.9 If we subtract the first row from every other row, then we have an upper triangular matrix B, with diagonal entries 1, 1, 2, 3 and 4. Then det(A) = det(B) = 24 by Algorithm 6.2.5b. 1 1 6.2.10 A = 1 1 1

1 2 3 4 5

1 3 6 10 15

1 4 10 20 35

1 5 15 35 70

−I −I → −I −I 273

Chapter 6

1 0 0 0 0

1 1 1 2 2 5 3 9 4 14

1 0 0 0 0

1 1 0 0 0

1 2 1 3 6

1 3 3 10 22

1 0 0 0 0

1 1 0 0 0

1 2 1 0 0

1 1 1 3 4 0 3 6 → B = 0 1 4 0 4 17 −4IV 0

1 3 9 19 34

1 4 14 −2II → 34 −3II 69 −4II 1 4 → 6 22 −3III 53 −6III 1 1 0 0 0

1 2 1 0 0

1 3 3 1 0

1 4 6 . 4 1

Now det(A) = det(B) = 1 by Algorithm 6.2.5b. 6.2.11 By Theorem 6.2.3a, the desired determinant is (−9)(8) = −72. 6.2.12 By Theorem 6.2.3b, the desired determinant is −8. 6.2.13 By Theorem 6.2.3b, applied twice, since there are two row swaps, the desired determinant is (−1)(−1)(8) = 8. 6.2.14 By Theorem 6.2.3c, the desired determinant is 8. 6.2.15 By Theorem 6.2.3c, the desired determinant is 8. 6.2.16 This determinant is 0, since the first row is twice the last.

3 0 2 6 , so that det(T ) = det(A) = 8. 0 2

−2 4 3 −12 , so that det(T ) = det(A) = 27. 0 9

1 6.2.19 The standard matrix of T is A = 0 0

0 0 −1 0 , so that det(T ) = det(A) = −1. 0 1

1 0 6.2.20 The standard matrix of L is M = 0 0

0 0 1 0

2 6.2.17 The standard matrix of T is A = 0 0 1 6.2.18 The standard matrix of T is A = 0 0

0 1 0 0

0 0 , so that det(L) = det(M ) = −1. 0 1 274

Section 6.2 1 0 6.2.21 The standard matrix of T is A = 0 0

0 −1 0 0

0 0 0 0 , so that det(T ) = det(A) = 1. 1 0 0 −1

6.2.22 Using Exercises 19 and 21 as a guide, we observe that the standard matrix A of T is diagonal, of size (n + 1) × (n + 1), with diagonal entries (−1)0 , (−1)1 , (−1)2 , . . . , (−1)n . Thus det(T ) = det(A) = (−1)1+2+···+n = (−1)n(n+1)/2 . 6.2.23 Consider the matrix M of T with respect to a basis consisting of n(n+1)/2 symmetric matrices and n(n−1)/2 skew-symmetric matrices (see Exercises 54 and 55 or Section 5.3). Matrix M will be diagonal, with n(n + 1)/2 entries 1 and n(n − 1)/2 entries -1 on the diagonal. Thus, det(T ) = det(M ) = (−1)n(n−1)/2 . 6.2.24 The standard matrix of T is A =

2 3

2 6.2.25 The standard matrix of T is A = 0 0

−3 , so that det(T ) = det(A) = 13. 2 0 0 2 3 , so that det(T ) = det(A) = 16. 0 4

6.2.26 The matrix of T with respect to the basis

1 0 0 , 0 0 1

1 0 , 0 0

det(A) = −16.

0 1

2 4 is A = 2 4 0 4

0 2 , so that det(T ) = 6

−b a 6.2.27 The matrix of T with respect to the basis cos(x), sin(x) is A = , so that det(T ) = det(A) = a2 +b2 . −a −b −3 −2 −6 −10 , so that det(T ) = det(A) = 14. 0 is A = 1 , 6.2.28 The matrix of T with respect to the basis 5 6 1 0

6.2.29 Expand down the first column, realizing that all but the first contribution are zero, since a21 = 0 and Ai1 has two equal rows for all i > 2. Therefore, det(Pn ) = det(Pn−1 ). Since det(P1 ) = 1, we can conclude that det(Pn ) = 1, for all n. 1 1 1 6.2.30 a f (t) = det a b t = (ab2 − a2 b) + (a2 − b2 )t + (b − a)t2 so f (t) is a quadratic function of t. The a2 b2 t2 2 coefficient of t is (b − a).

b In the cases t = a and t = b the matrix has two identical columns. It follows that f (t) = k(t − a)(t − b) with k = coefficient of t2 = (b − a). c The matrix is invertible for the values of t for which f (t) 6= 0, i.e., for t 6= a, t 6= b.

1 6.2.31 a If n = 1, then A = a0

1 , so det(A) = a1 − a0 (and the product formula holds). a1 275

Chapter 6 b Expanding the given determinant down the right-most column, we see that the coefficient k of tn is the n − 1 Vandermonde determinant which we assume is Y (ai − aj ). n−1≥i>j

Now f (a0 ) = f (a1 ) = · · · = f (an−1 ) = 0, since in each case the given matrix has two identical columns, hence its determinant equals zero. Therefore Y f (t) = (ai − aj ) (t − a0 )(t − a1 ) · · · (t − an−1 ) n−1≥i>j

and

det(A) = f (an ) =

Y

n≥i>j

(ai − aj ),

as required. 6.2.32 By Exercise 31, we need to compute

Q

i>j

Q

i>j

(ai − aj ) where a0 = 1, a1 = 2, a2 = 3, a3 = 4, a4 = 5 so

(ai − aj ) = (2 − 1)(3 − 1)(3 − 2)(4 − 1)(4 − 2)(4 − 3)(5 − 1)(5 − 2)(5 − 3)(5 − 4) = 288.

th

6.2.33 Think of the i

1 ai 2 a column of the given matrix as ai .i . .

1 a1 2 a be written as (a1 a2 · · · an ) det .1 . .

a1n−1 (see Exercise 6.2.31), and we get

1 a2 a22 .. .

··· ··· ···

1 an a2n .. .

a2n−1

···

ann−1

n Y

i=1

ai

Y

i>j

ain−1

, so, by Theorem 6.2.3a, the determinant can

. The new determinant is a Vandermonde determinant (ai − aj ).

6.2.34 a The hint pretty much gives it away. Since the columns of matrix B have [ In M ] = B − M = 0, and M = B, as claimed. −In

B −In

are in the kernel of [ In

. . b If B = A−1 we get rref[A..In ] = [In ..A−1 ] which tells us how to compute A−1 (see Theorem 2.4.5).

6.2.35

x1 x2

1 must satisfy det x1 x2

1 a1 a2

1 b1 = 0, i.e., must satisfy the linear equation b2 276

M ], we

Section 6.2 (a1 b2 − a2 b1 ) − x1 (b2 − a2 ) + x2 (b1 − a1 ) = 0. b x1 a1 x1 = 1 satisfy this equation, since the matrix has two identical columns and = We can see that b2 x2 a2 x2 in these cases. 6.2.36 Expanding down the first column we see that the equation has the form A − Bx1 + Cx2 − D(x21 + x22 ) = 0. If D 6= 0 this equation defines it is a line. From Exercise 35 acircle; otherwise a1 b1 c1 we know that D = 0 if and only if the three given points , , are collinear. Note that the circle a2 b2 c2 or line runs through the three given points. 6.2.37 Applying Theorem 6.2.6 to the equation AA−1 = In we see that det(A) det(A−1 ) = 1. The only way the product of the two integers det(A) and det(A−1 ) can be 1 is that they are both 1 or both -1. Therefore, det(A) = 1 or det(A) = −1. 6.2.38 det(AT A) = det(AT ) det(A) = [det(A)]2 = 9 ↑

↑

Theorem 6.2.6

Theorem 6.2.1 2

6.2.39 det(AT A) = det(AT ) det(A) = [det(A)] > 0 ↑

↑

Theorem 6.2.6

Theorem 6.2.1

6.2.40 By Exercise 38, det(AT A) = [det(A)]2 . Since A is orthogonal, AT A = In so that 1 = det(In ) = det (AT A) = [det(A)]2 and det(A) = ±1. 6.2.41 det(A) = det(AT ) = det(−A) = (−1)n (det A) = − det(A), so that det(A) = 0. We have used Theorems 6.2.1 and 6.2.3a. T 6.2.42 det AT A |{z} = det (QR) QR = det RT QT QR |{z} = det RT Im R Step 1 Step 3 Qm 2 2 T T = det R R |{z} = det R det(R) |{z} = (det R) |{z} = ( i=1 rii ) > 0. Step 5

Step 6

Step 7

In Step 1 we are using the definition of matrix A. Equation 3 holds since the column vectors of matrix Q are orthonormal. In Steps 5 and 6 we are using Theorems 6.2.6 and 6.2.1, respectively. Finally, Equation 7 holds since matrix R is triangular. T

6.2.43 det(A A) = det

~v T w ~T

~v · ~v [~v w] ~ = det ~v · w ~

~v · w ~ k~v k2 = det w ~ ·w ~ ~v · w ~

~v · w ~ kwk ~ 2

= k~v k2 kwk ~ 2 − (~v · w) ~ 2 ≥ 0 by the Cauchy-Schwarz inequality (Theorem 5.1.11). 6.2.44 a We claim that ~v2 ×~v3 ×· · ·×~vn 6= ~0 if and only if the vectors ~v2 , . . . , ~vn are linearly independent. If the vectors ~v2 , . . . , ~vn are linearly independent, then we can find a basis ~x, ~v2 , . . . , ~vn of Rn (any vector ~x that is not in span (~v2 , . . . , ~vn ) will do). Then ~x·(~v2 ×· · ·×~vn ) = det[~x ~v2 · · · ~vn ] 6= 0, so that ~v2 ×· · ·×~vn 6= ~0. Conversely, suppose that 277

Chapter 6 ~v2 ×~v3 ×· · ·×~vn 6= 0; say the ith component of this vector is nonzero. Then 0 6= ~ei ·(~v2 ×· · ·×~vn ) = det[~ei ~v2 · · · ~vn ], so that the vectors ~v2 , . . . , ~vn are linearly independent (being columns of an invertible matrix).

1 b ith component of ~e2 × ~e3 × · · · × ~en = det ~ei 1

1 ~e2 1

so ~e2 × ~e3 × · · · × ~en = ~e1 .

1 c ~vi · (~v2 × ~v3 × · · · × ~vn ) = det ~vi 1

1 ~v2 1

1 ~v3 1

1 n 1 · · · ~en = 0 1

if i = 1 if i > 1

1 · · · ~vn = 0 1

for any 2 ≤ i ≤ n since the above matrix has two identical columns.

d Compare the 1 1 det ~ei ~v2 1 1

ith components of the two vectors: 1 1 1 1 1 ~v3 · · · ~vn and det ~ei ~v3 ~v2 1 1 1 1 1

1 · · · ~vn · 1

The two determinants differ by a factor of −1 by Theorem 6.2.3b, so that ~v2 × ~v3 × · · · × ~vn = −~v3 × ~v2 × · · · × ~vn .

e det[~v2 × ~v3 × · · · × ~vn ~v2 ~v3 · · · ~vn ] = (~v2 × ~v3 × · · · × ~vn ) · (~v2 × ~v3 × · · · × ~vn ) = k~v2 × · · · × ~vn k2 f In Definition 6.1.1 we saw that the “old” cross product satisfies the defining equation of the “new” cross product: ~x · (~v2 × ~v3 ) = det [ ~x ~v2 ~v3 ]. 6.2.45 f (x) is a linear function, so f ′ (x) 1 0 that the coefficient of x is − det 0 0

a 3 6.2.46 a det b 3 c 3

is 2 2 0 0

the 3 3 3 0

coefficient of x (the slope). Expanding down the first column, we see 4 4 = −24, so f ′ (x) = −24. 4 4

a 1 d d e = 3 det b 1 e = 21. We have used Theorem 6.2.3. c 1 f f

a 1 a 2(1) d a 2(1) + 1 d a 3 d b det b 5 e = det b 2(2) + 1 e = det b 2(2) e +det b 1 c 1 c 2(3) f c 2(3) + 1 f c 7 f 2 · 11 + 7 = 29. We have used Theorem 6.2.2 and Theorem 6.2.3a.

6.2.47 Yes! For example, T

a 1 a 1 d d e = 2 det b 2 e +det b 1 c 1 c 3 f f

x b = dx + by is given by the matrix [d b], so that T is linear in the first column. y d

6.2.48 Since ~v2 , . . . , ~vn are linearly independent, T (~x) = 0 only if ~x is a linear combination of the ~vi ‘s, (otherwise the matrix [~x ~v2 · · · ~vn ] is invertible, and T (~x) 6= 0). Hence, the kernel of T is the span of ~v2 , . . . , ~vn , an (n − 1)dimensional subspace of Rn . The image of T is the real line R (since it must be 1-dimensional). 278

d e= f

Section 6.2 1 1 1 6.2.49 For example, we can start with an upper triangular matrix B with det(B) = 13, such as B = 0 1 1 . 0 0 13 Adding the first row of B to both the second and the third to make all entries nonzero, we end up with 1 1 1 A = 1 2 2 . Note that det(A) = det(B) = 13. 1 1 14

6.2.50 There are many ways to do this problem; here is one possible approach: Subtracting the second to last row from the last, we can make the last row into [0

0

···

0

1 ].

Now expanding along the last row we see that det(Mn ) = det(Mn−1 ). Since det(M1 ) = 1 we can conclude that det(Mn ) = 1 for all n. 6.2.51 Notice that it takes n row swaps (swap row i with n + i for each i between 1 and n) to turn A into I2n . So, det(A) = (−1)n det(I2n ) = (−1)n . 6.2.52 a We build B column-by-column: d −b −b d 0 1 . = = T T −c a a −c 1 0 b det(A) = ad − bc = det(B). The two determinants are equal.

d −b a b da − bc 0 c BA = = = (ad − bc)I2 . −c a c d 0 −cb + ad ad − bc 0 d −b a b = (ad − bc)I2 also. = AB = 0 −cb + da −c a c d a d −b a b a = = . We note that B + c2 d Any vector ~u in the image of A will be of the form c1 c −c a c d c ad − bc b = ~0. The same is true of B . Thus, anything in the image of A will be in the kernel of B. Since −ca + ac d both matrices have a rank of 1, the dimensions of the kernel and image of each will be exactly 1. So, it must be that im(A) = ker(B). −b a b −b d = . However, we see that +c2 Also, any vector ~u in the image of B will be of the form c1 a c d a −c −ab + ba d = ~0. The same is true for A . Thus, by the same reasoning as above, the image of B will equal −bc + ad −c the kernel of A. e A−1 =

1 ad−bc

d −b = −c a

1 ad−bc B.

6.2.53 a See Exercise 37. 279

Chapter 6

b If A =

a b , then A−1 = c d

1 det(A)

6.2.54 f (t) = (det(A + tB))2 − 1 =

d −b −c a

has integer entries.

a + tb1 det 1 a3 + tb3

a2 + tb2 a4 + tb4

2

− 1 assuming A =

a1 a3

b a2 ,B= 1 b3 a4

b2 . b4

Then the determinant above is a polynomial of degree ≤2 so its square is a polynomial of degree ≤4. Hence f (t) is a polynomial of degree ≤4. Since A, A + B, A + 2B, A + 3B, A + 4B are invertible and their inverses have integer entries, by Exercise 53a, it follows that their determinants are ±1. Hence f (0) = f (1) = f (2) = f (3) = f (4) = 0. Since f is a polynomial of degree ≤ 4 with at least 5 roots, it follows that f (t) = 0 for all t, in particular for t = 5, so det(A + 5B) = ±1. Hence A + 5B is an invertible 2 × 2 matrix whose inverse has integer entries by Exercise 53b. 6.2.55 We start out with a preliminary remark: If a square matrix A has two equal rows, then D(A) = 0. Indeed, if we swap the two equal rows and call the resulting matrix B, then B = A, so that D(A) = D(B) = −D(A), by property b, and D(A) = 0 as claimed. Next we need to understand how the elementary row operations affect D. Properties a and b tell us about how row multiplications and row swaps, but we still need to think about row additions. We will show that if B is obtained from A by adding k times the ith row to the j th , then D(B) = D(A). Let’s label the row vectors of A by ~v1 , . . . ~vn . By linearity of D in the j th row (property c) we have .. .. . . ~ v ~ v i i . . .. .. D(B) = D = D(A) + kD = D(A). ~ v + k~ v ~ v j i i .. .. . .

Note that in the last step we have used the preliminary remark. Now, using the terminology introduced on Page 265, we can write D(A) = (−1)s k1 k2 · · · kr D(rref A).

Next we observe that D(rref A) = det(rref A) for all square matrices A. Indeed, if A is invertible, then rref(A) = In , and D(In ) = 1 = det(In ) by property c of function D. If A fails to be invertible, then D(rref A) = 0 = det(rref A) by linearity in the last row. It follows that D(A) = (−1)s k1 k2 · · · kr D(rref A) = (−1)s k1 k2 · · · kr det(rref A) = det(A) for all square matrices, as claimed. 6.2.56 a We show first that D is linear in the i th row. ~v1 M ~v1 . . . . . . ~xM ~x . . 1 D .. = det M det .. ~vn M ~vn ↑ ↑ AM A 280

Section 6.2 The entries in the ith row of AM are linear combinations of the components xi of the vector ~x, while the other entries of AM are constants. Therefore, ) is a linear combination of the xi (expand along the ith row). det(AM ~v1 .. . Since det1M is a constant, we have D ~x = c1 x1 + c2 x2 + · · · + cn xn for some constants ci , as claimed. . .. ~vn c The property D(In ) = 1 is obvious. det(AM ) det(M )

It now follows from Exercise 41 that det(A) = D(A) = det(AM ) = det(A) det(M ).

and therefore

6.2.57 Note that matrix A1 is invertible, since det(A1 ) 6= 0. Now ~y ~y T = [A1 A2 ] = A1 ~y + A2 ~x = ~0 when A1 ~y = −A2 ~x, or, ~x ~x ~y = −A−1 x. This shows that for every ~x there is a unique ~y (that is, ~y is a function of ~x); furthermore, this 1 A2 ~ function is linear, with matrix M = −A−1 1 A2 .

1 2 1 2 6.2.58 Using the approach of Exercise 57, we have A1 = , A2 = , 3 7 4 3 1 −8 y1 1 −8 x1 and M = −A−1 A = . The function is = . 2 1 y2 −1 3 x2 −1 3 Alternatively, we can solve the linear system y1 + 2y2 + x1 + 2x2 3y1 + 7y2 + 4x1 + 3x2

=0 =0

Gaussian Elimination gives y1 − x1 + 8x2 = 0

and

y2 + x1 − 3x2 = 0

0 6.2.59 det 0

y1 = x1 − 8x2 y2 = −x1 + 3x2

0 1 0 = 0 = det , but these matrices fail to be similar. 0 0 0

6.2.60 We argue using induction on n. The base case (n = 2) is discussed in the text. Now we assume that B is obtained from the n × n matrix A by adding k times the pth row to the q th row. We will evaluate the determinant of B by expanding across the ith row (where i is neither p nor q). det(B) = Σnj=1 (−1)i+j bij det(Bij ) = Σnj=1 (−1)i+j aij det(Bij ) = Σnj=1 (−1)i+j aij det(Aij ) = det(A) 281

Chapter 6 Note that the (n − 1) × (n − 1) matrix Bij is obtained from Aij by adding k times some row to another row. Now, det(Bij ) = det(Aij ) by induction hypothesis. A B A B In 0 = 6.2.61 We follow the hint: −CA + AC −CB + AD C D −C A A B In 0 A B = . So, det = det(A) det(AD − CB). 0 AD − CB −C A C D A B = det(A) det(AD − CB), which leads to Thus, det(In ) det(A) det C D A B det = det(AD − CB), since det(A) 6= 0. C D

A B A B In 0 6.2.62 a We compute = . Since the matrix is invert−1 C D 0 −CA−1 B −CA +D In A B A B ible (its determinant is 1),the product will have the same rank as , namely, n. 0 −CA−1 B + D C D With A being invertible, this implies that −CA−1 B + D = 0, or CA−1 B = D, as claimed. In −CA−1

0 In

b Take determinants on both sides of the equation D = CA−1 B from part (a) to find that det(D) = det(C)(det A)−1 det(B), or det(A) det(D) − det(B) det(C) = 0, proving the claim. 6.2.63 Let Mn be the number of multiplications required to compute the determinant of an n × n matrix by Laplace expansion. We will use induction on n to prove that Mn > n!, for n ≥ 3. In the lowest applicable case, n = 3, we can check that M3 = 9 and 3! = 6. Now let’s do the induction step. If A is an n × n matrix, then det(A) = a11 det(A11 ) + · · · + (−1)n+1 an1 det(An1 ), by Theorem 6.2.10. We need to compute n determinants of (n − 1) × (n − 1) matrices, and then do n extra multiplications ai1 det(Ai1 ), so that Mn = nMn−1 + n. If n > 3, then Mn−1 > (n − 1)!, by induction hypothesis, so that Mn > n(n − 1)! + n > n!, as claimed. 6.2.64 To compute det(A) for an n × n matrix A by Laplace expansion, det(A) = a11 det(A11 ) − a21 det(A21 ) + · · · + (−1)n+1 an1 det(An1 ), we first need to compute the n minors, which requires nLn−1 operations; then we compute the n products ai1 det(Ai1 ); and finally we have to do n − 1 additions. Altogether, Ln = nLn−1 + n + (n − 1) = nLn−1 + 2n − 1. Now we can prove the formula

Ln n!

=1+1+

1 2!

+

1 3!

+ ··· +

1 (n−1)!

−

1 n!

by induction on n.

For n = 2, the formula gives L22 = 1 + 1 − 21 , or L2 = 3, which is correct: We have to perform 2 multiplications and 1 addition to compute the determinant of a 2 × 2 matrix. nLn−1 +2n−1 Ln−1 Ln = (n−1)! n! = n! Ln−1 2 1 Applying the induction hypothesis to the first summand, we find that Ln!n = (n−1)! + (n−1)! − n! 1 1 1 2 1 1 1 1 1 3! + · · · + (n−2)! − (n−1)! + (n−1)! − n! = 1 + 1 + 2! + 3! + · · · + (n−1)! − n! , as claimed.

For an n × n matrix A we can use the recursive formula derived above to see that 1 − n! . 1 + 2! +

2 (n−1)!

1+1

Now recall from the theory of Taylor series in calculus that e = e1 = 1 + 1 + 1 1 1 1 Thus Ln = (1 + 1 + 2! + 3! + · · · + (n−1)! − n! )n! < en!, as claimed. 282

1 2!

+

1 3!

+ ··· +

1 (n−1)!

+

1 n!

+ =

+ · · ·.

Section 6.3

6.2.65 a. Using Laplace expansion along the first row, we find dn = det (Mn ) = det (Mn−1 ) − det det (Mn−1 ) + det (Mn−2 ) = dn−1 + dn−2 . b. d1 = 1, d2 = 2, d3 = 3, d4 = 5, ..., d10 = 89 c. Since dn > 0 for all positive integers n, the matrix Mn is always invertible.

6.2.66 a. Using Laplace expansion along the first row, we find dn = det (Mn ) = det (Mn−1 ) − det

−1 ∗ 0 Mn−2

=

=

1 ∗ 0 Mn−2

det (Mn−1 ) − det (Mn−2 ) = dn−1 − dn−2 . b. d1 = 1, d2 = 0, d3 = −1, d4 = −1, d5 = 0, d6 = 1, d7 = 1, d8 = 0. c. Since d4 = −d1 and d5 = −d2 , the formula dn+3 = −dn holds for all positive integers n. (Give a formal proof by induction.) Now dn+6 = −dn+3 = − (−dn ) = dn for all positive integers n, meaning that the sequence dn has a period of six. 6.2.67 Let k be the number of pattern entries to the left and above aij . Then the number of pattern entries to the right and above aij is i − 1 − k, since there are i − 1 rows above aij , each of which contains exactly one pattern entry. Likewise, the number of pattern entries to the left and below aij is j − 1 − k. Now the number #(inversions involving aij ) of inversions involving aij is (i − 1 − k) + (j − 1 − k) = i + j − 2 − 2k. It follows that (−1) i+j−2−2k i+j = (−1) = (−1) . This means that the number of inversions involving aij is even if (and only if) i + j is even, as claimed. 6.2.68 Using Exercise 67 and the terminology introduced in the proof of Theorem 6.2.10, we have #(inversions in P ) #(inversions involving aij ) #(inversions not involving aij ) sgnP = (−1) = (−1) (−1) = (−1)

i+j

(−1)

#(inversions in Pij )

= (−1)

i+j

sgn (Pij ) .

Section 6.3 6.3.1 By Theorem 2.4.10, the area equals | det 6.3.2 By Theorem 2.4.10, Area =

6.3.3 Area of triangle = 21 | det

1 2

6 −2

3 8 | = | − 50| = 50. 7 2

3 8 det = 12 |−50| = 25 7 2 1 | = 13 (See Figure 6.1.) 4

b1 − a1 c1 − a1 6.3.4 Note that area of triangle = det . (See Figure 6.2.) b2 − a2 c2 − a2 a1 b1 − a1 c1 − a1 a1 b1 c1 On the other hand, det a2 b2 c2 = det a2 b2 − a2 c2 − a2 , by subtracting the first column from the 1 0 0 1 1 1 second and third. b1 − a1 c1 − a1 , by expanding across the bottom row. This, in turn, equals det b2 − a2 c2 − a2 a1 b1 c1 Therefore, area of triangle = 12 det a2 b2 c2 . 1 1 1 1 2

283

Chapter 6

Figure 6.1: for Problem 6.3.3.

Figure 6.2: for Problem 6.3.4. 6.3.5 The volume of the tetrahedron T0 defined by ~e1 , ~e2 , ~e3 is 13 (base)(height) = 16 . Here we are using the formula for the volume of a pyramid. (See Figure 6.3.)

Figure 6.3: for Problem 6.3.5. The tetrahedron T defined by ~v1 , ~v2 , ~v3 can be obtained by applying the linear transformation with matrix [~v1 ~v2 ~v3 ] to T0 . Now we have vol(T ) = | det[~v1 ~v2 ~v3 ]|vol(T0 ) = 16 | det[~v1 ~v2 ~v3 ]| = 16 V (~v1 , ~v2 , ~v3 ). ↑

↑

Theorem 6.3.7 and Page 283

6.3.6 From Exercise 5 we know that

area of triangle =

1 2

a1 det a2 1

b1 b2 1

Theorem 6.3.4 and Page 280

a1 b1 c1 volume of tetrahedron = 61 det a2 b2 c2 , and Exercise 4 tells us that 1 1 1 c1 c2 , so that area of tetrahedron = 31 (area of triangle). 1

284

Section 6.3 We can see this result more directly if we think of the tetrahedron as an inverted pyramid whose base is the triangle and whose height is 1. (See Figure 6.4.)

Figure 6.4: for Problem 6.3.6. c1 b1 a1 The three vertices of the shaded triangle are a2 , b2 , c2 . 1 1 1

6.3.7 Area =

1 2

10 det 11

−2 13

+

1 2

8 det 2

10 11

= 110. (See Figure 6.5.)

Figure 6.5: for Problem 6.3.7. 6.3.8 We need to show that both sides of the equation in Theorem 6.3.3 give zero. | det(A)| = 0 since A is not invertible. On the other hand, since A is not invertible, the ~vi will be linearly k dependent, i.e., one of the ~vi will be redundant. This implies that ~vi = ~vi and ~vi⊥ = ~0, so that the right-hand side of the equation is 0, as claimed. 6.3.9 Using linearity in the second column, we find that det det

~v1

~v2⊥ . Thus the two determinants are equal.

285

~v1

~v2

= det

h

~v1

i i h k ~v2 + ~v2⊥ = det ~v1 ~v2k + det ~v1 {z } | noninvertible

~v

Chapter 6 6.3.10 | det(A)| ≤ k~v1 kk~v2 k · · · k~vn k since | det(A)| = k~v1 kk~v2⊥ k · · · k~vn⊥ k and k~vi k ≥ k~vi⊥ k. The equality holds if k~vi k = k~vi⊥ k for all i, that is, if the ~vi ‘s are mutually perpendicular. 6.3.11 The matrix of the transformation T with respect to the basis ~v1 , ~v2 is B = det(B) = 12, by Theorem 6.2.7.

3 0 , so that det(A) = 0 4

6.3.12 Denote the columns by ~v1 , ~v2 , ~v3 , ~v4 . By Theorem 6.3.3 and Exercise 6.3.8 we know that | det(A)| ≤ k~v1 kk~v2 k k~v3√ kk~v4 k; equality holds if the columns are orthogonal. Since the entries of the ~vi are 0, 1, and −1, we have k~vi k ≤ 1 + 1 + 1 + 1 = 2. Therefore, | det A| ≤ 16. To build an example where det(A) = 16 we want all 1’s and −1’s as entries, and the columns need to be 1 1 1 1 1 −1 −1 1 orthogonal. A little experimentation produces A = (there are other solutions). Note that 1 −1 −1 1 1 −1 1 −1 we need to check that det(A) = 16 (and not −16). 6.3.13 By Theorem 6.3.6, the v u u 1 u 1 1 1 1 1 udet t 1 2 3 4 1 1

desired 2-volume is 1 s 2 4 = det 3 10 4

√ 10 = 20. 30

6.3.14 By Theorem 6.3.6, the desired 3-volume is v u 1 1 1 v u u 1 1 1 1 0 0 0 u u √ 0 1 2 u udet 4 10 = 6. 1 1 1 1 = tdet 1 t 0 1 3 1 10 30 1 2 3 4 0 1 4 6.3.15 If ~v1 , ~v2 , . . . , ~vm are linearly dependent and if A = [~v1 · · · ~vm ], then det(AT A) = 0 since AT A and A have equal and nonzero kernels (by Theorem 5.4.2), hence AT A fails to be invertible. On the other hand, since the ~vi are linearly dependent, at least one of them will be redundant. For such a k redundant ~vi , we will have ~vi = p ~vi and ~vi⊥ = ~0, so that V (~v1 , . . . , ~vm ) = 0, by Definition 6.3.5. This discussion shows that V (~v1 , . . . , ~vm ) = 0 = det(AT A) if the vectors ~v1 , . . . , ~vm are linearly dependent.

6.3.16 False

If T is given by A = 2I3 then | det(A)| = 8. But if Ω is the square defined by ~e1 , ~e2 in R3 (of area 1), then T (Ω) is the square defined by 2~e1 , 2~e2 and the area of T (Ω) is 4. 6.3.17 a Let w ~ = ~v1 ×~v2 ×~v3 . Note that w ~ is orthogonal to ~v1 , ~v2 and ~v3 , by Exercise 6.2.44c. Then V (~v1 , ~v2 , ~v3 , w) ~ = V (~v1 , ~v2 , ~v3 )kw ~ ⊥ k = V (~v1 , ~v2 , ~v3 )kwk. ~ ↑ by Definition 6.3.5. 286

Section 6.3 b By Exercise 6.2.44e, V (~v1 , ~v2 , ~v3 , ~v1 × ~v2 × ~v3 )

= | det [ ~v1 ~v2 ~v3 ~v1 × ~v2 × ~v3 ] | . = | det [ ~v1 × ~v2 × ~v3 ~v1 ~v2 ~v3 ] | = k~v1 × ~v2 × ~v3 k2

c By parts a and b, V (~v1 , ~v2 , ~v3 ) = k~v1 × ~v2 × ~v3 k. If the vectors ~v1 , ~v2 , ~v3 are linearly dependent, then both sides of the equation are 0, by Exercise 15 and Exercise 6.2.44a.

Figure 6.6: for Problem 6.3.18a. p 0 cos(t) p · cos(t) p 0 6.3.18 a (See Figure 6.6.) = , the ellipse with semi-axis ± and ± . 0 q sin(t) q · sin(t) 0 q (area of the ellipse) = | det(A)|(area of the unit circle) = pqπ b By Theorem 6.3.7, | det(A)| =

area of the ellipse = area of the unit circle

abπ π

= ab so | det(A)| = ab.

c The unit circle consists of all vectors of the form ~x = cos(t) √12 consisting of all vectors √ 1 √ 1 T (~x) = cos(t) 2 2 + sin(t) 2 1 −1 | {z } | {z } . (See Figure 6.7.) semi-major axis semi-minor axis

1 1 + sin(t) √12 ; its image is the ellipse 1 −1

6.3.19 det[~v1 ~v2 ~v3 ] = ~v1 ·(~v2 ×~v3 ) = k~v1 kk~v2 ×~v3 k cos θ where θ is the angle between ~v1 and ~v2 ×~v3 so det[~v1 ~v2 ~v3 ] > 0 if and only if cos θ > 0, i.e., if and only if θ is acute (0 ≤ θ ≤ π2 ). (See Figure 6.8.) 6.3.20 By Exercise 19, ~v1 , ~v2 , ~v3 constitute a positively oriented basis if and only if det[~v1 ~v2 ~v3 ] > 0. Assume that ~v1 , ~v2 , ~v3 is such a basis. We want to show that A~v1 , A~v2 , A~v3 is positively oriented if and only if det(A) > 0. We have det[A~v1 A~v2 A~v3 ] = det(A[~v1 ~v2 ~v3 ]) = det(A) det[~v1 ~v2 ~v3 ] so since det [~v1 ~v2 ~v3 ] > 0 by assumption, det [A~v1 A~v2 A~v3 ] > 0 if and only if det(A) > 0. Hence A is orientation preserving if and only if det(A) > 0. 6.3.21 a Reverses 287

Chapter 6

Figure 6.7: for Problem 6.3.18c. v2 × v3

v1 θ

v3

v2

Figure 6.8: for Problem 6.3.19. Consider ~v2 and ~v3 in the plane (not parallel), and let ~v1 = ~v2 × ~v3 ; then ~v1 , ~v2 , ~v3 is a positively oriented basis, but T (~v1 ) = −~v1 , T (~v2 ) = ~v2 , T (~v3 ) = ~v3 is negatively oriented. b Preserves Consider ~v2 and ~v3 orthogonal to the line (not parallel), and let ~v1 = ~v2 ×~v3 ; then ~v1 , ~v2 , ~v3 is a positively oriented basis, and T (~v1 ) = ~v1 , T (~v2 ) = −~v2 , T (~v3 ) = −~v3 is positively oriented as well. c Reverses The standard basis ~e1 , ~e2 , ~e3 is positively oriented, but T (~e1 ) = −~e1 , T (~e2 ) = −~e2 , T (~e3 ) = −~e3 is negatively oriented. 6.3.22 Here A = det

x=

"

1 7 3 11 5

#

1 3 7 , so by Theorem 6.3.8 , det(A) = 5, ~b = 3 4 11 "

det

= −2, y =

3 4 5

1 3

#

= 1. 288

Section 6.3

6.3.23 Here A =

1 5 −3 , so by Theorem 6.3.8 , det(A) = 17, ~b = 0 −6 7

1 −3 det 0 7 x1 = 17 2 3 6.3.24 Here A = 0 4 6 0 8 3 −1

x=

det

3 0 4 5 0 7

146

5 1 det −6 0 7 = , x2 = 17 17

=

6 . 17

8 0 5 , det(A) = 146, ~b = 3 , so by Theorem 6.3.8, −1 7

= 1, y =

2

8 0 3 5 6 −1 7

det 0

146

= 2, z =

2 3 4 6 0

det 0

146

8 3 −1

= −1.

6.3.25 By Theorem 6.3.9, the ij th entry of adj(A) is given by (−1)i+j det(Aji ), so since 1 0 1 0 1 0 3 2 = 1, and for i = 1, j = 2 we get (−1) det A = 0 1 0 for i = 1, j = 1, we get (−1) det 0 0 1 2 0 1 and so forth. 1 0 −1 Completing this process gives adj(A) = 0 −1 0 , hence by Theorem 6.3.9, −2 0 1 1 0 −1 −1 0 1 1 1 A−1 = det(A) adj(A) = −1 0 −1 0 = 0 1 0 . −2 0 1 2 0 −1

1 = 0, 1

1 adj(A), so if det(A) = 1, A−1 = adj(A). If A has integer entries then 6.3.26 By Theorem 6.3.9, A−1 = det(A) i+j (−1) det(Aji ) will be an integer for all 1 ≤ i, j ≤ n, hence adj(A) will have integer entries. Therefore, A−1 will also have integer entries.

x = det

1 0

a −b 1 , det(A) = a2 + b2 , ~b = , we get b a 0 −b a 1 a −b 1 1 = , y = det a2 +b2 a2 +b2 a2 +b2 = a2 +b2 , a b 0

6.3.27 By Theorem 6.3.8, using A =

so x is positive, y is negative (since a, b > 0), and x decreases as b increases.

s 6.3.28 Here A = m det

Y =

I◦ + G Ms − M ◦

−sh−ma

"

det

r=

"

◦ a I +G ~ , det(A) = −sh − ma, b = so, by Theorem 6.3.8 −h Ms + M ◦ a −h

s I◦ + G m Ms − M ◦ −sh−ma

#

#

=

=

−h(I ◦ +G)−a(Ms −M ◦ ) −sh−ma

s(Ms −M ◦ )−m(I ◦ +G) −sh−ma

=

=

h(I ◦ +G)+a(Ms −M ◦ ) , sh+ma

m(I ◦ +G)−s(Ms −M ◦ ) . sh+ma

289

Chapter 6 6.3.29 By Theorem 6.3.8,

dx1 =

det

dy1 =

det

dp =

det

0 0 −R2 de2 −R1 α R2

−R1 α R2

R1 1−α −R2 D

0 0 −R2 de2 D

R1 1−α −R2 D

−(1 − α) −(1 − α)2 2 − (1−α) α

−(1 − α) −(1 − α)2 2 − (1−α) α

=

R1 R2 (1−α)2 de2 −R2 (1−α)2 de2 D

0 0 −R2 de2

=

R2 de2 (R1 (1−α)2 +α(1−α)) D

>0

=

R1 R2 de2 D

> 0.

0 0 6 0. −5 3

−6 0 6.3.31 Using the procedure outlined in Exercise 25, we find adj(A) = −3 5 4 −5

1 −2 . 1

0 0 6.3.32 Using the procedure outlined in Exercise 25, we find that adj(A) = 0 −1

0 0 −1 −1 0 0 . 0 −1 0 0 0 0

18 6.3.30 Using the procedure outlined in Exercise 25, we find adj(A) = −12 −2

24 0 0 0 0 12 0 0 6.3.33 Using the procedure outlined in Exercise 25, we find that adj(A) = . Note that the matrix 0 0 8 0 0 0 0 6 adj(A) is diagonal, and the ith diagonal entry of adj(A) is the product of all ajj where j 6= i.

6.3.34 For an invertible n × n matrix A, Aadj(A) = A(det(A)A−1 ) = det(A)AA−1 = det(A)In . The same is true for adj(A)A. 6.3.35 det(adj(A)) = det(det(A)A−1 ). Taking the product det(A)A−1 amounts to multiplying each row of A−1 by 1 det(A), so that det(adj(A)) = (det A)n det(A−1 ) = (det A)n det(A) = (det A)n−1 . 6.3.36 adj(adjA) = adj(det(A)A−1 ) = det(det(A)A−1 )(det(A)A−1 )−1 = (det A)n det(A−1 )(det(A)A−1 )−1 1 (A−1 )−1 = (det A)n−1 (det(A)A−1 )−1 = (det A)n−1 det(A)

= (det A)n−2 A. 6.3.37 adj(A−1 ) = det(A−1 )(A−1 )−1 = (det A)−1 (A−1 )−1 = (adjA)−1 . 290

Section 6.3 6.3.38 adj(AB) = det(AB)(AB)−1 = det(A)(det(B)B −1 )A−1 = det(B)B −1 (det(A)A−1 ) = adj(B)adj(A). 6.3.39 Yes, let S be an invertible matrix such that AS = SB, or SB −1 = A−1 S. Multiplying both sides by det(A) = det(B), we find that S(det(B)B −1 ) = (det(A)A−1 )S, or, S(adjB) = (adjA)S, as claimed. 6.3.40 The ij th entry of the matrix B of T is (ith component of T (~ej )) = det(A~ej ,i ). Expand down the i′ th column to see that this is the ij th entry of adj(A). Thus B = adj(A). See Theorem 6.3.9. 6.3.41 If A has a nonzero minor det(Aij ), then the n − 1 columns of the invertible matrix Aij will be independent, so that the n − 1 columns of A, minus the j th , will be independent as well. Thus, the rank of A (the dimension of the image) is at least n − 1. Conversely, if rank(A) ≥ n−1, then we can find n−1 independent columns of A. The n×(n−1) matrix consisting of those n − 1 columns will have rank n − 1, so that there will be exactly one redundant row (compare with Exercises 3.3.69 through 3.3.71). Omitting this redundant row produces an invertible (n − 1) × (n − 1) submatrix of A, giving us a nonzero minor of A. 6.3.42 By Theorem 6.3.9, adj(A) = 0 if (and only if) all the minors Aji of A are zero. By Exercise 41, this is the case if (and only if) rank(A) ≤ n − 2. 6.3.43 A direct computation shows that A(adjA) = (adjA)A = (det A)(In ) for all square matrices. Thus we have A(adjA) = (adjA)A = 0 for noninvertible matrices, as claimed. Let’s write B = adj(A), and let’s verify the equation AB = (det A)(In ) for the diagonal entries; the verification for the off-diagonal entries is analogous. The ith diagonal entry of AB is n ith X aij bji . [ith row of A] column = ai1 b1i + · · · + ain bni = j=1 of B Since B is the adjunct of A, bji = (−1)j+i det(Aij ). So, our summation equals n X

aij (−1)i+j det(Aij )

j=1

which is our formula for Laplace expansion across the ith row, and equals det(A), proving our claim for the diagonal entries. 6.3.44 The equation A(adjA) = 0 from Exercise 43 means that im(adjA) is a subspace of ker(A). Thus rank(adjA) = dim(im(adjA)) ≤ dim(kerA) = n − rank(A) = n − (n − 1) = 1, implying that rank(adjA) ≤ 1. Since adj(A) 6= 0, by Exercise 42, we can conclude that rank(adjA) = 1. 291

Chapter 6 d −b a c a b . So, a = d and b = −c. Thus, the = . We want AT = adj(A), or b d c d −c a a b . equation AT = adj(A) holds for all matrices of the form −b a

6.3.45 Let A =

6.3.46 In the simple case when f (x, y) = 1 we have

Z

f (x, y)dA =

Ω2

Z

g(u, v)dA =

Ω1

Z

dA = area of Ω1 = 1, so that

Ω1

Z

Ω2

Z

Ω2

dA = area of Ω2 = | det M | and

f (x, y)dA = | det M | ·

Z

g(u, v)dA.

Ω1

This formula holds, in fact, for any continuous function f (x, y); see an introductory text in multivariable calculus for a justification.

x1 x2 is the area of the triangle OP1 P2 , where O denotes the origin. This is likewise y1 y2 true for one-half the second matrix. See Theorem 2.4.10. However, because of the reversal in orientation, x3 x4 1 is negative the area of the triangle OP3 P4 ; likewise for the last matrix. Finally, note that the 2 det y y4 3 area of the quadrilateral P1 P2 P3 P4 is equal to:

6.3.47 Note that

1 2

det

the area of triangle OP1 P2 + area of triangle OP2 P3 − area of triangle OP3 P4 − area of triangle OP4 P1 . 6.3.48 In what follows, we will freely use the fact that an invertible linear transformation L from R2 to R2 maps an ellipse into an ellipse (see Exercise 2.2.52). Now consider a linear transformation L that transforms our 3-4-5 right triangle R into an equilateral triangle T . 0 4 0 , and the vertices of the equilateral , , If we place the vertices of the right triangle R at the points 3 0 0 "1 # 1 1 2 0 2 3 1 √ , , triangle T at , then the transformation L has the matrix A = , with det(A) = 2√ . 3 0 0 3 0 √1 3

According to the hint, L will map the largest ellipse E inscribed √ into R into the circle C inscribed into T . The Figure 6.9 illustrates that the radius of C is tan(π/6) = 1/ 3, so that the area of C is π/3. Using the interpretation of the determinant as an expansion fact, we find that (area of C) = (det A)(area of E), or (area of C = √ 2π of E) = area ≈ 3.6 det(A) 3 6.3.49 We will use the terminology introduced in√ the solution of Exercise 48 throughout. Note that the trans√ 3 2 −2/ √ formation L−1 , with matrix A−1 = , maps the circle C (with radius 1/ 3) into the ellipse E. 0 3 cos θ of C, and find the maximal value M and the minimal value m of Now consider a radial vector ~v = √13 sin θ 2 25 8 1 4 (sin θ)(cos θ) = 18 (sin 2θ) (we are taking the square to facilitate kA−1~v k = 34 + 19 sin2 θ − 3√ − 18 (cos 2θ) − 3√ 3 √3 √ the computations). Then M and mq will be the lengths of the semi-axis of E. The function above is sinusoidal √ √ √ 25+ 193 193 16 1 and m = 25−18 193 , so that the with average value 25 18 and amplitude 182 + 27 = 18 . Thus M = 18 length of the semi-major axis of E is 292

True or False

1 2

[ 03 [

1 3 1 √3

A= 0 R

[ √13 [

L

2

–1

L

T 1

A

–1

=

2

−2 √3

0

√3 C

1

2

3

[ 40 [

π 6

r= 1 √3

[ 20 [

Figure 6.9: for Problem 6.3.48 and Problem 6.3.49. q √ √ M = 25+18 193 ≈ 1.47, and for the semi-minor axis we get q √ √ m = 25−18 193 ≈ 0.79.

True or False Ch 6.TF.1 T, by Definition 6.1.1 Ch 6.TF.2 F; We have det(4A) = 44 det(A), by Theorem 6.2.3a. Ch 6.TF.3 F; Let A = B = I5 , for example Ch 6.TF.4 T; We have det(−A) = (−1)6 det(A) = det(A), by Theorem 6.2.3a. Ch 6.TF.5 F; In fact, det(A) = 0, since A fails to be invertible Ch 6.TF.6 F; The matrix A fails to be invertible if det(A) = 0 by Theorem 6.2.4. Ch 6.TF.7 T, by Theorem 6.2.3a, applied to the columns. Ch 6.TF.8 T, by Theorem 6.2.6. Ch 6.TF.9 T, By theorem 6.1.4, a diagonal matrix is triangular as well. Ch 6.TF.10 T, by Theorem 6.2.3b. Ch 6.TF.11 T. Without computing its exact value, we will show that the determinant is positive. The pattern that contains all the entries 100 has a product of 1004 = 108 , with two inversions. Each of the other 4! − 1 = 23 patterns contains at most two entries 100, with the other entries being less than 10, so that the product of each of these patterns is less than 1002 · 102 = 106 . Thus the determinant in more than 108 − 23 · 106 > 0, so that the matrix in invertible. 293

Chapter 6 Ch 6.TF.12 F; The correct formula is det(A−1 ) =

1 , det(AT )

by Theorems 6.2.1 and 6.2.8.

Ch 6.TF.13 T; The matrix A is invertible. Ch 6.TF.14 T; Any nonzero noninvertible matrix A will do. Ch 6.TF.15 T, by Theorem 6.2.7. Ch 6.TF.16 F, by Theorem 6.3.1. The determinant can be −1. Ch 6.TF.17 T, by Theorem 6.2.6. Ch 6.TF.18 F; The second and the fourth column are linearly dependent. Ch 6.TF.19 T; The determinant is 0 for k = −1 or k = −2, so that the matrix is invertible for all positive k. Ch 6.TF.20 F. There is only one pattern with a nonzero product, containing all the 1’s. Since there are three inversions in this pattern, det A = −1.

1 1 Ch 6.TF.21 T; Let A = 1 1

1 1 1 −1 −1 1 −1 −1

0

8 Ch 6.TF.22 F; Let A = 0

1 2

1 −1 . The column vectors of A are orthogonal and they all have length 2. −1 1

1 , for example. and ~v = 0

Ch 6.TF.23 F; In fact, det(A) = det[~u ~v w] ~ = − det[~v ~u w] ~ = −~v · (~u × w). ~ We have used Theorem 6.2.3b and Definition 6.1.1.

1 0 Ch 6.TF.24 T; Let A = 0 1

1 and B = 0

1 Ch 6.TF.25 F; Note that det 0 1

0 , for example. −1

1 0 1 1 = 2. 0 1

Ch 6.TF.26 T, by Theorem 6.3.9. Ch 6.TF.27 T, by Theorem 6.3.3, since k~vi⊥ k ≤ k~vi k = 1 for all column vectors ~vi . Ch 6.TF.28 T; We have det(A) = det(rref A) = 0. 3 2 , for example. See Theorem 6.2.10. Ch 6.TF.29 F; Let A = 5 3

Ch 6.TF.30 F; Let A = 2I2 , for example 294

True or False Ch 6.TF.31 F; Note that det(S −1 AS) = det(A) but det(2A) = 23 (det A) = 8(det A). Ch 6.TF.32 F; Note that det(S T AS) = (det S)2 (det A) and det(−A) = −(det A) have opposite signs. Ch 6.TF.33 F; Let A = 2I2 , for example. 0 1 Ch 6.TF.34 F; Let A = 0 0

−1 0 0 0

0 0 0 1

0 0 , for example. −1 0

Ch 6.TF.35 F; Let A = I2 and B = −I2 , for example. Ch 6.TF.36 T; Note that det(B) = − det(A) < det(A), so that det(A) > 0. Ch 6.TF.37 T; Let’s do Laplace expansion along the first row, for example (see Theorem 6.2.10). Then det(A) =

n X j=1

(−1)1+j a1j det(A1j ) 6= 0. Thus det(A1j ) 6= 0 for at least one j, so that A1j is invertible.

Ch 6.TF.38 T; Note that det(A) and det(A−1 ) are both integers, and (det A)(det A−1 ) = 1. This leaves only the possibilities det(A) = det(A−1 ) = 1 and det(A) = det(A−1 ) = −1. Ch 6.TF.39 T, since adj(A) = (det A)(A−1 ), by Theorem 6.3.9. Ch 6.TF.40 F; Note that det(A2 ) = (det A)2 cannot be negative, but det(−I3 ) = −1. Ch 6.TF.41 T; The product associated with the diagonal pattern is odd, while the products associated with all other patterns are even. Thus the determinant of A is odd, so that A is invertible, as claimed.

2 Ch 6.TF.42 F; Let A = 1 1

1 1 2 1 , for example 5 2

a b 0 Ch 6.TF.43 T; Let A = . If a 6= 0, let B = c d 0 1 0 and if d 6= 0, let B = . 0 0

0 0 0 0 1 ; if b 6= 0, let B = ; if c 6= 0, let B = , 1 1 0 0 0

Ch 6.TF.44 T; Use Gaussian elimination for the first column only to transform A into a matrix of the form 1 ±1 ±1 ±1 ∗ ∗ 0 ∗ B= 0 ∗ ∗ ∗ 0 ∗ ∗ ∗ Note that det(B) = det(A) or det(B) = −(det A). The stars in matrix B all represent numbers (±1) ± (±1), so that they are 2, 0, or −2. Thus the determinant of the 3 × 3 matrix M containing the stars is divisible by 8, 295

Chapter 6 since each of the 6 terms in Sarrus’ rule is 8, 0 or -8. Now perform Laplace expansion down the first column of B to see that det(M ) = det(B) = +/ − det(A). Ch 6.TF.45 T; A(adjA) = A(det(A)A−1 ) = det(A)In = det(A)A−1 A = adj(A)A.

1 Ch 6.TF.46 T; Laplace expansion along the second row gives det(A) = −k det 8 0 constant C (we need not compute that C = −259). Thus A is invertible except 37 be 259 35 = 5 = 7.4).

1 Ch 6.TF.47 F; A = 0 0 AB 6= BA.

2 9 0 for

4 7 + C = 35k + C, for some 5 k = −C 35 (which turns out to

−1 0 0 0 0 0 −1 and B = 0 −1 0 are both orthogonal and det(A) = det(B) = 1. However, 0 0 1 1 0

296

Section 7.1

Chapter 7 Section 7.1 7.1.1 If ~v is an eigenvector of A, then A~v = λ~v . Hence A3~v = A2 (A~v ) = A2 (λ~v ) = A(Aλ~v ) = A(λA~v ) = A(λ2~v ) = λ2 A~v = λ3~v , so ~v is an eigenvector of A3 with eigenvalue λ3 . 7.1.2 We know A~v = λ~v so ~v = A−1 A~v = A−1 λ~v = λA−1~v , so ~v = λA−1~v or A−1~v = λ1 ~v . Hence ~v is an eigenvector of A−1 with eigenvalue

1 λ.

7.1.3 We know A~v = λ~v , so (A + 2In )~v = A~v + 2In~v = λ~v + 2~v = (λ + 2)~v , hence ~v is an eigenvector of (A + 2In ) with eigenvalue λ + 2. 7.1.4 We know A~v = λ~v , so 7A~v = 7λ~v , hence ~v is an eigenvector of 7A with eigenvalue 7λ. 7.1.5 Assume A~v = λ~v and B~v = β~v for some eigenvalues λ, β. Then (A + B)~v = A~v + B~v = λ~v + β~v = (λ + β)~v so ~v is an eigenvector of A + B with eigenvalue λ + β. 7.1.6 Yes. If A~v = λ~v and B~v = µ~v , then AB~v = A(µ~v ) = µ(A~v ) = µλ~v 7.1.7 We know A~v = λ~v so (A − λIn )~v = A~v − λIn~v = λ~v − λ~v = ~0 so a nonzero vector ~v is in the kernel of (A − λIn ) so ker(A − λIn ) 6= {~0} and A − λIn is not invertible. a b a b 1 1 a 5 7.1.8 We want all such that =5 hence = , i.e. the desired matrices must have c d c d 0 0 c 0 5 b the form . 0 d

a b c d

1 1 a λ =λ for any λ. Hence = , i.e., the desired matrices must have the form 0 0 c 0

7.1.9 We want λ b , they must be upper triangular. 0 d 7.1.10 We want

a c

a b c d

b d

5 − 2b 1 1 , i.e. the desired matrices must have the form =5 10 − 2d 2 2

2 −2 = . So, 2a + 3b = −2 and 2c + 3d = −3. Thus, b = 3 −3 a −2−2a 3 will fit. all matrices of the form c −3−2c 3

7.1.11 We want

7.1.12 Solving solving

2 3

t v1 2 0 v v1 = 2 1 we get = (with t 6= 0) and 3 4 v2 v2 v2 − 32 t v1 0 v1 v1 0 =4 we get (with t 6= 0). = 4 v2 v2 v2 t

297

−2−2a , 3

b . d and d =

−3−2c . 3

So

Chapter 7

7.1.13 Solving

−6 6 −15 13

v1 v2

=4

3 t v1 v1 (with t 6= 0). , we get = 5 v2 v2 t

7.1.14 We want to find all 4 × 4 matrices A such that A~e2 = λ~e2 , i.e. the second column of A must be of the form a 0 c d 0 e λ f g λ . , so A = h 0 i j 0 k 0 l m 0 7.1.15 Any vector on L is unaffected by the reflection, so that a nonzero vector on L is an eigenvector with eigenvalue 1. Any vector on L⊥ is flipped about L, so that a nonzero vector on L⊥ is an eigenvector with eigenvalue −1. Picking a nonzero vector from L and one from L⊥ , we obtain a basis consisting of eigenvectors. 7.1.16 Rotation by 180◦ is a flip about the origin so every nonzero vector is an eigenvector with the eigenvalue −1. Any basis for R2 consists of eigenvectors. 7.1.17 No (real) eigenvalues 7.1.18 Any nonzero vector in the plane is unchanged, hence is an eigenvector with the eigenvalue 1. Since any nonzero vector in V ⊥ is flipped about the origin, it is an eigenvector with eigenvalue −1. Pick any two non-collinear vectors from V and one from V ⊥ to form a basis consisting of eigenvectors. 7.1.19 Any nonzero vector in L is an eigenvector with eigenvalue 1, and any nonzero vector in the plane L⊥ is an eigenvector with eigenvalue 0. Form a basis consisting of eigenvectors by picking any nonzero vector in L and any two nonparallel vectors in L⊥ . 7.1.20 Any nonzero vector along the ~e3 -axis is unchanged, hence is an eigenvector with eigenvalue 1. No other (real) eigenvalues can be found. 7.1.21 Any nonzero vector in R3 is an eigenvector with eigenvalue 5. Any basis for R3 consists of eigenvectors. 7.1.22 Any nonzero scalar multiple of ~v is an eigenvector with eigenvalue 1. 7.1.23 a Since S = [~v1 · · · ~vn ], S −1~vi = S −1 (S~ei ) = ~ei . b i th column of S −1 AS = S −1 AS~ei = S −1 A~vi (by definition of S) = S −1 λi~vi (since ~vi is an eigenvector) = λi S −1~vi = λi~ei (by part a)

λ1 0 hence S −1 AS = ... 0

0 λ2

0 ··· 0 ···

0

0 ···

0 0 .

λn

298

Section 7.1 7.1.24 See Figure 7.1.

Figure 7.1: for Problem 7.1.24. 7.1.25 See Figure 7.2.

Figure 7.2: for Problem 7.1.25. 7.1.26 See Figure 7.3.

Figure 7.3: for Problem 7.1.26. 7.1.27 See Figure 7.4. 7.1.28 See Figure 7.5. 7.1.29 See Figure 7.6. 299

Chapter 7

Figure 7.4: for Problem 7.1.27.

Figure 7.5: for Problem 7.1.28.

Figure 7.6: for Problem 7.1.29. 7.1.30 Since the matrix is diagonal, ~e1 and ~e2 are eigenvectors. See Figure 7.7. 7.1.31 See Figure 7.8. 7.1.32 Since the matrix is diagonal, ~e1 and ~e2 are eigenvectors. See Figure 7.9. 1 t −1 , hence we know that the eigenvalues are 2 and 6 with corresponding +6 7.1.33 We are given that ~x(t) = 2 1 1 1 −1 −1 1 = respectively (see Theorem 7.1.3), so we want a matrix A such that A and eigenvectors 1 1 1 1 −1 1 −1 4 −2 2 −6 . . Multiplying on the right by , we get A = 1 1 −2 4 2 6 t

300

Section 7.1

Figure 7.7: for Problem 7.1.30.

Figure 7.8: for Problem 7.1.31.

Figure 7.9: for Problem 7.1.32. 7.1.34 (A2 + 2A + 3In )~v = A2~v + 2A~v + 3In~v = 42~v + 2 · 4~v + 3~v = (16 + 8 + 3)~v = 27~v so ~v is an eigenvector of A2 + 2A + 3In with eigenvalue 27. 7.1.35 Let λ be an eigenvalue of S −1 AS. Then for some nonzero vector ~v , S −1 AS~v = λ~v , i.e., AS~v = Sλ~v = λS~v so λ is an eigenvalue of A with eigenvector S~v . Conversely, if α is an eigenvalue of A with eigenvector w, ~ then Aw ~ = αw, ~ for some nonzero w. ~ Therefore, S −1 AS(S −1 w) ~ = S −1 Aw ~ = S −1 αw ~ = αS −1 w, ~ so S −1 w ~ is an eigenvector of S −1 AS with eigenvalue α. 3 15 1 10 3 1 15 10 7.1.36 We want A such that A = and A = , i.e. A = , so 1 5 2 20 1 2 5 20 301

Chapter 7

15 A= 5

10 20

3 1 1 2

−1

4 = −2

3 . 11

0.6 0.8 7.1.37 a A = 5 is a scalar multiple of an orthogonal matrix. By Theorem 7.1.2, the possible eigenvalues 0.8 −0.6 of the orthogonal matrix are ±1, so that the possible eigenvalues of A are ±5. In part b we see that both are indeed eigenvalues. b Solve A~v = ±5~v to get ~v1 =

1 1 2 1 −3 −1 = −2 = 2 −1 . The associated eigenvalue is 2. 2 −1 −2 −1

4 1 7.1.38 −5 0 −1 −1 7.1.39 a c So,

2 −1 , ~v2 = . 1 2

a We want c 0 1 =a d 0 0 1 0 , 1 0 0

b 0 0 0 =λ = . So b = 0, and d = λ (for any λ). Thus, we need matrices of the form d 1 1 λ 0 0 0 0 0 +c +d . 0 1 0 0 1 0 0 0 is a basis of V , and dim(V )= 3. , 0 1 0

7.1.40 We need all matrices A such that

a c

b d

1 1 λ =λ = . −3 −3 −3λ

Thus, a − 3b = λ and c − 3d = −3λ. Thus, c − 3d = −3(a − 3b) = −3a + 9b, or c = −3a + 9b + 3d. a b 1 0 0 1 0 0 So A must be of the form = a +b +d . Thus, a basis of V is −3 0 9 0 3 1 −3a + 9b + 3d d 0 0 0 1 1 0 , and the dimension of V is 3. , , 3 1 9 0 −3 0 7.1.41 We want 2λ2 = c + 2d.

a b c d

1 1 a , and = λ1 1 1 c

b d

1 1 . So, a + b = λ1 = c + d and a + 2b = λ2 and = λ2 2 2

So (a+2b)−(a+b) = λ2 −λ1 = b, a = λ1 −b = 2λ1 −λ = 2λ2−λ1 = d, c = λ1 −d = 2λ1 −2λ2 . (c+2d)−(c+d) 2 . Also, −1 1 2 −1 2λ1 − λ2 λ2 − λ1 . + λ2 = λ1 So A must be of the form: −2 2 2 −1 2λ1 − 2λ2 2λ2 − λ1 −1 1 2 −1 , and dim(V )= 2. , So a basis of V is −2 2 2 −1 1 7.1.42 We will do this in a slightly simpler manner than Exercise 40. Since A 0 is simply the first column of A, 0 the first column must be a multiple of ~e1 . Similarly, the third column must be a multiple of ~e3 . There are no 302

Section 7.1

other restrictions on the

0 0 0 0 + c0 1 0 0 0 1 Thus, a basis of V is 0 0 5.

0 1 b0 0 0 0

1 0 a b 0 form of A, meaning it can be any matrix of the form 0 c 0 = a 0 0 0 0 0 d e 0 0 0 0 0 0 0 0 + d 0 0 0 + e 0 0 0 . 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 , 0 0 0 , 0 1 0 , 0 0 0 , 0 0 0 and the dimension 0 0 1 0 1 0 0 0 0 0 0 0 0 0

7.1.43 A = AIn = A[ ~e1 . . . ~en ] = [ λ1~e1 . . . can be any diagonal matrix, and dim(V ) = n.

0 0 + 0

of V is

λn~en ], where the eigenvalues λ1 , . . . , λn are arbitrary. Thus A

7.1.44 We see that each of the columns 1 through m of A will have to be a multiple of its respective vector ~ei . Thus, there will be m free variables in the first m columns. The remaining n − m columns will each have n free variables. Thus, in total, the dimension of V is m + (n − m)n = m + n2 − nm. 7.1.45 Consider a vector w ~ that is not parallel to ~v . We want A[~v w] ~ = [λ~v a~v + bw], ~ where λ, a and b are arbitrary constants. Thus the matrices A in V are of the form A = [λ~v a~v + bw][~ ~ v w] ~ −1 . Using Summary 4.1.6, we see that [~v ~0][~v w] ~ −1 , [~0 ~v ][~v w] ~ −1 , [~0 w][~ ~ v w] ~ −1 is a basis of V , so that dim(V ) = 3. 7.1.46 a We need all matrices A such that

a b c d

k 1 1 . = =k 2k 2 2

Thus, a + 2b = k and c+ 2d = 2k. So, c + 2d = 2a + 4b, or c = −2d + 2a + 4b and A must be of the form a b 1 0 0 1 0 0 1 0 0 1 0 0 =a +b +d . So a basis of V is , , , and −2d + 2a + 4b d 2 0 4 0 −2 1 2 0 4 0 −2 1 the dimension of V is 3. 1 is a basis of the image of T by definition of V , so that the rank of T is 1. The kernel of T consists b Clearly 2 a b a b 1 of all matrices such that = ~0, or a + 2b = 0, c + 2d = 0. These are the matrices of the form c d c d 2 −2b b −2 1 0 0 −2 1 0 0 =b +d . Thus a basis of the kernel of T is , . −2d d 0 0 −2 1 0 0 −2 1 a b c Let’s find the kernel of L first. In part (a) we saw that the matrices in V are of the form A = . −2d + 2a + 4b d a b 1 A matrix A in V is in the kernel of L if = ~0, or a + 3b = 0, 2a + 4b + d = 0. This −2d + 2a + 4b d 3 −3b b = system simplifies to a = −3b and d = 2b, so that the matrices in the kernel of L are of the form −6b 2b −3 1 −3 1 forms a basis of the kernel of L. By the rank-nullity theorem, the rank of L . The matrix b −6 2 −6 2 is dim(V )−dim(ker L) = 3 − 1 = 2, and the image of L is all of R2 . 7.1.47 Suppose V is a one-dimensional A-invariant subspace of Rn , and ~v is a non-zero vector in V . Then A~v will be in V, so that A~v = λ~v for some λ, and ~v is an eigenvector of A. Conversely, if ~v is any eigenvector of A, then 303

Chapter 7 V = span(~v ) will be a one-dimensional A-invariant subspace. Thus the one-dimensional A-invariant subspaces V are of the form V = span(~v ), where ~v is an eigenvector of A. 7.1.48 a Since span(~e1 ) is an A-invariant subspace of R3 , it must be that ~e1 is an eigenvector of A, as revealed in a Exercise 47. Thus, the first column of A must be of the form 0 . Since span(~e1 , ~e2 ) is also an A-invariant 0 b subspace, it must be that A~e2 is in span(~e1 , ~e2 ). Thus, the second column of A must have the form c . The 0 1 1 1 third column may be any vector in R3 . Thus, we can choose A = 0 1 1 to maximize the number of non-zero 0 0 1 entries. b We see, from our construction a of all matrices of the form 0 0

above, that upper-triangular matrices fit this description. This space, V consists b d c e and has a dimension of 6. 0 f

100 7.1.49 The eigenvalues of the system are λ1 = 1.1, and λ2 = 0.9 and corresponding eigenvectors are ~v1 = 300 100 200 , we can see that ~x0 = 3~v1 − ~v2 . Therefore, by Theorem 7.1.3, , respectively. So if ~x0 = and ~v2 = 100 800 200 100 , i.e. c(t) = 300(1.1)t − 200(0.9)t and r(t) = 900(1.1)t − 100(0.9)t . − (0.9)t we have ~x(t) = 3(1.1)t 100 300 4 100 , and we see that A~v (0) = 7.1.50 a ~v (0) = 1 100 100 100 . = 2t At 100 100

−2 1

100 100

=

200 200

= 2

100 . So, ~v (t) = At~v (0) = 100

So c(t) = r(t) = 100(2)t . 4 200 , and we see that A~v (0) = b ~v (0) = 1 100 t 200 . 3 100

−2 1

200 600 200 t t 200 = . So, ~v (t) = A ~v (0) = A =3 = 100 100 300 100

So c(t) = 200(3)t and r(t) = 100(3)t . 200 100 600 . So, + . We can write this in terms of the previous eigenvectors as ~v (0) = 4 c ~v (0) = 100 100 500 200 100 200 100 . ~v (t) = At~v (0) = At 4 + (3)t = 4(2)t + At 100 100 100 100

So c(t) = 400(2)t + 200(3)t and r(t) = 400(2)t + 100(3)t . 304

Section 7.1 0 100 , and we see that A~v (0) = 7.1.51 a ~v (0) = −1.5 200 100 100 . = (1.5)t At 200 200

.75 2.25

100 150 100 . So, ~v (t) = At~v (0) = = 1.5 = 200 300 200

So c(t) = 100(1.5)t and r(t) = 200(1.5)t . 100 0 b ~v (0) = , and we see that A~v (0) = 100 −1.5 t 100 t 100 A = (.75) . 100 100

.75 2.25

100 100

=

75 75

= .75

100 . So, ~v (t) = At~v (0) = 100

So c(t) = 100(.75)t and r(t) = 100(.75)t .

500 100 100 c ~v (0) = . We can write this in terms of the previous eigenvectors as ~v (0) = 3 +2 . So, 700 100 200 100 100 100 100 . + 2(1.5)t = 3(.75)t + At 2 ~v (t) = At~v (0) = At 3 200 100 200 100 So c(t) = 300(.75)t + 200(1.5)t and r(t) = 300(.75)t + 400(1.5)t . 7.1.52 a and

0.978 0.004

0.978 0.004

g b ~x0 = 0 l0

−0.006 −1 −0.99 −1 = = 0.99 , 0.992 2 1.98 2 3 2.94 3 −0.006 . The eigenvalues are λ1 = 0.99 and λ2 = 0.98. = 0.98 = −1 −0.98 −1 0.992

100 −1 3 3 t t −1 = = 20 + 40 , hence + 40(0.98) so ~x(t) = 20(0.99) 0 2 −1 2 −1

g(t) = −20(0.99)t + 120(0.98)t and h(t) = 40(0.99)t − 40(0.98)t .

Figure 7.10: for Problem 7.1.52b. h(t) first rises, then falls back to zero. g(t) falls a little below zero, then goes back up to zero. See Figure 7.10. c We set g(t) = −20(0.99)t + 120(0.98)t = 0. Solving for t we get that g(t) = 0 for t ≈ 176 minutes. (After t = 176, g(t) < 0). 305

Chapter 7 1 1 1 6 7.1.53 a ~v (0) = 1 = 3 1 + 2 −1 + 0 . −1 0 1 2 1 1 1 So, ~v (t) = At~v (0) = At 3 1 + 2 −1 + 0 −1 0 1 1 1 1 1 1 1 = 3At 1 + 2At −1 + At 0 = 3λt1 1 + 2λt2 −1 + λt3 0 1 −1 0 0 1 −1 1 1 1 = 3 1 + 2(− 12 )t −1 + (− 21 )t 0 . 1 0 −1 So a(t) = 3 + 3(− 12 )t , b(t) = 3 − 2(− 12 )t and c(t) = 3 − (− 21 )t .

b a(365) = 3 + 3(− 12 )365 = 3 − c(365) = 3 − (− 12 )365 = 3 +

3 2365 ,

1 2365 .

b(365) = 3 − 2(− 12 )365 = 3 +

1 2364

and

So, Benjamin will have the most gold.

7.1.54 a We are given that n(t + 1) = 2a(t) a(t + 1) = n(t) + a(t), so that the matrix is A = 1 0 b A = 1 1

0 2 . 1 1

1 2 1 2 0 2 2 −2 2 = =2 and A = = =− , hence 2 and −1 are 1 2 1 1 −1 1 −1 1 −1 1 2 the eigenvalues associated with and respectively. 1 −1 2 1

2 2 n0 1 1 1 1 t 1 1 1 t + 3 (−1) (by Theorem 7.1.3), +3 , and ~x(t) = 3 2 c We are given ~x0 = = so ~x0 = 3 1 −1 1 −1 a0 0 hence n(t) = 31 2t + 23 (−1)t and a(t) = 31 2t − 13 (−1)t .

Section 7.2 7.2.1 λ1 = 1, λ2 = 3 by Theorem 7.2.2. 7.2.2 λ1 = 2 (Algebraic multiplicity 2) λ2 = 1 (Algebraic multiplicity 2), by Theorem 7.2.2.

5−λ 7.2.3 det(A − λI2 ) = det 2

−4 = λ2 − 4λ + 3 = (λ − 1)(λ − 3) = 0 so λ1 = 1, λ2 = 3. −1 − λ 306

Section 7.2 4 = −λ(4 − λ) + 4 = (λ − 2)2 = 0 so λ = 2 with algebraic multiplicity 2. 4−λ

7.2.4 det(A − λI2 ) = det

−λ −1

7.2.5 det(A − λI2 ) = det

11 − λ 6

1−λ 7.2.6 det(A − λI2 ) = det 3

−15 = λ2 − 4λ + 13 so det(A − λI2 ) = 0 for no real λ. −7 − λ 2 = λ2 − 5λ − 2 = 0 so λ1,2 = 4−λ

√ 5± 33 . 2

7.2.7 λ = 1 with algebraic multiplicity 3, by Theorem 7.2.2. 7.2.8 fA (λ) = −λ2 (λ + 3) so λ1 = 0 (Algebraic multiplicity 2) λ2 = −3. 7.2.9 fA (λ) = −(λ − 2)2 (λ − 1) so λ1 = 2 (Algebraic multiplicity 2) λ2 = 1. 7.2.10 fA (λ) = (1 + λ)2 (1 − λ) so λ1 = −1 (Algebraic multiplicity 2), λ2 = 1. 7.2.11 fA (λ) = −λ3 − λ2 − λ − 1 = −(λ + 1)(λ2 + 1) = 0 λ = −1 (Algebraic multiplicity 1). 7.2.12 fA (λ) = λ(λ + 1)(λ − 1)2 so λ1 = 0, λ2 = −1, λ3 = 1 (Algebraic multiplicity 2). 7.2.13 fA (λ) = −λ3 + 1 = −(λ − 1)(λ2 + λ + 1) so λ = 1 (Algebraic multiplicity 1). 7.2.14 fA (λ) = det(B − λI2 ) det(D − λI2 ) (see Theorem 6.1.5). The eigenvalues of A are the eigenvalues of B and D. The eigenvalues of C are irrelevant. 2

7.2.15 fA (λ) = λ − 2λ + (1 − k) = 0 if λ1,2 =

2±

√

4−4(1−k) 2

=1±

√ k

The matrix A has 2 distinct real eigenvalues when k > 0, no real eigenvalues when k < 0. 7.2.16 fA (λ) = λ2 − (a + c)λ + (ac − b2 ) The discriminant of this quadratic equation is (a + c)2 − 4(ac − b2 ) = a2 + 2ac + c2 − 4ac + 4b2 = (a − c)2 + 4b2 ; this quantity is always positive (since b 6= 0). There will always be two distinct real eigenvalues. √ 7.2.17 fA (λ) = λ2 − a2 − b2 = 0 so λ1,2 = ± a2 + b2 . The matrix A represents a reflection about a line followed by a scaling by 307

√

a2 + b2 , hence the eigenvalues.

Chapter 7 7.2.18 fA (λ) = λ2 − 2aλ + a2 − b2 so λ1,2 =

+2a±

√

4a2 −4(a2 −b2 ) 2

= a ± b.

Hence the eigenvalues are a ± b. 7.2.19 True, since fA (λ) = λ2 − tr(A)λ + det(A) and the discriminant [tr(A)]2 − 4 det(A) is positive if det(A) is negative. 7.2.20 The characteristic polynomial of A is fA (λ) = (λ − λ1 )(λ − λ2 ) = λ2 − (λ1 + λ2 )λ + λ1 λ2 . But from Theorem 7.2.4 we know that fA (λ) = λ2 − tr(A)λ + det(A). Comparing the coefficient of λ, we see that λ1 + λ2 = tr(A), as claimed. 7.2.21 If A has n eigenvalues, then fA (λ) = (λ1 − λ)(λ2 − λ) · · · (λn − λ). Then fA (λ) = (−λ)n + (λ1 + λ2 + · · · + λn )(−λ)n−1 + · · · + (λ1 λ2 · · · λn ). But, by Theorem 7.2.5, the coefficient of (−λ)n−1 is tr(A). So, tr(A) = λ1 + · · · + λn . 7.2.22 By Theorem 6.2.1, fA (λ) = det(A−λIn ) = det(A−λIn )T = det(AT −λIn ) = fAT (λ). Since the characteristic polynomials of A and AT are identical, the two matrices have the same eigenvalues, with the same algebraic multiplicities. 7.2.23 fB (λ) = det(B − λIn ) = det(S −1 AS − λIn ) = det(S −1 AS − λS −1 In S) = det(S −1 (A − λIn )S) = det(S −1 ) det(A − λIn ) det(S) = (det S)−1 det(A − λIn ) det(S) = det(A − λIn ) = fA (λ) Hence, since fA (λ) = fB (λ), A and B have the same eigenvalues. 7.2.24 λ1 = 0.25, λ2 = 1 b ab + cb (a + c)b b b = = = since a + c = b + d = 1; therefore, is an eigenvector with c cb + cd (b + d)c c c eigenvalue λ1 = 1. 1 a−b 1 1 Also, A = = (a−b) since a−b = −(c−d); therefore, is an eigenvector with eigenvalue −1 c−d −1 −1 λ2 = a − b. Note that |a − b| < 1; a possible phase portrait is shown in Figure 7.11.

7.2.25 A

7.2.26 Here

1 0.25 b with λ2 = a − b = 0.25. See Figure 7.12. with λ1 = 1 and = −1 0.5 c

7.2.27 a We know ~v1 = x1 (t) =

1 3

+

2 3

1 1 1 then ~x0 = 31 ~v1 + 23 ~v2 , so by Theorem 7.1.3, , λ2 = 14 . If ~x0 = , λ1 = 1 and ~v2 = 0 −1 2

1 t 4

t − 23 41 . 0 If ~x0 = then ~x0 = 13 ~v1 − 31 ~v2 , so by Theorem 7.1.3, 1 x2 (t) =

2 3

308

Section 7.2

Figure 7.11: for Problem 7.2.25.

Figure 7.12: for Problem 7.2.26. x1 (t) =

1 3

−

1 3

x2 (t) =

2 3

+

1 3

t

b A approaches

1 t 4

1 t 4 . 1 3

See Figure 7.13.

1 1 , as t → ∞. See part c for a justification. 2 2

c Let us think about the first column of At , which is At~e1 . We can use Theorem 7.1.3 to compute At~e1 . b 1 1 and c2 = Start by writing ~e1 = c1 + c2 ; a straightforward computation shows that c1 = b+c c −1 309

c b+c .

Chapter 7

Figure 7.13: for Problem 7.2.27a.

Now At~e1 =

1 b+c

b + c

c t b+c (λ2 )

1 , where λ2 = a − b. −1

Since |λ2 | < 1, the second summand goes to zero, so that lim (At~e1 ) = t→∞

Likewise, lim (At~e2 ) = t→∞

1 b+c

b , so that lim At = t→∞ c

1 b+c

b c

1 b+c

b . c

b . c

7.2.28 a w(t + 1) = 0.8w(t) + 0.1m(t) m(t + 1) = 0.2w(t) + 0.9m(t)

0.8 so A = 0.2

0.1 0.9

which is a regular transition matrix since its columns sum to 1 and its entries are positive.

b The eigenvectors of A are

~x0 =

0.1 0.2

or

1 1 with λ2 = 0.7. with λ1 = 1, and −1 2

1 1 1 1 1200 or + 800(0.7)t so ~x(t) = 400 + 800 = 400 −1 2 −1 2 0

w(t) = 400 + 800(0.7)t m(t) = 800 − 800(0.7)t . c As t → ∞, w(t) → 400 so Wipfs won’t have to close the store. 7.2.29 The ith entry of A~e is [ai1 ai2 · · · ain ]~e = to the eigenvector ~e.

n X

aij = 1, so A~e = ~e and λ = 1 is an eigenvalue of A, corresponding

j=1

310

Section 7.2 7.2.30 a Let vi be the largest component of the vector ~v , that is, vi ≥ vj for j = 1, . . . , n. Then the ith component of n n n X X X aij vi = vi aij vi = aij vj ≤ A ~v is λvi = j=1

j=1

↑

vj ≤ v i

j=1

n X

↑

aij = 1

j=1

We can conclude that λvi ≤ vi , and therefore λ ≤ 1, as claimed. Also note that if ~v is not a multiple of the n n X X aij vi and therefore aij vj < eigenvector ~e discussed in Exercise 29, then vj < vi for some index j, so that j=1

j=1

λ < 1.

b Let vi be the component of ~v with the largest absolute . . , n. Then the is, |vi | ≥ |vj | for j = 1, 2, . value, that n n n n X X X X aij |vi | = aij |vi | = aij |vj | ≤ aij vj ≤ absolute value of the ith component of A~v is |λkvi | = j=1

j=1

j=1

j=1

|vi | so that |λkvi | ≤ |vi | and |λ| ≤ 1, as claimed.

7.2.31 Since A and AT have the same eigenvalues (by Exercise 22), Exercise 29 states that λ = 1 is an eigenvalue of A,and Exercise 30 says that |λ| ≤ 1 for all eigenvalues λ. Vector ~e need not be an eigenvector of A; consider 0.9 0.9 A= . 0.1 0.1 7.2.32 fA (λ) = −λ3 + 3λ + k. The eigenvalues of A are the solutions of the equation −λ3 + 3λ + k = 0, or, λ3 − 3λ = k. Following the hint, we graph the function g(λ) = λ3 − 3λ as shown in Figure 7.14. We use the derivative f ′ (λ) = 3λ2 − 3 to see that g(λ) has a global minimum at (1, −2) and a global maximum at (−1, 2). To count the eigenvalues of A, we need to find out how many times the horizontal line y = k intersects the graph of g(λ). In Figure 7.14, we see that there are three solutions if k satisfies the inequality 2 > k > −2, two solutions if k = 2 or k = −2, and one solution if |k| > 2.

(−1, 2)

g(λ) = λ3 − 3λ (1, − 2)

Figure 7.14: for Problem 7.2.32. 7.2.33 a fA (λ) = det(A − λI3 ) = −λ3 + cλ2 + bλ + a 311

Chapter 7

0 b By part a, we have c = 17, b = −5 and a = π, so M = 0 π

1 0 0 1 . −5 17

7.2.34 Consider the possible graphs of fA (λ) assuming that it has 2 distinct real roots.

(1 – λ) (–2 – λ) (λ2 + 1)

Figure 7.15: for Problem 7.2.34. 1 0 Algebraic multiplicity of each eigenvalue is 1. Example: 0 0

0 −2 0 0

0 0 0 1

0 0 . See Figure 7.15. −1 0

(–2 – λ)2 (1 – λ)2

Figure 7.16: for Problem 7.2.34. −2 0 Algebraic multiplicity of each eigenvalue is 2. Example: 0 0

Algebraic multiplicity of λ1 is 1, and of λ2 is 3. −2 0 0 0 0 1 0 0 Example: . See Figure 7.17. 0 0 1 0 0 0 0 1 0 −1 0 1 7.2.35 A = 0 0 0 0

0 0 0 0 , with fA (λ) = (λ2 + 1)2 0 −1 1 0 312

0 −2 0 0

0 0 1 0

0 0 . See Figure 7.16. 0 1

Section 7.2

(– 2 – λ) (1 – λ)3

Figure 7.17: for Problem 7.2.34.

7.2.36 Let A =

B

0 B

0

..

. B

where B = 0 1

−1 , fA (λ) = (λ2 + 1)n . 0

7.2.37 We can write fA (λ) = (λ − λ0 )2 g(λ), for some polynomial g. The product rule for derivatives tells us that fA′ (λ) = 2(λ − λ0 )g(λ) + (λ − λ0 )2 g ′ (λ), so that fA′ (λ0 ) = 0, as claimed. 7.2.38 By Theorem 7.2.4, the characteristic polynomial of A is fA (λ) = λ2 − 5λ − 14 = (λ − 7)(λ + 2), so that the eigenvalues are 7 and -2. ae + bg − − − e f a b = ae + bg + cf + dh. =tr 7.2.39 tr(AB) =tr − − − cf + dh g h c d e f a b ea + f c − − − tr(BA) =tr =tr = ea + f c + gb + hd. So they are equal. g h c d − − − gb + hd 7.2.40 Let the entries of A be aij and the entries of B be bij . Now, tr(AB) = (a11 b11 + a12 b21 + · · · + a1n bn1 ) + (a21 b12 + · · · + a2n bn2 ) + · · · + (an1 b1n + · · · +ann bnn ). This is the sum of all products of the form aij bji . We see that tr(BA) = (b11 a11 + · · · + b1n an1 )+· · ·+(bn1 a1n + · · · + bnn ann ) , which also is the sum of all products of the form bji aij = aij bji . Thus, tr(AB) = tr(BA). 7.2.41 So there exists an invertible S such that B = S −1 AS, and tr(B) =tr(S −1 AS) =tr((S −1 A)S). By Exercise 40, this equals tr(S(S −1 A)) =tr(A). 7.2.42 tr (A + B)2 = tr(A2 + AB + BA + B 2 ) = tr(A2 ) + tr(AB) + tr(BA) + tr(B 2 ).. By Exercise 40, tr(AB) = tr(BA). Thus, tr (A + B)2 = tr(A2 ) + 2tr(BA) + tr(B 2 ) = tr(A2 ) + tr(B 2 ), since BA = 0.

7.2.43 tr(AB − BA) =tr(AB)−tr(BA) =tr(AB)−tr(AB) = 0, but tr(In ) = n, so no such A, B exist. We have used Exercise 40.

7.2.44 No, there are no such matrices A and B. We will argue indirectly, assuming that invertible matrices A and 313

Chapter 7 B with AB − BA = A do exist. Then AB = BA + A = (B + In )A, and ABA−1 = B + In . Using Exercise 41, we see that tr(B) = tr(ABA−1 ) = tr(B + In ) = tr(B) + n, a contradiction. 7.2.45 fA (λ) = λ2 −tr(A)λ+det(A) = λ2 − 2λ + (−3 − 4k). We want fA (5) = 25 − 10 − 3 − 4k = 0, or, 12 − 4k = 0, or k = 3. 7.2.46 a λ21 + λ22 = (λ1 + λ2 )2 − 2λ1 λ2 = (trA)2 − 2 det(A) = (a + d)2 − 2(ad − bc) = a2 + d2 + 2bc. b Based on part (a), we need to show that a2 + d2 + 2bc ≤ a2 + b2 + c2 + d2 , or 2bc ≤ b2 + c2 , or 0 ≤ (b − c)2 . But the last inequality is obvious. c By parts (a) and (b), the equality λ21 + λ22 = a2 + b2 + c2 + d2 holds if (and only if) 0 = (b − c)2 , or b = c. Thus equality holds for symmetric matrices A. 7.2.47 Let M = [ ~v1

~v2 ]. We want A[ ~v1

~v2 ] = [ ~v1

must be nonzero, 2 or 3 must be an eigenvalue of A.

2 0 , or, [ A~v1 ~v2 ] 0 3

A~v2 ] = [ 2~v1

3~v2 ]. Since ~v1 or ~v2

7.2.48 Let S = [~v1 ~v2 ]. Then AS = [A~v1 A~v2 ] and SD = [2~v1 3~v2 ], so that ~v1 must be an eigenvector with eigenvalue 2, and ~v2 must be an eigenvector with eigenvalue 3. Thus, both 2 and 3 must be eigenvalues of A. 7.2.49 As in problem 47, such an M will exist if A has an eigenvalue 2, 3 or 4. 7.2.50 a If f (x) = x3 + 6x − 20 then f ′ (x) = 3x2 + 6 so f ′ (x) > 0 for all x, i.e. f is always increasing, hence has only one real root. b If v 3 − u3 = 20 and vu = 2 then (v − u)3 + 6(v − u) = v 3 − 3v 2 u + 3vu2 − u3 + 6(v − u) = v 3 − u3 − 3vu(v − u) + 6(v − u) = 20 − 6(v − u) + 6(v − u) = 20 Hence x = v − u satisfies the equation x3 + 6x = 20. c The second equation tells us that u =

2 v

or u3 =

8 v3 .

Substituting into the first equation we find that

8 v3

= 20, or, (v 3 )2 − 8 = 20v 3 or (v 3 )2 − 20v 3 − 8 = 0, with solutions p √ √ √ √ 3 v 3 = 20± 400+32 = 10 ± 108 = 10 ± 6 3 and v = 10 ± 108. 2 p √ √ 3 Now u3 = v 3 − 20 = −10 ± 108 and u = −10 ± 108. v3 −

d Let v =

r 3

q 2

+

q

q 2 2

+

p 3 3

Then v 3 − u3 = q and vu = Since x = v − u we have

and u =

q 3

q 2 2

+

r 3

− 2q +

p 3 3

−

q

q 2 2

q 2 2

+

= p3 .

p 3 3 .

x3 + px = v 3 − 3v 2 u + 3vu2 − u3 + p(v − u) = v 3 − u3 − 3vu(v − u) + p(v − u) 314

Section 7.3 = q − p(v − u) + p(v − u) = q, as claimed. 2 3 If p is negative, the expression 2q + p3 may be negative. Also, the equation x3 + px = q may have more than one solution in this case. e Setting x = t −

a 3

we get t −

a 3 3

+a t−

a 2 3

+b t−

a 3

+ c = 0 or

t3 − at2 + at2 + (linear and constant terms) = 0 or t3 + (linear and constant terms) = 0, as claimed (bring the constant terms to the right-hand side).

Section 7.3

8 1 −2 8 4 = span , E9 = ker = span 2 0 0 0 1

0 7.3.1 λ1 = 7, λ2 = 9, E7 = ker 0 1 4 Eigenbasis: , 0 1 7.3.2 λ1 = 2, λ2 = 0, E2 = span

1 −1 , E0 = span 1 1

−1 1 , Eigenbasis: 1 1 1 3 , E9 = span 7.3.3 λ1 = 4, λ2 = 9, E4 = span 1 −2 1 3 , Eigenbasis: 1 −2

−1 7.3.4 λ1 = λ2 = 1, E1 = span 1

No eigenbasis 7.3.5 No real eigenvalues as fA (λ) = λ2 − 2λ + 2. 7.3.6 λ1,2 =

√ 7± 57 2

Eigenbasis:

3 3 3 3 ≈ , ≈ −2.27 λ2 − 2 5.27 λ1 − 2

7.3.7 λ1 = 1, λ2 = 2, λ3 = 3, eigenbasis: ~e1 , ~e2 , ~e3 1 1 1 7.3.8 λ1 = 1, λ2 = 2, λ3 = 3, eigenbasis: 0 , 1 , 2 0 0 1 315

Chapter 7 −1 0 1 7.3.9 λ1 = λ2 = 1, λ3 = 0, eigenbasis: 0 , 1 , 0 1 0 0 1 0 7.3.10 λ1 = λ2 = 1, λ3 = 0, E1 = span 0 , E0 = span 0 0 1 No eigenbasis

1 1 1 7.3.11 λ1 = λ2 = 0, λ3 = 3, eigenbasis: −1 , 0 , 1 1 −1 0

1 7.3.12 λ1 = λ2 = λ3 = 1, E1 = span 0 , no eigenbasis 0 1 1 0 7.3.13 λ1 = 0, λ2 = 1, λ3 = −1, eigenbasis: 1 , −3 , −1 2 1 0 0 1 0 7.3.14 λ1 = 0, λ1 = λ3 = 1, eigenbasis: 1 , −5 , 2 0 0 1 0 7.3.15 λ1 = 0, λ2 = λ3 = 1, E0 = span 1 . We can use Kyle Numbers to see that 0 1 −2 E1 = ker −3 −4

−1 0 −1 0

2 1 1 = span −1 . 1 2 2

There is no eigenbasis since the eigenvalue 1 has algebraic multiplicity 2, but the geometric multiplicity is only 1.

1 7.3.16 λ1 = 0 (no other real eigenvalues), with eigenvector −1 1 No real eigenbasis

7.3.17 λ1 = λ2 = 0, λ3 = λ4 = 1 0 0 0 1 0 −1 1 0 with eigenbasis , , , 0 0 1 0 1 0 0 0 316

Section 7.3 7.3.18 λ1 = λ2 = 0, λ3 = λ4 = 1, E0 = span(~e1 , ~e3 ), E1 = span(~e2 ) No eigenbasis 7.3.19 Since 1 is the only eigenvalue, with algebraic multiplicity 3, there exists an eigenbasis for A 0 the geometric multiplicity of the eigenvalue 1 is 3 as well, that is, if E1 = R3 . Now E1 = ker 0 0 (and only if) a = b = c = 0.

if (and only if) a b 0 c is R3 if 0 0

If a = b = c = 0 then E1 is 3-dimensional with eigenbasis ~e1 , ~e2 , ~e3 . If a 6= 0 and c 6= 0 then E1 is 1-dimensional and otherwise E1 is 2-dimensional. The geometric multiplicity of the eigenvalue 1 is dim(E1 ). 0 a 0 0 a b 7.3.20 For λ1 = 1, E1 = ker 0 0 c = ker 0 0 1 so if a = 0 then E1 is 2-dimensional, otherwise it is 0 0 0 0 0 1 1-dimensional. 1 −a −b 1 −c so E2 is 1-dimensional. For λ2 = 2, E2 = ker 0 0 0 0

Hence, there is an eigenbasis if a = 0.

1 4 2 2 1 1 , i.e. A = = 2 and A = 7.3.21 We want A such that A 2 6 3 3 2 2 −1 1 4 1 2 5 −2 . = 2 6 6 −2 2 3

2 3

=

1 4 2 6

so A =

The answer is unique.

7 0 7.3.22 We want A such that A~e1 = 7~e1 and A~e2 = 7~e2 hence A = . 0 7 7.3.23 λ1 = λ2 = 1 and E1 = span(~e1 ), hence there is no eigenbasis. The matrix represents a shear parallel to the x-axis. 2 2 a b a b , or 2a + b = 2, 2c + d = 1. This condition is satisfied = . First we want 7.3.24 Let A = 1 1 c d c d a 2 − 2a by all matrices of the form A = . Next, we want there to be no other eigenvalue, besides 1, so that c 1 − 2c 1 must have an algebraic multiplicity of 2. We want the characteristic polynomial to be (λ−1)2 = λ2−2λ+1, so that the trace must be 2, and a+(1−2c) = 2, 1 + 2c −4c . or, a = 1 + 2c. Thus we want a matrix of the form A = c 1 − 2c 2 instead of E1 = R2 . This means that we must exclude the case Finally, we have to make sure the E1 = span 1 1 + 2c −4c A = I2 . In order to ensure this, we state simply that A = , where c is any nonzero constant. c 1 − 2c 317

Chapter 7

−λ 1 7.3.25 If λ is an eigenvalue of A, then Eλ = ker(A − λI3 ) = ker 0 −λ a b

0 1 . c−λ

The second and third columns of the above matrix aren’t parallel, hence Eλ is always 1-dimensional, i.e., the geometric multiplicity of λ is 1.

7.3.26 Note that fA (0) = det(A − 0I6 ) = det(A) is negative. Since lim fA (λ) = ∞, there must be a positive root, λ→∞

by the Intermediate Value Theorem (see Exercise 2.2.47c). Therefore, the matrix A has a positive eigenvalue. See Figure 7.18.

Figure 7.18: for Problem 7.3.26. 7.3.27 By Theorem 7.2.4, we have fA (λ) = λ2 − 5λ + 6 = (λ − 3)(λ − 2) so λ1 = 2, λ2 = 3. 7.3.28 Since Jn (k) is triangular, its 0 0 . Ek = ker(Jn (k) − kIn ) = ker .. . . . 0

eigenvalues are its diagonal entries, hence its only eigenvalue is k. Moreover, 1 0 ··· 0 .. 0 1 . .. .. = span (~e1 ). . 0 . .. .. 1 . . 0 0 0

The geometric multiplicity of k is 1 while its algebraic multiplicity is n.

7.3.29 Note that r is the number of nonzero diagonal entries of A, since the nonzero columns of A form a basis of im(A). Therefore, there are n − r zeros on the diagonal, so that the algebraic multiplicity of the eigenvalue 0 is n − r. It is true for any n × n matrix A that the geometric multiplicity of the eigenvalue 0 is dim(ker(A)) = n − rank(A) = n − r. 7.3.30 Since A is triangular, fA (λ) = (a11 − λ)(a22 − λ) · · · (amm − λ)(0 − λ)n−m . Hence the algebraic multiplicity of λ = 0 is (n − m). Also note that the rank of A is at least m, since the first m columns of A are linearly independent. Therefore, the geometric multiplicity of the eigenvalue 0 is dim(ker(A)) = n − rank(A) ≤ n − m. 7.3.31 They must be the same. For if they are not, by Theorem 7.3.7, the geometric multiplicities would not add up to n. 318

Section 7.3 7.3.32 Recall that a matrix and its transpose have the same rank (Theorem 5.3.9c). The geometric multiplicity of λ as an eigenvalue of A is dim(ker(A − λIn )) = n − rank(A − λIn ). The geometric multiplicity of λ as an eigenvalue of AT is dim(ker(AT − λIn )) = dim(ker(A − λIn )T ) = n − rank(A − λIn )T = n − rank(A − λIn ). We can see that the two multiplicities are the same. 7.3.33 If S −1 AS = B, then S −1 (A − λIn )S = S −1 (AS − λS) = S −1 AS − λS −1 S = B − λIn . 7.3.34 a If ~x is in the kernel of B, then AS~x = SB~x = S~0 = ~0, so that S~x is in ker(A). b T is clearly linear, and the transformation R(~x) = S −1 ~x is the inverse of T (if ~x is in the kernel of B, then S −1 ~x is in the kernel of A, by part (a)). c The equation nullity(A) = nullity(B) follows from part (b); the equation rank(A) = rank(B) then follows from the rank-nullity theorem (Theorem 3.3.7). 7.3.35 No, since the two matrices have different eigenvalues (see Theorem 7.3.6c). 7.3.36 No, since the two matrices have different traces (see Theorem 7.3.6.d) 7.3.37 a A~v · w ~ = (A~v )T w ~ = (~v T AT )w ~ = (~v T A)w ~ = v T (Aw) ~ = ~v · Aw ~ ↑ A symmetric b Assume A~v = λ~v and Aw ~ = αw ~ for λ 6= α, then (A~v ) · w ~ = (λ~v ) · w ~ = λ(~v · w), ~ and ~v · Aw ~ = ~v · αw ~ = α(~v · w). ~ By part a, λ(~v · w) ~ = α(~v · w) ~ i.e., (λ − α)(~v · w) ~ = 0. Since λ 6= α, it must be that ~v · w ~ = 0, i.e., ~v and w ~ are perpendicular. 7.3.38 Note that fA (0) = det(A − 0I3 ) = det(A) = 1. Since lim fA (λ) = −∞, the polynomial fA (λ) must have a positive root λ0 , by the Intermediate Value Theorem λ→∞

in single variable calculus. In other words, the matrix A will have a positive eigenvalue λ0 . Since A is orthogonal, this eigenvalue λ0 will be 1, by Theorem 7.1.2. This means that there is a nonzero vector ~v in R3 such that A~v = 1~v = ~v , as claimed. See Figure 7.19. 7.3.39 a There are two eigenvalues, λ1 = 1 (with E1 = V ) and λ2 = 0 (with E0 = V ⊥ ). Now geometric multiplicity(1) = dim(E1 ) = dim(V ) = m, and geometric multiplicity(0) = dim(E0 ) = dim(V ⊥ ) = n − dim(V ) = n − m. Since geometric multiplicity(λ) ≤ algebraic multiplicity(λ), by Theorem 7.3.7, and the algebraic multiplicities cannot add up to more than n, the geometric and algebraic multiplicities of the eigenvalues are the same here. 319

Chapter 7

fA (λ) 1 λ0 = 1

Figure 7.19: for Problem 7.3.38. b Analogous to part a: E1 = V and E−1 = V ⊥ . geometric multiplicity(1) = algebraic multiplicity(1) = dim(V ) = m, and geometric multiplicity(−1) = algebraic multiplicity(−1) = dim(V ⊥ ) = n − m.

a b 7.3.40 The matrix of the dynamical system is A = b a

Hence, λ1,2 = a ± b, and the respective eigenvectors are Since ~x(0) =

3 1 2

=

7 4

1 − 1

5 4

so fA (λ) = (a − λ)2 − b2 . 1 −1 and . 1 1

1 −1 5 7 t 1 t + 4 (a − b) . , by Theorem 7.1.3, ~x(t) = 4 (a + b) 1 −1 1

Note that a − b is between 0 and 1, so that the second summand in the formula above goes to ~0 as t goes to infinity. Qualitatively different outcomes occur depending on whether a + b exceeds 1, equals 1, or is less than 1. See Figure 7.20.

Figure 7.20: for Problem 7.3.40. 1 2 9 7.3.41 The eigenvalues of A are 1.2, −0.8, −0.4 with eigenvectors 6 , −2 , −2 . 2 1 2 320

Section 7.3 2 9 1 2 9 Since ~x0 = 50 6 + 50 −2 + 50 −2 we have ~x(t) = 50(1.2)t 6 + 50(−0.8)t −2 + 1 2 2 1 2 1 50(−0.4)t −2 , so, as t goes to infinity, j(t) : n(t) : a(t) approaches the proportion 9 : 6 : 2. 2 50 1 0.8 10 C(t + 1) , . A has eigenvectors ,A = = 1 0 0 1 1 C(0) 0 0 1 50 corresponding to λ1 = 0.8 and λ2 = 1. Since = , and = −50 + , we have C(t) = 1 1 1 0 1 t −50(0.9) + 50, hence in the long run, there will be 50 spectators. The graph of C(t) looks similar to the graph in Figure 7.21.

7.3.42 C(t + 1) = 0.8C(t) + 10 so if A

C(t) 1

Figure 7.21: for Problem 7.3.42.

0 1 7.3.43 a A = 12 1 0 1 1

1 1 0

7 7.6660156 b After 10 rounds, we have A10 11 ≈ 7.6699219 . 5 7.6640625 7.66666666667 7 After 50 rounds, we have A50 11 ≈ 7.66666666667 . 7.66666666667 5 c The eigenvalues of A are 1 and − 21 with −1 0 1 E1 = span 1 and E− 12 = span 1 , −1 2 −1 1 1 −1 0 t t so ~x(t) = 1 + c30 1 + − 21 1 + − 21 c30 −1 . 1 2 −1 After 1001 rounds, Alberich will be ahead of Brunnhilde by

1 1001 2

, so that Carl needs to beat Alberich to 1001 win the game. A straightforward computation shows that c(1001) − a(1001) = 21 (1 − c0 ); Carl wins if this quantity is positive, which is the case if c0 is less than 1. Alternatively, observe that the ranking of the players is reversed in each round: whoever is first will be last after the next round. Since the total number of rounds is odd (1001), Carl wants to be last initially to win the game; he wants to choose a smaller number than both Alberich and Brunnhilde. 321

Chapter 7 7.3.44 a a11 = 0.7 means that only 70% of the pollutant present in Lake Sils at a given time is still there a week later; some is carried down to Lake Silvaplana by the river Inn, and some is absorbed or evaporates. The other diagonal entries can be interpreted analogously: a21 = 0.1 means that 10% of the pollutant present in Lake Sils at any given time can be found in Lake Silvaplana a week later, carried down by the river Inn. The significance of the coefficient a32 = 0.2 is analogous; a31 = 0 means that no pollutant is carried down from Lake Sils to Lake St. Moritz in just one week. The matrix is lower triangular since no pollutant is carried from Lake Silvaplana to Lake Sils. The river Inn would have to flow the other way. b The eigenvalues of A are 0.8, 0.6, 0.7, with corresponding eigenvectors 1 0 0 0 , 1 , 1 . −2 −1 1 100 0 0 1 ~x(0) = 0 = 100 0 − 100 1 + 100 1 , 0 1 −1 −2 1 0 0 so ~x(t) = 100(0.8)t 0 − 100(0.6)t 1 + 100(0.7)t 1 or −2 −1 1 x1 (t) = 100(0.7)t

x2 (t) = 100(0.7)t − 100(0.6)t x3 (t) = 100(0.8)t + 100(0.6)t − 200(0.7)t . See Figure 7.22.

Figure 7.22: for Problem 7.3.44b. Using calculus, we find that the function x2 (t) = 100(0.7)t − 100(0.6)t reaches its maximum at t ≈ 2.33. Keep in mind, however, that our model holds for integer t only.

0.1 7.3.45 a A = 0.4 A ~b b B= 0 1

1 0.2 ~ ,b = 2 0.3

322

Section 7.3

c The eigenvalues of A are 0.5 and −0.1 with associated eigenvectors

1 1 . and −1 2

~v ~v ~v The eigenvalues of B are 0.5, −0.1, and 1. If A~v = λ~v then B =λ so is an eigenvector of B. 0 0 0 2 −(A − I2 )−1~b Furthermore, 4 is an eigenvector of B corresponding to the eigenvalue 1. Note that this vector is . 1 1 2 1 1 x1 (0) d Write ~y (0) = x2 (0) = c1 2 + c2 −1 + c3 4 . 1 0 0 1

Note that c3 = 1.

2 2 1 1 2 t→∞ t→∞ t t Now ~y (t) = c1 (0.5) 2 + c2 (−0.1) −1 + 4 −→ 4 so that ~x(t) −→ . 4 1 1 0 0 7.3.46 a T1 (t + 1) = 0.6T1 (t) + 0.1T2 (t) + 20 T2 (t + 1) = 0.1T1 (t) + 0.6T2 (t) + 0.1T3 (t) + 20 T3 (t + 1) = 0.1T2 (t) + 0.6T3 (t) + 40 20 0.6 0.1 0 so A = 0.1 0.6 0.1 , ~b = 20 40 0 0.1 0.6 b B=

A ~b 0 1

0 70.86 0 93.95 c ~y (10) = B 10 ≈ 0 120.56 1 1 0 74.989 0 99.985 ~y (30) = B 30 ≈ 0 124.989 1 1 75 100 ~y (t) seems to approach as t → ∞ 125 1

d The eigenvalues of A are λ1 ≈ 0.45858, λ2 = 0.6, λ3 ≈ 0.74142 so the eigenvalues of B are λ1 ≈ 0.45858, λ2 = 0.6, λ3 ≈ 0.74142, λ4 = 1. 323

Chapter 7 ~v1 ~v ~v If ~v1 , ~v2 , ~v3 are eigenvectors of A (with A~vi = λi~vi ), then , 2 , 3 are corresponding eigenvectors 0 0 0 75 100 of B. Furthermore, is an eigenvector if B with eigenvalue 1. Since λ1 , λ2 , λ3 are all less than 1, 125 1 75 lim ~x(t) = 100 , as in Exercise 45. t→∞ 125 1 r(t) 2 7.3.47 a If ~x(t) = p(t) , then ~x(t + 1) = A~x(t) with A = 12 w(t) 0

1 4 1 2 1 4

0 1 2 1 2

.

1 1 1 The eigenvalues of A are 0, 12 , 1 with eigenvectors −2 , 0 , 2 . 1 −1 1 1 1 1 1 1 t Since ~x(0) = 0 = 14 −2 + 21 0 + 41 2 , ~x(t) = 12 21 0 + 1 −1 1 −1 0

1 1 2 for t > 0. 4 1

b As t → ∞ the ratio is 1 : 2 : 1 (since the first term of ~x(t) drops out). 7.3.48 a We are told that a(t + 1) = a(t) + j(t)

1 1 j(t + 1) = a(t), so that A = . 1 0 λ2 λ1 . and 1 1 λ λ 1 λ1 λ2 = √15 Since ~x(0) = − √15 we have ~x(t) = √15 (λ1 )t 1 − √15 (λ2 )t 2 , i.e. 1 1 0 1 1

b fA (λ) = λ(λ − 1) − 1 = λ2 − λ − 1 so λ1,2 =

a(t) =

√1 ((λ1 )t+1 5

j(t) =

√1 ((λ1 )t 5

c As t → ∞,

a(t) j(t)

√ 1± 5 2

with eigenvectors

− (λ2 )t+1 )

− (λ2 )t ). → λ1 =

√ 1+ 5 2 ,

since |λ2 | < 1.

7.3.49 This “random” matrix A = [~0 ~v2 · · · ~vn ] is unlikely to have any zeros above the diagonal. In this case, the columns ~v2 , . . . , ~vn will be linearly independent (none of them is redundant), so that rank(A) = n − 1 and geometric multiplicity(0) = dim(ker(A)) = n − rank(A) = 1. Alternatively, you can argue in terms of rref(A). √ 1 −1 7.3.50 a fA (λ) = (2 − λ)2 − 3 so λ1,2 = 2 ± 3 (or approximately 3.73 and 0.27) with eigenvectors √ and √ . 3 3 See Figure 7.23. 324

Section 7.3

Figure 7.23: for Problem 7.3.50a.

b The trajectory starting at

0 0 = is above the line Eλ1 , so that At 1 1

(second column of At ) has a slope of more than

√ √ 3, for all t. Applying this to t = 6 gives the estimate 3 <

1351 780 .

1 t 1 Likewise, the trajectory starting at = is below Eλ1 , so that A 1 1 (sum of the two columns of √ At ) has a slope of less than 3. Applying this to t = 4 gives

265 153

<

√ 3.

c det(A6 ) = (det A)6 = 1 and det(A6 ) = 13512 − 780 · 2340, so that 13512 − 780 · 2340 = 1. Dividing both sides by 1351 · 780 we obtain Now note that Therefore

1351 780

2340 1351

−

d The slope of A6

is the slope of A6

√ 3<

1351 780

−

2340 1351

1351 780

−

=

1 780·1351

< 10−6 .

√ 1 , which is less than 3. 0

< 10−6 , as claimed.

√ 2131 1 is less than 3, i.e. = 3691 1

2340 1351

−λ 7.3.51 fA (λ) = det(A − λI3 ) = det 1 0

3691 2131

<

√

3.

0 a −λ b = −λ3 + cλ2 + bλ + a. 1 c−λ

n 7.3.52 The characteristic polynomial is fA (λ) = (−1) λn − an−1 λn−1 − ... − a1 λ − a0 . We can prove this by induction, using Laplace expansion along the first row: n+1 fA (λ) = det (A − λIn ) = −λ det (A − λIn )11 + (−1) a0 det (A − λIn )1n n−1 n+1 = −λ (−1) λn−1 − an−1 λn−2 − ... − a2 λ − a1 + (−1) a0 n n n−1 = (−1) λ − an−1 λ − ... − a1 λ − a0 . Note that (A − λIn )1n is upper triangular with all 1’s on the diagonal, so that det (A − λIn )1n = 1. 325

Chapter 7

7.3.53 a B =

0 1 0 0 0

0 a ∗ 0 b ∗ 1 c ∗ 0 0 ∗ 0 0 ∗

∗ ∗ ∗ ∗ ∗

.

0 0 a b Let B11 = 1 0 b and B22 be the diagonal blocks of the matrix B we found in part (a). Since A is similar 0 1 c to B, we have fA (λ) = fB (λ) = fB22 (λ) fB11 (λ) = fB22 (λ) −λ3 + cλ2 + bλ + a , by Exercise 51. Now we observe that h (λ) = fB22 (λ)is a quadratic polynomial, completing our proof. c From part (b) we know that fA (A) = h(A) −A3 + cA2 + bA + aI5 . The given equation A3~v = a~v + bA~v + cA2~v can be written as −A3 + cA2 + bA + aI5 ~v = ~0, implying that fA (A) ~v = h(A) −A3 + cA2 + bA + aI5 ~v = ~0, as claimed.

7.3.54 a. B11

=

0 1 0 ... 0 0

0 0 1 ... 0 0

0 0 0 ... 0 0

... ... ... ... ... ...

0 a0 0 a1 0 a2 ... ... 0 am−2 1 am−1

, B21 = 0.

b. We find fA (λ) = fB (λ) = fB22 (λ) fB11 (λ) as in part (b) of Exercise 53. Now matrix B11 has the form discussed in Exercise 52, where we show that m that det (B11 ) = (−1) λm − am−1 λm−1 − ... − a1 λ − a0 . It follows m fA (λ) = (−1) fB22 (λ) λm − am−1 λm−1 − ... − a1 λ − a0 , as claimed. c. Using the same approach as in Exercise 53, part (c ), we can write the given equation for Am~v as m m−1 ~ A − am−1 A − ... − a1 A − a0 In ~v = 0. Thus m fA (A) ~v = (−1) fB22 (A) Am − am−1 Am−1 − ... − a1 A − a0 In ~v = ~0, as claimed. d. Our work in parts (a) through (c) shows that fA (A) ~v = ~0 for all ~v in Rn , meaning that fA (A) = 0, the zero matrix.

Section 7.4 7.4.1 Matrix A is diagonal already, so it’s certainly diagonalizable. Let S = I2 . 7.4.2 Diagonalizable. The eigenvalues are 2,3, with associated eigenvectors 2 0 −1 . S AS = D = 0 3 326

1 1 1 1 , . If we let S = , then 0 1 0 1

Section 7.4

7.4.3 Diagonalizable. The eigenvalues are 0,3, with associated eigenvectors 0 0 . then S −1 AS = D = 0 3 7.4.4 Diagonalizable. The eigenvalues are 0,7, with associated eigenvectors 0 0 then S −1 AS = D = . 0 7

−1 1 1 −1 , . If we let S = , 1 2 2 1

−2 1 −2 1 , . If we let S = , 1 3 1 3

7.4.5 Fails to be diagonalizable. There is only one eigenvalue, 1, with a one-dimensional eigenspace. 7.4.6 Fails to be diagonalizable. There is only one eigenvalue, 2, with a one-dimensional eigenspace. 4 −1 4 −1 7.4.7 Diagonalizable. The eigenvalues are 2,−3, with associated eigenvectors , . If we let S = , 1 1 1 1 2 0 then S −1 AS = D = . 0 −3 7.4.8 Diagonalizable. The eigenvalues are 4,−2, with associated eigenvectors 4 0 −1 . then S AS = D = 0 −2

1 −1 −1 1 , . If we let S = , 1 1 1 1

7.4.9 Fails to be diagonalizable. There is only one eigenvalue, 1, with a one-dimensional eigenspace. 7.4.10 Fails to be diagonalizable. There is only one eigenvalue, 1, with a one-dimensional eigenspace. 7.4.11 Fails to be diagonalizable. The eigenvalues are 1,2,1, and the eigenspace E1 = ker(A − I3 ) = span(~e1 ) is only one-dimensional. 1 0 −1 7.4.12 Diagonalizable. The eigenvalues are 2,1,1, with associated eigenvectors 0 , 1 , 0 . If we let S = 0 0 1 2 0 0 1 0 −1 0 1 0 , then S −1 AS = D = 0 1 0 . 0 0 1 0 0 1 7.4.13 Diagonalizable. The eigenvalues are

1 0 0

1 0 1 1 1 2 , then S −1 AS = D = 0 2 0 0 0 1

1 1 1 1,2,3, with associated eigenvectors 0 , 1 , 2 . If we let S = 0 0 1 0 0 . 3

0 −1 1 7.4.14 Diagonalizable. The eigenvalues are 3,2,1, with associated eigenvectors 0 , 1 , −1 . If we let 1 0 0 327

Chapter 7

1 −1 1 S = 0 0 0 7.4.15

2 1 0

3 0 0 −1 , then S −1 AS = D = 0 2 0 0 1

2 1 0 Diagonalizable. The eigenvalues are 1, −1, 1, with associated eigenvectors 1 , 1 , 0 . If we let S = 0 0 1 1 0 0 1 0 1 0 , then S −1 AS = D = 0 −1 0 . 0 0 1 0 1

7.4.16 Diagonalizable. The eigenvalues are

2 0 1

0 0 . 1

1 0 3 0 0 1 , then S −1 AS = D = 0 2 1 0 0 0

2 1 0 3,2,1, with associated eigenvectors 0 , 0 , 1 . If we let S = 1 1 0 0 0 . 1

7.4.17 Diagonalizable. 1 1 1 1 1 The eigenvalues are 0,3,0, with associated eigenvectors −1 , 1 , 0 . If we let S = −1 1 0 1 −1 1 0 0 0 0 S −1 AS = D = 0 3 0 . 0 0 0

1 0 , then −1

−1 1 0 7.4.18 Diagonalizable. The eigenvalues are 0,2,1, with associated eigenvectors 0 , 0 , 1 . If we let S = 1 1 0 0 0 0 −1 1 0 0 0 1 , then S −1 AS = D = 0 2 0 . 0 0 1 1 1 0 7.4.19 Fails to be diagonalizable. The eigenvalues are 1,0,1, and the eigenspace E1 = ker(A − I3 ) = span(~e1 ) is only one-dimensional. 0 1 −1 7.4.20 Diagonalizable. The eigenvalues are 1,2,0, with associated eigenvectors 1 , 2 , 0 . If we let S = 0 1 1 0 1 −1 1 0 0 1 2 0 , then S −1 AS = D = 0 2 0 . 0 1 1 0 0 0 7.4.21 Diagonalizable for all values of a, since there are always two distinct eigenvalues, 1 and 2. See Theorem 7.4.3. 7.4.22 Diagonalizable except if b = 1 and a 6= 0. (In that case we have only one eigenvalue, 1, with a one-dimensional eigenspace.). 328

Section 7.4 7.4.23 Diagonalizable for positive a. The characteristic polynomial is (λ − 1)2 − a, so that the eigenvalues are √ λ = 1 ± a. If a is positive, then we have two distinct real eigenvalues, so that the matrix is diagonalizable. If a is negative, then there are no real eigenvalues. If a is 0, then 1 is the only eigenvalue, with a one-dimensional eigenspace. 7.4.24 Diagonalizable for all values polynomial is λ2 − (a + c)λ + ac − b2 , so that √ √ of a, b, and c. The characteristic a+c± (a−c)2 +4b2 a+c± (a+c)2 −4(ac−b2 ) = . Note that the expression whose square root the eigenvalues are λ = 2 2 we take (the “discriminant”) is always positive or 0, since it is the sum of two squares. If the discriminant is positive, then we have two distinct real eigenvalues, and everything is fine. The discriminant is 0 only if a = c and b = 0. In that case the matrix is diagonal already, and certainly diagonalizable as well. 7.4.25 Diagonalizable for all values of a, b, and c, since we have three distinct eigenvalues, 1, 2, and 3. 7.4.26 The eigenvalues are 1, 2, 1, and thematrix is diagonalizable if (and only if) the eigenspace E1 is two0 a b 0 1 c dimensional. Now E1 = ker(A − I3 ) = ker 0 1 c = ker 0 0 b − ac is two-dimensional if (and only 0 0 0 0 0 0 if) b − ac = 0. Thus the matrix is diagonalizable if and only if b = ac. 7.4.27 Diagonalizable only if a = b = c = 0. Since 1 is the only eigenvalue, it is required that E1 = R3 , that is, the matrix must be the identity matrix. 7.4.28 Diagonalizable for positive values of a. The characteristic polynomial is −λ3 + aλ = −λ(λ2 − a). If a is √ positive, then we have three distinct real eigenvalues, 0, ± a, so that the matrix will be diagonalizable. If a is negative or 0, then 0 is the only real eigenvalue, and the matrix fails to be diagonalizable. 7.4.29 Not diagonalizable for any a. The characteristic polynomial is −λ3 + a, so that there is only one real √ eigenvalue, 3 a, for all a. Since the corresponding eigenspace isn’t all of R3 , the matrix fails to be diagonalizable.

0 0 a 7.4.30 First we observe that all the eigenspaces of A = 1 0 3 are one-dimensional, regardless of the value of 0 1 0 1 0 ∗ a, since rref(A − λI3 ) is of the form 0 1 ∗ for all λ. Thus A is diagonalizable if and only if there are three 0 0 ∗ distinct real eigenvalues. The characteristic polynomial of A is −λ3 + 3λ + a. Thus the eigenvalues of A are the solutions of the equation λ3 − 3λ = a. See Figure 7.24 with the function f (λ) = λ3 − 3λ; using calculus, we find the local maximum f (−1) = 2 and the local minimum f (1) = −2. To count the distinct eigenvalues of A, we have to examine how many times the horizontal line y = a intersects the graph of f (λ). The answer is three if |a| < 2, two if a = ±2, and one if |a| > 2. Thus A is diagonalizable if and only if |a| < 2, that is, −2 < a < 2. 1 2 are −1 and 5, with associated 7.4.31 In Example 2 of Section 7.3 we see that the eigenvalues of A = 4 3 1 1 1 1 −1 0 −1 eigenvectors and . If we let S = , then S AS = D = . −1 2 −1 2 0 5

329

Chapter 7

Thus A = SDS

−1

t

t

and A = SD S

−1

=

1 3

=

1 3

1 −1

1 2

(−1)t 0

2(−1)t + 5t 2(5t ) − 2(−1)t

0 55

2 1

−1 1

(−1)t+1 + 5t 2(5t ) + (−1)t

1 2 4 −2 . If we let S = and are 3 and 2, with associated eigenvectors 7.4.32 The eigenvalues of A = 1 1 1 t 1 2 1 3 0 2 1 3 0 1 −1 , then S −1 AS = D = . Thus A = SDS −1 and At = SDt S −1 = = 0 2 1 1 0 2t −1 2 1 t1 2(3 ) − 2t 2t+1 − 2(3t ) . 3t − 2t 2t+1 − 3t

1 2 −2 1 7.4.33 The eigenvalues of A = are 0 and 7, with associated eigenvectors and . If we let S = 3 6 1 3 0 0 −3 1 0 0 −2 1 1 −2 1 −1 t t −1 −1 = . Thus A = SDS and A = SD S = 7 , then S AS = D = 0 7t 1 2 1 3 0 7 1 t3 t 7 2(7 ) 1 t−1 A. We can find the same result more directly by observing that A2 = 7A. 7 3(7t ) 6(7t ) = 7

−1 1 −1 1 7.4.34 The eigenvalues of A are 1/4 and 1, with associated eigenvectors and . If we let S = , 1 2 1 2 −2 1 (1/4)t 0 −1 1 1/4 0 = . Thus A = SDS −1 and At = SDt S −1 = 13 then S −1 AS = D = 1 1 0 1 1 2 0 1 t 1 − (1/4)t 1 1 + 2(1/4) · 3 2 − 2(1/4)t 2 + (1/4)t −1 6 7.4.35 Matrix has the eigenvalues 3 and 2. If ~v and w ~ are associated eigenvectors, and if we let S = [~v w], ~ −2 6 3 0 −1 6 3 0 −1 6 . is indeed similar to , so that matrix S= then S −1 0 2 −2 6 0 2 −2 6

−1 −2 3 to the diagonal matrix 0 Theorem 3.4.6.

7.4.36 Yes. The matrices

2 both have the eigenvalues 3 and 2, so that each of them is similar 4 0 −1 6 1 2 , by Algorithm 7.4.4. Thus is similar to , by parts b and c of 2 −2 6 −1 4 1 6 and −1 6

7.4.37 Yes. Matrices A and B have the same characteristic polynomial, λ2 − 7λ + 7, so that they have the same √ λ1 0 7± 21 two distinct real eigenvalues λ1,2 = 2 . Thus both A and B are similar to the diagonal matrix , by 0 λ2 Algorithm 7.4.4. Therefore A is similar to B, by parts b and c of Theorem 3.4.6.

2 0 7.4.38 No. As a counterexample, consider A = 0 2

2 1 and B = . 0 2

7.4.39 The eigenfunctions with eigenvalue λ are the nonzero functions f (x) such that T (f (x)) = f ′ (x) − f (x) = λf (x), or f ′ (x) = (λ + 1)f (x). From calculus we recall that those are the exponential functions of the form 330

Section 7.4 f (x) = Ce(λ+1)x , where C is a nonzero constant. Thus all real numbers are eigenvalues of T , and the eigenspace Eλ is one-dimensional, spanned by e(λ+1)x . 7.4.40 The eigenfunctions with eigenvalue λ are the nonzero functions f (x) such that T (f (x)) = 5f ′ (x) − 3f (x) = λf (x), or f ′ (x) = λ+3 5 f (x). From calculus we recall that those are the exponential functions of the form (λ+3)x/5 f (x) = Ce , where C is a nonzero constant. Thus all real numbers are eigenvalues of T , and the eigenspace Eλ is one-dimensional, spanned by e(λ+3)x/5 . 7.4.41 The nonzero symmetric matrices are eigenmatrices with eigenvalue 2, since L(A) = A + AT = 2A in this T case. The nonzero skew-symmetric matrices have 0, eigenvalue since L(A) = A+ A = A − A = 0. Yes, L 1 0 0 1 0 0 0 1 is diagonalizable, since we have the eigenbasis , , , (three symmetric matrices, 0 0 1 0 0 1 −1 0 and one skew-symmetric one). 7.4.42 The nonzero symmetric matrices are eigenmatrices with eigenvalue 0, since L(A) = A − AT = A − A = 0 in T this case. The nonzero skew-symmetric matrices have eigenvalue 2, since L(A) = A− A = A + A = 2A. Yes, L 1 0 0 1 0 0 0 1 is diagonalizable, since we have the eigenbasis , , , (three symmetric matrices, 0 0 1 0 0 1 −1 0 and one skew-symmetric one). 7.4.43 The nonzero real numbers are “eigenvectors” with eigenvalue 1, and the nonzero imaginary numbers (of the form iy) are “eigenvectors” with eigenvalue −1. Yes, T is diagonalizable, since we have the eigenbasis 1,i. 7.4.44 The nonzero sequence (x0 , x1 , x2 , . . .) is an eigensequence with eigenvalue λ if T (x0 , x1 , x2 , . . .) = (x2 , x3 , x4 , . . .) = λ(x0 , x1 , x2 , . . .) = (λx0 , λx1 , λx2 , . . .). This means that x2 = λx0 , x3 = λx1 , . . . , xn+2 = λxn , . . .. These are the sequences of the form (a, b, λa, λb, λ2 a, λ2 b, . . .), where at least one of the first two terms, a and b, is nonzero. Thus all real numbers λ are eigenvalues of T , and the eigenspace Eλ is two-dimensional, with basis (1, 0, λ, 0, λ2 , 0, . . .), (0, 1, 0, λ, 0, λ2 , . . .). 7.4.45 The nonzero sequence (x0 , x1 , x2 , . . .) is an eigensequence with eigenvalue λ if T (x0 , x1 , x2 , . . .) = (0, x0 , x1 , x2 , . . .) = λ(x0 , x1 , x2 , . . .) = (λx0 , λx1 , λx2 , . . .). This means that 0 = λx0 , x0 = λx1 , x1 = λx2 , . . . , xn = λxn+1 , . . . . If λ is nonzero, then these equations imply that x0 = λ1 0 = 0, x1 = 1 1 λ x0 = 0, x2 = λ x1 = 0, . . . , so that there are no eigensequences in this case. If λ = 0, then we have x0 = λx1 = 0, x1 = λx2 = 0, x2 = λx3 = 0, . . . , so that there aren’t any eigensequences either. In summary: There are no eigenvalues and eigensequences for T . 7.4.46 The nonzero sequence (x0 , x1 , x2 , . . .) is an eigensequence with eigenvalue λ if T (x0 , x1 , x2 , . . .) = (x0 , x2 , x4 , . . .) = λ(x0 , x1 , x2 , . . .) = (λx0 , λx1 , λx2 , . . .). This means that x0 = λx0 , x2 = λx1 , x4 = λx2 , . . . , x2n = λxn , . . . . For each λ, there are lots of eigensequences: we can choose the terms xk for odd k freely and then fix the xk for even k according to the formula x2n = λxn . For example, eigenspace E3 consists of the sequences of the form (x0 = 0, x1 , x2 = 3x1 , x3 , x4 = 9x1 , x5 , x6 = 3x3 , x7 , x8 = 27x1 , x9 , . . .), where x1 , x3 , x5 , x7 , x9 , . . . are arbitrary. Note that all the eigenspaces are infinite-dimensional. The condition x0 = λx0 implies that x0 = 0, except for λ = 1, in which case x0 is arbitrary. 7.4.47 The nonzero even functions, of the form f (x) = a+cx2 , are eigenfunctions with eigenvalue 1, and the nonzero odd functions, of the form f (x) = bx, have eigenvalue −1. Yes, T is diagonalizable, since the standard basis, 331

Chapter 7 1, x, x2 , is an eigenbasis for T . 7.4.48 Apply T to the standard basis: T (1) = 1, T (x) = 2x, and T (x2 ) = (2x)2 = 4x2 . This gives the eigenvalues 1, 2, and 4, with corresponding eigenfunctions 1, x, x2 . Yes, T is diagonalizable, since the standard basis is an eigenbasis for T . 1 −1 1 3 −6 . The eigenvalues of B are 7.4.49 The matrix of T with respect to the standard basis 1, x, x2 is B = 0 0 0 9 1 −1 1 1, 3, 9, with corresponding eigenvectors 0 , 2 , −4 . The eigenvalues of T are 1,3,9, with corresponding 4 0 0 eigenfunctions 1, 2x − 1, 4x2 − 4x + 1 = (2x − 1)2 . Yes, T is diagonalizable, since the functions 1, 2x − 1, (2x − 1)2 from an eigenbasis.

1 −3 9 7.4.50 The matrix of T with respect to the standard basis 1, x, x2 is B = 0 1 −6 . The only eigenvalue of B 0 0 1 1 is 1, with corresponding eigenvector 0 . The only eigenvalue of T is 1 as well, with corresponding eigenfunction 0 f (x) = 1. T fails to be diagonalizable, since there is only one eigenvalue, with a one-dimensional eigenspace. 7.4.51 The nonzero constant functions f (x) = b are the eigenfunctions with eigenvalue 0. If f (x) is a polynomial of degree ≥ 1, then the degree of f (x) exceeds the degree of f ′ (x) by 1 (by the power rule of calculus), so that f ′ (x) cannot be a scalar multiple of f (x). Thus 0 is the only eigenvalue of T , and the eigenspace E0 consists of the constant functions. 7.4.52 Let f (x) = a0 + a1 x + a2 x2 + · · · + an xn , with an 6= 0, be an eigenfunction of T with eigenvalue λ. Then T (f (x)) = x(a1 + 2a2 x + · · · + nan xn−1 ) = a1 x + 2a2 x2 + · · · + nan xn = λ(a0 + a1 x + a2 x2 + · · · + an xn ) = λa0 + λa1 x + λa2 x2 + · · · + λan xn . This means that λa0 = 0, λa1 = a1 , λa2 = 2a2 , . . . , λan = nan . Since we assumed that an 6= 0, we can conclude that λ = n. Now it follows that a0 = a1 = · · · = an−1 = 0, so that the eigenfunctions with eigenvalue n are the nonzero scalar multiples of xn , of the form f (x) = an xn . This makes good sense, since T (xn ) = x(nxn−1 ) = nxn . In summary: The eigenvalues are the integers n = 0, 1, 2, . . ., and the eigenspace En is span (xn ) 7.4.53 Suppose basis D consists of f1 , . . . , fn . We are told that the D-matrix D of T is diagonal; let λ1 , λ2 , . . . , λn be the diagonal entries of D. By Theorem 4.3.3., we know that [T (fi )]D = (ith column of D) = λi~ei , for i = 1, 2, . . . , n, so that T (fi ) = λi fi , by definition of coordinates. Thus f1 , . . . , fn is an eigenbasis for T , as claimed. 7.4.54 Note that A2 = 0, but B 2 6= 0. Since A2 fails to be similar to B 2 , matrix A isn’t similar to B (see Example 7 of Section 3.4). 7.4.55 Let A =

0 1 0 0

and B =

1 0

0 , for example. 0

AB 7.4.56 The hint shows that matrix M = B

0 0

0 is similar to N = B 332

0 ; thus matrices M and N have the BA

Section 7.4 AB − λIn 0 = (−λ)n det(AB − B −λIn λIn ) = (−λ)n fAB (λ). To understnd the second equality, consider Theorem 6.1.5. Likewise, fN (λ) = (−λ)n fBA (λ). It follows that (−λ)n fAB (λ) = (−λ)n fBA (λ) and therefore fAB (λ) = fBA (λ), as claimed.

same characteristic polynomial, by Theorem 7.3.6a. Now fM (λ) = det

AB B

0 0

Im 0

A In

Im 0

A In

7.4.57 Modifying the hint in Exercise 56 slightly, we can write = 0 0 AB 0 0 0 . Thus matrix M = is similar to N = . By Theorem 7.3.6a, matrices M B BA B 0 B BA and N have the same characteristic polynomial. AB − λIm 0 Now fM (λ) = det = (−λ)n det(AB − λIm ) = (−λ)n fAB (λ). To understnd the second B −λIn equality, consider Theorem 6.1.5. Likewise, fN (λ) −λIm 0 = det = (−λ)m fBA (λ). B BA − λIn It follows that (−λ)n fAB (λ) = (−λ)m fBA (λ). Thus matrices AB and BA have the same nonzero eigenvalues, with the same algebraic multiplicities. If mult(AB) and mult(BA) are the algebraic multiplicities of 0 as an eigenvalue of AB and BA, respectively, then the equation (−λ)n fAB (λ) = (−λ)m fBA (λ) implies that n + mult(AB) = m + mult(BA). 7.4.58 a. If ~v is in the image of A, then ~v = Aw ~ for some vector w. ~ Now A~v = A2 w ~ = ~0, showing that ~v is in the kernel of A. b. From part (a) we know that dim imA ≤ dim ker A. Also, dim imA > 0 since A is nonzero, and dim ker A + dim imA = 3 by the rank-nullity theorem. This leaves us with only one possibility, namely, dim imA = 1 and dim ker A = 2. c. Consider a relation c1~v1 + c2~v2 + c3~v3 = ~0. Multiplying both sides with A from the left, keeping in mind that ~v1 and ~v3 are in the kernel, we find c2 A~v2 = c2~v1 = ~0, so c2 = 0. Now c1 = c3 = 0 since ~v1 and ~v3 are independent by construction. 0 1 0 d. We have B = 0 0 0 since A~v1 = ~0, A~v2 = ~v1 and A~v3 = ~0. 0 0 0 0 1 0 7.4.59 Yes, A is similar to B, since both A and B are similar to 0 0 0 , by Exercise 58. 0 0 0

7.4.60 We will use the method outlined in Exercise 58. We start

1 Since ~v1 = (first column of A) = A~e1 , we can let ~v2 = ~e1 = 0 0 333

1 with the vector ~v1 = 2 in the image of A. 3 2 . Finally, we let ~v3 = 1 , another vector in 0

Chapter 7

1 the kernel (use Kyle numbers!). Thus S = 2 3

1 2 0 1 will do the job, but there are other answers. 0 0

7.4.61 First we need to verify that the vectors ~v1 , ..., ~vr , w ~ 1 , ..., w ~ r , ~u1 , ..., ~um are linearly independent. Consider a relation a1~v1 + ... + ar ~vr + b1 w ~ 1 + ... + br w ~ r + c1 ~u1 + ... + cm ~um = ~0. Multiplying both sides with A, we find b1~v1 + ... + br ~vr = ~0, so that b1 = ... = br = 0 since the ~vi are independent by construction. Now the ai and the ci must be 0 as well, since the vectors ~v1 , ..., ~vr , ~u1 , ..., ~um are independent by construction. Secondly, we need to show that 2r + m = n, meaning that we have the right number of vectors to form a basis of Rn. Indeed. ~ i gives us a block n = dim ker A + dim imA = (r + m) + r = 2r + m. Each of the r pairs ~vi , w 0 1 J= along the diagonal of B, since A~vi = ~0 and Aw ~ i = ~vi , while all other entries of matrix B are zero. 0 0 7.4.62 A nonzero function f is an eigenfunction of T with eigenvalue λ if T (f ) = f ′′ + af ′ + bf = λf , or, f ′′ + af ′ + (b − λ)f = 0. By Theorem 4.1.7, this differential equation has a two-dimensional solution space. Thus all real numbers are eigenvalues of T , and all the eigenspaces are two-dimensional. 7.4.63 a We need to solve the differential equation f ′′ (x) = f (x). As in Example 18 of Section 4.1, we will look for exponential solutions. The function f (x) = ekx is a solution if k 2 = 1, or k = ±1. Thus the eigenspace E1 is the span of functions ex and e−x . b We need to solve the differential equation f ′′ (x) = 0. Integration gives f ′ (x) = C, a constant. If we integrate again, we find f (x) = Cx + c, where c is another arbitrary constant. Thus E0 = span(1, x). c The solutions of the differential equation f ′′ (x) = −f (x) are the functions f (x) = a cos(x) + b sin(x), so that E−1 = span(cos x, sin x). See the introductory example of Section 4.1 and Exercise 4.1.58. d Modifying part c, we see that the solutions of the differential equation f ′′ (x) = −4f (x) are the functions f (x) = a cos(2x) + b sin(2x), so that E−4 = span(cos(2x), sin(2x)). 1 1 . Arguing as in Exercise 65, we and 7.4.64 The eigenvalues of A are 1 and 3, with associated eigenvectors 1 0 1 0 0 1 find the basis , for V, so that dim(V ) = 2. 0 0 0 1 7.4.65 Let’s write S in terms of its columns, as S = [ ~v w ~ ]. 5 0 , or, [ A~v Aw ~ ] = [ 5~v −w ~ ] , that is, we want ~v to be in the eigenspace We want A [ ~v w ~ ] = [ ~v w ~] 0 −1 1 1 , so that S must be of the form and E−1 = span E5 , and w ~ in E−1 . We find that E5 = span −1 2 0 1 1 0 0 1 1 0 1 1 b a , and dim(V ) = 2. , . Thus, a basis of the space V is +b =a 0 −1 2 0 0 −1 2 0 −1 2 1 0 1 ~ 7.4.66 For A we find the eigenspaces E1 = span 0 , 1 and E2 = span 1 . If we write S = [~u ~v w], 0 −1 0 334

Section 7.4

1 0 then we want A[~u ~v w] ~ = [~u ~v w] ~ 0 1 0 0 w ~ must be in E2 . The matrices S we seek

1 and a basis of V is 0 0 five.

0 0 0 0 0 0, 1 0 0 0 −1 0

0 ~ = [~u ~v 2w], ~ that is, 0 , or [A~u A~v Aw] 2 0 1 are of the form S = a 0 + b 1 −1 0 0 0 1 0 0 0 0 0 0 0,0 0 0,0 1 0,0 0 0 0 0 0 0 −1 0 0 0

~u and ~v must be in E1 , and 1 0 1 c 0 + d 1 e 1 , 0 −1 0 1 1 . The dimension of V is 0

~ 1, w ~ 2 ). As in Exercise 65, we can see that S must be of the 7.4.67 Let Eλ1 = span(~v1 , ~v2 , ~v3 ) and Eλ2 = span(w form [ ~x1 ~x2 ~x3 ~x4 ~x5 ] where ~x1 , ~x2 and ~x3 are in Eλ1 and ~x4 and ~x5 are in Eλ2 . Thus, we can write ~x1 = c1~v1 + c2~v2 + c3~v3 , for example, or ~x5 = d1 w ~ 1 + d2 w ~ 2. Using Summary 4.1.6, we find a basis: [ ~v1 ~0 ~0 ~0 ~0 ] , [ ~v2 ~0 ~0 ~0 ~0 ] , [ ~v3 ~0 ~0 ~0 ~0 ] , [ ~0 ~v1 ~0 ~0 ~0 ] , [ ~0 ~v2 ~0 ~0 ~0 ] , [ ~0 ~v3 ~0 ~0 ~0 ] , [ ~0 ~0 ~v1 ~0 ~0 ] , [ ~0 ~0 ~v2 ~0 ~0 ] , [ ~0 ~0 ~v3 ~0 ~0 ] , [ ~0 ~0 ~0 w ~ 1 ~0 ] , [ ~0 ~0 ~0 w ~ 2 ~0 ] , [ ~0 ~0 ~0 ~0 w ~ 1 ] , [ ~0 ~0 ~0 ~0 w ~2 ] . Thus, the dimension of the space of matrices S is 3 + 3 + 3 + 2 + 2 = 13. 7.4.68 Let ~v1 , . . . , ~vn be an eigenbasis for A, with A~vi = λi~vi . Arguing as in Exercises 64 through 67, we see that the ith column of S must be in Eλi , so that it must be of the form ci~vi for some scalar ci . The matrices S we seek are of the form S = [c1~v1 . . . cn~vn ], involving the n arbitrary constants c1 , . . . , cn , so that the dimension of V is n. 7.4.69 a B is diagonalizable since it has three distinct eigenvalues, so that S −1 BS is diagonal for some invertible S. But S −1 AS = S −1 I3 S = I3 is diagonal as well. Thus A and B are indeed simultaneously diagonalizable. b There is an invertible S such that S −1 AS = D1 and S −1 BS = D2 are both diagonal. Then A = SD1 S −1 and B = SD2 S −1 , so that AB = (SD1 S −1 )(SD2 S −1 ) = SD1 D2 S −1 and BA = (SD2 S −1 )(SD1 S −1 ) = SD2 D1 S −1 . These two results agree, since D1 D2 = D2 D1 for the diagonal matrices D1 and D2 . c Let A be In and B a nondiagonalizable n × n matrix, for example, A =

1 0

0 1

and B =

1 0

1 . 1

d Suppose BD = DB for a diagonal D with distinct diagonal entries. The ij-th entry of the matrix BD = DB is bij djj = dii bij . For i 6= j this implies that bij = 0. Thus B must be diagonal. e Since A has n distinct eigenvalues, A is diagonalizable, that is, there is an invertible S such that S −1 AS = D is a diagonal matrix with n distinct diagonal entries. We claim that S −1 BS is diagonal as well; by part d it suffices to show that S −1 BS commutes with D = S −1 AS. This is easy to verify: (S −1 BS)D = (S −1 BS)(S −1 AS) = S −1 BAS = S −1 ABS = (S −1 AS)(S −1 BS) = D(S −1 BS) .

335

Chapter 7 7.4.70 Consider an n × n matrix A with m distinct eigenvalues λ1 , ..., λm . If ~v is an eigenvector of A with eigenvalue λm , then (A − λm In ) ~v = ~0, so that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) ~v = ~0. Since the factors in the product (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) commute, we can conclude that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) ~v = ~0 for all eigenvectors ~v of A. Now assume that A is diagonalizable. Then there exists a basis consisting of eigenvectors, so that (A − λ1 In ) (A − λ2 In )· · · (A − λm In ) ~v = ~0 for all vectors in Rn . We can conclude that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) = 0. Conversely, assume that (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) = 0. Then n = dim ker ((A − λ1 In ) (A − λ2 In ) · · · (A − λm In )) ≤ dim ker (A − λ1 In ) + ... + dim ker (A − λm In ) by Exercise 4.2.83. This means that the geometric multiplicities of the eigenvalues add up to n (we know that this sum cannot exceed n), so that A is diagonalizable, by Theorem 7.3.4b. 7.4.71 The eigenvalues are 1 and 2, and (A − I3 ) (A − 2I3 ) = 0. Thus A is diagonalizable.

0 7.4.72 The eigenvalues of this upper triangular matrix are 0 and 1. Now A (A − I3 ) = 0 0 A is diagonalizable if (and only if) b = −ac.

0 b + ac . Matrix 0 0 0 0

7.4.73 If an n × n matrix A has m distinct eigenvalues λ1 , ..., λm , then we can write its characteristic polynomial as fA (λ) = (λ − λ1 ) (λ − λ2 ) · · · (λ − λm ) g (λ) for some polynomial g(λ) of degree n − m. Now fA (A) = (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) g (A) = 0 for a diagonalizable matrix A since (A − λ1 In ) (A − λ2 In ) · · · (A − λm In ) = 0, as we have seen in Exercise 70. (In Exercise 7.3.54 we prove the Cayley-Hamilton theorem for all n × n matrices A, but that proof is a bit longer.) 7.4.74 a For a diagonalizable n × n matrix A with only two distinct eigenvalues, λ1 and λ2 , we have (A − λ1 In )(A − λ2 In ) = 0, by Exercise 70. Thus the column vectors of A − λ2 In are in the kernel of A − λ1 In , that is, they are eigenvectors of A with eigenvalue λ1 (or else they are ~0). Conversely, the column vectors of A − λ1 In are eigenvectors of A with eigenvalue λ2 (or else they are ~0). b If A is a 2 × 2 matrix with distinct eigenvalues λ1 and λ2 , then the nonzero columns of A − λ1 I2 are eigenvectors λ1 0 of A with eigenvalue λ2 , as we observed in part (a). Since the matrices A − and A − λ1 I2 have the 0 λ2 λ1 0 same first column, the first column of A − will be an eigenvector of A with eigenvalue λ2 as well (or 0 λ2 λ 0 it is zero). Likewise, the second column of A − 1 will be an eigenvector of A with eigenvalue λ1 (or it is 0 λ2 zero).

Section 7.5 p √ 7.5.1 z = 3 − 3i so |z| = 32 + (−3)2 = 18 and arg(z) = − π4 , √ so z = 18 cos − π4 + i sin − π4 . 336

Section 7.5 7.5.2 If z = r(cos θ + i sin θ) then z 4 = r4 (cos 4θ + i sin 4θ). z 4 = 1 if r = 1, cos 4θ = 1 and sin 4θ = 0 so 4θ = 2kπ for an integer k, and θ = kπ 2 , + i sin kπ i.e. z = cos kπ 2 2 , k = 0, 1, 2, 3. Thus z = 1, i, −1, −i. See Figure 7.24.

Figure 7.24: for Problem 7.5.2. 7.5.3 If z = r(cos θ + i sin θ), then z n = rn (cos(nθ) + i sin(nθ)). z n = 1 if r = 1, cos(nθ) = 1, sin(nθ) = 0 so nθ = 2kπ for an integer k, and θ = + i sin 2kπ i.e. z = cos 2kπ n n , k = 0, 1, 2, . . . , n − 1. See Figure 7.25.

2kπ n ,

Figure 7.25: for Problem 7.5.3. 7.5.4 Let z = r(cos θ + i sin θ) then w =

√

7.5.5 Let z = r(cos θ + i sin θ) then w =

√ n

r cos r cos

θ+2πk 2

θ+2πk n

+ i sin

θ+2πk 2

+ i sin

θ+2πk n

, k = 0, 1.

, k = 0, 1, 2, . . . , n − 1.

7.5.6 If we have z = r(cos θ + i sin θ) then z1 must have the property that z · z1 = 1 = cos 0 + i sin 0 i.e. |z| · z1 = 1 and arg z · z1 = arg(z) + arg z1 = 0 so z1 = 1r (cos(−θ) + i sin(−θ)) = 1 1 ¯. See Figure 7.26. r (cos θ − i sin θ) (since cosine is even, sine odd). Hence z is a real scalar multiple of z √ 7.5.7 |T (z)| = |z| 2 and arg(T (z)) = arg(1 − i) + arg(z) = − π4 + arg(z) so T is a clockwise rotation by √ by a scaling of 2. 7.5.8 By Theorem 7.5.1, (cos 3θ + i sin 3θ) = (cos θ + i sin θ)3 , i.e. cos 3θ + i sin 3θ = cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 φ 337

π 4

followed

Chapter 7

Figure 7.26: for Problem 7.5.6. = (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ). Equating real and imaginary parts, we get cos 3θ = cos3 θ − 3 cos θ sin2 θ sin 3θ = 3 cos2 θ sin θ − sin3 θ. 7.5.9 |z| =

√

0.82 + 0.72 =

√ 1.15, arg(z) = arctan − 0.7 0.8 ≈ −0.72. See Figure 7.27.

Figure 7.27: for Problem 7.5.9. The trajectory spirals outward, in the clockwise direction. 7.5.10 Let p(x) = ax3 + bx2 + cx + d, where a 6= 0. Since p must have a real root, say λ1 , we can write p(x) = a(x − λ1 )g(x) where g(x) is of the form g(x) = x2 + px + q. On page 350, we see that g(x) = (x − λ2 )(x − λ3 ), so that p(x) = a(x − λ1 )(x − λ2 )(x − λ3 ), as claimed. 7.5.11 Notice that f (1) = 0 so λ = 1 is a root of f (λ). Hence f (λ) = (λ − 1)g(λ), where g(λ) = Setting g(λ) = 0 we get λ = 1 ± 2i so that f (λ) = (λ − 1)(λ − 1 − 2i)(λ − 1 + 2i).

f (λ) λ−1

= λ2 − 2λ + 5.

7.5.12 We will use the facts: i) z + w = z¯ + w ¯ and ii) (z n ) = z¯n which are easy to check. Assume λ0 is a complex root of f (λ) = an λn + · · · + a1 λ + a0 , where the coefficients ai are real. Since λ0 is a root of f , we have an λn0 + an−1 λ0n−1 + · · · + a1 λ0 + a0 = 0. 338

Section 7.5 Taking the conjugate of both sides we get an λn0 + an−1 λ0n−1 + · · · + a1 λ0 + a0 = ¯0 so by fact i), and factoring the real constants we get an λn0 + an−1 λ0n−1 + · · · + a1 λ0 + a0 = 0. Now, by fact ii), an (λ0 )n + an−1 (λ0 )n−1 + · · · + a1 λ0 + a0 = 0, i.e. λ0 is also a root of f , as claimed. 2 2 0 0 ~ ~v = , +i is an eigenvector with eigenvalue 2i, so that we can let S = w 0 0 1 1 | {z } | {z } w ~ ~v 0 −2 −1 with S AS = . 2 0

7.5.13

2i 1

1+i 1

=

1 1 1 1 ~ ~v = 7.5.14 = +i is an eigenvector with eigenvalue i, so that we can let S = w , 1 0 0 1 | {z } | {z } w~ ~v 0 −1 −1 . with S AS = 1 0 1 +i 2 | {z } | ~ v 0 1 , with S −1 AS = 1 2

7.5.15

1 2+i

1 1+i

=

0 ~ ~v = is an eigenvector with eigenvalue 2 + i, so that we can let S = w 1 {z } w ~ 2 −1 . 1 2

0 ~ ~v = is an eigenvector with eigenvalue 4 + i, so that we can let S = w 1 {z } w ~ 4 −1 . 1 4

1 7.5.16 = +i 1 | {z } | ~ v 0 1 −1 , with S AS = 1 1

2 +i −1 | {z } | ~ v 0 2 3 , with S −1 AS = 2 −1 4

7.5.17

2 −1 + 2i

=

0 ~ ~v = is an eigenvector with eigenvalue 3 + 4i, so that we can let S = w 2 {z } w ~ −4 . 3

7.5.18 Let ~v1 , ~v2 be two eigenvectors of A. They define a parallelogram of area S = | det[~v1 ~v2 ]|. Now A~v1 = λ1~v1 and A~v2 = λ2~v2 define a parallelogram of area S1 = | det[λ1~v1 λ2~v2 ]| = |λ1 λ2 det[~v1 ~v2 ]| so SS1 = |λ1 λ2 | = | det(A)|. Hence | det(A)| = |λ1 λ2 |, as claimed. In R3 , a similar argument holds if we replace areas by volumes. See Figure 7.28. 7.5.19 a Since A has eigenvalues 1 and 0 associated with V and V ⊥ respectively and since V is the eigenspace of λ = 1, by Theorem 7.5.5, tr(A) = m, det(A) = 0. b Since B has eigenvalues 1 and −1 associated with V and V ⊥ respectively and since V is the eigenspace associated with λ = 1, tr(A) = m − (n − m) = 2m − n, det B = (−1)n−m . 7.5.20 fA (λ) = (3 − λ)(−3 − λ) + 10 = λ2 + 1 so λ1,2 = ±i. 339

Chapter 7

Figure 7.28: for Problem 7.5.18. 7.5.21 fA (λ) = (11 − λ)(−7 − λ) + 90 = λ2 − 4λ + 13 so λ1,2 = 2 ± 3i. 7.5.22 fA (λ) = (1 − λ)(10 − λ) + 12 = λ2 − 11λ + 22 so λ1,2 =

√ 11± 33 . 2

7.5.23 fA (λ) = −λ3 + 1 = −(λ − 1)(λ2 + λ + 1) so λ1 = 1, λ2,3 =

√ −1± 3i . 2

7.5.24 fA (λ) = −λ3 + 3λ2 − 7λ + 5 so λ1 = 1, λ2,3 = 1 ± 2i. (See Exercise 11.) 7.5.25 fA (λ) = λ4 − 1 = (λ2 − 1)(λ2 + 1) = (λ − 1)(λ + 1)(λ − i)(λ + i) so λ1,2 = ±1 and λ3,4 = ±i 7.5.26 fA (λ) = (λ2 − 2λ + 2)(λ2 − 2λ) = (λ2 − 2λ + 2)(λ − 2)λ = 0, so λ1,2 = 1 ± i, λ3 = 2, λ4 = 0. 7.5.27 By Theorem 7.5.5, tr(A) = λ1 + λ2 + λ3 , det(A) = λ1 λ2 λ3 but λ1 = λ2 6= λ3 by assumption, so tr(A) = 1 = 2λ2 + λ3 and det(A) = 3 = λ22 λ3 . Solving for λ2 , λ3 we get −1, 3 hence λ1 = λ2 = −1 and λ3 = 3. (Note that the eigenvalues must be real; why?) 7.5.28 Suppose the complex eigenvalues are z = a+ib and z¯ = a−ib. By Theorem 7.5.5, we have tr(A) = 2+z + z¯ = 2 + 2a = 8, so that a = 3. Furthermore , det(A) = 2z z¯ = 2(a2 + b2 ) = 2(9 + b2 ) = 50, so that b = 4. Hence the complex eigenvalues are 3 ± 4i. 7.5.29 tr(A) = 0 so λ1 + λ2 + λ3 = 0. Also, we can compute det(A) = bcd > 0 since b, c, d > 0. Therefore, λ1 λ2 λ3 > 0. Hence two of the eigenvalues must be negative, and the largest one (in absolute value) must be positive. 7.5.30 a The ith entry of A~x is

n X

aik xk , so that the sum of all the entries of A~x is

k=1 n n X X

i=1 k=1

aik xk =

n n X X

k=1 i=1

aik xk =

n n X X

k=1

aik

i=1

↑

!

xk =

n X

xk = 1.

k=1

1 340

Section 7.5 b As we do some computer experiments, At appears to approach a matrix with identical columns, with column sum 1. Let ~v1 , ~v2 , . . . , ~vn be an eigenbasis with λ1 = 1 and |λj | < 1 for j = 2, . . . , n. For a fixed i, write ~ei = c1~v1 + c2~v2 + · · · + cn~vn , so that (ith column of At ) = At~ei = c1~v1 + [c2 λt2~v2 + · · · + cn λtn~vn ]. (The term in square brackets goes to zero as t goes to infinity.) Therefore, lim (ith column of At ) = lim (At~ei ) = c1~v1 . t→∞

t→∞

Furthermore, the entries of At~ei add up to 1, for all t, by part a. Therefore, the same is true for the limit (since the limit of a sum is the sum of the limits). It follows that lim (At ) exists and has identical columns, with column sum 1, as claimed. t→∞

1 A is a regular transition matrix, so that lim 7.5.31 No matter how we choose A, 15

t→∞

t

t 1 15 A

is a matrix with identical

columns by Exercise 30. Therefore, the columns of A “become more and more alike” as t approaches infinity, in ij th entry of At = 1 for all i, j, k. the sense that lim ik t→∞ th entry of At 0.6 0.6a(t) + 0.1m(t) + 0.5s(t) a(t) 7.5.32 a ~x(t) = m(t) = 0.2a(t) + 0.7m(t) + 0.1s(t) so A = 0.2 0.2 0.2a(t) + 0.2m(t) + 0.4s(t) s(t)

0.1 0.7 0.2

Note that A is a regular transition matrix.

0.5 0.1 . 0.4

b By Exercise 30, lim (At ) = [~v ~v ~v ], where ~v is the unique eigenvector of A with eigenvalue 1 and column sum 1. t→∞ 0.4 We find that ~v = 0.35 . 0.25 Now lim ~x(t) = lim (At ~x0 ) = lim At ~x0 = [~v ~v ~v ]~x0 = ~v , since the components of ~x0 add up to 1. The market t→∞

t→∞

t→∞

shares approach 40%, 35%, and 25%, respectively, regardless of the initial shares.

7.5.33 a C is obtained from B by dividing each column of B by its first component. Thus, the first row of C will consist of 1’s. b We observe that the columns of C are almost identical, so that the columns of B are “almost parallel” (that is, almost scalar multiples of each other). c Let λ1 , λ2 , . . . , λ5 be the eigenvalues. Assume λ1 real and positive and λ1 > |λj | for 2 ≤ j ≤ 5. Let ~v1 , . . . , ~v5 be corresponding eigenvectors. For a fixed i, write ~ei =

5 X

cj ~vj ; then

j=1

(ith column of At ) = At~ei = c1 λt1~v1 + · · · + c5 λt5~v5 . But in the last expression, for large t, the first term is dominant, so the ith column of At is almost parallel to ~v1 , the eigenvector corresponding to the dominant eigenvalue. 341

Chapter 7 d By part c, the columns of B and C are almost eigenvectors of A associated with the largest eigenvalue, λ1 . Since the first row of C consists of 1’s, the entries in the first row of AC will be close to λ1 . 7.5.34 a The eigenvalues of A − λIn are λ1 − λ, λ2 − λ, . . . , λn − λ, and we were told that |λ1 − λ| < |λi − λ| for i = 2, . . . , n. We may assume that λ1 6= λ (otherwise we are done). The eigenvalues of (A−λIn )−1 are (λ1 −λ)−1 , (λ2 −λ)−1 , . . . , (λn −λ)−1 , and (λ1 −λ)−1 has the largest modulus. The matrices A, A − λIn , and (A − λIn )−1 have the same eigenvectors. For large t, the columns of the tth power of (A − λIn )−1 will be almost eigenvectors of A. If ~v is such a column, compare ~v and A~v to find an approximation of λ1 . b See Figure 7.29.

λ 3 ≈ 17 λ 1 ≈ −1

λ2 ≈ 0 (not to scale)

Figure 7.29: for Problem 7.5.34b. Let λ = −1.

2 A − λI3 = 4 7

9 2 3 6 6 , N = (A − λI3 )−1 = −1 −5 8 11

1 −3 0.5 0 , and B = N 20 . −1 2

Obtain C from B as in Exercise 33: 1 1 1 C ≈ −0.098922005729 −0.098922005729 −0.098922005729 −0.569298722688 −0.569298722688 −0.569298722688 −0.905740179522 −0.905740179522 −0.905740179522 AC ≈ ∗ ∗ ∗ ∗ ∗ ∗

The entries in the first row of AC give us a good approximation for λ1 , and the columns of C give us a good approximation for a corresponding eigenvector.

7.5.35 We have fA (λ) = (λ1 − λ)(λ2 − λ) · · · (λn − λ) = (−λ)n + (λ1 + λ2 + · · · + λn )(−λ)n−1 + · · · + (λ1 λ2 · · · λn ). But, by Theorem 7.2.5, the coefficient of (−λ)n−1 is tr(A). So, tr(A) = λ1 + · · · + λn . 7.5.36 a The entries in the first row are age-specific birth rates and the entries just below the diagonal are age-specific survival rates. For example, the entry 1.6 in the first row tells us that during the next 15 years the people who 342

Section 7.5 are 15–30 years old today will on average have 1.6 children (3.2 per couple) who will survive to the next census. The entry 0.53 tells us that 53% of those in the age group 45–60 today will still be alive in 15 years (they will then be in the age group 60–75). b Using technology, we find the largest eigenvalue λ1 = 1.908 with associated eigenvector 0.574 0.247 0.115 ~v1 ≈ . 0.047 0.014 0.002

The components of ~v1 give the distribution of the population among the age groups in the long run, assuming that current trends continue. λ1 gives the factor by which√the population will grow in the long run in a period of 15 years; this translates to an annual growth factor of 15 1.908 ≈ 1.044, or an annual growth of about 4.4%. 7.5.37 a Use that w + z = w + z and wz = wz. w1 −z 1 w2 −z 2 w1 + w2 −(z1 + z2 ) + = is in H. z1 z2 w1 w2 z1 + z2 w1 + w2

w1 z1

−z 1 w1

w2 z2

−z 2 w2

w1 w2 − z 1 z2 = z1 w2 + w1 z2

−(z1 w2 + w1 z2 ) w1 w2 − z 1 z2

is in H.

b If A in H is nonzero, then det(A) = ww + zz = |w|2 + |z|2 > 0, so that A is invertible. −z , then A−1 = w

w c Yes; if A = z

d For example, if A = 0 0 2 7.5.38 a C4 = 1 0

0 0 0 1

1 0 0 0

i 0

0 −i

1 |w|2 +|z|2

and B =

0 0 1 3 0 , C4 = 0 0 1 0

1 0 0 0

w −z

z w

is in H.

0 −1 0 , then AB = 1 0 −i 0 1 0 0

−i 0

and BA =

0 0 4 4+k , C4 = I4 , then C4 = C4k . 1 0

Figure 7.30 illustrates how C4 acts on the basis vectors ~ei .

Figure 7.30: for Problem 7.5.38a.

343

0 i

i . 0

Chapter 7 λ3k 2 λ b The eigenvalues are λ1 = 1, λ2 = −1, λ3 = i, and λ4 = −i, and for each eigenvalue λk , ~vk = k is an associated λk 1 eigenvector.

c M = aI4 + bC4 + cC42 + dC43 If ~v is an eigenvector of C4 with eigenvalue λ, then M~v = a~v + bλ~v + cλ2~v + dλ3~v = (a + bλ + cλ2 + dλ3 )~v , so that ~v is an eigenvector of M as well, with eigenvalue a + bλ + cλ2 + dλ3 . The eigenbasis for C4 we found in part b is an eigenbasis for all circulant 4 × 4 matrices. 7.5.39 Figure 7.31 illustrates how Cn acts on the standard basis vectors ~e1 , ~e2 , . . . , ~en of Rn .

Figure 7.31: for Problem 7.5.39. 7.5.39 a Based on Figure 7.31, we see that Cnk takes ~ei to ~ei+k “modulo n,” that is, if i + k exceeds n then Cnk takes ~ei to ~ei+k−n (for k = 1, . . . , n − 1). To put it differently: Cnk is the matrix whose ith column is ~ei+k if i + k ≤ n and ~ei+k−n if i + k > n (for k = 1, . . . , n − 1). b The characteristic polynomial is 1 − λn , so that the eigenvalues are the n distinct solutions of the λn = 1 equation 2πk 2πk (the so-called nth roots of unity), equally spaced points along the unit circle, λk = cos n + i sin n , for k = 0, 1, . . . , n − 1 (compare with Exercise 5 and Figure 7.7.). For each eigenvalue λk , n−1 λk . .. ~vk = λ2 is an associated eigenvector. k λk 1 c The eigenbasis ~v0 , ~v1 , . . . , ~vn−1 for Cn we found in part b is in fact an eigenbasis for all circulant n × n matrices. 7.5.40 In Exercise 7.2.50 we derived the formula x =

r 3

for the solution of the equation x3 + px = q. Here r r q q 2 3 3 q q 2 −p 3 + 2q − i − 2q + x= 2 +i − 2 + 3

q 2

+

q 2 2

q

+

−p 3 . 3

344

q 2 2

p 3 3

+

p 3 3

+

r 3

q 2

−

q

q 2 2

+

p 3 3

is negative, and we can write

Section 7.5 Let us write this solution in polar coordinates: q q 3/2 3/2 3 3 (cos α + i sin α) + (cos α − i sin α) x= − p3 − p3 q −p p α+2πk + cos α+2πk + i sin − i sin α+2πk = − p3 cos α+2πk 3 3 3 3 3

p = 2 − p3 cos

α+2πk 3

, k = 0, 1, 2. See Figure 7.32.

Figure 7.32: for Problem 7.5.40. Answer: x1,2,3

q = 2 −p 3 cos

α+2πk 3

, k = 0, 1, 2, where α = arccos q

q

−p 3 ,2

−p 3

Note that x is on the interval q q −p , when k = −1 (Think about it!). − −p 3 3 7.5.41 Substitute ρ = 14 x2

+

12 x3

1 x

(

q 2 p 3/2 −3

)

.

q q −p when k = 0, on −2 3 , − −p when k = 1 and on 3

into 14ρ2 + 12ρ3 − 1 = 0;

−1=0

14x + 12 − x3 = 0 x3 − 14x = 12 Now use the formula derived in Exercise 40 to find x, with p = −14 and q = 12. There is only one positive solution, x ≈ 4.114, so that ρ = x1 ≈ 0.243. 7.5.42 a We will use the fact that for any two complex numbers z and w, z + w = z + w and zw = zw. The ij th entry of AB is

p X

k=1

aik bkj =

p X

k=1

aik bkj =

p X

aik bkj , which is the ij th entry of AB, as claimed.

k=1

b Use part a, where B is the n × 1 matrix ~v + iw. ~ We are told that AB = λB, where λ = p + iq. Then ¯ B, ¯ = AB = λB = λ ¯ or A(~v − iw) AB = A¯ B ~ = (p − iq)(~v − iw). ~ 7.5.43 Note that f (z) is not the zero polynomial, since f (i) = det(S1 + iS2 ) = det(S) 6= 0, as S is invertible. A nonzero polynomial has only finitely many zeros, so that there is a real number x such that f (x) = det(S1 +xS2 ) 6= 345

Chapter 7 0, that is, S1 + xS2 is invertible. Now SB = AS or (S1 + iS2 )B = A(S1 + iS2 ). Considering the real and the imaginary part, we can conclude that S1 B = AS1 and S2 B = AS2 and therefore (S1 + xS2 )B = A(S1 + xS2 ). Since S1 + xS2 is invertible, we have B = (S1 + xS2 )−1 A(S1 + xS2 ), as claimed. 7.5.44 Let A be a complex 2 × 2 matrix. Let λ be a complex eigenvalue of A, and consider an associated eigenvector ~v , so that A~v = λ~v . Now let P be an invertible 2 ×2 matrix ~ (the first column of P is of the form P = [~v w] λ ∗ −1 , so that we have found an upper triangular matrix our eigenvector ~v ). Then P AP will be of the form 0 ∗ similar to A (compare with the proof of Theorem 7.4.1). Yes, any complex square matrix is similar to an upper triangular matrix, although the proof is challenging at this stage of the course. Following the hint, we will assume that the claim holds for n × n matrices, and we will prove it for an (n + 1) × (n + 1) matrix A. As inthe case of a 2 × 2 matrix discussed above, we can find an invertible λ w ~ P such that P −1 AP is of the form for some scalar λ, a row vector w ~ with n components, and an n × n 0 B matrix B (just make the first column of P an eigenvector of A). By the induction hypothesis, B is similar to some upper triangular matrix T , that is, R−1 BR = T for some invertible R. 1 ~0 Now let S = P , an invertible (n + 1) × (n + 1) matrix. Then 0 R ~0 ~0 1 1 0 1 λ w ~ 1 0 λ wR ~ −1 S −1 AS = P AP = = , an upper triangular matrix, 0 R−1 0 R 0 R−1 0 B 0 R 0 T showing that A is indeed similar to an upper triangular matrix. You will see an analogous proof in Section 8.1 (proof of Theorem 8.1.1, Page 371). 7.5.45 If a 6= 0, thenthere are two distinct eigenvalues, 1 ± 1 1 1 1 = fails to be diagonalizable. a 1 0 1

√

a, so that the matrix is diagonalizable. If a = 0, then

7.5.46 6= 0,then there are two distinct eigenvalues, ±ia, so that the matrix is diagonalizable. If a = 0, then If a 0 0 0 −a is diagonalizable as well. Thus the matrix is diagonalizable for all a. = 0 0 a 0 7.5.47 Ifa 6= 0, 0 0 then 1 0 0 1

then there 0 0 a = 1 0 0

√ are three distinct eigenvalues, 0, ± a, so that the matrix is diagonalizable. If a = 0, 0 0 0 0 fails to be diagonalizable. 1 0

7.5.48 The characteristic polynomial is f (λ) = −λ3 + 3λ + a. We need to find the values a such that this polynomial has multiple roots. Now λ is a multiple root if (and only if) f (λ) = f ′ (λ) = 0 (see Exercise 7.2.37). Since f ′ (λ) = −3λ2 + 3 = −3(λ − 1)(λ + 1), the only possible multiple roots are 1 and −1. Now 1 is a multiple root if f (1) = 2 + a = 0, or, a = −2, and −1 is a multiple root if a = 2. Thus, if a is neither 2 nor −2, then the matrix is diagonalizable. Conversely, if a = 2 or a = −2, then the matrix fails to be diagonalizable, since all the eigenspaces will be one-dimensional (verify this!). 7.5.49 The eigenvalues are 0, 1, a − 1. If a is neither 1 nor 2, then there are three distinct eigenvalues, so that the matrix is diagonalizable. Conversely, if a = 1 or a = 2, then the matrix fails to be diagonalizable, since all the eigenspaces will be one-dimensional (verify this!). 346

Section 7.6 0 1 7.5.50 The eigenvalues are 0, 0, 1. Since the kernel is always two-dimensional, with basis 1 , 1 , the matrix is 1 0 diagonalizable for all values of constant a. 7.5.51 Yes, Q is a field. Check the axioms on page 350. 7.5.52 No, Z is not a field since multiplicative inverses do not exist, i.e. division within Z is not possible. (Axiom 8 does not hold). 0 ). 1

7.5.53 Yes, check the axioms on page 350. (additive identity

1 0 0 , multiplicative identity 0 0 0

7.5.54 Yes, check the axioms on page 350. (additive identity

1 0 0 0 ). Also notice , multiplicative identity 0 1 0 0

that rotation-scaling matrices commute when multiplied.

7.5.55 No, since multiplication is not commutative; Axiom 5 does not hold.

Section 7.6 7.6.1 λ1 = 0.9, λ2 = 0.8, so, by Theorem 7.6.2, ~0 is a stable equilibrium. 7.6.2 λ1 = −1.1, λ2 = 0.9, so by Theorem 7.6.2, ~0 is not a stable equilibrium. (|λ1 | > 1) 7.6.3 λ1,2 = 0.8 ± (0.7)i so |λ1 | = |λ2 | =

√

7.6.4 λ1,2 = −0.9 ± (0.4)i so |λ1 | = |λ2 | =

0.64 + 0.49 > 1 so ~0 is not a stable equilibrium.

√

0.81 + 0.16 < 1 so ~0 is a stable equilibrium.

7.6.5 λ1 = 0.8, λ2 = 1.1 so ~0 is not a stable equilibrium. 7.6.6 λ1,2 = 0.8 ± (0.6)i so |λ1 | = |λ2 | = 7.6.7 λ1,2 = 0.9 ± (0.5)i so |λ1 | = |λ2 | =

√

0.64 + 0.36 = 1 and ~0 is not a stable equilibrium.

√

0.81 + 0.25 > 1 and ~0 is not a stable equilibrium.

7.6.8 λ1 = 0.9, λ2 = 0.8 so ~0 is a stable equilibrium. 7.6.9 λ1,2 = 0.8 ± (0.6)i, λ3 = 0.7, so |λ1 | = |λ2 | = 1 and ~0 is not a stable equilibrium. 7.6.10 λ1,2 = 0, λ3 = 0.9 so ~0 is a stable equilibrium. 7.6.11 λ1 = k, λ2 = 0.9 so ~0 is a stable equilibrium if |k| < 1. 7.6.12 λ1,2 = 0.6 ± ik so ~0 is a stable equilibrium if |λ1 | = |λ2 | = 347

√

0.36 + k 2 < 1 i.e. if k 2 < 0.64 or |k| < 0.8.

Chapter 7 7.6.13 Since λ1 = 0.7, λ2 = −0.9, ~0 is a stable equilibrium regardless of the value of k. 7.6.14 λ1 = 0, λ2 = 2k so ~0 is a stable equilibrium if |2k| < 1 or |k| < 12 . 7.6.15 λ1,2 = 1 ±

1 10

√

k

If k ≥ 0 then λ1 = 1 + for any real k.

1 10

√

k ≥ 1. If k < 0 then |λ1 | = |λ2 | > 1. Thus, the zero state isn’t a stable equilibrium

√ √ 1+30k 1 7.6.16 λ1,2 = 2± 10 so |2 ± 1 + 30k| must be less than 10. λ1,2 are real if k ≥ − 30 . In this case it is required √ √ √ 21 that 2 + 1 + 30k < 10 and −10 < 2 − 1 + 30k, which means that 1 + 30k < 8 or k < 10 . 1 97 λ1,2 are complex if k < − 30 . Here it is required that 4 + (−1 − 30k) < 100 or k > − 30 . Overall, ~0 is a stable 21 97 equilibrium if − 30 < k < 10 .

7.6.17 λ1,2 = 0.6 ± (0.8)i = 1(cos θ ± i sin θ), where 0.8 4 θ = arctan 0.8 0.6 = arctan 0.6 = arctan 3 ≈ 0.927. −0.8i −0.8 −1 0 −1 Eλ1 = ker = span so w ~= , ~v = . 0.8 −0.8i i 1 0 0 ~x0 = = 1w ~ + 0~v , so a = 1 and b = 0. Now we use Theorem 7.6.3: 1 0 −1 cos(θt) − sin(θt) 1 0 −1 cos θt − sin θt ~x(t) = = = , where 1 0 sin(θt) cos(θt) 0 1 0 sin θt cos θt θ = arctan 34 ≈ 0.927. The trajectory is the circle shown in Figure 7.33.

Figure 7.33: for Problem 7.6.17.

7.6.18 λ1,2 =

√ −4±2 3i 5

= r(cos θ ± i sin θ), where r ≈ 1.058 and θ = π − arctan 348

2

√ 3 5 4 5

≈ 2.428 (second quadrant).

Section 7.6

Eλ1 = span

√ 3 2

i

√ 3 0 , ~v = 2 . , so w ~= 1 0

~x0 = 1w ~ + 0~v , so a = 1, b = 0. cos(θt) − sin(θt) 0 0.866 1 cos(θt) ~x(t) = rt [ w ~ ~v ] ≈ (1.058)t sin(θt) cos(θt) 1 0 0 sin(θt) 0.866 · sin(2.428t) . See Figure 7.34. ≈ (1.058)t cos(2.428t)

Figure 7.34: for Problem 7.6.18. Spirals slowly outwards (plot the first few points). √ 13, and θ = arctan 32 ≈ 0.98, so √ 0 −1 a 1 ~ ~v ] = , = and ~x(t) λ1 ≈ 13(cos(0.98) + i sin(0.98)), [w 1 0 b 0 √ t − sin(0.98t) . ≈ 13 cos(0.98t)

7.6.19 λ1,2 = 2 ± 3i, r =

The trajectory spirals outwards; see Figure 7.35. 7.6.20 λ1,2 = 4 ± 3i, r = 5, θ = arctan 34 ≈ 0.64, so λ1 ≈ 5(cos(0.64) + i sin(0.64)), [w ~ ~v ] = sin(0.64t) . See Figure 7.36. and ~x(t) ≈ 5t cos(0.64t) Spirals outwards (rotation-dilation). 7.6.21 λ1,2 = 4 ± i, r =

√ 17, θ = arctan

1 4

≈ 0.245 so √ 1 a 0 5 = , ~ ~v ] = λ1 ≈ 17(cos(0.245) + i sin(0.245)), [w 0 b 1 3 √ t 5 sin(0.245t) and ~x(t) ≈ 17 cos(0.245t) + 3 sin(0.245t) 349

0 1 a 1 , = 1 0 b 0

Chapter 7

Figure 7.35: for Problem 7.6.19.

Figure 7.36: for Problem 7.6.20.

Figure 7.37: for Problem 7.6.21.

The trajectory spirals outwards; see Figure 7.37. 7.6.22 λ1,2 = −2 ± 3i, r =

√

13, θ ≈ 2.16 (in second quadrant) 350

Section 7.6

[w ~ ~v ] =

0 1

√ t −5 sin(θt) 1 a −5 , where θ ≈ 2.16. so ~x(t) = 13 = , cos(θt) − 3 sin(θt) 0 b −3

Spirals outwards, as in Figure 7.38.

Figure 7.38: for Problem 7.6.22. 7.6.23 λ1,2 = 0.4 ± 0.3i, r = 12 , θ = arctan 0 5 a 1 [w ~ ~v ] = , = so ~x(t) = 1 3 b 0

0.3 0.4

≈ 0.643 5 sin(θt) 1 t . 2 cos(θt) + 3 sin θ(t)

The trajectory spirals inwards as shown in Figure 7.39.

Figure 7.39: for Problem 7.6.23. 7.6.24 λ1,2 = −0.8 ± 0.6i, r = 1, θ = π − arctan .6 .8 ≈ 2.5 (second quadrant) 0 −5 a 1 −5 sin(θt) [w ~ ~v ] = , = so ~x(t) = , an ellipse, as shown in Figure 7.40. 1 −3 b 0 cos(θt) − 3 sin(θt) 7.6.25 Not stable since if λ is an eigenvalue of A, then

1 λ

is an eigenvalue of A−1 and

351

1 λ

=

1 |λ|

> 1.

Chapter 7

Figure 7.40: for Problem 7.6.24. 7.6.26 Stable since A and AT have the same eigenvalues. 7.6.27 Stable since if λ is an eigenvalue of −A, then −λ is an eigenvalue of −A and | − λ| = |λ|. 7.6.28 Not stable, since if λ is an eigenvalue of A, then (λ − 2) is an eigenvalue of (A − 2In ) and |λ − 2| > 1. 1

2

0

3

2

0

, then A + I2 is 7.6.29 Cannot tell; for example, if A = 1 0 23 1 1 0 2 0 0 −2 A= then A + I2 = 2 1 and the zero state is stable. 0 − 21 0 2

and the zero state is not stable, but if

7.6.30 Consider the dynamical systems ~x(t + 1) = A2 ~x(t) and ~y (t + 1) = A~y (t) with equal initial values, ~x(0) = ~y (0). Then ~x(t) = ~y (2t) for all positive integers t. We know that lim ~y (t) = ~0; thus lim ~x(t) = ~0, proving that the t→∞

zero state is a stable equilibrium of the system ~x(t + 1) = A2 ~x(t).

t→∞

7.6.31 We need to determine for which values of det(A) and tr(A) the modulus of both eigenvalues is less than 1. We will first think about the border line case and examine when one of the moduli is exactly 1: If one of the eigenvalues is 1 and the other is λ, then tr(A) = λ + 1 and det(A) = λ, so that det(A) = tr(A) − 1. If one of the eigenvalues is −1 and the other is λ, then tr(A) = λ − 1 and det(A) = −λ, so that det(A) = −tr(A) − 1. If the eigenvalues are complex conjugates with modulus 1, then det(A) = 1 and |tr(A)| < 2 (think about it!). It is convenient to represent these conditions in the tr-det plane, where each 2 × 2 matrix A is represented by the point (trA, detA), as shown in Figure 7.41. If tr(A) = det(A) = 0, then both eigenvalues of A are zero. We can conclude that throughout the shaded triangle in Figure 7.42 the modulus of both eigenvalues will be less than 1, since the modulus of the eigenvalues changes continuously with tr(A) and det(A) (consider the quadratic formula!). Conversely, we can choose sample points to show that in all the other four regions in Figure 7.42 the modulus of at least one of the eigenvalues exceeds one; consider 0 −2 2 0 −2 0 2 0 . , and , , the matrices 2 0 0 −2 0 0 0 0 ↑ ↑ ↑ ↑ in (I) in (II) in (III) in (IV) 352

Section 7.6

Figure 7.41: for Problem 7.6.31. It follows that throughout these four regions, (I), (II), (III), and (IV), at least one of the eigenvalues will have a modulus exceeding one. The point (trA, detA) is in the shaded triangle if det(A) < 1, det(A) > tr(A) − 1 and det(A) > −tr(A) − 1. This means that |trA| − 1 < det(A) < 1, as claimed. 7.6.32 Take conjugates of both sides of the equation ~x0 = c1 (~v + iw) ~ + c2 (~v − iw): ~ ~x0 = ~x0 = c1 (~v + iw) ~ + c2 (~v − iw) ~ = c¯1 (~v − iw) ~ + c¯2 (~v + iw) ~ = c¯2 (~v + iw) ~ + c¯1 (~v − iw). ~ The claim that c2 = c¯1 now follows from the fact that the representation of ~x0 as a linear combination of the linearly independent vectors ~v + iw ~ and ~v − iw ~ is unique. 7.6.33 Take conjugates of both sides of the equation ~x0 = c1 (~v + iw) ~ + c2 (~v − iw): ~ ~ + c2 (~v − iw) ~ = c¯1 (~v − iw) ~ + c¯2 (~v + iw) ~ = c¯2 (~v + iw) ~ + c¯1 (~v − iw). ~ ~x0 = ~x0 = c1 (~v + iw) The claim that c2 = c¯1 now follows from the fact that the representation of ~x0 as a linear combination of the linearly independent vectors ~v + iw ~ and ~v − iw ~ is unique. 7.6.34 a If | det A| = |λ1 λ2 · · · λn | = |λ1 λ2 | · · · |λn | > 1 then at least one eigenvalue is greater than one in modulus and the zero state fails to be stable. b If | det A| = |λ1 kλ2 | · · · |λn | < 1 we cannot conclude anything about the stability of ~0. |2k0.1| < 1 and |0.2k0.1| < 1 but in the first case we would not have stability, in the second case we would. 7.6.35 a Let ~v1 , . . . , ~vn be an eigenbasis for A. Then ~x(t) =

n X

ci λti~vi and

i=1

k~x(t)k = |

n X i=1

ci λti~vi | ≤

n X i=1

kci λti~vi k =

n X i=1

|λi |t kci~vi k ≤

n X i=1

↑ ≤1 353

kci~vi k.

Chapter 7

The last quantity,

n X i=1

kci~vi k, gives the desired bound M .

t 0 k+1 k 1 1 is = , so that ~x(t) = At = represents a shear parallel to the x-axis, with A 1 1 1 1 0 1 not bounded. This does not contradict part a, since there is no eigenbasis for A.

b A=

7.6.36 If the zero state is stable, then lim (ith column of At ) = lim (At~ei ) = ~0, so that all columns and therefore t→∞

all entries of At approach 0.

t→∞

Conversely, if lim At = 0, then lim (At ~x0 ) = t→∞

t→∞

lim At ~x0 = ~0 for all ~x0 (check the details).

t→∞

7.6.37 a Write Y (t + 1) = Y (t) = Y, C(t + 1) = C(t) = C, I(t + 1) = I(t) = I. Y = C + I + G0 → Y = γY G+0 G0 C = γY Y = 1−γ I=0 Y =

G0 1−γ , C

γG0 1−γ , I

=

b y(t) = Y (t) −

=0

G0 1−γ , c(t)

= C(t) −

γG0 1−γ , i(t)

= I(t)

Substitute to verify the equations. C(t + 1) γ γ c(t) = i(t + 1) αγ − α αγ i(t) c A=

0.2 0.2 −4 1

eigenvalues 0.6 ± 0.8i

not stable d A=

γ γ−1

γ e A= αγ − α

γ , trA = 2γ, detA = γ, stable (use Exercise 31) γ γ trA = γ(1 + α) > 0, detA = αγ αγ

Use Exercise 31; stable if det(A) = αγ < 1 and trA − 1 = αγ + γ − 1 < αγ. The second condition is satisfied since γ < 1. Stable if γ < α1 eigenvalues are real if γ ≥

4α (1+α)2

7.6.38 a T (~v ) = A~v + ~b = ~v if ~v − A~v = ~b or (In − A)~v = ~b. In − A is invertible since 1 is not an eigenvalue of A. Therefore, ~v = (In − A)−1 ~b is the only solution. 354

Section 7.6 b Let ~y (t) = ~x(t) − ~v be the deviation of ~x(t) from the equilibrium ~v . Then ~y (t + 1) = ~x(t + 1) − ~v = A~x(t) + ~b − ~v = A(~y (t) + ~v ) + ~b − ~v = A~y (t) + A~v + ~b − ~v = A~y (t), so that ~y (t) = At ~y (0), or ~x(t) = ~v + At (~x0 − ~v ). lim ~x(t) = ~v for all ~x0 if lim At (~x0 − ~v ) = ~0. This is the case if the modulus of all the eigenvalues of A is less t→∞ than 1. t→∞

7.6.39 Use Exercise 38: ~v = (I2 − A)−1~b =

0.9 −0.4

−0.2 0.7

−1 2 1 . = 4 2

2 is a stable equilibrium since the eigenvalues of A are 0.5 and −0.1. 4 7.6.40 a AT A =

BT −C

b By part a, A−1 =

CT B

B C

−C T BT

1 T p2 +q 2 +r 2 +s2 A

= (p2 + q 2 + r2 + s2 )I4

if A 6= 0.

c (det A)2 = (p2 + q 2 + r2 + s2 )4 , by part a, so that det A = ±(p2 + q 2 + r2 + s2 )2 . Laplace Expansion along the first row produces the term +p4 , so that det(A) = (p2 + q 2 + r2 + s2 )2 . d Consider det(A − λI4 ). Note that the matrix A − λI4 has the same “format” as A, with p replaced by p − λ and q, r, s remaining unchanged. By part c, det(A − λI4 ) = ((p − λ)2 + q 2 + r2 + s2 )2 = 0 when (p − λ)2 = −q 2 − r2 − s2 p p − λ = ±i q 2 + r2 + s2 p λ = p ± i q 2 + r2 + s2

Each of these eigenvalues has algebraic multiplicity 2 (if q = r = s = 0 then λ = p has algebraic multiplicity 4).

p e By part a we can write A = p2 + q 2 + r2 + s2

1 |

p

p2 + q 2 + r2 + s2 {z

!

A , where S is orthogonal.

S

p p Therefore, kA~xk = k p2 + q 2 + r2 + s2 (S~x)k = p2 + q 2 + r2 + s2 k~xk.

3 −3 3 3 f Let A = 4 −5 5 4

}

−4 −5 1 −39 5 −4 2 13 and ~x = ; then A~x = . 3 3 4 18 −3 3 4 13 2

By part e, kA~xk2 = (32 + 32 + 42 + 52 )k~xk , or 392 + 132 + 182 + 132 = (32 + 32 + 42 + 52 )(12 + 22 + 42 + 42 ), as desired. 355

Chapter 7 g Any positive integer m can be written as m = p1 p2 . . . pn . Using part f repeatedly we see that the numbers p1 , p1 p2 , p1 p2 p3 , . . . , p1 p2 p3 · · · pn−1 , and finally m = p1 · · · pn can be expressed as the sums of four squares. 7.6.41 Find the 2 × 2 matrix A

8 6

−3 −8 −3 = 4 −6 4

8 −3 −3 −8 A that transforms into and into : 6 4 4 −6 −1 36 −73 8 −3 −3 −8 1 . = 50 and A = 52 −36 6 4 4 −6

There are many other correct answers. 7.6.42 a x(t + 1) = x(t) − ky(t) y(t + 1) = kx(t) + y(t) = kx(t) + (1 − k 2 )y(t) so

x(t + 1) 1 = y(t + 1) k

−k 1 − k2

x(t) . y(t)

b fA (λ) = λ2 − (2 − k 2 )λ + 1 = 0 The discriminant is (2 − k 2 )2 − 4 = −4k 2 + k 4 = k 2 (k 2 − 4), which is negative if k is a small positive number (k < 2). Therefore, the eigenvalues are complex. By Theorem 7.6.4 the trajectory will be an ellipse, since det(A) = 1.

True or False Ch 7.TF.1 F; If

1 1 , then eigenvalue 1 has geometric multiplicity 1 and algebraic multiplicity. 2. 0 1

Ch 7.TF.2 T, by Theorem 7.4.3 Ch 7.TF.3 T; A = AIn = A[~e1 . . . ~en ] = [λ1~e1 . . . λn~en ] is diagonal. Ch 7.TF.4 T; If A~v = λ~v , then A3~v = λ3~v . Ch 7.TF.5 T; Consider a diagonal 5 × 5 matrix with only two distinct diagonal entries. Ch 7.TF.6 F, by Theorem 7.2.7. Ch 7.TF.7 T, by Summary 7.1.5 Ch 7.TF.8 T, by Theorem 7.2.4 Ch 7.TF.9 T, by Theorem 7.2.2 Ch 7.TF.10 T, by Definition 7.2.3 Ch 7.TF.11 T, by Example 6 of Section 7.5 Ch 7.TF.12 T; The geometric multiplicity of eigenvalue 0 is dim(kerA) = n − rank(A). 356

True or False Ch 7.TF.13 T; If S −1 AS = D, then S T AT (S T )−1 = D.

2 Ch 7.TF.14 F; Let A = 0 0

1 0 0 3 0 and B = 0 0 0 0

0 0 4 0 , for example. 0 0

1 1 Ch 7.TF.15 F; Consider A = . 0 1 Ch 7.TF.16 F; Let A = 8 0 . AB = 0 15

2 0 4 0 , α = 2, B = , β = 5, for example. Then αβ = 10 isn’t an eigenvalue of 0 3 0 5

Ch 7.TF.17 T; If A~v = 3~v , then A2~v = 9~v . Ch 7.TF.18 T; Construct an eigenbasis by concatenating a basis of V with a basis of V ⊥ . Ch 7.TF.19 T, by Theorem 7.5.5

1 1 Ch 7.TF.20 F; Let A = , for example. 0 −1 Ch 7.TF.21 T; If S −1 AS = D, then S −1 A−1 S = D−1 is diagonal. Ch 7.TF.22 F; the equation det(A) = det(AT ) holds for all square matrices, by Theorem 6.2.1. Ch 7.TF.23 T; The sole eigenvalue, 7, must have geometric multiplicity 3. Ch 7.TF.24 F; Let A =

1 1 0 0

and B =

0 1 1 , with A + B = 0 1 0

2 , for example. 1

Ch 7.TF.25 F; Consider the zero matrix. Ch 7.TF.26 T; If A~v = α~v and B~v = β~v , then (A + B)~v = A~v + B~v = α~v + β~v = (α + β)~v .

0 1 0 0 2 Ch 7.TF.27 F; Consider A = , with A = . 0 0 0 0 Ch 7.TF.28 T, by Theorem 7.5.5 1 0 , for example. Ch 7.TF.29 F; Let A = 1 1

1 1 Ch 7.TF.30 F; Let A = 0 0

0 0 1 , with AB = and B = 0 0 1

357

2 , for example. 0

Chapter 7 Ch 7.TF.31 T, Consider the proof of Theorem 7.3.4a. Ch 7.TF.32 T; An eigenbasis for A is an eigenbasis for A + 4I4 as well. Ch 7.TF.33 F; Consider the identity matrix.

1 Ch 7.TF.34 T; Both A and B are similar to 0 0 Ch 7.TF.35 F; Let A =

Ch 7.TF.36 F; Consider

Ch 7.TF.37 F; Let A =

1 1 0 1

1 1 0 1

and ~v =

0 0 2 0 , by Theorem 7.4.1 0 3

1 , for example. 0

2 0 1 0 , ~v = , and w ~= , for example. 0 3 0 1

Ch 7.TF.38 T; A nonzero vector on L and a nonzero vector on L⊥ form an eigenbasis. Ch 7.TF.39 T; The eigenvalues are 3 and −2. Ch 7.TF.40 T, We will us Theorem 7.3.7 throughout: The geometric multiplicity of an eigenvalue is ≤ its algebraic multiplicity. Now let’s show the contrapositive of the given statement: If the geometric multiplicity of some eigenvalue is less than its algebraic multiplicity, then the matrix A fails to be diagonalizable. Indeed, in this case the sum of the geometric multiplicities of all the eigenvalues is less than the sum of their algebraic multiplicities, which in turn is ≤ n (where A is an n × n matrix). Thus the geometric multiplicities do not add up to n, so that A fails to be diagonalizable, by Theorem 7.3.4b. Ch 7.TF.41 F; Consider a rotation through π/2. Ch 7.TF.42 T; Suppose

Ch 7.TF.43 F; Consider

A A ~v A(~v + w) ~ λ~v ~v = = for a nonzero vector . If w ~ is nonzero, then it 0 A w ~ Aw ~ λw ~ w ~ is an eigenvector of A with eigenvalue λ; otherwise ~v is such an eigenvector. 1 0 0 1

and

1 1 . 0 1

Ch 7.TF.44 T; Note that S −1 AS = D, so that D4 = S −1 A4 S = S −1 0S = 0, and therefore D = 0 (since D is diagonal) and A = SDS −1 = 0. Ch 7.TF.45 T; There is an eigenbasis ~v1 , . . . , ~vn , and we can write ~v = c1~v1 + · · · + cn~vn . The vectors ci~vi are either eigenvectors or zero. Ch 7.TF.46 T; If A~v = α~v and B~v = β~v , then AB~v = αβ~v . 358

True or False Ch 7.TF.47 T, by Theorem 7.3.6a

1 0 Ch 7.TF.48 F; Let A = , for example. 0 0 Ch 7.TF.49 T; Recall that the rank is the dimension of the image. If ~v is in the image of A, then A~v is in the image of A as well, so that A~v is parallel to ~v . Ch 7.TF.50 F; Consider

0 1 . 0 0

Ch 7.TF.51 T; If A~v = λ~v for a nonzero ~v , then A4~v = λ4~v = ~0, so that λ4 = 0 and λ = 0.

1 1 Ch 7.TF.52 F; Let A = 0 0

0 1 , for example. and B = 0 1

~ Ch 7.TF.53 T; If the eigenvalue associated with ~v is λ = 0, then A~v = 0, so that ~v is in the kernel of A; otherwise ~v = A λ1 ~v , so that ~v is in the image of A. Ch 7.TF.54 T; either there are two distinct real eigenvalues, or the matrix is of the form kI2 . Ch 7.TF.55 T; Either A~u = 3~u or A~u = 4~u. Ch 7.TF.56 T; Note that (~u~uT )~u = k~uk2 ~u. Ch 7.TF.57 T; Suppose A~vi = αi~vi and B~vi = βi~vi , and let S = [~v1 . . . ~vn ]. Then ABS = BAS = [α1 β1~v1 . . . αn βn~vn ], so that AB = BA. p a b ap + bq Ch 7.TF.58 T; Note that a nonzero vector ~v = is an eigenvector of A = if (and only if) A~v = q c d cp + dq p p ap + bq is parallel to ~v = , that is, if det = 0. Check that this is the case if (and only if) ~v is an q q cp + dq eigenvector of adj(A) (use the same criterion).

359

Chapter 8

Chapter 8 Section 8.1 8.1.1 ~e1 , ~e2 is an orthonormal eigenbasis. 8.1.2

√1 2

1 √1 1 , 2 is an orthonormal eigenbasis. 1 −1

8.1.3

√1 5

2 √1 −1 , 5 is an orthonormal eigenbasis. 1 2

1 1 1 8.1.4 √13 1 , √12 −1 , √16 1 is an orthonormal eigenbasis. −1 0 2

8.1.5 Eigenvalues −1, −1, 2 −1 Choose ~v1 = √12 1 in E−1 and ~v2 = 0 2 8.1.6 31 2 , 1 8.1.7

√1 2

1 , 1

2 1 −1 , 3 −2

√1 2

1 −1

1 √1 1 in E2 and let ~ v3 = ~v1 × ~v2 = 3 1

1 √1 1 . 6 −2

1 1 −2 is an orthonormal eigenbasis. 3 2

is an orthonormal eigenbasis, so S =

√1 2

1 1

1 −1

5 and D = 0

0 . 1

3 −1 1 √ 8.1.8 , 10 is an orthonormal eigenbasis, with λ1 = 4 and λ2 = −6, so S = 3 1 4 0 . D= 0 −6 √1 10

√1 10

1 −1 0 8.1.9 √12 0 , √12 0 , 1 is an orthonormal eigenbasis, with λ1 = 3, λ2 = −3, and λ3 = 2, so 1 1 0 3 0 0 1 −1 √0 S = √12 0 2 and D = 0 −3 0 . 0 0 0 2 1 1 0 8.1.10 λ1 = λ2 = 0 and λ3 = 9. 2 ~v1 = √15 1 is in E0 and ~v2 = 0

1 1 −2 is in E9 . 3 2 360

3 −1 1 3

and

Section 8.1 2 1 Let ~v3 = ~v1 × ~v2 = 3√ −4 ; then ~v1 , ~v2 , ~v3 is an orthonormal eigenbasis. 5 −5

S=

√2 5 √1 5

2 √ 3 5 4 − 3√ 5 √ 5 − 3

1 3

− 32

0

2 3

0 0 and D = 0 9 0 0

0 0 . 0

0 1 −1 8.1.11 √12 0 , √12 0 , 1 is an orthonormal eigenbasis, with λ1 = 2, λ2 = 0, and λ3 = 1, so S = 0 1 1 1 −1 0 2 0 0 √ √1 0 0 2 and D = 0 0 0 . 2 0 0 1 1 1 0 1 8.1.12 a E1 = span 0 and E−1 = (E1 )⊥ . An orthonormal eigenbasis is 2

1 b Use Theorem 7.4.1: B = 0 0 c A = SBS −1

−0.6 = 0 0.8

0 1 √1 0 , 1 , 5 0 2

2 √1 0 . 5 −1

0 0 −1 0 . 0 −1

1 √ 0 0.8 5 0 , where S = −1 0 0 0.6 √2 5

0 1

√2 5

0 .

0 − √15

8.1.13 Yes; if ~v is an eigenvector of A with eigenvalue λ, then ~v = I3~v = A2~v = λ2~v , so that λ2 = 1 and λ = 1 or λ = −1. Since A is symmetric, E1 and E−1 will be orthogonal complements, so that A represents the reflection about E1 . 0 0 0 8.1.14 Let S be as in Example 3. Then S −1 AS = 0 0 0 . 0 0 3 0 0 0 a. This matrix is 2A so that S −1 (2A)S = 0 0 0 . 0 0 6

−3 0 0 b. This is A − 3I3 , so that S −1 (A − 3I3 )S = S −1 AS − 3I3 = 0 −3 0 . 0 0 0 1 0 −2 c. This is 12 (A − I3 ), so that S −1 12 (A − I3 ) S = 21 (S −1 AS − I3 ) = 0 − 12 0

361

0

0 0 . 1

Chapter 8 8.1.15 Yes, if A~v = λ~v , then A−1~v = λ1 ~v , so that an orthonormal eigenbasis for A is also an orthonormal eigenbasis for A−1 (with reciprocal eigenvalues). 8.1.16 a ker(A) is four-dimensional, so that the eigenvalue 0 has multiplicity 4, and the remaining eigenvalue is tr(A) = 5. b B = A + 2I5 , so that the eigenvalues are 2, 2, 2, 2, 7. c det(B) = 24 · 7 = 112 (product of eigenvalues) 8.1.17 If A is the n × n matrix with all 1’s, then the eigenvalues of A are 0 (with multiplicity n − 1) and n. Now B = qA + (p − q)In , so that the eigenvalues of B are p − q (with multiplicity n − 1) and qn + p − q. Thus det(B) = (p − q)n−1 (qn + p − q). p 8.1.18 By Theorem 6.3.6, the volume is |det A| = det(AT A). Now ~vi · ~vj = k~vi kk~vj kcos(θ) = 12 , so that AT A has n−1 1 1 all 1’s on the diagonal and 21 ’s outside. By Exercise 17 with p = 1 and q = 12 , det(AT A) = 21 2n + 2 = p n/2 √ 1 n n + 1. (n + 1), so that the volume is det(AT A) = 21 2

8.1.19 Let L(~x) = A~x. Then AT A is symmetric, since (AT A)T = AT (AT )T = AT A, so that there is an orthonormal eigenbasis ~v1 , . . . , ~vm for AT A. Then the vectors A~v1 , . . . , A~vm are orthogonal, since A~vi · A~vj = (A~vi )T A~vj = ~viT AT A~vj = ~vi · (AT A~vj ) = ~vi · (λj ~vj ) = λj (~vi · ~vj ) = 0 if i 6= j.

8.1.20 By Exercise 19, there is an orthonormal basis ~v1 , . . . , ~vm of Rm such that T (~v1 ), . . . , T (~vm ) are orthogonal. 1 vi ) for Suppose that T (~v1 ), . . . , T (~vr ) are nonzero and T (~vr+1 ), . . . , T (~vm ) are zero. Then let w ~ i = kT (~ vi )k T (~ ⊥ i = 1, . . . , r and choose an orthonormal basis w ~ r+1 , . . . , w ~ n of [span(w ~ 1, . . . , w ~ r )] . Then w ~ 1, . . . , w ~ n does the job. 8.1.21 For each eigenvalue there are two unit eigenvectors: ±~v1 , ±~v2 , and ±~v3 . We have 6 choices for the first column of S, 4 choices remaining for the second column, and 2 for the third. Answer: 6 · 4 · 2 = 48. 8.1.22 a If we let k = 2 then A is symmetric and therefore (orthogonally) diagonalizable. b If we let k = 0 then 0 is the only eigenvalue (but A 6= 0), so that A fails to be diagonalizable. 8.1.23 The eigenvalues are real (by Theorem 8.1.3), so that the only possible eigenvalues are ±1. Since A is symmetric, E1 and E−1 are orthogonal complements. Thus A represents a reflection about E1 . 8.1.24 Note that A is symmetric and orthogonal, so that the eigenvalues are 1 and −1 (see Exercise 23). 1 1 0 0 0 1 0 1 E1 = span , and E−1 = span , , so that 0 0 −1 1 −1 1 0 0 1 0 1 0 0 1 0 1 √1 , √1 , √1 , √12 is an orthonormal eigenbasis. 2 2 2 0 1 0 −1 1 0 −1 0 362

Section 8.1 8.1.25 Note that A is symmetric an orthogonal, so that the eigenvalues of A are 1 and −1. 1 0 0 1 0 0 1 0 0 1 E1 = span 0 , 0 , 1 , E−1 = span 0 , 0 0 1 0 0 −1 1 0 0 −1 0 1 0 0 1 0 0 1 0 1 √0 2 0 0 is one possible choice. The columns of S must form an eigenbasis for A : S = √12 0 0 0 1 0 0 −1 1 0 0 −1 0 8.1.26 Since Jn is both orthogonal and symmetric, the eigenvalues are 1 and −1. If n is even, then both have multiplicity n2 (as in Exercise 24). If n is odd, then the multiplicities are n+1 for 1 and n−1 for −1 (as in 2 2 Exercise 25). One way to see this is to observe that tr(Jn ) is 0 for even n, and 1 for odd n (recall that the trace is the sum of the eigenvalues). 8.1.27 If n is even, then this matrix is Jn + In , for the Jn introduced in Exercise 26, so that the eigenvalues are 0 and 2, with multiplicity n2 each. E2 is the span of all ~ei + ~en+1−i , for i = 1, . . . , n2 , and E0 is spanned by all ~ei −~en+1−i . If n is odd, then E2 is spanned by all ~ei +~en+1−i , for i = 1, . . . , n−1 ei −~en+1−i , 2 ; E0 is spanned by all ~ n+1 . for i = 1, . . . , n−1 , and E is spanned by ~ e 1 2 2

8.1.28 For λ 6= 0 −λ fA (λ) = det 0

1

−λ

1 = λ det

−λ

0

−λ

1

0

.

··· 0

..

.

0 0

..

···

| | | | | | −λ | | | 1 |

| | | | | | −λ | | | 0 |

1 1 .. . 1 1−λ 1 1 .. .

=

1 −λ2 + λ + 12

= −λ11 (λ2 − λ − 12) = −λ11 (λ − 4)(λ + 3)

−λ

1 det λ

Eigenvalues are 0 (with multiplicity 11), 4 and −3. Eigenvalues for 0 are ~e1 − ~ei (i = 2, . . . , 12), 1 1 1 1 . . . . E4 = span . (12 ones), E−3 = span . (12 ones) 1 1 4 −3 363

0

−λ

..

.

0 λ

λ

···

| | | | | | −λ | | | λ |

1 1 .. . 1 λ − λ2

Chapter 8 so 1 1 0 −1 0 −1 0 0 0 0 0 0 S= 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 0 0 0 0 −1 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 −1 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 −1 0 0 0 0 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 −1 0 0

1 1 1 1 1 1 1 1 1 1 1 1 4

1 1 1 1 1 1 1 1 1 1 1 1 −3

diagonalizes A, and D = S −1 AS will have all zeros as entries except d12,

12

= 4 and d13,

13

= −3.

8.1.29 By Theorem 5.4.1 (im A)⊥ = ker(AT ) = ker(A), so that ~v is orthogonal to w. ~ 8.1.30 The columns ~v , ~v2 , . . . , ~vn of R form an orthogonal eigenbasis for A = ~v ~v T , with eigenvalues 1, 0, 0, . . . , 0(n − 1 zeros), since A~v = ~v v T ~v = ~v (~v · ~v ) = ~v , (since ~v · ~v = 1) and A~vi = ~v v T ~vi = ~v (~v · ~vi ) = ~0 (since ~v · ~vi = 0). 1 0 ··· 0 0 0 ··· 0 . Therefore we can let S = R, and D = ... 0

0 ···

0

8.1.31 True; A is diagonalizable, that is, A is similar to a diagonal matrix D; then A2 is similar to D2 . Now rank(D) = rank(D2 ) is the number of nonzero entries on the diagonal of D (and D2 ). Since similar matrices have the same rank (by Theorem 7.3.6b) we can conclude that rank(A) = rank(D) = rank(D2 ) = rank(A2 ). 8.1.32 By Exercise 17, det(A) = (1 − q)n−1 (qn + 1 − q). A is invertible if det(A) 6= 0, that is, if q 6= 1 and q 6= 8.1.33 The angles must add up to 2π, so θ =

2π 3

= 120◦ . (See Figure 8.1.)

v1 θ

v2

θ

unit circle

θ

v3

Figure 8.1: for Problem 8.1.33. 364

1 1−n .

Section 8.1 Algebraically, we can see this as follows: let A = [ ~v1 ~v2 ~v3 ], a 2 × 3 matrix. 1 cos θ cos θ 1 cos θ is a noninvertible 3 × 3 matrix, so that cos θ = Then AT A = cos θ cos θ cos θ 1 ◦ = 120 . and θ = 2π 3 8.1.34 Let ~v1 , ~v2 , ~v3 , ~v4 be such vectors. Form A = [ ~v1

~v2

~v3

1 1−3

= − 21 , by Exercise 32,

~v4 ], a 3 × 4 matrix.

1 cos θ cos θ cos θ 1 cos θ cos θ cos θ Then AT A = is noninvertible, so that cos θ = cos θ cos θ 1 cos θ 1 cos θ cos θ cos θ θ = arccos − 31 ≈ 109.5◦ . See Figure 8.2.

1 1−4

= − 13 , by Exercise 32, and

Figure 8.2: for Problem 8.1.34. The tips of ~v1 , ~v2 , ~v3 , ~v4 form a regular tetrahedron. 8.1.35 Let ~v1 , . . . , ~vn+1 be these vectors. Form A = [ ~v1 · · · ~vn+1 ], an n × (n + 1) matrix. 1 cos θ · · · cos θ cos θ 1 · · · cos θ is a noninvertible (n + 1) × (n + 1) matrix with 1’s on the diagonal and Then AT A = .. ... . cos θ

cos θ outside, so that cos θ =

···

1 1−n ,

1

by Exercise 32, and θ = arccos

1 1−n

.

8.1.36 If ~v is an eigenvector with eigenvalue λ, then λ~v = A~v = A2~v = λ2~v , so that λ = λ2 and therefore λ = 0 or λ = 1. Since A is symmetric, E0 and E1 are orthogonal complements, so that A represents the orthogonal projection onto E1 . 8.1.37 In Example 4 we see that the image of the unit circle is an ellipse with semi-axes 2 and 3. Thus ||A~u|| takes all values in the interval [2, 3]. 8.1.38 The spectral theorem tells us that there exists an orthonormal eigenbasis ~v1 , ~v2 for A, with associated eigenvalues -2 and 3. Consider a unit vector ~u = c1~v1 + c2~v2 in R2 , with c21 + c22 = 1. Then ~u · A~u = (c1~v1 + c2~v2 )·(−2c1~v1 + 3c2~v2 ) = −2c21 +3c22 , which takes all values on the interval [−2, 3] since −2 = −2c21 −2c22 ≤ −2c21 + 3c22 ≤ 3c21 + 3c22 = 3. 8.1.39 The spectral theorem tells us that there exists an orthonormal eigenbasis ~v1 , ~v2 , ~v3 for A, with associated eigenvalues -2, 3 and 4. Consider a unit vector ~u = c1~v1 + c2~v2 + c3~v3 in R3 , with c21 + c22 + c23 = 1. Then 365

Chapter 8 ~u · A~u = (c1~v1 + c2~v2 + c3~v3 ) · (−2c1~v1 + 3c2~v2 + 4c3~v3 ) = −2c21 + 3c22 + 4c23 , which takes all values on the interval [−2, 4] since −2 = −2c21 − 2c22 − 2c23 ≤ −2c21 + 3c22 + 4c23 ≤ 4c21 + 4c22 + 4c23 = 4. 8.1.40 Using the terminology introducedp in Exercise 8.1.39, we have kA~uk = k−2c1~v1 + 3c2~v2 + 4c3~v3 k = 4c21 + 9c22 + 16c22 , which takes all values on the interval [2, 4]. Geometrically, the image of the unit sphere under A is the ellipsoid with semi-axes 2, 3, and 4. 8.1.41 The spectral theorem tells us that there exists an orthogonal matrix S such that S −1 AS = D is diagonal. Let D1 be the diagonal matrix such that D13 = D; the diagonal entries of D1 are the cube roots of those of D. 3 Now B = SD1 S −1 does the job, since B 3 = SD1 S −1 = SD13 S −1 = SDS −1 = A.

8.1.42 We will use the strategy outlined in Exercise 8.1.41. An orthogonal matrix that diagonalizes A = 15 2 0 8 0 1 −2 6 and B = SD1 S −1 = 51 . Now D1 = , with S −1 AS = D = is S = √15 0 1 0 1 2 1 2

12 14 14 33 2 . 9

1 1 1 8.1.43 There is an orthonormal eigenbasis ~v1 = √12 −1 , ~v2 = √16 1 , ~v3 = √13 1 with associated 0 −2 1 eigenvalues -9, -9, 24. We are looking for a nonzero vector ~v = c1~v1 + c2~v2 + c3~v3 such that ~v · A~v = (c1~v1 + c2~v2 + c3~v3 ) · (−9c1~v1 − 9c2~v2 + 24c3~v3 ) = −9c21 −9c22 + 24c23 = 0 or −3c21 − 3c22 + 8c23 = 0. One possible 3 √ √ √ solution is c1 = 8 = 2 2, c2 = 0, c3 = 3, so that ~v = −1 . 1 8.1.44 Use Exercise 8.1.43 as a guide. Consider an orthonormal eigenbasis ~v1 , ...., ~vn for A, with associated eigenvalues λ1 ≤ λ2 ≤ .... ≤ λn , listed in ascending order. If ~v = c1~v1 + ... + cn~vn is any nonzero vector in Rn , then ~v · A~v = (c1~v1 + ... + cn~vn ) · (λ1 c1~v1 + ... + λn cn~vn ) = λ1 c21 + ... + λn c2n . If all the eigenvalues are positive, then ~v · A~v will be positive. Likewise, if all the eigenvalues are negative, then ~v · A~v will be negative. However, if A has positive as well as negative eigenvalues, meaning√that λ1 < √ 0 < λn (as in Example 8.1.43) , then there exist nonzero vectors ~v with ~v · A~v = 0, for example, ~v = λn~v1 + −λ1~vn . 8.1.45 a If S −1 AS is upper triangular then the first column of S is an eigenvector of A. Therefore, any matrix 0 −1 . without real eigenvectors fails to be triangulizable over R, for example, 1 0 b Proof by induction on n: For an n × n matrix A we can choose a complex invertible n × n matrix P whose first λ ~v −1 column is an eigenvector for A. Then P AP = . B is triangulizable, by induction hypothesis, that is, 0 B there is an invertible that Q−1 BQ = triangular. Q such (n − 1) × (n −1) matrix T is upper λ ~ v 1 0 1 0 λ ~ v 1 0 λ ~v Q Now let R = R= . Then R−1 = is upper triangular. 0 B 0 Q−1 0 B 0 Q 0 T 0 Q λ ~v R = R−1 P −1 AP R = S −1 AS, where S = P R, proving our claim. R−1 0 B 8.1.46 a By definition of an upper triangular matrix, ~e1 is in ker U , ~e2 is in ker(U 2 ), . . . , ~en is in ker(U n ), so that all ~x in Cn are in ker(U n ), that is, U n = 0.

366

Section 8.2 b By Exercise 45b, there exists an invertible S such that S −1 AS = U is upper triangular. The diagonal entries of U are all zero, since A and U have the same eigenvalues; therefore U n = 0 by part a. Now A = SU S −1 and An = SU n S −1 = 0, as claimed. 8.1.47 a For all i, j,

"

n X

k=1

#

aik bkj ≤

n X

k=1

|aik bkj | =

n X

k=1

|aik ||bkj |

↑ triangle inequality b By induction on t: |At | = |At−1 A| ≤ |At−1 ||A| ≤ |A|t−1 |A| = |A|t ↑ part a

↑ by induction hypothesis

t t t t 2 n−1 n 8.1.48 If t ≥ n − 1 then (In + U ) = In + U+ U +···+ U , since U = 0. Now ≤ tn 1 2 n−1 k for k = 1, . . . , n − 1, so that (In + U )t ≤ tn (In + U + · · · + U n−1 ), as claimed. Check that the formula holds for t < n − 1 as well. t

8.1.49 Let λ be the largest |rii |; note that λ < 1. Then |R| =

|r11 |

..

∗ .

0 |rnn | λ(In + U ), and |Rt | ≤ |R|t ≤ λt (In + U )t ≤ λt tn (In + U + · · · + U n−1 ).

≤

λ

..

.

0

∗ λ

= λ

1 0

..

.

∗ 1

=

We learn in Calculus that lim (λt tn ) = 0, so that lim (Rt ) = 0. t→∞

t→∞

8.1.50 a From Exercise 45b we know that there is an invertible S and an upper triangular R such that S −1 AS = R, and |rii | < 1 for all i, since the diagonal entries of R are the eigenvalues of A. Now lim Rt = 0 by Exercise 49. Note that A = SRS −1 and At = SRt S −1 , so that lim At = 0, as claimed.

t→∞

t→∞

b See the remark after Definition 7.6.1.

Section 8.2 8.2.1 We have a11 = coefficient of x21 = 6, a22 = coefficient of x22 = 8, a12 = a21 = 6 − 27 So, A = − 72 8 0

1 2

1 2

0

3 8.2.3 A = 0

0 4

3

7 2

5

8.2.2 A =

3

7 2

367

1 2(

coefficient of x1 x2 ) = − 72 .

Chapter 8

8.2.4 A =

6 2 , positive definite 2 3

1 2 , indefinite (since det(A) < 0) 8.2.5 A = 2 1

2 3 , indefinite (since det(A) < 0) 8.2.6 A = 3 4

0 0 2 8.2.7 A = 0 3 0 , indefinite (eigenvalues 2, −2, 3) 2 0 0

8.2.8 If S −1 AS = D is diagonal, then S −1 A2 S = D2 , so that all eigenvalues of A2 are ≥0. So A2 is positive semi-definite; it is positive definite if and only if A is invertible. 8.2.9 a (A2 )T = (AT )2 = (−A)2 = A2 , so that A2 is symmetric. b q(~x) = ~x T A2 ~x = ~x T AA~x = −~x T AT A~x = −(A~x) · (A~x) = −kA~xk2 ≤ 0 for all ~x, so that A2 is negative semi-definite. The eigenvalues of A2 will be ≥ 0. c If ~v is a complex eigenvector of A with eigenvalue λ, then A2~v = λ2~v , and λ2 ≤ 0, by part b. Therefore, λ is imaginary, that is, λ = bi for a real b. Thus, the zero matrix is the only skew-symmetric matrix that is diagonalizable over R. 8.2.10 L(~x) = (~x+~v )T A(~x + ~v ) − ~x T A~x − ~v T A~v =~x T A~x + ~x T A~v + ~v T A~x + ~v T A~v − ~x T A~x − ~v T A~v =~x T A~v + ~v T A~x = ~v T A~x + ~v T A~x = (2~v T A)~x, ↑ note that ~x T A~v is a scalar so that ~x T A~v = (~x T A~v )T = ~v T AT ~x = ~v T A~x if A is symmetric. So L is linear with matrix 2~v T A. 8.2.11 The eigenvalues of A−1 are the reciprocals of those of A, so that A and A−1 have the same definiteness. 8.2.12 det(A) is the product of the two (real) eigenvalues. q is indefinite if an only if those have different signs, that is, their product is negative. 8.2.13 q(~ei ) = ~ei · A~ei = aii > 0 8.2.14 If det(A) is positive then both eigenvalues have the same sign, so that A is positive definite or negative definite. Since ~e1 · A~e1 = a > 0, A is in fact positive definite.

2 ; eigenvalues λ1 = 7 and λ2 = 2 3 2 −1 1 1 √ √ orthonormal eigenbasis ~v1 = 5 , ~v2 = 5 1 2

6 8.2.15 A = 2

368

Section 8.2 λ1 c21 + λ2 c22 = 1 or 7c21 + 2c22 = 1. (See Figure 8.3.)

E7

v2 1 v 2

1 v 1

√2

√7

– 1 v1 √7

v1 1

E2 – 1 v2 √2

Figure 8.3: for Problem 8.2.15.

8.2.16 A =

0 1 2

1 2

0

; eigenvalues λ1 = 12 , and λ2 = − 21

orthonormal eigenbasis ~v1 = 1 2 2 c1

√1 2

1 and ~v2 = 1

√1 2

1 −1

− 21 c22 = 1. (See Figure 8.4.)

2 , eigenvalues λ1 = 4, λ2 = −1 0 2 −1 orthonormal eigenbasis ~v1 = √15 , ~v2 = √15 1 2

8.2.17 A =

3 2

4c21 − c22 = 1 (hyperbola) (See Figure 8.5.) −2 , eigenvalues λ1 = 10, λ2 = 5 6 2 −1 , ~v2 = √15 orthonormal eigenbasis ~v1 = √15 1 2

9 8.2.18 A = −2

10c21 + 5c22 = 1. This is an ellipse, as shown in Figure 8.6. 2 ; eigenvalues λ1 = 5, λ2 = 0 4 1 −2 1 1 √ √ eigenvectors ~v1 = 5 , ~v2 = 5 2 1

1 8.2.19 A = 2

369

Chapter 8

E1 2

√2 v1 v1

1 v2 –√2 v1 E– 1 2

Figure 8.4: for Problem 8.2.16.

E4 v2 1 2 v1

v1

1 – 12 v1

E–1

Figure 8.5: for Problem 8.2.17. 5c21 = 1 (a pair of lines) (See Figure 8.7.) Note that (x21 + 4x1 x2 + 4x22 ) = (x1 + 2x2 )2 = 1, so that x1 + 2x2 = ±1, and the two lines are x2 =

1−x1 2

and x2 =

−1−x1 . 2

370

Section 8.2

E10 v2 1 v √5 2

– 1 v1

1 v √10 1

v1

– 1 v2

1

√5

√10

E5

Figure 8.6: for Problem 8.2.18.

v1

v2 1 v √5 1

– 1 v1 √5

1

E0 E5

Figure 8.7: for Problem 8.2.19.

3 ; eigenvalues λ1 = 6 and λ2 = −4 5 1 −3 , ~v2 = √110 orthonormal eigenbasis ~v1 = √110 3 1

−3 8.2.20 A = 3

6c21 − 4c22 = 1. This is a hyperbola, as shown in Figure 8.8. 8.2.21 a In each case, it is informative to think about the intersections with the three coordinate planes: x1 − x2 , x1 − x3 , and x2 − x3 . 371

Chapter 8

v1 v2

1 v √6 1

– 1 v1 √6

E–4

E6

Figure 8.8: for Problem 8.2.20. •] For the surface x21 + 4x22 + 9x23 = 1, all these intersections are ellipses, and the surface itself is an ellipsoid . 0 1 This surface is connected and bounded; the points closest to the origin are ± 0 , and those farthest ± 0 . 1 0 3 (See Figure 8.9.) x3

x2 x1

Figure 8.9: for Problem 8.2.21a: x21 + 4x22 + 9x23 = 1, an ellipsoid (not to scale). •] In the case of x21 + 4x22 − 9x23 = 1, the intersection with the x1 − x2 plane is an ellipse, and the two other intersections are hyperbolas. The surface is connected and not bounded; the points closest to the origin are 372

Section 8.2 0 ± 12 . (See Figure 8.10.) 0 x3

x2

x1

Figure 8.10: for Problem 8.2.21a: x21 + 4x22 − 9x23 = 1, a hyperboloid of one sheet (not to scale). •] In the case −x21 −4x22 +9x23 = 1, the intersection with the x1 −x2 plane is empty, and the two other intersections 0 are hyperbolas. The surface consists of two pieces and is unbounded. The points closest to the origin are ± 0 . 1 3

(See Figure 8.11.)

x3

x2

x1

Figure 8.11: for Problem 8.2.21a: −x21 − 4x22 + 9x23 = 1, a hyperboloid of two sheets (not to scale). 373

Chapter 8 1

1 2

1

b A = 12 1

2

3 2

3 2

3

is positive definite, with three positive eigenvalues λ1 , λ2 , λ3 .

The surface is given by λ1 c21 + λ2 c22 + λ3 c23 = 1 with respect to the principal axis, an ellipsoid. To find the points closest to and farthest from the origin, use technology to find the eigenvalues and eigenvectors:

eigenvalues: λ1 ≈ 0.56, λ2 ≈ 4.44, λ3 = 1 0.86 0.31 unit eigenvectors: ~v1 ≈ 0.19 , ~v2 ≈ 0.54 , ~v3 = −0.47 0.78 Equation: 0.56c21 + 4.44c22 + c23 = 1

1 √1 −2 6 1

1 Farthest points when c1 = ± √0.56 and c2 = c3 = 0 1 and c1 = c3 = 0 Closest points when c2 = ± √4.44

0.86 1.15 1 0.19 ≈ ± 0.26 Farthest points ≈ ± √0.56 −0.47 −0.63 0.31 0.15 1 0.54 ≈ ± 0.26 Closest points ≈± √4.44 0.78 0.37

−1 0 8.2.22 A = 0 1 5 0

5 0 ; eigenvalues λ1 = 4, λ2 = −6, λ3 = 1 −1

Equation with respect to principal axis: 4c21 − 6c22 + c23 = 1, a hyperboloid of one sheet (see Figure 8.10). Closest to origin when c1 = ± 12 , c2 = c3 = 0.

1 1 0.35 A unit eigenvector for eigenvalue 4 is ~v = √12 0 , so that the desired points are ± 21 √12 0 ≈ ± 0 . 1 1 0.35 8.2.23 Yes; M = 21 (A + AT ) is symmetric, and ~x T M~x = 12 ~x T A~x + 12 ~x T AT ~x = 21 ~x T A~x + 12 ~x T A~x = ~x T A~x Note that ~x T A~x is a 1 × 1 matrix, so that ~x T A~x = (~x T A~x)T = ~x T AT ~x. 8.2.24 q(~e1 ) = ~e1 · A~e1 = ~e1 · (first column of A) = a11 8.2.25 q(~v ) = ~v · A~v = ~v · (λ~v ) = λ(~v · ~v ) = λ ↑ ~v is a unit vector 374

Section 8.2

8.2.26 False; If A =

0 1 1 0

then q

0 1 1 0 1 1 1 = 0. · = · = 1 0 0 1 0 0 0

8.2.27 Let ~v1 , . . . , ~vn be an orthonormal eigenbasis for A with A~vi = λi~vi . We know that q(~vi ) = λi (see Exercise 25), so that q(~v1 ) = λ1 and q(~vn ) = λn are in the image. We claim that all numbers between λn and λ1 are in the image as well. To see this, apply the Intermediate Value π Theorem to the continuous function f (t) = q((cos t)~ v + (sin t)~ v ) on 0, vn ) = λn and n 1 2 (note that f (0) = q(~ f π2 = q(~v1 ) = λ1 ). (See Figure 8.12.) λ1 v1

(cos t)vn + (sin t)v1, a unit vector

f(t) λn 0

π 2

t

vn

Figure 8.12: for Problem 8.2.27. The Intermediate Value Theorem tells us that for any c between λn and λ1 , there is a t0 such that f (t0 ) = q((cos t0 )~vn + (sin t0 )~v1 ) = c. Note that (cos t0 )~vn + (sin t0 )~v1 is a unit vector. Now we will show that, conversely, q(~v ) is on [λn , λ1 ] for all unit vectors ~v . Write ~v = c1~v1 + · · · + cn~vn and note that k~v k2 = c21 + · · · + c2n = 1. Then q(~v ) = λ1 c21 + λ2 c22 + · · · + λn c2n ≤ λ1 c21 + λ1 c22 + · · · + λ1 c2n = λ1 . Likewise, q(~v ) ≥ λn . We have shown that the image of S n−1 under q is the closed interval [λn , λ1 ]. 8.2.28 The hint almost gives it away. Since D is a diagonal matrix with positive diagonal entries, we can write D = D12 , where D1 is diagonal with positive diagonal entries (the square roots of the entries of D). Now A = SDS T = SD1 D1 S T = SD1 (SD1 )T = BB T where B = SD1 . The columns of B are scalar multiples of the corresponding columns of S, so that they are orthogonal. 8.2.29 From Example 1 we have S = 6 2 √1 . 5 −3 4

√1 5

2 1 −1 2

and D =

9 0

3 0 . Let D1 = 0 4

0 2

and B = SD1 =

8.2.30 Define D1 as in Exercise 28. Then A = SDS −1 = SD1 D1 S −1 = (SD1 S −1 )(SD1 S −1 ) = B 2 , where B = SD1 S −1 . B is positive definite, since S −1 BS = D1 is diagonal with positive diagonal entries. 8.2.31 S =

√1 5

2 −1

1 2

and D1 =

3 0 0 2

(see Exercise 29), so that B = SD1 S −1 =

8.2.32 Recall that a = q(~e1 ) > 0 and det A = ac − b2 = λ1 λ2 > 0.

x x 0 a b = 0 y z b c

y z

x2 = xy

xy y2 + z2

x2 = a means that xy = b y2 + z2 = c 375

2.8 −0.4

−0.4 . 2.2

Chapter 8 It is required that x and z be positive. This system has the unique solution √ x= a y=

b x

z=

p

=

√b a

q

c − y2 =

c−

b2 a

=

q

ac−b2 a

8.2.33 Use the formulas for x, y, z derived in Exercise 32. √ √ √ x= a= 8=2 2 y= z=

√b a

q

2 = − √12 = − 2√ 2

ac−b2 a

√ 2 2 L= − √12

=

q

0

√3 2

36 8

=

√3 , 2

so that

.

8.2.34 (i) implies (ii): See the hint at the end of the exercise. (ii) implies (iii): det A(m) is the product of the (positive) eigenvalues. (iii) implies (iv): T A(n−1) ~v B 0 B ~x BB T = = T T ~v k ~x 1 0 t ~x T B T B~x = ~v has the unique solution The system ~x T ~x + t = k A=

B~x ~x T ~x + t

~x = B −1~v t = k − ~x T ~x = k − kB −1~v k2 . (n)

Note that t is positive since 0 < det(A

B ) = det(A) = det ~x T

(iv) ⇒ (i)

T B 0 det 0 1

~x = (det B)2 · t. t

~x T A~x = ~x T LLT ~x = (LT ~x)T (LT ~x) = kLT ~xk2 > 0 if ~x 6= ~0, since L is invertible.

4 8.2.35 Solve the system −4 8

x2 = 4, so x = 2 2y = −4, so y = −2 2z = 8, so z = 4 4 + w2 = 13, so w = 3 −8 + 3t = 1, so t = 3 16 + 9 + s2 = 26, so s = 1

−4 8 x 0 13 1 = y w 1 26 z t 2 0 0 L = −2 3 0 4 3 1

0 x y 00 w s 0 0

z t s

8.2.36 If A = QR, then AT A = (QR)T QR = RT QT QR = RT R = LLT , L = RT . 376

Section 8.2 2

∂q ∂x1

2

2

∂q ∂ q ∂ q ∂ q = 2ax1 + bx2 and ∂x = bx1 + 2cx2 , so that q11 = ∂x = b, and 2 = 2a, q22 = ∂x2 = 2c, and q12 = ∂x ∂x 1 2 1 2 2 2a b q q D = det 11 12 = det = 4ac − b2 > 0. q12 q22 b 2c a 2b The matrix A = b of q is positive definite, since a > 0 and det(A) = 14 D > 0. This means, by definition, c 2 that q has a minimum at ~0, since q(~x) > 0 = q(~0) for all ~x 6= ~0.

8.2.37

8.2.38 The eigenvalues of B are p − q and nq + p − q = p + (n − 1)q, so that B is positive definite if p − q > 0 and p + (n − 1)q > 0. 8.2.39 If ~v1 , . . . , ~vn is such a basis consisting of unit vectors, and we let A = [~v1 · · · ~vn ], then 1 cos θ · · · cos θ .. . cos θ 1 cos θ AT A = . is positive definite, so that, by Exercise 38, 1−cos θ > 0 and 1+(n−1) cos θ > 0 .. .. .. .. . . . cos θ cos θ · · · 1 1 1 or 0 < θ < arccos 1−n . or 1 > cos θ > 1−n

1

cos θ Conversely, if θ is in this range, then the matrix . ..

cos θ 1 .. .

··· .. . .. .

cos θ

cos θ is positive definite, so that it has a .. .

cos θ cos θ · · · 1 Cholesky factorization LLT . The columns of LT give us a basis with the desired property.

8.2.40 Let λ be the smallest eigenvalue of A. If we let k = 1 − λ, then the smallest eigenvalue of the matrix A + kIn will be λ + k = 1, so that all the eigenvalues of A + kIn will be positive. Thus matrix A + kIn will be positive definite, by Theorem 8.2.4. 8.2.41 The functions x21 , x1 x2 , x22 form a basis of Q2 , so that dim(Q2 ) = 3. 8.2.42 The functions xi xj form a basis of Qn , where 1 ≤ i ≤ j ≤ n. A little combinatorics shows that there are 1 + 2 + 3 + · · · + n = n(n + 1)/2 such functions, so that dim(Qn ) = n(n + 1)/2 8.2.43 Note that T (ax21 + bx1 x2 + cx22 ) = ax21 (we let x2 = 0). Thus im(T ) = span(x21 ), rank(T ) = 1, ker(T ) = span(x1 x2 , x22 ), nullity(T ) = 2. 8.2.44 Note that T (ax21 +bx1 x2 +cx22 ) = ax21 +bx1 +c (we let x2 = 1). Thus im(T ) = P2 , rank(T ) = 3, ker(T ) = {0}, nullity(T ) = 0 (T is an isomorphism). 8.2.45 Note that T (ax21 + bx22 + cx23 + dx1 x2 + ex1 x3 + f x2 x3 ) = ax21 + b + c + dx1 + ex1 + f (we let x2 = x3 = 1). Thus im(T ) = P2 and rank(T ) = 3. The kernel of T consists of the quadratic forms with a = 0, d + e = 0, and b + c + f = 0 (consider the coefficients of x21 , x1 , and 1). The general element of the kernel is q(x1 , x2 , x3 ) = (−c − f )x22 + cx23 − ex1 x2 + ex1 x3 + f x2 x3 = c(x23 − x22 ) + e(x1 x3 − x1 x2 ) + f (x2 x3 − x22 ). Thus ker(T ) = span(x23 − x22 , x1 x3 − x1 x2 , x2 x3 − x22 ) and nullity(T ) = 3. 8.2.46 Note that T (ax21 + bx22 + cx23 + dx1 x2 + ex1 x3 + f x2 x3 ) = ax21 + bx22 + cx21 + dx1 x2 + ex21 + f x1 x2 (we let x3 = x1 ). Thus im(T ) = Q2 and rank(T ) = 3. The kernel of T consists of the quadratic forms with 377

Chapter 8 a + c + e = 0, b = 0, and d + f = 0 (consider the coefficients of x21 , x22 , and x1 x2 ). The general element of the kernel is q(x1 , x2 , x3 ) = (−c − e)x21 + cx23 − f x1 x2 + ex1 x3 + f x2 x3 = c(x23 − x21 ) + e(x1 x3 − x21 ) + f (x2 x3 − x1 x2 ). Thus ker(T ) = span(x23 − x21 , x1 x3 − x21 , x2 x3 − x1 x2 ) and nullity(T ) = 3. 8.2.47 T (A + B)(~x) = ~x T (A + B)~x = ~x T A~x + ~x T B~x equals (T (A) + T (B))(~x) = T (A)(~x) + T (B)(~x) = ~x T A~x + ~x T B~x. The verification of the second axiom of linearity is analogous. By definition of a quadratic form, im(T ) = Qn : For every quadratic form q in Qn there is a symmetric n × n matrix A such that q = T (A). Thus, the rank of T is dim(Qn ) = n(n + 1)/2 (see Exercise 42). By the rank nullity theorem, nullity(T ) = dim(Rn×n ) − rank(T ) = n2 −

n(n + 1) n(n − 1) = 2 2

Next, let’s think about the kernel of T . In our solution to Exercise 23 we observed that T (A) = T ( 12 (A + AT )); note that matrix 21 (A + AT ) is symmetric. Now 21 (A + AT ) = 0 if (and only if) AT = −A, that is, if A is skew-symmetric. Thus the skew-symmetric matrices are in the kernel of T . Since the space of skew-symmetric matrices has the same dimension as ker(T ), namely, n(n − 1)/2, we can conclude that ker(T ) consists of all skew-symmetric n × n matrices. 0 0 1 8.2.48 The matrix of T with respect to the basis x21 , x1 x2 , x22 is A = 0 1 0 , with the eigenvalues 1,1, −1 and 1 0 0 0 1 1 corresponding eigenvectors 1 , 0 , 0 . Thus x1 x2 and x21 + x22 are eigenfunctions with eigenvalue 1, and 0 1 −1 x21 − x22 has eigenvalue −1. Yes, T is diagonalizable, since there is an eigenbasis.

1 0 0 8.2.49 The matrix of T with respect to the basis x21 , x1 x2 , x22 is A = 0 2 0 , with the eigenvalues 1,2,4 and 0 0 4 1 0 0 corresponding eigenvectors 0 , 1 , 0 . Thus x21 , x1 x2 , x22 are eigenfunctions with eigenvalues 1, 2, and 4, 0 0 1 respectively. Yes, T is diagonalizable, since there is an eigenbasis.

0 8.2.50 The matrix of T with respect to the basis x21 , x1 x2 , x22 is A = 2 0 1 1 1 corresponding eigenvectors 0 , 2 , −2 . −1 1 1

1 0 0 2 , with the eigenvalues 0,2, −2 and 1 0

Thus x21 − x22 , x21 + 2x1 x2 + x22 , x21 − 2x1 x2 + x22 are eigenfunctions with eigenvalues 0, 2, and −2, respectively. Yes, T is diagonalizable, since there is an eigenbasis.

8.2.51 If B is negative definite, then A = −B is positive definite, so that the determinants of all principal submatrices A(m) are positive. Thus det(B (m) ) = det(−A(m) ) = (−1)m det(A(m) ) is positive for even m and negative for odd m. 378

Section 8.2 8.2.52 Because aij = ~eTj A~ei , we have q(~ ei ) = aii . Further, using linearity q(~ei + ~ej ) = (~ei + ~ej )T A(~ei + ~ej ) = T T T T ~ei A~ei + ~ei A~ej + ~ej A~ei + ~ej A~ej = q(~ei ) + q(~ej ) + 2aij . Solving for aij gives aij = 12 (q(~ei + ~ej ) − q(~ei ) − q(~ej )). T

8.2.53 a. Because p (x, y) = q (x~ei + y~ej ) = (x~ei + y~ej ) A (x~ei + y~ej ) = aii x2 + aij xy + aji yx + ajj y 2 , this is a quadratic form with matrix B. b. If q is positive definite and (x, y) 6= (0, 0) , then p (x, y) = q (x~ei + y~ej ) > 0. c. If q is positive semidefinite, then p (x, y) = q (x~ei + y~ej ) ≥ 0 for all x, y. d. If q (x1 , x2 , x3 ) = x21 + x22 − x23 and we let i = 1, j = 2, then p(x, y) = q(x, y, 0) = x2 + y 2 is positive definite. 8.2.54 The entries a1j = aj1 must all be 0. To see that a1j = 0, consider the function p(x, y) = q(x~e1 + y~ej )defined a11 a1j 0 a1j in Exercise 8.2.53. By Exercise 53a, the symmetric matrix of p will be = . This a a a ajj j1 jj j1 0 a1j matrix is positive semidefinite, by Exercise 53c, implying that det = −a21j ≥ 0. Thus a1j = 0, as aj1 ajj claimed. 8.2.55 As the hint suggests, it suffices to prove that aij < aii or aij < ajj , implying that for every entry off the T diagonal there exists a larger entry on the diagonal. Now q (~ei − ~ej ) = (~ei − ~ej ) A (~ei − ~ej ) = aii −2aij +ajj > 0, or, aii + ajj > 2aij , proving the claim. 8.2.56 Let λ1 ≥ λ2 ≥ ... ≥ λn be the eigenvalues of A. In Exercise 8.2.27 we see that the range of q on unit vectors is the interval [λn , λ1 ]. Since a11 = q (~e1 ) is in that range, we must have a11 ≤ λ1 , as claimed. 8.2.57 Working in coordinates with respect to an orthonormal eigenbasis for A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 + λ3 c23 = 1, where the eigenvalues λ1 , λ2 , λ3 are positive. This level surface is an ellipsoid. 8.2.58 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 = 1, where the eigenvalues λ1 and λ2 are positive. This level surface is a cylinder. 8.2.59 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation√ q (~x) = 1 as λ1 c21 = 1, where the eigenvalue λ1 is positive. This level surface is a pair of parallel planes, c1 = ±1 λ1

8.2.60 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 + λ3 c23 = 1, where the eigenvalue λ1 is positive, while λ2 and λ3 are negative. This level surface is a hyperboloid of two sheets. 8.2.61 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 1 as λ1 c21 + λ2 c22 + λ3 c23 = 1, where the eigenvalues λ1 and λ2 are positive, while λ3 is negative. This level surface is a hyperboloid of one sheet. 8.2.62 Working in coordinates with respect to an orthonormal eigenbasis of A, we can write the equation q (~x) = 0 as λ1 c21 + λ2 c22 + λ3 c23 = 0, where the eigenvalues λ1 and λ2 are positive, while λ3 is negative. This level surface is a cone. 8.2.63 Note that w ~i · w ~ i = λ1i . Now q (c1 w ~ 1 + ... + cn w ~ n) = (c1 w ~ 1 + ... + cn w ~ n ) · A (c1 w ~ 1 + ... + cn w ~ n )= (c1 w ~ 1 + ... + cn w ~ n ) · (λ1 c1 w ~ 1 + ... + λn cn w ~ n )= λ1 c21 λ11 + ... + 379

Chapter 8 λn c2n λ1n = c21 + ... + c2n , as claimed.

8 −2 −2 5

, with

8.2.64 We will use the strategy outlined in Exercise 8.2.63. The symmetric matrix of q is A = 2 1 , ~v2 = √15 , with associated eigenvalues λ1 = 9 and λ2 = 4. an orthonormal eigenbasis ~v1 = √15 −1 2 2 1 1 1 1 √ √ Thus the orthogonal basis w ~ 1 = √1λ ~v1 = 3√ , w ~ = has the required property. (See ~ v = 2 λ2 2 5 2 5 1 −1 2 Figure 8.13.)

y 1 √ , √15 2 5

1 2

w2

x

− 21

w1

1 2 2 1 √ , − 3√ 3 5 5

− 21 Figure 8.13: for Problem 8.2.64.

8.2.65 Working in coordinates c1 , c2 with respect to an orthonormal eigenbasis ~v1 , ~v2 for A, we can write q (~x) = λ1 c21 +λ2 c22 , where the eigenvalue λ1 is positive while λ2 is negative. We define the orthogonal vectors w ~ 1 = √1λ ~v1 1 1 1 1 ~2 · w ~ 2 = (−λ2 ) . Now q (c1 w ~ 1 + c2 w ~ 2 ) = (c1 w ~ 1 + c2 w ~ 2) · ~1 · w ~ 1 = λ1 and w and w ~ 2 = √−λ ~v2 . Note that w 2 1 (λ1 c1 w ~ 1 + λ2 c2 w ~ 2 ) = λ1 c21 λ11 + λ2 c22 (−λ = c21 − c22 , as claimed. 2)

3 −5 −5 3

, with

8.2.66 We will use the strategy outlined in Exercise 8.2.65. The symmetric matrix of q is A = 1 1 , ~v2 = √12 an orthonormal eigenbasis ~v1 = √12 , with associated eigenvalues λ1 = 8 and λ2 = −2. −1 1 1 1 1 , w ~ 2 = √−λ has the required property that Thus the orthogonal basis w ~ 1 = √1λ ~v1 = 41 ~v2 = 21 1 2 −1 1 q (c1 w ~ 1 + c2 w ~ 2 ) = c21 − c22 . (See Figure 8.14.) 8.2.67 Consider an orthonormal eigenbasis ~v1 , ..., ~vn for A with associated eigenvalues λ1 , ..., λn , such that the eigenvalues λ1 , ..., λp are positive, λp+1 , ..., λr are negative, and the remaining eigenvalues are 0. Define a 380

Section 8.2

y

1

1 1 2, 2

w2 w1

-1

x

1 1, 1 4 −4

-1

Figure 8.14: for Problem 8.2.66. new orthogonal eigenbasis w ~ 1 , ..., w ~ n by setting w ~i =

.p |λi | ~vi for i = 1, ..., r and w ~ i = ~vi for i = 1

r + 1, ..., n. Note that w ~i · w ~ i = 1/λi for i = 1, ..., p and w ~i · w ~ i = 1/ (−λi ) for i = p + 1, ..., r. Now q (c1 w ~ 1 + ... + cp w ~ p + ... + cr w ~ r + ... + cn w ~ n ) = (c1 w ~ 1 + ... + cp w ~ p + ... + cr w ~ r + ... + cn w ~ n )·(λ1 c1 w ~ 1 + ... + λp cp w ~ p + ... + λ 1 2 2 2 2 = λ1 c21 λ11 + ... + λp c2p λ1p − ... − λr c2r (−λ = c + ... + c − c − ... − c , as claimed. p r 1 p+1 r) T 8.2.68 p (~x) = q (L (~x)) = q (R~x) = (R~x) A (R~x) = ~xT RT AR ~x, proving that p is a quadratic form with symmetric matrix RT AR. T T 8.2.69 If A is positive definite, then ~xT RT AR ~x = n(R~ ox) A (R~x) ≥ 0 for all ~x, meaning that R AR is positive T semidefinite. If A is positive definite and ker R = ~0 , then ~xT RT AR ~x = (R~x) A (R~x) > 0 for all nonzero T T ~x, meaning that RT AR is positive definite. n o Conclusion: R AR is always positive semidefinite, and R AR is positive definite if (and only if) ker R = ~0 , meaning that the rank of R is m.

8.2.70 Since A is indefinite, there exist vectors ~v1 and ~v2 in Rn such that ~v1T A~v1 > 0 and ~v2T A~v2 < 0. Since the n×m matrix R has rank n, we know that the image of R is all of Rn , so that there exist vectors w ~ 1 and w ~ 2 in Rm with T T T T T T ~ 2 = ~v2T A~v2 < 0, ~ 2 R ARw ~ 1 = (Rw ~ 1 ) A (Rw ~ 1 ) = ~v1 A~v1 > 0 and w Rw ~ 1 = ~v1 and Rw ~ 2 = ~v2 . Now w ~ 1 R ARw T proving that matrix R AR is indefinite. 8.2.71 Anything can happen. Consider the example A =

381

1 0 0 −1

1 0 , R1 = I2 , R2 = and R3 = . Then 0 1

Chapter 8 R1T AR1 = A is indefinite, R2T AR2 = [1] is positive definite, and R3T AR3 = [−1] is negative definite.

Section 8.3 8.3.1 σ1 = 2, σ2 = 1 8.3.2 The image of the unit circle is the unit circle, since the transformation defined by A preserves length. Thus σ1 = σ2 = 1 by Theorem 8.3.2. 8.3.3 AT A = In ; the eigenvalues of AT A are all 1, so that the singular values of A are all 1. q √ √ 1 1 , with eigenvalues λ1,2 = 3±2 5 . The singular values of A are σ1 = 3+2 5 = 8.3.4 AT A = 1 2 q √ √ 3− 5 σ2 = = −1+2 5 ≈ 0.62. 2

√ 1+ 5 2

≈ 1.62 and

2 p + q2 0 , with eigenvalues λ1 = λ2 = p2 + q 2 . The singular values of A are σ1 = σ2 = 8.3.5 AT A = 0 p2 + q 2 p p p2 + q 2 . A represents a rotation combined with a scaling, with a scaling factor of p2 + q 2 , so that the image p 2 2 of the unit circle is a circle with radius p + q . 8.3.6 The eigenvalues of AT A are λ1 = 25 and λ2 = 0, so that the singular values of A are σ1 = 5 and σ2 = 0 (these are also the eigenvalues of A; compare with Exercise 24). 1 1 1 works. The image of the unit circle is the line segment connecting the tips E5 = span , so that ~v1 = √5 2 2 of A~v1 = 5~v1 and A(−~v1 ) = −5~v1 . See Figure 8.15. 5 v1 v1

A

– v1 –5 v1

Figure 8.15: for Problem 8.3.6.

1 8.3.7 A A = 0 T

0 4

λ1 = 4, λ2 = 1; σ1 = 2, σ2 = 1 382

Section 8.3 1 0 , ~u1 = , ~v2 = eigenvectors of AT A : ~v1 = 0 1 0 1 2 0 . ,V = Σ= 1 0 0 1

1 v1 ) σ1 (A~

=

0 , ~u2 = −1

1 v2 ) σ2 (A~

8.3.9 AT A =

0 1 1 , , so that U = −1 0 0

p2 + q 2 0

p 0 2 2 p2 + q 2 2 2 ; λ1 = λ2 = p + q ; σ 1 = σ 2 = p +q 0 1 T , ~u1 = σ11 A~v1 = , ~v2 = eigenvectors of A A : ~v1 = 1 0 p √ 21 2 −q , so that U = √ 21 2 p −q , Σ = ( p2 + q 2 )I2 , V = I2 . p +q p +q p q p

8.3.8 AT A =

=

5 10

10 20

√ 21 2 p +q

p , ~u2 q

=

1 v2 σ2 A~

=

(See Exercise 6)

λ1 = 25, λ2 = 0; σ1 = 5, σ2 = 0; eigenvectors of AT A : 1 −2 1 1 1 1 1 √ √ √ , ~v2 = 5 , ~u1 = σ1 A~v1 = 5 , ~u2 = a unit vector orthogonal to ~v1 = 5 2 1 2 −2 1 −2 5 0 ~u1 = √15 so that U = V = √15 ,Σ= . 1 2 1 0 0 8.3.10 In Example 4 we found sides: 6 −7 = 2 6

√1 5

2 1 −1 2 ↑ U

6 −7

10 0 0 5

2 = 6

√1 5

√1 5

↑ Σ

1 2

1 2 −2 1

−2 1

10 0

0 5

√1 5

2 −1 1 2

; now take the transpose of both

.

↑ VT

1 0 8.3.11 A A = ; λ1 = 4, λ2 = 1; σ1 = 2, σ2 = 1 eigenvectors of AT A : 0 4 0 0 1 1 0 , ~u1 = σ11 A~v1 = 1 , ~u2 = σ12 A~v2 = 0 , ~u3 = 0 , , ~v2 = ~v1 = 0 1 1 0 0 0 1 0 2 0 0 1 U = 1 0 0 , Σ = ,V = . 0 1 1 0 0 0 1 T

8.3.12 In Example 5 we see that

0 1 1 1

1 = 0

√1 2

1 1

−1 1

Now take the transpose of both sides. 383

√

3 0 0 1

0 0

√1 6 √1 2 √1 3

√2 6

0 − √13

√1 6 − √12 √1 3

.

Chapter 8

0 1 1

1 √ 6 1 2 √ 1 = 6 0 √1 6

√1 2

0 − √12

√1 3 − √13 √1 3

↑ U

√ 3 0 0

0 1 1 √12 −1 0

↑ Σ

1 1

↑ VT

√ √ 37 16 8.3.13 A A = ; λ1 = 45, λ2 = 5; σ1 = 3 5, σ2 = 5 eigenvectors of AT A : 16 13 2 −1 1 0 , ~v2 = √15 , ~u1 = σ11 A~v1 = ~v1 = √15 , ~v2 = σ12 A~v2 = , so that 1 2 0 1 √ 2 −1 1 0 3 5 √0 1 √ . ,V = 5 U= ,Σ= 1 2 5 0 0 1 T

4 6 ; λ1 = 16, λ2 = 1; σ1 = 4, σ2 = 1 8.3.14 A A = 6 13 1 −2 1 1 T eigenvectors of A A : ~v1 = √5 , ~v2 = √5 , ~u1 = 2 1 −1 , so that = √15 2 2 −1 1 −2 4 0 1 1 √ √ U= 5 ,Σ= . ,V = 5 1 2 2 1 0 1 T

8.3.15 If A~v1 = σ1 ~u1 and A~v2 = σ2 ~u2 , then A−1 ~u1 = are the reciprocals of the singular values of A.

1 v1 σ1 ~

1 v1 σ1 A~

=

√1 5

2 , ~u2 = 1

and A−1 ~u2 =

1 v2 , σ2 ~

1 v2 σ2 A~

so that the singular values of A−1

2 8.3.16 If A = U ΣV T then A−1 = V Σ−1 U T and (A−1 )T A−1 = U Σ−1 U −1 . Thus (A−1 )T A−1 is similar to 2 Σ−1 , so that the eigenvalues of (A−1 )T A−1 are the squares of the reciprocals of the singular values of A. It follows that the singular values of A−1 are the reciprocals of those of A. 8.3.17 We need to check that A But A

~b·~ u1 v1 σ1 ~

+ ··· +

~b·~ um vm σm ~

~b·~ u1 v1 σ1 ~

=

+ ··· +

~b·~ u1 v1 σ1 A~

~b·~ um vm σm ~

+ ··· +

= projimA~b. (see Page 223).

~b·~ um vm σm A~

= (~b · ~u1 )~u1 + · · · + (~b · ~um )~um

= projimA~b, since ~u1 , . . . , ~um is an orthonormal basis of im(A) (see Theorem 5.1.5). 1 1 1 u2 = 2 1 , ~ 1 −0.1 ~b·~ ~b·~ u1 u2 ∗ . ~x = σ1 ~v1 + σ2 ~v2 = −3.2

1 2 8.3.18 ~b = , ~u1 = 3 4

1 1 1 v1 = 2 −1 , ~ −1

1 5

3 , ~v2 = −4

384

1 5

4 , σ1 = 2, σ2 = 1, so that 3

Section 8.3 8.3.19 ~x = c1~v1 + · · · + cm~vm is a least-squares solution if A~x = c1 A~v1 + · · · + cm A~vm = c1 σ1 ~u1 + · · · + cr σr ~ur = projimA~b. But projimA~b = (~b·~u1 )~u1 +· · ·+(~b·~ur )~ur , since ~u1 , . . . , ~ur is an orthonormal basis of im(A). Comparing ~ ui the coefficients of ~ui above we find that it is required that ci σi = ~b · ~ui or ci = b·~ σi , for i = 1, . . . , r, while no u1 v1 + · · · + condition is imposed on cr+1 , . . . , cm . The least-squares solutions are of the form ~x∗ = b·~ σ1 ~ cr+1~vr+1 + · · · cm~vm , where cr+1 , . . . , cm are arbitrary (see Exercise 17 for a special case).

~b·~ ur vr σr ~

+

8.3.20 a A = U ΣV T = U V T V ΣV T = QS, where Q = U V T and S = V ΣV T . Note that Q is orthogonal, being the product of orthogonal matrices; S is symmetric as S T = (V T )T ΣT V T = V ΣV T = S; and S is similar to Σ, so that the eigenvalues of S are the (nonnegative) diagonal entries of Σ. b Yes, write A = U ΣV T = U ΣU T U V T = S1 Q1 where S1 = U ΣU T and Q1 = U V T . 1 1 2 −1 10 0 1 2 √ √ 8.3.21 A = 2 0 5 5 −2 1 5 1 {z } | {z } | {z } | Σ U VT 1 1 2 −1 1 1 2 −1 2 1 10 0 1 2 √ √ √ √ = 2 2 0 5 5 −2 1 5 1 5 −1 2 5 1 {z }| {z }| {z } | {z } | {z } | Σ U V VT VT 1 4 3 9 −2 = −2 6 5 −3 4 | {z }| {z } Q S

8.3.22 a. T1 is the orthogonal projection onto the plane perpendicular to the vector ~v . T2 scales by the length of the vector ~v and T3 is a rotation about the line through the origin spanned by ~v by a rotation angle π/2. Because Q = A3 is orthogonal and S = A2 A1 is symmetric this is a polar decomposition: A = QS. b. Here, A1 represents the orthogonal projection onto the xz plane, A2 represents a scaling by a factor of 2, and A3 represents a rotation about the y axis through an angle of π/2, counterclockwise as viewed from the positive y axis:

0 0 0 0 , A2 0 1 0 Thus Q = A3 = 0 −1 1 A1 = 0 0

2 0 = 0 2 0 0 0 1 1 0 ,S 0 0

0 0 1 0 0 , A3 = 0 1 0 . −1 0 0 2 0 2 0 0 = A2 A1 = 0 0 0 , and A = QS = 0 −2 0 0 2

0 2 0 0 . 0 0

8.3.23 AAT U = U ΣV T V ΣT U T U = U ΣSigmaT , since V T V = Im and U T U = In , so that 2 σi ~ui for i = 1, . . . , r T AA ~ui = ~0 for i = r + 1, . . . , n The nonzero eigenvalues of AT A and AAT are the same. 8.3.24 The eigenvalues of AT A = A2 are the squares of the eigenvalues of A, so that the singular values of A are the absolute values of the eigenvalues of A. 385

Chapter 8 8.3.25 See Figure 8.16.

→

A unit circle

Au

→

σ2

u

σ1

Figure 8.16: for Problem 8.3.25. Algebraically: Write ~u = c1~v1 + c2~v2 and note that k~uk2 = c21 + c22 = 1. Then A~u = c1 σ1 ~u1 + c2 σ2 ~u2 , so that kA~uk2 = c21 σ12 + c22 σ22 ≥ c21 σ22 + c22 σ22 = σ22 and kA~uk ≥ σ2 . Likewise kA~uk ≤ σ1 . 8.3.26 Write ~v = c1~v1 + · · · + cm~vm and note that k~v k2 = c21 + · · · + c2m . Then A~v = c1 σ1 ~u1 + · · · + cr σr ~ur and kA~v k2 = c21 σ12 + c22 σ22 + · · · + c2r σr2 ≤ c21 σ12 + c22 σ12 + · · · + c2r σ12 ≤ σ12 kvk2 so that kA~v k ≤ σ1 k~v k. Likewise, kA~v k ≥ σm k~v k. 8.3.27 Let ~v be a unit eigenvector with eigenvalue λ and use Exercise 26. 8.3.28 If λ1 , . . . , λn are the eigenvalues of AT A, then (det A)2 = det(AT A) = λ1 · · · λn = σ12 · · · σn2 , so that | det A| = σ1 · · · σn . For a 2 × 2 matrix:

A unit circle

σ1 σ2 Ω

A(Ω)

Figure 8.17: for Problem 8.3.28. | det(A)| = expansion factor =

8.3.29 A = U ΣV T

area of ellipse A(Ω) = area of unit circle Ω

= [~u1 · · · ~ur · · ·]

σ1

..

. σr ..

0

.

πσ1 σ2 π

0 T ~v .1 . . T ~vr . .. 0 386

= σ1 σ2 . See Figure 8.17.

σ ~v T 1 1 .. = [~u1 · · · ~ur · · ·] . σr ~vrT 0

True or False = σ1 ~u1 ~v1T + · · · + σr ~ur ~vrT 1 2 2 −1 1 10 0 6 2 1 1 1 √ √ √1 [2 − 1] = 10 8.3.30 = √5 5 1 5 −2 5 −2 1 2 0 5 −7 6 2 4 −2 2 4 √1 [ 1 2 ] = +5 √15 + 5 1 −8 4 1 2

8.3.31 The formula A = σ1 ~u1 ~v1T + · · · + σr ~ur ~vrT gives such a representation. 8.3.32 (SAR)T SAR = RT AT S T SAR = RT AT AR is similar to AT A, so that the matrices AT A and (SAR)T SAR have the same eigenvalues. Thus A and SAR have the same singular values. 8.3.33 Yes; since AT A is diagonalizable and has only 1 as an eigenvalue, we must have AT A = In . 8.3.34 A = U ΣU T means that U T AU = U −1 AU = Σ, i.e., A is orthogonally diagonalizable, with eigenvalues ≥ 0. This is the case if and only if A is symmetric and positive semidefinite. 8.3.35 We will freely use the diagram on Page 393 (with r = m). We have AT A~vi = AT (σi ~ui ) = σi2~vi and therefore (AT A)−1~vi = σ12 ~vi for i = 1, . . . , m. Then (AT A)−1 AT ~ui = (AT A)−1 (σi~vi ) = σ1i ~vi for i = 1, . . . , m i and (AT A)−1 AT ~ui = ~0 for i = m + 1, . . . , n since ~ui is in ker(AT ) in this case. Note that (AT A)−1 AT ~ui is the least-squares solution of the equation A~x = ~ui ; for i = 1, . . . , m this is the exact solution since ~ui is in im(A). 8.3.36 We will freely use the diagram on Page 393. By construction of the ~vi as eigenvectors of AT A we have AT A~vi = λi~vi = σi2~vi , or (AT A)−1~vi = σ12 ~vi . Then A(AT A)−1 AT ~ui = A(AT A)−1 (σi~vi ) = A σ1i ~vi = σ1i A~vi = i T −1 T ~0 for i = m + 1, . . . , n since ~ui is in ker(AT ) in this case. The fact that ~ui for i = 1, . . . , m and A(A A) A ~ u = i ~ u if i = 1, . . . , m i A(AT A)−1 AT ~ui = means that the matrix A(AT A)−1 AT represents the orthogonal ~0 if i = m + 1, . . . , n projection onto im(A) = span(~u1 , . . . , ~um ).

True or False Ch 8.TF.1 F. The orthogonal matrix A =

0 −1 1 0

fails to be diagonalizable (over R).

3 , then the eigenvalue of AT A = [ 3 Ch 8.TF.2 T. If A = 4 √ A is σ = λ = 5.

3 = [25] is λ = 25, so that the singular value of 4] 4

Ch 8.TF.3 F. The last term, 5x2 , does not have the form required in Definition 8.2.1 Ch 8.TF.4 F. The singular values of A are the square roots of the eigenvalues of AT A, by Definition 8.3.1. Ch 8.TF.5 T, by Theorem 8.2.4. Ch 8.TF.6 T, by Definition 8.2.1 387

Chapter 8 2 λ1 . λ1 . 0 Ch 8.TF.7 T. If D = . . . , then DT D = D2 = . . 0 . 0 . λn p p and the singular values of D are λ21 = |λ1 |, . . . , λ2n = |λn |.

Ch 8.TF.8 F, since det

2 5 2

5 2

3

0 . . The eigenvalues of DT D are λ21 , . . . , λ2n , λ2n

= − 14 < 0 (see Theorem 8.2.7).

Ch 8.TF.9 T, by the spectral theorem (Theorem 8.1.1) Ch 8.TF.10 T. Note that [ 1

0]

a b b c

1 = a > 0, by Definition 8.2.3. 0

Ch 8.TF.11 T, by Theorem 8.3.2 Ch 8.TF.12 T. All four eigenvalues are negative, so that their product, the determinant, is positive. Ch 8.TF.13 T, by Theorem 8.1.2 Ch 8.TF.14 F, since the determinant is 0, so that 0 is an eigenvalue. Ch 8.TF.15 F. Consider A =

0 −1 . 1 0

Ch 8.TF.16 T, since AAT is symmetric (use the spectral theorem) Ch 8.TF.17 T, by Theorem 8.2.4: all the eigenvalues are positive. Ch 8.TF.18 T, since the matrix is symmetric. Ch 8.TF.19 F. Consider the shear matrix A = values fail to be 1, 1.

1 0

1 . The unit circle isn’t mapped into itself, so that the singular 1

Ch 8.TF.20 F. In general, (AT A)T = AT A 6= AAT Ch 8.TF.21 F. Let A = I2 , ~v =

1 1 . ,w ~= 1 0

P P Ch 8.TF.22 T. Consider the singular value decomposition A = U V T , or P AV = U , where V is orthogonal (see Theorem 8.3.5). We can let S = V , since the columns of AS = AV = U are orthogonal, by construction.

Ch 8.TF.23 T. By the spectral theorem, A is diagonalizable: S −1 AS = D for some invertible S and a diagonal D. Now Dn = S −1 An S = S −1 0S = 0, so that D = 0 (since D is diagonal). Finally, A = SDS −1 = S0S −1 = 0, as claimed. Ch 8.TF.24 F. If k is negative, then kq(~x) will be negative definite. 388

True or False Ch 8.TF.25 T. The eigenvalues λ1 , . . . , λn of A are nonzero, since A is invertible, so that the eigenvalues λ21 , . . . , λ2n of A2 are positive. Now use Theorem 8.2.4. Ch 8.TF.26 T, by Theorem 8.3.2, since ~v = A~e1 and w ~ = A~e2 are the principal semi-axis of the image of the unit circle. Ch 8.TF.27 F. As a counterexample, consider A = S = 2In . Ch 8.TF.28 T, since ~eiT A~ei = aii < 0. Ch 8.TF.29 T. By Theorem 7.3.6, matrices A and B have the same eigenvalues. Now use Theorem 8.2.4. Ch 8.TF.30 T. The spectral theorem guarantees that there is an orthogonal R such that RT AR is diagonal. Now let S = RT . Ch 8.TF.31 F. Let A =

0 1 0 0

and B =

1 0 . Then 1 is a singular value of BA but not of AB . 0 0

Ch 8.TF.32 T, since A + A−1 = A + AT is symmetric. Ch 8.TF.33 F. For example, (x21 )(x2 x3 ) fails to be a quadratic form. Ch 8.TF.34 T. We can write q(~x) = ~x

−1 0 Ch 8.TF.35 F. Consider A = 0 1 0 0

T

1

5 2

5 2

4

~x.

0 0 , which is indefinite. 1

Ch 8.TF.36 T, by Definition 8.2.3: ~x T (A + B)~x = ~x T A~x + ~x T B~x > 0 for all nonzero ~x. Ch 8.TF.37 T, since ~x · A~x is positive, so that cos θ is positive, where θ is the angle enclosed by ~x and A~x. Ch 8.TF.38 T. Preliminary remark: If σ is the largest singular value of an n×m matrix M , then kM~v k ≤ σk~v k for all ~v in Rm (see Exercise 8.3.26). Now let σ1 , σ2 be the singular values of matrix AB , with σ1 ≥ σ2 , and let ~v1 be a unit vector in R2 such that kAB~v1 k = σ1 (see Theorem 8.3.3). Now σ2 ≤ σ1 = kA(B~v1 )k ≤ 3kB~v1 k ≤ 3 · 5k~v1 k = 15, proving our claim; note that we have used the preliminary remark twice. 0 1 . Ch 8.TF.39 F. Consider A = 0 0

Ch 8.TF.40 T. If λ is the smallest eigenvalue of A, let k = 1 − λ. Then the smallest eigenvalue of A + kIn is λ + k = 1, so that all the eigenvalues of A + kIn are positive. Now use Theorem 8.2.4.

Ch 8.TF.41 T. The quadratic form q(x1 , x2 ) = [ x1

a 0 x2 ] b c 389

b d e

c x1 e 0 = ax21 + 2cx1 x2 + f x22 is positive x2 f

Chapter 8

definite. The matrix of this quadratic form is A = Thus af > c2 , as claimed.

c , and det(A) = af − c2 > 0 since A is positive definite. f

a c

1 −1 Ch 8.TF.42 F. Consider the positive definite matrix A = . −1 2

0 . −1

1 Ch 8.TF.43 F. Consider the indefinite matrix A = 0

cos x Ch 8.TF.44 T. By Theorem 8.3.2., the continuous function f (x) = A sin x

has the global maximum 5 and the cos x . By the global minimum 3. Note that the image of the unit circle consists of all vectors of the form A sin x cos c intermediate value theorem, f (c) = 4 for some c. Let ~u = (draw a sketch!). sin c

Ch 8.TF.45 T, since ~x T A2 ~x = −~x T AT A~x = −(A~x)T A~x = −kA~xk2 ≤ 0 for all ~x. Ch 8.TF.46 T. If λ1 , . . . , λn are the eigenvalues of AT A, then λ1 λ2 . . . λn = det(AT A) = (det A)2 . If √ √ σ1 = λ1 , . . . , σn = λn are the singular values of A, then √ σ1 σ2 . . . σn = λ1 λ2 . . . λn = | det A|, as claimed. Ch 8.TF.47 F. Note that the columns of S must be unit eigenvectors of A. There are two distinct real eigenvalues, λ1 , λ2 , and for each of them there are two unit eigenvectors, ±~v1 (for λ1 ) and ±~v2 (for λ2 ). (Draw a sketch!) Thus there are 8 matrices S, namely S = [ ±~v1 ±~v2 ] and S = [ ±~v2 ±~v1 ] Ch 8.TF.48 T. See the remark following Definition 8.2.1. Ch 8.TF.49 F. Some eigenvalues of A may be negative. Ch 8.TF.50 F. Consider the similar matrices A = and 3, while those of B are 0 and 5.

0 0

0 3

and B =

0 4 . Matrix A has the singular values 0 0 3

Ch 8.TF.51 T. Let ~v1 , ~v2 be an orthonormal eigenbasis, with A~v1 = ~v1 and A~v2 = 2~v2 . Consider a nonzero vector ~x = c1~v1 + c2~v2 ; then A~x = c1~v1 + 2c2~v2 . If c1 = 0, then ~x = c2~v2 and A~x = 2c2~v2 are parallel, and we are all set. Now consider the case when c1 6= 0. Then the angle between ~x and A~x is arctan(2c2 /c1 ) − arctan(c2 /c1 ); to see this, subtract the angle between ~v1 and ~x from the angle between ~v1 and A~x (draw a sketch). Let m = c2 /c1 and use calculus to see that the function f (m) = arctan(2m) − arctan(m) assumes its global maximum at m = √12 . √ √ The maximal angle between ~x and A~x is arctan( 2) − arctan(1/ 2) < 0.34 < π/6. √ a b 1 a Ch 8.TF.52 T. Let A = . By Theorem 8.3.2, A = = a2 + c2 < 5 (since the length of the c d 0 c semi-major axis of the image of the unit circle is less than 5). Thus a < 5 and c < 5. Likewise, b < 5 and d < 5.

390

True or False Ch 8.TF.53 T. We need to show that each entry aij = aji off the diagonal is smaller than some entry on the diagonal. Now (~ei −~ej )T A(~ei −~ej ) = aii + ajj − 2aij > 0, so that aii + ajj > 2aij . Thus the larger of the diagonal entries aii and ajj must exceed aij . Ch 8.TF.54 T. Let λ1 , ..., λm be the Pmdistinct eigenvalues of A, with the associated eigenspaces EA,λi . Since A is diagonalizable, we know that k=1 dim (EA,λi ) = n. By definition of an eigenvector ,EA,λi is a subspace of Pm EA3 ,λ3i , Since k=1 dim EA3 ,λ3i cannot exceed n, we must have EA,λi =EA3 ,λ3i for all eigenvalues. Applying the same reasoning to B and B 3 , we can conclude that EA,λi =EA3 ,λ3i = EB 3 ,λ3i = EB,λi . Since the diagonalizable matrices A and B have the same eigenvectors with the same eigenvalues, they must be equal.

391

Chapter 9

Chapter 9 Section 9.1 9.1.1 x(t) = 7e5t , by Theorem 9.1.1. 9.1.2 x(t) = −e · e−0.71t = −e1−0.71t , by Theorem 9.1.1. 9.1.3 P (t) = 7e0.03t , by Theorem 9.1.1. 2

9.1.4 This is just an antiderivative problem: y(t) = 0.8 t2 + C = 0.4t2 + C, and C = −0.8, so that y(t) = 0.4t2 − 0.8. 9.1.5 y(t) = −0.8e0.8t , by Theorem 9.1.1. 9.1.6 x dx = dt x2 2

= t + C, and

1 2

= 0 + C, so that

x2 2

=t+

1 2

x2 = 2t + 1 √ x(t) = 2t + 1 9.1.7 x−2 dx = dt −x−1 = t + C − x1 = t + C, and −1 = 0 + C, so that − x1 = t − 1 x(t) =

1 1−t ;

note that lim x(t) = ∞. x→1−

9.1.8 x−1/2 dx = dt √ 2x1/2 = t + C, and 2 4 = 0 + C, so that 2x1/2 = t + 4. 2 x(t) = 2t + 2 for t ≥ −4.

9.1.9 x−k dx = dt 1 1−k 1−k x

= t + C, and

1 1−k 1−k x

=t+

1 1−k

= C, so that

1 1−k

x1−k = (1 − k)t + 1 x(t) = ((1 − k)t + 1)1/1−k . 9.1.10 cos x dx = dt sin x = t + C, and C = 0. x(t) = arcsin(t) for |t| < 1. 392

Section 9.1 9.1.11

dx 1+x2

= dt

arctan(x) = t + C and C = 0. x(t) = tan(t) for |t| <

π 2.

9.1.12 We want ekt = 3t or ek = 3 or k = ln(3) :

dx dt

= ln(3)x.

9.1.13 a The debt in millions is 0.45(1.06)212 ≈ 104, 245, or about 100 billion dollars. b The debt in millions is 0.45e0.06·212 ≈ 150, 466, or about 150 billion dollars. t

9.1.14 a x(t) = e− 8270 , by Theorem 9.1.1 T

If T is the half-life, then e− 8270 =

1 2

T or − 8270 = ln

1 2

The half-life is about 5732 years.

or T = −8270 ln

1 2

≈ 5732.

t

t = ln(0.53) or t = −8270 ln(0.53) ≈ 5250. The b We want to find t such that e− 8270 = 1 − 0.47 = 0.53 or − 8270 Iceman died about 5000 years before A.D. 1991, or about 3000 B.C. The Austrian expert was wrong.

9.1.15 If a quantity P grows at an instantaneous rate of k %, then P (t) = P0 ekt/100 , by Theorem 9.1.1. The 2) doubling time T is given by P (T ) = P0 ekT /100 = 2P0 , or ekT /100 = 2 or kT /100 = ln 2 or k = 100(ln ≈ 69 k k since ln 2 ≈ 0.69. 9.1.16 See Figure 9.1. x2

x1

Figure 9.1: for Problem 9.1.16. 9.1.17 See Figure 9.2. 9.1.18 See Figure 9.3. 9.1.19 See Figure 9.4.

0 −1 9.1.20 A~x = 1 0

x1 x2

−x2 = . See Figure 9.5. x1 393

Chapter 9

Figure 9.2: for Problem 9.1.17.

Figure 9.3: for Problem 9.1.18.

Figure 9.4: for Problem 9.1.19. 1 cos(t) we will trace out the unit circle ~x(t) = . 0 sin(t) − sin(t) 0 −1 − sin(t) , as claimed. ~x(t) = equals = cos(t) 1 0 cos(t)

It appears that the trajectories will be circles. If we start at We can verify that

9.1.21 A~x =

0 1 0 0

d~ x dt

x1 x2

=

x2 0

(see Figure 9.6).

p , then the horizontal velocity will be q, so that The trajectories will be horizontal lines. If we start at q x1 (t) p + qt q 0 1 q d~ x ~x(t) = = . We can verify that dt = equals ~x(t) = , as claimed. x2 (t) q 0 0 0 0 394

Section 9.1 x2

(1, 0)

x1

Figure 9.5: for Problem 9.1.20. x2

x1

Figure 9.6: for Problem 9.1.21. x1 9.1.22 We are told that d~ x1 and dt = A~ A(~x1 + ~x2 ) = A~x, as claimed.

9.1.23 We are told that claimed.

d~ x1 dt

d~ x2 dt

= A~x2 . Let ~x(t) = ~x1 (t) + ~x2 (t). Then

= A~x1 . Let ~x(t) = k~x1 (t). Then

d~ x dt

=

d~ c dt

=

d kt x) dt (e ~

x x. Let ~c(t) = ekt ~x(t). Then 9.1.24 We are told that d~ dt = A~ (A + kIn )(ekt ~x) = (A + kIn )~c, as claimed.

d x1 ) dt (k~

=

d~ x dt

=

d~ x1 dt

+

d~ x2 dt

= A~x1 + A~x2 =

x1 = k d~ x1 = A(k~x1 ) = A~x, as dt = kA~

d kt dt e

x kt ~x + ekt d~ x + ekt A~x = dt = ke ~

x 9.1.25 We are told that d~ x. Let ~c(t) = ~x(kt). Using the chain rule we find that dt = A~ kA(~x(kt)) = kA~c(t), as claimed.

d~ c dt

=

d x(kt)) dt (~

x = k d~ dt ]kt =

To get the vector field kA~c we scale the vectors of the field A~x by k. 1 −2 −2t −2 3t 1 9.1.26 λ1 = 3, λ2 = −2; ~v1 = , ~v2 = . −e , c1 = 5, c2 = −1, so that ~x(t) = 5e 1 3 1 3 9.1.27 Use Theorem 9.1.3.

−4 3 2 −3

are λ1 = −6 and λ2 = −1, with associated eigenvectors ~v1 = The eigenvalues of A = 1 1 ~v2 = . The coordinates of ~x(0) = with respect to ~v1 and ~v2 are c1 = − 15 and c2 = 52 . 1 0 395

−3 2

and

Chapter 9 By Theorem 9.1.3 the solution is ~x(t) = − 51 e−6t

−3 1 + 25 e−t . 2 1

−3 1 1 5 1 2t −3 5 10t 1 9.1.28 λ1 = 2, λ2 = 10; ~v1 = , ~v2 = ; c1 = − 8 , c2 = 8 , so that ~x(t) = − 8 e + 8e . 2 2 2 2

−2 1 , ~v2 = ; c1 = −2, c2 = 1, 1 2 1 4 1 −2 . + e5t = + e5t so that ~x(t) = −2 2 −2 2 1

9.1.29 λ1 = 0, λ2 = 5; ~v1 =

−2 1 −2 2 9.1.30 λ1 = 0, λ2 = 5, ~v1 = , ~v2 = ; c1 = −1, c2 = 0, so that ~x(t) = − . = 1 2 −1 1

1 9.1.31 λ1 = 1, λ2 = 6, λ3 = 0; ~v1 = −2 . Since ~x(0) = ~v1 we need not find ~v2 and ~v3 . 1 1 c1 = 1, c2 = c3 = 0, so that ~x(t) = et −2 . 1

In Exercises 32 to 35, find the eigenvalues and eigenspaces. Then determine the direction of the flow along the eigenspaces (outward if λ > 0 and inward if λ < 0). Use Figure 11 of Section 9.1 as a guide to sketch the other trajectories.

9.1.32 See Exercise 26 and Figure 9.7. E3

E–2

Figure 9.7: for Problem 9.1.32. 9.1.33 See Exercise 27 and Figure 9.8. 9.1.34 See Exercise 28 and Figure 9.9. 9.1.35 See Exercise 29 and Figure 9.10. In Exercises 36 to 39, find the eigenvalues and eigenspaces (the eigenvalues will always be positive). Then determine the direction of the flow along the eigenspaces (outward if λ > 1 and inward if 1 > λ > 0). Use Figure 11 of Section 7.1 as a guide to sketch the other trajectories. 396

Section 9.1

E–1

E–6

Figure 9.8: for Problem 9.1.33. E10

E2

Figure 9.9: for Problem 9.1.34. E5

E0

Figure 9.10: for Problem 9.1.35.

9.1.36 See Figure 9.11. 9.1.37 See Figure 9.12. 9.1.38 See Figure 9.13. 9.1.39 See Figure 9.14. 2 3t 3 +e 9.1.40 ~x(t) = e 4 3 2t

397

Chapter 9 E1.3

E0.8

Figure 9.11: for Problem 9.1.36. E1.6

E1.1

Figure 9.12: for Problem 9.1.37.

E0.7

E0.9

Figure 9.13: for Problem 9.1.38.

E1 E1.4

Figure 9.14: for Problem 9.1.39.

398

Section 9.1 2 and We want a 2 × 2 matrix A with eigenvalues λ1 = 2 and λ2 = 3 and associated eigenvectors ~v1 = 3 −1 11 −6 −4 3 4 9 2 3 4 9 4 9 2 3 3 . = = or A = = ; that is A ~v2 = 12 −6 3 −2 6 12 3 4 6 12 6 12 3 4 4 9.1.41 The trajectories are of the form ~x(t) = c1 eλ1 t~v1 + c2 eλ2 t~v2 = c1~v1 + c2 eλ2 t~v2 . See Figure 9.15. span (v2)

span (v1)

Figure 9.15: for Problem 9.1.41. 9.1.42 a The term 0.8x in the second equation indicates that species y is helped by x, while species x is hindered by y (consider the term −1.2y in the first equation). Thus y preys on x. b See Figure 9.16. y E1 = span

1 2

E1 = span

3 1

x

Figure 9.16: for Problem 9.1.42b. y(t) t→∞ x(t)

c If

y(0) x(0)

< 2 then both species will prosper, and lim

If

y(0) x(0)

≥ 2 then both species will die out.

= 31 .

9.1.43 a These two species are competing as each is hindered by the other (consider the terms −y and −2x). b Although only the first quadrant is relevant for our model, it is useful to consider the phase portrait in the other quadrants as well. See Figure 9.17. y(0) x(0) y(t) x(t) =

c If

> 2 then species y wins (x will die out); if

y(0) x(0)

< 2 then x wins. If

2 for all t. 399

y(0) x(0)

= 2 then both will prosper and

Chapter 9 y E3 = span

1 2

x

E6 = span

1 –1

Figure 9.17: for Problem 9.1.43b. 9.1.44 a The two species are in symbiosis: Each is helped by the other (consider the terms 4y and 2x). b See Figure 9.18. y

E3 = span

2 1

x

E–3 = span

1 –1

Figure 9.18: for Problem 9.1.44b. y(t) t→∞ x(t)

c Both populations will prosper and lim

= 21 , regardless of the initial populations.

9.1.45 a Species y has the more vicious fighters, since they kill members of species x at a rate of 4 per time unit, while the fighters of species x only kill at a rate of 1. b See Figure 9.19. c If

y(0) x(0)

<

1 2

then x wins; if

y(0) x(0)

>

1 2

then y wins; if

y(0) x(0)

=

1 2

nobody will survive the battle.

9.1.46 Look at the phase portrait in Figure 9.20. 9.1.47 a The two species are in symbiosis: Each is helped by the other (consider the positive terms kx and ky). b λ1,2 =

√ −5± 9+4k2 2

Both eigenvalues are negative if

√

9 + 4k 2 < 5 or 9 + 4k 2 < 25 or 4k 2 < 16 or k < 2 (recall that k is positive). 400

Section 9.1 y E–2 = span

2 1

x

E2 = span

2 –1

Figure 9.19: for Problem 9.1.45b. y E–√pq = span

√p √q

x

E√pq = span

√p –√q

Figure 9.20: for Problem 9.1.46. If k = 2 then the eigenvalues are −5 and 0. If k > 2 then there is a positive and a negative eigenvalue. c See Figure 9.21.

k=1

k=3

k=2 E0

both species die out

E–5 system approaches an equilibrium state

both species prosper

Figure 9.21: for Problem 9.1.47c.

9.1.48 a Symbiosis 401

Chapter 9 b λ1,2 =

√ −5± 9+4k 2

Both eigenvalues are negative if

√

9 + 4k < 5 or 9 + 4k < 25 or 4k < 16 or k < 4.

If k = 4 then the eigenvalues are −5 and 0. If k > 4 then there is a positive and a negative eigenvalue. c k = 1: See corresponding figure in Exercise 47 and Figure 9.22. k=4

k = 10

E0

E1

E–6 E–5 system approaches an equilibrium state

system approaches an equilibrium state

Figure 9.22: for Problem 9.1.48c. −1 −0.2 , λ1 = −0.4, λ2 = −0.8 9.1.49 A = 0.6 −0.2 −1 1 E−0.4 = span , E−0.8 = span 3 −1 g(0) −1 1 = 15 + 45 , so that c1 = 15, c2 = 45. h(0) 3 −1 1 g(t) −0.8t −0.4t −1 , so that + 45e = 15e −1 3 h(t)

g(t) = −15e−0.4t + 45e−0.8t h(t) = 45e−0.4t − 45e−0.8t . See Figure 9.23. 9.1.50 We want both eigenvalues λ1 and λ2 to be negative, so that tr(A) = λ1 + √λ2 < 0 and det(A) = λ1 λ2 > 0. tr(A)± (trA)2 −4 det(A) are both negative. Conversely, if tr(A) < 0 and det(A) > 0, then the two eigenvalues λ1,2 = 2 Answer: tr(A) < 0 and det(A) > 0. 9.1.51 ith component of

d x) dt (S~

=

d dt (si1 x1 (t)

+ si2 x2 (t) + · · · + sin xn (t))

dx2 dxn 1 = si1 dx dt + si2 dt + · · · + sin dt

402

Section 9.1

30 0

E–0.4

E–0.8

Figure 9.23: for Problem 9.1.49. x = ith component of S d~ dt

0 1 p + qt p = ~x are of the form , where ~x(0) = , by Exercise 21. Since 9.1.52 The solutions of 0 0 q q 0 1 λ 1 λt p + qt , by Exercise 24. , the solutions of the given system are of the form ~x(t) = e = λI2 + q 0 0 0 λ The zero state is a stable equilibrium solution if and only if λ < 0. The case λ = 0 is discussed in Exercise 21. See Figure 9.24. d~ x dt

λ>0

λ<0

Figure 9.24: for Problem 9.1.52.

1 0 −1 cos(t) x 9.1.53 For the initial value , the system d~ = ~ x has the solution ~ x (t) = , by Exercise 20; the dt 0 sin(t) 0 1 cos(qt) 0 −q p −q d~ x d~ x , by Exercise 25; and the system dt = has the solution ~x(t) = ~x has system dt = sin(qt) q 0 q p cos(qt) p −q 0 −q , by Exercise 24 write the solution ~x(t) = ept = pI2 + . See Figure 9.25. sin(qt) q p q 0 9.1.54 A =

0 −2

1 1 1 , λ1 = −1, λ2 = −2; ~v1 = and ~v2 = . See Figure 9.26. −3 −1 −2

In the case of trajectory 3 the door will slam: Initially the door is opened just a little (θ is small) and given a θ(0) strong push to close it (ω is large negative). More generally, the door will slam if the point representing ω(0) 403

Chapter 9

p=0

p>0

p<0

Figure 9.25: for Problem 9.1.53. ω

θ trajectory 1 E–1 trajectory 2

door slams trajectory 3

E–2

Figure 9.26: for Problem 9.1.54.

the initial state is located below the line E−2 = span

1 , that is, if −2

ω(0) θ(0)

< −2.

p 0 1 , λ1,2 = 12 −q ± q 2 − 4p ; note that both eigenvalues are negative. −p −q 1 1 . and Eλ2 = = span λ2 λ1

9.1.55 A = Eλ1

See Figure 9.27. In the case of trajectory 3 the door will slam: Initially the door is opened just a little (θ is small) and given a strong push to close it (ω is large negative). More generally, the door will slam if the point θ(0) 1 representing the initial state is located below the line Eλ2 = span , that is, if ω(0) θ(0) < λ2 . ω(0) λ2

Section 9.2 9.2.1 By Euler’s formula (Theorem 9.2.2), e2πi = cos(2π) + i sin(2π) = 1. 1

9.2.2 By Euler’s formula (Theorem 9.2.2), e 2 πi = cos 9.2.3 r =

p

(−1)2 + 12 =

√ 2

π 2

+ i sin

404

π 2

= i.

Section 9.2 ω

θ trajectory 1 Eλ 1 trajectory 2

door slams

Eλ

trajectory 3

2

Figure 9.27: for Problem 9.1.55. –1 + i

i r

θ

–1

1

Figure 9.28: for Problem 9.2.3. θ=

3π 4 ,

so that z =

√

3

2e 4 πi . See Figure 9.28.

9.2.4 e3it = cos(3t) + i sin(3t) t=

π 6

i t=

π 3

i –i

t = 0,

2π 3

–i t=

π 2

Figure 9.29: for Problem 9.2.4. Period T given by 3T = 2π or T =

2π 3 .

See Figure 9.29.

9.2.5 e−0.1t−2it = e−0.1t e−2it = e−0.1t (cos(2t) − i sin(2t)) spirals inward, in clockwise direction. See Figure 9.30. 9.2.6 λ1,2

3−i = ±i; Ei = ker 5

3−i 3+i −2 . and E−i = span = span 5 5 −3 − i

General solution: 405

Chapter 9

t=

3π 4

–1 t=

π 2

i

t=π 3π t= 2

1 t=0

t= –i

π 4

Figure 9.30: for Problem 9.2.5.

~x(t) = c1 eit

3−i 3+i + c2 e−it 5 5

6 cos(t) − 2 sin(t) 3−i 3+i . = + (cos(t) − i sin(t)) If c1 = c2 = 1 then ~x(t) = (cos(t)+i sin(t)) 10 cos(t) 5 5

9.2.7 det(A) = −2, so that the zero state is not stable, by Theorem 9.2.5. 9.2.8 Recall that all eigenvalues of a symmetric matrix are real. The zero state is stable if (and only if) all eigenvalues are negative (by Theorem 9.2.4); this is the case if (and only if) the matrix is negative definite (by Theorem 8.2.2). 9.2.9 The eigenvalues are conjugate complex, λ1,2 = p ± iq, and tr(A) = 2p < 0, so that p is negative. By Theorem 9.2.4, the zero state is stable. 9.2.10 a The matrix of the system is

2a 2b

2b , which is 2A, so that 2c

d~ x dt

= grad(q) = 2A~x.

b By Theorem 8.2.7, the level curves are ellipses. From multivariable calculus we know that grad (q) is perpendicular to the level curve, pointing inwards, so that all trajectories will approach the origin: The zero state is stable. See Figure 9.31. c By Theorem 8.2.7, the level curves are hyperbolas, as shown in Figure 9.32. λ 1 > 0 > λ2 x d The zero state is a stable equilibrium solution of the system d~ x if (and only if) the eigenvalues dt = grad(q) = 2A~ of 2A (and A) are negative. This means that the quadratic form q(~x) = ~x · A~x is negative definite.

9.2.11 a q(~x) = 2ai1 xi x1 + 2ai2 xi x2 + · · · + aii x2i + · · · + 2ain xi xn + terms not involving xi , so that x x. 2ai2 x2 + · · · + 2aii xi + · · · + 2ain xn and d~ dt = grad(q) = 2A~

∂q ∂xi

= 2ai1 x1 +

The matrix of the system is B = 2A. x x if (and only if) all the eigenvalues b The zero state is a stable equilibrium solution of the system d~ dt = grad(q) = 2A~ of 2A are negative. This means that the quadratic form q(~x) = ~x · A~x is negative definite.

406

Section 9.2

Eλ1

q = –1 q = –4 q = –9

Eλ2

Figure 9.31: for Problem 9.2.10b.

q=4 q = –4

q=1

q = –1

q = –1 q = –4 q=1

Eλ2

q=4

Eλ1

Figure 9.32: for Problem 9.2.10c. 9.2.12 We will show that the real parts of all the eigenvalues are negative, so that the zero state is a stable equilibrium solution. Now the characteristic polynomial of A is fA (λ) = −λ3 − 2λ2 − λ − 1. It is convenient to get rid of all these minus signs: The eigenvalues are the solutions of the equation g(λ) = λ3 + 2λ2 + λ + 1 = 0. Since g(−1) = 1 and g(−2) = −1, there will be an eigenvalue λ1 between -2 and -1. Using calculus (or a graphing calculator), we see that the equation g(λ) = 0 has no other real solutions. Thus there must be two complex 1 conjugate eigenvalues p ± iq. Now the sum of the eigenvalues is λ1 + 2p = tr(A) = −2, and p = −2−λ will be 2 negative , as claimed. The graph of g(λ) is shown in Figure 9.33. 9.2.13 Recall that the zero state is stable if (and only if) the real parts of all eigenvalues are negative. Now the eigen 1 −1 . = pp−iq values of A are the reciprocals of those of A; the real parts have the same sign if λ = p + iq, then λ1 = p+iq 2 +q 2 407

Chapter 9

g(λ) = λ3 + 2λ2 + λ + 1

(–1, 1)

(0, 1) λ1 (–2, –1)

Figure 9.33: for Problem 9.2.12. i 9.2.14 a For i > 1, dx dt = −ki xi + xi−1 . This means that in the absence of quantity xi−1 (t), the quantity xi (t) will decay exponentially, but the presence of xi−1 helps xi to grow.

For i = 1, the beginning of the loop, of x1 .

dx1 dt

= −k1 x1 − bxn , so that the presence of xn contributes to the decrease

−k1 b If n = 2 then the matrix of the system is A = 1 so that the zero state is stable, by Theorem 9.2.5.

−b −k2

with tr(A) = −k1 −k2 < 0 and det(A) = k1 k2 +b > 0,

−1 0 −b c No; consider the case k1 = k2 = k3 = 1 for simplicity. Then the matrix of the system is A = 1 −1 0 0 1 −1 √ 3 3 and fA (λ) = − (λ + 1) − b. Matrix A has the real eigenvalue λ1 = − b − 1 and the √complex . eigenvalues √ λ2,3 = 3 b e±πi/3 − 1. (Compare with Exercise 7.5.5.) The real part of λ2 and λ3 is 3 b 2 − 1, which is positive if b exceeds 8. The zero state fails to be stable in that case. 9.2.15 The eigenvalues are λ1 = tr(A) > 0 and λ2 = 0. See Figure 9.34.

E0

Eλ1

Figure 9.34: for Problem 9.2.15.

0 1 a b both negative.

9.2.16 If A =

then tr(A) = b and det(A) = −a. By Theorem 9.2.5, the zero state is stable if a and b are

408

Section 9.2 −1 k then tr(A) = −2 and det(A) = 1 − k 2 . By Theorem 9.2.5, the zero state is stable if k −1 det(A) = 1 − k 2 > 0, that is, if |k| < 1.

9.2.17 If A =

9.2.18 If λ1 , λ2 , λ3 are real and negative, then tr(A) = λ1 + λ2 + λ3 < 0 and det(A) = λ1 λ2 λ3 < 0. If λ1 is real and negative and λ2,3 = p ± iq, where p is negative, then tr(A) = λ1 + 2p < 0 and det(A) = λ1 (p2 + q 2 ) < 0. Either way, both trace and determinant are negative.

1 0 0 9.2.19 False, consider A = 0 2 0 . 0 0 −4 9.2.20 Use Theorem 9.2.6, with p = 0, q = π; a = 1, b = 0. 1 cos(πt) − sin(πt) = (cos(πt))w ~ + (sin(πt))~v . See Figure 9.35. ~x(t) = [ w ~ ~v ] 0 sin(πt) cos(πt)

( )

x 12 = v x(0) = x(2) = w

x(1) = –w –v

Figure 9.35: for Problem 9.2.20.

9.2.21 a

db

dt ds dt

= 0.05b+ =

s 0.07s

b λ1 = 0.07, λ2 = 0.05; ~v1 =

and

b(0) 1, 000 = s(0) 1, 000

1 50 ; ~x(0) = 1, 000~v1 − 49, 000~v2 ; so that , ~v2 = 0 1

b(t) = 50, 000e0.07t − 49, 000e0.05t and s(t) = 1, 000e0.07t . 0 1 , E0.5 = span 9.2.22 λ1 = 3, λ2 = 0.5; E3 = span 1 −1

System is discrete so choose VII. 9.2.23 λ1,2 = − 12 ± i, r > 1, so that trajectory spirals outwards. Choose II. 9.2.24 λ1 = 3, λ2 = 0.5, E3 =

0 1 . , E0.5 = 1 −1

System is continuous, so choose I. 1 9.2.25 negative so that trajectories spiral inwards in the counterclockwise direction λ1,2 = − 2 ± i; realpart is −1.5 1 x . Choose IV. then d~ if ~x = dt = 2 0

409

Chapter 9

9.2.26 λ1 = 1, λ2 = −2; E1 = span

1 0 . , E−2 = span −1 1

System is continuous so choose V. 9.2.27 λ1,2 = ±3i, E3i = span

1 0 0 1 +i , so that p = 0, q = 3, w ~= , ~v = . 0 −1 −1 0

Now use Theorem 9.2.6: 0 1 cos(3t) ~x(t) = e0t −1 0 sin(3t)

− sin(3t) cos(3t)

a sin(3t) = b − cos(3t)

2 0 9.2.28 λ1,2 = ±6i, E6i = span +i , so that 0 3 0 2 cos(6t) − sin(6t) a 2 sin(6t) ~x(t) = = 3 0 sin(6t) cos(6t) b 3 cos(6t)

cos(3t) sin(3t)

2 cos(6t) −3 sin(6t)

1 0 +i , so that 0 1 sin(4t) a − sin(4t) = e2t cos(4t) b cos(4t)

a b

a . b

9.2.29 λ1,2 = 2 ± 4i, E2+4i = span ~x(t) = e2t

0 1 1 0

cos(4t) sin(4t)

cos(4t) − sin(4t)

a . b

0 5 , so that +i 1 3 5 sin(3t) a − sin(3t) = e−2t cos(3t) + 3 sin(3t) b cos(3t)

9.2.30 λ1,2 = −2 ± 3i, E−2+3i = span ~x(t) = e−2t

0 5 1 3

cos(3t) sin(3t)

9.2.31 λ1,2 = −1 ± 2i, E−1+2i

5 cos(3t) − sin(3t) + 3 cos(3t)

1 0 0 = span +i , so that p = −1, q = 2, w ~= , 0 −1 −1

1 1 ~v = . Now = ~x(0) = w ~ + ~v , so that a = 1 and b = 1. 0 −1 0 1 cos(2t) − sin(2t) sin(2t) + cos(2t) 1 Then ~x(t) = e−t . = e−t −1 0 sin(2t) cos(2t) sin(2t) − cos(2t) 1 See Figure 9.36. 1 0 0 1 9.2.32 λ1,2 = ±2i, E2i = span +i , ~x(0) = 0 +1 , so that a = 0 and b = 1. 0 2 2 0 0 1 cos(2t) − sin(2t) 0 cos(2t) ~x(t) = = . See Figure 9.37. 2 0 sin(2t) cos(2t) 1 −2 sin(2t) 1 0 9.2.33 λ1,2 = ±i, Ei = span +i 1 1 0 1 cos(t) a = 1, b = 0, so that ~x(t) = 1 1 sin(t)

− cos(t) cos(t)

1 sin(t) = 0 sin(t) + cos(t)

410

a . b

Section 9.2

1

1

Figure 9.36: for Problem 9.2.31.

1 x (0) = x(π) = 1

1 0

–1 x π2 = 0

( )

0 x π4 = –2

( )

Figure 9.37: for Problem 9.2.32.

= cos(t)

0 1 + sin(t) . See Figure 9.38. 1 1

3 1 +i −2 0 1 3 cos(2t) − sin(2t) cos(2t) + 3 sin(2t) 1 a = 1, b = 0, so that ~x(t) = et . See Figure 9.39. = et 0 −2 sin(2t) cos(2t) −2 sin(2t) 0

9.2.34 λ1,2 = 1 ± 2i, E1+2i = span

9.2.35 If z = f + ig and w = p + iq then zw = (f p − gq) + i(f q + gp), so that (zw)′ = (f ′ p + f p′ − g ′ q − gq ′ ) + i(f ′ q + f q ′ + g ′ p + gp′ ). Also z ′ w = (f ′ + ig ′ )(p + iq) = (f ′ p − g ′ q) + i(f ′ q + g ′ p) and zw′ = (f + ig)(p′ + iq ′ ) = (f p′ − gq ′ ) + i(gp′ + f q ′ ). We can see that (zw)′ = z ′ w + zw′ , as claimed. 9.2.36 A =

0 −b

1 −c

and fA (λ) = λ2 + cλ + b, with eigenvalues λ1,2 = 411

√ −c± c2 −4b . 2

Chapter 9

x (0) = x(2π) =

0 1

1 x π2 = 1

( )

x(π) =

0 –1

Figure 9.38: for Problem 9.2.33.

Figure 9.39: for Problem 9.2.34. √ 9.2.36 a If c = 0 then λ1,2 = ±i b. The trajectories are ellipses. See Figure 9.40. v

x

Figure 9.40: for Problem 9.2.36a. The block oscillates harmonically, with period

2π √ . b

The zero state fails to be asymptotically stable. 412

Section 9.2 b λ1,2 =

√ −c±i 4b−c2 2

The trajectories spiral inwards, since Re(λ1 ) = Re(λ2 ) = − 2c < 0. This is the case of a damped oscillation. The zero state is asymptotically stable. See Figure 9.41.

Figure 9.41: for Problem 9.2.36b. c This case is discussed in Exercise 9.1.55. The zero state is stable here. is differentiable when z(t) 6= 0, since both the real and the imaginary parts are differentiable if z = p + iq then ′ ′ ′ ′ To find z1 , apply the product rule to the equation z z1 = 1: z ′ z1 + z z1 = 0, so that z1 = − zz2 .

9.2.37 a

b

z ′ w

9.2.38 a

1 z(t)

′

= z ′ w1 + z

=

z ′ 1 z2 −z1 z ′ 2 z22

= z w1

z1 z2

′

1 ′ w

=

=

z′ w

−

zw′ w2

λz1 z2 −λz1 z2 z22

=

z ′ w−zw′ w2

= 0, so that

z1 (t) z2 (t)

= k, a constant.

Now z1 (t) = kz2 (t); substituting t = 0 gives 1 = z1 (0) = kz2 (0) = k, so that z1 (t) = z2 (t), as claimed. b Let z2 (t) = ept (cos(qt) + i sin(qt)) be the solution constructed in the text (see Page 413). Since z2 (t) 6= 0 for all t, this is the only solution, by Part a.

λ 1 9.2.39 Let A = 0 λ 0 0 3 c3 (t), dc = 0. dt

0 1 . We first solve the system λ

c3 (t) = k3 , a constant, so that

dc2 dt

d~ c dt

0 = (A − λI3 )~c = 0 0

= k3 and c2 (t) = k3 t + k2 . Likewise c1 (t) =

1 0 0 1 ~c, or 0 0 k3 2 2 t

dc1 dt

2 = c2 (t), dc dt =

+ k2 t + k1 .

Applying Exercise 9.1.24, with k = −λ, we find that ~c(t) = e−λt ~x(t) or ~x(t) = eλt~c(t) k1 + k2 t + k23 t2 where k1 , k2 , k3 are arbitrary constants. The zero state is stable if (and only if) the real k2 + k3 t = eλt k3 part of λ is negative. 9.2.40 a B(t) = 1000(1 + 0.05i)t = 1000(r(cos θ + i sin θ))t = 1000rt (cos(θt) + i sin(θt)), where √ r = 1 + 0.052 > 1 and θ = arctan(0.05) ≈ 0.05. See Figure 9.42. 413

1 z

=

Chapter 9 b B(t) = 1000e0.05i = 1000(cos(0.05t) + i sin(0.05t)). See Figure 9.42.

1000 1000

1000

trajectory to part a slowly spiral outwards

trajectory to part b circle 2π ≅ period = 126 (years) 0.05

Figure 9.42: for Problem 9.2.40. c We would choose an account with annual compounding, since the modulus of the balance grows in this case. In the case of continuous compounding the modulus of the balance remains unchanged.

Section 9.3 9.3.1 The characteristic polynomial of this differential equation is λ − 5, so that λ1 = 5. By Theorem 9.3.8 the general solution is f (t) = Ce5t , where C is an arbitrary constant. −3t , where C is an arbitrary constant, and the differential 9.3.2 The solutions of dx dt +3x = 0 are of the form x(t) = Ce dx equation dt + 3x = 7 has the particular solution xp (t) = 73 , so that the general solution is x(t) = Ce−3t + 37 (where C is a constant). Alternatively, we could use Theorem 9.3.13.

9.3.3 Use Theorem 9.3.13, where a = −2 and g(t) = e3t : Z Z 1 5t 1 f (t) = e−2t e2t e3t dt = e−2t e5t dt = e−2t e + C = e3t + Ce−2t , where C is a constant. 5 5 9.3.4 We can look for a sinusoidal solution xp (t) = P cos(3t) + Q sin(3t), as in Example 7. P and Q needto be −2P + 3Q = 1 with chosen in such a way that −3P sin(3t) + 3Q cos(3t) − 2P cos(3t) − 2Q sin(3t) = cos(3t) or −3P − 2Q = 0 3 dx 2 2t solution P = − 13 and Q = 13 . Since the general solution of dt − 2x = 0 is x(t) = Ce , the general solution of dx 2 3 2t dt − 2x = cos(3t) is x(t) = Ce − 13 cos(3t) + 13 sin(3t), where C is an arbitrary constant. t

9.3.5 Using Theorem 9.3.13, f (t) = e stant.

Z

2t

9.3.6 Using Theorem 9.3.13, f (t) = e

Z

e−t t dt = et (−te−t − e−t + C) = Cet − t − 1, where C is an arbitrary con-

−2t 2t

e

2t

e dt = e

Z

414

dt = e2t (t + C), where C is an arbitrary constant.

Section 9.3 9.3.7 By Definition 9.3.6, pT (λ) = λ2 + λ − 12 = (λ + 4)(λ − 3). Since pT (λ) has distinct roots λ1 = −4 and λ2 = 3, the solutions of the differential equation are of the form f (t) = c1 e−4t + c2 e3t , where c1 and c2 are arbitrary constants (by Theorem 9.3.8). 9.3.8 pT (λ) = λ2 + 3λ − 10 = (λ + 5)(λ − 2) = 0 x(t) = c1 e−5t + c2 e2t , where c1 , c2 are arbitrary constants. 9.3.9 pT (λ) = λ2 − 9 = (λ − 3)(λ + 3) = 0 f (t) = c1 e3t + c2 e−3t , where c1 , c2 are arbitrary constants. 9.3.10 pT (λ) = λ2 + 1 = 0 has roots λ1,2 = ±i. By Theorem 9.3.9, f (t) = c1 cos(t) + c2 sin(t), where c1 , c2 are arbitrary constants. 9.3.11 pT (λ) = λ2 − 2λ + 2 = 0 has roots λ1,2 = 1 ± i. By Theorem 9.3.9, x(t) = et (c1 cos(t) + c2 sin(t)), where c1 , c2 are arbitrary constants. 9.3.12 pT (λ) = λ2 − 4λ + 13 = 0 has roots λ1,2 = 2 ± 3i. By Theorem 9.3.9, f (t) = e2t (c1 cos(3t) + c2 sin(3t)), where c1 , c2 are arbitrary constants. 9.3.13 pT (λ) = λ2 + 2λ + 1 = (λ + 1)2 = 0 has the double root λ = −1. Following Summary 9.3.15 in the text, we find f (t) = e−t (c1 t + c2 ), where c1 , c2 are arbitrary constants. 9.3.14 pT (λ) = λ2 + 3λ = λ(λ + 3) = 0 has roots λ1 = 0, λ2 = −3. 9.3.15 By integrating twice we find f (t) = c1 + c2 t, where c1 , c2 are arbitrary constants. 9.3.16 By Theorem 9.3.10, the differential equation has a particular solution of the form fp (t) = P cos(t) + Q sin(t). Plugging fp into the equation we find (−P cos(t) − Q sin(t)) + 4(−P sin(t) + Q cos(t)) + 13(P cos(t) + Q sin(t)) = cos(t) or 12P + 4Q = 1 , so −4P + 12Q = 0 P =

3 40

Q=

1 40 .

Therefore, fp (t) =

3 40

cos(t) +

1 40

sin(t).

Next we find a basis of the solution space of f ′′ (t) + 4f ′ (t) + 13f (t) = 0. pT (λ) = λ2 + 4λ + 13 = 0 has roots −2 ± 3i. By Theorem 9.3.9, f1 (t) = e−2t cos(3t) and f2 (t) = e−2t sin(3t) is a basis of the solution space. By Theorem 9.3.4, the solutions of the original differential equation are of the form f (t) = c1 f1 (t) + c2 f2 (t) + 1 3 cos(t) + 40 sin(t), where c1 , c2 are arbitrary constants. fp (t) = c1 e−2t cos(3t) + c2 e−2t sin(3t) + 40 415

Chapter 9 9.3.17 By Theorem 9.3.10, the differential equation has a particular solution of the form fp (t) = P cos(t) + Q sin(t). Plugging fp into we find (−P cos(t) − Q sin(t)) + 2(−P sin(t) + Q cos(t)) + P cos(t) + Q sin(t) = sin(t) the equation P = − 12 2Q = 0 . or , so −2P = 1 Q=0 Therefore, fp (t) = − 21 cos(t). Next we find a basis of the solution space of f ′′ (t) + 2f ′ (t) + f (t) = 0. In Exercise 13 we see that f1 (t) = e−t , f2 (t) = te−t is such a basis. By Theorem 9.3.4, the solutions of the original differential equation are of the form f (t) = c1 f1 (t) + c2 f2 (t) + fp (t) = c1 e−t + c2 te−t − 12 cos(t), where c1 , c2 are arbitrary constants. 9.3.18 We follow the approach outlined in Exercises 16 and 17. • Particular solution fp =

1 10

cos(t) +

3 10

sin(t)

• Solutions of f ′′ (t) + 3f ′ (t) + 2f (t) = 0 are f1 (t) = e−t and f2 (t) = e−2t . • The solutions of the original differential equation are of the form f (t) = c1 e−t + c2 e−2t + where c1 and c2 are arbitrary constants.

1 10

cos(t) +

3 10

sin(t),

9.3.19 We follow the approach outlined in Exercise 17. • Particular solution xp (t) = cos(t) • Solutions of

d2 x dt2

√ √ + 2x = 0 are x1 (t) = cos( 2t) and x2 (t) = sin( 2t).

√ √ • The solutions of the original differential equation are of the form x(t) = c1 cos( 2t) + c2 sin( 2t) + cos(t), where c1 and c2 are arbitrary constants. 9.3.20 pT (λ) = λ3 − 3λ2 + 2λ = λ(λ − 1)(λ − 2) = 0 has roots λ1 = 0, λ2 = 1, λ3 = 2. By Theorem 9.3.8, the general solution is f (t) = c1 + c2 et + c3 e2t , where c1 , c2 , c3 are arbitrary constants. 9.3.21 pT (λ) = λ3 + 2λ2 − λ − 2 = (λ − 1)(λ + 1)(λ + 2) = 0 has roots λ1 = 1, λ2 = −1, λ3 = −2. By Theorem 9.3.8, the general solution is f (t) = c1 et + c2 e−t + c3 e−2t , where c1 , c2 , c3 are arbitrary constants. 9.3.22 pT (λ) = λ3 − λ2 − 4λ + 4 = (λ − 1)(λ − 2)(λ + 2) = 0 has roots λ1 = 1, λ2 = 2, λ3 = −2. By Theorem 9.3.8, the general solution is f (t) = c1 et + c2 e2t + c3 e−2t , where c1 , c2 , c3 are arbitrary constants. 9.3.23 General solution f (t) = Ce5t Plug in: 3 = f (0) = Ce0 = C, so that f (t) = 3e5t . 9.3.24 General solution x(t) = Ce−3t +

7 3

(see Exercise 2).

Plug in: 0 = x(0) = C + 37 , so that C = − 37 and x(t) = − 73 e−3t + 37 . 9.3.25 General solution f (t) = Ce−2t Plug in: 1 = f (1) = Ce−2 , so that C = e2 and f (t) = e2 e−2t = e2−2t . 416

Section 9.3 9.3.26 General solution f (t) = c1 e3t + c2 e−3t (see Exercise 9), with f ′ (t) = 3c1 e3t − 3c2 e−3t Plug in: 0 = f (0) = c1 + c2 and 1 = f ′ (0) = 3c1 − 3c2 , so that c1 = 61 , c2 = − 16 , and f (t) = 16 e3t − 61 e−3t . 9.3.27 General solution f (t) = c1 cos(3t) + c2 sin(3t) (Theorem 9.3.9) Plug in: 0 = f (0) = c1 and 1 = f π2 = −c2 , so that c1 = 0, c2 = −1, and f (t) = − sin(3t). 9.3.28 General solution f (t) = c1 e−4t + c2 e3t , with f ′ (t) = −4c1 e−4t + 3c2 e3t

Plug in: 0 = f (0) = c1 + c2 and 0 = f ′ (0) = −4c1 + 3c2 , so that c1 = c2 = 0 and f (t) = 0. 9.3.29 General solution f (t) = c1 cos(2t) + c2 sin(2t) + (use the approach outlined in Exercise 17)

1 3

sin(t), so that f ′ (t) = −2c1 sin(2t) + 2c2 cos(2t) +

Plug in: 0 = f (0) = c1 and 0 = f ′ (0) = 2c2 + 13 , so that c1 = 0, c2 = − 61 , and f (t) = − 61 sin(2t) +

1 3

1 3

cos(t)

sin(t).

9.3.30 a k is a positive constant that depends on the rate of cooling of the coffee (it varies with the material of the cup, for example). A is the room temperature. b T ′ (t) + kT (t) = kA Constant particular solution: Tp (t) = A General solution of T ′ (t) + kT (t) = 0 is T (t) = Ce−kt . General solution of the original differential equation: T (t) = Ce−kt + A Plug in: T0 = T (0) = C + A, so that C = T0 − A and T (t) = (T0 − A)e−kt + A. 9.3.31

dv dt

+

k mv

=g

constant particular solution: vp = General solution of

dv dt

+

k mv

mg k k

= 0 is v(t) = Ce− m t . k

General solution of the original differential equation: v(t) = Ce− m t + mg k k mg mg t −m Plug in: 0 = v(0) = C + mg 1 − e , so that C = − and v(t) = k k k lim v(t) =

t→∞

mg k

(the “terminal velocity”). See Figure 9.43.

Figure 9.43: for Problem 9.3.31.

417

Chapter 9 dB dt

9.3.32

= kB − r or

dB dt

− kB = −r

↑ ↑ interest withdrawals constant particular solution Bp = General solution of

dB dt

r k

− kB = 0 is B(t) = Cekt

General solution of the original differential equation: B(t) = Cekt + Plug in: B0 = B(0) = C + kr , so that C = B0 −

r k

and B(t) = B0 −

r k r k

ekt +

r k

if B0 >

r k

then interest will exceed withdrawals and balance will grow.

if B0 <

r k

then withdrawals will exceed interest and account will eventually be depleted.

if B0 =

r k

then the balance will remain the same.

The graph in Figure 9.44 shows the three possible scenarios.

Figure 9.44: for Problem 9.3.32. 9.3.33 By Theorem 9.3.9, x(t) = c1 cos

L

√

2 = P = 2π √Lg or L =

√ pg pg 2π √ √L . It is required that g = 2π L t + c2 sin L t , with period P = g

g π2

≈ 0.994 (meters).

9.3.34 a We will take downward forces as positive. Let g = acceleration due to gravity, ρ = density of block a = length of edge of block Then (weight of block) = (mass of block) · g = (density of block)(volume of block) g buoyancy = (weight of displaced water) = (mass of displaced water) · g = (density of water) (volume of displaced water) g = 1a2 x(t)g = a2 gx(t). b Newton’s Second Law of Motion tells us that 2

m ddt2x = F = weight − buoyancy = ρa3 g − a2 gx(t), where m = ρa3 is the mass of the block. 2

ρa3 ddt2x = ρa3 g − a2 gx(t) 418

=

ρa3 g

Section 9.3 d2 x dt2

=g−

d2 x dt2

+

g ρa x

g ρa x(t)

=g

constant solution xp = ρa general solution (use Theorem 9.3.9): x(t) = c1 cos Now c2 = 0 since block is at rest at t = 0.

q

g ρa t

+ c2 sin

q

g ρa t

+ ρa

Plug in: a = x(0) = c1 + ρa, so that c1 = a − ρa and q g x(t) = (a − ρa) cos ρa t + ρa ≈ 2 cos(11t) + 8 (measured in centimeters) c The period is P = √2πg = ρa

√ 2π ρa √ g .

Thus the period increases as ρ or a increases (denser wood or larger block), or

as g decreases (on the moon). The period is independent of the initial state. 9.3.35 a pT (λ) = λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0 with roots λ1 = −1 and λ2 = −2, so x(t) = c1 e−t + c2 e−2t . b x′ (t) = −c1 e−t − 2c2 e−2t Plug in: 1 = x(0) = c1 + c2 and 0 = x′ (0) = −c1 − 2c2 , so that c1 = 2, c2 = −1 and x(t) = 2e−t − e−2t . See Figure 9.45.

Figure 9.45: for Problem 9.3.35b. c Plug in: 1 = x(0) = c1 + c2 and −3 = x′ (0) = −c1 − 2c2 , so that c1 = −1, c2 = 2, and x(t) = −e−t + 2e−2t . See Figure 9.46.

Figure 9.46: for Problem 9.3.35c. d The oscillator in part (b) never reaches the equilibrium, while the oscillator in part (c) goes through the equilibrium once, at t = ln(2). Take another look at Figures 9.45 and 9.46. 419

Chapter 9 9.3.36 fT (λ) = λ2 + 2λ + 101 = 0 has roots λ1,2 = −1 ± 20i. By Theorem 9.3.9, x(t) = e−t (c1 cos(20t) + c2 sin(20t)). Any nonzero solution goes through the equilibrium infinitely many times. See Figure 9.47.

Figure 9.47: for Problem 9.3.36. 9.3.37 fT (λ) = λ2 + 6λ + 9 = (λ + 3)2 has roots λ1,2 = −3. Following the method of Example 10, we find the general solution x(t) = e−3t (c1 + c2 t) with x′ (t) = e−3t (c2 − 3c1 − 3c2 t). Plug in: 0 = x(0) = c1 , and 1 = x′ (0) = c2 − 3c1 , so that c1 = 0, c2 = 1, and x(t) = te−3t . See Figure 9.48.

Figure 9.48: for Problem 9.3.37. The oscillator does not go through the equilibrium at t > 0. 9.3.38 a (D − λ)(p(t)eλt ) = [p(t)eλt ]′ − λp(t)eλt = p′ (t)eλt + λp(t)eλt − λp(t)eλt = p′ (t)eλt , as claimed. b Applying the result from part (a) m times we find (D − λ)m (p(t)eλt ) = p(m) (t)eλt = 0, since p(m) (t) = 0 for a polynomial of degree less than m. c By Theorem 9.3.3, we are looking for m linearly independent functions. By part (b), the functions eλt , teλt , t2 eλt , . . . , tm−1 eλt do the job. d Note that the kernel of (D − λi )mi is contained in the kernel of (D − λ1 )m1 · · · (D − λr )mr , for any 1 ≤ i ≤ r. Therefore, we have the following basis: eλ1 t , teλ1 t , . . . , tm1 −1 eλ1 t , eλ2 t , teλ2 t , . . . , tm2 −1 eλ2 t , . . . eλr t , teλr t , . . . , tmr −1 eλr t . 420

Section 9.3 9.3.39 fT (λ) = λ3 + 3λ2 + 3λ + 1 = (λ + 1)3 = 0 has roots λ1,2,3 = −1. In other words, we can write the differential equation as (D + 1)3 f = 0. By Exercise 38, part (c), the general solution is f (t) = e−t (c1 + c2 t + c3 t2 ). 9.3.40 fT (λ) = λ3 + λ2 − λ − 1 = (λ + 1)2 (λ − 1) = 0 has roots λ1,2 = −1, λ3 = 1. In other words, we can write the differential equation as (D + 1)2 (D − 1) = 0. By Exercise 38, part (d), the general solution is x(t) = e−t (c1 + c2 t) + c3 et . 9.3.41 We are looking for functions x such that T (x) = λx, or T (x) − λx = 0. Now T (x) − λx is an nth-order linear differential operator, so that its kernel is n-dimensional, by Theorem 9.3.3. Thus λ is indeed an eigenvalue of T , with an n-dimensional eigenspace. 2

9.3.42 a We need to solve the second-order differential equation T x = D2 x = ddt2x = λx. This differential equation has a two-dimensional solution space Eλ for any λ, so that all λ are eigenvalues of T . √ √ if λ > 0 then Eλ = span e λt , e− λt if λ = 0 then Eλ = span(1, t)

if λ < 0 then Eλ = span sin

√ √ −λt , cos −λt

b Among the eigenfunctions of T we found in part (a), we seek those of period 1. In the case λ < 0 the shortest . Now 1 is a period if P = √2π = k1 for some positive integer k, or, λ = −4π 2 k 2 . Then period is P = √2π −λ −λ Eλ = span(cos(2πkt), (sin(2πkt)). In the case λ > 0 there are no periodic solutions. In the case λ = 0 we have the constant solutions, so that λ = 0 is an eigenvalue with E0 = span(1). Summary: λ = −4π 2 k 2 is an eigenvalue, for k = 1, 2, 3, . . ., with Eλ = span(cos(2πkt), (sin(2πkt)). λ = 0 is an eigenvalue, with E0 = span(1). 9.3.43 a Using the approach of Exercise 17, we find x(t) = c1 e−2t + c2 e−3t + b For large t, x(t) ≈

1 10

cos t +

1 10

1 10

cos t +

1 10

sin t.

sin t.

9.3.44 a Using the approach of Exercises 16 and 17 we find x(t) = e−2t (c1 cos t + c2 sin t) − 1 b For large t, x(t) ≈ − 40 cos(3t) +

9.3.45 We can write the system as

3 40

sin(3t).

dx1 dt

= x1 + 2x2 dx2 dt = x2

with x1 (0) = 1, x2 (0) = −1.

The solution of the second equation, with the given initial value, is x2 (t) = −et . Now the first equation takes the form

dx1 dt

− x1 = −2et .

Using Example 9 (with a = 1 and c = −2) we find x1 (t) = et (−2t + C). 421

1 40

cos(3t) +

3 40

sin(3t).

Chapter 9

plug in: 1 = x1 (0) = C, so that x1 (t) = et (1 − 2t) and ~x(t) = et dx1

9.3.46 We can write the system as

dt dx2 dt dx3 dt

= = =

We solve for x2 and x3 as in Exercise 45:

1 − 2t . −1

2x1 + 3x2 + x3 x2 + 2x3 with x1 (0) x3

=

2, x2 (0)

=

1, x3 (0)

= −1 .

x2 (t) = et (1 − 2t) x3 (t) = −et Now the first equation takes the form

dx1 dt

− 2x1 = 3et (1 − 2t) − et = et (2 − 6t), x1 (0) = 2.

We use Theorem 9.3.13 to solve this differential equation: Z Z x1 (t) = e2t e−2t et (2 − 6t) dt = e2t (2e−t − 6te−t ) dt = e2t [−2e−t + 6te−t + 6e−t + C] plug in: 2 = x1 (0) = (−2 + 6 + c), so that c = −2 and x1 (t) = e2t (4e−t + 6te−t − 2) = 4et + 6tet − 2e2t . t 4e + 6tet − 2e2t et − 2tet ~x(t) = t −e 9.3.47 a We start with a preliminary remark that will be useful below: If f (t) = p(t)eλt , where p(t) is a polynomial, then f (t) has an antiderivative of the form q(t)eλt , where q(t) is another polynomial. We leave this remark as a calculus exercise. The function xn (t) satisfies the differential equation sired form.

dxn dt

= ann xn , so that xn = Ceann t , which is of the de-

Now we will show that xk is of the desired form, assuming that xk+1 , . . . , xn have this form. xk satisfies the dxk k differential equation dx dt = akk xk + ak,k+1 xk+1 + · · · + akn xn or dt − akk xk = ak,k+1 xk+1 + · · · + akn xn . Note that, by assumption, the function on the right-hand side has the form p1 (t)eλ1 t + · · · + pm (t)eλm t . If we set λ1 t k akk = a for simplicity, we can write dx + · · · + pm (t)eλm t . dt − axk = p1 (t)e By Theorem 9.3.13, the solution is Z xk (t) = eat e−at (p1 (t)eλ1 t + · · · + pm (t)eλm t ) dt = eat

Z

(p1 (t)e(λ1 −a)t + · · · + pm (t)e(λm −a)t ) dt

= eat (q1 (t)e(λ1 −a)t + · · · + qm (t)e(λm −a)t + C) = q1 (t)eλ1 t + · · · + qm (t)eλm t + Ceat as claimed (note that a is one of the λi ). The constant C is determined by xk (0). Note that we used the preliminary remark in the second to last step. b It is shown in introductory calculus classes that lim (tm eλt ) = 0 if and only if λ is negative (here m is a fixed t→∞

positive integer). In light of part (a), this proves the claim. 422

Section 9.3 9.3.48 a By Exercise 47, the system solution ~x(t) = S~u(t).

d~ u dt

= B~u has a unique solution ~u(t). Then the system

b It suffices to note that lim ~x(t) = ~0 if and only if lim ~u(t) = ~0, where ~u = S −1 ~x. t→∞

t→∞

423

d~ x dt

= A~x has the unique

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