Elementary Linear Algebra, Students Solutions Manual (e-only)

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Author: Stephen Andrilli | David Hecker

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Academic Press is an imprint of Elsevier 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA 525 B Street, Suite 1900, San Diego, California 92101-4495, USA 84 Theobald’s Road, London WC1X 8RR, UK c 2010 Elsevier Inc. All rights reserved. Copyright No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. ISBN: 978-0-12-381455-5 For information on all Academic Press publications visit our Web site at www.elsevierdirect.com

Dedication To all the students who have used the various editions of our book over the years

Table of Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Chapter 1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Chapter 3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Chapter 4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Chapter 5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Chapter 6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Chapter 7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 Chapter 8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Chapter 9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 Appendix B Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 Appendix C Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295

iii

Preface This Student Solutions Manual is designed to accompany the fourth edition of Elementary Linear Algebra, by Andrilli and Hecker. It contains detailed solutions for all of the exercises in the textbook marked with a star () or a triangle (). In the triangle exercises, the student is typically asked to prove a theorem that appears in the textbook. The solutions presented are generally self-contained, except that a comment may appear at the beginning of an exercise that applies to all the solutions for each part of the exercise. For example, the solution to Exercise 1 in Section 1.1 begins with a description of the general strategy used to solve problems in each of parts (a) and (c). Then the solution is given to each part separately, following the previously stated strategy. Therefore, you should always check for a comment at the heading of a problem before jumping to the part in which you are interested. We hope you find these solutions helpful. You can find other useful information at the web site at which you obtained this manual. Stephen Andrilli David Hecker August 2009

iv

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition

Section 1.1

Solutions to Selected Exercises Chapter 1 Section 1.1 (1) In each part, to find the vector, subtract the corresponding coordinates of the initial vector from the terminal vector. (a) In this case, [5, −1] − [−4, 3] = [9, −4]. And so, the desired vector is [9, −4]. The distance between √ the two points = [9, −4] = 92 + (−4)2 = 97. (c) In this case, [0, −3, 2, −1, −1] − [1, −2, 0, 2, 3] = [−1, −1, 2, −3, −4]. And so, the desired vector is [−1, −1, 2, −3, −4]. The distance between the two points = [−1, −1, 2, −3, −4] = √ (−1)2 + (−1)2 + 22 + (−3)2 + (−4)2 = 31 (2) In each part, the terminal point is found by adding the coordinates of the given vector to the corresponding coordinates of the initial point (1, 1, 1). (a) [1, 1, 1] + [2, 3, 1] = [3, 4, 2] (see Figure 1), so the terminal point = (3, 4, 2). (c) [1, 1, 1] + [0, −3, −1] = [1, −2, 0] (see Figure 2), so the terminal point = (1, −2, 0).

Figure 1

Figure 2

(3) In each part, the initial point is found by subtracting the coordinates of the given vector from the corresponding coordinates of the given terminal point. (a) [6, −9] − [−1, 4] = [7, −13], so the initial point is (7, −13). (c) [2, −1, −1, 5, 4] − [3, −4, 0, 1, −2] = [−1, 3, −1, 4, 6], so the initial point = (−1, 3, −1, 4, 6).

1


Section 1.1

(4) (a) First, we find the vector v having the given initial and terminal point by subtracting: [10, −10, 11] − [−4, 7, 2] = [14, −17, 9] = v. Next, the desired point is found by adding 23 v (which 2 3

the length of v) to the vector for the initial point (−4, 7, 2): 16 13 34 [−4, 7, 2] + 23 v = [−4, 7, 2] + 23 [14, −17, 9] = [−4, 7, 2] + 28 3 ,− 3 ,6 = 3 ,− 3 ,8 , 16 13 so the desired point is 3 , − 3 , 8 . is

(5) Let v represent the given vector. Then the desired unit vector u equals (a) u =

1 v v

=

1 [3,−5,6] [3, −5, 6]

=√

1 [3, −5, 6] 32 +(−5)2 +62

u is shorter than v since u = 1 ≤ (c) u =

1 v v

=

1 [0.6,−0.8] [0.6, −0.8]

=

√

70 = v. 1 = √ 2

(0.6) +(−0.8)2

1 v v.

√3 , − √5 , √6 70 70 70

;

[0.6, −0.8] = [0.6, −0.8]. Neither vector is

longer because u = v. (6) Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. (a) [12, −16] and [9, −12] are parallel because 34 [12, −16] = [9, −12]. (c) [−2, 3, 1] and [6, −4, −3] are not parallel. To show why not, suppose they are. Then there would be a c ∈ R such that c[−2, 3, 1] = [6, −4, −3]. Comparing first coordinates shows that −2c = 6, or c = −3. However, comparing second coordinates shows that 3c = −4, or c = − 43 instead. But c cannot have both values. (7) (a) 3[−2, 4, 5] = [3(−2), 3(4), 3(5)] = [−6, 12, 15] (c) [−2, 4, 5] + [−1, 0, 3] = [(−2 + (−1)), (4 + 0), (5 + 3)] = [−3, 4, 8] (e) 4[−1, 0, 3] − 5[−2, 4, 5] = [4(−1), 4(0), 4(3)] − [5(−2), 5(4), 5(5)] = [−4, 0, 12] − [−10, 20, 25] = [(−4 − (−10)), (0 − 20), (12 − 25)] = [6, −20, −13] (8) (a) x+y = [−1, 5]+[2, −4] = [(−1+2), (5−4)] = [1, 1], x−y = [−1, 5]−[2, −4] = [(−1−2), (5−(−4))] = [−3, 9], y − x = [2, −4] − [−1, 5] = [(2 − (−1)), ((−4) − 5)] = [3, −9] (see Figure 3, next page) (c) x + y = [2, 5, −3] + [−1, 3, −2] = [(2 + (−1)), (5 + 3), ((−3) + (−2))] = [1, 8, −5], x − y = [2, 5, −3] − [−1, 3, −2] = [(2 − (−1)), (5 − 3), ((−3) − (−2))] = [3, 2, −1], y − x = [−1, 3, −2] − [2, 5, −3] = [((−1) − 2), (3 − 5), ((−2) − (−3))] = [−3, −2, 1] (see Figure 4, next page) (10) In each part, consider the center of the clock to be the origin. (a) At 12 PM, the tip of the minute hand is at (0, 10). At 12:15 PM, the tip of the minute hand is at (10, 0). To find the displacement vector, we subtract the vector for the initial point from the vector for the terminal point, yielding [10, 0] − [0, 10] = [10, −10]. (b) At 12 PM, the tip of the minute hand is at (0, 10). At 12:40 PM, the minute hand makes a 210◦ angle with the positive x-axis. So, as shown in Figure 1.10 in the textbook, hand makes the minute √

the vector v = [v cos θ, v sin θ] = [10 cos(210◦ ), 10 sin(210◦ )] = 10 − 23 , 10 − 12 = √ [−5 3, −5]. To find the displacement vector, √ for the initial point from the √ we subtract the vector vector for the terminal point, yielding [−5 3, −5] − [0, 10] = [−5 3, −15].

2


Figure 3

Section 1.1

Figure 4

(13) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2 , v = [v cos θ, v sin θ], where θ is the angle v makes with the positive x-axis. Now “southwest” corresponds to 225◦ , “east” corresponds to 0◦ , and northwest corresponds to 135◦ . Hence, if v1 , v2 , and v3 are the vectors corresponding to the three given movements, then v1 = [1 cos(225◦ ), 1 sin(225◦ )] = −

√ √ 2 2 2 ,− 2

, v2 = [0.5 cos(0◦ ), 0.5 sin(0◦ )] = [0.5, 0], and √

√ v3 = [0.2 cos(135◦ ), 0.2 sin(135◦ )] = 0.2 − 22 , 0.2 22 . Hence, the result of all three putts is √ √

√ √ √ √ v1 +v2 +v3 = − 22 , − 22 +[0.5, 0]+ 0.2 − 22 , 0.2 22 = [0.5−0.6 2, −0.4 2] ≈ [−0.3485, −0.5657].

(15) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2 , v = [v cos θ, v sin θ], where θ is the angle v makes with the positive x-axis. Now “northwestward” corresponds to 135◦ , and “southward” rower, and v2 corresponds to the current, then corresponds to 270◦ . Hence, if v1 corresponds the √ √ to √ √ 2 2 ◦ ◦ v1 = [4 cos(135 ), 4 sin(135 )] = 4 − 2 , 4 2 = −2 2, 2 2 and v2 = [3 cos(270◦ ), 3 sin(270◦ )] = √ √ √ √ [0, −3]. Therefore, the rower’s net velocity = v1 +v2 = −2 2, 2 2 +[0, −3] = [−2 2, −3+2 2]. The √ √ √ √ 2 √ resultant speed = [−2 2, −3 + 2 2] = (−2 2)2 + −3 + 2 2 = 25 − 12 2 ≈ 2.83 km/hr. (17) If v1 corresponds to the rower, and v2 corresponds to the current, and v3 corresponds to the resultant velocity, then v3 = v1 + v2 , or v1 = v3 − v2 . Now, as shown in Figure 1.10 in the textbook, for any vector v ∈ R2 , v = [v cos θ, v sin θ], where θ is the angle v makes with the positive x-axis. Also, corresponds to 45◦ . Thus, “westward” corresponds to 180◦, and “northeastward” √ √ √ √ 2, 2 and v3 = [8 cos(180◦ ), 8 sin(180◦ )] = [−8, 0]. v2 = [2 cos(45◦ ), 2 sin(45◦ )] = 2 22 , 2 22 = √ √ √ √ Hence, v1 = v3 − v2 = [−8, 0] − 2, 2 = [−8 − 2, − 2]. (18) Let fR and aR represent the resultant force and the resultant acceleration, respectively. Then fR =

3


Section 1.1

48 16 [3, −12, 4] = 12 13 , − 13 , 13 , [0,−4,−3] 2 f2 = 2 [0,−4,−3] =√2 [0, −4, −3] = 25 [0, −4, −3] = 0, − 85 , − 65 , and 0 +(−4)2 +(−3)2 [0,0,1] 48 16 8 6 f3 = 6 [0,0,1 = 61 [0, 0, 1] = [0, 0, 6]. Therefore, fR = f1 +f2 +f3 = 12 13 , − 13 , 13 + 0, − 5 , − 5 +[0, 0, 6] = 12 344 392 1 1 12 344 392 13 , − 65 , 65 . Finally, fR = maR . Since m = 20, acceleration = aR = m fR = 20 [ 13 , − 65 , 65 ] ≈

[3,−12,4] f1 + f2 + f3 . Now, f1 = 4 [3,−12,4] =√

4

32 +(−12)2 +42

[3, −12, 4] =

4 13

[0.0462, −0.2646, 0.3015]. (21) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2 , v = [v cos θ, v sin θ], where θ is the angle v makes with the positive x-axis. Thus, if we let r = a and s = b, then √ √ √ a = [r cos(135◦ ), r sin(135◦ )] = −r 22 , r 22 and b = [s cos(60◦ ), s sin(60◦ )] = s 12 , s 23 . Now, since √ √ √ the weight is not moving, a + b + g = 0. That is, −r 22 , r 22 + s 12 , s 23 + [0, −mg] = [0, 0]. Using √ √ the first coordinates gives us −r 22 + s 12 = 0. Hence, s = 2r. Using the second coordinates gives us √ √ √ √ √ √ r 22 + s 23 − mg = 0. Substituting in s = 2r yields r 22 + 2r 23 − mg = 0. Solving for r produces √ 2mg √ . Plugging these values in to the formulas for a and b gives √ , and s = r = √2 2mg 2r = 1+ 3 (1+ 3) √ −mg mg mg mg√ 3 √ √ √ a = [ 1+ 3 , 1+ 3 ] and b = [ 1+ 3 , 1+ 3 ]. (23) Let x= [x1 , x2 , . . . , xn ]. Then cx = (cx1 )2 + · · · + (cxn )2 = c2 (x21 + · · · + x2n ) = |c| x21 + · · · + x2n = |c| x. (24) In each part, let x = [x1 , x2 , . . . , xn ], y = [y1 , y2 , . . . , yn ], and c ∈ R. (b) x + (−x) = [x1 , . . . , xn ] + [−x1 , . . . , −xn ] = [(x1 + (−x1 )), . . . , (xn + (−xn ))] = [0, . . . , 0] = 0. Also, (−x) + x = x + (−x) (by part (1) of Theorem 1.3) = 0, by the above. (c) c(x+y) = c([(x1 +y1 ), . . . , (xn +yn )]) = [c(x1 +y1 ), . . . , c(xn +yn )] = [(cx1 +cy1 ), . . . , (cxn +cyn )] = [cx1 , . . . , cxn ] + [cy1 , . . . , cyn ] = cx + cy. (25) If c = 0, we are done. Otherwise, cx = 0 ⇒ ( 1c )(cx) = 1c (0) ⇒ ( 1c · c)x = 0 (by part (7) of Theorem 1.3) ⇒ x = 0. Thus, either c = 0 or x = 0. (27) (a) False. The length of [a1 , a2 , a3 ] is a21 + a22 + a23 . The formula given in the problem is missing the square root. So, for example, the length of [0, 3, 4] is actually 5, not 02 + 32 + 42 = 25. (b) True. This is easily proved using parts (1) and (2) of Theorem 1.3. (c) True. [2, 0, −3] = 2[1, 0, 0] + (−3)[0, 0, 1]. (d) False. Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. To show why [3, −5, 2] and [6, −10, 5] are not parallel, suppose they are. Then there would be a c ∈ R such that c[3, −5, 2] = [6, −10, 5]. Comparing first coordinates shows that 3c = 6, or c = 2. But this value of c must work in all coordinates. However, it does not work in the third coordinate, since (2)(2) = 5. (e) True. Multiply both sides of dx = 0 by d1 . (f) False. Parallel vectors can be in opposite directions. For example, [1, 0] and [−2, 0] are parallel but are not in the same direction. (g) False. The properties of vectors in Theorem 1.3 are independent of the location of the initial points of the vectors. 4


Section 1.2

Section 1.2 (1) In each part, we use the formula cos θ = (a) cos θ =

x·y xy

=

[−4,3]·[6,−1] [−4,3][6,−1]

x·y xy .

= √

(−4)(6)+(3)(−1) (−4)2 +32

√

62 +(−1)2

=

−27 √ . 5 37

And so, using a calculator yields

27 ), or about 152.6◦ , or 2.66 radians. θ = arccos(− 5√ 37

(c) cos θ =

x·y xy

=

[7,−4,2]·[−6,−10,1] [7,−4,2][−6,−10,1]

θ = arccos(0), which is 90◦ , or

π 2

= √

(7)(−6)+(−4)(−10)+(2)(1) 72 +(−4)2 +22

√

(−6)2 +(−10)2 +12

=

√ 0 √ 69 137

= 0. And so,

radians.

(4) When a force f moves an object by a displacement d, the work performed is f · d. (b) First, we need to compute f . A unit vector in the direction of f is u =

1 [−2,4,5]

1 1 √ [−2, 4, 5] = 3√ [−2, 4, 5]. Thus, since f = 26, f = 26u = 2+ 42 +52 5 (−2) √ √ √ −52 5 104 5 26 5 , where we have rationalized the denominators. 15 , 15 , 3

26 √ 3 5

[−2, 4, 5] = [−2, 4, 5] =

1 [−1, 2, 2] = Next, we need to compute d. A unit vector in the direction of d is v = [−1,2,2] 1 20 20 √ [−1, 2, 2] = − 13 , 23 , 23 . Thus, since d = 10, d = 10v = − 10 3 , 3 , 3 . 2 2 2 (−1) +2 +2

Hence, the work performed is √ √ √ √ 104√5 20 26√5 20 5 104 5 26 5 5 20 20 · − 10 − 10 = −52 + + f · d = −52 15 , 15 , 3 3 , 3 , 3 15 3 15 3 3 3 √ √ √ √ 1040 5 , or 258.4 joules. = 19 104 5 + 416 5 + 520 5 = 9 (6) In all parts, let x = [x1 , x2 , . . . , xn ] , y = [y1 , y2 , . . . , yn ] , and z = [z1 , z2 , . . . , zn ] . Part (1): x · y = [x1 , x2 , . . . , xn ] · [y1 , y2 , . . . , yn ] = x1 y1 + · · · + xn yn = y1 x1 + · · · + yn xn = [y1 , y2 , . . . , yn ] · [x1 , x2 , . . . , xn ] = y · x Part (2): x · x = [x1 , x2 , . . . , xn ] · [x1 , x2 , . . . , xn ] = x1 x1 + · · · + xn xn = x21 + · · · + x2n . Now x21 + · · · + x2n is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and so

2 2 x21 + · · · + x2n = x . its square root is defined. Thus, 0 ≤ x · x = x21 + · · · + x2n = Part (3): Suppose x · x = 0. From part (2), 0 = x21 + · · · + x2n ≥ x2i , for each i, since the sum of the remaining squares (without x2i ) is nonnegative. Hence, 0 ≥ x2i for each i. But x2i ≥ 0, because it is a square. Hence, each xi = 0. Therefore, x = 0. Next, suppose x = 0. Then x · x = [0, . . . , 0] · [0, . . . , 0] = (0)(0) + · · · + (0)(0) = 0. Part (4): c(x · y) = c ([x1 , x2 , . . . , xn ] · [y1 , y2 , . . . , yn ]) = c (x1 y1 + · · · + xn yn ) = cx1 y1 + · · · + cxn yn = [cx1 , cx2 , . . . , cxn ] · [y1 , y2 , . . . , yn ] = (cx) · y. Next, c(x · y) = c(y · x) (by part (1)) = (cy) · x (by the above) = x · (cy), by part (1). Part (6): (x + y) · z = ([x1 , x2 , . . . , xn ] + [y1 , y2 , . . . , yn ]) · [z1 , z2 , . . . , zn ] = [x1 + y1 , x2 + y2 , . . . , xn + yn ] · [z1 , z2 , . . . , zn ] = (x1 + y1 )z1 + (x2 + y2 )z2 + · · · + (xn + yn )zn = (x1 z1 + x2 z2 + · · · + xn zn ) + (y1 z1 + y2 z2 + · · · + yn zn ). Also, (x · z) + (y · z) = ([x1 , x2 , . . . , xn ] · [z1 , z2 , . . . , zn ]) + ([y1 , y2 , . . . , yn ] · [z1 , z2 , . . . , zn ]) = (x1 z1 + x2 z2 + · · · + xn zn ) + (y1 z1 + y2 z2 + · · · + yn zn ). Hence, (x + y) · z = (x · z) + (y · z). (7) No. Consider x = [1, 0], y = [0, 1], and z = [1, 1]. Then x · z = [1, 0] · [1, 1] = (1)(1) + (0)(1) = 1 and y · z = [0, 1] · [1, 1] = (0) (1) + (1)(1) = 1, so x · z = y · z. But x = y. 5


Section 1.2

[a,b,c]·[1,0,0] a(1)+b(0)+c(0) a √ √ = √a2 +b . 2 +c2 [a,b,c][1,0,0] = a2 +b2 +c2 12 +02 +02 a(0)+b(1)+c(0) [a,b,c]·[0,0,1] b √ √ = √a2 +b2 +c2 , and cos θ3 = [a,b,c][0,0,1] = a2 +b2 +c2 02 +12 +02

(13) θ1 is the angle between x and [1, 0, 0]. So cos θ1 = Similarly, cos θ2 = a(0)+b(0)+c(1) √ √ a2 +b2 +c2 02 +02 +12

[a,b,c]·[0,1,0] [a,b,c][0,1,0]

=

=

c √ . a2 +b2 +c2

(14) Position the cube so that one of its corners is at the origin and the three edges from that corner are along the positive x-, y-, and z-axes. Then the corner diagonally opposite from the origin is at the point (s, s, s). The vector representing this diagonal is d = [s, s, s]. √ √ √ (a) The length of the diagonal = d = s2 + s2 + s2 = 3s2 = 3s. (b) The angle d makes with a side of the cube is equal between d and [s, 0, 0], which to the angle 2

√ d·[s,0,0] s(s)+s(0)+s(0) is arccos d[s,0,0] = arccos √3s √s2 +02 +02 = arccos √s3s2 = arccos( 33 ) ≈ 54.7◦ , or ( )( ) 0.955 radians.

(2)(1)+(1)(4)+(5)(−3) a·b (15) (a) proja b = a a = [2,1,5]·[1,4,−3] [2, 1, 5] = [2, 1, 5] = −9 2 2 2 +12 +52 2 30 [2, 1, 5] [2,1,5] 3 3 3 3 3 3 8 43 3 = − 5 , − 10 , − 2 . b − proja b = [1, 4, −3] − − 5 , − 10 , − 2 = 5 , 10 , − 2 . proja b is orthogonal 3 3 , − 32 · 85 , 43 to (b − proja b) because proja b · (b − proja b) = − 35 , − 10 10 , − 2 = 3 8 3 43 3 3 − 5 5 + − 10 10 + − 2 − 2 = 0.

a·b (c) proja b = a a = [1,0,−1,2]·[3,−1,0,−1] [1, 0, −1, 2] = 2 2 [1,0,−1,2]

(1)(3)+(0)(−1)+(−1)(0)+(2)(−1) [1, 0, −1, 2] = 16 [1, 0, −1, 2] = 16 , 0, − 16 , 13 . 12 +02 +(−1)2 +22 1 4 b−proja b = [3, −1, 0, −1]− 16 , 0, − 16 , 13 = 17 6 , −1, 6 , − 3 . proja b is orthogonal to (b − proja b) 1 4 because proja b · (b − proja b) = 16 , 0, − 16 , 13 · 17 6 , −1, 6 , − 3 = 1 17 1 1 1 4 − 3 = 0. 6 6 + (0) (−1) + − 6 6 + 3

i·a 1a+0b+0c (17) Let a = [a, b, c]. Then proji a = i i = [1,0,0]·[a,b,c] [1, 0, 0] = [1, 0, 0] = ai. Similarly, 2 2 2 2 2 1 +0 +0 [1,0,0]

[0,1,0]·[a,b,c] j·a 0a+1b+0c k·a a = j = [0, 1, 0] = [0, 1, 0] = bj, and proj k = projj a = j 2 2 2 2 2 2 k 0 +1 +0 k

[0,1,0]

[0,0,1]·[a,b,c] [0, 0, 1] = 00a+0b+1c [0, 0, 1] = ck. 2 +02 +12 [0,0,1]2 (18) In each part, if v is the given vector, then by Theorem 1.10, x = projv x + (x − projv x), where projv x is parallel to v and (x − projv x) is orthogonal to v.

v·x v= (a) In this case, x = [−6, 2, 7] and v = [2, −3, 4]. Then projv x = v 2

10 [2,−3,4]·[−6,2,7] (2)(−6)+(−3)(2)+(4)(7) 30 40 [2, −3, 4] = [2, −3, 4] = 29 [2, −3, 4] = [ 20 22 +(−3)2 +42 29 , − 29 , 29 ]. [2,−3,4]2 And so projv x is clearly parallel to v since projv x = 10 29 v. Also, 30 40 194 88 163 (x − projv x) = [−6, 2, 7] − [ 20 29 , − 29 , 29 ] = [− 29 , 29 , 29 ]. We can easily check that 30 40 194 88 163 projv x + (x − projv x) = [ 20 29 , − 29 , 29 ] + [− 29 , 29 , 29 ] = [−6, 2, 7] = x, and that projv x 30 40 194 88 163 and (x − projv x) are orthogonal, since projv x · (x − projv x) = [ 20 29 , − 29 , 29 ] · [− 29 , 29 , 29 ] 20 194 30 88 40 163 = 0. = 29 − 29 + − 29 29 + 29 29

6


Section 1.3

v·x v= (c) In this case, x = [−6, 2, 7] and v = [3, −2, 6]. Then projv x = v 2

60 40 120 [3,−2,6]·[−6,2,7] [3, −2, 6] = (3)(−6)+(−2)(2)+(6)(7) [3, −2, 6] = 20 32 +(−2)2 +62 49 [3, −2, 6] = 49 , − 49 , 49 . [3,−2,6]2 And so, projv x is clearly parallel to v since projv x = 20 49 v. Also, 60 40 120 354 138 223 (x − projv x) = [−6, 2, 7] − 49 , − 49 , 49 = − 49 , 49 , 49 . We can easily check that 354 138 223 40 120 = [−6, 2, 7] = x, and that projv x projv x + (x − projv x) = 60 49 , − 49 , 49 + − 49 , 49 , 49 354 138 223 40 120 and (x − projv x) are orthogonal, since projv x · (x − projv x) = 60 49 , − 49 , 49 · − 49 , 49 , 49 354 40 138 120 223 − 49 + − 49 = 0. = 60 49 49 + 49 49 (23) (a) True by part (4) of Theorem 1.5. (b) True. Theorem 1.6 (the Cauchy-Schwarz Inequality) states that |x · y| ≤ x y. Since x · y ≤ |x · y|, we have x · y ≤ x y. But then, because x > 0, we can divide both sides by x to obtain the desired inequality. (c) False. For example, if x = [1, 0, 0] and y = [0, 1, 0], then x − y = [1, −1, 0] = √ 12 + (−1)2 + 02 = 2, and x − y = [1, 0, 0] − [0, 1, 0 = √ √ √ 12 + 02 + 02 − 02 + 12 + 02 = 1 − 1 = 0. However, 2 0. (d) False. Theorem 1.8 shows that if θ >

π 2,

then x · y < 0. (Remember, θ is defined to be ≤ π.)

(e) True. If i = j, then ei · ej = 0(0) + · · · + 1(0) + · · · + 0(1) + 0(0) + · · · + 0(0) = 0 + · · · + 0 = 0. (f) False. proja b =

a·b a2

ith terms

jth terms

a, and so proja b is parallel to a, since it is a scalar multiple of a. Thus, if

example, suppose a = [1, 0] proja b = b, a and b are parallel, not perpendicular.

For a particular

[1,0]·[2,0] 1(2)+0(0) a·b √ [1, 0] = 2[1, 0] = [2, 0] = and b = [2, 0]. Then proja b = a2 a = [1,0] [1, 0] = 12 +02 b. However, a = [1, 0] and b = [2, 0] are parallel (since b = 2a) and not perpendicular (since a · b = [1, 0] · [2, 0] = 1(2) + 0(0) = 2 = 0.

Section 1.3 (1) (b) Let m = max{|c|, |d|}. Then cx ± dy ≤ m(x + y). The proof is as follows: cx ± dy = ≤ = ≤ =

cx + (±dy) cx + ±dy by the Triangle Inequality |c| x + | ± d| y by Theorem 1.1 m x + m y because |c| ≤ m and | ± d| ≤ m m (x + y)

(2) (b) The converse is: If an integer has the form 3k + 1, then it also has the form 6j − 5, where j and k are integers. For a counterexample, consider that the number 4 = 3k + 1 with k = 1, but there is no integer j such that 4 = 6j − 5, since this equation implies j = 32 , which is not an integer. 2

2

(5) (a) Consider x = [1, 0, 0] and y = [1, 1, 0]. Then x · y = [1, 0, 0] · [1, 1, 0] = 1 = [1, 0, 0] = x , but x = y. 7


Section 1.3

(b) If x = y, then x · y = x2 . (c) Yes. Using the counterexample from part (a), x = y, but x · y = x2 . (8) (a) Contrapositive: If x = 0, then x is not a unit vector. Converse: If x is nonzero, then x is a unit vector. Inverse: If x is not a unit vector, then x = 0. (c) (Let x, y be nonzero vectors.) Contrapositive: If projy x = 0, then projx y = 0. Converse: If projy x = 0, then projx y = 0. Inverse: If projx y = 0, then projy x = 0. (10) (b) Converse: Let x and y be vectors in Rn . If x + y ≥ y, then x · y = 0. The original statement is true, but the converse is false in general. Proof of the original statement: x + y2

= (x + y) · (x + y) by part (2) of Theorem 1.5 2 2 = x + 2(x · y) + y using Theorem 1.5 = x2 + y2 since x · y = 0 ≥

y2

by part (2) of Theorem 1.5

Taking the square root of both sides yields x + y ≥ y. √ Counterexample √ to converse: let x = [1, 0] and y = [1, 1]. Then x + y = [2, 1] = 5 and y = [1, 1] = 2. Hence, x + y ≥ y. However, x · y = [1, 0] · [1, 1] = 1 = 0. (18) Step 1 cannot be reversed because y could equal −(x2 + 2) and then the converse would not hold. Step 2 cannot be reversed because y 2 could equal x4 +4x2 +c with c > 4. For a specific counterexample √ to the converse, set y = x4 + 4x2 + 5. 3 dy +8x = 4x 2y by 2y. Step 3 is reversible. To prove the converse, multiply both sides of dx Step 4 cannot be reversed because in general y does not have to equal x2 + 2. In particular, y could equal x2 + 5. 2 +2) Step 5 can be reversed by multiplying 2x by 2(x 2(x2 +2) . Step 6 cannot be reversed, since

dy dx

could equal 2x + c for some c = 0.

(19) (a) Negation: For every unit vector x in R3 , x · [1, −2, 3] = 0. (Note that the existential quantifier has changed to a universal quantifier.) (c) Negation: x = 0 or x + y = y, for all vectors x and y in Rn . (The “and” has changed to an “or,” and the existential quantifier has changed to a universal quantifier.) (e) Negation: There is an x ∈ R3 such that for every nonzero y ∈ R3 , x · y = 0. (Note that the universal quantifier on x has changed to an existential quantifier, and the existential quantifier on y has changed to a universal quantifier.) (20) (a) When negating the conclusion “x = 0 or x − y > y” of the original statement, we must change the “or” to “and.” Contrapositive: If x = 0 and x − y ≤ y, then x · y = 0. Converse: If x = 0 or x − y > y, then x · y = 0. Inverse: If x · y = 0, then x = 0 and x − y ≤ y. 8


Section 1.4

(25) (a) False. We must also verify that the reversed steps are truly reversible. That is, we must also supply valid reasons for the forward steps thus generated, as in the proof of Result 2. (b) True. “If not B then not A” is the contrapositive of “If A then B.” A statement and its contrapositive are always logically equivalent. (c) True. We saw that “A only if B” is an equivalent form for “If A then B,” whose converse is “If B then A.” (d) False. “A is a necessary condition for B” is an equivalent form for “If B then A.” As such, it does not include the statement “If A then B” that is included in the “A if and only if B” statement. (e) False. “A is a necessary condition for B” and “B is a sufficient condition for A” are both equivalent forms for “If B then A.” As such, neither includes the statement “If A then B” that is included in the “A if and only if B” statement. (f) False. The inverse of a statement is the contrapositive of the converse. Hence, the converse and the inverse are logically equivalent and always have the same truth values. (g) False. The problem describes only the inductive step. The base step of the proof is also required. (h) True. This rule is given in the section for negating statements with quantifiers. (i) False. The “and” must change to an “or.” Thus, the negation of “A and B” is “not A or not B.”

Section 1.4 ⎡

−4 (1) (a) ⎣ 0 6 ⎡ 2 ⎣ = 2 9 ⎡ −4 (c) 4 ⎣ 0 6

⎤ ⎡ ⎤ ⎡ ⎤ 2 3 6 −1 0 (−4 + 6) (2 + (−1)) (3 + 0) 5 −1 ⎦ + ⎣ 2 2 −4 ⎦ = ⎣ (0 + 2) (5 + 2) ((−1) + (−4)) ⎦ 1 −2 3 −1 1 (6 + 3) (1 + (−1)) ((−2) + 1) ⎤ 1 3 7 −5 ⎦ 0 −1 ⎤ ⎡ ⎤ ⎡ ⎤ 2 3 4(−4) 4(2) 4(3) −16 8 12 5 −1 ⎦ = ⎣ 4(0) 4(5) 4(−1) ⎦ = ⎣ 0 20 −4 ⎦ 1 −2 4(6) 4(1) 4(−2) 24 4 −8

(e) Impossible. C is a 2 × 2 matrix, F is a 3 × 2 matrix, and E is a 3 × 3 matrix. Hence, 3F is 3 × 2 and −E is 3 × 3. Thus, C, 3F, and −E have different sizes. However, if we want to add matrices, they must be the same size. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −4 2 3 3 −3 5 6 −1 0 0 −2 ⎦ − ⎣ 2 2 −4 ⎦ = (g) 2 ⎣ 0 5 −1 ⎦ − 3 ⎣ 1 6 1 −2 6 7 −2 3 −1 1 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ −6 −(−1) −0 2(−4) 2(2) 2(3) −3(3) −3(−3) −3(5) ⎣ 2(0) 2(5) 2(−1) ⎦ + ⎣ −3(1) −2 −(−4) ⎦ −3(0) −3(−2) ⎦ + ⎣ −2 −3(6) −3(7) −3(−2) −3 −(−1) −1 2(6) 2(1) 2(−2) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −8 4 6 −9 9 −15 −6 1 0 0 6 ⎦ + ⎣ −2 −2 4 ⎦ = ⎣ 0 10 −2 ⎦ + ⎣ −3 12 2 −4 −18 −21 6 −3 1 −1 ⎡ ⎤ ⎡ ⎤ (−8) + (−9) + (−6) 4+9+1 6 + (−15) + 0 −23 14 −9 10 + 0 + (−2) (−2) + 6 + 4 ⎦ = ⎣ −5 8 8 ⎦ = ⎣ 0 + (−3) + (−2) 12 + (−18) + (−3) 2 + (−21) + 1 (−4) + 6 + (−1) −9 −18 1 9

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡

⎤T ⎡ ⎤T ⎡ ⎤ ⎡ −4 2 3 3 −3 5 −4 0 6 3 1 0 −2 ⎦ = ⎣ 2 5 1 ⎦ + ⎣ −3 0 (i) ⎣ 0 5 −1 ⎦ + ⎣ 1 6 1 −2 6 7 −2 3 −1 −2 5 −2 ⎡ ⎤ ⎡ ⎤ ((−4) + 3) (0 + 1) (6 + 6) −1 1 12 ⎦ = ⎣ −1 (5 + 0) (1 + 7) 5 8 ⎦ = ⎣ (2 + (−3)) (3 + 5) ((−1) + (−2)) ((−2) + (−2)) 8 −3 −4

Section 1.4 ⎤ 6 7 ⎦ −2

(l) Impossible. Since C is a 2 × 2 matrix, both CT and 2CT are 2 × 2. Also, since F is a 3 × 2 matrix, −3F is 3 × 2. Thus, 2CT and −3F have different sizes. However, if we want to add matrices, they must be the same size. ⎡ ⎤ ⎡ ⎤ 10 2 −3 3 1 6 0 7 ⎦. (n) (B − A)T = ⎣ −3 −3 −2 ⎦ and ET = ⎣ −3 −3 −3 3 5 −2 −2 ⎡ ⎤ ⎡ ⎤ 13 3 3 13 −6 2 T = ⎣ 3 −3 −5 ⎦. Thus, (B − A)T + ET = ⎣ −6 −3 5 ⎦, and so (B − A)T + ET 2 −5 1 3 5 1 (2) For a matrix to be square, it must have the same number of rows as columns. A matrix X is diagonal if and only if it is square and xij = 0 for i = j. A matrix X is upper triangular if and only if it is square and xij = 0 for i > j. A matrix X is lower triangular if and only if it is square and xij = 0 for i < j. A matrix X is symmetric if and only if it is square and xij = xji for all i, j. A matrix X is skew-symmetric if and only if it is square and xij = −xji for all i, j. Notice that this implies that if X is skew-symmetric, then xii = 0 for all i. Finally, the transpose of an m × n matrix X is the n × m matrix whose (i, j)th entry is xji . We now check each matrix in turn. A is a 3 × 2 matrix, so it is not square. Therefore, it cannot be diagonal, upper triangular, lower −1 0 6 T . triangular, symmetric, or skew-symmetric. Finally, A = 4 1 0 B is a 2 × 2 matrix, so it is square. Because b12 = b21 = 0, B is diagonal, upper triangular, lower triangular, and symmetric. B is not skew-symmetric because b11 = 0. Finally, BT = B. C is a 2×2 matrix, so it is square. Because c12 = 0, C is neither diagonal nor lower triangular. Because c21 = 0, C is not upper triangular. C is not symmetric because c12 = c21 . C is not skew-symmetric −1 −1 T . because c11 = 0. Finally, C = 1 1 D is a 3 × 1 matrix, so it is not square. Therefore, it cannot be diagonal, upper triangular, lower triangular, symmetric, or skew-symmetric. Finally, DT = −1 4 2 . E is a 3×3 matrix, so it is square. Because e13 = 0, E is neither diagonal nor lower triangular. Because e31 = 0, E is not upper triangular. E is not symmetric because e31 = e13 . E is not skew-symmetric ⎡ ⎤ 0 0 −6 0 ⎦. because e22 = 0. Finally, ET = ⎣ 0 −6 6 0 0 F is a 4 × 4 matrix, so it is square. Because f14 = 0, F is neither diagonal nor lower triangular. Because f41 = 0, F is not upper triangular. F is symmetric because fij = fji for every i, j. F is not skew-symmetric because f32 = −f23 . Finally, FT = F. G is a 3 × 3 matrix, so it is square. Because gij = 0 whenever i = j, G is diagonal, lower triangular, upper triangular, and symmetric. G is not skew-symmetric because g11 = 0. Finally, GT = G. 10


Section 1.4

H is a 4×4 matrix, so it is square. Because h14 = 0, H is neither diagonal nor lower triangular. Because h41 = 0, H is not upper triangular. H is not symmetric because h12 = h21 . H is skew-symmetric because hij = −hji for every i, j. Finally, HT = −H. J is a 4 × 4 matrix, so it is square. Because j12 = 0, J is neither diagonal nor lower triangular. Because j21 = 0, J is not upper triangular. J is symmetric because jik = jki for every i, k. J is not skew-symmetric because j32 = −j23 . Finally, JT = J. K is a 4×4 matrix, so it is square. Because k14 = 0, K is neither diagonal nor lower triangular. Because k41 = 0, K is not upper triangular. K is not symmetric because k12 = k21 . K is not skew-symmetric ⎡ ⎤ 1 −2 −3 −4 ⎢ 2 1 −5 −6 ⎥ ⎥. because k11 = 0. Finally, KT = ⎢ ⎣ 3 5 1 −7 ⎦ 4 6 7 1 L is a 3 × 3 matrix, so it is square. Because l13 = 0, L is neither diagonal nor lower triangular. Because lij = 0 whenever i˙ > j, L is upper triangular. L is not symmetric because l31 = l13 . L is not ⎡ ⎤ 1 0 0 skew-symmetric because l11 = 0 (or because l21 = −l12 ). Finally, LT = ⎣ 1 1 0 ⎦. 1 1 1 M is a 3 × 3 matrix, so it is square. Because m31 = 0, M is neither diagonal nor upper triangular. Because mij = 0 whenever i˙ < j, M is lower triangular. M is not symmetric because m31 = m13 . ⎡ ⎤ 0 1 1 M is not skew-symmetric because m31 = −m13 . Finally, MT = ⎣ 0 0 1 ⎦. 0 0 0 N is a 3 × 3 matrix, so it is square. Because nij = 0 whenever i = j, N is diagonal, lower triangular, upper triangular, and symmetric. N is not skew-symmetric because n11 = 0. Finally, NT = N = I3 . P is a 2×2 matrix, so it is square. Because p12 = 0, P is neither diagonal nor lower triangular. Because p21 = 0, P is not upper triangular. P is symmetric because p12 = p21 . P is not skew-symmetric because p12 = −p21 . Finally, PT = P. Q is a 3 × 3 matrix, so it is square. Because q31 = 0, Q is neither diagonal nor upper triangular. Because qij = 0 whenever i˙ < j, Q is lower triangular. Q is not symmetric because q31 = q13 . Q is ⎡ ⎤ −2 4 −1 2 ⎦. not skew-symmetric because q11 = 0 (or because q31 = −q13 ). Finally, QT = ⎣ 0 0 0 0 3 R is a 3 × 2 matrix, so it is not square. Therefore, it cannot be diagonal, upper triangular, lower 6 3 −1 T . triangular, symmetric, or skew-symmetric. Finally, R = 2 −2 0 (3) Just after Theorem 1.13 in the textbook, the desired decomposition is described as A = S + V, where S = 12 (A + AT ) is symmetric and V = 12 (A − AT ) is skew-symmetric. ⎛⎡

⎤ ⎡ 3 −1 4 3 ⎥ ⎢ 1 1 ⎜⎢ T 2 5 ⎦ + ⎣ −1 (a) S = 2 (A + A ) = 2 ⎝⎣ 0 1 −3 2 4

11

0 2 5

⎤⎞ 1 ⎥⎟ −3 ⎦⎠ = 2

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎛⎡

(3 + 3) 1 ⎝⎣ (0 + (−1)) 2 (1 + 4) ⎡ 1 2

6 −1 ⎣ −1 4 5 2

((−1) + 0) (2 + 2) ((−3) + 5) ⎤

⎡

5 ⎢ 2 ⎦=⎣ 4

⎤⎞ (4 + 1) (5 + (−3)) ⎦⎠ = (2 + 2)

1 2 (6)

1 2 (−1) 1 2 (4) 1 2 (2)

1 2 (−1) 1 2 (5)

1 2 (5) 1 2 (2) 1 2 (4)

⎤

⎡

⎥ ⎢ ⎦=⎢ ⎣ −

sij = sji for all i, j. ⎛⎡

⎡

(3 + 0) ⎢ 1 1 ⎣ −2 + 2 5 3 2 + −2

1

2

5 1

− 12 + − 2 (2 + 0)

(1 + (−4))

⎤ 3

−

3 2

⎡ 3 ⎥ ⎣ (1 + 4) ⎦ = 0 1 (2 + 0) 2

+

2

−4 −1 2 −3

5 2

⎤

3

− 12

1 2

2

⎥ 1 ⎥ ⎦, which is symmetric, since

5 2

1

2

⎤ ⎡ ⎤⎞ 3 −1 4 3 0 1 2 5 ⎦ − ⎣ −1 2 −3 ⎦⎠ V = 12 (A − AT ) = 12 ⎝⎣ 0 1 −3 2 4 5 2 ⎛⎡ ⎤⎞ (3 − 3) ((−1) − 0) (4 − 1) (2 − 2) (5 − (−3)) ⎦⎠ = = 12 ⎝⎣ (0 − (−1)) (1 − 4) ((−3) − 5) (2 − 2) ⎡ ⎡ ⎤ ⎡ 1 (0) 1 (−1) 1 (3) ⎤ 0 − 12 2 2 2 0 −1 3 ⎢ ⎥ ⎢ 1 1 1 1 ⎣ ⎢ 1 0 8 ⎦ = ⎣ 12 (1) 0 2 (0) 2 (8) ⎦ = ⎣ 2 2 1 1 1 −3 −8 0 − 32 −4 2 (−3) 2 (−8) 2 (0) since vij = −vji for all i, j. ⎡ ⎤ ⎤ ⎡ 3 − 12 52 0 − 12 32 ⎥ ⎢ ⎥ ⎢ 1 1 ⎢ 2 1 ⎥ 0 4 ⎥ Note that S + V = ⎢ 2 ⎣ −2 ⎦= ⎦+⎣ 5 2

Section 1.4

3 2

⎤

⎥ 4 ⎥ ⎦, which is skew-symmetric,

0

0 ⎤ 4 5 ⎦ = A. 2

(5) (d) The matrix (call it A) must be a square zero matrix; that is, On , for some n. First, since A is diagonal, it must be square (by definition of a diagonal matrix). To prove that aij = 0 for all i, j, we consider two cases: First, if i = j, then aij = 0 since A is diagonal. Second, if i = j, then aij = aii = 0 since A is skew-symmetric (aii = −aii for all i, implying aii = 0 for all i). Hence, every entry aij = 0, and so A is a zero matrix. (11) Part (1): Let B = AT and C = (AT )T . Then cij = bji = aij . Since this is true for all i and j, we have C = A; that is, (AT )T = A. Part (3): Let B = c(AT ), D = cA, and F = (cA)T . Then fij = dji = caji = bij . Since this is true for all i and j, we have F = B; that is, (cA)T = c(AT ). (14) (a) Trace (B) = b11 + b22 = 2 + (−1) = 1, trace (C) = c11 + c22 = (−1) + 1 = 0, trace (E) = e11 + e22 + e33 = 0 + (−6) + 0 = −6, trace (F) = f11 + f22 + f33 + f44 =1 + 0 + 0 + 1 = 2, trace (G) = g11 + g22 + g33 = 6 + 6 + 6 = 18, trace (H) = h11 + h22 + h33 + h44 = 0+0+0+0 = 0, trace (J) = j11 +j22 +j33 +j44 = 0+0+1+0 = 1, trace (K) = k11 +k22 +k33 +k44 = 1 + 1 + 1 + 1 = 4, trace (L) = l11 + l22 + l33 = 1 + 1 + 1 = 3, trace (M) = m11 + m22 + m33 = 0 + 0 + 0 = 0, trace (N) = n11 + n22 + n33 = 1 + 1 + 1 = 3, trace (P) = p11 + p22 = 0 + 0 = 0, trace (Q) = q11 + q22 + q33 = (−2) + 0 + 3 = 1. 12


Section 1.5

(c) Not necessarily true when n > 1. Consider matrices L and N in Exercise 2. From part (a) of this exercise, trace(L) = 3 = trace(N). However, L = N. (15) (a) False. The main diagonal entries of a 5 × 6 matrix A are a11 , a22 , a33 , a44 , and a55 . Thus, there are 5 entries on the main diagonal. (There is no entry a66 because there is no 6th row.) (b) True. Suppose A is lower triangular. Then aij = 0 whenever i < j. If B = AT , then bij = aji = 0, if i > j, and so B is upper triangular. (c) False. The square zero matrix On is both skew-symmetric and diagonal. (d) True. This follows from the definition of a skew-symmetric matrix.

T T T

(by part (3) of Theorem 1.12) = c AT + BT (e) True. c AT + B = c AT + B (by part (2) of Theorem 1.12) = c A + BT (by part (1) of Theorem 1.12) = cA + cBT (by part (5) of Theorem 1.11) = cBT + cA (by part (1) of Theorem 1.11).

Section 1.5

⎡

⎤⎡ −5 3 6 −2 3 ⎣ ⎦ ⎣ 3 8 0 6 5 (1) (b) BA = −2 0 4 1 −4 ⎡ ((−5) (−2) + (3) (6) + (6) (1)) ⎣ ((3) (−2) + (8) (6) + (0) (1)) ((−2) (−2) + (0) (6) + (4) (1))

⎤ ⎦= ⎤ ⎡ ((−5) (3) + (3) (5) + (6) (−4)) 34 ((3) (3) + (8) (5) + (0) (−4)) ⎦ = ⎣ 42 ((−2) (3) + (0) (5) + (4) (−4)) 8

⎤ −24 49 ⎦ −22

(c) Impossible; (number of columns of J) = 1 = 2 = (number of rows of M) ⎡ ⎤ 8 (e) RJ = −3 6 −2 ⎣ −1 ⎦ = [(−3) (8) + (6) (−1) + (−2) (4)] = [−38] 4 ⎤ ⎡ ⎤ ⎡ ⎡ (8) (−3) (8) (6) (8) (−2) −24 8 3 (f) JR = ⎣ −1 ⎦ −3 6 −2 = ⎣ (−1) (−3) (−1) (6) (−1) (−2) ⎦ = ⎣ 4 (4) (−3) (4) (6) (4) (−2) −12 (g) Impossible; (number of columns of R) = 3 = 1 = (number of rows of T) (j) Impossible; (number of columns of ⎡ 1 ⎢ 1 (l) E3 = E(E2 ). Let W = E2 = ⎢ ⎣ 0 1 w11 w12 w13 w14 w21 w22 w23 w24 w31 w32

F) = 2 = 4 = (number of rows of F) ⎤2 1 0 1 0 1 0 ⎥ ⎥ . Then 0 0 1 ⎦ 0 1 0

= (1) (1) + (1) (1) + (0) (0) + (1) (1) = 3 = (1) (1) + (1) (0) + (0) (0) + (1) (0) = 1 = (1) (0) + (1) (1) + (0) (0) + (1) (1) = 2 = (1) (1) + (1) (0) + (0) (1) + (1) (0) = 1 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 = (1) (1) + (0) (0) + (1) (0) + (0) (0) = 1 = (1) (0) + (0) (1) + (1) (0) + (0) (1) = 0 = (1) (1) + (0) (0) + (1) (1) + (0) (0) = 2 = (0) (1) + (0) (1) + (0) (0) + (1) (1) = 1 = (0) (1) + (0) (0) + (0) (0) + (1) (0) = 0 13

48 −6 24

⎤ −16 2 ⎦ −8


w33 w34 w41 w42 w43 w44

= (0) (0) + (0) (1) + (0) (0) + (1) (1) = 1 = (0) (1) + (0) (0) + (0) (1) + (1) (0) = 0 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 = (1) (1) + (0) (0) + (1) (0) + (0) (0) = 1 = (1) (0) + (0) (1) + (1) (0) + (0) (1) = 0 = (1) (1) + (0) (0) + (1) (1) + (0) (0) = 2

Let Y = E3 ⎡ 1 1 0 ⎢ 1 0 1 ⎢ ⎣ 0 0 0 1 0 1

= EE2 = EW = ⎤⎡ 1 3 1 2 1 ⎢ 1 1 0 2 0 ⎥ ⎥⎢ 1 ⎦⎣ 1 0 1 0 0 1 1 0 2

⎤ ⎥ ⎥. So, ⎦

y11 y12 y13 y14 y21 y22 y23 y24 y31 y32 y33 y34 y41 y42 y43 y44

= (1) (3) + (1) (1) + (0) (1) + (1) (1) = 5 = (1) (1) + (1) (1) + (0) (0) + (1) (1) = 3 = (1) (2) + (1) (0) + (0) (1) + (1) (0) = 2 = (1) (1) + (1) (2) + (0) (0) + (1) (2) = 5 = (1) (3) + (0) (1) + (1) (1) + (0) (1) = 4 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 = (1) (2) + (0) (0) + (1) (1) + (0) (0) = 3 = (1) (1) + (0) (2) + (1) (0) + (0) (2) = 1 = (0) (3) + (0) (1) + (0) (1) + (1) (1) = 1 = (0) (1) + (0) (1) + (0) (0) + (1) (1) = 1 = (0) (2) + (0) (0) + (0) (1) + (1) (0) = 0 = (0) (1) + (0) (2) + (0) (0) + (1) (2) = 2 = (1) (3) + (0) (1) + (1) (1) + (0) (1) = 4 = (1) (1) + (0) (1) + (1) (0) + (0) (1) = 1 = (1) (2) + (0) (0) + (1) (1) + (0) (0) = 3 = (1) (1) + (0) (2) + (1) (0) + (0) (2) = 1 ⎡ ⎤ 5 3 2 5 ⎢ 4 1 3 1 ⎥ ⎥ Hence, E3 = Y = ⎢ ⎣ 1 1 0 2 ⎦. 4 1 3 1 ⎤ ⎞ ⎛⎡ 9 −3 ⎜⎢ 5 −4 ⎥ 2 1 −5 ⎟ ⎥ ⎟= ⎢ (n) D(FK) = D ⎜ ⎝⎣ 2 7 ⎠ 0 ⎦ 0 2 8 −3 ⎡ ((9) (2) + (−3) (0)) ((9) (1) + (−3) (2)) ((9) (−5) + (−3) (7)) ⎢ ((5) (2) + (−4) (0)) ((5) (1) + (−4) (2)) ((5) (−5) + (−4) (7)) D⎢ ⎣ ((2) (2) + (0) (0)) ((2) (1) + (0) (2)) ((2) (−5) + (0) (7)) ((8) (2) + (−3) (0)) ((8) (1) + (−3) (2)) ((8) (−5) + (−3) (7)) ⎡ ⎤ ⎤ 18 ⎡ 3 −66 −1 4 3 7 ⎢ 10 −3 −53 ⎥ ⎥ = Z. Then, 5 ⎦⎢ =⎣ 2 1 7 ⎣ 4 2 −10 ⎦ 0 5 5 −2 16 2 −61 z11 = (−1) (18) + (4) (10) + (3) (4) + (7) (16) = 146 z12 = (−1) (3) + (4) (−3) + (3) (2) + (7) (2) = 5 14

⎤ ⎥ ⎥ ⎦

Section 1.5


Section 1.5

= (−1) (−66) + (4) (−53) + (3) (−10) + (7) (−61) = −603 = (2) (18) + (1) (10) + (7) (4) + (5) (16) = 154 = (2) (3) + (1) (−3) + (7) (2) + (5) (2) = 27 = (2) (−66) + (1) (−53) + (7) (−10) + (5) (−61) = −560 = (0) (18) + (5) (10) + (5) (4) + (−2) (16) = 38 = (0) (3) + (5) (−3) + (5) (2) + (−2) (2) = −9 = (0) (−66) + (5) (−53) + (5) (−10) + (−2) (−61) = −193 ⎡ ⎤ 146 5 −603 Hence, D(FK) = Z = ⎣ 154 27 −560 ⎦. 38 −9 −193 10 9 7 −1 ((10) (7) + (9) (11)) ((10) (−1) + (9) (3)) (2) (a) No. LM = = 8 7 11 3 ((8) (7) + (7) (11)) ((8) (−1) + (7) (3)) 169 17 7 −1 10 9 = , but ML = 133 13 11 3 8 7 ((7) (10) + (−1) (8)) ((7) (9) + (−1) (7)) 62 56 = = . Hence, LM = ML. ((11) (10) + (3) (8)) ((11) (9) + (3) (7)) 134 120 z13 z21 z22 z23 z31 z32 z33

(c) No. Since A is a 3 × 2 matrix and K is a 2 × 3 matrix, AK is a 3 × 3 matrix, while KA is a 2 × 2 matrix. Therefore, AK cannot equal KA, since they are different sizes. 0 0 3 −1 ((0) (3) + (0) (4)) ((0) (−1) + (0) (7)) 0 0 (d) Yes. NP = = = . 0 0 4 7 ((0) (3) + (0) (4)) ((0) (−1) + (0) (7)) 0 0 3 −1 0 0 ((3) (0) + (−1) (0)) ((3) (0) + (−1) (0)) 0 0 Also, PN = = = . 4 7 0 0 ((4) (0) + (7) (0)) ((4) (0) + (7) (0)) 0 0 Hence, NP = PN, and so N and P commute. (3) (a) (2nd row of BG) = (2nd row of B)G =

3

8

0

⎡

5 ⎣ 0 1

⎤ 1 0 −2 −1 ⎦. The entries obtained from 0 3

this are: (1st column entry) = ((3) (5) + (8) (0) + (0) (1)) = 15, (2nd column entry) = ((3) (1) + (8) (−2) + (0) (0)) = −13, (3rd column entry) = ((3) (0) + (8) (−1) + (0) (3)) = −8. Hence, (2nd row of BG) = [15, −13, −8].

⎡

(c) (1st column of SE) = S(1st column of E) =

6 −4

3

⎤ 1 ⎢ 1 ⎥ ⎥ 2 ⎢ ⎣ 0 ⎦ 1

= [(6) (1) + (−4) (1) + (3) (0) + (2) (1)] = [4] (4) (a) Valid, by Theorem 1.14, part (1). (b) Invalid. The equation claims that all pairs of matrices that can be multiplied in both orders commute. However, parts (a) and (c) of Exercise 2 illustrate two different pairs of matrices that do not commute (see above). (c) Valid, by Theorem 1.14, part (1). (d) Valid, by Theorem 1.14, part (2). 15


Section 1.5

(e) Valid, by Theorem 1.16. (f) Invalid. For a counterexample, consider the matrices L and M from Exercises 1 through 3. 10 9 62 56 Using our computation of ML from Exercise 2(a), above, L(ML) = = 8 7 134 120 ((10) (62) + (9) (134)) ((10) (56) + (9) (120)) 1826 1640 = , ((8) (62) + (7) (134)) ((8) (56) + (7) (120)) 1434 1288

10 9 10 9 2 but L M = (LL)M = M 8 7 8 7 ((10) (10) + (9) (8)) ((10) (9) + (9) (7)) 172 153 7 −1 = M= ((8) (10) + (7) (8)) ((8) (9) + (7) (7)) 136 121 11 3 ((172) (7) + (153) (11)) ((172) (−1) + (153) (3)) 2887 287 = = . ((136) (7) + (121) (11)) ((136) (−1) + (121) (3)) 2283 227 (g) Valid, by Theorem 1.14, part (3). (h) Valid, by Theorem 1.14, part (2). (i) Invalid. For a counterexample, consider the matrices A and K from Exercises 1 through 3. Note that A is a 3 × 2 matrix and K is a 2 × 3 matrix. Therefore, AK is a 3 × 3 matrix. Hence, (AK)T is a 3 × 3 matrix. However, AT is a 2 × 3 matrix and KT is a 3 × 2 matrix. Thus, AT KT is a 2 × 2 matrix and so cannot equal (AK)T . The equation is also false in general for square matrices, where it is not the sizes of the matrices 1 1 1 −1 that cause the problem. For example, let A = , and let K = . Then, 0 0 0 0

T T 1 1 1 −1 ((1) (1) + (1) (0)) ((1) (−1) + (1) (0)) = = (AK)T = 0 0 0 0 ((0) (1) + (0) (0)) ((0) (−1) + (0) (0)) T 1 0 1 0 1 0 1 −1 T T = , while A K = −1 0 1 0 −1 0 0 0 ((1) (1) + (0) (−1)) ((1) (0) + (0) (0)) 1 0 = = . ((1) (1) + (0) (−1)) ((1) (0) + (0) (0)) 1 0 However, there are some rare instances for which the equation is true, such as when A = K, or when either A or K equals In . Another example for which the equation holds is for A = G and K = H, where G and H are the matrices used in Exercises 1, 2, and 3. (j) Valid, by Theorem 1.14, part (3), and Theorem 1.16. (5) To find the total in salaries paid by Outlet 1, we must compute the total amount paid to executives, the total amount paid to salespersons, and the total amount paid to others, and then add these amounts. But each of these totals is found by multiplying the number of that type of employee working at Outlet 1 by the salary for that type of employee. Hence, the total salary paid at Outlet 1 is (3) (30000) + (7) (22500) + (8) (15000) = 367500 . Executives

Salespersons

Others

Salary Total

Note that this is the (1, 1) entry obtained when multiplying the two given matrices. A similar analysis shows that multiplying the two given matrices gives all the desired salary and fringe benefit totals. In 16


particular, if ⎡ 3 7 ⎢ 2 4 W=⎢ ⎣ 6 14 3 6 w11 w12 w21 w22 w31 w32 w41 w42

⎤ 8 ⎡ 30000 5 ⎥ ⎥ ⎣ 22500 18 ⎦ 15000 9

Section 1.5

⎤ 7500 4500 ⎦, then 3000

= (3) (30000) + (7) (22500) + (8) (15000) = 367500, = (3) (7500) + (7) (4500) + (8) (3000) = 78000, = (2) (30000) + (4) (22500) + (5) (15000) = 225000, = (2) (7500) + (4) (4500) + (5) (3000) = 48000, = (6) (30000) + (14) (22500) + (18) (15000) = 765000, = (6) (7500) + (14) (4500) + (18) (3000) = 162000, = (3) (30000) + (6) (22500) + (9) (15000) = 360000, = (3) (7500) + (6) (4500) + (9) (3000) = 76500.

Hence, we have: W =

Outlet Outlet Outlet Outlet

Salary Fringe Benefits ⎡ ⎤ 1 $367500 $78000 2 ⎢ $48000 ⎥ ⎢ $225000 ⎥. ⎣ 3 $765000 $162000 ⎦ 4 $360000 $76500

(7) To compute the tonnage of a particular chemical applied to a given field, for each type of fertilizer we must multiply the percent concentration of the chemical in the fertilizer by the number of tons of that fertilizer applied to the given field. After this is done for each fertilizer type, we add these results to get the total tonnage of the chemical that was applied. For example, to compute the number of tons of potash applied to field 2, we compute 0.05

% Potash in Fert. 1

·

2

Tons of Fert. 1

+

·

0.05

% Potash in Fert. 2

1 Tons of Fert. 2

+

0.20

% Potash in Fert. 3

·

1

Tons of Fert. 3

=

0.35

.

Tons of Potash

Note that this is the dot product (3rd column of A)·(2nd column of B). Since this is the dot product of two columns, it is not in the form of an entry of a matrix product. To turn this into a form that looks like an entry of a matrix product, we consider the transpose of AT and express it as (3rd row of AT )·(2nd column of B). Thus, this is the (3, 2) entry of AT B. A similar analysis shows that computing all of the entries of AT B produces all of the information requested in the problem. Hence, to solve the problem, we compute ⎡ ⎤⎡ ⎤ 0.10 0.25 0.00 5 2 4 Y = AT B = ⎣ 0.10 0.05 0.10 ⎦ ⎣ 2 1 1 ⎦ . 0.05 0.05 0.20 3 1 3 Solving for each entry produces y11 y12 y13 y21 y22 y23

= (0.10) (5) + (0.25) (2) + (0.00) (3) = 1.00, = (0.10) (2) + (0.25) (1) + (0.00) (1) = 0.45, = (0.10) (4) + (0.25) (1) + (0.00) (3) = 0.65, = (0.10) (5) + (0.05) (2) + (0.10) (3) = 0.90, = (0.10) (2) + (0.05) (1) + (0.10) (1) = 0.35, = (0.10) (4) + (0.05) (1) + (0.10) (3) = 0.75, 17


Section 1.5

y31 = (0.05) (5) + (0.05) (2) + (0.20) (3) = 0.95, y32 = (0.05) (2) + (0.05) (1) + (0.20) (1) = 0.35, y33 = (0.05) (4) + (0.05) (1) + (0.20) (3) = 0.85. Field 1 Field 2 Field 3 ⎤ ⎡ 1.00 0.45 0.65 Nitrogen 0.90 0.35 0.75 ⎦(in tons). Hence, Y = Phosphate ⎣ 0.95 0.35 0.85 Potash 1 1 1 1 1 1 2 (9) (a) An example: A = . Note that A = 0 −1 0 −1 0 −1 ((1) (1) + (1) (0)) ((1) (1) + (1) (−1)) = = I2 . ((0) (1) + (−1) (0)) ((0) (1) + (−1) (−1)) ⎡ ⎡ ⎤ 1 1 1 1 0 (b) One example: Suppose A = ⎣ 0 −1 0 ⎦. Let W = A2 = ⎣ 0 −1 0 0 1 0 0 Computing each entry yields w11 w12 w13 w21 w22 w23 w31 w32 w33

⎤⎡ 0 1 0 ⎦⎣ 0 1 0

1 −1 0

⎤ 0 0 ⎦. 1

= (1) (1) + (1) (0) + (0) (0) = 1, = (1) (1) + (1) (−1) + (0) (0) = 0, = (1) (0) + (1) (0) + (0) (1) = 0, = (0) (1) + (−1) (0) + (0) (0) = 0, = (0) (1) + (−1) (−1) + (0) (0) = 1, = (0) (0) + (−1) (0) + (0) (1) = 0, = (0) (1) + (0) (0) + (1) (0) = 0, = (0) (1) + (0) (−1) + (1) (0) = 0, = (0) (0) + (0) (0) + (1) (1) = 1.

Hence, A2 = W = I3 . ⎤ ⎡ ⎡ 0 0 0 1 (c) Consider A = ⎣ 1 0 0 ⎦. Let Y = A2 = ⎣ 1 0 0 1 0

0 0 1

⎤⎡ 1 0 0 ⎦⎣ 1 0 0

0 0 1

⎤ 1 0 ⎦. First, we compute each 0

entry of Y. y11 y12 y13 y21 y22 y23 y31 y32 y33

= (0) (0) + (0) (1) + (1) (0) = 0, = (0) (0) + (0) (0) + (1) (1) = 1, = (0) (1) + (0) (0) + (1) (0) = 0, = (1) (0) + (0) (1) + (0) (0) = 0, = (1) (0) + (0) (0) + (0) (1) = 0, = (1) (1) + (0) (0) + (0) (0) = 1, = (0) (0) + (1) (1) + (0) (0) = 1, = (0) (0) + (1) (0) + (0) (1) = 0, = (0) (1) + (1) (0) + (0) (0) = 0. ⎡ ⎤ ⎡ 0 1 0 0 And so Y = A2 = ⎣ 0 0 1 ⎦. Hence, A3 = A2 A = YA = ⎣ 0 1 0 0 1 Letting this product equal Z, we compute each entry as follows: z11 = (0) (0) + (1) (1) + (0) (0) = 1, z12 = (0) (0) + (1) (0) + (0) (1) = 0, 18

1 0 0

⎤⎡ 0 0 1 ⎦⎣ 1 0 0

0 0 1

⎤ 1 0 ⎦. 0


z13 z21 z22 z23 z31 z32 z33

Section 1.5

= (0) (1) + (1) (0) + (0) (0) = 0, = (0) (0) + (0) (1) + (1) (0) = 0, = (0) (0) + (0) (0) + (1) (1) = 1, = (0) (1) + (0) (0) + (1) (0) = 0, = (1) (0) + (0) (1) + (0) (0) = 0, = (1) (0) + (0) (0) + (0) (1) = 0, = (1) (1) + (0) (0) + (0) (0) = 1.

Therefore, A3 = Z = I3 . (10) (a) This is the dot product of the 3rd row of A with the 4th column of B. Hence, the result is the 3rd row, 4th column entry of AB. (c) This is (2nd column of A)·(3rd row of B) = (3rd row of B)·(2nd column of A) = 3rd row, 2nd column entry of BA. (11) (a) ((3, 2) entry of AB) = (3rd row of A)·(2nd column of B) =

!n k=1

a3k bk2

(12) (a) 3v1 − 2v2 + 5v3 = 3[4, 7, −2] − 2[−3, −6, 5] + 5[−9, 2, −8] = [12, 21, −6] + [6, 12, −10] + [−45, 10, −40] = [−27, 43, −56] ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 7 −2 (b) 2w1 + 6w2 − 3w3 = 2 ⎣ −3 ⎦ + 6 ⎣ −6 ⎦ − 3 ⎣ 5 ⎦ −9 2 −8 ⎡

⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8 42 6 56 = ⎣ −6 ⎦ + ⎣ −36 ⎦ + ⎣ −15 ⎦ = ⎣ −57 ⎦ −18 12 24 18 (15) Several crucial steps in the following proofs rely on Theorem 1.5 for their validity. Proof of Part (2): The (i, j) entry of A(B + C) = = = = =

(ith row of A)·(jth column of (B + C)) (ith row of A)·(jth column of B + jth column of C) (ith row of A)·(jth column of B) + (ith row of A)·(jth column of C) (i, j) entry of AB + (i, j) entry of AC (i, j) entry of (AB + AC).

Proof of Part (3): The (i, j) entry of (A + B)C = (ith row of (A + B))·(jth column of C) = ((ith row of A) + (ith row of B)) · (jth column of C) = (ith row of A)·(jth column of C) + (ith row of B)·(jth column of C) = (i, j) entry of AC + (i, j) entry of BC = (i, j) entry of (AC + BC). For the first equation in part (4), the (i, j) entry of c(AB) = c ((ith row of A)·(jth column of B)) = (c(ith row of A))·(jth column of B) = (ith row of cA)·(jth column of B) = (i, j) entry of (cA)B. 19


Section 1.5

Similarly, the (i, j) entry of c(AB) = c ((ith row of A)·(jth column of B)) = (ith row of A)·(c(jth column of B)) = (ith row of A)·(jth column of cB) = (i, j) entry of A(cB). (20) Proof of Part (1): We use induction on the variable t. Base Step: As+0 = As = As I = As A0 . Inductive Step: Assume As+t = As At for some t ≥ 0. We must prove As+(t+1) = As At+1 . But As+(t+1) = A(s+t)+1 = As+t A = (As At )A (by the inductive hypothesis) = As (At A) = As At+1 . Proof of Part (2): Again, we use induction on the variable t. Base Step: (As )0 = I = A0 = As0 . Inductive Step: Assume (As )t = Ast for some integer t ≥ 0. We must prove (As )t+1 = As(t+1) . But (As )t+1 = (As )t As (by definition) = Ast As (by the inductive hypothesis) = Ast+s (by part (1)) = As(t+1) . Finally, reversing the roles of s and t in the proof above shows that (At )s = Ats , which equals Ast . 1 0 1 0 1 0 2 (27) (a) Consider any matrix A of the form . Then A = = x 0 x 0 x 0 ((1) (1) + (0) (x)) ((1) (0) + (0) (0)) 1 0 = A. So, for example, is idempotent. ((x) (1) + (0) (x)) ((x) (0) + (0) (0)) 1 0 ⎤ ⎡ ⎤ ⎡ 1 −2 1 −2 1 2 −1 ⎣ 1 2 −1 0 1 ⎦= 1 ⎦ . Then AB = (28) (b) Consider A = and B = ⎣ 0 2 4 −2 2 4 −2 1 0 1 0 ((1) (1) + (2) (0) + (−1) (1)) ((1) (−2) + (2) (1) + (−1) (0)) = O2 . ((2) (1) + (4) (0) + (−2) (1)) ((2) (−2) + (4) (1) + (−2) (0)) (29) A 2 × 2 matrix A that commutes with every other 2 × 2 matrix must have the form A = cI2 . To prove 0 1 this, note that if A commutes with every other 2×2 matrix, then AB = BA, where B = . But 0 0 0 a11 0 1 ((a11 ) (0) + (a12 ) (0)) ((a11 ) (1) + (a12 ) (0)) a11 a12 = = and AB = a21 a22 0 a21 ((a21 ) (0) + (a22 ) (0)) ((a21 ) (1) + (a22 ) (0)) 0 0 a21 a22 ((0) (a11 ) + (1) (a21 )) ((0) (a12 ) + (1) (a22 )) 0 1 a11 a12 = . = BA = a21 a22 0 0 ((0) (a11 ) + (0) (a21 )) ((0) (a12 ) + (0) (a22 )) 0 0 c a12 . Setting these equal shows that a21 = 0 and a11 = a22 . Let c = a11 = a22 . Then A = 0 c 0 0 0 0 c a12 = Let D = . Then AD = 0 c 1 0 1 0 ((c) (0) + (a12 ) (1)) ((c) (0) + (a12 ) (0)) a12 0 0 0 c a12 = and DA = = 0 c ((0) (0) + (c) (1)) ((0) (0) + (c) (0)) c 0 1 0 0 0 ((0) (c) + (0) (0)) ((0) (a12 ) + (0) (c)) . = c a12 ((1) (c) + (0) (0)) ((1) (a12 ) + (0) (c)) Setting AD = DA shows that a12 = 0, and so A = cI2 .

20


Chapter 1 Review

Finally, we must show that cI2 actually does commute with every 2 × 2 matrix. If M is a 2 × 2 matrix, then (cI2 )M = c(I2 M) = cM. Similarly, M(cI2 ) = c(MI2 ) = cM. Hence, cI2 commutes with M. (31) (a) True. This is a “boxed” fact given in the section. (b) True. D (A + B) = DA+DB (by part (2) of Theorem 1.14) = DB+DA (by part (1) of Theorem 1.1) (c) True. This is part (4) of Theorem 1.14. 1 1 1 −2 1 −2 1 1 (d) False. Let D = , and let E = . Now DE = = 0 −1 0 1 0 −1 0 1 ((1) (1) + (1) (0)) ((1) (−2) + (1) (1)) 1 −1 = . Hence, ((0) (1) + (−1) (0)) ((0) (−2) + (−1) (1)) 0 −1 1 −1 1 −1 ((1) (1) + (−1) (0)) ((1) (−1) + (−1) (−1)) 2 = = I2 . (DE) = 0 −1 0 −1 ((0) (1) + (−1) (0)) ((0) (−1) + (−1) (−1)) 1 1 1 1 ((1) (1) + (1) (0)) ((1) (1) + (1) (−1)) = = I2 . Also, But D2 = 0 −1 0 −1 ((0) (1) + (−1) (0)) ((0) (1) + (−1) (−1)) 1 −2 1 −2 ((1) (1) + (−2) (0)) ((1) (−2) + (−2) (1)) 1 −4 E2 = = = . 0 1 0 1 ((0) (1) + (1) (0)) ((0) (−2) + (1) (1)) 0 1 Hence, D2 E2 = I2 E2 = E2 = I2 . Thus, (DE)2 = D2 E2 . The problem here is that D and E do not commute. For, if they did, we would have (DE)2 = (DE) (DE) = D(ED)E = D(DE)E = (DD)(EE) = D2 E2 . (e) False. See the answer for part (i) of Exercise 4 for a counterexample. We know from Theorem 1.16 that (DE)T = ET DT . So if DT and ET do not commute, then (DE)T = DT ET . (f) False. See the answer for part (b) of Exercise 28 for a counterexample. DE = O whenever every row of D is orthogonal to every column of E, but neither matrix is forced to be zero for DE = O to be true.

Chapter 1 Review Exercises 1 2

√ 3 2 3 2 1 394 9 9 394 + − + = + + = 4 5 4 16 25 16 16·25 = 20 ≈ 0.99247. x 5 Thus, u = x ≈ [0.2481, −0.5955, 0.7444]. This is slightly longer than x = √394 , − √12 , √15 394 394 because we have divided x by a scalar with absolute value less than 1, which amounts to multiplying x by a scalar having absolute value greater than 1.

(2) x =

1 f . For the force (4) We will use Newton’s Second Law, which states that or equivalently, a = m f = ma,

[6,17,−6] [6,17,−6] [6,17,−6] 1 √ = 19 = 19 √ 2 2 = [6, 17, −6]. Similarly, for the f1 , a1 = 7 133 [6,17,−6] 361 6 +17 +(−6)2

[−8,−4,8] √ force f2 , a2 = 17 168 [−8,−4,8] = 24 [−8,−4,8] = 24 √ [−8,−4,8] = [−16, −8, 16]. Hence, 2 2 2 144 (−8) +(−4) +8

a = a1 + a2 = [6, 17, −6] + [−16, −8, 16] = [−10, 9, 10], in m/ sec2 .

−69 x·y −69 ≈ cos−1 (−0.7237) ≈ 136◦ . (6) The angle θ = cos−1 xy ≈ cos−1 (10.05)(9.49) ≈ cos−1 95.34

21


[15,−8] [15,−8]

[−7,24] · 75 [−7,24] = 34 √ [15,−8] 2

Chapter 1 Review

· 75 √ [−7,24] 2 2

(8) Work = f · d = 34 = 15 +(−8) (−7) +242

· 75 [−7,24] = (2 [15, −8])·(3 [−7, 24]) = [30, −16]·[−21, 72] = (30)(−21)+(−16)(72) 34 [15,−8] 17 25 = −1782 joules. (10) First, x = 0, or else projx y is not defined. Also, y = 0, since that would imply proj x y = y. Now,

x·y x·cx assume xy. Then, there is a scalar c = 0 such that y = cx. Hence, projx y = x x = x 2 x2

2 x = cx = y, contradicting the assumption that y = projx y. = cx x2

5 3

(11) (a) 3A − 4CT = 3

15 −6 9 −3

−1 4

⎡

−1 4

3 5

−4

−2 4

−4 3

=

12 −8 20 16 ⎡ 2 5 −2 −1 ⎣ −4 AB = 3 −1 4 3 (5)(2) + (−2)(−4) + (−1)(3) (5)(−3) + (−2)(5) + (−1)(−4) (5)(−1) + (−2)(−2) + (−1)(3) = (3)(2) + (−1)(−4) + (4)(3) (3)(−3) + (−1)(5) + (4)(−4) (3)(−1) + (−1)(−2) + (4)(3) =

−3 12

−2 −1

⎤T 3 5 5 −2 − 4 ⎣ −2 4 ⎦ = 3 3 −1 −4 3 −16 3 2 13 = ; 12 −11 −19 0 ⎤ −3 −1 5 −2 ⎦ −4 3

15 −21 22 −30

−

−4 11

;

BA is not defined because B which is 2; ⎡ 3 5 −2 −1 ⎣ −2 AC = 3 −1 4 −4

has 3 columns, which does not match the number of rows of A, ⎤ 5 4 ⎦ 3

(5)(3) + (−2)(−2) + (−1)(−4) (5)(5) + (−2)(4) + (−1)(3) (3)(3) + (−1)(−2) + (4)(−4) (3)(5) + (−1)(4) + (4)(3) ⎡ ⎤ 3 5 5 −2 −1 ⎣ ⎦ CA = −2 4 3 −1 4 −4 3

=

=

23 −5

14 23

⎡

⎤ ⎡ (3)(5) + (5)(3) (3)(−2) + (5)(−1) (3)(−1) + (5)(4) 30 2 = ⎣ (−2)(5) + (4)(3) (−2)(−2) + (4)(−1) (−2)(−1) + (4)(4) ⎦ = ⎣ (−4)(5) + (3)(3) (−4)(−2) + (3)(−1) (−4)(−1) + (3)(4) −11 A3 is not defined. (Powers of matrices are defined only for square matrices.); ⎛⎡

2 B3 = (B2 )(B). Now B2 = ⎝⎣ −4 3

−3 5 −4

⎤⎡ ⎤⎞ −1 2 −3 −1 −2 ⎦ ⎣ −4 5 −2 ⎦⎠ . 3 3 −4 3

22

;

⎤ −11 17 0 18 ⎦; 5 16


Chapter 1 Review

The (1, 1) entry of B2 = (2)(2) + (−3)(−4) + (−1)(3) = 13; the (1, 2) entry of B2 = (2)(−3) + (−3)(5) + (−1)(−4) = −17; the (1, 3) entry of B2 = (2)(−1) + (−3)(−2) + (−1)(3) = 1; the (2, 1) entry of B2 = (−4)(2) + (5)(−4) + (−2)(3) = −34; the (2, 2) entry of B2 = (−4)(−3) + (5)(5) + (−2)(−4) = 45; the (2, 3) entry of B2 = (−4)(−1) + (5)(−2) + (−2)(3) = −12; the (3, 1) entry of B2 = (3)(2) + (−4)(−4) + (3)(3) = 31; the (3, 2) entry of B2 = (3)(−3) + (−4)(5) + (3)(−4) = −41; the (3, 3) entry of B2 = (3)(−1) + (−4)(−2) + (3)(3) = 14. ⎡

13 Thus, B2 = ⎣ −34 31 ⎡

−17 45 −41

13 B3 = B2 (B) = ⎣ −34 31 the the the the the the the the the

(1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (3, 3)

entry entry entry entry entry entry entry entry entry

of of of of of of of of of

B3 B3 B3 B3 B3 B3 B3 B3 B3

⎤ 1 −12 ⎦ . Hence, 14

−17 45 −41

⎤⎡ ⎤ 1 2 −3 −1 −12 ⎦ ⎣ −4 5 −2 ⎦ . Thus, 14 3 −4 3

= (13)(2) + (−17)(−4) + (1)(3) = 97; = (13)(−3) + (−17)(5) + (1)(−4) = −128; = (13)(−1) + (−17)(−2) + (1)(3) = 24; = (−34)(2) + (45)(−4) + (−12)(3) = −284; = (−34)(−3) + (45)(5) + (−12)(−4) = 375; = (−34)(−1) + (45)(−2) + (−12)(3) = −92; = (31)(2) + (−41)(−4) + (14)(3) = 268; = (31)(−3) + (−41)(5) + (14)(−4) = −354; = (31)(−1) + (−41)(−2) + (14)(3) = 93.

⎡

⎤ 97 −128 24 375 −92 ⎦ . Hence, B3 = ⎣ −284 268 −354 93 (b) Third row of BC =(3rd row of B)C =

3

−4

3

⎡

3 ⎣ −2 −4

⎤ 5 4 ⎦ 3

= [(3)(3) + (−4)(−2) + (3)(−4) (3)(5) + (−4)(4) + (3)(3)] = [5 8]. T (13) (a) Now, (3(A − B)T )T = 3 (A − B)T (by part (3) of Theorem 1.12) = 3(A − B) (by part (1) of Theorem 1.12). Also, −(3(A − B)T ) = 3(−1)(AT − BT ) (parts (2) and (3) of Theorem 1.12) = 3(−1)((−A) − (−B)) (definition of skew-symmetric) = 3(A − B). Hence, (3(A − B)T )T = −(3(A − B)T ), and so 3(A − B)T is skew-symmetric. ⎡ ⎤ 5200 4300 20 15 (14) Let A = , the matrix of costs vs. commodities (steel and iron), and B = ⎣ 6300 5100 ⎦, 3 2 4600 4200 the matrix of companies vs. commodities. In order for the units to line up properly for matrix multiplication, the “commodities” label, the shared unit, must represent the columns of the first matrix in the product as well as the rows of the second matrix. One way to do this is to compute the product BAT . Now, 23


⎡

5200 ⎢ BAT = ⎣ 6300 4600

⎤ 4300 ⎥ 20 5100 ⎦ 15 4200

3 2

⎡

(5200)(20) + (4300)(15) ⎢ = ⎣ (6300)(20) + (5100)(15) (4600)(20) + (4200)(15)

Chapter 1 Review

⎤ (5200)(3) + (4300)(2) ⎥ (6300)(3) + (5100)(2) ⎦ (4600)(3) + (4200)(2)

Price Shipping Cost ⎤ $168500 $24200 Company I ⎢ ⎥ $29100 ⎦ . = Company II ⎣ $202500 $155000 $22200 Company III ⎡

T T (15) Take the transpose of both sides of AT BT = BT AT to get AT BT = BT AT , which by Theorem 1.16 and part (1) of Theorem 1.12 implies that BA = AB. Then, (AB)2 = (AB)(AB) = A(BA)B = A(AB)B = A2 B2 . (17) If A = O22 , then some row of A, say the ith row, is nonzero. Let x = (ith row of A). Then, by Result 1 5 in Section 1.3, either x · [1, 0] = 0 or x · [0, 1] = 0. But x · [1, 0] is the (i, 1) entry of A , and 0 0 1 x · [0, 1] is the (i, 1) entry of A . Thus, either at least one of the entries of A is nonzero or 1 0 0 at least one of the entries of A is nonzero. 1 (19) (a) Let A and B be n × n matrices having the properties given in the exercise. Let C = AB. Then we know that aij = 0 for all i < j, bij = 0 for all i > j, cij = 0 for all i = j, and that aii = 0 and bii = 0 for all i. We need to prove that aij = 0 for all i > j.!We use induction on j. ! n n Base Step ! (j = 1): Let i > j = 1. Then 0 = ci1 = k=1 aik bk1 = ai1 b11 + k=2 aik bk1 n = ai1 b11 + k=2 aik · 0 = ai1 b11 . Since b11 = 0, this implies that ai1 = 0, completing the Base Step. Inductive Step: Assume for j = 1, 2, . . . , m − 1 that aij = 0 for all i > j. That is, assume we have already proved that, for some m ≥ 2, the first m − 1 columns of A have all zeroes below the main diagonal. To complete the inductive step, we need to prove that the mth column of A has all zeroes below the main! diagonal. That!is, we must prove that a! im = 0 for all n m−1 n i > m. Let i > m. Then, 0 = cim = k=1 aik bkm = k=1 aik bkm + aim bmm + k=m+1 aik bkm !m−1 !n = k=1 0 · bkm + aim bmm + k=m+1 aik · 0 = aim bmm . But, since bmm = 0, we must have aim = 0, completing the proof. (20) (a) False. If A is a nonzero matrix, there must exist some i and j such that aij = 0. For this i and j, the (i, j) entry of cA, which equals caij , must be nonzero. (b) True. A nonzero vector x is parallel to the unit vector u =

1 x x.

(c) False. For example, [1, 4, 3] − [2, 5, 4] = [−1, −1, −1] is a linear combination of [1, 4, 3] and [2, 5, 4] having negative entries. (d) False. By definition, an angle between vectors has a measure from 0 to π. But measure of the angle between [1, 0] and [0, −1] is π2 .

3π 2

> π. The

(e) False. projx y could have the opposite direction from x. For example, if x = [1, 0] and y = [−1, 1], then projx y = [−1, 0].

24


Chapter 1 Review

(f) True. Apply the Triangle Inequality twice: x + y + z = x + (y + z) + z ≤ !≤ x + y !k k x+y+z. Note that a proof by induction can be used to prove that i=1 xi ≤ i=1 xi (see Exercise 14 in Section 1.3). (g) False. Recall that a set {v1 , . . . , vn } of vectors is mutually orthogonal if and only if for every pair of distinct vectors vi , vj in the set, vi · vj = 0. When we negate statements involving a universal quantifier, such as this, the universal quantifier must change to an existential quantifier. The correct negation is: There is a pair of distinct vectors vi , vj in {v1 , . . . , vn } such that vi · vj = 0. (h) False. Disproving a statement involving a universal quantifier typically involves finding a counterexample. To disprove a statement involving an existential quantifier, we must prove its negation, which is a universally quantified statement. Thus, the negation must be proven in all cases. For example, to disprove the statement “There is a 2 × 2 matrix A such that AO22 = O22 ,” we must show that all 2 × 2 matrices A have the property AO22 = O22 . It is not enough to provide just one specific matrix A for which AO22 = O22 . 1 1 As a further illustration, consider the following statement: “Let A = . Then there is 1 1 a nonzero 2×2 matrix Bsuch that AB = O22 .” This statement is actually true, since the nonzero 1 1 matrix B = does satisfy AB = O22 . Now it is possible to find other matrices, such −1 −1 as B = I2 , for which AB = O22 . However, the existence of such possibilities for B does not prove the original existential statement false. The moral here is that finding a “counterexample” to an existential statement is not enough to ensure that the statement is false. 1 1 2 0 3 1 (i) False. For a counterexample, note that + = (not symmetric). 2 2 2 3 0 1 (j) True. Let A be a skew-symmetric matrix. Then aij = −aji , and so it must be true that aii = −aii . Hence, all of the main diagonal entries of A are zero. Since the trace of a matrix is the sum of its main diagonal entries, trace(A) = 0. (k) True. If A ∈ Un ∩ Ln , then A is both upper and lower triangular. Because A is upper triangular, aij = 0 for all i˙ > j. Because A is lower triangular, aij = 0 for all i < j. Hence, all of the entries of A off the main diagonal equal zero. Therefore, A ∈ Dn . This proves that Un ∩ Ln ⊆ Dn . It is also clear that every diagonal matrix is both upper and lower triangular, since all of its entries both below and above the main diagonal equal zero. Hence, Dn ⊆ Un ∩ Ln . Thus, because we have subset inclusion in both directions, Un ∩ Ln = Dn . (l) True. This is easily proven using parts (2) and (3) of Theorem 1.12 and an induction argument. (m) True. First, because A is m×n and B is n×1, AB is an m×1 matrix. Also, if aj = [a1j , . . . , anj ]T represents the jth column of A, then AB = b11 a1 + · · · + bn1 an , which shows that AB is a linear combination of the columns of A. (To verify this, note that, by definition, the (i, 1) entry of AB equals ai1 b11 + · · · + ain bn1 , while the (i, 1) entry of b11 a1 + · · · + bn1 an = b11 ai1 + · · · + bn1 ain , and so AB = b11 a1 + · · · + bn1 an .) (n) False. In fact, using an argument similar to the one used in part(m), AD is the matrix whose ith 1 2 3 0 3 10 column is the ith column of A multiplied by dii . For example, = . 3 4 0 5 9 20 (o) False. If A is m × n and B is n × m, then both AB and BA are defined, with AB being m × m and BA being n × n. If m = n, then, clearly, AB = BA. For a specific counterexample, choose A = O23 and B = O32 .

25


(p) False. (A + B)2 = A2 + 2AB + B2 ifand 1 in Section 1.5). In particular, if A = 1 5 4 but A2 + 2AB + B2 = . 4 5 (q) False. For a counterexample, note that

Chapter 1 Review

only 21 if A and B commute (see part (b) of Exercise 1 1 2 7 6 2 and B = , then (A + B) = , 1 0 −1 2 3

0 2 −2 0

0 3

−3 0

=

(r) True. This follows immediately from part (2) of Theorem 1.15.

26

6 0

0 6

.


Section 2.1

Chapter 2 Section 2.1 (1) In each part, we first set up the augmented matrix corresponding to the given system of linear equations. Then we perform the row operations designated by the Gaussian elimination method. Finally, we use the final augmented matrix obtained to give the solution set for the original system. " ⎤ ⎡ −5 −2 2 "" 14 1 −1 "" −8 ⎦. The first pivot is the (1,1) entry. We make (a) First augmented matrix: ⎣ 3 2 2 −1 " −3 this a “1” by performing the following type (I) row ⎡ 1 25 − 25 1 ⎣ Type (I) operation: 1 ← − 5 1 : 3 1 −1 2 2 −1

operation: " ⎤ " − 14 5 " " −8 ⎦ " " −3

We now target the (2,1) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 2 ← (−3) × 1 + 2 : Side Calculation (−3)× (row 1) −3 − 65 (row 2) (sum)

6 5

42 5

3

1

−1

−8

0

− 15

1 5

2 5

Resultant Matrix " ⎤ 2 1 − 25 "" − 14 5 5 ⎢ 1 " 2 ⎥ ⎣ 0 − 15 5 "" 5 ⎦ " 2 2 −1 −3 ⎡

Next, we target the (3,1) entry. Type (II) operation: 3 ← (−2) × 1 + 3 : Side Calculation (−2)× (row 1) −2 − 45 (row 3) (sum)

4 5

28 5

2

2

−1

−3

0

6 5

− 15

13 5

Resultant Matrix " ⎤ 2 1 − 25 "" − 14 5 5 ⎢ 1 " 2 ⎥ ⎣ 0 − 15 5 "" 5 ⎦ 6 1 " 13 0 −5 5 5 ⎡

Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation. " ⎤ ⎡ 1 25 − 25 "" − 14 5 " ⎥ ⎢ 1 −1 " −2 ⎦ Type (I) operation: 2 ← −5 2 : ⎣ 0 " 13 0 6 −1 " 5

Now we target the (3,2) entry. Type (II) operation: 3 ← (− 65 ) × 2 + 3 : Side Calculation (− 65 )× (row 2) 0 − 65 (row 3)

0

6 5

6 5 − 15

(sum)

0

0

1

12 5 13 5

5

5

5

Resultant Matrix " ⎤ 1 25 − 25 "" − 14 5 " ⎥ ⎢ 1 −1 " −2 ⎦ ⎣ 0 " 0 0 1 " 5 ⎡

The pivot moves to the (3,3) entry. However, the (3,3) entry already equals 1. Since there are no rows below the third row, we are finished performing row operations. The final augmented

27


matrix corresponds to the following linear system: ⎧ 2 ⎨ x1 + 5 x2 − x2 − ⎩

2 5 x3

x3 x3

Section 2.1

= − 14 5 = =

−2 5

The last equation clearly gives x3 = 5. Substituting this value into the second equation yields x2 −5 = −2, or x2 = 3. Substituting both values into the first equation produces x1 + 25 (3)− 25 (5) = − 14 5 , which leads to x1 = −2. Hence, the solution set for the system is {(−2, 3, 5)}. " ⎤ ⎡ 3 −2 4 "" −54 20 ⎦. The first pivot is the (1,1) entry. We make 1 −2 "" (c) First augmented matrix: ⎣ −1 " −83 5 −4 8 this a “1” by performing the following type (I) row operation: ⎤ ⎡ 4 "" 1 − 23 3 " −18 Type (I) operation: 1 ← 13 1 : ⎣ −1 1 −2 "" 20 ⎦ 5 −4 8 " −83 We now target the (2,1) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 2 ← (1) × 1 + 2 : Side Calculation (1)× (row 1) 1 − 23 (row 2)

4 3

−18

−1

1

−2

20

0

1 3

− 23

2

(sum)

Resultant Matrix " ⎤ 4 " 1 − 23 3 " −18 ⎥ " ⎢ 1 2 ⎦ − 23 " ⎣ 0 3 " 5 −4 8 " −83 ⎡

Next, we target the (3,1) entry. Type (II) operation: 3 ← (−5) × 1 + 3 : Side Calculation 10 − 20 (−5)× (row 1) −5 3 3 (row 3) (sum)

90

5

−4

8

−83

0

− 23

4 3

7

Resultant Matrix " ⎤ 4 " 1 − 23 3 " −18 ⎥ " ⎢ 1 2 ⎦ − 23 " ⎣ 0 3 " 4 " 7 0 − 23 3 ⎡

Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation. " ⎤ ⎡ 4 " 1 − 23 3 " −18 ⎥ " ⎢ 1 −2 " 6 ⎦ Type (I) operation: 2 ← 3 2 : ⎣ 0 " 4 " 7 0 − 23 3 Now we target the (3,2) entry. Type (II) operation: 3 ← ( 23 ) × 2 + 3 : Side Calculation 2 ( 23 )× (row 2) 0 − 43 3

4

(row 3)

0

− 23

4 3

7

(sum)

0

0

0

11

Resultant Matrix ⎤ 4 "" 1 − 23 3 " −18 ⎣ 0 1 −2 "" 6 ⎦ 0 0 0 " 11 ⎡

The pivot moves to the (3,3) entry. However, the (3,3) entry equals 0. Because there are no rows below the third row, we cannot switch with a row below the third row to produce a nonzero pivot. Also, since there are no more columns before the augmentation bar, we are finished performing 28


Section 2.1

row operations. The final augmented matrix corresponds to the following linear system: ⎧ 2 4 ⎨ x1 − 3 x2 + 3 x3 = −18 x2 − 2x3 = 6 ⎩ 0 = 11 The last equation states that 0 = 11. Since this equation cannot be satisfied, the original linear system has no solutions. The solution set is empty. " ⎤ ⎡ 6 −12 −5 16 −2 "" −53 29 ⎦. The first pivot is the (1,1) 6 3 −9 1 "" (e) First augmented matrix: ⎣ −3 33 −4 8 3 −10 1 " entry. We make this a “1” by performing the following type (I) row operation: " ⎤ ⎡ 8 1 −2 − 56 − 13 " − 53 3 6 " Type (I) operation: 1 ← 16 1 : ⎣ −3 6 3 −9 1 "" 29 ⎦ " −4 8 3 −10 1 33 We now target the (2,1) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 2 ← (3) × 1 + 2 : 8

−1

− 53 2

6

3

−9

1

29

0

1 2

−1

0

5 2

3

−6

(row 2)

−3

Side Calculation:

(sum) ⎡ Resultant Matrix:

− 52

(3) × (row 1)

⎢ ⎣

0

" ⎤ − 13 "" − 53 6 " 5 ⎥ −1 0 " 2 ⎦ " " −10 1 33 8 3

− 56

1

−2

0

0

1 2

−4

8

3

Next, we target the (3,1) entry. Type (II) operation: 3 ← (4) × 1 + 3 : 32 3

8

3

0

− 13

4

−8

(row 3)

−4

Side Calculation:

(sum) ⎡ Resultant Matrix:

− 10 3

(4) × (row 1)

1

⎢ ⎣ 0 0

−2 0 0

0 8 3

− 56

1 2 − 13

−1 2 3

− 43

− 106 3

−10

1

33

2 3

− 13

− 73

" ⎤ − 13 "" − 53 6 " 5 ⎥ 0 " 2 ⎦ " − 13 " − 73

We attempt to move the pivot to the (2,2) entry; however, there is a “0” there. Also, the entry below this is also 0, so we cannot switch rows to make the (2,2) entry nonzero. Thus, we move the pivot horizontally to the next column – to the (2,3) entry. We make this pivot a “1” using the following type (I) row operation: " ⎤ ⎡ 8 − 13 "" − 53 1 −2 − 56 3 6 " ⎥ ⎢ 1 −2 0 0 " 5 ⎦ Type (I) operation: 2 ← 2 2 : ⎣ 0 " 2 0 0 − 13 − 13 " − 73 3 Now we target the (3,3) entry. 29


Type (II) operation: 3 ← ( 13 ) × 2 + 3 : ( 13 ) × (row 2)

0

(row 3)

0

(sum)

0

Side Calculation: ⎡ Resultant Matrix:

1 −2 ⎢ 0 ⎣ 0 0 0

0 0 0

− 56

8 3

1

−2

0

0

1 3 − 13

− 23 2 3

0 0 " ⎤ 53 " −6 " " ⎥ 0 " 5 ⎦ " − 13 " − 23

0 − 13 − 13

Section 2.1

5 3 − 73 − 23

− 13

We attempt to move the pivot diagonally to the (3,4) entry; however, that entry is also 0. Since there are no rows below with which to swap, we must move the pivot horizontally to the (3,5) entry. We make this pivot a “1” by performing the following type (I) row operation: " ⎤ ⎡ 8 − 13 "" − 53 1 −2 − 56 3 6 " ⎥ ⎢ 0 1 −2 0 " 5 ⎦ Type (I) operation: 3 ← −3 3 : ⎣ 0 " 1 " 0 0 0 0 2 At this point, we are finished performing row operations. The linear system corresponding to the final augmented matrix is ⎧ ⎨ x1 ⎩

− 2x2

−

5 6 x3 x3

+ −

8 3 x4 2x4

−

1 3 x5

x5

= − 53 6 = 5 = 2

The last equation clearly gives us x5 = 2. The variable x4 corresponds to a nonpivot column and hence is an independent variable. Let x4 = d. Then the second equation yields x3 − 2d = 5, or x3 = 2d + 5. The variable x2 also corresponds to a nonpivot column, making it an independent variable as well. We let x2 = b. Finally, plugging all these values into the first equation produces x1 − 2b − 56 (2d + 5) + 83 d − 13 (2) = − 53 6 . Solving this equation for x1 gives x1 = 2b − d − 4. Hence, the general solution set for the original system is {(2b − d − 4, b, 2d + 5, d, 2) | b, d ∈ R}. To find three particular solutions, we plug in three different sets of values for b and d. Using b = 0, d = 0 yields the solution (−4, 0, 5, 0, 2). Letting b = 1, d = 0 gives us (−2, 1, 5, 0, 2). Finally, setting b = 0, d = 1 produces (−5, 0, 7, 1, 2). " ⎤ ⎡ 5 4 −2 −7 "" ⎢ −6 5 10 "" −11 ⎥ ⎥. The first pivot is the (1,1) entry. We make (g) First augmented matrix: ⎢ ⎣ −2 3 4 "" −3 ⎦ −3 2 5 " −5 this a “1” by performing the following type (I) row operation: " ⎡ 5 ⎤ 1 − 12 − 74 " 4 " " −11 ⎥ ⎢ −6 5 10 1 ⎢ " ⎥ Type (I) operation: 1 ← 4 1 : ⎣ −2 3 4 "" −3 ⎦ −3 2 5 " −5 We now target the (2,1) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 2 ← (6) × 1 + 2 :

30


Side Calculation (6)× (row 1)

6

−3

− 21 2

15 2

(row 2)

−6

5

10

−11

(sum)

0

2

− 12

− 72

Matrix " ⎡ Resultant 5 1 − 12 − 74 "" 4 " ⎢ 2 − 12 " − 72 ⎢ 0 " ⎢ ⎣ −2 3 4 "" −3 −3 2 5 " −5

Section 2.1

⎤ ⎥ ⎥ ⎥ ⎦

Next, we target the (3,1) entry. Type (II) operation: 3 ← (2) × 1 + 3 : Side Calculation − 72

5 2

(2)× (row 1)

2

−1

(row 3)

−2

3

4

−3

2

1 2

− 12

(sum)

0

Matrix " ⎡ Resultant 5 1 − 12 − 74 "" 4 " ⎢ ⎢ 0 2 − 12 " − 72 ⎢ " ⎢ 0 1 " 1 2 ⎣ 2 " −2 " −3 2 5 " −5

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

To finish the first column, we now target the (4,1) entry. Type (II) operation: 4 ← (3) × 1 + 4 : Side Calculation − 32

− 21 4

15 4

(3)× (row 1)

3

(row 4)

−3

2

5

−5

0

1 2

− 14

− 54

(sum)

Matrix " ⎡ Resultant 5 1 − 12 − 74 "" 4 " ⎢ 1 " ⎢ 0 2 −2 − 72 " ⎢ " ⎢ 0 1 1 2 ⎣ 2 " −2 " 1 0 −1 " −5 2

4

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

4

Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation. " ⎤ ⎡ 5 1 − 12 − 74 "" 4 " ⎥ ⎢ ⎢ 0 1 − 14 " − 74 ⎥ 1 " ⎥ ⎢ Type (I) operation: 2 ← 2 2 : ⎢ 1 " 1 ⎥ 2 ⎣ 0 2 " −2 ⎦ " 1 0 − 14 " − 54 2 Now we target the (3,2) entry. Type (II) operation: 3 ← (−2) × 2 + 3 : Side Calculation (−2)× (row 2)

0

−2

(row 3)

0

2

(sum)

0

0

1 2 1 2

7 2 − 12

1

3

Matrix " ⎡ Resultant 5 1 − 12 − 74 "" 4 " ⎢ 1 ⎢ 0 1 − 4 " − 74 " ⎢ ⎢ 0 0 1 "" 3 ⎣ " 1 1 " 5 0 −4 −4 2

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

To finish the second column, we target the (4,2) entry. Type (II) operation: 4 ← (− 12 ) × 2 + 4 : Side Calculation (− 12 )×

(row 2)

0

(row 4)

0

− 12 1 2

(sum)

0

0

1 8 − 14 − 18

7 8 − 54 − 38

Matrix " ⎡ Resultant 5 1 − 12 − 74 "" 4 " ⎢ ⎢ 0 1 − 14 " − 74 " ⎢ ⎢ 0 0 1 "" 3 ⎣ " 1 " 3 0 0 −8 −8

31

⎤ ⎥ ⎥ ⎥ ⎥ ⎦


Section 2.1

The pivot now moves to the (3,3) entry. Luckily, the pivot already equals 1. Therefore, we next target the (4,3) entry. Type (II) operation: 4 ← ( 18 ) × 3 + 4 : Matrix " ⎡ Resultant 5 1 − 12 − 74 "" 4 " ⎢ 1 ⎢ 0 1 − 4 " − 74 ⎢ " ⎢ 0 1 "" 0 3 ⎣ " 0 0 0 " 0

Side Calculation ( 18 )×

(row 3)

0

0

(row 4)

0

0

(sum)

0

0

1 8 − 18

3 8 − 38

0

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

This finishes the third column. Since there are no more columns before the augmentation bar, we are finished performing row operations. The linear system corresponding to the final matrix is ⎧ x1 ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

−

1 2 x2

−

x2

−

7 4 x3 1 4 x3

=

x3 0

= =

=

5 4 − 74

3 0

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of x1 , x2 , and x3 . Thus, we can ignore it. The third equation clearly gives x3 = 3. Plugging this value into the second equation yields x2 − 14 (3) = − 74 , or x2 = −1. Substituting the values we have for x2 and x3 into the first equation produces x1 − 12 (−1) − 74 (3) = 54 , or x1 = 6. Hence, the original linear system has a unique solution. The full solution set is {(6, −1, 3)}. (2) (a) The system of linear equations corresponding to the given augmented matrix is ⎧ x1 − 5x2 + 2x3 + 3x4 − 2x5 = −4 ⎪ ⎪ ⎨ x2 − x3 − 3x4 − 7x5 = −2 5 x4 + 2x5 = ⎪ ⎪ ⎩ 0 = 0 The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of x1 , x2 , x3 , x4 , and x5 . So, we can ignore it. The column corresponding to the variable x5 is not a pivot column, and so x5 is an independent variable. Let x5 = e. Substituting e for x5 into the third equation yields x4 + 2e = 5, or x4 = −2e + 5. The column corresponding to the variable x3 is not a pivot column, and so x3 is another independent variable. Let x3 = c. Substituting the expressions we have for x3 , x4 , and x5 into the second equation produces x2 − c − 3(−2e + 5) − 7e = −2. Solving for x2 and simplifying gives x2 = c + e + 13. Finally, we plug all these expressions into the first equation to get x1 − 5(c + e + 13) + 2c + 3(−2e + 5) − 2e = −4, which simplifies to x1 = 3c + 13e + 46. Hence, the complete solution set is {(3c + 13e + 46, c + e + 13, c, −2e + 5, e) | c, e ∈ R}. (c) The system of linear equations corresponding to the given augmented matrix is ⎧ x1 + 4x2 − 8x3 − x4 + 2x5 − 3x6 = −4 ⎪ ⎪ ⎨ x2 − 7x3 + 2x4 − 9x5 − x6 = −3 . 2 x5 − 4x6 = ⎪ ⎪ ⎩ 0 = 0 32


Section 2.1

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of x1 , x2 , x3 , x4 , x5 , and x6 . So, we can ignore it. The column corresponding to the variable x6 is not a pivot column, and so x6 is an independent variable. Let x6 = f . Substituting f for x6 into the third equation yields x5 − 4f = 2, or x5 = 4f + 2. The columns corresponding to the variables x3 and x4 are not pivot columns, and so x3 and x4 are also independent variables. Let x3 = c and x4 = d. Substituting the expressions we have for x3 , x4 , x5 , and x6 into the second equation produces x2 − 7c + 2d − 9(4f + 2) − f = −3. Solving for x2 and simplifying gives x2 = 7c − 2d + 37f + 15. Finally, we plug all these expressions into the first equation to get x1 + 4(7c − 2d + 37f + 15) − 8c − d + 2(4f + 2) − 3f = −4, which simplifies to x1 = −20c + 9d − 153f − 68. Hence, the complete solution set is {(−20c + 9d − 153f − 68, 7c − 2d + 37f + 15, c, d, 4f + 2, f ) | c, d, f ∈ R}. (3) Let x represent the number of nickels, y represent the number of dimes, and z represent the number of quarters. The fact that the total value of all of the coins is $16.50 gives us the following equation: 0.05x + 0.10y + 0.25z = 16.50, or 5x + 10y + 25z = 1650. Next, using the fact that there are twice as many dimes as quarters gives us y = 2z, or y − 2z = 0. Finally, given that the total number of nickels and quarters is 20 more than the number of dimes produces x + z = y + 20, or x − y + z = 20. Putting these three equations into a single system yields ⎧ y + z = 20 ⎨ x − y − 2z = 0 , ⎩ 5x + 10y + 25z = 1650 where we have arranged the equations in an order augmented matrix for this system is ⎡ 1 −1 1 ⎣ 0 1 −2 5 10 25

that will make the row operations easier. The " ⎤ " 20 " " 0 ⎦. " " 1650

The (1,1) entry is the first pivot. Since this already equals 1, we continue by targeting nonzero entries in the first column. We get a zero in the (3,1) entry by performing a type (II) operation: Type (II) operation: 3 ← (−5) × 1 + 3 : Side Calculation (−5)× (row 1) −5 5 −5 (row 3) 5 10 25 (sum) 0 15 20

−100 1650 1550

" ⎤ ⎡ Resultant Matrix 1 −1 20 1 "" ⎣ 0 0 ⎦ 1 −2 "" " 1550 0 15 20

Next, the pivot moves to the (2,2) entry. Since this already equals 1, we continue with the second column by targeting the (3,2) entry. Type (II) operation: 3 ← (−15) × 2 + 3 : Side Calculation (−15)× (row 2) 0 −15 30 (row 3) 0 15 20 (sum) 0 0 50

0 1550 1550

" ⎤ ⎡ Resultant Matrix 20 1 −1 1 "" ⎣ 0 1 −2 "" 0 ⎦ 0 0 50 " 1550

The pivot now moves to the (3,3) entry. We must a “1.” ⎡ 1 −1 1 1 1 −2 Type (I) operation: 3 ← 50 3 : ⎣ 0 1 0 0 33

perform a type (I) operation to turn the pivot into " ⎤ " 20 " " 0 ⎦ " " 31


Section 2.1

This ends the row operations. The final matrix corresponds to the linear system ⎧ ⎨ x − y + z = 20 y − 2z = 0 . ⎩ z = 31 The third equation in this system yields z = 31. Plugging this value into the second equation produces y − 2(31) = 0, or y = 62. Finally, we substitute these values into the first equation to get x − 62 + 31 = 20, or x = 51. Hence, there are 51 nickels, 62 dimes, and 31 quarters. (4) First, plug each of the three given linear system: ⎧ ⎨ 18 = 9 = ⎩ 13 =

points into the equation y = ax2 + bx + c. This yields the following

9a + 3b + c ←− using (3, 18) 4a + 2b + c ←− using (2, 9) . 4a − 2b + c ←− using (−2, 13) " ⎤ ⎡ 9 3 1 "" 18 2 1 "" 9 ⎦. The first pivot is the (1,1) entry. We The augmented matrix for this system is ⎣ 4 4 −2 1 " 13 make this a “1” by performing the following type (I) row operation: ⎤ ⎡ 1 "" 1 1 2 3 9 " Type (I) operation: 1 ← 19 1 : ⎣ 4 2 1 "" 9 ⎦ 4 −2 1 " 13

We now target the (2,1) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 2 ← (−4) × 1 + 2 : Side Calculation (−4)× (row 1) −4 − 43 − 49 (row 2) (sum)

−8

4

2

1

9

0

2 3

5 9

1

Resultant Matrix " ⎤ ⎡ 1 " 1 1 2 3 9 " ⎥ ⎢ 5 " 2 1 ⎦ ⎣ 0 3 9 "" 4 −2 1 " 13

Next, we target the (3,1) entry. Type (II) operation: 3 ← (−4) × 1 + 3 : Side Calculation (−4)× (row 1) −4 − 43 − 49 (row 3) (sum)

−8

4

−2

1

13

0

− 10 3

5 9

5

Resultant Matrix " ⎤ ⎡ 1 1 " 1 3 9 " 2 ⎥ ⎢ 5 " 2 ⎣ 0 3 9 "" 1 ⎦ 5 " 5 0 − 10 3 9

Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation. " ⎡ ⎤ 1 1 " 1 3 9 " 2 " ⎥ ⎢ 1 56 " 32 ⎦ Type (I) operation: 2 ← 32 2 : ⎣ 0 " 5 " 5 0 − 10 3 9 Now we target the (3,2) entry. Type (II) operation: 3 ← ( 10 3 ) × 2 + 3 :

34


Side Calculation 10 ( 10 )× (row 2) 0 3 3 (row 3)

0

− 10 3

(sum)

0

0

25 9 5 9 10 3

5 5 10

Section 2.1

Resultant Matrix " ⎡ ⎤ 1 " 2 1 13 9 " ⎢ 3 ⎥ 5 " 1 ⎣ 0 2 ⎦ 6 "" " 10 0 0 10 3

Finally, we move the pivot to the (3,3) entry and perform the following row operation to turn that entry into a “1.” " ⎡ ⎤ 1 " 1 13 9 " 2 " ⎢ ⎥ 3 Type (I) operation: 3 ← 10 3 : ⎣ 0 1 56 " 32 ⎦ " 1 " 3 0 0 This matrix corresponds to the following linear system: ⎧ 1 1 ⎪ ⎨ a + 3b + 9c = b + 56 c = ⎪ ⎩ c =

2 3 2

.

3

The last equation clearly gives c = 3. Plugging c = 3 into the second equation yields b + 56 (3) = 32 , or b = −1. Substituting these values into the first equation produces a + 13 (−1) + 19 (3) = 2, or a = 2. Hence, the desired quadratic equation is y = 2x2 − x + 3. (6) First, plug each of the three given following linear system: ⎧ 2 ⎨ (6 + 82 ) (82 + 42 ) ⎩ 2 (3 + 92 )

points into the equation x2 + y 2 + ax + by = c. This yields the + + +

6a + 8a + 3a +

8b 4b 9b

= c ←− using (6, 8) = c ←− using (8, 4) . = c ←− using (3, 9)

Rearranging terms leads to the following augmented matrix: " ⎤ ⎡ 6 8 −1 "" −100 ⎣ 8 4 −1 " −80 ⎦ . " 3 9 −1 " −90 The first pivot is the (1,1) entry. We make this a “1” by performing the following type (I) row operation: " ⎤ ⎡ 1 43 − 16 " − 50 3 " Type (I) operation: 1 ← 16 1 : ⎣ 8 4 −1 "" −80 ⎦ 3 9 −1 " −90 We now must target the (2,1) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 2 ← (−8) × 1 + 2 : Side Calculation (−8)× (row 1) −8 − 32 3 (row 2) (sum)

4 3

400 3

8

4

−1

−80

0

− 20 3

1 3

160 3

Resultant Matrix " ⎤ 4 1 − 16 "" − 50 3 3 ⎢ 1 " 160 ⎥ ⎣ 0 − 20 3 3 "" 3 ⎦ " 3 9 −1 −90 ⎡

Next, we target the (3,1) entry. 35


Section 2.1

Type (II) operation: 3 ← (−3) × 1 + 3 : Side Calculation (−3)× (row 1) −3 −4

1 2

50

(row 3)

3

9

−1

−90

(sum)

0

5

− 12

−40

Resultant Matrix " ⎤ 4 1 − 16 "" − 50 3 3 ⎢ 160 ⎥ 1 " ⎣ 0 − 20 3 3 "" 3 ⎦ 0 5 − 12 " −40 ⎡

Now we move the pivot to the (2,2) entry. We make ⎡ 1 43 − 16 ⎢ 1 3 1 − 20 Type (I) operation: 2 ← − 20 2 : ⎣ 0 0 5 − 12

this entry a “1” using a type (I) row operation. " ⎤ " − 50 3 " " ⎥ " −8 ⎦ " " −40

Now we target the (3,2) entry. Type (II) operation: 3 ← (−5) × 2 + 3 : Side Calculation (−5)× (row 2) 0 −5 (row 3)

0

5

(sum)

0

0

1 4 − 12 − 14

40 −40 0

Resultant Matrix " ⎤ − 16 "" − 50 1 43 3 ⎥ ⎢ 1 " 1 − 20 " −8 ⎦ ⎣ 0 " 0 0 − 14 " 0 ⎡

Finally, we move the pivot to the (3,3) entry and perform the following row operation to turn that entry into a “1.” " ⎤ ⎡ − 16 "" − 50 1 43 3 ⎥ ⎢ 1 " Type (I) operation: 3 ← (−4) 3 : ⎣ 0 1 − 20 " −8 ⎦ " 1 " 0 0 0 This matrix corresponds to the following linear system: ⎧ 4 1 50 ⎪ 6c = − 3 ⎨ a + 3b − 1 b − 20 c = −8 . ⎪ ⎩ c = 0 1 The last equation clearly gives c = 0. Plugging c = 0 into the second equation yields b − 20 (0) = −8, 4 1 or b = −8. Substituting these values into the first equation produces a + 3 (−8) − 6 (0) = − 50 3 , or a = −6. Hence, the desired equation of the circle is x2 + y 2 − 6x − 8y = 0, or (x − 3)2 + (y − 4)2 = 25. ⎡ ⎤ 26 15 −6 ⎢ 6 4 1 ⎥ ⎥. To find R(AB) we perform the type (II) row (7) (a) First, we compute AB = ⎢ ⎣ 18 6 15 ⎦ 10 4 −14

operation R : 3 ← (−3) × 2 + 3 : Side Calculation (−3)× (row 2) (row 3) (sum)

−18 18 0

−12 6 −6

−3 15 12

Resultant Matrix ⎤ ⎡ 26 15 −6 ⎢ 6 4 1 ⎥ ⎥ R(AB) = ⎢ ⎣ 0 −6 12 ⎦ 10 4 −14

36


⎡

2 ⎢ 0 Next we compute R(A). Using A = ⎢ ⎣ −2 3 Side Calculation (−3)× (row 2) (row 3) (sum)

0 −2 −2

−3 1 −2

−3 5 2

3 1 1 0

Section 2.2

⎤ 4 1 ⎥ ⎥ and R : 3 ← (−3) × 2 + 3 : 5 ⎦ 1

Resultant Matrix ⎡ 2 3 4 ⎢ 0 1 1 ⎢ R(A) = ⎣ −2 −2 2 3 0 1

⎤ ⎥ ⎥ ⎦

Multiplying R(A) by the given matrix B produces the same result computed above for R(AB). (8) (a) To save space, we use the notation C i for the ith row of a matrix C. Also, recall the following hint from the textbook, which we can rewrite as follows in the notation we have just defined: (Hint: Use the fact from Section 1.5 that AB k = A k B.) For the Type (I) operation R : i ←− c i : Now, R(AB) i = c AB i = c A i B (by the hint) = R(A) i B = R(A)B i . But, if k = i, R(AB) k = AB k = A k B (by the hint) = R(A) k B = R(A)B k . For the Type (II) operation R : i ←− c j + i : Now, R(AB) i = c AB j + AB i = c A j B + A i B (by the hint) = (c A j + A i )B = R(A) i B = R(A)B i . But, if k = i, R(AB) k = AB k = A k B (by the hint) = R(A) k B = R(A)B k . For the Type (III) operation R : i ←→ j : Now, R(AB) i = AB j = A j B (by the hint) = R(A) i B = R(A)B i . Similarly, R(AB) j = AB i = A i B (by the hint) = R(A) j B = R(A)B j . And, if k = i and k = j, R(AB) k = AB k = A k B (by the hint) = R(A) k B = R(A)B k . (11) (a) True. The augmented matrix contains all of the information necessary to solve the system completely. (b) False. A linear system has either no solutions, one solution, or infinitely many solutions. Having exactly three solutions is not one of the possibilities. (c) False. A linear system is consistent if it has at least one solution. It could have either just one or infinitely many solutions. (d) False. Type (I) operations are used to convert nonzero pivot entries to 1. When following the standard algorithm for Gaussian elimination, type (II) row operations are never used for this purpose. Type (II) row operations are used only to convert a nonzero target to a zero. (e) True. In the standard algorithm for Gaussian elimination, this is the only instance in which a type (III) row operation is used. (f) True. This is the statement of part (1) of Theorem 2.1 expressed in words rather than symbols.

Section 2.2 (1) The matrix in (e) is in reduced row echelon form, as it satisfies all four conditions. The matrices in (a), (b), (c), (d), and (f) are not in reduced row echelon form. The matrix in (a) fails condition 2 of the definition, since the first nonzero entry in row 2 is in the third column, while the first nonzero entry in row 3 is only in the second column.

37


Section 2.2

The matrix in (b) fails condition 4 of the definition, as the second row contains all zeroes, but later rows do not. The matrix in (c) fails condition 1 of the definition because the first nonzero entry in the second row equals 2, not 1. The matrix in (d) fails conditions 1, 2, and 3 of the definition. Condition 1 fails because the first nonzero entry in row 3 equals 2, not 1. Condition 2 fails because the first nonzero entries in rows 2 and 3 appear in the same column. Condition 3 fails because the (3,3) entry below the first nonzero entry of the second column is nonzero. The matrix in (f) fails condition 3 of the definition since the entries in the fifth column above the 1 in the third row do not equal zero. (2) (a) First, convert the pivot (the (1,1) entry) to “1.” " 11 ⎤ ⎡ " −5 1 4 − 18 5 " 1 ⎣ 1 ← ( 5 ) 1 3 12 −14 "" 3 ⎦ " −4 −16 13 13 Next, target the (2,1) and (3,1) entries. ⎡ 1 4 2 ← (−3) × 1 + 2 ⎢ ⎣ 0 0 3 ← (4) × 1 + 3 0 0

− 18 5

− 16 5

− 75

" 11 " −5 " " 48 " 5 " " 21

⎤ ⎥ ⎦

5

We try to move the pivot to the second column but cannot due to the zeroes in the (2,2) and (3,2) entries. Hence, the pivot moves to the (2,3) entry. We set that equal to 1. " 11 ⎤ ⎡ " −5 1 4 − 18 5 " " ⎥ ⎢ 5 1 " −3 ⎦ 2 ← (− 16 ) 2 ⎣ 0 0 " 0 0 − 75 " 21 5 Finally, we target the (1,3) and (3,3) ⎡ 1 ) × 2 + 1 1 ← ( 18 5 ⎣ 0 3 ← ( 75 ) × 2 + 3 0

entries. 4 0 0

0 1 0

" ⎤ " −13 " " −3 ⎦ " " 0

This is the desired reduced row echelon form matrix. (b) First, convert the pivot (the (1,1) entry) ⎡ 1 − 12 − 12 ⎢ 6 −1 −2 ⎢ 1 ← (− 12 ) 1 ⎣ 1 −1 −1 −5 −5 −5

to “1.” ⎤ − 15 2 −36 ⎥ ⎥ −11 ⎦ −14

Next, target the (2,1), (3,1), and (4,1) entries. ⎡ 1 − 12 − 12 − 15 2 ⎢ 2 ← (−6) × 1 + 2 ⎢ 0 2 1 9 ⎢ 3 ← (−1) × 1 + 3 ⎢ 0 −1 1 7 −2 −2 ⎣ 2 4 ← (5) × 1 + 4 15 15 0 − 2 − 2 − 103 2 The pivot now moves to the (2,2) entry.

38

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ 2 ← ( 12 ) 2

1

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

− 12 1 − 12

− 15 2

− 12

1 2 − 12 − 15 2

− 15 2

9 2 − 72 − 103 2

Section 2.2

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

Next, target the (1,2), (3,2), and (4,2) entries. ⎡ 1 0 − 14 1 1 ← ( 2 ) × 2 + 1 ⎢ 1 ⎢ 0 1 2 ⎢ 3 ← ( 12 ) × 2 + 3 ⎢ 0 1 0 − ⎣ 4 4 ← ( 15 ) × 2 + 4 2 0 0 − 15 4

− 21 4

9 2 − 54 − 71 4

The pivot now moves to the (3,3) entry. ⎡ 1 0 − 14 − 21 4 ⎢ 9 1 ⎢ 0 1 2 2 ⎢ 3 ← (−4) 3 ⎢ 0 0 1 5 ⎣ 15 71 0 0 −4 −4

⎤

Next, target the (1,3), (2,3), and (4,3) ⎡ 1 1 ← ( 14 ) × 3 + 1 ⎢ ⎢ 0 2 ← (− 12 ) × 3 + 2 ⎣ 0 4 ← ( 15 0 4 ) × 3 + 4

⎤ 0 −4 0 2 ⎥ ⎥ 1 5 ⎦ 0 1

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

⎥ ⎥ ⎥ ⎥ ⎦

entries. 0 1 0 0

The pivot moves to the (4,4) entry. Since (1,4), (2,4), and (3,4) entries. ⎡ 1 0 1 ← (4) × 4 + 1 ⎢ 0 1 ⎢ 2 ← (−2) × 4 + 2 ⎣ 0 0 3 ← (−5) × 4 + 3 0 0

this already equals 1, we now need to target only the 0 0 1 0

⎤ 0 0 ⎥ ⎥ 0 ⎦ 1

Thus, I4 is the desired reduced row echelon form matrix. (c) First, convert the pivot (the (1,1) entry) to “1.” ⎡ 17 "" 19 1 −2 5 5 " −4 ⎢ 6 −11 −11 "" 14 ⎢ −3 1 ← (− 15 ) 1 ⎣ −7 14 −26 −25 "" 31 9 −18 34 31 " −37 Next, target the (2,1), (3,1), and (4,1) entries. ⎡ 1 −2 19 5 ⎢ 2 ← (3) × 1 + 2 2 ⎢ 0 0 5 ⎢ 3 ← (7) × 1 + 3 ⎢ 0 3 0 ⎣ 5 4 ← (−9) × 1 + 4 0 0 − 15

17 5 − 45 − 65 2 5

⎤ ⎥ ⎥ ⎦

" ⎤ " −4 " ⎥ " " 2 ⎥ ⎥ " " 3 ⎥ " ⎦ " " −1

We try to move the pivot to the second column but cannot due to the zeroes in the (2,2), (3,2), and (4,2) entries. Hence, the pivot moves to the (2,3) entry. We set that equal to 1.

39


1

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

−2

19 5

0

1

0

3 5 − 15

0

Section 2.2

" ⎤ " −4 " ⎥ " −2 " 5 ⎥ ⎥ " − 65 "" 3 ⎥ ⎦ " 2 " −1 5 17 5

Next, target the (1,3), (3,3), and (4,3) entries. ⎡ 1 −2 0 1 ← (− 19 ) × 2 + 1 5 ⎢ 1 0 0 ⎢ 3 ← (− 35 ) × 2 + 3 ⎣ 0 0 0 4 ← ( 15 ) × 2 + 4 0 0 0

11 −2 0 0

" ⎤ " −23 " " 5 ⎥ ⎥ " " 0 ⎦ " " 0

This is the desired reduced row echelon form matrix. (e) First, convert the pivot (the (1,1) entry) to “1.” " ' ( 5 1 −2 13 0 "" 53 3 1 1 ← (− 3 ) 1 " −1 2 3 −5 10 " 5 Next, target the (2,1) entry. ' 2 ← (1) × 1 + 2

1

−2

0

0

1 3 10 3

5 3 − 10 3

" 0 "" " 10 "

5 3 20 3

(

We try to move the pivot to the (2,2) entry, but it equals zero. There are no rows below it to provide a nonzero pivot to be swapped up, and so the pivot moves to the (2,3) entry. We set that equal to 1. " ' ( 5 1 " 5 1 −2 0 " 3 3 3 3 2 ← ( 10 ) 2 " 1 −1 3 " 2 0 0 Finally, we target the (1,3) entry to obtain the desired reduced row echelon form matrix. " 1 −2 0 2 −1 "" 1 . 1 ← (− 13 ) × 2 + 1 1 −1 3 " 2 0 0 (3) (a) The final matrix obtained in Exercise 1(a) in Section 2.1 is ⎡

1

⎣ 0 0

2 5

1 0

" ⎤ − 25 " − 14 5 " −1 "" −2 ⎦ . 1 " 5

We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2,2) entry as the pivot and target the (1,2) entry. " ⎡ ⎤ 1 0 0 "" −2 ⎣ 0 1 −1 "" −2 ⎦ 1 ← (− 25 ) × 2 + 1 0 0 1 " 5 Next, we pivot at the (3,3) entry zero.) ⎡ 1 ⎣ 0 2 ← (1) × 3 + 2 0

and target the (2,3) entry. (Note that the (1,3) entry is already 0 1 0

0 0 1

" ⎤ " −2 " " 3 ⎦ " " 5

40


This matrix corresponds to the linear system ⎧ ⎨ x1 x2 ⎩

x3

Section 2.2

= −2 = 3 . = 5

Hence, the unique solution to the system is (−2, 3, 5). (e) The final matrix obtained in Exercise 1(e) in Section 2.1 is ⎡

1

⎣ 0 0

−2

− 56

0 0

1 0

" ⎤ − 13 " − 53 6 " −2 0 "" 5 ⎦. " 0 1 2 8 3

We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2,3) entry as the pivot and target the (1,3) entry. " ⎤ ⎡ 1 −2 0 1 − 13 " − 14 3 " ⎣ 0 1 ← ( 56 ) × 2 + 1 1 −2 0 0 "" 5 ⎦ " 0 0 0 0 1 2 Next, we pivot at the (3,5) entry and target the zero.) ⎡ 1 −2 0 1 ⎣ 0 0 1 −2 1 ← ( 13 ) × 3 + 1 0 0 0 0 This matrix corresponds to the linear system ⎧ ⎨ x1 − 2x2 x3 ⎩

(1,5) entry. (Note that the (2,5) entry is already 0 0 1

" ⎤ " −4 " " 5 ⎦ " " 2

+ x4 − 2x4 x5

= −4 = 5 . = 2

Columns 2 and 4 are not pivot columns, so x2 and x4 are independent variables. Let x2 = b and x4 = d. Then the first equation gives x1 − 2b + d = −4, or x1 = 2b − d − 4. The second equation yields x3 − 2d = 5, or x3 = 2d + 5. The last equation states that x5 = 2. Hence, the complete solution set is {(2b − d − 4, b, 2d + 5, d, 2) | b, d ∈ R}. (g) The final matrix obtained in Exercise 1(g) in Section 2.1 is " ⎡ 5 ⎤ 1 − 12 − 74 "" 4 " ⎥ ⎢ 1 − 14 " − 74 ⎥ ⎢ 0 " ⎢ ⎥. ⎣ 0 0 1 "" 3 ⎦ 0 0 0 " 0 We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2,2) entry as the pivot and target the (1,2) entry. " ⎡ 3 ⎤ " 1 0 − 15 8 " 8 " ⎥ ⎢ 1 − 14 " − 74 ⎥ ⎢ 0 1 1 ← ( 2 ) × 2 + 1 ⎢ ⎥ " ⎣ 0 0 1 "" 3 ⎦ 0 0 0 " 0 41


Section 2.2

Next, we use the (3,3) entry as a pivot and target the (1,3) and (2,3) entries. " ⎡ ⎤ 1 0 0 "" 6 ⎢ 0 1 0 " −1 ⎥ 1 ← ( 15 8 ) × 3 + 1 ⎢ ⎥ " ⎣ 0 0 1 "" 3 ⎦ 2 ← ( 14 ) × 3 + 2 0 0 0 " 0 Ignoring the last row, which gives the equation 0 = 0, this matrix yields the linear system ⎧ = 6 ⎨ x1 = −1 , x2 ⎩ 3 x3 = producing the unique solution (6, −1, 3).

⎡

−2 −3 2 4 (4) (a) The augmented matrix for the system is ⎣ −4 −7 1 2 −1 The (1,1) entry is the first pivot. We convert ⎡ 3 13 1 −1 2 2 1 ⎣ −4 −7 1 ← (− ) 1 4 −29 2

1

2

it " " " " " 8 "

−1

−13 −29 8

" ⎤ " 0 " " 0 ⎦. " " 0

to 1. ⎤ 0 0 ⎦ 0

Next, we target the (2,1) and (3,1) entries. " ⎤ ⎡ 3 " 1 −1 13 2 2 " 0 2 ← (4) × 1 + 2 ⎥ " ⎢ 0 −3 " 0 ⎦ ⎣ 0 −1 3 ← (−1) × 1 + 3 " 1 3 " 0 0 0 2 2 The (2,2) entry is the second pivot. We convert it to 1. " ⎤ ⎡ " 1 32 −1 13 2 " 0 ⎥ " ⎢ 1 0 3 " 0 ⎦ 2 ← (−1) 2 ⎣ 0 " 3 " 0 0 1 0 2

Next, we target the (1,2) and (3,2) ⎡ 1 ← (− 32 ) × 2 + 1 ⎣ 3 ← (− 12 ) × 2 + 3

2

entries. 1 0 0

" ⎤ 0 −1 2 "" 0 1 0 3 "" 0 ⎦ 0 0 0 " 0

This matrix is in reduced row echelon form. It corresponds ⎧ − x3 + 2x4 ⎨ x1 + 3x4 x2 ⎩ 0

to the linear system = = =

0 0 . 0

We can ignore the last equation, since it provides us with no information regarding the values of the variables. Columns 3 and 4 of the matrix are nonpivot columns, and so x3 and x4 are independent variables. Let x3 = c and x4 = d. Substituting these into the first equation yields x1 − c + 2d = 0, or x1 = c − 2d. The second equation gives us x2 + 3d = 0, or x2 = −3d. Therefore. the complete solution set is {(c − 2d, −3d, c, d) | c, d ∈ R}. Letting c = 1 and d = 2 produces the particular nontrivial solution (−3, −6, 1, 2). 42


(c) The augmented matrix for the system is " ⎡ 7 28 4 −2 10 19 "" 0 ⎢ −9 −36 −5 3 −15 −29 "" 0 ⎢ ⎣ 3 12 2 0 6 11 "" 0 6 24 3 −3 10 20 " 0 The (1,1) entry is the first pivot. We convert it to 1. ⎡ 10 19 "" 4 1 4 − 27 7 7 7 " 0 ⎢ 3 −15 −29 "" 0 ⎢ −9 −36 −5 1 ← ( 17 ) 1 ⎣ 3 12 2 0 6 11 "" 0 6 24 3 −3 10 20 " 0 Next, we target the (2,1), (3,1), and ⎡ 1 ⎢ 2 ← (9) × 1 + 2 ⎢ 0 ⎢ 3 ← (−3) × 1 + 3 ⎢ 0 ⎣ 4 ← (−6) × 1 + 4 0

0 0 0

4 7 1 7 2 7 − 37

10 7 − 15 7 12 7 10 7

− 27

3 7 6 7 − 97

⎤ ⎥ ⎥. ⎦

⎤ ⎥ ⎥ ⎦

(4,1) entries. 4

Section 2.2

19 7 − 32 7 20 7 26 7

" " " " " " " " " "

0

⎤

⎥ 0 ⎥ ⎥ 0 ⎥ ⎦ 0

We try to move the pivot to the second column but cannot due to the zeroes in the (2,2), (3,2), and (4,2) entries. Hence, the pivot moves to the (2,3) entry. We convert that to 1. " ⎡ ⎤ 10 19 " 4 1 4 − 27 7 7 7 " 0 ⎢ ⎥ " ⎢ 0 0 1 3 −15 −32 " 0 ⎥ ⎢ ⎥ " 2 ← (7) 2 ⎢ 0 0 ⎥ 6 12 20 " 2 ⎣ 7 7 7 7 " 0 ⎦ " 10 26 " 0 0 0 −3 −9 7

7

Next, we target the (1,3), (3,3), and (4,3) ⎡ 1 4 1 ← (− 47 ) × 2 + 1 ⎢ ⎢ 0 0 3 ← (− 27 ) × 2 + 3 ⎣ 0 0 4 ← ( 37 ) × 2 + 4 0 0

7

7

entries. 0 −2 1 3 0 0 0 0

10 21 −15 −32 6 12 −5 −10

" " " " " " " "

⎤ 0 0 ⎥ ⎥ 0 ⎦ 0

We try to move the pivot to the third column but cannot due to the zeroes in the (3,4) and (4,4) entries. Hence, the pivot moves to the (3,5) entry. We convert that to 1. " ⎤ ⎡ 1 4 0 −2 10 21 "" 0 ⎢ 0 0 1 3 −15 −32 "" 0 ⎥ ⎥ ⎢ 3 ← ( 16 ) 3 ⎣ 0 0 0 1 0 2 "" 0 ⎦ 0 0 0 0 −5 −10 " 0 Next, we target the (1,5), (2,5), and (4,5) ⎡ 1 4 1 ← (−10) × 3 + 1 ⎢ 0 0 ⎢ 2 ← (15) × 3 + 2 ⎣ 0 0 4 ← (5) × 3 + 4 0 0

entries.

" 0 −2 0 1 "" 0 1 3 0 −2 "" 0 1 0 0 2 "" 0 0 0 0 0 " 0

43

⎤ ⎥ ⎥ ⎦


This matrix is in reduced row echelon form. It corresponds to the ⎧ x1 + 4x2 − 2x4 + x6 ⎪ ⎪ ⎨ − 2x6 x3 + 3x4 x5 + 2x6 ⎪ ⎪ ⎩ 0

Section 2.2

linear system = = = =

0 0 . 0 0

We can ignore the last equation, since it provides us with no information regarding the values of the variables. Columns 2, 4, and 6 of the matrix are nonpivot columns, and so x2 , x4 , and x6 are independent variables. Let x2 = b, x4 = d, and x6 = f . Substituting these into the first equation yields x1 +4b−2d+f = 0, or x1 = −4b+2d−f . The second equation gives us x3 +3d−2f = 0, or x3 = −3d + 2f . The third equation produces x5 + 2f = 0, or x5 = −2f . Therefore, the complete solution set is {(−4b + 2d − f, b, −3d + 2f, d, −2f, f ) | b, d, f ∈ R}. Letting b = 1, d = 2, and f = 3 produces the particular nontrivial solution (−3, 1, 0, 2, −6, 3). " ⎤ ⎡ −2 1 8 "" 0 (5) (a) The augmented matrix for the system is ⎣ 7 −2 −22 "" 0 ⎦. 3 −1 −10 " 0 The (1,1) entry is the first pivot. We convert it to 1. " ⎤ ⎡ 1 − 12 −4 " 0 " ⎣ 7 −2 −22 " 0 ⎦ 1 ← (− 12 ) 1 " 3 −1 −10 " 0 Next, we target the (2,1) and (3,1) entries. ⎡ 1 − 12 2 ← (−7) × 1 + 2 ⎢ 3 ⎣ 0 2 3 ← (−3) × 1 + 3 1 0 2

" ⎤ −4 "" 0 ⎥ " 6 " 0 ⎦ " 2 " 0

The (2,2) entry is the second pivot. ⎡ 1 − 12 −4 ⎢ 1 4 2 ← ( 23 ) 2 ⎣ 0 0

1 2

We convert it to 1. " ⎤ " 0 " ⎥ " " 0 ⎦ " 2 " 0

Next, we target the (1,2) and (3,2) ⎡ 1 ← ( 12 ) × 2 + 1 ⎣ 3 ← (− 12 ) × 2 + 3

entries.

" ⎤ 1 0 −2 "" 0 1 0 4 "" 0 ⎦ 0 0 0 " 0 This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ − 2x3 = 0 ⎨ x1 x2 + 4x3 = 0 . ⎩ 0 = 0

We can ignore the last equation, since it provides us with no information regarding the values of the variables. Column 3 of the matrix is a nonpivot column, and so x3 is an independent variable. Let x3 = c. Substituting this into the first equation yields x1 − 2c = 0, or x1 = 2c. The second equation gives us x2 + 4c = 0, or x2 = −4c. Therefore, the complete solution set is {(2c, −4c, c) | c ∈ R} = {c(2, −4, 1) | c ∈ R}. 44


2 ⎢ 1 (c) The augmented matrix for the system is ⎢ ⎣ 2 3

6 4 8 10

13 10 20 21

1 1 1 2

" " " " " " " "

Section 2.2

⎤ 0 0 ⎥ ⎥. 0 ⎦ 0

The (1,1) entry is the first pivot. We convert it to 1. ⎡ ⎤ 1 "" 1 3 13 2 2 " 0 ⎢ 1 4 10 1 "" 0 ⎥ ⎢ ⎥ 1 ← ( 12 ) 1 ⎣ 2 8 20 1 "" 0 ⎦ 3 10 21 2 " 0 Next, we target the (2,1), (3,1), and (4,1) entries. " ⎤ ⎡ 1 " 1 3 13 2 2 " 0 ⎥ ⎢ 2 ← (−1) × 1 + 2 7 1 "" ⎥ ⎢ 0 1 2 2 " 0 ⎥ ⎢ 3 ← (−2) × 1 + 3 " ⎢ 0 2 7 0 " 0 ⎥ ⎦ ⎣ 4 ← (−3) × 1 + 4 " 1 " 3 0 0 1 2 2 The (2,2) entry is the second pivot. It already equals 1, so now we target the (1,2), (3,2), and (4,2) entries. " ⎡ ⎤ 1 0 −4 −1 "" 0 1 ← (−3) × 2 + 1 ⎢ ⎥ 7 1 " 1 ⎢ 0 2 2 "" 0 ⎥ 3 ← (−2) × 2 + 3 ⎢ ⎥ ⎣ 0 0 0 −1 "" 0 ⎦ 4 ← (−1) × 2 + 4 0 0 −2 0 " 0 The (3,3) entry equals 0. into the (3,3) position. ⎡ 1 ⎢ ⎢ 0 3 ←→ 4 ⎢ ⎣ 0 0

However, we can switch the 3rd and 4th rows to bring a nonzero number 0 −4 1

7 2

0 −2 0 0

" −1 "" 1 " 2 "" 0 "" −1 "

0

⎤

⎥ 0 ⎥ ⎥ 0 ⎦ 0

We now convert the pivot (the (3,3) entry) to 1. " ⎤ ⎡ 1 0 −4 −1 "" 0 ⎥ ⎢ 1 " 7 ⎢ 0 1 2 2 "" 0 ⎥ 3 ← (− 12 ) 3 ⎥ ⎢ ⎣ 0 0 1 0 "" 0 ⎦ 0 0 0 −1 " 0 Next, we target the (1,3) and (2,3) entries. The (4,3) entry already equals 0. " ⎤ ⎡ 1 0 0 −1 "" 0 ⎥ ⎢ 1 " 1 ← (4) × 3 + 1 ⎢ 0 1 0 2 "" 0 ⎥ ⎥ ⎢ ⎣ 0 0 2 ← (− 72 ) × 3 + 2 1 0 "" 0 ⎦ 0 0 0 −1 " 0 The (4,4) entry is the last pivot. We convert it to 1.

45

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ 4 ← (−1) 4

1

0

0

⎢ ⎢ 0 1 0 ⎢ ⎣ 0 0 1 0 0 0

" −1 "" 1 " 2 "" 0 "" 1 "

Finally, we target the (1,4) and (2,4) ⎡ 1 ⎢ 0 1 ← (1) × 4 + 1 ⎢ ⎣ 0 2 ← (− 12 ) × 4 + 2 0

0

Section 2.2

⎤

⎥ 0 ⎥ ⎥ 0 ⎦ 0

entries. The " 0 0 0 "" 1 0 0 "" 0 1 0 "" 1 " 0 0

(3,4) entry is already 0. ⎤ 0 0 ⎥ ⎥ 0 ⎦ 0

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ x1 = 0 ⎪ ⎪ ⎨ = 0 x2 . = 0 x ⎪ 3 ⎪ ⎩ x4 = 0 Clearly, this system has only the trivial solution. The solution set is {(0, 0, 0, 0)}. (6) (a) First, we find the system of linear equations corresponding to the given chemical equation by considering each element separately. This produces ⎧ = c ← Carbon equation ⎨ 6a 6a = 2d ← Hydrogen equation . ⎩ 2b = 2c + d ← Oxygen equation Moving all variables to the left side and ⎡ 6 ⎣ 6 0 The (1,1) entry is the first ⎡ 1 1 ⎣ 1 ← ( 6 ) 1 6 0

then creating an augmented matrix yields " ⎤ 0 −1 0 "" 0 0 0 −2 "" 0 ⎦ . 2 −2 −1 " 0

pivot. We convert it to 1. " ⎤ 0 − 16 0 " 0 " 0 0 −2 "" 0 ⎦ 2 −2 −1 " 0

Next, we target the (2,1) entry. (The (3,1) entry is already 0.) " ⎤ ⎡ 1 0 − 16 0 " 0 " ⎣ 0 0 2 ← (−6) × 1 + 2 1 −2 "" 0 ⎦ 0 2 −2 −1 " 0 The pivot moves to the (2,2) entry, which is 0. So, we switch the 2nd and 3rd rows to move a nonzero number into the pivot position. " ⎤ ⎡ 1 0 − 16 0 " 0 " ⎣ 0 2 ←→ 3 2 −2 −1 "" 0 ⎦ 0 0 1 −2 " 0 Now we convert the (2,2) entry to 1. 46


1

0

⎢ 1 ⎣ 0 0 0

Section 2.2

" ⎤ 0 "" 0 " ⎥ −1 − 12 " 0 ⎦ " 1 −2 " 0

− 16

The pivot moves to the (3,3) entry. entries. ⎡ 1 1 1 ← ( 6 ) × 3 + 1 ⎢ ⎣ 0 2 ← (1) × 3 + 2 0

Since this already equals 1, we target the (1,3) and (2,3) " ⎤ 0 0 − 13 "" 0 " ⎥ 1 0 − 52 " 0 ⎦ " 1 −2 " 0 0

The matrix is now in reduced row echelon form. It corresponds to the following linear system: ⎧ − 13 d = 0 ⎪ ⎨ a b − 52 d = 0 , ⎪ ⎩ c − 2d = 0 or a = 13 d, b = 52 d, and c = 2d. Setting d = 6 eliminates all fractions and provides the smallest solution containing all positive integers. Hence, the desired solution is a = 2, b = 15, c = 12, d = 6. (c) First, we find the system of linear equations corresponding to the given chemical equation by considering each element separately. This produces ⎧ a = c ← Silver equation ⎪ ⎪ ⎨ a = e ← Nitrogen equation , 2d + 3e ← Oxygen equation ⎪ 3a + b = ⎪ ⎩ 2b = e ← Hydrogen equation Moving all variables to the left side and then creating an augmented matrix yields " ⎤ ⎡ 1 0 −1 0 0 "" 0 ⎢ 1 0 0 0 −1 "" 0 ⎥ ⎥. ⎢ ⎣ 3 1 0 −2 −3 "" 0 ⎦ 0 2 0 0 −1 " 0 The (1,1) entry is the first pivot. Since it already equals 1, we target the (2,1) and (3,1) entries. (The (4,1) entry is already 0.) ⎡ " ⎤ 1 0 −1 0 0 "" 0 ⎢ 0 0 2 ← (−1) × 1 + 2 1 0 −1 "" 0 ⎥ ⎥ ⎢ ⎣ 3 ← (−3) × 1 + 3 0 1 3 −2 −3 "" 0 ⎦ 0 2 0 0 −1 " 0 The pivot moves to the (2,2) entry. However, it move a nonzero number into the pivot position. " ⎡ 1 0 −1 0 0 "" 0 ⎢ 0 1 3 −2 −3 "" 0 ⎢ 2 ←→ 3 ⎣ 0 0 1 0 −1 "" 0 0 2 0 0 −1 " 0

equals 0. So, we switch the 2nd and 3rd rows to ⎤ ⎥ ⎥ ⎦

The (2,2) entry already equals 1, so we target the (4,2) entry. (The (1,2) and (3,2) entries already equal 0, and so do not need to be targeted.) 47


1 ⎢ 0 ⎢ ⎣ 0 0

4 ← (−2) × 2 + 4

The pivot moves to the (3,3) entry. (4,3) entries. ⎡ 1 1 ← (1) × 3 + 1 ⎢ 0 ⎢ 2 ← (−3) × 3 + 2 ⎣ 0 4 ← (6) × 3 + 4 0

0 1 0 0

−1 3 1 −6

" 0 0 "" 0 −2 −3 "" 0 0 −1 "" 0 4 5 " 0

Section 2.2

⎤ ⎥ ⎥ ⎦

Since this already equals 1, we target the (1,3), (2,3), and " ⎤ 0 0 0 −1 "" 0 1 0 −2 0 "" 0 ⎥ ⎥ 1 0 0 −1 "" 0 ⎦ 0 0 4 −1 " 0

The pivot moves to the (4,4) ⎡ 1 0 ⎢ 0 1 ⎢ 4 ← ( 14 ) 4 ⎣ 0 0 0

entry. We convert this to 1. " ⎤ 0 0 −1 " 0 " 0 −2 0 " 0 ⎥ ⎥ " 1 0 −1 " 0 ⎦ " 1 − 14 " 0 0 0

Finally, we target the (2,4) entry. ⎡ 1 ⎢ ⎢ 0 ⎢ 2 ← (2) × 4 + 2 ⎢ 0 ⎣ 0

0

0

1

0

0

1

0

0

" −1 "" " 0 − 12 " " 0 −1 "" " 1 − 14 " 0

0

⎤

⎥ 0 ⎥ ⎥ 0 ⎥ ⎦ 0

The matrix is now in reduced row echelon form. It corresponds to the following linear system: ⎧ a − e = 0 ⎪ ⎪ ⎨ b − 12 e = 0 , c − e = 0 ⎪ ⎪ ⎩ d − 14 e = 0 or a = e, b = 12 e, c = e, and d = 14 e. Setting e = 4 eliminates all fractions and provides the smallest solution containing all positive integers. Hence, the desired solution is a = 4, b = 2, c = 4, d = 1, e = 4. (7) (a) Combining the fractions on the right side of the given equation over a common denominator produces B C A(x − 3)(x + 4) + B(x − 1)(x + 4) + C(x − 1)(x − 3) A + + = (x − 1) (x − 3) (x + 4) (x − 1)(x − 3)(x + 4) 2 (A + B + C) x + (A + 3B − 4C) x + (−12A − 4B + 3C) . = (x − 1)(x − 3)(x + 4) Now we set the coefficients of x2 and x and the constant term equal to the corresponding coefficients in the numerator of the left side of the equation given in the problem. Doing this yields the following linear system: ⎧ A + B + C = 5 ← coefficients of x2 ⎨ A + 3B − 4C = 23 ← coefficients of x . ⎩ −12A − 4B + 3C = −58 ← constant terms 48


1 1 The augmented matrix for the system is ⎣ −12 The (1,1) entry is the first pivot. Since ⎡ 1 2 ← (−1) × 1 + 2 ⎣ 0 3 ← (12) × 1 + 3 0

1 3 −4

1 −4 3

Section 2.2

" ⎤ " 5 " " 23 ⎦. " " −58

it already equals 1, we target the (2,1) and (3,1) entries. " ⎤ 1 1 "" 5 2 −5 "" 18 ⎦ 8 15 " 2

The (2,2) entry is the second pivot. We convert it to 1. " ⎤ ⎡ 1 1 1 "" 5 ⎥ " ⎢ 1 − 52 " 9 ⎦ 2 ← ( 12 ) 2 ⎣ 0 " 0 8 15 " 2 Next, we target the (1,2) and (3,2) entries. ⎡ 7 1 0 2 1 ← (−1) × 2 + 1 ⎢ 1 − 52 0 ⎣ 3 ← (−8) × 2 + 3 0 0 35

" ⎤ " −4 " ⎥ " 9 ⎦ " " " −70

The (3,3) entry is the last pivot. We convert it to 1. " ⎤ ⎡ 7 " 1 0 2 " −4 ⎥ " ⎢ 1 3 ← ( 35 ) 3 ⎣ 0 1 − 52 " 9 ⎦ " 1 " −2 0 0 Finally, we target the (1,3) and (2,3) ⎡ 1 1 ← (− 72 ) × 3 + 1 ⎣ 0 2 ← ( 52 ) × 3 + 2 0

entries. 0 1 0

0 0 1

" ⎤ " 3 " " 4 ⎦ " " −2

This matrix is in reduced row echelon form. It corresponds to the linear system ⎧ = 3 ⎨ A B = 4 , ⎩ C = −2 which gives the unique solution to the problem. (8) We set up the augmented matrix having two columns to the right of the augmentation bar: " ⎤ ⎡ 9 2 2 "" −6 −12 ⎣ 3 −3 ⎦ . 2 4 "" 0 8 27 12 22 " 12 The (1,1) entry is the first pivot. We convert it to 1. ⎤ ⎡ 2 "" 2 2 1 − 43 9 9 " −3 ⎣ 3 2 1 ← ( 19 ) 1 4 "" 0 −3 ⎦ 27 12 22 " 12 8 We next target the (2,1) and (3,1) entries.

49

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ 2 ← (−3) × 1 + 2 3 ← (−27) × 1 + 3

1

⎢ ⎣ 0 0

2 9 4 3

6

" 2 " −3 " " " 2 " 16 " 30 2 9 10 3

− 43

Section 2.2

⎤

⎥ 1 ⎦ 44

The (2,2) entry is the second pivot. We convert it to 1. " 2 ⎤ ⎡ 2 " − 43 1 29 9 " −3 ⎢ 3 ⎥ 5 " 3 1 2 ← ( 34 ) 2 ⎣ 0 2 "" 2 4 ⎦ " 0 6 16 30 44 Next, we target the (1,2) and (3,2) entries. ⎡ 1 0 − 13 2 1 ← (− 9 ) × 2 + 1 ⎢ 5 1 ⎣ 0 2 3 ← (−6) × 2 + 3 0 0 1

" " −1 " " 3 " 2 " " 21

− 32

3 4 79 2

⎤ ⎥ ⎦

The (3,3) entry is the last pivot. Since it already equals 1, we target the (1,3) and (2,3) entries. " ⎡ 35 ⎤ 1 0 0 "" 6 3 1 1 ← ( 3 ) × 3 + 1 " ⎥ ⎢ 0 1 0 −51 −98 " ⎦ ⎣ " 2 ← (− 52 ) × 3 + 2 79 " 1 0 0 21 2 This matrix is in reduced row echelon form. It gives the unique solution for each of the two original linear systems. In particular, the solution for the system AX = B1 is (6, −51, 21), and the solution for 79 the system AX = B2 is ( 35 3 , −98, 2 ).

(11) (b) Any nonhomogeneous system with two equations and two unknowns that has a unique solution will serve as a counterexample. For instance, consider ) x + y = 1 . x − y = 1 This system has a unique solution: (1, 0). Let (s1 , s2 ) and (t1 , t2 ) both equal (1, 0). Then the sum of solutions is not a solution in this case. Also, if c = 1, then the scalar multiple of a solution by c is not a solution. (14) (a) True. The Gaussian elimination method puts the augmented matrix for a system into row echelon form. The statement gives a verbal description of the definition of row echelon form given in the section. (b) True. The Gaussian elimination method puts the augmented matrix for a system into row echelon form, while the Gauss-Jordan method puts the matrix into reduced row echelon form. The statement describes the difference between those two forms for a matrix. (c) False. A column is skipped if the pivot entry that would usually be used is zero, and all entries below that position in that column are zero. For example, the reduced row echelon form matrix 1 2 0 has no pivot in the second column. Rows, however, are never skipped over, although 0 0 1 rows of zeroes may occur at the bottom of the matrix. (d) True. A homogeneous system always has the trivial solution. (e) False. This statement reverses the roles of the “independent (nonpivot column)” and “dependent (pivot column)” variables. Note, however, that the statement does correctly define the terms “independent” and “dependent.” 50


Section 2.3

(f) False. For a counterexample, consider the system ⎧ = 0 ⎨ x y = 0 , ⎩ x + y = 0 which clearly has only the trivial solution. The given statement is the reverse of a related true statement, which says: If a homogeneous system has more variables than equations, then the system has a nontrivial solution.

Section 2.3 (1) Recall that a matrix A is row equivalent to a matrix B if a finite sequence of row operations will convert A into B. In parts (a) and (c), below, only one row operation is necessary, although longer finite sequences could be used. (a) A row operation of type (I) converts A to B: 2 ←− −5 2 . (c) A row operation of type (II) converts A to B: 2 ←− 3 + 2 . (2) (b) We must first compute B by putting A into reduced row echelon form. Performing the following sequence of row operations converts A to B = I3 : (I): (II): (II): (III): (II):

1 ← 14 1 2 ← 2 1 + 2 3 ← −3 1 + 3 2 ↔ 3 1 ← 5 3 + 1

To convert B back to A, we use the inverses of these row operations in the reverse order: (II): 1 ← −5 3 + 1 (III): 2 ↔ 3 (II): 3 ← 3 1 + 3 (II): 2 ← −2 1 + 2 (I): 1 ← 4 1

the the the the the

inverse inverse inverse inverse inverse

of of of of of

(II): 1 ← 5 3 + 1 (III): 2 ↔ 3 (II): 3 ← −3 1 + 3 (II): 2 ← 2 1 + 2 (I): 1 ← 14 1

(3) (a) Both A and B row reduce to I3 . To convert A to I3 , use the following sequence of row operations: (II): 3 ←− 2 2 + 3 (I): 3 ←− −1 3 (II): 1 ←− −9 3 + 1 (II): 2 ←− 3 3 + 2 To convert B to I3 , use these row operations: (I): 1 ←− − 15 1 (II): 2 ←− 2 1 + 2 (II): 3 ←− 3 1 + 3 (I): 2 ←− −5 2 (II): 1 ←− (II): 3 ←−

3 5 9 5

2 + 1 2 + 3

(b) First, we convert A to I3 using the sequence of row operations we computed in part (a): (II): 3 ←− 2 2 + 3 51


Section 2.3

(I): 3 ←− −1 3 (II): 1 ←− −9 3 + 1 (II): 2 ←− 3 3 + 2 Next, we continue by converting I3 to B. To do this, we use the inverses of the row operations that converted B to I3 (from part (a)), in the reverse order: (II): 3 ← − 95 2 + 3

the inverse of (II): 3 ←

(II): 1 ←

2 + 1

the inverse of (II): 1 ←

2

the inverse of (I): 2 ← −5 2

(I): 2 ←

− 35 − 15

9 5 3 5

2 + 3 2 + 1

(II): 3 ← −3 1 + 3

the inverse of (II): 3 ← 3 1 + 3

(II): 2 ← −2 1 + 2

the inverse of (II): 2 ← 2 1 + 2

(I): 1 ← −5 1

the inverse of (I): 1 ← − 15 1

(5) The rank of a matrix is the number of nonzero rows in its corresponding reduced row echelon form matrix. ⎡ ⎤ 1 0 2 (a) The given matrix reduces to ⎣ 0 1 −1 ⎦, which has two nonzero rows. Hence, its rank is 2. 0 0 0 ⎡ ⎤ 1 0 0 (c) The given matrix reduces to ⎣ 0 0 1 ⎦, which has two nonzero rows. Hence, its rank is 2. 0 0 0 ⎡ ⎤ 1 0 0 1 ⎢ 0 1 0 −1 ⎥ ⎢ ⎥ (e) The given matrix reduces to ⎢ 3 ⎥, which has 3 nonzero rows. Hence, its rank is 3. ⎣ 0 0 1 2 ⎦ 0 0 0 0 (6) (a) Corollary 2.6 cannot be used because there are matrix for this system is " ⎡ ⎤ ⎡ 1 −2 6 3 "" 0 ⎢ 0 " 0 ⎥ ⎢ 5 −9 −4 ⎥, which reduces to ⎢ " ⎢ ⎣ 0 ⎣ 4 −8 −3 "" 0 ⎦ 0 6 −11 −5 " 0

more equations than variables. The augmented 0 1 0 0

0 0 1 0

" " " " " " " "

⎤ 0 0 ⎥ ⎥. Since there are three nonzero rows 0 ⎦ 0

in the reduced row echelon form matrix, the rank = 3. Theorem 2.5 predicts that the system has only the trivial solution. The reduced row echelon form matrix corresponds to the linear system ⎧ x1 = 0 ⎪ ⎪ ⎨ = 0 x2 , x3 = 0 ⎪ ⎪ ⎩ 0 = 0 which clearly has the solution set {(0, 0, 0)}, agreeing with the prediction of Theorem 2.5. (7) In each case, we give the smallest and largest rank possible, with a particular example of a linear system and the reduced row echelon form of its corresponding augmented matrix for each of the two ranks. 52


(a)

Smallest rank = 1 ⎧ x1 + 2x2 + 3x3 ⎪ ⎪ ⎨ x1 + 2x2 + 3x3 x1 + 2x2 + 3x3 ⎪ ⎪ ⎩ x1 + 2x2 + 3x3 ⎡

1 ⎢ 0 ⎢ ⎣ 0 0 (c)

2 0 0 0

3 0 0 0

" " " " " " " "

Largest rank = 4 ⎧ x1 ⎪ ⎪ ⎨ x2 x3 ⎪ ⎪ ⎩ 0

=4 =4 =4 =4

⎤ 4 0 ⎥ ⎥ 0 ⎦ 0

⎡

1 ⎢ 0 ⎢ ⎣ 0 0

Smallest rank = 2 ⎧ ⎨ x1 + x2 + x3 + x4 = 0 x1 + x2 + x3 + x4 = 1 ⎩ x1 + x2 + x3 + x4 = 0 ⎡

1 ⎣ 0 0

1 0 0

1 0 0

1 0 0

" ⎤ " 0 " " 1 ⎦ " " 0

Section 2.3

0 1 0 0

= = = = 0 0 1 0

0 0 0 1 " " " " " " " "

⎤ 0 0 ⎥ ⎥ 0 ⎦ 1

Largest rank = 3 ⎧ ⎨ x1 + x3 + x4 = 0 x2 + x3 + x4 = 0 ⎩ x1 + x3 + x4 = 1 ⎡

1 ⎣ 0 0

0 1 0

1 1 0

1 1 0

" ⎤ " 0 " " 0 ⎦ " " 1

(8) In each part, to express the vector x as a linear combination of the vectors a1 , . . . , an , we need to solve the linear system whose augmented matrix has the vectors a1 , . . . , an as columns to the left of the augmentation bar, and has the vector x as a column to the right of the augmentation bar. A solution of this linear system gives the coefficients for the desired linear combination. (a) Following 1 −2 4 3

the procedure described above, we reduce the matrix " ' ( " " −3 1 0 "" − 21 11 " to obtain . The unique solution to this system provides the " " −6 6 0 1 " 11

coefficients for x as a linear combination of a1 and a2 , and so x = − 21 11 a1 + (c) Following ⎡ 3 2 ⎣ 6 10 2 −4

the procedure described ⎡ " ⎤ 1 " 2 " " −1 ⎦ to obtain ⎢ ⎣ 0 " " 4 0

6 11 a2 .

above, we reduce the matrix " ⎤ 0 "" 11 9 " ⎥ 1 " − 56 ⎦. " 0 " − 16 9

The last row of the reduced matrix shows that the system is inconsistent. Therefore, it is not possible to express x as a linear combination of a1 and a2 . (e) Following the procedure described above, we " ⎤ ⎡ ⎡ 1 0 −1 1 5 4 "" 7 ⎣ −2 −2 0 " 2 ⎦ to obtain ⎣ 0 1 1 " 0 0 0 3 6 3 " 3

reduce the matrix " ⎤ " −3 " " 2 ⎦. The linear system has the infinite solution " " 0

set {(c − 3, −c + 2, c) | c ∈ R}, and hence, there are many sets of coefficients b1 , b2 , b3 for which x = b1 a1 + b2 a2 + b3 a3 . Choosing the particular solution when c = 0 yields x = −3a1 + 2a2 + 0a3 . Other choices for c produce alternate linear combinations.

53


(g) Following the procedure described above, we reduce " " ⎡ ⎤ ⎡ 1 0 0 "" 2 3 −2 6 "" 2 ⎢ " ⎢ 2 0 1 "" 3 ⎥ ⎥ to obtain ⎢ 0 1 0 " −1 ⎢ ⎣ 0 0 1 " −1 ⎣ −2 1 2 "" −7 ⎦ " 0 0 0 " 0 4 −3 8 " 3

Section 2.3

the matrix ⎤ ⎥ ⎥. The unique solution to this system pro⎦

vides the coefficients for x as a linear combination of a1 , a2 , and a3 , and so x = 2a1 − a2 − a3 . (9) To determine whether a given vector X is in the row space of matrix A, we row reduce the augmented matrix AT |X . If the corresponding linear system is consistent, the solution set gives coefficients for expressing X as a linear combination of the rows of A. If the system is inconsistent, then X is not in the row space of A. (a) Following the procedure described above, " ⎤ ⎡ ⎡ 1 0 3 2 2 "" 7 ⎣ 6 10 −1 " 1 ⎦ to obtain ⎣ 0 1 " 0 0 2 −4 4 " 18

we reduce the matrix " ⎤ 0 "" 5 0 "" −3 ⎦. Since the system has the unique solution 1 " −1

(5, −3, −1), X is in the row space of A, and X = 5(row 1) − 3(row 2) − 1(row 3). (c) Following the procedure described above, " ⎤ ⎡ ⎡ 1 0 4 −2 6 "" 2 ⎣ −1 3 1 "" 2 ⎦ to obtain ⎣ 0 1 0 0 2 5 9 " −3

we reduce the matrix " ⎤ 1 2 "" 1 ⎦. The last row of the reduced matrix shows 1 "" 0 " −10

that the system is inconsistent. Hence, X is not in the row space of A. (e) Following the procedure described above, " ⎡ ⎤ ⎡ 1 2 7 3 "" 1 ⎢ 0 " 11 ⎥ ⎢ −4 −1 7 ⎢ ⎥ " ⎢ ⎣ 1 −1 −3 " −4 ⎦ to obtain ⎣ 0 " 0 −3 2 8 " 11

we reduce the matrix " ⎤ 0 −2 "" −3 1 1 "" 1 ⎥ ⎥. 0 0 "" 0 ⎦ 0 0 " 0

The linear system has the infinite solution set {(2c − 3, −c + 1, c) | c ∈ R}, and hence, there are many sets of coefficients b1 , b2 , b3 for which [1, 11, −4, 11] = b1 (row 1) + b2 (row 2) + b3 (row 3). Choosing the particular solution when c = 0 yields [1, 11, −4, 11] = −3(row 1) + 1(row 2) + 0(row 3). Other choices for c produce alternate linear combinations. (10) (a) To express the vector [13, −23, 60] as a linear combination of vectors q1 , q2 , and q3 , we need to solve the linear system whose augmented matrix has the vectors q1 , q2 , and q3 as columns to the left of the augmentation bar, and has the vector [13, −23, 60] as a column to the right of the augmentation bar. Hence, we row reduce " " ⎡ ⎤ ⎡ ⎤ −1 −10 7 "" 13 1 0 0 "" −2 ⎣ −5 3 −12 "" −23 ⎦ to obtain ⎣ 0 1 0 "" 1 ⎦. 11 −8 30 " 60 0 0 1 " 3 The unique solution (−2, 1, 3) for the system corresponding to this matrix gives us the coefficients for the desired linear combination. Hence, [13, −23, 60] = −2q1 + q2 + 3q3 . (b) As in part (a), for each qi , we need to solve a linear system whose augmented matrix of the form [r1 r2 r3 | qi ], where the vectors listed represent columns of the matrix. But using the method for solving several systems simultaneously, we can solve all three problems together by row reducing

54


[r1 r2 r3 | q1 q2 ⎡ 3 2 4 ⎣ −2 1 −1 4 −3 2

q3 ]. Hence, we row reduce " ⎤ ⎡ " −1 −10 7 1 " " −5 ⎦ to get ⎣ 0 3 −12 " " 11 −8 30 0

0 1 0

0 0 1

" " 3 " " −1 " " −2

2 2 −5

Section 2.3

⎤ 1 −6 ⎦. 4

Reading the three solutions (3, −1, −2), (2, 2, −5), and (1, −6, 4) from the reduced matrix gives the coefficients needed for the desired linear combinations. Hence, q1 = 3r1 − r2 − 2r3 , q2 = 2r1 + 2r2 − 5r3 , and q3 = r1 − 6r2 + 4r3 . (c) Combining the answers from parts (a) and (b) yields [13, −23, 60] = −2q1 + q2 + 3q3 = −2 (3r1 − r2 − 2r3 ) + (2r1 + 2r2 − 5r3 ) + 3 (r1 − 6r2 + 4r3 ) = −6r1 + 2r2 + 4r3 + 2r1 + 2r2 − 5r3 + 3r1 − 18r2 + 12r3 = −r1 − 14r2 + 11r3 . ⎡

1 (11) (a) (i) A row reduces to B = ⎣ 0 0

0 1 0

−1 3 0

⎤ 2 2 ⎦. 0

(ii) To check that the ith row, bi , of B is in the row space of A, we must solve the system whose augmented matrix is [AT |bi ], where bi is a column. But we can solve this problem for the two nonzero rows of B simultaneously by row reducing the matrix [AT |b1 b2 ]. Performing this row reduction produces the matrix " ⎤ ⎡ " 1 0 − 38 " − 78 14 " ⎥ ⎢ ⎢ 0 1 1 " 1 0 ⎥ ⎥. ⎢ 2 " 2 " ⎥ ⎢ " ⎣ 0 0 0 " 0 0 ⎦ 0 0 0 " 0 0 Both systems have infinitely many solutions. Hence, there are many ways to express the two nonzero rows of B as linear combinations of the rows of A. The two solution sets are {( 38 c − 78 , − 12 c + 12 , c) | c ∈ R} for b1 and {( 38 c + 14 , − 12 c, c) | c ∈ R} for b2 . If we find particular solutions by choosing c = 0, we get [1, 0, −1, 2] = − 78 [0, 4, 12, 8] + 12 [2, 7, 19, 18] + 0[1, 2, 5, 6] and [0, 1, 3, 2] = 14 [0, 4, 12, 8] + 0[2, 7, 19, 18] + 0[1, 2, 5, 6]. (iii) In this part, we perform the same operations as in part (ii), except that we reverse the roles of A and B. Hence, we row reduce the matrix [b1 b2 |AT ], which yields " ⎡ ⎤ 1 0 "" 0 2 1 ⎢ 0 1 " 4 7 2 ⎥ ⎢ " ⎥ ⎣ 0 0 " 0 0 0 ⎦. " 0 0 " 0 0 0 In this case, each of the three linear systems has a unique solution. These three solutions are (0, 4), (2, 7), and (1, 2). Each gives the coefficients for expressing a row of A as a linear combination of the two nonzero rows of B. Hence, we have [0, 4, 12, 8] = 0[1, 0, −1, 2] + 4[0, 1, 3, 2], [2, 17, 19, 18] = 2[1, 0, −1, 2] + 7[0, 1, 3, 2], and [1, 2, 5, 6] = 1[1, 0, −1, 2] + 2[0, 1, 3, 2]. (13) (a) Suppose we are performing row operations on an m × n matrix A. Throughout this part, we use the notation B i for the ith row of a matrix B. 55


Section 2.3

For the Type (I) operation R : i ←− c i : Now R−1 is i ←− 1c i . Clearly, R and R−1 change only the ith row of A. We want to show that R−1 R leaves A i unchanged. But R−1 (R(A)) i = 1c R(A) i = 1c (c A i ) = A i . For the Type (II) operation R : i ←− c j + i : Now R−1 is i ←− −c j + i . Again, R and R−1 change only the ith row of A, and we need to show that R−1 R leaves A i unchanged. But R−1 (R(A)) i = −c R(A) j + R(A) i = −c A j + R(A) i = −c A j + c A j + A i = A i . For the Type (III) operation R : i ←→ j : Now, R−1 = R. Also, R changes only the ith and jth rows of A, and these get swapped. Obviously, a second application of R swaps them back to where they were, proving that R is indeed its own inverse. (b) An approach similar to that used for Type (II) operations in the abridged proof of Theorem 2.3 in the text works just as easily for Type (I) and Type (III) operations. For Type (I) operations: ⎧ a11 x1 ⎪ ⎪ ⎪ ⎨ a21 x1 .. ⎪ . ⎪ ⎪ ⎩ am1 x1

Suppose that the original system has the form + +

a12 x2 a22 x2 .. .

+ am2 x2

a13 x3 a23 x3 .. .

+ ··· + ··· .. .

+ +

+ am3 x3

+ ···

+ amn xn

+ +

a1n xn a2n xn .. .

= =

b1 b2 .. .

= bm

and that the row operation used is R : i ← c i . When R is applied to the corresponding augmented matrix, all rows except the ith row remain unchanged. The new ith equation then has the form (cai1 )x1 +(cai2 )x2 +· · ·+(cain )xn = cbi . We must show that any solution (s1 , s2 , . . . , sn ) of the original system is a solution of the new one. Now, since (s1 , s2 , . . . , sn ) is a solution of the ith equation in the original system, we have ai1 s1 + ai2 s2 + · · · + ain sn = bi . Multiplying this equation by c yields (cai1 )s1 + (cai2 )s2 + · · · + (cain )sn = cbi , and so (s1 , s2 , . . . , sn ) is also a solution of the new ith equation. Now, since none of the other equations in the system have changed, (s1 , s2 , . . . , sn ) is still a solution for each of them. Therefore, (s1 , s2 , . . . , sn ) is a solution to the new system formed. For Type (III) operations: A type (III) row operation merely changes the order of the rows of a matrix. Since the rows of the augmented matrix correspond to equations in the system, performing a type (III) row operation changes only the order in which the equations are written, without making any changes to the actual equations at all. Hence, any solution to the first system of equations must also be a solution to the second system of equations, since none of the actual equations themselves have changed. There is, however, an alternate approach to this problem that manages to prove the result for all three types of row operations at once: Suppose R is a row operation, and let X satisfy AX = B. Multiplying both sides of this matrix equation by the matrix R(I) yields R(I)AX = R(I)B, implying R(IA)X = R(IB), by Theorem 2.1. Thus, R(A)X = R(B), showing that X is a solution to the new linear system obtained from AX = B after the row operation R is performed. You might wonder what motivates the second method of proof used here. The technique is based on the topic of Elementary Matrices, which is covered in Section 8.6 of the textbook. You will have sufficient background to read that section after you finish studying Chapter 2. (14) The zero vector is a solution to AX = O, but it is not a solution for AX = B. Hence, these two systems have different solution sets and thus cannot be equivalent systems. (15) Consider the systems

)

x + y x + y

= =

1 0

) and

56

x − y x − y

= =

1 . 2


The augmented matrices for these systems are, respectively, " 1 −1 1 1 "" 1 and 1 −1 1 1 " 0

" " 1 " " 2 .

The reduced row echelon matrices for these are, respectively, " 1 1 "" 0 1 −1 and 0 0 " 1 0 0

" " 0 " " 1 ,

Section 2.3

where we have continued the row reduction beyond the augmentation bar. Thus, the original augmented matrices are not row equivalent, since their reduced row echelon forms are different. However, the reduced row echelon form matrices reveal that both systems are inconsistent. Therefore, they both have the same solution set, namely, the empty set. Hence, the systems are equivalent. (19) (a) As in the abridged proof of Theorem 2.8 in the text, let a1 , . . . , am represent the rows of A, and let b1 , . . . , bm represent the rows of B. For the Type (I) operation R : i ←− c i : Now bi = 0a1 + 0a2 + · · · + cai + 0ai+1 + · · · + 0am , and, for k = i, bk = 0a1 + 0a2 + · · · + 1ak + 0ak+1 + · · · + 0am . Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. For the Type (II) operation R : i ←− c j + i : Now bi = 0a1 + 0a2 + · · · + caj + 0aj+1 + · · · + ai + 0ai+1 + · · · + 0am , where our notation assumes i > j. (An analogous argument works for i < j.) And, for k = i, bk = 0a1 + 0a2 + · · · + 1ak + 0ak+1 + · · · + 0am . Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. For the Type (III) operation R : i ←→ j : Now, bi = 0a1 + 0a2 + · · · + 1aj + 0aj+1 + · · · + 0am , bj = 0a1 + 0a2 + · · · + 1ai + 0ai+1 + · · · + 0am , and, for k = i, k = j, bk = 0a1 + 0a2 + · · · + 1ak + 0ak+1 + · · · + 0am . Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. (b) If x is in the row space of B, then x is a linear combination of the rows of B. By part (a), each row of B is a linear combination of the rows of A. Hence, by Lemma 2.7, x is a linear combination of the rows of A as well. Therefore, x is in the row space of A. Therefore, since every member of the row space of B is also in the row space of A, we get that the row space of B is contained in the row space of A. (20) Let k be the number of matrices between A and B when performing row operations to get from A to B. Use a proof by induction on k. Base Step: If k = 0, then there are no intermediary matrices, and Exercise 19 shows that the row space of B is contained in the row space of A. Inductive Step: Given the chain A → D1 → D2 → · · · → Dk → Dk+1 → B, we must show that the row space B is contained in the row space of A. The inductive hypothesis shows that the row space of Dk+1 is in the row space of A, since there are only k matrices between A and Dk+1 in the chain. Thus, each row of Dk+1 can be expressed as a linear combination of the rows of A. But by Exercise 19, each row of B can be expressed as a linear combination of the rows of Dk+1 . Hence, by Lemma 2.7, each row of B can be expressed as a linear combination of the rows of A and therefore is in the row space of A. By Lemma 2.7 again, the row space of B is contained in the row space of A. (22) (a) True. This is the statement of Theorem 2.3. 57


Section 2.4

(b) True. In general, if two matrices are row equivalent, then they are both row equivalent to the same reduced row echelon form matrix. Hence, A and B are both row equivalent to the same matrix C in reduced row echelon form. Since the ranks of A and B are both the number of nonzero rows in their common reduced echelon form matrix C, A and B must have the same rank. (c) False. The inverse of the type (I) row operation i ← c i is the type (I) row operation i ←

1 c

i .

(d) False. The statement is true for homogeneous systems but is false in general. If the system in question is not homogeneous, it could be inconsistent and thus have no solutions at all, and hence, ) x+y+z =1 no nontrivial solutions. For a particular example, consider the system . This x+y+z =2 system has three variables, but the rank of its augmented matrix is ≤ 2, since the matrix has only two rows. However, the system has no solutions because it is impossible for x + y + z to equal both 1 and 2. (e) False. This statement directly contradicts part (2) of Theorem 2.5. (f) True. By the definition of row space, x is clearly in the row space of A. But Theorem 2.8 shows that A and B have the same row space. Hence, x is also in the row space of B.

Section 2.4 (2) To find the rank of a matrix, find its corresponding reduced row echelon form matrix and count its nonzero rows. An n × n matrix is nonsingular if and only if its rank equals n. (a) The given 2 × 2 matrix row reduces to I2 , which has two nonzero rows. Hence, the original matrix has rank = 2. Therefore, the matrix is nonsingular. (c) The given 3 × 3 matrix row reduces to I3 , matrix has rank = 3. Therefore, the matrix ⎡ 1 ⎢ 0 (e) The given 4 × 4 matrix row reduces to ⎢ ⎣ 0 0

which has three nonzero rows. Hence, the original is nonsingular. ⎤ 0 0 0 1 0 0 ⎥ ⎥, 0 1 −2 ⎦ 0 0 0

which has three nonzero rows. Hence, the rank of the original matrix is 3. But this is less than the number of rows in the matrix, and so the original matrix is singular. (3) Use the formula for the inverse of a 2 × 2 matrix given in Theorem 2.13. (a) First, we compute δ = (4)(−3) − (2)(9) = −30. Since δ = 0, the given matrix is nonsingular, and ' 1 1 ( 10 15 −3 −2 1 its inverse is −30 = . 2 3 −9 4 − 15 10 (c) First, we compute δ = (−3)(−8) − (5)(−12) = 84. Since δ = 0, the given matrix is nonsingular, ' −2 −5 ( 21 84 −8 −5 1 and its inverse is 84 . = 1 1 12 −3 − 7 28 (e) First, we compute δ = (−6)(−8) − (12)(4) = 0. Since δ = 0, the given matrix is singular; it has no inverse.

58


Section 2.4

(4) To find the inverse of an n × n matrix A, row reduce [A|In ] to obtain [In |A−1 ]. If row reduction does not produce In to the left of the augmentation bar, then A does not have an inverse. " " ⎡ ⎤ ⎡ ⎤ −4 7 6 "" 1 0 0 1 0 0 "" 1 3 2 2 ⎦. Therefore, (a) We row reduce ⎣ 3 −5 −4 "" 0 1 0 ⎦ to obtain ⎣ 0 1 0 "" −1 0 " " −2 4 3 0 0 1 0 0 1 2 2 −1 ⎡ ⎤ 1 3 2 ⎣ −1 0 2 ⎦ is the inverse of the original matrix. 2 2 −1 " ⎡ " ⎤ ⎡ 1 ⎤ 1 0 0 "" 32 0 2 2 −2 3 "" 1 0 0 ⎥ " ⎢ 9 "" 0 1 0 ⎦ to obtain ⎣ 0 1 0 " −3 12 − 12 ⎦. Therefore, (c) We row reduce ⎣ 8 −4 " " 0 0 1 −4 6 −9 0 0 1 " − 83 13 − 23 ⎤ ⎡ 3 1 0 2 2 ⎥ ⎢ ⎢ − 3 1 − 1 ⎥ is the inverse of the original matrix. 2 2 ⎦ ⎣ −

8 3

1 3

− 23 ⎡

2 0 −1 3 ⎢ 1 −2 3 1 (e) We row reduce ⎢ ⎣ 4 1 0 −1 1 3 −2 −5

" " " " " " " "

1 0 0 0

0 1 0 0

0 0 1 0

⎤

⎡

1

0 ⎢ ⎢ 0 0 ⎥ ⎥ to ⎢ 0 ⎦ ⎢ ⎣ 0 1 0

0

0

1

0

0

1

0

0

" 1 " " 5 " 4 " −5 " " −3 " 5 " 0 " 1

8 15 − 47 15 − 29 15

1 15 4 − 15 2 15

2 15 7 15 4 15

1

−1

0

⎤

⎥ 0 ⎥ ⎥. 0 ⎥ ⎦ 1

Therefore, since I4 is not obtained to the left of the augmentation bar, the original matrix does not have an inverse. (Note: If you use a calculator to perform the row reduction above, it might continue to row reduce beyond the augmentation bar.) (5) (c) To find the inverse of a diagonal matrix, merely find the diagonal matrix whose main diagonal entries are the reciprocals of the main diagonal entries of the original matrix. Hence, when this matrix is multiplied by the original, the product will be a diagonal matrix with all 1’s on main diagonal –⎤that is, the identity matrix. Therefore, the inverse matrix we need is ⎡ the 1 0 ··· 0 a ⎢ 11 ⎥ 1 ⎢ 0 ··· 0 ⎥ a22 ⎥ ⎢ . ⎢ . .. .. ⎥ .. ⎢ .. . . . ⎥ ⎦ ⎣ 0

0

···

1 ann

(6) (a) Using Theorem 2.13, δ = (cos θ)(cos θ) − (sin θ)(−sin θ) = cos2 θ + sin2 θ = 1. Therefore, the cos θ sin θ general inverse is . Plugging in the given values for θ yields −sin θ cos θ

59


θ

Original Matrix '

√ 3 2

π 6

1 2

'

√ 2 2 √ 2 2

π 4

π 2

0 1

(b) It is easy to ⎡ a11 a12 ⎣ a21 a22 0 0

Inverse Matrix

(

− 12

'

√ 3 2

√ 3 2

−

− (

√ 2 2 √ 2 2

'

1 2

(

√ 2 2 √ 2 2

0 −1

see that if A and ⎤⎡ b11 b12 0 0 ⎦ ⎣ b21 b22 1 0 0

(

1 2 √ 3 2

√ 2 2 √ 2 2

−

−1 0

Section 2.4

1 0

B are 2 × 2 matrices with AB = C, then ⎤ ⎡ ⎤ c11 c12 0 0 0 ⎦ = ⎣ c21 c22 0 ⎦. (Check it out!) Therefore, to find the 1 0 0 1

inverse of the matrix given in part (b), we merely use the associated 2 × 2 inverses we found in part (a), glued into the upper corner of a 3 × 3 matrix, as illustrated. Hence, the general inverse ⎡ ⎤ cos θ sin θ 0 is ⎣ − sin θ cos θ 0 ⎦ . Substituting the given values of θ produces 0 0 1 θ

Original Matrix ⎡

π 6

⎢ ⎢ ⎣ ⎡

π 4

√ 3 2

⎢ ⎢ ⎣

− 12

1 2

√ 3 2

0

0

√ 2 2 √ 2 2

−

√ 2 2 √ 2 2

0 ⎡ π 2

0 ⎣ 1 0

0

⎤

⎥ 0 ⎥ ⎦

Inverse Matrix ⎡

√ 3 2

⎢ ⎢ − ⎣

1 0

⎤

⎥ 0 ⎥ ⎦

⎡ ⎢ ⎢ − ⎣

0 1 −1 0 0

⎤ 0 0 ⎦ 1

⎡

0

1 2

1 2 √ 3 2

0

0

1

√ 2 2 √ 2 2

√ 2 2 √ 2 2

0

0

0 ⎣ −1 0

1 0 0

⎤

⎥ 0 ⎥ ⎦

0

⎤

⎥ 0 ⎥ ⎦ 1

⎤ 0 0 ⎦ 1

(7) By Theorem 2.15, if A is a nonsingular matrix, then the unique solution to AX = B is A−1 B. (a) The given system corresponds to the matrix equation 5 −1 x1 20 5 −1 = . We use Theorem 2.13 to compute the inverse of . x2 −7 2 −31 −7 2 ' 2 1 ( 3 3 2 1 . The unique First, δ = (5)(2) − (−7)(−1) = 3. Hence, the inverse = 13 = 5 7 7 5 3

60

3

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ' solution to the system is thus system is {(3, −5)}.

2 3

1 3

7 3

5 3

(

20 −31

=

⎡

0 ⎢ −7 (c) The system corresponds to the matrix equation ⎢ ⎣ 5 −3

3 −5

Section 2.4

. That is, the solution set for the

⎤⎡ −2 5 1 x1 ⎢ x2 −4 5 22 ⎥ ⎥⎢ 3 −4 −16 ⎦ ⎣ x3 −1 0 9 x4

We find the inverse of the coefficient matrix by row reducing " ⎡ ⎤ ⎡ 1 0 0 0 −2 5 1 "" 1 0 0 0 ⎢ 0 1 0 " 0 1 0 0 ⎥ ⎢ −7 −4 5 22 ⎥ to obtain ⎢ " ⎢ ⎣ 0 0 1 ⎣ 5 3 −4 −16 "" 0 0 1 0 ⎦ 0 0 0 −3 −1 0 9 " 0 0 0 1

0 0 0 1

⎤ 25 ⎢ ⎥ ⎥ ⎥ = ⎢ −15 ⎥. ⎣ ⎦ 9 ⎦ −16 ⎤

⎡

" ⎤ " 1 −13 −15 5 " " −3 3 0 −7 ⎥ ⎥. The " " −1 2 1 −3 ⎦ " " 0 −4 −5 1

inverse of the coefficient matrix is the 4 × 4 matrix to the right of the augmentation bar in the row reduced matrix. Hence, the unique solution to the system is ⎡ ⎤ ⎤⎡ ⎤ ⎡ 1 −13 −15 5 5 25 ⎢ −3 ⎥ ⎢ ⎥ ⎢ 3 0 −7 ⎥ ⎢ ⎥ ⎢ −15 ⎥ = ⎢ −8 ⎥. ⎣ −1 ⎦ ⎦ ⎣ ⎣ 2 1 −3 2 ⎦ 9 −1 0 −4 −5 1 −16 That is, the solution set for the linear system is {(5, −8, 2, −1)}. 0 1 (8) (a) Through trial and error, we find the involutory matrix . 1 0 (b) Using the answer to part (a), the comments at the beginning of the answer to Section 2.4, Exercise ⎤ ⎡ 0 1 0 6(b) in this manual, suggest that we consider the matrix ⎣ 1 0 0 ⎦. Squaring this matrix 0 0 1 verifies that it is, in fact, involutory. (c) If A is involutory, then A2 = AA = In . Hence, A itself satisfies the definition of an inverse for A, and so A−1 = A. (10) (a) Since A is nonsingular, A−1 exists. Hence, AB = On ⇒ A−1 (AB) = A−1 On ⇒ (A−1 A)B = On ⇒ In B = On ⇒ B = On . Thus, B must be the zero matrix. (b) No. AB = In implies B = A−1 , and so A is nonsingular. We can now apply part (a), substituting C where B appears, proving that AC = On ⇒ C = On . (11) Now A4 = In ⇒ A3 A = In ⇒ A−1 = A3 . Also, since Ikn = In for every integer k, we see that A4k = (A4 )k = Ikn = In for every integer k. Hence, for every integer k, A4k+3 = A4k A3 = In A3 = A3 = A−1 . Thus, all n × n matrices of the form A4k+3 equal A−1 . These are . . . , A−9 , A−5 , A−1 , A3 , A7 , A11 , . . .. Now, could any other powers of A also equal A−1 ? Suppose that Am = A−1 . Then, Am+1 = Am A = A−1 A = In . Now, if m + 1 = 4l for some integer l, then m = 4(l − 1) + 3, and so m is already on our list of powers of A such that Am = A−1 . On the other hand, if m + 1 is not a multiple of 4, then, dividing m + 1 by 4 yields an integer k, with a remainder r of either 1, 2, or 3. Then m + 1 = 4k + r, and so A4k+r = In implying A4k Ar = In . But A4k = In from above, and so Ar = In . But, since r equals either 1, 2, or 3, this contradicts the given conditions of the problem, and so m + 1 cannot equal 4k + r. Therefore, the only powers of A that equal A−1 are of the form A4k+3 .

61


Section 2.4

(12) By parts (1) and (3) of Theorem 2.11, B−1 A is the inverse of A−1 B. (Proof: (A−1 B)−1 = B−1 (A−1 )−1 = B−1 A.) Thus, if A−1 B is known, simply compute its inverse to find B−1 A. (14) (a) No step in the row reduction process will alter the column of zeroes, and so the unique reduced row echelon form for the matrix must contain a column of zeroes and so cannot equal In . (15) (a) Part (1): Since AA−1 = I, we must have (A−1 )−1 = A. Part (2): For k > 0, to show (Ak )−1 = (A−1 )k , we must show that Ak (A−1 )k = I. Proceed by induction on k. Base Step: For k = 1, clearly AA−1 = I. Inductive Step: Assume Ak (A−1 )k = I. Prove Ak+1 (A−1 )k+1 = I. Now, Ak+1 (A−1 )k+1 = AAk (A−1 )k A−1 = AIA−1 = AA−1 = I. This concludes the proof for k > 0. We now show Ak (A−1 )k = I for k ≤ 0. For k = 0, clearly A0 (A−1 )0 = I I = I. The case k = −1 is covered by part (1) of the theorem. For k ≤ −2, (Ak )−1 = ((A−1 )−k )−1 (by definition) = ((A−k )−1 )−1 (by the k > 0 case) = A−k (by part (1)). (17) (a) Let p = −s, q = −t. Then p, q > 0. Now, As+t = A−(p+q) = (A−1 )p+q = (A−1 )p (A−1 )q (by Theorem 1.15) = A−p A−q = As At . (21) (a) Suppose that n > k. Then, by Corollary 2.6, there is a nontrivial X such that BX = O. Hence, (AB)X = A(BX) = AO = O. But (AB)X = In X = X. Therefore, X = O, which gives a contradiction. (b) Suppose that k > n. Then, by Corollary 2.6, there is a nontrivial Y such that AY = O. Hence, (BA)Y = B(AY) = BO = O. But (BA)Y = Ik Y = X. Therefore, Y = O, which gives a contradiction. (c) Parts (a) and (b) combine to prove that n = k. Hence, A and B are both square matrices. They are nonsingular with A−1 = B by the definition of a nonsingular matrix, since AB = In = BA. (22) (a) False. Many n × n matrices are singular, having no inverse. For example, the 2 × 2 matrix 6 3 has no inverse, by Theorem 2.13, because δ = (6)(4) − (3)(8) = 0. 8 4 (b) True. This follows directly from Theorem 2.9. (c) True. ((AB)T )−1 = ((AB)−1 )T (by part (4) of Theorem 2.11) = (B−1 A−1 )T (by part (3) of Theorem 2.11) = (A−1 )T (B−1 )T (by Theorem 1.16).

(d) False. This statement contradicts Theorem 2.13, which states that

a c

b d

is nonsingular if and

only if δ = ad − bc = 0. (e) False. This statement contradicts the Inverse Method given in Section 2.4 as well as Theorem 2.14. To correct the statement, change the word “nonsingular” to “singular.” (f) True. This follows directly from combining Theorems 2.14 and 2.15.

62


Chapter 2 Review Exercises ⎡

2 ⎢ 1 (1)(i) (a) First augmented matrix: ⎢ ⎣ −3 −2

5 −3 4 3

−4 2 7 −1

Chap 2 Review

" ⎤ " 48 " " −40 ⎥ ⎥. The first pivot is the (1,1) entry. We make " " 15 ⎦ " " 41

this a “1” by performing the following type (I) row operation: " ⎤ ⎡ 5 1 24 −2 " 2 " ⎢ 1 −3 2 "" −40 ⎥ ⎥ Type (I) operation: 1 ← 12 1 : ⎢ ⎣ −3 15 ⎦ 4 7 "" 41 −2 3 −1 " We now target the (2,1), (3,1), and (4,1) entries and make them “0.” We do this using the following type (II) row operations: Type (II) operation: 2 ← (−1) × 1 + 2 : Type (II) operation: 3 ← (3) × 1 + 3 : Type (II) operation: 4 ← (2) × 1 + 4 : This results in the matrix " ⎤ ⎡ 5 1 24 −2 "" 2 ⎥ ⎢ " ⎢ 0 − 11 4 " −64 ⎥ 2 ⎥ ⎢ " ⎥ " ⎢ 0 23 87 1 " ⎦ ⎣ 2 " 89 0 8 −5 " Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation. " ⎤ ⎡ 5 1 −2 "" 24 2 ⎢ 8 "" 128 ⎥ ⎥ ⎢ 0 1 − 11 2 ⎢ " 11 ⎥ Type (I) operation: 2 ← − 11 2 : ⎢ " 23 1 " 87 ⎥ ⎦ ⎣ 0 2 " " 89 0 8 −5 Now we target the (3,2) and (4,2) entries. Type (II) operation: 3 ← (− 23 2 ) × 2 + 3 : Type (II) operation: 4 ← (−8) × 2 + 4 : The resultant matrix is " ⎤ ⎡ 24 1 52 −2 "" ⎢ 128 ⎥ 8 "" ⎥ ⎢ 0 1 − 11 11 ⎥ " ⎢ ⎢ 0 515 ⎥ 103 " 0 ⎣ 11 " − 11 ⎦ " 9 " − 45 0 0 11 11 Now we move the pivot to the (3,3) entry. We make this entry a “1” using a type (I) row operation. " ⎤ ⎡ 24 1 52 −2 "" ⎢ 128 ⎥ 8 "" ⎥ ⎢ 0 1 − 11 11 ⎥ 11 " ⎢ Type (I) operation: 3 ← 103 3 : ⎢ " 1 " −5 ⎥ ⎦ ⎣ 0 0 " 45 9 " − 0 0 11

63

11


Chap 2 Review

9 We target the (4,3) entry using the following type (II) operation: 4 ← (− 11 ) × 3 + 4 : " ⎡ ⎤ 5 " 1 2 −2 " 24 ⎢ 8 "" 128 ⎥ ⎢ 0 1 − 11 ⎥ ⎢ " 11 ⎥ ⎢ 0 0 " 1 " −5 ⎥ ⎣ ⎦ " " 0 0 0 0

This finishes the third column. Since there are no more columns before the augmentation bar, we are finished performing row operations. The linear system corresponding to the final matrix is ⎧ x1 ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

+

5 2 x2

−

x2

−

2x3 8 11 x3

=

24

=

128 11

x3 0

= =

−5 0

The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of x1 , x2 , and x3 . Thus, we can ignore it. The third equation 8 (−5) = 128 clearly gives x3 = −5. Plugging this value into the second equation yields x2 − 11 11 , or x2 = 8. Substituting the values we have for x2 and x3 into the first equation produces x1 + 52 (8) − 2(−5) = 24, or x1 = −6. Hence, the original linear system has a unique solution. The full solution set is {(−6, 8, −5)}. " ⎤ ⎡ 4 3 −7 5 "" 31 ⎢ −2 −3 5 −1 "" −5 ⎥ ⎥. The first pivot is the (1,1) entry. We (b) First augmented matrix: ⎢ ⎣ 2 −6 −2 3 "" 52 ⎦ 6 −21 −3 12 " 16 make this a “1” by performing the following type (I) row operation: " 31 ⎤ ⎡ 5 " 3 1 − 74 4 4 " 4 " ⎢ ⎥ " −5 ⎥ ⎢ −2 −3 1 5 −1 Type (I) operation: 1 ← 4 1 : ⎢ " ⎥ ⎣ 2 −6 −2 3 "" 52 ⎦ 6 −21 −3 12 " 16 We now target the (2,1), (3,1), and (4,1) entries and make them “0.” We do this using the following type (II) row operations: Type (II) operation: 2 ← (2) × 1 + 2 : Type (II) operation: 3 ← (−2) × 1 + 3 : Type (II) operation: 4 ← (−6) × 1 + 4 : This results in the matrix " ⎡ ⎤ 3 31 1 − 74 54 "" 4 4 " ⎢ ⎥ 21 ⎥ 3 3 " ⎢ 0 −3 2 2 2 " 2 ⎥ ⎢ ⎥ ⎢ " 3 1 " 73 ⎥ ⎢ 0 − 15 2 2 2 " 2 ⎦ ⎣ " 15 9 " 0 − 51 − 61 2 2 2 2 Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation.

64

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ Type (I) operation: 2 ←

− 23

2 :

3 4

− 74

1

−1

− 15 2

3 2

− 51 2

15 2

1

⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣ 0

" 31 " 4 " " " −1 " −7 " 1 " 73 2 " 2 " 9 " 61 −2 2 5 4

Chap 2 Review

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Now we target the (3,2) and (4,2) entries. Type (II) operation: 3 ← ( 15 2 ) × 2 + 3 : Type (II) operation: 4 ← ( 51 2 ) × 2 + 4 : The resultant matrix is " ⎡ ⎤ 5 " 31 1 34 − 74 4 " 4 " ⎢ ⎥ ⎢ 0 1 −1 −1 "" −7 ⎥ ⎢ ⎥ " ⎢ ⎥ ⎣ 0 0 −6 −7 "" −16 ⎦ 0 0 −18 −21 " −209 Now we move the pivot to the (3,3) entry. We make this entry a “1” using a type (I) row operation. " ⎤ ⎡ 5 " 31 1 34 − 74 4 " 4 " ⎥ ⎢ ⎢ 0 1 −1 −1 "" −7 ⎥ ⎥ ⎢ 1 Type (I) operation: 3 ← − 6 3 : ⎢ ⎥ " 7 " 8 ⎥ ⎢ 0 0 1 6 " 3 ⎦ ⎣ " 0 0 −18 −21 " −209 We target the (4,3) entry using the following type (II) operation: 4 ← (18) × 3 + 4 : " ⎡ ⎤ 5 " 31 1 34 − 74 4 " 4 ⎢ " ⎥ ⎢ 0 1 −1 −1 " −7 ⎥ ⎢ ⎥ " ⎢ ⎥ " 7 " 8 ⎥ ⎢ 0 0 1 3 ⎦ 6 " ⎣ " 0 0 0 0 " −161 The last row of this augmented matrix corresponds to the equation 0 = −161. Since this equation cannot be satisfied for any values of x1 , x2 , and x3 , the original system of linear equations has no solutions. " ⎤ ⎡ 6 −2 2 −1 −6 "" −33 13 ⎦. The first pivot is the (1,1) entry. 1 0 2 −1 "" (c) First augmented matrix: ⎣ −2 " −24 4 −1 2 −3 1 We make this a “1” by performing the following type (I) row operation: " ⎡ ⎤ 1 − 13 13 − 16 −1 "" − 11 2 " ⎢ ⎥ Type (I) operation: 1 ← 16 1 : ⎣ −2 13 ⎦ 1 0 2 −1 "" 4 −1 2 −3 1 " −24 We now target the (2,1) and (3,1) entries and make them “0.” We do this using the following type (II) row operations: Type (II) operation: 2 ← (2) × 1 + 2 : Type (II) operation: 3 ← (−4) × 1 + 3 : This results in the matrix 65


1

⎢ ⎢ 0 ⎣ 0

− 13

1 3

− 16

1 3

2 3

5 3

1 3

2 3

− 73

Chap 2 Review

" ⎤ −1 "" − 11 2 ⎥ " −3 "" 2 ⎥ ⎦ " 5 " −2

Now we move the pivot to the (2,2) entry. We make this entry a “1” using a type (I) row operation. " ⎡ ⎤ 1 − 13 13 − 16 −1 "" − 11 2 " ⎢ ⎥ 1 2 5 −9 "" 6 ⎥ Type (I) operation: 2 ← 3 2 : ⎢ ⎣ 0 ⎦ " 2 1 7 " −2 0 − 5 3 3 3 Now we target the (3,2) entry. Type (II) operation: 3 ← (− 13 ) × 2 + 3 : The resultant matrix is " ⎡ ⎤ 1 − 13 13 − 16 −1 "" − 11 2 " ⎢ ⎥ ⎣ 0 1 2 6 ⎦ 5 −9 "" 0 0 0 −4 8 " −4 We attempt to move the pivot to the (3,3) entry. However, that entry is a “0.” There are no entries below it in the matrix, and so a type (III) operation will not resolve the issue. Thus, we move the pivot horizontally to the (3,4) position. We make this entry a “1” using a type (I) row operation. " ⎡ ⎤ 1 − 13 13 − 16 −1 "" − 11 2 ⎢ " ⎥ Type (I) operation: 3 ← − 14 3 : ⎣ 0 1 2 6 ⎦ 5 −9 "" 1 0 0 0 1 −2 " This finishes the fourth column and the third row. Since there are no more rows in the matrix, we are finished performing row operations. The linear system corresponding to the final matrix is ⎧ ⎪ ⎨ x1

−

⎪ ⎩

1 3 x2

+

1 3 x3

−

1 6 x4

−

x2

+

2x3

+

5x4 x4

− 9x5 − 2x5

x5

= − 11 2 = =

6 1

The variables x3 and x5 are independent variables because there were no pivots in the third and fifth columns. We will let x3 = c and x5 = e. The last equation gives x4 = 1 + 2x5 = 1 + 2e. Plugging this expression and the values of x3 and x5 into the second equation yields x2 + 2c + 5(1 + 2e) − 9e = 6, which simplifies to x2 = 1 − 2c − e. Substituting the expressions we have for x2 through x5 into the first equation produces x1 − 13 (1−2c−e)+ 13 (c)− 16 (1+2e)−e = − 11 2 , which simplifies to x1 = −5 − c + e. Hence, the original linear system has an infinite solution set. The full solution set is {[−5 − c + e, 1 − 2c − e, c, 1 + 2e, e] | c, e ∈ R}. (2) First, plug each of the three given following linear system: ⎧ 120 = −27a ⎪ ⎪ ⎨ 51 = −8a −24 = 27a ⎪ ⎪ ⎩ −69 = 64a

points into the equation y = ax3 + bx2 + cx + d. This yields the + + + +

9b − 3c 4b − 2c 9b + 3c 16b + 4c 66

+ + + +

d ←− d ←− d ←− d ←−

using using using using

(−3, 120) (−2, 51) . (3, −24) (4, −69)


−27 ⎢ −8 The augmented matrix for this system is ⎢ ⎣ 27 64

9 4 9 16

" −3 1 "" 120 51 −2 1 "" 3 1 "" −24 4 1 " −69

Chap 2 Review

⎤ ⎥ ⎥. The first pivot is the (1,1) ⎦

entry. We make this a “1” by performing the following type (I) row operation: ⎡ ⎤ 1 1 "" 1 − 13 − 27 − 40 9 9 " ⎢ −8 4 −2 1 "" 51 ⎥ 1 ⎥. Type (I) operation: 1 ← − 27 1 : ⎢ " ⎣ 27 9 3 1 " −24 ⎦ 64 16 4 1 " −69 We now target the (2,1), (3,1), and (4,1) entries and make them “0.” We do this using the following type (II) row operations: Type (II) operation: 2 ← (8) × 1 + 2 : Type (II) operation: 3 ← (−27) × 1 + 3 : Type (II) operation: 4 ← (−64) × 1 + 4 : This results in the matrix " ⎤ ⎡ 1 1 " 40 1 − 13 − 27 " −9 9 ⎢ 4 19 "" 139 ⎥ ⎥ ⎢ 0 − 10 3 9 ⎥ 9 27 " ⎢ . " ⎢ 0 18 0 2 " 96 ⎥ ⎦ ⎣ " 91 " 1939 0 112 − 28 3 9 27 9 The second pivot is the (2,2) entry. operation: ⎡ Type (I) operation: 2 ←

3 4

2 :

We make this a “1” by performing the following type (I) row 1

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

− 13 1

1 9 − 56

18

0

112 3

− 28 9

" 1 " 40 − 27 " −9 " 139 19 " 36 " 12 2 "" 96 " 91 " 1939 27

⎤ ⎥ ⎥ ⎥. ⎥ ⎦

9

We now target the (3,2) and (4,2) entries and make them “0.” We do this using the following type (II) row operations: Type (II) operation: 3 ← (−18) × 2 + 3 : Type (II) operation: 4 ← (− 112 3 ) × 2 + 4 : This results in the matrix " ⎤ ⎡ 1 1 " 40 1 − 13 − 27 " −9 9 ⎢ 19 "" 139 ⎥ ⎥ ⎢ 0 1 − 56 36 " 12 ⎥ ⎢ . ⎢ 0 15 " 225 ⎥ 0 15 − 2 " − 2 ⎦ ⎣ " " −217 0 0 28 − 49 3 The third pivot is the (3,3) entry. operation:

Type (I) operation: 3 ←

1 15

3 :

We make this a “1” by performing the following type (I) row ⎡

1

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

− 13 1

" 1 " 40 − 27 " −9 " 19 " 139 36 " 12 1 − 12 "" − 15 2 " " −217 28 − 49 3

1 9 − 56

0 0

67

⎤ ⎥ ⎥ ⎥. ⎥ ⎦


Chap 2 Review

We now target the (4,3) entry and make it “0.” We do this using the following type (II) row operation: Type (II) operation: 4 ← (−28) × 3 + 4 : This results in the matrix " ⎤ ⎡ 1 1 " 40 1 − 13 − 27 " −9 9 ⎢ 139 ⎥ 19 "" ⎢ 0 ⎥ 1 − 56 36 " 12 ⎥ ⎢ . ⎢ 0 1 " 15 ⎥ 1 0 −2 " − 2 ⎦ ⎣ " 0 0 0 − 73 " −7 The last pivot is the (4,4) entry. operation:

Type (I) operation: 4 ←

− 37

4 :

We make this a “1” by performing the following type (I) row ⎡

1

⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

− 13 1

1 9 − 56

0

1

0

0

" 1 " 40 − 27 " −9 " 139 19 " 36 " 12 − 12 "" − 15 2 " 1 " 3

⎤ ⎥ ⎥ ⎥. ⎥ ⎦

This matrix is in row echelon form. We solve the system using back substitution. The last equation produces d = 3. Plugging this value into the third equation yields c− 32 = − 15 2 , or c = −6. Substituting 139 = , which results in b = 5. Finally, we use all of these into the second equation gives us b + 5 + 19 12 12 , or a = −2. Thus, the desired cubic polynomial values in the first equation to get a − 53 − 23 − 19 = − 40 9 is y = −2x3 + 5x2 − 6x + 3. (4) First, we find the system of linear equations corresponding to the given chemical equation by considering each element separately. This produces ⎧ = c ← Nitrogen equation ⎨ a 3a = 2d ← Hydrogen equation . ⎩ 2b = 2c + d ← Oxygen equation Moving all variables to the left side and then creating an augmented matrix yields " ⎤ ⎡ 1 0 −1 0 "" 0 ⎣ 3 0 0 −2 "" 0 ⎦ . 0 2 −2 −1 " 0 The (1,1) entry is the first pivot. Since it already equals 1, we do not need a type (I) operation. Next, we target the (2,1) entry. (The (3,1) entry is already 0.) " ⎤ ⎡ 1 0 −1 0 "" 0 ⎣ 0 0 3 −2 "" 0 ⎦ 2 ← (−3) × 1 + 2 0 2 −2 −1 " 0 The pivot moves to the (2,2) entry, which is zero. So, we switch the 2nd and 3rd rows to move a nonzero number into the pivot position. " ⎡ ⎤ 1 0 −1 0 "" 0 ⎣ 0 2 −2 −1 " 0 ⎦ 2 ←→ 3 " 0 0 3 −2 " 0 Now we convert the (2,2) entry to 1.

68


1

0 −1

⎢ 1 ⎣ 0 0 0

−1 3

Chap 2 Review

" ⎤ 0 "" 0 ⎥ " − 12 " 0 ⎦ " −2 " 0

The pivot moves to the (3,3) entry. We convert the (3,3) entry to 1. " ⎤ ⎡ 1 0 −1 0 "" 0 " ⎥ ⎢ 3 ← ( 13 ) 3 ⎣ 0 1 −1 − 12 " 0 ⎦ " 1 −2 " 0 0 0 3

Next, we target the (1,3) and (2,3) entries: ⎡ 1 0 0 − 23 1 ← (1) × 3 + 1 ⎢ ⎣ 0 1 0 − 76 2 ← (1) × 3 + 2 1 − 23 0 0

" ⎤ " 0 " " ⎥ " 0 ⎦ " " 0

The matrix is now in reduced row echelon form. It corresponds to the following linear system: ⎧ − 23 d = 0 ⎪ ⎨ a b − 76 d = 0 , ⎪ ⎩ c − 23 d = 0 or a = 23 d, b = 76 d, and c = 23 d. Setting d = 6 eliminates all fractions and provides the smallest solution containing all positive integers. Hence, the desired solution is a = 4, b = 7, c = 4, d = 6. ⎡ ⎤ ⎡ ⎤ 2 −5 3 1 0 −1 −3 4 ⎦ row reduces to ⎣ 0 1 −1 ⎦. (8) (a) A = ⎣ −1 7 −12 5 0 0 0 Since the reduced row echelon form matrix has 2 nonzero rows, rank(A) = 2. ⎡ ⎤ 0 0 −2 −1 ⎢ −2 0 −1 0 ⎥ ⎥ B=⎢ ⎣ −1 −2 −1 −5 ⎦ row reduces to I4 . 0 1 1 3 Since the reduced row echelon form matrix has 4 nonzero rows, rank(B) = 4. ⎡ ⎤ ⎡ ⎤ 3 −1 −5 −6 1 0 −1 0 4 8 −2 ⎦ reduces to ⎣ 0 1 2 0 ⎦. C=⎣ 0 −2 −3 −4 0 0 0 0 1 Since the reduced row echelon form matrix has 3 nonzero rows, rank(C) = 3. (b) We apply Theorem 2.5. The homogeneous system AX = 0 has 3 variables (because A has 3 columns). Since rank(A) = 2 < 3, Theorem 2.5 implies that AX = 0 has nontrivial solutions. Hence, AX = 0 has infinitely many solutions. The homogeneous system BX = 0 has 4 variables (because B has 4 columns). Since rank(B) = 4, Theorem 2.5 tells us that BX = 0 has only the trivial solution. Hence, BX = 0 has exactly one solution. The homogeneous system CX = 0 has 4 variables (because C has 4 columns). Since rank(C) = 3 < 4, Theorem 2.5 indicates that CX = 0 has nontrivial solutions. Hence, CX = 0 has infinitely many solutions. 69


Chap 2 Review

(10) (a) To express the vector [−34, 29, −21] as a linear combination of the vectors x1 = [2, −3, −1], x2 = [5, −2, 1], and x3 = [9, −8, 3], we need to solve the linear system whose augmented matrix has the vectors x1 , x2 , and x3 as columns to the left of the augmentation bar, and has the vector [−34, 29, −21] as a column to the right of the augmentation bar. A solution of this linear system gives the coefficients for the desired linear combination. Hence, we row reduce " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 5 2 5 9 "" −34 ⎣ −3 −2 −8 " 29 ⎦, obtaining ⎣ 0 1 0 " 2 ⎦. " " 0 0 1 " −6 −1 1 3 " −21 Hence [−34, 29, −21] = 5[2, −3, −1] + 2[5, −2, 1] − 6[9, −8, 3]. We see that [−34, 29, −21] is a linear combination of x1 , x2 , and x3 . (b) A vector is in the row space of a matrix if and only if the vector can be expressed as a linear combination of the rows of the matrix. Since the rows of the given matrix A are x1 , x2 , and x3 , and since, in part (a), we showed how to express [−34, 29, −21] as a linear combination of x1 , x2 , and x3 , it follows that [−34, 29, −21] is in the row space of A. (12) To find the inverse of an n × n matrix B, row reduce [B|In ] to obtain [In |B−1 ]. If row not produce In to the left of the augmentation bar, then B does not have an inverse. " " ⎤ ⎡ ⎡ 1 0 0 2 "" 5 −1 3 4 3 5 "" 1 0 0 0 ⎢ 0 1 0 −1 " −2 ⎢ 4 5 5 0 8 "" 0 1 0 0 ⎥ " ⎥ ⎢ (b) We row reduce ⎢ " −2 ⎣ 7 9 8 13 " 0 0 1 0 ⎦ to ⎣ 0 0 1 1 1 " " 0 0 0 0 " −1 −1 2 3 2 3 " 0 0 0 1

reduction does 0 0 0 1

⎤ −5 3 ⎥ ⎥. 1 ⎦ 0

Therefore, since I4 is not obtained to the left of the augmentation bar, the original matrix does not have an inverse. (Note: If you use a calculator to perform the row reduction above, it might continue to row reduce beyond the augmentation bar.) ⎤ ⎡ ⎤ ⎡ ⎤⎡ 17 4 −6 1 x1 2 −1 ⎦ ⎣ x2 ⎦ = ⎣ −14 ⎦. We find the (15) The system corresponds to the matrix equation ⎣ −1 x3 23 3 −5 1 inverse of the coefficient " ⎡ 4 −6 1 "" 1 0 ⎣ −1 2 −1 "" 0 1 3 −5 1 " 0 0

matrix by row reducing ⎤ ⎡ 0 1 0 0 ⎦ to obtain ⎣ 0 1 1 0 0

0 0 1

" " 3 " " 2 " " 1

−1 −1 −2

⎤ −4 −3 ⎦. The inverse of the coefficient −2

matrix is the 3 × 3 matrix to the right of the augmentation bar in the row reduced matrix. Hence, the unique solution to the system is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 3 −1 −4 17 −27 ⎣ 2 −1 −3 ⎦ ⎣ −14 ⎦ = ⎣ −21 ⎦. 1 −2 −2 23 −1 That is, the solution set for the linear system is {(−27, −21, −1)}. (17) (a) False. Theorem 2.3 shows that each of these methods produces the precise solution set for the system. (b) False. Theorem 2.1 states instead that R(AB) = R(A)B. For a counterexample to R(A)B = 0 1 −1 0 and AR(B), consider A = , B = I2 , and R = 1 ↔ 2 . Then R(A)B = −1 0 0 1 0 −1 AR(B) = . 1 0 70


1 (c) False. Consider the system associated with the augmented matrix ⎣ 0 0

Chap 2 Review

2 0 0

3 0 0

" ⎤ " 0 " " 1 ⎦. " " 0

(d) True. The augmented matrices for the two systems are row equivalent. (Multiply each row of the augmented matrix for the second system by c to get the augmented matrix for the first system.) Hence, the systems have the same solution set by Theorem 2.3. ⎡ ⎤ 1 0 0 0 ⎢ 0 0 0 1 ⎥ ⎥ (e) False. Consider ⎢ ⎣ 0 0 1 1 ⎦. 0 0 0 1 (f) True. This follows from conditions (2) and (3) for reduced row echelon form: Each successive row has its first nonzero entry in a later column, and all values below those entries are zero. (g) True. The row operations for solving any system are those that put the coefficient matrix A into reduced row echelon form. The constants on the right side of the equation have no effect in determining the row operations used. (h) True. If X = 0 is a solution to the system AX = B, then A0 = B, implying 0 = B. (i) False. We show that if both statements are true, we obtain a contradiction. Suppose X1 is a nontrivial solution to AX = 0, and X2 is the unique solution to AX = B. Then A(X1 + X2 ) = AX1 + AX2 = 0 + B = B. Hence, (X1 + X2 ) is a solution to AX = B that is different from X2 , because X1 = 0. This contradicts the uniqueness of X2 as a solution to AX = B. (j) True. These are listed in Table 2.1 near the beginning of Section 2.3. ⎡ ⎤ ⎡ ⎤ 1 0 −1 1 0 −2 (k) False. Consider A = ⎣ 0 1 −1 ⎦ and B = ⎣ 0 1 −1 ⎦ . The solution set for AX = 0 is 0 0 0 0 0 0 {[c, c, c] | c ∈ R}, while the solution set for BX = 0 is {[2c, c, c] | c ∈ R}. For a smaller-sized example, consider A = 1 0 and B = 0 1 . Both matrices have rank 1. However, the solution set for AX = 0 is {[0, y] | y ∈ R}, while the solution set for BX = 0 is {[x, 0] | x ∈ R}. (Note, however, if rank(A) = rank(B) = n, then both systems will have only the trivial solution, and so their solution sets will be identical.) (l) False. The rank equals the number of nonzero rows in the reduced row echelon form for the matrix. If the rank is greater than zero, the number of vectors in the row space is infinite, since every scalar multiple of any of the nonzero rows is in the row space. (m) True. By Theorem 2.14, an n × n matrix A is nonsingular if and only if rank(A) = n. But if A is nonsingular, then A−1 is also nonsingular by part (1) of Theorem 2.11. Hence, rank(A−1 ) = n as well. (n) True. A−1 = B(AB)−1 because A(B(AB)−1 ) = (AB)(AB)−1 = In . (Alternately, from Exercise 18(d) in Section 2.3, we know that rank(AB) ≤ rank (A), and since rank(AB) = n here, we must have rank(A) = n, and hence A is nonsingular.) ⎡ ⎤ 1 0 1 0 0 ⎣ 1 0 ⎦ 0 1 = (o) False. Consider . 0 1 0 0 1 0 0 (p) True. First, note that since the two given systems are homogeneous, both have, at least, the trivial solution. If the determinants of both A and B are zero, then both A and B are singular (Theorem 2.13), and by part (2) of Theorem 2.15, both have an infinite number of solutions. 71


Chap 2 Review

If the determinants of A and B are both nonzero, then both A and B are nonsingular (Theorem 2.13), and by part (1) of Theorem 2.15, both have a unique solution. (q) True. This is the essence of the Inverse Method, which is justified in Section 2.4. (r) True. This follows from part (2) of Theorem 2.11 and part (2) of Theorem 2.12. (s) True. (A−1 BT C)−1 = C−1 (BT )−1 (A−1 )−1 = C−1 (B−1 )T A, where we used a generalized form of part (3) of Theorem 2.11 for the first equality, and parts (1) and (4) of Theorem 2.11 for the second equality.

72


Section 3.1

Chapter 3 Section 3.1

" " " −2 5 " " " = (−2)(1) − (5)(3) = −17 (1) (a) " 3 1 " " " " 6 −12 " " " = (6)(8) − (−12)(−4) = 0 (c) " −4 8 " (e) To use basketweaving, we form a new array by taking the given matrix and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern. 2

0 ×

−4

1 ×

0

5 × 7 ×

× −3

3

2 ×

0 ×

−4

1 ×

0

3

Hence, the determinant equals (2)(1)(−3) + (0)(7)(0) + (5)(−4)(3) − (5)(1)(0) − (2)(7)(3) − (0)(−4)(−3) = −108. (g) To use basketweaving, we form a new array by taking the given matrix and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern. 5

0 ×

3 × −1

0 ×

−2

0 ×

8

5 × 3 ×

× −1

4

0 ×

−2 8

Hence, the determinant equals (5)(−2)(4) + (0)(0)(−1) + (0)(3)(8) − (0)(−2)(−1) − (5)(0)(8) − (0)(3)(4) = −40. (i) To use basketweaving, we form a new array by taking the given matrix and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving

73


Section 3.1

pattern.

3

1

× −1

4 ×

3

×

−2

× −2

1

3 ×

5 ×

1

× −1

4 ×

3

1

Hence, the determinant equals (3)(4)(−2) + (1)(5)(3) + (−2)(−1)(1) − (−2)(4)(3) − (3)(5)(1) − (1)(−1)(−2) = 0. (j) The determinant of a 1 × 1 matrix is defined to be the (1,1) entry of the matrix. Therefore, the determinant equals −3. (2) Recall that the submatrix Aij is found by deleting the ith row and the jth column from A. The (i, j) minor is the determinant of this submatrix. " " " 4 3 " " = (4)(4) − (3)(−2) = 22. " (a) |A21 | = " −2 4 " " " " −3 0 5 "" " (c) |C42 | = "" 2 −1 4 "". We will use basketweaving to compute this determinant. To do this, we " 6 4 0 " form a new array by taking C42 and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern.

−3

0 ×

2 × 6

5 × −1

× 4

×

4

−3

0

× 6

×

2 ×

0

−1 4

Hence, |C42 | = (−3)(−1)(0) + (0)(4)(6) + (5)(2)(4) − (5)(−1)(6) − (−3)(4)(4) − (0)(2)(0) = 118. (3) The cofactor Aij is defined to be (−1)i+j |Aij |, where |Aij | is the (i, j) minor – that is, the determinant of the (i, j) submatrix, obtained by deleting the ith row and the jth column from A. " "

" 4 −3 " " = (1) (4)(−7) − (−3)(9) = −1. (a) A22 = (−1)2+2 |A22 | = (−1)4 "" 9 −7 " " " " −5 2 13 "" " 2 22 "" = (−1)(−222) = 222, where we have computed (c) C43 = (−1)4+3 |C43 | = (−1)7 "" −8 " −6 −3 −16 " the 3 × 3 determinant |C43 | to be −222 using basketweaving as follows: 74


Section 3.1

To compute |C43 |, we form a new array by taking C43 and then adding a second copy of columns 1 and 2 as the new columns 4 and 5. We then form terms using the basketweaving pattern.

−5 −8

2 ×

13

2

× 22

× −3

×

×

−6

× −16

−5 −8

2 × 2 ×

−6

−3

Hence, |C43 | = (−5)(2)(−16) + (2)(22)(−6) + (13)(−8)(−3) −(13)(2)(−6) − (−5)(22)(−3) − (2)(−8)(−16) = −222. " "

" " 1+2 3" x−4 x−3 " = (−1) (x − 4)(x + 2) − (x − 3)(x − 1) (d) D12 = (−1) |D12 | = (−1) " x−1 x+2 "

= (−1) x2 − 2x − 8 − x2 − 4x + 3 = −2x + 11. (4) In this problem, we are asked to use only the formal definition (cofactor expansion) to compute determinants. Therefore, we will not use basketweaving for 3 × 3 determinants or the simple a11 a22 − a12 a21 formula for 2 × 2 determinants. Of course, the cofactor expansion method still produces the same results obtained in Exercise 1. −2 5 (a) Let A = . Then |A| = a21 A21 + a22 A22 = a21 (−1)2+1 (a12 ) + a22 (−1)2+2 (a11 ) 3 1 = (3)(−1)(5) + (1)(1)(−2) = −17. 6 −12 (c) Let A = . Then |A| = a21 A21 + a22 A22 = a21 (−1)2+1 (a12 ) + a22 (−1)2+2 (a11 ) −4 8 = (−4)(−1)(−12) + (8)(1)(6) = 0. ⎡ ⎤ 2 0 5 7 ⎦ . Then |A| = a31 A31 + a32 A32 + a33 A33 (e) Let A = ⎣ −4 1 0 3 −3 " " " " " " " " 2 5 " " 2 3+1 " 0 5 " 3+2 " 3+3 " " = (0)(−1) " 1 7 " + (3)(−1) " −4 7 " + (−3)(−1) " −4

0 1

" " ". Now, for a 2 × 2 matrix "

B, |B| = b21 B21 + b22 B22 = b21 (−1)2+1 (b12 ) + b22 (−1)2+2 (b11 ). " " " " " 2 5 " " 2 0 " " " " " = (−4)(−1)(0) + (1)(1)(2) = 2. Hence, " = (−4)(−1)(5) + (7)(1)(2) = 34 and " " −4 7 " −4 1 " " " 0 5 " " because its coefficient in the determinant formula is zero. It is not necessary to compute "" 1 7 " Hence, |A| = 0 + (3)(−1)(34) + (−3)(1)(2) = −108. ⎡ ⎤ 5 0 0 (g) Let A = ⎣ 3 −2 0 ⎦ . Then |A| = a31 A31 + a32 A32 + a33 A33 −1 8 4 75


3+1

= (−1)(−1)

" " 0 " " −2

" " " 0 "" 3+2 " 5 + (8)(−1) " " 3 0

" " " 0 "" 3+3 " 5 + (4)(−1) " " 3 0

Section 3.1

" 0 "" . Now, for a 2 × 2 matrix −2 "

B, |B| = b21 B21 + b22 B22 = b21 (−1)2+1 (b12 ) + b22 (−1)2+2 (b11 ). " " " " " 0 0 " " 5 0 " " " " " = (3)(−1)(0) + (0)(1)(5) = 0 Hence, " = (−2)(−1)(0) + (0)(1)(0) = 0, " −2 0 " 3 0 " " " " 5 0 "" = (3)(−1)(0) + (−2)(1)(5) = −10. Hence, |A| = (−1)(1)(0) + (8)(−1)(0) + and "" 3 −2 " (4)(1)(−10) = −40. ⎡ ⎤ 3 1 −2 5 ⎦ . Then |A| = a31 A31 + a32 A32 + a33 A33 (i) Let A = ⎣ −1 4 3 1 −2 " " " " " " " 1 −2 " " " " + (1)(−1)3+2 " 3 −2 " + (−2)(−1)3+3 " 3 = (3)(−1)3+1 "" " " " " −1 4 5 −1 5

" 1 "" . Now, for a 2 × 2 4 "

matrix B, |B| = b21 B21 + b22 B22 = b21 (−1)2+1 (b12 ) + b22 (−1)2+2 (b11 ). " " " " " " " 1 −2 " " = (4)(−1)(−2) + (5)(1)(1) = 13, " 3 −2 " = (−1)(−1)(−2) + (5)(1)(3) = Hence, "" " " −1 5 " 4 5 " " " 3 1 " " = (−1)(−1)(1) + (4)(1)(3) = 13. Hence, |A| = (3)(1)(13) + (1)(−1)(13) + 13, and "" −1 4 " (−2)(1)(13) = 0. (j) By definition, the determinant of a 1 × 1 matrix is defined to be the (1,1) entry of the matrix. Thus, the determinant of [−3] equals −3. ⎡ ⎤ 5 2 1 0 ⎢ −1 3 5 2 ⎥ ⎥ (5) (a) Let A = ⎢ ⎣ 4 1 0 2 ⎦ . Then |A| = a41 A41 + a42 A42 + a43 A43 + a44 A44 0 2 3 0 = (0)(−1)4+1 |A41 | + (2)(−1)4+2 |A42 | + (3)(−1)4+3 |A43 | + (0)(−1)4+4 |A44 | ⎡ ⎤ 5 1 0 = 0 + 2|A42 | − 3|A43 | + 0. Now, let B = A42 = ⎣ −1 5 2 ⎦ . Then |A42 | = |B| = 4 0 2 " " " " " " " " " 5 0 " " 5 1 " 3+1 " 1 0 " 3+2 " 3+3 " " " b31 B31 + b32 B32 + b33 B33 = (4)(−1) " 5 2 " + (0)(−1) " −1 2 " + (2)(−1) " −1 5 " ⎡ ⎤ 5 2 0

= (4)(1) (1)(2) − (0)(5) +0+2(1) (5)(5) − (1)(−1) = 60. Also, if C = A43 = ⎣ −1 3 2 ⎦. 4 1 2 Then |A43 | = |C| = c31 C31 + c32 C32 + c33 C33 " " " " " " " " " 2 0 " " " " + (1)(−1)3+2 " 5 0 " +(2)(−1)3+3 " 5 2 " = (4)(−1)3+1 "" " " " " 3 2 −1 2 −1 3 "

= (4)(1) (2)(2) − (3)(0) + (1)(−1) (5)(2) − (0)(−1) + 2(1) (5)(3) − (2)(−1) = 40. Hence, |A| = 2|A42 | − 3|A43 | = 2(60) − 3(40) = 0. 76


Section 3.1

(d) Let A be the given 5 × 5 matrix. Then |A| = a51 A51 + a52 A52 + a53 A53 + a54 A54 + a55 A55 = (0)(−1)5+1 |A51 | + (3)(−1)5+2 |A52 | + (0)(−1)5+3 |A53 | + (0)(−1)5+4 |A54 | + (2)(−1)5+5 |A55 | = (0) + (3)(−1)|A52 | + (0) + (0) + (2)(1)|A55 |. ⎡ ⎤ 0 1 3 −2 ⎢ 2 3 −1 0 ⎥ ⎥ . Then, |A52 | = |B| = b41 B41 + b42 B42 + b43 B43 + b44 B44 Let B = A52 = ⎢ ⎣ 3 2 −5 1 ⎦ 1 −4 0 0 = (1)(−1)4+1 |B41 | + (−4)(−1)4+2 |B42 | + (0)(−1)4+3 |B43 | + (0)(−1)4+4 |B44 | = ⎡ ⎤ 1 3 −2 0 ⎦. We compute |B41 | using (1)(−1)|B41 | + (−4)(1)|B42 | + (0) + (0). Now, B41 = ⎣ 3 −1 2 −5 1 basketweaving. 1

3 ×

3 ×

−1

−2

1 ×

0 ×

−5

2

×

3 ×

1

3 × ×

−5

2

−1

This yields |B41 | = (1)(−1)(1) + (3)(0)(2) + (−2)(3)(−5) − (−2)(−1)(2) − (1)(0)(−5) − (3)(3)(1) = ⎡ ⎤ 0 3 −2 0 ⎦. We compute |B42 | using basketweaving. 16. Next, B42 = ⎣ 2 −1 3 −5 1

0

3

× 2 ×

−1

−2

0

3

2 ×

−1 −5

3

×

× 1

×

0 ×

−5

3

×

This yields |B42 | = (0)(−1)(1) + (3)(0)(3) + (−2)(2)(−5) − (−2)(−1)(3) − (0)(0)(−5) − (3)(2)(1) = 8. Therefore, |A52 | = (1)(−1)|B41 | + (−4)(1)|B42 | = (1)(−1)(16) + (−4)(1)(8) = −48. Now, ⎡ ⎤ 0 4 1 3 ⎢ 2 2 3 −1 ⎥ ⎥. Then |A55 | = |C| = c41 C41 + c42 C42 + c43 C43 + c44 C44 let C = A55 = ⎢ ⎣ 3 1 2 −5 ⎦ 1 0 −4 0 = (1)(−1)4+1 |C41 | + (0)(−1)4+2 |C42 | +(−4)(−1)4+3 |C43 | + (0)(−1)4+4 |C44 |

77


= (1)(−1)4+1 |C41 | + (0) + (−4)(−1)4+3 |C43 | + (0). Now, C41

4 =⎣ 2 1

|C41 | using basketweaving.

4

1

× 2

3 ×

3 ×

1

×

4 ×

−1

⎤ 1 3 3 −1 ⎦. We compute 2 −5

1

× 2

×

−5

2

Section 3.1

3 ×

1

2

This yields |C41 | = (4)(3)(−5)+(1)(−1)(1)+(3)(2)(2)−(3)(3)(1)−(4)(−1)(2)−(1)(2)(−5) = −40. ⎡ ⎤ 0 4 3 Similarly, C43 = ⎣ 2 2 −1 ⎦. We compute |C43 | using basketweaving. 3 1 −5

0

4

× 2

3 ×

2 ×

3

×

0

× −1

4

2

2 ×

3

×

×

−5

1

1

This yields |C43 | = (0)(2)(−5) + (4)(−1)(3) + (3)(2)(1) − (3)(2)(3) − (0)(−1)(1) − (4)(2)(−5) = 16. Therefore, |A55 | = (1)(−1)4+1 |C41 | + (−4)(−1)4+3 |C43 | = (1)(−1)(−40) + (−4)(−1)(16) = 104. Finally, |A| = (0) + (3)(−1)|A52 | + (2)(1)|A55 | = (3)(−1)(−48) + (2)(1)(104) = 352. " " " 1 1 " 1 1 1 0 " = (1)(1) − (1)(1) = 0 and |B| = " (7) Let A = , and let B = . Then |A| = " 1 1 " 1 1 0 1 " " " " " " " 1 0 " " = (1)(1) − (0)(0) = 1. Hence, |A| + |B| = 1. But, |A + B| = " 2 1 " = (2)(2) − (1)(1) = 3. " " 1 2 " " 0 1 " Hence, |A + B| = |A| + |B|. (9) According to part (1) of Theorem 3.1, the area of the parallelogram is the determinant of the matrix whose rows are the two given vectors. Be careful! negative, we must take the absolute value to find the area. " " 3 (a) The area of the parallelogram is the absolute value of the determinant "" 4 equals (3)(5) − (2)(4) = 7. Hence, the area is 7.

78

absolute value of the If the determinant is " 2 "" . This determinant 5 "


Section 3.1

" " 5 (c) The area of the parallelogram is the absolute value of the determinant "" −3 minant equals (5)(3) − (−1)(−3) = 12. Hence, the area is 12.

" −1 "" . This deter3 "

(10) Let x = [x1 , x2 ] and y = [y1 , y2 ]. Consider the hint in the

text. Then the area of the parallelogram is x·y x h, where h = y − projx y. Now, projx y = x2 x =

x·y 1 1 2 x2 [(x1 y1 + x2 y2 )x1 , (x1 y1 + x2 y2 )x2 ]. Hence, h = y − projx y = x2 (x y) − x2 x = 1 1 2 2 2 2 x2 [(x1 + x2 )y1 , (x1 + x2 )y2 ] − x2 [(x1 y1 + x2 y2 )x1 , (x1 y1 + x2 y2 )x2 ] 1 2 2 2 = x2 [x1 y1 + x2 y1 − x1 y1 − x1 x2 y2 , x21 y2 + x22 y2 − x1 x2 y1 − x22 y2 ] = 1 1 2 2 x2 [x2 y1 −x1 x2 y2 , x1 y2 −x1 x2 y1 ] = x2 [x2 (x2 y1 −x1 y2 ), x1 (x1 y2 −x2 y1 )]

=

x1 y2 −x2 y1 [−x2 , x1 ]. x"2 "

" x 2 −x2 y1 | x y − projx y = x |x1 yx x22 + x21 = |x1 y2 − x2 y1 | = absolute value of "" 1 2 y1

Thus,

x2 "" . y2 "

(11) According to part (2) of Theorem 3.1, the volume of the parallelepiped is the absolute value of the determinant of the matrix whose rows are the three given vectors. Be careful! If the determinant is negative, we must take the absolute value to find the volume. " " " −2 3 1 " " " (a) The volume of the parallelepiped is the absolute value of the determinant "" 4 2 0 "". We use " −1 3 2 " basketweaving to compute this.

−2

3

×

4

1 ×

2 ×

−1

× 0

× 3

−2

2 ×

−1

2

3 ×

4 ×

3

The determinant equals (−2)(2)(2)+(3)(0)(−1)+(1)(4)(3)−(1)(2)(−1)−(−2)(0)(3)−(3)(4)(2) = −18. The volume is the absolute value of this result. Hence, the volume equals 18. " " " −3 4 0 "" " (c) The volume of the parallelepiped is the absolute value of the determinant "" 6 −2 1 "". We " 0 −3 3 " use basketweaving to compute this.

−3

4 ×

6 ×

0

× −2

× 1

×

−3

0

79

−3

4

×

−2 −3

0

×

6 ×

3


Section 3.1

The determinant equals (−3)(−2)(3) + (4)(1)(0) + (0)(6)(−3) − (0)(−2)(0) − (−3)(1)(−3) − (4)(6)(3) = −63. The volume is the absolute value of this result. Hence, the volume equals 63. (12) First, " " x1 " " y1 " " z1

read the hint in the textbook. Then note that " x2 x3 "" y2 y3 "" = x1 y2 z3 + x2 y3 z1 + x3 y1 z2 − x3 y2 z1 − x1 y3 z2 − x2 y1 z3 = z2 z 3 "

(x2 y3 − x3 y2 )z1 + (x3 y1 − x1 y3 )z2 + (x1 y2 − x2 y1 )z3 = (x × y)·z (from the definition in Exercise 8). Also note that the formula (x2 y3 − x3 y2 )2 + (x1 y3 − x3 y1 )2 + (x1 y2 − x2 y1 )2 given in the hint for this exercise for the area of the parallelogram (verified below) equals x × y. Hence, the volume of " " " z·(x×y) " the parallelepiped equals proj(x×y) z x × y = " x×y " x × y = |z·(x × y)| = absolute value of " " " x1 x2 x3 " " " " y1 y2 y3 ". " " " z1 z2 z 3 " Now let us verify the formula A = (x2 y3 − x3 y2 )2 + (x1 y3 − x3 y1 )2 + (x1 y2 − x2 y1 )2 for the area of the parallelogram determined by x and y. As in the solution for Exercise 10, above, the area of this parallelogram equals x y − projx y. We must show that A = x y − projx y. We can verify this by a tedious, brute force, argument. (Algebraically expand and simplify x2 y − projx y2 to get (x2 y3 − x3 y2 )2 + (x1 y3 − x3 y1 )2 + (x1 y2 − x2 y1 )2 .) An alternate approach is the following: Now, A2 = (x2 y3 − x3 y2 )2 + (x1 y3 − x3 y1 )2 + (x1 y2 − x2 y1 )2 = x22 y32 − 2x2 x3 y2 y3 + x23 y22 + x21 y32 − 2x1 x3 y1 y3 + x23 y12 + x21 y22 − 2x1 x2 y1 y2 + x22 y12 . Using some algebraic manipulation, we can express this as 2 2 A2 = (x21 +x22 +x23 )(y12 +y22 +y32 )−(x1 y1 +x2 y2 +x3 y3 ). (Verify this!) Therefore, A2 = x y −(x·y)2 . Now, 2 x·y 2 2 2 x y − projx y = x y − 2 x x * 2

= x

y−

(x · y) x x

2

⎛

*

2 = x ⎝(y · y) − 2

* 2

= x

2

2

x (y · y) x

2

+ * ·

y−

x·y x

x

* (x · y) +

2

+

2

+

* −2

(x · y) x

2

(x · y) x

2

+

x·y x

(x · x)⎠

2

2

+

(x · y) x

⎞

+2

+

2

2

= x y − (x · y)2 = A2 . Hence, A = x y − projx y. " " " x 2 "" (15) (a) "" = 0 ⇒ x(x + 3) − (2)(5) = 0 ⇒ x2 + 3x − 10 = 0 ⇒ (x + 5)(x − 2) = 0 ⇒ x = −5 or 5 x+3 " x = 2. 80


Section 3.2

(c) First, we compute the given determinant using basketweaving.

x−3

5 ×

0 × 0

× x−1

−19

×

6 ×

x−3

5

× 0

×

0 ×

x−2

0

x−1 0

The determinant equals (x − 3)(x − 1)(x − 2) + (5)(6)(0) + (−19)(0)(0) − (−19)(x − 1)(0) − (x − 3)(6)(0) − (5)(0)(x − 2) = (x − 3)(x − 1)(x − 2). Setting this determinant equal to zero yields x = 3, x = 1, or x = 2. (16) (b) The given matrix is a 3 × 3 Vandermonde matrix with a = 2,

b= 3, and c = −2. Thus,

when we use part (a), the determinant is (a − b)(b − c)(c − a) = 2 − 3 3 − (−2) (−2) − 2 = 20. (18) (a) False. The basketweaving technique only works to find determinants of 3 × 3 matrices. For larger matrices, the only method we have learned at this point is the cofactor expansion along the last row, although alternate methods are given in Sections 3.2 and 3.3. " " " x1 x2 " " = x1 y2 − x2 y1 . Part (1) of Theorem 3.1 assures us that the absolute " (b) True. Note that " y 1 y2 " value of this determinant gives the area of the desired parallelogram. (c) False in general. (True for n = 2.) An n × n matrix has n2 cofactors – one corresponding to each entry in the matrix. (d) False in general. (True in the special case in which B23 = 0.). By definition, B23 = (−1)2+3 |B23 | = −|B23 |, not |B23 |. (e) True. The given formula is the cofactor expansion of A along the last row. By definition, this equals the determinant of A.

Section 3.2 (1) (a) The row operation used is (II): 1 ← −3 2 + 1 . By Theorem 3.3, performing a type (II) operation does not change the determinant of a matrix. Thus, since |I3 | = 1, the determinant of the given matrix also equals 1. (c) The row operation used is (I): 3 ← −4 3 . By Theorem 3.3, performing a type (I) operation multiplies the determinant by the constant used in the row operation. Thus, since |I3 | = 1, the determinant of the given matrix equals (−4)(1) = −4. (f) The row operation used is (III): 1 ↔ 2 . By Theorem 3.3, performing a type (III) row operation changes the sign of the determinant. Thus, since |I3 | = 1, the determinant of the given matrix equals −1. (2) In each part, we use row operations to put the given matrix into upper triangular form. Notice in the solutions, below, that we stop performing row operations as soon as we obtain an upper triangular 81


Section 3.2

matrix. We give a chart indicating the row operations used, keeping track of the variable P , as described in Example 5 in Section 3.2 of the textbook. Recall that P = 1 at the beginning of the process. We then use the final upper triangular matrix obtained and the value of P to compute the desired determinant. (a)

(I): 1 ←

1 10

1

Multiply P by

(II): 3 ← 5 1 + 3 (I): 2 ←

− 14

P

Effect

Row Operations

1 10

No change Multiply P by − 14

2

(II): 3 ← (−1) 2 + 3

No change

1 10 1 10 1 − 40 1 − 40

⎡

1

⎢ The upper triangular matrix obtained from these operations is B = ⎣ 0 0 3.2, |B| original (c)

= (1)(1)(− 34 ) matrix is P1 ×

=

− 34 .

2 5

1 0

21 10 − 34 − 34

⎤ ⎥ ⎦. By Theorem

Hence, as in Example 5 in the textbook, the determinant of the

|B| = (−40)(− 34 ) = 30.

Row Operations (II): 2 ← 2 1 + 2 (II): 3 ← 3 1 + 3 (II): 4 ← (−2) 1 + 4 (I): 2 ← (−1) 2 (II): 3 ← 2 + 3 (II): 4 ← (−1) 2 + 4 (III): 3 ↔ 4

Effect No change No change No change Multiply P by −1 No change No change Multiply P by −1

P 1 1 1 −1 −1 −1 1

⎡

1 −1 5 ⎢ 0 1 −3 ⎢ The upper triangular matrix obtained from these operations is B = ⎣ 0 0 2 0 0 0

⎤ 1 −3 ⎥ ⎥. By 2 ⎦ −2

Theorem 3.2, |B| = (1)(1)(2)(−2) = −4. Hence, as in Example 5 in the textbook, the determinant of the original matrix is P1 × |B| = 1(−4) = −4. (e)

Row Operations (I): 1 ←

1 5

1

Multiply P by

(II): 2 ← (− 15 2 ) 1 + 2

No change

(II): 3 ←

No change

5 2

1 + 3

(II): 4 ← (−10) 1 + 4 (I): 2 ←

(− 14 ) 2

P

Effect 1 5

No change Multiply P by − 14

(II): 3 ← (−3) 2 + 3

No change

(II): 4 ← 9 2 + 4

No change

(I): 3 ← 4 3

Multiply P by 4

(II): 4 ←

3 4

3 + 4

No change

82

1 5 1 5 1 5 1 5 1 − 20 1 − 20 1 − 20 − 15 − 15


Section 3.2

1

⎢ ⎢ 0 The upper triangular matrix obtained from these operations is B = ⎢ ⎣ 0 0

3 5

1 0 0

4 5 13 4

− 85

⎤

⎥ ⎥ ⎥. By 1 −27 ⎦ 0 −7

− 11 4

Theorem 3.2, |B| = (1)(1)(1)(−7) = −7. Hence, as in Example 5 in the textbook, the determinant of the original matrix is P1 × |B| = (−5)(−7) = 35. (3) In this problem, first compute the determinant by any convenient method. " " " 5 6 "" " (a) " = (5)(−4) − (6)(−3) = −2. Since the determinant is nonzero, Theorem 3.5 implies −3 −4 " that the given matrix is nonsingular. (c) Using basketweaving on the given matrix produces the value (−12)(−1)(−8) + (7)(2)(3) + (−27)(4)(2) − (−27)(−1)(3) − (−12)(2)(2) − (7)(4)(−8) = −79 for the determinant. Since the determinant is nonzero, Theorem 3.5 implies that the given matrix is nonsingular. ⎡ ⎤ −6 3 −22 4 −31 ⎦. Using basketweaving (4) (a) The coefficient matrix for the given system is A = ⎣ −7 11 −6 46 yields |A| = (−6)(4)(46) + (3)(−31)(11) + (−22)(−7)(−6) −(−22)(4)(11) − (−6)(−31)(−6) − (3)(−7)(46) = −1. Because |A| = 0, Corollary 3.6 shows that rank(A) = 3. Hence, by Theorem 2.5, the system has only the trivial solution. (6) Perform the following type (III) row operations on A:

⎡

⎧ ⎢ ⎢ ⎨ (III) : 1 ↔ 6 ⎢ (III) : 2 ↔ 5 . This produces the matrix B = ⎢ ⎢ ⎩ ⎢ (III) : 3 ↔ 4 ⎣

a61 0 0 0 0 0

a62 a52 0 0 0 0

a63 a53 a43 0 0 0

a64 a54 a44 a34 0 0

a65 a55 a45 a35 a25 0

a66 a56 a46 a36 a26 a16

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

Since B is upper triangular, Theorem 3.2 gives |B| = a61 a52 a43 a34 a25 a16 . But applying a type (III) row operation to a matrix changes the sign of its determinant. Since we performed three type (III) row operations, |B| = −(−(−|A|)), or |A| = −|B| = −a61 a52 a43 a34 a25 a16 . (16) (a) False in general, although Theorem 3.2 shows that it is true for upper triangular matrices. For a 1 1 counterexample to the general statement, consider the matrix A = whose determinant 1 1 is (1)(1) − (1)(1) = 0. However, the product of the main diagonal entries of A is (1)(1) = 1. (b) True. Part (3) of Theorem 3.3 shows that performing a type (III) row operation changes the sign of the determinant. Therefore, performing two type (III) row operations in succession changes the sign twice. Hence, there is no overall effect on the determinant. (c) False in general. If A is a 4 × 4 matrix, Corollary 3.4 shows that |3A| = 34 |A| = 81|A|. This equals 3|A| only in the exceptional case in which |A| = 0. The matrix A = I4 provides a specific counterexample to the original statement.

83


Section 3.3

(d) False. If A is a matrix having (row i) = (row j), then |A| = 0 = 1. To prove this, note that performing the type (III) row operation i ↔ j on A results in A. Thus, by part (3) of Theorem 3.3, |A| = −|A|. Hence, |A| = 0. For a specific counterexample to the original statement, consider On , for any n ≥ 2. (e) False. This statement contradicts Theorem 3.5. (f) True. This statement is the contrapositive of Corollary 3.6.

Section 3.3 (1) (a) |A| = a31 A31 + a32 A32 + a33 A33 + a34 A34 = a31 (−1)3+1 |A31 | + a32 (−1)3+2 |A32 | + a33 (−1)3+3 |A33 | + a34 (−1)3+4 |A34 | = a31 |A31 | − a32 |A32 | + a33 |A33 | − a34 |A34 |. (c) |A| = a14 A14 + a24 A24 + a34 A34 + a44 A44 = a14 (−1)1+4 |A14 | + a24 (−1)2+4 |A24 | + a34 (−1)3+4 |A34 | + a44 (−1)4+4 |A44 | = −a14 |A14 | + a24 |A24 | − a34 |A34 | + a44 |A44 |. Then |A| = a21 A21 + a22 A22 + a23 A23 " " " " 2 " −1 4 4 "" 2+2 2+3 " + (3) "" |A22 | + a23 (−1) |A23 | = −(0) " a22 (−1) " 5 −3 −2 −3

0 + (3) (2)(−3) − (4)(5) − (−2) (2)(−2) − (−1)(5) = −76.

(2) (a) Let A be the given matrix.

= a21 (−1)2+1 |A21 | + " " " " " " " − (−2) " 2 −1 " = " 5 −2 " "

(c) Let C be the given matrix. Then |C| = c11 C11 + c21 C21 + c31 C31 = c11 (−1)1+1 |C11 | " " " " " " −2 " −1 −2 " " −2 3 " 3 2+1 3+1 " " " " + (3) "" |C21 | + c31 (−1) |C31 | = (4) " − (5) " + c21 (−1) −1 −2 3 2 " 3 2 "

(4) (−1)(2) − (−2)(3) − (5) (−2)(2) − (3)(3) + (3) (−2)(−2) − (3)(−1) = 102. (3) (a) Let A represent the given matrix. First, " " " 0 −3 " 1+1 " " A11 = (−1) " −2 −33 " = −6, " " " −1 −21 " " = 9, A21 = (−1)2+1 "" −2 −33 " " " " " 3+1 " −1 −21 " A31 = (−1) " 0 −3 " = 3, " " " 2 0 "" A13 = (−1)1+3 "" = −4, 20 −2 "

we compute each " " 1+2 " A12 = (−1) " " " A22 = (−1)2+2 "" " " 3+2 " A32 = (−1) "

" " " = "

cofactor: 2 20 14 20 14 2

" " 14 A23 = (−1)2+3 "" 20 " " 14 A33 = (−1)3+3 "" 2

" −3 "" = 6, −33 " " −21 "" = −42, −33 " " −21 "" = 0, −3 " " −1 "" = 8, −2 " " −1 "" = 2. 0 "

⎡

−6 The adjoint matrix A is the 3 × 3 matrix whose (i, j) entry is Aji . Hence, A = ⎣ 6 −4

⎤ 9 3 −42 0 ⎦. 8 2

Next, to find the determinant, we perform a cofactor expansion along the second column, since a22 = 0. Using the cofactors we have already computed, |A| = a12 A12 + a22 A22 + a32 A32 =

84


Section 3.3

(−1)(6) + (0)(−42) + (−2)(0) = −6. Finally, by Corollary 3.12, ⎡ ⎤ ⎡ 1 −3 −1 ⎤ 2 2 −6 9 3 ⎥ ⎢ 1 1 ⎣ 6 −42 0 ⎦ = ⎣ −1 7 0 ⎦. A = (−6) A−1 = |A| 2 −4 8 2 − 43 − 13 3 (c) Let A represent the given matrix. First, we compute each cofactor. Each of the 3×3 determinants can be calculated using basketweaving. " " " " " −4 " 1 1 4 "" 0 −1 "" " " A21 = (−1)2+1 "" 11 −2 −8 "" = 0, A11 = (−1)1+1 "" 11 −2 −8 "" = −3, " 10 −2 −7 " " 10 −2 −7 " " " " " " 1 " 1 0 −1 "" 0 −1 "" " " 1 4 "" = 3, 1 4 "" = −3, A31 = (−1)3+1 "" −4 A41 = (−1)4+1 "" −4 " 10 −2 −7 " " 11 −2 −8 " " " " " " " −2 7 1 4 "" 0 −1 "" " " A22 = (−1)2+2 "" −14 −2 −8 "" = 0, A12 = (−1)1+2 "" −14 −2 −8 "" = 0, " −12 −2 −7 " " −12 −2 −7 " " " " " " −2 " −2 0 −1 "" 0 −1 "" " " 7 1 4 "" = 0, 7 1 4 "" = 0, A42 = (−1)4+2 "" A32 = (−1)3+2 "" " −12 −2 −7 " " −14 −2 −8 " " " " " " " −2 7 −4 4 "" 1 −1 "" " " A23 = (−1)2+3 "" −14 11 −8 "" = 0, A13 = (−1)1+3 "" −14 11 −8 "" = −3, " −12 10 −7 " " −12 10 −7 " " " " " " −2 " −2 1 −1 "" 1 −1 "" " " 7 −4 4 "" = 3, 7 −4 4 "" = −3, A33 = (−1)3+3 "" A43 = (−1)4+3 "" " −12 10 −7 " " −14 11 −8 " " " " " " " −2 7 −4 1 "" 1 0 "" " " A14 = (−1)1+4 "" −14 11 −2 "" = 6, A24 = (−1)2+4 "" −14 11 −2 "" = 0, " −12 10 −2 " " −12 10 −2 " " " " " " −2 " −2 1 0 "" 1 0 "" " " 7 −4 1 "" = −6, 7 −4 1 "" = 6. A44 = (−1)4+4 "" A34 = (−1)3+4 "" " −12 10 −2 " " −14 11 −2 " ⎡ ⎤ −3 0 3 −3 ⎢ 0 0 0 0 ⎥ ⎥. The adjoint matrix A is the 4 × 4 matrix whose (i, j) entry is Aji . So, A = ⎢ ⎣ −3 0 3 −3 ⎦ 6 0 −6 6 Next, we use a cofactor expansion along the first row to compute the determinant of A because a13 = 0. Using the cofactors we calculated above: |A| = a11 A11 + a12 A12 + a13 A13 + a14 A14 = (−2)(−3) + (1)(0) + (0)(−3) + (−1)(6) = 0. Finally, since |A| = 0, A has no inverse, by Theorem 3.5. (e) Let A represent the given matrix. First, we compute each cofactor:

85


A11 A21 A31 A13

" " " −3 2 "" " = (−1) " 0 −1 " = 3, " " " 0 "" 2+1 " −1 = (−1) " 0 −1 " = −1, " " " " 3+1 " −1 0 " = (−1) " −3 2 " = −2, " " " 0 −3 " " = 0, = (−1)1+3 "" 0 0 " 1+1

A12 A22 A32

" " " = (−1) " " " 2+2 " = (−1) " " " 3+2 " = (−1) " 1+2

0 0 3 0 3 0

" " 3 A23 = (−1)2+3 "" 0 " " 3 A33 = (−1)3+3 "" 0

" 2 "" = 0, −1 " " 0 "" = −3, −1 " " 0 "" = −6, 2 " " −1 "" = 0, 0 " " −1 "" = −9. −3 "

Section 3.3

⎡

3 The adjoint matrix A is the 3 × 3 matrix whose (i, j) entry is Aji . Hence, A = ⎣ 0 0

−1 −3 0

⎤ −2 −6 ⎦. −9

Next, since A is upper triangular, |A| = (3)(−3)(−1) = 9. Finally, by Corollary 3.12, ⎡ 1 ⎤ ⎡ ⎤ − 19 − 29 3 3 −1 −2 ⎥ ⎢ 1 2 ⎥ 1 A = 19 ⎣ 0 −3 −6 ⎦ = ⎢ A−1 = |A| ⎣ 0 − 3 − 3 ⎦. 0 0 −9 0 0 −1 ⎡

⎤ 3 −1 −1 (4) (a) Let A be the coefficient matrix ⎣ 2 −1 −2 ⎦. The matrices A1 , A2 , and A3 are formed −9 1 0 by removing, respectively, the first, second, and third columns, from A, and replacing them with ⎡ ⎤ ⎡ ⎤ −8 −1 −1 3 −8 −1 3 −2 ⎦ , [−8, 3, 39] (as a column vector). Hence, A1 = ⎣ 3 −1 −2 ⎦, A2 = ⎣ 2 39 1 0 −9 39 0 ⎡ ⎤ 3 −1 −8 3 ⎦. Basketweaving, or any other convenient method, can be used to and A3 = ⎣ 2 −1 −9 1 39 compute the determinant of each of these matrices, yielding |A| = −5, |A1 | = 20, |A2 | = −15, and |A3 | = 35. Therefore, Cramer’s Rule states that the unique solution to the system is given by x1 =

|A1 | |A|

=

20 (−5)

set is {(−4, 3, −7)}.

= −4, x2 =

|A2 | |A|

=

(−15) (−5)

= 3, and x3 =

|A3 | |A|

=

35 (−5)

= −7. The full solution

⎡

⎤ −5 2 −2 1 ⎢ 2 −1 2 −2 ⎥ ⎥. The matrices A1 , A2 , A3 , and A4 (d) Let A be the coefficient matrix ⎢ ⎣ 5 −2 3 −1 ⎦ −6 2 −2 1 are formed by removing, respectively, the first, second, third, and fourth columns, from A, and replacing them with [−10, −9, 7, −14] (as a column vector). Hence,

86


Section 3.3

⎡

⎡ ⎡ ⎤ ⎤ ⎤ −10 2 −2 1 −5 −10 −2 1 −5 2 −10 1 ⎢ −9 −1 ⎢ ⎢ 2 −2 ⎥ −9 2 −2 ⎥ −9 −2 ⎥ ⎥, A2 = ⎢ 2 ⎥ , A3 = ⎢ 2 −1 ⎥ A1 = ⎢ ⎣ ⎣ 5 ⎣ 5 −2 7 −2 3 −1 ⎦ 7 3 −1 ⎦ 7 −1 ⎦ −14 2 −2 1 −6 −14 −2 1 −6 2 −14 1 ⎡ ⎤ −5 2 −2 −10 ⎢ 2 −1 2 −9 ⎥ ⎥. Row reduction or cofactor expansion can be used to compute and A4 = ⎢ ⎣ 5 −2 3 7 ⎦ −6 2 −2 −14 the determinant of each of these matrices, yielding |A| = 3, |A1 | = 12, |A2 | = −3, |A3 | = −9, and |A4 | = 18. Therefore, Cramer’s Rule states that the unique solution to the system is given by x1 =

|A1 | |A|

=

12 3

= 4, x2 =

|A2 | |A|

=

(−3) 3

= −1, x3 =

|A3 | |A|

=

(−9) 3

The full solution set is {(4, −1, −3, 6)}.

(8) (b) Try n = 2. All 2 × 2 skew-symmetric matrices are of the form 0 −1 A= . Note that |A| = 1 = 0. 1 0

= −3, and x4 = 0 −a a 0

⎡

|A4 | |A|

=

18 3

= 6.

. Letting a = 1 yields

0 (9) (b) Use trial and error, rearranging the rows of I3 . For example, A = ⎣ 1 0

1 0 0

⎤ 0 0 ⎦ works, since 1

AAT = AA (because A is symmetric) = I3 . (Note: All of the four other matrices obtained by rearranging the rows of I3 are also orthogonal matrices!) (13) (b) Simply choose any nonsingular 2 × 2 matrix P and compute B = P−1 AP. For example, using 1 −1 2 1 (see Theorem 2.13), yielding P= , we get P−1 = −1 2 1 1 1 −1 1 2 2 1 −6 −4 2 −5 = . Similarly, letting P = B = P−1 AP = −1 2 3 4 1 1 16 11 −1 3 3 5 1 2 2 −5 10 −12 produces B = = . 1 2 3 4 −1 3 4 −5 1 1 A, and B−1 = |B| B. Hence, by part (3) of Theorem 2.11, (14) By Corollary 3.12, A−1 = |A|

1 1 −1 −1 −1 (AB) = B A = |B| B |A| A = BA/(|A| |B|).

(18) (b) If we try a skew-symmetric 2 × 2 matrix as a possible example, we find that the adjoint is ⎡ ⎤ 0 1 1 0 1 ⎦. skew-symmetric. So we consider a skew-symmetric 3 × 3 matrix, such as A = ⎣ −1 −1 −1 0 Then each of the nine minors |Aij | is easily seen to equal 1. Hence, the (i, j) entry of A is ⎡ ⎤ 1 −1 1 1 −1 ⎦, which is not Aji = (−1)j+i |Aji | = (−1)j+i (1) = (−1)j+i . Therefore, A = ⎣ −1 1 −1 1 skew-symmetric.

87


(22) (a) True. Corollary 3.8 shows that |A−1 | =

1 |A| .

Section 3.3

But Theorem 3.9 asserts that |A| = |AT |. Combining

these two equations produces the desired result. (b) True. According to Theorem 3.10, both cofactor expansions produce |A| as an answer. (The size of the square matrix is not important.) (c) False. Type (III) column operations applied to square matrices change the sign of the determinant, just as type (III) row operations do. Hence, |B| = −|A|. The equation |B| = |A| can be true only in the special case in which |A| = 0. For a particular counterexample to |B| = |A|, consider 0 1 . Note that A = I2 and use the column operation col . 1 ↔ col . 2 . Then B = 1 0 |B| = −1, but |A| = 1. (d) True. The (i, j) entry of the adjoint is defined to be Aji = (−1)j+i |Aji |, which clearly equals (−1)i+j |Aji |. (e) False. Theorem 3.11 states that AA = |A| I. Hence, the equation AA = I can only be true when |A| = 1. For a specific counterexample, consider A = 2I2 . A very short computation shows that A = 2I2 as well. Hence, AA = (2I2 ) (2I2 ) = 4I2 = I2 . (f) True. This follows directly from Cramer’s Rule. The coefficient matrix A for the given system is upper triangular, and so |A| is easily found to be (4)(−3)(1) = −12 (see Theorem 3.2). Notice ⎡ ⎤ 4 −6 −1 1 2| 5 4 ⎦. Cramer’s Rule then states that x2 = |A that A2 = ⎣ 0 |A| = − 12 |A2 |. 0 3 1 (23) (a) Let R be the type (III) row operation k ←→ k − 1 , let A be an n × n matrix, and let B = R(A). Then, by part (3) of Theorem 3.3, |B| = (−1)|A|. Next, notice that the submatrix A(k−1)j = Bkj because the (k − 1)st row of A becomes the kth row of B, implying the same row is being eliminated from both matrices, and since all other rows maintain their original relative positions (notice the same column is being eliminated in both cases). Hence, a(k−1)1 A(k−1)1 + a(k−1)2 A(k−1)2 + · · · + a(k−1)n A(k−1)n = bk1 (−1)(k−1)+1 |A(k−1)1 | + bk2 (−1)(k−1)+2 |A(k−1)2 | + · · · + bkn (−1)(k−1)+n |A(k−1)n | (because a(k−1)j = bkj for 1 ≤ j ≤ n) = bk1 (−1)k |Bk1 | + bk2 (−1)k+1 |Bk2 | + · · · + bkn (−1)k+n−1 |Bkn | = (−1)(bk1 (−1)k+1 |Bk1 | + bk2 (−1)k+2 |Bk2 | + · · · + bkk (−1)k+n |Bkn |) = (−1)|B| (by applying Theorem 3.10 along the kth row of B) = (−1)|R(A)| = (−1)(−1)|A| = |A|, finishing the proof. (b) The definition of the determinant is the Base Step for an induction proof on the row number, counting down from n to 1. Part (a) is the Inductive Step. (24) Suppose row i of A equals row j of A. Let R be the type (III) row operation i ←→ j : Then, clearly, A = R(A). Thus, by part (3) of Theorem 3.3, |A| = −|A|, or 2|A| = 0, implying |A| = 0. (25) Let A, i, j, and B be as given in the exercise and its hint. Then by Exercise 24, |B| = 0, since its ith and jth rows are the same. Also, since every row of A equals the corresponding row of B, with the exception of the jth row, the submatrices Ajk and Bjk are equal for 1 ≤ k ≤ n. Hence, Ajk = Bjk for 1 ≤ k ≤ n. Now, computing the determinant of B using a cofactor expansion along the jth row (allowed by Exercise 23) yields 0 = |B| = bj1 Bj1 + bj2 Bj2 + · · · + bjn Bjn = ai1 Bj1 + ai2 Bj2 + · · · + ain Bjn (because the jth row of B equals the ith row of A) = ai1 Aj1 + ai2 Aj2 + · · · + ain Ajn , completing the proof. (26) Using the fact that the general (k, m) entry of A is Amk , we see that the (i, j) entry of AA equals ai1 Aj1 + ai2 Aj2 + · · · + ain Ajn . If i = j, Exercise 23 implies that this sum equals |A|, while if i = j, 88


Section 3.3

Exercise 25 implies that the sum equals 0. Hence, AA equals |A| on the main diagonal and 0 off the main diagonal, yielding (|A|)In .

1 A = In . (27) Suppose A is nonsingular. Then |A| = 0 by Theorem 3.5. Thus, by Exercise 26, A |A|

1 Therefore, by Theorem 2.9, |A| A A = In , implying AA = (|A|)In . (28) Using the fact that the general (k, m) entry of A is Amk , we see that the (j, j) entry of AA equals a1j A1j + a2j A2j + · · · + anj Anj . But all main diagonal entries of AA equal |A| by Exercise 27. (29) Let A be a singular n × n matrix. Then |A| = 0 by Theorem 3.5. By part (4) of Theorem 2.11, AT is also singular (or else A = (AT )T is nonsingular). Hence, |AT | = 0 by Theorem 3.5, and so |A| = |AT | in this case. (30) Case 1: Assume 1 ≤ k < j and 1 ≤ i < m. Then the (i, k) entry of (Ajm )T = (k, i) entry of Ajm = (k, i) entry of A = (i, k) entry of AT = (i, k) entry of (AT )mj . Case 2: Assume j ≤ k < n and 1 ≤ i < m. Then the (i, k) entry of (Ajm )T = (k, i) entry of Ajm = (k + 1, i) entry of A = (i, k + 1) entry of AT = (i, k) entry of (AT )mj . Case 3: Assume 1 ≤ k < j and m < i ≤ n. Then the (i, k) entry of (Ajm )T = (k, i) entry of Ajm = (k, i + 1) entry of A = (i + 1, k) entry of AT = (i, k) entry of (AT )mj . Case 4: Assume j ≤ k < n and m < i ≤ n. Then the (i, k) entry of (Ajm )T = (k, i) entry of Ajm = (k + 1, i + 1) entry of A = (i + 1, k + 1) entry of AT = (i, k) entry of (AT )mj . Hence, the corresponding entries of (Ajm )T and (AT )mj are all equal, proving that the matrices themselves are equal. (31) Suppose A is a nonsingular n × n matrix. Then, by part (4) of Theorem 2.11, AT is also nonsingular. We prove |A| = |AT | by induction on n. Base Step: Assume n = 1. Then A = [a11 ] = AT , and so their determinants must be equal. Inductive Step: Assume that |B| = |BT | for any (n − 1) × (n − 1) nonsingular matrix B, and prove that |A| = |AT | for any n × n nonsingular matrix A. Now, first note that, by Exercise 30, |(AT )ni | = |(Ain )T |. But, |(Ain )T | = |Ain |, either by the inductive hypothesis, if Ain is nonsingular, or by Exercise 29, if Ain is singular. Hence, |(AT )ni | = |Ain |. Let D = AT . So, |Dni | = |Ain |. Then, by Exercise 28, |A| = a1n A1n + a2n A2n + · · · + ann Ann = dn1 (−1)1+n |A1n | + dn2 (−1)2+n |A2n | + · · · + dnn (−1)n+n |Ann | = dn1 (−1)1+n |Dn1 | + dn2 (−1)2+n |Dn2 | + · · · + dnn (−1)n+n |Dnn | = |D|, since this is the cofactor expansion for |D| along the last row of D. This completes the proof. (32) Let A be an n × n singular matrix. Let D = AT . D is also singular by part (4) of Theorem 2.11 (or else A = DT is nonsingular). Now, Exercise 30 shows that |Djk | = |(Akj )T |, which equals |Akj |, by Exercise 29 if Akj is singular, or by Exercise 31 if Akj is nonsingular. So, |A| = |D| (by Exercise 29) = dj1 Dj1 + dj2 Dj2 + · · · + djn Djn (by Exercise 23) = a1j (−1)j+1 |Dj1 | + a2j (−1)j+2 |Dj2 | + · · · + anj (−1)j+n |Djn | = a1j (−1)1+j |A1j | + a2j (−1)2+j |A2j | + · · · + anj (−1)n+j |Anj | = a1j A1j + a2j A2j + · · · + anj Anj , and we are finished. (33) If A has two identical columns, then AT has two identical rows. Hence, |AT | = 0 by Exercise 24. Thus, |A| = 0, by Theorem 3.9 (which is proven in Exercises 29 and 31). (34) Let B be the n × n matrix, all of whose entries are equal to the corresponding entries in A, except that the jth column of B equals the ith column of A (as does the ith column of B). Then |B| = 0 by Exercise 33. Also, since every column of A equals the corresponding column of B, with the exception of the jth column, the submatrices Akj and Bkj are equal for 1 ≤ k ≤ n. Hence, Akj = Bkj for 1 ≤ k ≤ n. Now, computing the determinant of B using a cofactor expansion along the jth column (allowed by part (2) of Theorem 3.10, proven in Exercises 28 and 32) yields 0 = |B| = b1j B1j + b2j B2j + · · · + bnj Bnj 89


Section 3.4

= a1i B1j + a2i B2j + · · · + ani Bnj (because the jth column of B equals the ith column of A) = a1i A1j + a2i A2j + · · · + ani Anj , completing the proof. (35) Using the fact that the general (k, m) entry of A is Amk , we see that the (i, j) entry of AA equals a1j A1i + a2j A2i + · · · + anj Ani . If i = j, Exercise 32 implies that this sum equals |A|, while if i = j, Exercise 34 implies that the sum equals 0. Hence, AA equals |A| on the main diagonal and 0 off the main diagonal, and so AA = (|A|)In . (Note that, since A is singular, |A| = 0, and so, in fact, AA = On .) (36) (a) By Exercise 27, 1 AB. X = |A|

1 |A| AA

= In .

Hence, AX = B ⇒

1 |A| AAX

=

1 |A| AB

⇒ In X =

1 |A| AB

⇒

(b) Using part (a) and the fact that the general (k, m)th entry of A is Amk , we see that the kth entry 1 1 AB = |A| (b1 A1k + · · · + bn Ank ). of X equals the kth entry of |A| (c) The matrix Ak is defined to equal A in every entry, except in the kth column, which equals the column vector B. Thus, since this kth column is ignored when computing (Ak )ik , the (i, k) submatrix of Ak , we see that (Ak )ik = Aik for each 1 ≤ i ≤ n. Hence, computing |Ak | by performing a cofactor expansion down the kth column of Ak (by part (2) of Theorem 3.10) yields |Ak | = b1 (−1)1+k |(Ak )1k | + b2 (−1)2+k |(Ak )2k | + · · · + bn (−1)n+k |(Ak )nk | = b1 (−1)1+k |A1k | + b2 (−1)2+k |A2k | + · · · + bn (−1)n+k |Ank | = b1 A1k + b2 A2k + · · · + bn Ank . (d) Theorem 3.13 claims that the kth entry of the solution X of AX = B is |Ak |/|A|. Part (b) gives 1 (b1 A1k + · · · + bn Ank ) for this kth entry, and part (c) replaces the expression |A| (b1 A1k + · · · + bn Ank ) with |Ak |.

Section 3.4 (1) In each part, let A represent the given matrix. " " " x−3 −1 "" = (x − 3)(x − 4) − (2)(−1) = x2 − 7x + 14. (a) pA (x) = |xI2 − A| = "" 2 x−4 " " " " x−2 −1 1 "" " x − 6 0 "". At this point, we could use basketweaving, but instead, (c) pA (x) = |xI3 −A| = "" 6 " −3 0 x " we will use a cofactor expansion on the last column. Hence, " " " " " 6 x−6 " " −1 "" " + (0) + x(−1)3+3 " x − 2 pA (x) = (1)(−1)1+3 "" " 6 −3 0 " x−6 "

= (1)(0 + 3(x − 6)) + x (x − 2)(x − 6) + 6 = x3 − 8x2 + 21x − 18. " " " x 1 0 −1 "" " " 5 x−2 1 −2 "" . Following the hint, we do a cofactor expansion (e) pA (x) = |xI4 − A| = "" −1 x − 1 0 "" " 0 " −4 1 −3 x "" " " x 0 −1 "" " 1 −2 "" along the third row: pA (x) = (0) + (−1)(−1)3+2 "" 5 " −4 −3 x " " " " x 1 −1 "" " + (x − 1)(−1)3+3 "" 5 x − 2 −2 "" + (0). Using basketweaving produces pA (x) = " −4 1 x " 90


Section 3.4

(1) (x)(1)(x) + (0)(−2)(−4) + (−1)(5)(−3) −(−1)(1)(−4) − (x)(−2)(−3) − (0)(5)(x) +

(x − 1) (x)(x − 2)(x) + (1)(−2)(−4) + (−1)(5)(1) −(−1)(x − 2)(−4) − (x)(−2)(1) − (1)(5)(x) , which, after algebraic simplification, equals x4 − 3x3 − 4x2 + 12x. (2) In each part, let A represent the given matrix. Then, to find the eigenspace Eλ , we form the matrix [(λIn − A)|0] and row reduce to solve the associated linear system. " " 1 −1 "" 0 1 −1 "" 0 . This row reduces to , (a) Since λ = 2, [(λI2 −A)|0] = [(2I2 −A)|0] = 2 −2 " 0 0 0 " 0 ) x1 − x2 = 0 which corresponds to the system . The independent variable is x2 . Let 0 = 0 x2 = b. The first equation then yields x1 = b as well. Hence, E2 , the solution set for this system, is {b[1, 1] | b ∈ R}. A fundamental eigenvector for λ = 2 is [1, 1]. " ⎤ ⎡ 4 −2 0 "" 0 (c) Since λ = −1, [(λI3 − A)|0] = [((−1)I3 − A)|0] = ⎣ 8 −4 0 "" 0 ⎦ . This reduces to −4 2 0 " 0 ⎧ " ⎤ ⎡ 1 − 12 0 " 0 ⎨ x1 − 12 x2 = 0 " " ⎦ ⎣ 0 0 = 0 . The second and 0 0 " 0 , which corresponds to the system ⎩ 0 = 0 0 0 0 " 0 third columns are nonpivot columns, and so x2 and x3 are independent variables. Let x2 = b and x3 = c. The first equation then yields x1 =

1 2 b.

We can eliminate fractions by letting b = 2t,

giving x1 = t and x2 = 2t. Hence, E−1 , the solution set for this system, is {[t, 2t, c] | c, t ∈ R} = {t[1, 2, 0] + c[0, 0, 1] | c, t ∈ R}. Fundamental eigenvectors for λ = −1 are [1, 2, 0] and [0, 0, 1]. (3) In each part, let A represent the given matrix. We will follow the first three steps of the Diagonalization Method, as described in Section 3.4 of the textbook. " " " x−1 −3 "" " = (x − 1)2 . The eigenvalues of A are the roots of pA (x). (a) pA (x) = |xI2 − A| = " 0 x−1 " Clearly, λ = 1 is the only root of (x − 1)2 . Since the linear factor (x − 1) appears to the 2nd power in pA (x), this eigenvalue has algebraic multiplicity 2. Next, we solve for the eigenspace E1 by solving the homogeneous system whose augmented matrix is [(1I2 − A)|0]. Now [(1I2 − A)|0] = " " ) x2 = 0 0 1 "" 0 0 −3 "" 0 . The associated linear system is , which reduces to . 0 0 " 0 0 = 0 0 0 " 0 Since column 1 is not a pivot column, x1 is an independent variable. Let x1 = a. Then the solution set is E1 = {[a, 0] | a ∈ R} = {a[1, 0] | a ∈ R}. A fundamental eigenvector for λ = 1 is [1, 0]. " " " x−1 0 −1 "" " x−2 3 "" = (x − 1)(x − 2)(x + 5), where we have used the (c) pA (x) = |xI3 − A| = "" 0 " 0 0 x+5 " fact that |xI3 − A| is upper triangular to calculate the determinant quickly, using Theorem 3.2. The eigenvalues of A are the roots of pA (x). These are clearly λ1 = 1, λ2 = 2, and λ3 = −5. 91


Section 3.4

Since each of the linear factors (x − 1), (x − 2), and (x + 5) appears in pA (x) raised to the 1st power, each of these eigenvalues has algebraic multiplicity 1. Next, we solve for each eigenspace. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. Now " " ⎤ ⎡ ⎤ ⎡ 0 1 0 "" 0 0 0 −1 "" 0 3 "" 0 ⎦ , which reduces to ⎣ 0 0 1 "" 0 ⎦. The associated linear [(1I3 −A)|0] = ⎣ 0 −1 0 0 6 " 0 0 0 0 " 0 ⎧ = 0 ⎨ x2 x3 = 0 . Since column 1 is not a pivot column, x1 is an independent variable. system is ⎩ 0 = 0 Let x1 = a. Then the solution set is E1 = {[a, 0, 0] | a ∈ R} = {a[1, 0, 0] | a ∈ R}. A fundamental eigenvector for λ1 = 1 is [1, 0, 0]. For λ2 = 2, we solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 0 1 0 −1 "" 0 3 "" 0 ⎦ , which reduces to ⎣ 0 0 1 "" 0 ⎦. The associated linear [(2I3 − A)|0] = ⎣ 0 0 0 0 0 " 0 0 0 7 " 0 ⎧ = 0 ⎨ x1 x3 = 0 . Since column 2 is not a pivot column, x2 is an independent variable. system is ⎩ 0 = 0 Let x2 = b. Then the solution set is E2 = {[0, b, 0] | b ∈ R} = {b[0, 1, 0] | a ∈ R}. A fundamental eigenvector for λ2 = 2 is [0, 1, 0]. For λ3 = −5, we solve the homogeneous system whose augmented matrix is [((−5)I3 − A)|0]. " ⎤ ⎡ " ⎡ ⎤ 1 " 1 0 6 " 0 −6 0 −1 "" 0 ⎥ " ⎢ 3 "" 0 ⎦ , which reduces to ⎣ 0 1 − 37 " 0 ⎦. The Now [((−5)I3 − A)|0] = ⎣ 0 −7 " 0 0 0 " 0 0 0 0 " 0 ⎧ + 16 x3 = 0 ⎪ ⎨ x1 x2 − 37 x3 = 0 . Since column 3 is not a pivot column, associated linear system is ⎪ ⎩ 0 = 0 x3 is an independent variable. Let x3 = c. Then x1 = − 16 x3 = − 16 c and x2 = 37 x3 = 37 c. , - , Hence, the solution set is E−5 = − 16 c, 37 c, c | c ∈ R = c − 16 , 37 , 1 | c ∈ R . A fundamental eigenvector for λ3 = −5 is − 16 , 37 , 1 . We can eliminate fractions by multiplying by 42 to obtain the fundamental eigenvector [−7, 18, 42] instead. " " " x−4 0 2 "" " x−2 6 "" . Cofactor expansion along the second column (e) pA (x) = |xI3 − A| = "" −6 " −4 0 x+2 " " "

" 2 "" 2+2 " x − 4 + (0) = (x − 2) (x − 4)(x + 2) − (−4)(2) = yields pA (x) = (0) + (x − 2)(−1) " −4 x+2 " (x − 2)(x2 − 2x) = x(x − 2)2 . The eigenvalues of A are the roots of pA (x). These are clearly λ1 = 0, and λ2 = 2. Since the linear factor (x − 0) = x appears in pA (x) raised to the 1st power, λ1 = 0 has algebraic multiplicity 1. Similarly, since the linear factor (x − 2) appears to the 2nd power in pA (x), λ2 = 2 has algebraic multiplicity 2. Next, we solve for each eigenspace.

92


Section 3.4

For λ1 = 0, we solve the homogeneous system whose augmented matrix is [(0I3 − A)|0] = [−A|0]. " ⎤ ⎡ " ⎡ ⎤ 1 0 − 12 "" 0 −4 0 2 "" 0 " ⎥ ⎢ Now [−A|0] = ⎣ −6 −2 6 "" 0 ⎦ , which reduces to ⎣ 0 1 − 32 " 0 ⎦. The associated " −4 0 2 " 0 0 0 0 " 0 ⎧ − 12 x3 = 0 ⎪ ⎨ x1 x2 − 32 x3 = 0 . Since column 3 is not a pivot column, x3 is an linear system is ⎪ ⎩ 0 = 0 independent variable. Let x3 = 2c, where the coefficient “2” is used to eliminate fractions. Then x1 =

1 2 x3

= c, and x2 =

3 2 x3

= 3c. Hence, the solution set is E0 = {[c, 3c, 2c] | c ∈ R} =

{c[1, 3, 2] | c ∈ R}. A fundamental eigenvector for λ1 = 0 is [1, 3, 2]. For λ2 = 2, we solve the ⎡ −2 0 [(2I3 −A)|0] = ⎣ −6 0 −4 0 ⎧ ⎨ x1 − x3 0 system is ⎩ 0

homogeneous system whose augmented matrix is [(2I3 − A)|0]. Now " " ⎤ ⎤ ⎡ 1 0 −1 "" 0 2 "" 0 0 "" 0 ⎦. The associated linear 6 "" 0 ⎦ , which reduces to ⎣ 0 0 0 0 0 " 0 4 " 0 = 0 = 0 . Since columns 2 and 3 are not pivot columns, x2 and x3 are = 0

independent variables. Let x2 = b and x3 = c. Then x1 = x3 = c, and the solution set is E2 = {[c, b, c] | b, c ∈ R} = {b[0, 1, 0] + c[1, 0, 1] | b, c ∈ R}. A pair of fundamental eigenvectors for λ2 = 2 is [0, 1, 0] and [1, 0, 1]. " " " x−3 1 −4 1 "" " " 0 x−3 3 −3 "" . We use a cofactor expansion along the first (h) pA (x) = |xI4 − A| = "" 6 −2 x + 8 −2 "" " " 6 4 2 x+4 " " " " x−3 3 −3 "" " x+8 −2 "" + (0) column. This gives pA (x) = (x − 3)(−1)1+1 "" −2 " 4 2 x+4 " " " " " " 1 " 1 −4 1 "" −4 1 "" " " −3 "" + (6)(−1)4+1 "" x − 3 3 −3 "". Using basketweaving to +(6)(−1)3+1 "" x − 3 3 " " 4 " 2 x+4 −2 x + 8 −2 " find each of the 3 × 3 determinants, we get pA (x) = (x − 3) (x − 3)(x + 8)(x + 4) + (3)(−2)(4)

+(−3)(−2)(2) − (−3)(x + 8)(4) − (x − 3)(−2)(2) − (3)(−2)(x + 4) + 6 (1)(3)(x + 4)

+(−4)(−3)(4) + (1)(x − 3)(2) − (1)(3)(4) − (1)(−3)(2) − (−4)(x − 3)(x + 4) − 6 (1)(3)(−2)

+(−4)(−3)(−2) + (1)(x − 3)(x + 8) − (1)(3)(−2) − (1)(−3)(x + 8) − (−4)(x − 3)(−2) = (x − 3)(x3 + 9x2 + 18x) + 6(4x2 + 9x) − 6(x2 ) = x4 + 6x3 + 9x2 = x2 (x2 + 6x + 9) = x2 (x + 3)2 . The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 0 and λ2 = −3. Since each of the linear factors (x − 0) = x and (x + 3) appears in pA (x) raised to the 2nd power, each of these eigenvalues has algebraic multiplicity 2. Next, we solve for each eigenspace.

93


For λ1 = 0, we solve the homogeneous ⎡ −3 1 −4 ⎢ 0 −3 3 [−A|0]. Now [−A|0] = ⎢ ⎣ 6 −2 8 6 4 2 ⎧ x1 + ⎪ ⎪ ⎨ x2 − associated linear system is ⎪ ⎪ ⎩

system whose augmented " ⎡ ⎤ 1 "" 0 ⎢ −3 "" 0 ⎥ ⎥ , reducing to ⎢ " ⎣ ⎦ −2 " 0 " 4 0 x3 x3

+ x4 0 0

= = = =

Section 3.4

matrix is [(0I4 − A)|0] = " ⎤ 1 0 1 0 "" 0 0 1 −1 1 "" 0 ⎥ ⎥. The 0 0 0 0 "" 0 ⎦ 0 0 0 0 " 0

0 0 . Since columns 3 and 4 are not 0 0

pivot columns, x3 and x4 are independent variables. Let x3 = c and x4 = d. Thus, x1 = −x3 = −c and x2 = x3 − x4 = c − d. Hence, the solution set is E0 = {[−c, c − d, c, d] | c, d ∈ R} = {c[−1, 1, 1, 0]+d[0, −1, 0, 1] | c, d ∈ R}. A pair of fundamental eigenvectors for λ1 = 0 is [−1, 1, 1, 0] and [0, −1, 0, 1]. For λ2 = −3, we solve the homogeneous system whose augmented matrix is [((−3)I4 − A)|0]. " ⎡ ⎡ ⎤ ⎤ 1 "" 1 0 0 −6 1 −4 1 "" 0 2 " 0 ⎢ ⎢ 0 −6 3 −3 "" 0 ⎥ 0 "" 0 ⎥ ⎥ , reducing to ⎢ 0 1 0 ⎥. The Now [((−3)I4 − A)|0] = ⎢ " ⎣ ⎣ 6 −2 ⎦ 5 −2 " 0 0 0 1 −1 "" 0 ⎦ 6 4 2 1 " 0 0 0 0 0 " 0 ⎧ x1 + 12 x4 = 0 ⎪ ⎪ ⎨ x2 = 0 . Since column 4 is not a pivot associated linear system is ⎪ − x = 0 x 3 4 ⎪ ⎩ 0 = 0 column, x4 is an independent variable. Let x4 = 2d, where we have included the factor “2” to eliminate fractions. Thus, x1 = − 12 x4 = −d, x2 = 0, and x3 = x4 = 2d. Hence, the solution set is E−3 = {[−d, 0, 2d, 2d] | d ∈ R} = {d[−1, 0, 2, 2] is [−1, 0, 2, 2]. " " x − 19 48 (4) (a) Step 1: pA (x) = |xI2 − A| = "" −8 x + 21 (x − 3)(x + 5).

| d ∈ R}. A fundamental eigenvector for λ2 = −3 " " " = (x − 19)(x + 21) − (48)(−8) = x2 + 2x − 15 = "

Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 3 and λ2 = −5. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 3, we solve the −16 [(3I2 − A)|0] = −8 ) x1 − 3x2 system is 0

homogeneous system whose augmented matrix is [(3I2 − A)|0]. Now " " 1 −3 "" 0 48 "" 0 . The associated linear , which reduces to 0 0 " 0 24 " 0 = 0 . Since column 2 is not a pivot column, x2 is an independent = 0

variable. Let x2 = 1. Then x1 = 3x2 = 3. This yields the fundamental eigenvector [3, 1]. For λ2 = −5, we solve the homogeneous system whose augmented matrix is [((−5)I2 −A)|0]. Now

94

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition [((−5)I2 − A)|0] = ) system is

x1

" −24 48 "" 0 1 , which reduces to −8 16 " 0 0

Section 3.4

" −2 "" 0 . The associated linear 0 " 0

− 2x2 = 0 . Since column 2 is not a pivot column, x2 is an independent 0 = 0

variable. Let x2 = 1. Then x1 = 2x2 = 2. This yields the fundamental eigenvector [2, 1]. Step 4: Since n = 2, and since we have found 2 fundamental eigenvectors, A is diagonalizable. Step 5: We form the2 × 2 matrix whose columns are the two fundamental eigenvectors we have 3 2 found: P = . 1 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is 1 −2 3 0 . It is easy to verify for D= . Also, using Theorem 2.13, we get P−1 = −1 3 0 −5 these matrices that D = P−1 AP. " " x − 13 34 (c) Step 1: pA (x) = |xI2 − A| = "" −5 x + 13

" " " = (x − 13)(x + 13) − (34)(−5) = x2 + 1. "

Step 2: The eigenvalues of A are the roots of pA (x). But pA (x) = x2 + 1 has no real roots. Therefore, A has no eigenvalues. Step 3: Because A has no eigenvalues, there are no associated eigenvectors, fundamental or otherwise. Step 4: Since there are no eigenvectors for A, A is not diagonalizable. " " " x + 13 3 −18 "" " x+4 −26 "". Using basketweaving gives pA (x) = (d) Step 1: pA (x) = |xI3 − A| = "" 20 " 14 3 x − 19 " (x + 13)(x + 4)(x − 19) + (3)(−26)(14) + (−18)(20)(3) −(−18)(x + 4)(14) − (x + 13)(−26)(3) − (3)(20)(x − 19) = x3 − 2x2 − x + 2 = (x2 − 1)(x − 2) = (x − 1)(x + 1)(x − 2). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are three eigenvalues, λ1 = 1 λ2 = −1, and λ3 = 2. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. Now " ⎡ ⎤ " ⎤ ⎡ 1 0 − 65 "" 0 14 3 −18 "" 0 ⎢ ⎥ " [(1I3 − A)|0] = ⎣ 20 5 −26 "" 0 ⎦ , which reduces to ⎣ 0 1 − 25 " 0 ⎦. The associated " " 14 3 −18 0 0 0 0 " 0 ⎧ − 65 x3 = 0 ⎪ ⎨ x1 x2 − 25 x3 = 0 . Since column 3 is not a pivot column, x3 is an linear system is ⎪ ⎩ 0 = 0 independent variable. We choose x3 = 5 to eliminate fractions. Then x1 = x2 = 25 x3 = 2. This yields the fundamental eigenvector [6, 2, 5].

6 5 x3

= 6 and

For λ2 = −1, we solve the homogeneous system whose augmented matrix is [((−1)I3 −A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 0 −1 "" 0 12 3 −18 "" 0 [((−1)I3 − A)|0] = ⎣ 20 3 −26 "" 0 ⎦ , which reduces to ⎣ 0 1 −2 "" 0 ⎦. The associated 0 0 0 " 0 14 3 −20 " 0 95

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎧ ⎨ x1 linear system is

⎩

x2

− x3 − 2x3 0

= = =

Section 3.4

0 0 . Since column 3 is not a pivot column, x3 is an 0

independent variable. Let x3 = 1. Then x1 = x3 = 1 and x2 = 2x3 = 2. This yields the fundamental eigenvector [1, 2, 1]. For λ3 = 2, we solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 0 −1 "" 0 15 3 −18 "" 0 [(2I3 − A)|0] = ⎣ 20 6 −26 "" 0 ⎦ , which reduces to ⎣ 0 1 −1 "" 0 ⎦. The associated 0 0 0 " 0 14 3 −17 " 0 ⎧ − x3 = 0 ⎨ x1 x2 − x3 = 0 . Since column 3 is not a pivot column, x3 is an linear system is ⎩ 0 = 0 independent variable. Let x3 = 1. Then x1 = x3 = 1 and x2 = x3 = 1. This yields the fundamental eigenvector [1, 1, 1]. Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable. Step 5: We form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have found: ⎤ ⎡ 6 1 1 P = ⎣ 2 2 1 ⎦. 5 1 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the ⎡ ⎡ ⎤ 1 0 1 0 0 1 D = ⎣ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ 3 0 0 2 −8 −1 to verify for these matrices that D = P−1 AP. " " x−5 8 12 " x−3 −4 (f) Step 1: pA (x) = |xI3 − A| = "" 2 " −4 6 x+9

correct order) is ⎤ −1 −4 ⎦. It is easy 10

" " " ". Using basketweaving gives us pA (x) = " "

(x − 5)(x − 3)(x + 9) + (8)(−4)(−4) + (12)(2)(6) −(12)(x − 3)(−4) − (x − 5)(−4)(6) − (8)(2)(x + 9) = x3 + x2 − x − 1 = (x2 − 1)(x + 1) = (x − 1)(x + 1)2 . Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 1 and λ2 = −1. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. Now " " ⎡ ⎤ ⎡ ⎤ −4 8 12 "" 0 1 0 −1 "" 0 1 "" 0 ⎦. The associated [(1I3 − A)|0] = ⎣ 2 −2 −4 "" 0 ⎦ , which reduces to ⎣ 0 1 " −4 6 10 0 0 0 0 " 0 ⎧ − x3 = 0 ⎨ x1 x2 + x3 = 0 . Since column 3 is not a pivot column, x3 is an linear system is ⎩ 0 = 0 independent variable. Let x3 = 1. Then x1 = x3 = 1 and x2 = −x3 = −1. This yields the fundamental eigenvector [1, −1, 1]. 96


Section 3.4

For λ2 = −1, we solve the homogeneous system whose augmented matrix is [((−1)I3 −A)|0]. Now " " ⎡ ⎤ ⎡ ⎤ −6 8 12 "" 0 1 0 −2 "" 0 0 "" 0 ⎦. The associated [((−1)I3 − A)|0] = ⎣ 2 −4 −4 "" 0 ⎦ , reducing to ⎣ 0 1 " −4 6 8 0 0 0 0 " 0 ⎧ − 2x3 = 0 ⎨ x1 = 0 . Since column 3 is not a pivot column, x3 is an x2 linear system is ⎩ 0 = 0 independent variable. Let x3 = 1. Then x1 = 2x3 = 2 and x2 = 0. This yields the fundamental eigenvector [2, 0, 1]. Step 4: Since n = 3, and the process has produced only 2 fundamental eigenvectors, A is not diagonalizable. " " " x−2 0 0 "" " x−4 −1 "" . Using a cofactor expansion along the first (g) Step 1: pA (x) = |xI3 − A| = "" 3 " −3 2 x−1 " " "

" −1 "" 1+1 " x − 4 row gives us pA (x) = (x − 2)(−1) = (x − 2) (x − 4)(x − 1) − (−1)(2) = " 2 x−1 " (x − 2)(x2 − 5x + 6) = (x − 2)2 (x − 3). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 2 and λ2 = 3. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 2, we solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 − 23 − 13 " 0 0 0 0 "" 0 " [(2I3 − A)|0] = ⎣ 3 −2 −1 "" 0 ⎦ , which reduces to ⎣ 0 0 0 "" 0 ⎦. The associated −3 2 1 " 0 0 0 0 " 0 ⎧ 2 1 ⎨ x1 − 3 x2 − 3 x3 = 0 linear system is 0 = 0 . Since columns 2 and 3 are not pivot columns, ⎩ 0 = 0 x2 and x3 are independent variables. First, we choose x2 = 3 (to eliminate fractions) and x3 = 0. Then x1 =

2 3 x2

+ 13 x3 = 2, producing the fundamental eigenvector [2, 3, 0]. Second, we choose

x2 = 0 and x3 = 3 (to eliminate fractions). This gives x1 = fundamental eigenvector [1, 0, 3].

2 3 x2

+ 13 x3 = 1, yielding the

For λ2 = 3, we solve the homogeneous system whose augmented matrix is [(3I3 − A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 0 1 0 0 "" 0 [(3I3 − A)|0] = ⎣ 3 −1 −1 "" 0 ⎦ , which reduces to ⎣ 0 1 1 "" 0 ⎦. The associated 0 0 0 " 0 −3 2 2 " 0 ⎧ = 0 ⎨ x1 x2 + x3 = 0 . Since column 3 is not a pivot column, x3 is an linear system is ⎩ 0 = 0 independent variable. Let x3 = 1. Then x1 = 0 and x2 = −x3 = −1. This yields the fundamental eigenvector [0, −1, 1].

97


Section 3.4

Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable. Step 5: We form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have found: ⎡ ⎤ 2 1 0 P = ⎣ 3 0 −1 ⎦. 0 3 1 Step 6: The ⎡ 2 0 D=⎣ 0 2 0 0

matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎤ 1 − 13 − 13 0 ⎢ 2 ⎥ 2 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −1 3 3 ⎦. It is easy to 3 3 −2 −1

verify for these matrices that D = P−1 AP. " " " x − 3 −1 6 2 "" " " −4 x 6 4 "" . Using a cofactor expansion along the (i) Step 1: pA (x) = |xI4 − A| = "" −2 0 x + 3 2 "" " " 0 −1 2 x−1 " " " " " " x−3 " x − 3 −1 2 " 6 2 "" " " " 6 4 "" + (2)(−1)4+3 "" −4 x 4 "" last row gives us pA (x) = (0) + (−1)(−1)4+2 "" −4 " −2 " −2 x+3 2 " 0 2 " " " " x − 3 −1 6 "" " x 6 "" . Using basketweaving on each 3 × 3 determinant pro+(x − 1)(−1)4+4 "" −4 " −2 0 x+3 " duces pA (x) = − (x − 3)(6)(2) + (6)(4)(−2) + (2)(−4)(x + 3) − (2)(6)(−2) − (x − 3)(4)(x + 3)

−(6)(−4)(2) −2 (x − 3)(x)(2) + (−1)(4)(−2) + (2)(−4)(0) − (2)(x(−2) − (x − 3)(4)(0)

−(−1)(−4)(2) +(x − 1) (x − 3)(x)(x + 3) + (−1)(6)(−2) + (6)(−4)(0) − (6)(x)(−2)

−(x − 3)(6)(0) − (−1)(−4)(x + 3) = −(−4x2 + 4x) − 2(2x2 − 2x) + (x − 1)(x3 − x) = 4x(x − 1) − 4x(x − 1) + (x − 1)x(x2 − 1) = (x − 1)2 (x + 1)x. Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are three eigenvalues, λ1 = 1, λ2 = −1, and λ3 = 0. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(1I4 − A)|0]. Now " " ⎤ ⎡ ⎤ ⎡ 1 0 −2 −1 "" 0 −2 −1 6 2 "" 0 ⎢ ⎢ −4 0 "" 0 ⎥ 1 6 4 "" 0 ⎥ ⎥. The associ⎥ , reducing to ⎢ 0 1 −2 [(1I4 − A)|0] = ⎢ ⎣ 0 0 ⎣ −2 0 0 "" 0 ⎦ 0 4 2 "" 0 ⎦ 0 0 0 0 " 0 0 −1 2 0 " 0 ⎧ x1 − 2x3 − x4 = 0 ⎪ ⎪ ⎨ = 0 x2 − 2x3 ated linear system is . Since columns 3 and 4 are not 0 = 0 ⎪ ⎪ ⎩ 0 = 0 pivot columns, x3 and x4 are independent variables. First, we let x3 = 1 and x4 = 0. Then

98


Section 3.4

x1 = 2x3 + x4 = 2 and x2 = 2x3 = 2. This yields the fundamental eigenvector [2, 2, 1, 0]. Second, we let x3 = 0 and x4 = 1. Then x1 = 2x3 + x4 = 1 and x2 = 2x3 = 0, producing the fundamental eigenvector [1, 0, 0, 1]. For λ2 = −1, we solve the homogeneous system whose augmented matrix " ⎡ ⎤ ⎡ 1 0 −4 −1 6 2 "" 0 ⎢ 0 1 " 0 ⎥ ⎢ −4 −1 6 4 ⎥ , reducing to ⎢ " Now [((−1)I3 − A)|0] = ⎢ ⎣ 0 0 ⎣ −2 0 2 2 "" 0 ⎦ 0 0 0 −1 2 −2 " 0 ⎧ x1 − x3 = 0 ⎪ ⎪ ⎨ = 0 x2 − 2x3 associated linear system is . Since column = 0 x ⎪ 4 ⎪ ⎩ 0 = 0

is [((−1)I4 − A)|0]. " ⎤ −1 0 "" 0 −2 0 "" 0 ⎥ ⎥. The 0 1 "" 0 ⎦ 0 0 " 0

3 is not a pivot col-

umn, x3 is an independent variable. Let x3 = 1. Then x1 = x3 = 1, x2 = 2x3 = 2, and x4 = 0. This yields the fundamental eigenvector [1, 2, 1, 0]. For λ3 = 0, we solve the homogeneous system whose augmented matrix is [(0I4 − A)|0] = [−A|0]. " " ⎤ ⎡ ⎤ ⎡ 1 0 0 −1 "" 0 −3 −1 6 2 "" 0 ⎢ ⎢ −4 1 "" 0 ⎥ 0 6 4 "" 0 ⎥ ⎥. The associated ⎥ , reducing to ⎢ 0 1 0 Now [−A|0] = ⎢ " ⎣ ⎦ ⎣ −2 0 0 1 0 "" 0 ⎦ 0 3 2 " 0 0 0 0 0 " 0 0 −1 2 −1 " 0 ⎧ x1 − x4 = 0 ⎪ ⎪ ⎨ + x4 = 0 x2 linear system is . Since column 4 is not a pivot column, x4 is an = 0 x3 ⎪ ⎪ ⎩ 0 = 0 independent variable. Let x4 = 1. Then x1 = x4 = 1, x2 = −x4 = −1, and x3 = 0. This yields the fundamental eigenvector [1, −1, 0, 1]. Step 4: Since n = 4, and we have found 4 fundamental eigenvectors, A is diagonalizable. Step 5: We form the 4 × 4 matrix whose columns are the four fundamental eigenvectors we have found: ⎡ ⎤ 2 1 1 1 ⎢ 2 0 2 −1 ⎥ ⎥. P=⎢ ⎣ 1 0 1 0 ⎦ 0 1 0 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎡ ⎤ 1 0 0 0 1 0 −1 −1 ⎢ 0 1 ⎢ 0 0 0 ⎥ 1 −2 1 ⎥ −1 ⎥ ⎢ ⎥. It D=⎢ ⎣ 0 0 −1 0 ⎦. Also, by row reducing [P|I4 ], we get P = ⎣ −1 0 2 1 ⎦ 0 0 0 0 0 −1 2 0 is easy to verify for the given matrices that D = P−1 AP. (5) In each part, we use the Diagonalization Method to find a matrix P and a diagonal matrix D such that A = PDP−1 . Then An = PDn P−1 . For a diagonal matrix D, Dn is easily computed as the diagonal matrix whose (i, i) entry is dnii .

99

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition " " x−4 (a) First, we diagonalize A. Step 1: pA (x) = |xI2 − A| = "" −3

Section 3.4

" 6 "" x+5 "

= (x − 4)(x + 5) − (6)(−3) = x2 + x − 2 = (x − 1)(x + 2). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 1 and λ2 = −2. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(1I2 − A)|0]. Now " " −3 6 "" 0 1 −2 "" 0 , which reduces to . The associated linear system [(1I2 − A)|0] = −3 6 " 0 0 0 " 0 ) x1 − 2x2 = 0 is . Since column 2 is not a pivot column, x2 is an independent variable. 0 = 0 Let x2 = 1. Then x1 = 2x2 = 2. This yields the fundamental eigenvector [2, 1]. For λ2 = −2, we solve the homogeneous system whose augmented matrix is [((−2)I2 −A)|0]. Now " " 1 −1 "" 0 −6 6 "" 0 . The associated linear , which reduces to [((−2)I2 − A)|0] = 0 0 " 0 −3 3 " 0 ) x1 − x2 = 0 system is . Since column 2 is not a pivot column, x2 is an independent 0 = 0 variable. Let x2 = 1. Then x1 = x2 = 1. This yields the fundamental eigenvector [1, 1]. Step 4: Since n = 2, and we have found 2 fundamental eigenvectors, A is diagonalizable. Step 5: We form the2 × 2 matrix whose columns are the two fundamental eigenvectors we have 2 1 found: P = . 1 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is 1 −1 1 0 . Thus, D= . Also, using Theorem 2.13, we get P−1 = −1 2 0 −2 ( ' 15 1 0 2 1 1 −1 32770 −65538 15 −1 15 A = PD P = = . 1 1 −1 2 32769 −65537 0 (−2)15 " " " x − 11 6 12 "" " x+6 16 "". Using bas(c) First, we diagonalize A. Step 1: pA (x) = |xI3 − A| = "" −13 " −5 3 x+5 " ketweaving gives pA (x) = (x − 11)(x + 6)(x + 5) + (6)(16)(−5) + (12)(−13)(3) −(12)(x + 6)(−5) − (x − 11)(16)(3) − (6)(−13)(x + 5) = x3 − x = x(x2 − 1) = x(x − 1)(x + 1). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are three eigenvalues, λ1 = 0, λ2 = 1, and λ3 = −1. Normally, at this point, we solve for fundamental eigenvectors. However, as we will see, we do not need to compute the actual matrix P in this case. Note that we get (at least) one fundamental eigenvector from each eigenvalue, so we obtain 3 fundamental Hence, A is ⎡ eigenvectors. ⎤ 0 0 0 0 ⎦ is the diagonal diagonalizable, and A = PDP−1 for some matrix P, where D = ⎣ 0 1 0 0 −1

100


Section 3.4

0

0

0

⎤

⎥ ⎢ 0 matrix whose diagonal entries are the eigenvalues of A. Now D49 = ⎣ 0 149 ⎦ = D. 49 0 0 (−1) Thus, A49 = PD49 P−1 = PDP−1 = A. " " " x−7 −9 12 "" " 22 "". Using bas(e) First, we diagonalize A. Step 1: pA (x) = |xI3 − A| = "" −10 x − 16 " −8 −12 x + 16 " ketweaving gives pA (x) = (x − 7)(x − 16)(x + 16) + (−9)(22)(−8) + (12)(−10)(−12) −(12)(x − 16)(−8) − (x − 7)(22)(−12) − (−9)(−10)(x + 16) = x3 − 7x2 + 14x − 8 = (x − 1)(x − 2)(x − 4). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are three eigenvalues, λ1 = 1, λ2 = 2, and λ3 = 4. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(1I3 − A)|0]. Now " " ⎤ ⎡ ⎤ ⎡ 1 32 0 " 0 −6 −9 12 "" 0 " [(1I3 − A)|0] = ⎣ −10 −15 22 "" 0 ⎦ , which reduces to ⎣ 0 0 1 "" 0 ⎦. The associated −8 −12 17 " 0 0 0 0 " 0 ⎧ 3 = 0 ⎨ x1 + 2 x2 linear system is x3 = 0 . Since column 2 is not a pivot column, x2 is an ⎩ 0 = 0 independent variable. We choose x2 = 2 to eliminate fractions. Then x1 = − 32 x2 = −3 and x3 = 0, producing the fundamental eigenvector [−3, 2, 0]. For λ2 = 2, we solve the homogeneous system whose augmented matrix is [(2I3 − A)|0]. Now " ⎤ ⎡ " ⎡ ⎤ 1 0 − 32 "" 0 −5 −9 12 "" 0 ⎥ ⎢ " [(2I3 − A)|0] = ⎣ −10 −14 22 "" 0 ⎦ , which reduces to ⎣ 0 1 − 12 " 0 ⎦. The associ" " −8 −12 18 0 0 0 0 " 0 ⎧ − 32 x3 = 0 ⎪ ⎨ x1 x2 − 12 x3 = 0 . Since column 3 is not a pivot column, x3 is ated linear system is ⎪ ⎩ 0 = 0 an independent variable. We choose x3 = 2 to eliminate fractions. Then x1 = x2 =

1 2 x3

3 2 x3

= 3 and

= 1, yielding the fundamental eigenvector [3, 1, 2].

For λ3 = 4, we solve the homogeneous system whose augmented matrix is [(4I3 − A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 0 −1 "" 0 −3 −9 12 "" 0 [(4I3 − A)|0] = ⎣ −10 −12 22 "" 0 ⎦ , which reduces to ⎣ 0 1 −1 "" 0 ⎦. The associated 0 0 0 " 0 −8 −12 20 " 0 ⎧ − x3 = 0 ⎨ x1 x2 − x3 = 0 . Since column 3 is not a pivot column, x3 is an indelinear system is ⎩ 0 = 0 pendent variable. Let x3 = 1. Then x1 = x3 = 1 and x2 = x3 = 1. This yields the fundamental eigenvector [1, 1, 1]. 101


Section 3.4

Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable. Step 5: We found: ⎡ −3 P=⎣ 2 0

form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have 3 1 2

⎤ 1 1 ⎦. 1

Step 6: The ⎡ 1 0 D=⎣ 0 2 0 0

matrix whose diagonal entries are the eigenvalues of A ⎤ ⎡ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −2 4 4 ⎤⎡ ⎤ ⎡ 110 ⎡ 0 0 −1 −1 2 −3 3 1 ⎢ ⎥ 5 0 ⎦ ⎣ −2 −3 A10 = PD10 P−1 = ⎣ 2 1 1 ⎦ ⎣ 0 210 4 6 −9 0 2 1 10 0 0 4 ⎡ ⎤ 4188163 6282243 −9421830 = ⎣ 4192254 6288382 −9432060 ⎦. 4190208 6285312 −9426944

(in the correct order) is ⎤ −1 2 −3 5 ⎦. Thus, 6 −9 ⎤ ⎦

−1 (7) (b) If A has all eigenvalues nonnegative, then A has a square root. √ Suppose A2 = PDP for some diagonal matrix D. Let C be the diagonal matrix with cii = dii . Thus, C = D. Let B = PCP−1 . Then B2 = (PCP−1 )2 = (PCP−1 )(PCP−1 ) = PC(P−1 P)CP−1 = PCIn CP−1 = PC2 P−1 = PDP−1 = A. ⎡ ⎤ 15 −14 −14 16 17 ⎦. We will use the Diagonalization Method to find a matrix P and a diag(8) Let B = ⎣ −13 20 −22 −23 √ onal matrix D such that B = PDP−1 . If C is the diagonal matrix with cii = 3 dii , then C3 = D. So, −1 if A = PCP−1 , then A3 = (PCP−1 )3 = (PCP−1 )(PCP−1 )(PCP−1 ) = PC(P P)C(P−1 P)CP−1 −1 3 −1 −1 = PCIn CIn CP = PC P = PDP = B. Hence, to solve this problem, we first need to find P and D. To do this, we diagonalize B. " " " x − 15 14 14 "" " x − 16 −17 "". Step 1: pB (x) = |xI3 − B| = "" 13 " −20 22 x + 23 "

Using basketweaving, we obtain pB (x) = (x − 15)(x − 16)(x + 23) + (14)(−17)(−20) + (14)(13)(22) −(14)(x − 16)(−20) − (x − 15)(−17)(22) − (14)(13)(x + 23) = x3 − 8x2 − x + 8 = (x2 − 1)(x − 8) = (x − 1)(x + 1)(x − 8). Step 2: The eigenvalues of B are the roots of pB (x). Hence, there are three eigenvalues, λ1 = 1, λ2 = −1, and λ3 = 8. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented " ⎤ ⎡ ⎡ 1 0 −14 14 14 "" 0 [(1I3 −B)|0] = ⎣ 13 −15 −17 "" 0 ⎦ , which reduces to ⎣ 0 1 0 0 −20 22 24 " 0

102

matrix " 1 "" 0 2 "" 0 0 " 0

is [(1I3 − B)|0]. ⎤

Now

⎦. The associated linear

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎧ ⎨ x1 system is

⎩

x2

+ +

x3 2x3 0

= = =

Section 3.4

0 0 . Since column 3 is not a pivot column, x3 is an independent 0

variable. Let x3 = 1. Then x1 = −x3 = −1 and x2 = −2x3 = −2. This yields the fundamental eigenvector [−1, −2, 1]. For λ2 = −1, we solve the homogeneous system whose augmented matrix is [((−1)I3 − B)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 0 −16 14 14 "" 0 [((−1)I3 − B)|0] = ⎣ 13 −17 −17 "" 0 ⎦ , which reduces to ⎣ 0 1 1 "" 0 ⎦. The associated 0 0 0 " 0 −20 22 22 " 0 ⎧ = 0 ⎨ x1 x2 + x3 = 0 . Since column 3 is not a pivot column, x3 is an indepenlinear system is ⎩ 0 = 0 dent variable. Let x3 = 1. Then x1 = 0 and x2 = −x3 = −1, producing the fundamental eigenvector [0, −1, 1]. For λ3 = 8, we solve the homogeneous system whose augmented matrix is [(8I3 − B)|0]. Now " ⎤ ⎡ " ⎤ ⎡ 1 0 −1 "" 0 −7 14 14 "" 0 ⎥ ⎢ 1 " [(8I3 − B)|0] = ⎣ 13 −8 −17 "" 0 ⎦ , which reduces to ⎣ 0 1 2 "" 0 ⎦. The associated −20 22 31 " 0 0 0 0 " 0 ⎧ − x3 = 0 ⎪ ⎨ x1 1 x2 + 2 x3 = 0 . Since column 3 is not a pivot column, x3 is an indelinear system is ⎪ ⎩ 0 = 0 pendent variable. We choose x3 = 2 to eliminate fractions. Then x1 = x3 = 2 and x2 = − 12 x3 = −1, yielding the fundamental eigenvector [2, −1, 2]. Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, B is diagonalizable. Step ⎡5: We form the 3⎤× 3 matrix whose columns are the 3 fundamental eigenvectors we have found: −1 0 2 P = ⎣ −2 −1 −1 ⎦. 1 1 2 Step 6: The matrix whose diagonal entries are the eigenvalues of B (in the correct order) is ⎡ ⎤ ⎡ ⎤ 1 −2 −2 1 0 0 4 5 ⎦. D = ⎣ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −3 1 −1 −1 0 0 8 Next, we compute the diagonal matrix C whose main diagonal entries are the cube roots of the eigen⎡ ⎤ 1 0 0 values of B. That is, C = ⎣ 0 −1 0 ⎦ . Then 0 0 2 ⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤ −1 0 2 1 0 0 1 −2 −2 3 −2 −2 4 5 ⎦ = ⎣ −7 10 11 ⎦. Direct A = PCP−1 = ⎣ −2 −1 −1 ⎦ ⎣ 0 −1 0 ⎦ ⎣ −3 1 1 2 0 0 2 1 −1 −1 8 −10 −11 calculation of A3 verifies that A3 = B.

103

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition (10) (b) Consider " " x 1 " " −1 x

0 1

−1 0

Section 3.4

. Now A has no eigenvalues because pA (x) = |xI2 − A| =

the matrix A = " " " = x2 +1, which has no real roots. However, A4 = I2 has 1 as an eigenvalue. Thus, with " √ 4 k = 4, λ = 1 is an eigenvalue for A4 , but neither solution for λ = ± 4 1 = ±1 is an eigenvalue for A.

(24) (a) True. This is the definition of what it means for 5 to be an eigenvalue of A. (b) False. The eigenvalues of A are the solutions of pA (x) = |xIn − A| = 0. The statement given in the text is missing the determinant symbol and also has a zero matrix on the right side of the equation. (c) True. This is part of Theorem 3.14. (d) True. This fact is stated in Step 6 of the Diagonalization Method. (e) False. If every column of P is a multiple of the same eigenvector, then |P| = 0, and P is singular. 1 1 Consider A = , which has eigenvalues λ1 = 1 and λ2 = 2. (Do you see why?) The 0 2 Diagonalization Method produces fundamental eigenvectors [1, 0] for λ1 and [1, 1] for λ2 . But note that [2, 0] is also an eigenvector for λ1 . If we use the eigenvectors [1, 0] and [2, 0] as columns 1 2 for P, instead of [1, 0] and [1, 1] as we should, we get P = . Note that the columns of P 0 0 are eigenvectors for A, but P is singular (|P| = 0). (f) True. The given characteristic polynomial indicates that the algebraic multiplicity of the eigenvalue (−1) is 1. We stated in Section 3.4 (without proof) that the number of fundamental eigenvectors produced for a given eigenvalue in Step 3 of the Diagonalization Method is always less than or equal to the algebraic multiplicity of the eigenvalue. (This is proven in Chapter 5.) Hence, no more than one fundamental eigenvector for (−1) can be produced in the Diagonalization Method. (Note that since the Diagonalization Method always produces at least one fundamental eigenvector for each eigenvalue, in this case the method must produce exactly one fundamental eigenvector for the eigenvalue (−1).) (g) True. Since Step 3 of the Diagonalization Method always produces at least one fundamental eigenvector for each eigenvalue of A, the method must yield (at least) three fundamental eigenvectors. Step 4 then assures us that A is diagonalizable. (h) False. In the discussion of using eigenvalues and eigenvectors to compute large powers of a matrix (in Section 3.4), we noted that, in general, A = PDP−1 implies An = PDn P−1 (where D is a diagonal matrix). Neither P nor P−1 should be raised to the nth power. For a specific counterex 19 −48 n n n −1 n , ample to the equation A = P D (P ) given in the problem, consider A = 8 −21 from Exercise 4(a). In the solution to that exercise (see above), we found that D = P−1 AP, and 3 2 3 0 −1 so A = PDP , for the matrices P = and D = . Now, a short computation 1 1 0 −5 −23 96 −503 1408 , but P2 D2 (P−1 )2 = instead. Note, however, shows that A2 = −16 57 −192 537 that PD2 P−1 = A2 .

104


Chapter 3 Review Exercises (1) (b) The (3,4) cofactor = A34

" " 4 " 3+4 = (−1) |A34 | = − "" −6 " −7

−5 1 0

2 −2 −1

Chap 3 Review

" " " ". We calculate this determinant " "

using basketweaving:

4 × −6 −7

−5

2 ×

1 ×

×

4 ×

−2

×

−1

0

× −6 −7

−5 1

× 0

" " " 4 −5 −3 " " " " −6 1 −4 "" = 4 × 1 × (−1) + (−5) × (−2) × (−7) + 2 × (−6) × 0 − 2 × 1 × (−7) − 4 × (−2) × 0 " " 3 −8 2 " − (−5) × (−6) × (−1) = −4 − 70 + 0 + 14 + 0 + 30 = −30. Hence, A34 = −(−30) = 30. (d) |A| = a12 A12 + a22 A22 + a32 A32 + a42 A42 " " " " −6 −2 −4 " " 4 " " " 5 2 "" +1(−1)2+2 "" 3 = (−5)(−1)1+2 "" 3 " −7 −1 "−7 9 "

2 5 −1

−3 2 9

" " " " 4 " " " +(−8)(−1)3+2 " −6 " " " " −7

" 2 −3 "" −2 −4 "" + 0. −1 9 "

Using cofactor expansion along the first row for each 3 × 3 matrix yields " " " " " "

" 3 " 3 2 " " 5 2 " 5 "" " " " " " + (−4) " − (−2) " |A| = 5 (−6) " −7 −1 " −7 9 " −1 9 " " " " " "

" " 3 " 3 2 " " 5 2 " 5 "" " " " " " + (−3) " − 2" + 1 4" −7 −1 " −7 9 " −1 9 " " " " " "

" " " " " " −2 −4 " " − 2 " −6 −4 " + (−3) " −6 −2 " + 8 4 "" " " " " −7 −1 " −7 9 −1 9 = 5 ((−6) × 47 + 2 × 41 − 4 × 32) + (4 × 47 − 2 × 41 − 3 × 32) + 8 (4 × (−22) − 2 × (−82) − 3 × (−8)) = 5 × (−328) + 10 + 8 × 100 = −1640 + 10 + 800 = −830. (3) We use row operations to put A into upper triangular form. Notice in the solution below that we stop performing row operations as soon as we obtain an upper triangular matrix. We give charts indicating the row operations used, keeping track of the variable P , as described in Example 5 in Section 3.2 of the textbook. Recall that P = 1 at the beginning of the process. We then use the final upper triangular matrix obtained and the value of P to compute the desired determinant. ⎡ ⎤ 2 0 −3 3 ⎢ 4 −2 −1 3 ⎥ ⎥ . Pivoting at the (1,1) entry: Initial Matrix: A = ⎢ ⎣ 1 −1 0 −2 ⎦ 2 1 −2 1

105


Row Operations (I): 1 ←

1 2

1

P

Effect Multiply P by

(II): 2 ← (−4) 1 + 2

No change

(II): 3 ← (−1) 1 + 3

No change

(II): 4 ← (−2) 1 + 4

No change

Chap 3 Review

1 2

1 2 1 2 1 2 1 2

Performing these operations results in the matrix ⎤ ⎡ 3 1 0 − 32 2 ⎥ ⎢ ⎢ 0 −2 5 −3 ⎥ ⎥ . We move to the second column: ⎢ ⎢ 0 −1 3 − 72 ⎥ ⎦ ⎣ 2 0 1 1 −2 Row Operations (I): 2 ←

− 12

2

P

Effect Multiply P by

(II): 3 ← (1) 2 + 3

No change

(II): 4 ← (−1) 2 + 4

No change

− 12

− 14 − 14 − 14

Performing these operations results in the matrix ⎤ ⎡ 3 1 0 − 32 2 ⎢ 3 ⎥ ⎥ ⎢ 0 1 − 52 2 ⎥ ⎢ ⎢ 0 0 −1 −2 ⎥ . We move on to the third column: ⎦ ⎣ 7 7 0 0 − 2 2 Effect

P

(I): 3 ← (−1) 3

Multiply P by −1

(II): 4 ← (− 72 ) 3 + 4

No change

1 4 1 4

Row Operations

Performing these operations results in the matrix ⎤ ⎡ 3 1 0 − 32 2 ⎢ 3 ⎥ ⎥ ⎢ 0 1 − 52 2 ⎥ ⎢ = B, the desired matrix in upper triangular form. The final value of P is ⎢ 0 0 1 2 ⎥ ⎣ ⎦ 0 0 0 − 21 2 1 4.

=

1 P

21 By Theorem 3.2, |B| = (1)(1)(1)(− 21 2 ) = − 2 . Hence, as in Example 5 in Section 3.2, |A| 21 × |B| = 4 × (− 2 ) = −42.

(5) (a) By part (1) of Theorem 3.3, |B| = (−4) × |A| = (−4) × (−15) = 60. (b) By part (2) of Theorem 3.3, performing a type (II) row operation does not change the determinant of a matrix. Hence, |B| = |A| = −15. (c) By part (3) of Theorem 3.3, performing a type (III) row operation changes the sign of the determinant of a matrix. Hence, |B| = −|A| = −(−15) = 15. (7) |−3AT B−1 | = (−3)3 |AT B−1 | (by Corollary 3.4) = (−27)|AT | |B−1 | (by Theorem 3.7) = (−27)|A| |B−1 | 1 (by Corollary 3.8) = (−27)(−7)(1/(1/2)) = (−27)(−7)(2) = 378. (by Theorem 3.9) = (−27)|A| |B| 106


⎤ ⎡ 2 −3 2 4 3 ⎦ and B = ⎣ (10) We need to solve the system AX = B, where A = ⎣ 3 −1 2 1 Cramer’s Rule, we first construct ⎡ ⎤ ⎡ ⎤ ⎡ 11 −3 2 2 11 2 2 −3 4 3 ⎦ , A2 = ⎣ 3 −9 3 ⎦ , and A3 = ⎣ 3 4 A1 = ⎣ −9 3 2 1 −1 3 1 −1 2

Chap 3 Review ⎤ 11 −9 ⎦. Following 3 ⎤ 11 −9 ⎦ . 3

Computing the required determinants results in |A| = 34, |A1 | = −136, |A2 | = −102, and |A3 | = 170. Hence, x1 =

−136 −102 170 |A1 | |A2 | |A3 | = = −4, x2 = = = −3, and x3 = = = 5. |A| 34 |A| 34 |A| 34

Therefore, the solution set for the system is {(−4, −3, 5)}. (11) (a) Using Theorem 3.7 three times shows that for any square matrix A, |A4 | = |A|4 . Hence, |A| = 4 |A4 |. Now, using basketweaving, or any other method, we see that " " " 5 −4 −2 " " " " −8 −3 3 "" = −289. " " −2 4 7 " √ Thus, if such a matrix A exists, then |A4 | = −289, implying |A| = 4 −289. However, negative numbers do not have real fourth roots. Therefore, such a matrix A, having real entries, cannot exist. ⎡ ⎤ 3 −2 5 1 4 ⎦. A short computation shows that |B| = 0. But, if B = A−1 , (b) Suppose that B = ⎣ −1 1 0 13 then, by part (1) of Theorem 2.11, B = A−1 is nonsingular, and so |B| = 0 by Theorem 3.5. This gives a contradiction. (12) B similar to A implies there is a matrix P such that B = P−1 AP. (b) |BT | = |B| (by Theorem 3.9) = |P−1 AP| = |P−1 ||A||P| (by using Theorem 3.7 twice) = (by Corollary 3.8) = |A| = |AT | (by Theorem 3.9).

1 |P| |A||P|

(e) B + In = P−1 AP + In = P−1 AP + P−1 In P = P−1 (A + In )P. Hence, B + In is similar to A + In . (13) Assume in what follows that A is an n × n matrix. Let Ai be the ith matrix (as defined in Theorem 3.13) for the matrix A. (a) Let C = R([A|B]). We must show that for each i, 1 ≤ i ≤ n, the matrix Ci (as defined in Theorem 3.13) for C is identical to R(Ai ). First, we consider the columns of Ci other than the ith column. If 1 ≤ j ≤ n with j = i, then the jth column of Ci is the same as the jth column of C = R([A|B]). But since the jth column of [A|B] is identical to the jth column of Ai , it follows that the jth column of R([A|B]) 107


Chap 3 Review

is identical to the jth column of R(Ai ). Therefore, for 1 ≤ j ≤ n with j = i, the jth column of Ci is the same as the jth column of R(Ai ). Finally, we consider the ith column of Ci . Now, the ith column of Ci is identical to the last column of R([A|B]) = [R(A)|R(B)], which is R(B). But since the ith column of Ai equals B, it follows that the ith column of R(Ai ) = R(B). Thus, the ith column of Ci is identical to the ith column of R(Ai ). Therefore, since Ci and R(Ai ) agree in every column, we have Ci = R(Ai ). (b) Consider each type of operation in turn. First, if R is the type (I) operation k ← c k for c|Ai | |Ai | i )| some c = 0, then by part (1) of Theorem 3.3, |R(A |R(A)| = c|A| = |A| . If R is any type (II) operation, then by part (2) of Theorem 3.3, |R(Ai )| |R(A)|

=

(−1)|Ai | (−1)|A|

=

|Ai | |A| .

|R(Ai )| |R(A)|

=

|Ai | |A| .

If R is any type (III) operation, then

(c) In the special case when n = 1, we have A = [1]. Let B = [b]. Hence, A1 (as defined in Theorem b 1| 3.13) = B = [b]. Then the formula in Theorem 3.13 gives x1 = |A |A| = 1 = b, which is the correct solution to the equation AX = B in this case. For the remainder of the proof, we assume n > 1. If A = In , then the solution to the system is X = B. Therefore, we must show that, for each i, 1 ≤ i ≤ n, the formula given in Theorem 3.13 yields xi = bi (the ith entry of B). First, note that |A| = |In | = 1. Now, the ith matrix Ai (as defined in Theorem 3.13) for A is identical to In except in its ith column, which equals B. Therefore, the ith row of Ai has bi as its ith entry, and zeroes elsewhere. Thus, a cofactor expansion along the ith row of Ai yields |Ai | = bi (−1)i+i |In−1 | bi i| = bi . Hence, for each i, the formula in Theorem 3.13 produces xi = |A |A| = 1 = bi , completing the proof. (d) Because A is nonsingular, [A|B] row reduces to [In |X], where X is the unique solution to the system. Let [C|D] represent any intermediate augmented matrix during this row reduction pro|Ai | i| cess. Now, by part (a) and repeated use of part (b), the ratio |C |C| is identical to the ratio |A| obtained from the original augmented matrix, for each i, 1 ≤ i ≤ n. But part (c) proves that for the final augmented matrix, [In |X], this common ratio gives the correct solution for xi , for each i, 1 ≤ i ≤ n. Since all of the systems corresponding to these intermediate matrices have the same unique solution, the formula in Theorem 3.13 gives the correct solution for the original system, [A|B], as well. Thus, Cramer’s Rule is validated. " " " x−5 −16 16 "" " x + 67 −64 "". Using basketweaving gives us pA (x) = (14) (b) Step 1: pA (x) = |xI3 − A| = "" 32 " 32 64 x − 61 " (x − 5)(x + 67)(x − 61) + (−16)(−64)(32) + (16)(32)(64) −(16)(x + 67)(32) − (x − 5)(−64)(64) − (−16)(32)(x − 61) = x3 + x2 − 21x − 45 = (x − 5)(x2 + 6x + 9) = (x − 5)(x + 3)2 . Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 5 and λ2 = −3. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 5, we solve the homogeneous system whose augmented matrix " ⎤ ⎡ ⎡ 1 1 0 0 −16 16 "" 0 4 72 −64 "" 0 ⎦ , which reduces to ⎣ 0 1 −1 [(5I3 − A)|0] = ⎣ 32 32 64 −56 " 0 0 0 0

108

is [(5I3 − A)|0]. Now " ⎤ " 0 " " 0 ⎦. The associated " " 0

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎧ ⎨ x1 linear system is

⎩

x2

Chap 3 Review

+

1 4 x3

=

0

−

x3 0

= =

0 . Since column 3 is not a pivot column, x3 is an 0

independent variable. Let x3 = 4 (to eliminate fractions). Then x1 = − 14 x3 = −1 and x2 = x3 = 4. This yields the fundamental eigenvector [−1, 4, 4]. The eigenspace E5 = {a[−1, 4, 4] | a ∈ R}. For λ2 = −3, we solve the homogeneous system whose augmented matrix is [((−3)I3 −A)|0]. Now " " ⎤ ⎤ ⎡ ⎡ 1 2 −2 "" 0 −8 −16 16 "" 0 0 "" 0 ⎦. The associated 64 −64 "" 0 ⎦ , reducing to ⎣ 0 0 [((−3)I3 − A)|0] = ⎣ 32 " 0 0 0 0 " 0 32 64 −64 ⎧ ⎨ x1 + 2x2 − 2x3 = 0 0 = 0 . Since columns 2 and 3 are not pivot columns, linear system is ⎩ 0 = 0 x2 and x3 are independent variables. Letting x2 = 1 and x3 = 0 produces x1 = −2, thus yielding the fundamental eigenvector [−2, 1, 0]. Letting x2 = 0 and x3 = 1 produces x1 = 2, thus yielding the fundamental eigenvector [2, 0, 1]. The eigenspace E−3 = {a[−2, 1, 0] + b[2, 0, 1] | a, b ∈ R}. Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable. Step 5: We form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have found: ⎤ ⎡ −1 −2 2 1 0 ⎦. P=⎣ 4 4 0 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the correct order) is ⎡ ⎤ ⎡ ⎤ 5 0 0 −1 −2 2 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ 4 9 −8 ⎦. It is easy D = ⎣ 0 −3 0 0 −3 4 8 −7 to verify for these matrices that D = P−1 AP. (15) (b) From the hint in the text, pA (x) = x4 + 6x3 + 9x2 , which factors as x2 (x + 3)2 . Thus, we have two eigenvalues: λ1 = 0 and λ2 = −3. Each of these eigenvalues has algebraic multiplicity 2. We continue from Step 3 of the Diagonalization Method, looking for fundamental eigenvectors: For λ1 = 0, we solve the homogeneous system whose augmented matrix is [(0I4 − A)|0] = [−A|0]. " " ⎡ ⎤ ⎤ ⎡ 1 12 0 − 94 " 0 468 234 754 −299 "" 0 " ⎢ ⎢ −324 −162 −525 204 "" 0 ⎥ 1 "" 0 ⎥ ⎥ , reducing to ⎢ 0 0 1 ⎥. The Now [−A|0] = ⎢ ⎣ 0 0 0 ⎣ −144 −72 −231 93 "" 0 ⎦ 0 "" 0 ⎦ 108 54 174 −69 " 0 0 0 0 0 " 0 ⎧ x1 ⎪ ⎪ ⎨

associated linear system is

+

1 2 x2

x3

⎪ ⎪ ⎩

−

9 4 x4

=

0

+

x4 0 0

= = =

0 . Since columns 2 and 4 are 0 0

not pivot columns, x2 and x4 are independent variables. First, we let x2 = 1 and x4 = 0. Then x1 = − 12 x2 + 94 x4 = − 12 and x3 = −x4 = 0. Multiplying these results by 2 to eliminate fractions yields the fundamental eigenvector [−1, 2, 0, 0]. Second, we let x2 = 0 and x4 = 1. Then

109


Chap 3 Review

x1 = − 12 x2 + 94 x4 = 94 and x3 = −x4 = −1. Multiplying these results by 4 to eliminate fractions produces the fundamental eigenvector [9, 0, −4, 4]. For λ2 = −3, we solve the homogeneous system whose augmented matrix is [(−3I4 − A)|0]. Now " ⎤ ⎡ " " ⎡ ⎤ 1 0 0 − 13 3 " 0 465 234 754 −299 "" 0 ⎥ ⎢ 5 "" ⎥ ⎢ 0 1 0 ⎢ −324 −165 −525 204 "" 0 ⎥ 2 " 0 ⎥ ⎢ ⎢ ⎥ , reducing to ⎢ [(−3I4 − A)|0] = ⎣ ⎥. 3 " −144 −72 −234 93 "" 0 ⎦ ⎣ 0 0 1 2 " 0 ⎦ " 108 54 174 −72 " 0 0 0 0 0 " 0 ⎧ x − 13 = 0 ⎪ ⎪ 3 x4 ⎪ 1 ⎪ 5 ⎨ x2 + = 0 2 x4 The associated linear system is . Since column 4 is not a pivot 3 ⎪ x3 = 0 ⎪ 2 x4 ⎪ ⎪ ⎩ 0 = 0 13 column, x4 is an independent variable. Setting x4 = 1 in the system produces x1 = 13 3 x4 = 3 , 5 5 3 3 x2 = − 2 x4 = − 2 , and x3 = − 2 x4 = − 2 . Multiplying these results by 6 to eliminate fractions produces the fundamental eigenvector [26, −15, −9, 6].

Step 4: Since n = 4, and the process has produced only 3 fundamental eigenvectors, A is not diagonalizable. We actually did more work than necessary to solve this problem. We only really needed to know how many fundamental eigenvectors we were going to get, not what the fundamental eigenvectors actually are! This could be determined by the number of independent variables in the systems corresponding to [(0I4 − A)|0] and [(−3I4 − A)|0]. The number of these independent variables is 2 and 1, respectively, which means we only get a total of 3 fundamental eigenvectors. In fact, if we had started by working on the eigenvalue λ2 = −3, we would have quickly discovered that there would only be one fundamental eigenvector for this eigenvalue, even though its algebraic multiplicity is two. In general, we can conclude that a matrix is not diagonalizable if any eigenvalue has fewer fundamental eigenvectors than its algebraic multiplicity. " " " x + 21 −22 −16 "" " x − 29 −20 "". Using basketweaving (16) First, we diagonalize A. Step 1: pA (x) = |xI3 −A| = "" 28 " −8 8 x+5 " gives pA (x) = (x + 21)(x − 29)(x + 5) + (−22)(−20)(−8) + (−16)(28)(8) −(−16)(x − 29)(−8) − (x + 21)(−20)(8) − (−22)(28)(x + 5) = x3 − 3x2 − x + 3 = (x − 3)(x2 − 1) = (x − 3)(x − 1)(x + 1). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are three eigenvalues, λ1 = 1, λ2 = −1, and λ3 = 3. Step 3: Now we solve for fundamental eigenvectors. For λ1 = 1, we solve ⎡ 22 [(1I3 − A)|0] = ⎣ 28 −8 ⎧ ⎨ x1 linear system is ⎩

the homogeneous system whose augmented matrix is [(1I3 − A)|0]. Now " " ⎤ ⎡ ⎤ −22 −16 "" 0 1 −1 0 "" 0 −28 −20 "" 0 ⎦ , which reduces to ⎣ 0 0 1 "" 0 ⎦. The associated 8 6 " 0 0 0 0 " 0 − x2 x3 0

= = =

0 0 . Since column 2 is not a pivot column, x2 is an inde0

110


Chap 3 Review

pendent variable. We choose x2 = 1. Then x1 = x2 = 1 and x3 = 0, producing the fundamental eigenvector [1, 1, 0]. For λ2 = −1, we solve ⎡ 20 [(−1I3 − A)|0] = ⎣ 28 −8 ⎧ ⎨ x1 ear system is x2 ⎩

the homogeneous system whose augmented matrix " " ⎤ ⎡ 1 0 52 " −22 −16 "" 0 " −30 −20 "" 0 ⎦ , which reduces to ⎣ 0 1 3 "" 8 4 " 0 0 0 0 "

is [(−1I3 − A)|0]. Now ⎤ 0 0 ⎦. The associated lin0

+

5 2 x3

=

0

+

3x3 0

= =

0 . Since column 3 is not a pivot column, x3 is an independent 0

variable. We choose x3 = 2 to eliminate fractions. Then x1 = − 52 x3 = −5 and x2 = −3x3 = −6, producing the fundamental eigenvector [−5, −6, 2]. For λ3 = 3, we solve the homogeneous system whose augmented matrix is [(3I3 − A)|0]. Now " " ⎤ ⎡ ⎤ ⎡ 1 0 3 "" 0 24 −22 −16 "" 0 [(3I3 − A)|0] = ⎣ 28 −26 −20 "" 0 ⎦ , which reduces to ⎣ 0 1 4 "" 0 ⎦. The associated linear 0 0 0 " 0 −8 8 8 " 0 ⎧ + 3x3 = 0 ⎨ x1 x2 + 4x3 = 0 . Since column 3 is not a pivot column, x3 is an independent system is ⎩ 0 = 0 variable. We choose x3 = 1. Then x1 = −3x3 = −3 and x2 = −4x3 = −4, producing the fundamental eigenvector [−3, −4, 1]. Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable. Step 5: We form the 3 × 3 matrix whose columns are the 3 fundamental eigenvectors we have found: ⎡ ⎤ 1 −5 −3 P = ⎣ 1 −6 −4 ⎦. 0 2 1 Step 6: The matrix whose diagonal entries are the eigenvalues of ⎡ ⎡ ⎤ 2 1 0 0 ⎣ 0 −1 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = ⎣ −1 2 0 0 3 ⎤⎡ ⎡ ⎤ ⎡ 113 0 0 2 1 −5 −3 ⎢ ⎥ 0 ⎦ ⎣ −1 A13 = PD13 P−1 = ⎣ 1 −6 −4 ⎦ ⎣ 0 (−1)13 2 0 2 1 0 0 313 ⎡ ⎤ −9565941 9565942 4782976 6377300 ⎦. = ⎣ −12754588 12754589 3188648 −3188648 −1594325

A (in the correct order) is D = ⎤ −1 2 1 1 ⎦. Thus, −2 −1 ⎤ −1 2 1 1 ⎦ −2 −1

(17) Because D = P−1 AP, with D being a diagonal matrix, A is diagonalizable with its eigenvalues appearing on the main diagonal of D. The columns of P are fundamental eigenvectors corresponding to these eigenvalues. (a) The eigenvalues of A are the main diagonal entries of D. In particular, they are λ1 = 2, λ2 = −1, and λ3 = 3. 111


Chap 3 Review

(b) Because λ1 = 2 is the (1,1) entry of D, the first column of P is a fundamental eigenvector for A corresponding to λ1 . Hence, E2 = {a[1, −2, 1, 1] | a ∈ R}. Next, λ2 = −1 appears as both the (2,2) and (3,3) entries of D. Therefore, both the second and third columns of P are fundamental eigenvectors for A corresponding to λ2 . Hence, E−1 = {a[1, 0, 0, 1] + b[3, 7, −3, 2] | a, b ∈ R}. Finally, because λ3 = 3 is the (4,4) entry of D, the fourth column of P is a fundamental eigenvector for A corresponding to λ3 . Hence, E3 = {a[2, 8, −4, 3] | a ∈ R}. (c) The matrices D and A are similar to each other. Therefore, by part (g) of Exercise 13 in Section 3.3, |A| = |D|. But, since D is diagonal, it is also upper triangular. Hence, by Theorem 3.2, |D| = 2 × (−1) × (−1) × 3 = 6, the product of its main diagonal entries. Therefore, |A| = 6. 1 0 2 3 (18) (a) False. For example, and both have determinant 1 but are not equal as 0 1 1 2 matrices. (b) True. Since A is symmetric, A = AT , and so A (the adjoint of A) equals the adjoint of AT . But by Exercise 17(a) in Section 3.3, the adjoint of AT equals AT . Hence, A = AT . Thus, Aij (the (j, i) entry of A) = Aji (the (j, i) entry of AT ). An alternate proof is as follows: Exercise 30 in Section 3.3 (whose complete solution appears in this manual) shows that for any n × n matrix, (Ajm )T and (AT )mj are equal. Also, since A is symmetric, A = AT . Hence, Aij = (−1)i+j |Aij | = (−1)j+i |(Aij )T | (by Theorem 3.9) = (−1)j+i |(AT )ji | (by Exercise 30 in Section 3.3) = (−1)j+i |Aji | (because A = AT ) = Aji . (c) False. If A is an n × n matrix with n ≥ 3, Aij is a matrix, while (−1)i+j Aij is a real number. For a specific counterexample, suppose A = I3 . Then A11 = I2 , a 2 × 2 matrix, while (−1)1+1 A11 = (−1)1+1 (−1)1+1 a22 = 1. (d) False. By part (1) of Theorem 3.1, to get the area of the parallelogram, we need to take the absolute value of this determinant in case the determinant is negative. For example, consider the square whose corners are at (0, 0), (1, 0), (0, 1), and (1, 1), " whose " area is 1. The vectors [1, 0] and " 0 1 " " = −1, which does not equal the [0, 1] determine this square. However, the determinant "" 1 0 " area of the square. Note that if this problem had said “in some order” rather than “in either order,” the answer would have been true. The reason is that switching the two rows (that is, switching the order of the two vectors) reverses the sign of the determinant. (e) True. Part (2) of Theorem 3.1 tells us that the volume of the parallelepiped is the absolute value of the determinant of the matrix whose rows are the given vectors. However, Theorem 3.9 assures us that the determinant of its transpose, that is, the matrix whose columns determine the parallelepiped, has the same determinant. (f) True. See the comments in the text just after Example 2 in Section 3.3. (A formal proof is as follows: If A is lower triangular, then AT is upper triangular. By Theorem 3.2, |AT | is the product of its main diagonal entries. But these are the same as the main diagonal entries of A, one of which is zero. Hence, |AT | equals zero. Thus, |A| = |AT | (by Theorem 3.9) = 0. Theorem 3.5 then asserts that A is singular.) (g) True. First note that changing the order of the rows of a matrix can be accomplished by a sequence of type (III) row operations, each of which changes the sign of the determinant (part (3) of Theorem 3.3). The comments in the text just after Example 2 in Section 3.3 assert that a sequence of type (III) column operations affects the determinant in a similar manner. (The reason for this is that a column swap can be performed by first taking the transpose (which 112


Chap 3 Review

does not change the determinant, by Theorem 3.9), then doing a row swap, and then taking the transpose again.) Such a sequence of column operations effectively changes the sign of the original determinant once for each row operation involved. If the number of sign changes is even, the original and final determinants will be equal. If the number of sign changes is odd, the original and final determinants will have opposite signs. (h) True. Since Ae1 = 0, e1 is a nontrivial solution to the homogeneous system AX = 0. By Table 3.1 (at the end of Section 3.2), rank(A) < n. Hence, by Corollary 3.6, |A| = 0. (i) False. As the text says at the beginning of Section 3.2, using row reduction to compute determinants for large matrices is, in general, more efficient than cofactor expansion. (It can be shown that for an n × n matrix, row reduction takes less than a constant times n3 steps, while cofactor expansion takes more than a constant times n! steps. Because of the comparative rates of increase between n3 and n!, for large values of n, row reduction takes much less effort.) (j) True. Suppose |A| = |B| = 0. By Theorem 3.5, both A and B are nonsingular. Hence, by the Inverse Method for finding the inverse of a matrix, both A and B are row equivalent to In . Therefore, since they are both row equivalent to the same matrix, they are row equivalent to each other, by an argument similar to that in Exercise 3(b) in Section 2.3. 0 0 1 1 (k) False. For example, A = and B = both have determinant 0, but they are 0 0 1 1 not row equivalent because they have different ranks. It is also easy to find counterexamples in ⎡ ⎤ 1 0 0 which the matrices have the same rank. Consider, for example, A = ⎣ 0 1 0 ⎦ = I3 and 0 0 1 ⎡ ⎤ 1 0 1 B = ⎣ 0 1 1 ⎦, where A and B have the same determinant (= 1) and the same rank (= 3), 0 0 1 but A cannot be obtained from B using a single type (II) row operation. (l) False. In fact, if the coefficient matrix has a nonzero determinant, Theorems 3.5 and 2.15 imply that the system has only one solution. Since the system is homogeneous, that solution is the trivial solution. (m) True. By Theorem 3.7, |AB| = 0 implies that |A| |B| = 0, which implies that |A| = 0 or |B| = 0. n 1 1 A| (by Corollary 3.12) = |A| |A| (n) False. If A is an n × n matrix, then in general, |A−1 | = | |A| 1 2 (by Corollary 3.4). For a particular counterexample, consider A = , for which A−1 = 3 4 ( ' −2 1 4 −2 |A| = −2 , A = , and |A−1 | = − 12 , but |A| −2 = 1. 3 1 −3 1 − 2 2 " " " 2 0 " 1 0 1 0 " = 2, (o) False. For example, consider A = and B = . Then |A + B| = "" 0 1 " 0 0 0 1 but |B| = 1. " " " 0 1 " " = −1. (p) False. For example, "" 1 0 " (q) False. If λ is an eigenvalue for an n × n matrix A, and v is any nonzero eigenvector for A corresponding to λ, then every distinct scalar multiple of v is also an eigenvector for A corresponding to

113


Chap 3 Review

λ. Hence, Eλ contains an infinite number of elements. However, the Diagonalization Method will produce at most a finite number of fundamental eigenvectors: one for each independent variable encountered when solving the homogeneous system (λIn − A)X = 0. (r) True. This follows from the definition of the algebraic multiplicity of an eigenvalue and the fact that an nth degree polynomial cannot have more than n roots, counting multiplicities. (See the comments right before Example 2 in Section 3.4.) (s) True. See part (b) of Exercise 23 in Section 3.4. (t) False. In general, if A = P−1 BP, then B = PAP−1 , not P−1 AP. For a particular counterexam 1 0 1 −1 0 1 1 0 1 −1 ple, consider B = and P = . Then A = = 0 0 1 0 −1 1 0 0 1 0 0 0 0 1 0 0 1 −1 0 1 , and P−1 AP = = = B. −1 1 −1 1 −1 1 1 0 0 1 (u) False. The only n × n matrix similar to In is In itself. The reason is that, for any nonsingular n × n matrix P, P−1 In P = P−1 (In P) = P−1 P = In . For a specific counterexample, consider any nonsingular matrix other than In . (v) True. The discussion that precedes Theorem 3.14 in the text shows that if λ is a root of pA (x), then the system (λIn − A)X = 0 has a nontrivial solution. Hence, there is at least one nonzero vector X such that AX = λX. (w) True. First, D−1 = R (see Exercise 5 in Section 2.4). Next, since D = P−1 AP, A = PDP−1 . Hence, by parts (3) and (1) of Theorem 2.11, A−1 = (P−1 )−1 D−1 P−1 = PRP−1 . (x) True. Any nontrivial solution to (λIn − A)X = 0 would be an eigenvector for A, making λ an eigenvalue for A. 1 0 0 1 1 1 (y) False. It is easy to show that and are both diagonalizable, but is 0 0 0 1 0 1 not. 1 1 2 1 (z) False. It is easy to show that and are both diagonalizable (see Exercise 16 in 0 2 0 1 2 2 Section 3.4), but is not. 0 2

114


Section 4.1

Chapter 4 Section 4.1 (5) The set is not closed under addition. For example, 1 1 1 2 2 3 1 1 1 2 + = . Both and are singular (their determinants are 1 1 2 4 3 5 1 1 2 4 2 3 zero), but is nonsingular (its determinant is 1 = 0). 3 5 (8) Properties (2), (3), and (6) are not satisfied. Property (4) makes no sense without Property (3). The following is a counterexample for Property (2): 3 ⊕ (4 ⊕ 5) = 3 ⊕ 18 = 42, but (3 ⊕ 4) ⊕ 5 = 14 ⊕ 5 = 38. While this single counterexample for just one of the properties is enough to prove that the set is not a vector space, we will also illustrate why properties (3) and (6) do not hold. For Property (3), suppose z ∈ R is the zero vector. Then, for all x ∈ R, x ⊕ z = x ⇒ 2(x + z) = x ⇒ z = − x2 . But then different values of x would produce different values of z. For if x = 2, z = −1, and if x = 4, z = −2. However, the value of z must be unique. For a counterexample to Property (6), note that (1 + 2)(3) = (3)(3) = 9, but (1)(3) ⊕ (2)(3) = 3 ⊕ 6 = 18. (20) (a) False. The two closure properties and the eight remaining properties must be true for all elements in Rn in order for it to be a vector space under some “addition” and “scalar multiplication.” For example, Example 10 in Section 4.1 illustrates operations on R2 for which R2 is not a vector space. Also, Exercise 8, whose answer appears above, provides a counterexample in R. (b) False. This set is not closed under addition. Both x7 + x and −x7 are in this set, but their sum is x, which does not have degree 7. (Also note that the zero polynomial is not in the set.) (c) True. This is the vector space P7 described in Example 4 of Section 4.1. (d) True. This statement is logically equivalent to part (4) of Theorem 4.1. Note that part (4) of Theorem 4.1 is stated in the form “If A then B or C,” with A = “cx = 0,” B = “c = 0,” and C = “x = 0” (where we use c and x instead of a and v). Now, in Section 1.3 we found that “If A then B or C” is logically equivalent to “If A is true and B is false, then C.” Here, B is given to be false (since c = 0), and so A implies C. That is, when c = 0, “cx = 0” implies “x = 0.” (e) False. Scalar multiplication by the zero scalar always results in the zero vector, not a scalar. This is part (2) of Theorem 4.1. (f) True. This is the statement of part (3) of Theorem 4.1. (g) True. Suppose V represents the given set of functions. Then V is closed under addition because if f , g ∈ V, then h = f + g is a real valued function, and h(1) = f (1) + g(1) = 0 + 0 = 0. Similarly, V is closed under scalar multiplication because if f ∈ V and c ∈ R, then h = cf is a real valued function, and h(1) = cf (1) = c0 = 0. The proofs of Properties (1), (2), (5), (6), (7), and (8) are exactly identical to those in Example 6 of Section 4.1 of the textbook. For Property (3), z(x) = 0 is in V because z(1) = 0. The same argument for Property (3) in Example 6 then shows that z works as the identity element for addition. Similarly, for Property (4): if f ∈ V, then the function −f defined by [−f ](x) = −(f (x)) is in V because [−f ](1) = −(f (1)) = −(0) = 0. Thus, Example 6 finishes the proof that Property (4) holds.

115


Section 4.2

Section 4.2 (1) In each part, we will refer to the given set as V. If V is a subspace of R2 , we will prove it using Theorem 4.2 by showing that it is nonempty and that it is closed under both vector addition and scalar multiplication. If V is not a subspace, we will show that it lacks at least one of these properties. Or, since every subspace of R2 contains the zero vector [0, 0], we can also show that V is not a subspace by proving that it does not contain the zero vector. In that case, it is not necessary to also prove that one of the closure properties fails. (a) This set V is not a subspace of R2 because the zero vector, [0, 0], is not a unit vector, and so is not in V. (Note: It can also be shown that V is not closed under either operation.) (c) This set V is a subspace of R2 . V is nonempty since [0, 0] ∈ V because it is of the form [a, 2a] with a = 0. V is closed under addition because [a, 2a] + [b, 2b] = [(a + b), 2(a + b)] = [A, 2A] ∈ V, where A = (a + b). Finally, V is closed under scalar multiplication since c[a, 2a] = [ca, 2(ca)] = [B, 2B] ∈ V, where B = ca. (e) This set V is not a subspace of R2 since the zero vector, [0, 0], does not equal [1, 2], and so is not in V. (Note: It can also be shown that V is not closed under either operation.) (g) This set V is not a subspace of R2 because it is not closed under addition. The counterexample [1, 1] + [1, −1] = [2, 0] ∈ / V (since |2| = |0|) proves this. (j) This set V is not a subspace of R2 because it is not closed under addition. To prove this, consider the following counterexample: [1, 1] + [2, 4] = [3, 5] ∈ / V, since 5 = 32 . (Note: It can also be shown that V is not closed under scalar multiplication.) (l) This set V is not a subspace of R2 because it is not closed under scalar multiplication. To prove this, consider the following counterexample: 2[0.75, 0] = [1.5, 0] ∈ / V. (Note: It can also be shown that V is not closed under addition.) (2) In each part, we will refer to the given set as V. If V is a subspace of M22 , we will prove it using Theorem 4.2 by showing that it is nonempty and that it is closed under both matrix addition and scalar multiplication. If V is not a subspace, we will show that it lacks at least one of these properties. Or, since every subspace of M22 contains the zero matrix O22 , we can also show that V is not a subspace by proving that it does not contain O22 . In that case, it is not necessary to also prove that one of the closure properties fails. (a) This set V is a subspace of M22 . V is nonempty since setting a = b = 0 shows that O22 ∈ V. V is closed under matrix addition since a −a c −c (a + c) −(a + c) A −A + = = ∈ V, where A = (a + c) and b 0 d 0 (b + d) 0 B 0 B = (b + d). Similarly, V is closed under scalar multiplication because a −a (ca) −(ca) C −C c = = ∈ V, where C = (ca) and D = (cb). b 0 (cb) 0 D 0 (c) This set V is a subspace of M22 . V is clearly nonempty since O22 is symmetric, and hence in V. a b d e (a + d) (b + e) V is closed under matrix addition since + = , which is in b c e f (b + e) (c + f ) V, since it is symmetric. Similarly, V is closed under scalar multiplication because a b (ca) (cb) c = , which is in V, since it is symmetric. b d (cb) (cd) 116


Section 4.2

(e) This set V is a subspace of M22 . Since the sum of the entries of O22 is zero, O22 ∈ V. Thus, V a b is nonempty. Note that a typical element of V has the form , where d = −(a + b + c). c d a b d e Now, V is closed under addition because + c −(a + b + c) f −(d + e + f ) a+d b+e a+d b+e , which is = = c + f −(a + b + c) − (d + e + f ) c + f −((a + d) + (b + e) + (c + f )) in V, since its entries sum to zero. Finally, V is closed under scalar multiplication since a b ka kb ka kb k = = , c −(a + b + c) kc k(−(a + b + c)) kc −((ka) + (kb) + (kc)) which is, again, in V, since its entries sum to zero. 1 3 (g) This set V is a subspace of M22 . Let B = . Because O22 B = O22 , O22 ∈ V, and −2 −6 so V is nonempty. Next, we show that V is closed under addition. If A, C ∈ V, then (A + C)B = AB + CB = O22 + O22 = O22 . Hence, (A + C) ∈ V. Finally, we show that V is closed under scalar multiplication. If A ∈ V, then (cA)B = c(AB) = cO22 = O22 , and so (cA) ∈ V. (h) This set V is not a subspace of M22 since it is not closed under addition. This is proven by 1 1 0 0 1 1 the following counterexample: + = ∈ / V (because the product 0 0 1 1 1 1 (1)(1)(1)(1) of its entries is not zero). (3) In each part, we will refer to the given set as V. If V is a subspace of P5 , we will prove it using Theorem 4.2 by showing that it is nonempty and that it is closed under both polynomial addition and scalar multiplication. If V is not a subspace, we will show that it lacks at least one of these properties. Or, since every subspace of P5 contains z, the zero polynomial, we can also show that V is not a subspace by proving that it does not contain z. In that case, it is not necessary to also prove that one of the closure properties fails. (a) This set V is a subspace of P5 . Clearly, z ∈ V because both the fifth degree term and the first degree term of z are zero. Thus, V is nonempty. Next, we show that V is closed under addition. (ax5 + bx4 + cx3 + dx2 + ax + e) + (rx5 + sx4 + tx3 + ux2 + rx + w) = (a + r)x5 + (b + s)x4 + (c + t)x3 + (d + u)x2 + (a + r)x + (e + w), which is in V since the coefficient of x5 equals the coefficient of x. Finally, we show that V is closed under scalar multiplication. t(ax5 + bx4 + cx3 + dx2 + ax + e) = (ta)x5 + (tb)x4 + (tc)x3 + (td)x2 + (ta)x + (te), which is in V, since the coefficient of x5 equals the coefficient of x. (b) This set V is a subspace of P5 . Now z(3) = 0, so z ∈ V. Hence, V is nonempty. Suppose p, q ∈ V. Then V is closed under addition because (p + q)(3) = p(3) + q(3) = 0 + 0 = 0, proving (p + q) ∈ V. Also, V is closed under scalar multiplication because (cp)(3) = cp(3) = c(0) = 0, which shows that (cp) ∈ V. / V, since the degree of z is 0. (Note: It can also (e) This set V is not a subspace of P5 . Note that z ∈ be shown that V is not closed under either operation.) (g) This set V is a subspace of P5 . Now z (4) = 0, so z ∈ V. Hence, V is nonempty. Suppose p, q ∈ V. Then V is closed under addition because (p + q) (4) = p (4) + q (4) = 0 + 0 = 0, proving 117


Section 4.3

(p + q) ∈ V. Also, V is closed under scalar multiplication because (cp) (4) = cp (4) = c(0) = 0, which shows that (cp) ∈ V. (12) (e) No; if |A| = 0 and c = 0, then |cA| = |On | = 0. (For a specific counterexample, consider A = In .) (15) If λ is not an eigenvalue for A, then, by the definition of an eigenvalue, there are no nonzero vectors X such that AX = λX. Thus, no nonzero vector can be in S. However, A0 = 0 = λ0, hence 0 ∈ S. Therefore, S = {0}, the trivial subspace of Rn . (22) (a) False. By Theorem 4.2, a nonempty subset W of a vector space V must be closed under addition and scalar multiplication in order to be a subspace of V. For example, in Example 5 in the textbook, the first quadrant in R2 is a nonempty subset of R2 that is not a subspace of R2 . (b) True. Every vector space has itself as a subspace. The trivial subspace {0} is also a subspace of every vector space. (Note that if V = {0}, these two subspaces are equal.) (c) False. The plane W must pass through the origin, or else 0 ∈ / W, and so W is not a subspace. For example, the plane x + y + z = 1 is not a subspace for this reason. (d) True. Let V be the set of lower triangular 5 × 5 matrices. To show that V is a subspace of M55 , we must show that it is nonempty and that it is closed under both addition and scalar multiplication. Since O5 is lower triangular, O5 ∈ V, and so V is nonempty. Next, let A, B ∈ V. Then aij = bij = 0 for all i < j. If C = A + B, then for i < j, cij = aij + bij = 0 + 0 = 0, and so C ∈ V. Hence, V is closed under addition. Similarly, if D = rA, then for i < j, dij = r(aij ) = r(0) = 0, proving D ∈ V. Therefore, V is also closed under scalar multiplication. (e) True. Let V be the given subset of R4 . To show that V is a subspace of R4 , we must show that it is nonempty and that it is closed under both matrix addition and scalar multiplication. Letting a = b = 0 shows that [0, 0, 0, 0] ∈ V, hence, V is nonempty. V is closed under addition because [0, a, b, 0] + [0, c, d, 0] = [0, (a + c), (b + d), 0] = [0, A, B, 0] ∈ V, where A = (a + c) and B = (b + d). Finally, V is closed under scalar multiplication since c[0, a, b, 0] = [0, ca, cb, 0] = [0, C, D, 0] ∈ V, where C = ca and D = cb. (f) False. While containing the zero vector is a necessary condition for a subset of a vector space to be a subspace, it is not a sufficient condition. For example, the subset W = {[0, 0], [1, 0]} of R2 contains the zero vector but is not a subspace of R2 . It is not closed under scalar multiplication because 2[1, 0] = [2, 0] ∈ / W. (Note that W also fails to be closed under addition.) (g) True. This is the statement of Theorem 4.3. (h) True. Theorem 4.4 asserts that the eigenspace corresponding to any eigenvalue of any n×n matrix is a subspace of Rn .

Section 4.3 (1) In each part we follow the three steps of 1 (a) Step 1: Form the matrix A = 2 1 Step 2: A row reduces to C = 0

the Simplified Span Method. 1 0 . −3 −5 0 −1 . 1 1

Step 3: span(S) = {a[1, 0, −1] + b[0, 1, 1] | a, b ∈ R} = {[a, b, −a + b] | a, b ∈ R}.

118


Section 4.3

⎡

⎤ 1 −1 1 3 ⎦. (c) Step 1: Form the matrix A = ⎣ 2 −3 0 1 −1 ⎡ ⎤ 1 0 0 Step 2: A row reduces to C = ⎣ 0 1 −1 ⎦. 0 0 0 Step 3: Using only the nonzero rows of C gives span(S) = {a[1, 0, 0] + b[0, 1, −1] | a, b ∈ R} = {[a, b, −b] | a, b ∈ R}. ⎡ ⎤ 1 3 0 1 ⎢ 0 0 1 1 ⎥ ⎥ (e) Step 1: Form the matrix A = ⎢ ⎣ 0 1 0 1 ⎦. 1 5 1 4 ⎤ ⎡ 1 0 0 −2 ⎢ 0 1 0 1 ⎥ ⎥. Step 2: A row reduces to C = ⎢ ⎣ 0 0 1 1 ⎦ 0 0 0 0 Step 3: Using only the nonzero rows of C gives span(S) = {a[1, 0, 0, −2] + b[0, 1, 0, 1] + c[0, 0, 1, 1] | a, b, c ∈ R} = {[a, b, c, −2a + b + c] | a, b, c ∈ R}. (2) (a) First, we convert the polynomials in S into vectors in R4 : (x3 − 1) → [1, 0, 0, −1], (x2 − x) → [0, 1, −1, 0], and (x − 1) → [0, 0, 1, −1]. Now, we use the Simplified Span Method on these vectors. To do this, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ 1 0 0 −1 1 0 0 −1 0 ⎦ to obtain C = ⎣ 0 1 0 −1 ⎦. A = ⎣ 0 1 −1 0 0 1 −1 0 0 1 −1 We must convert the nonzero rows of C to polynomial form: [1, 0, 0, −1] → (x3 − 1), [0, 1, 0, −1] → (x2 − 1), and [0, 0, 1, −1] → (x − 1). Hence, span(S) is the set of linear combinations of these 3 polynomials. That is, span(S) = {a(x3 − 1) + b(x2 − 1) + c(x − 1) | a, b, c ∈ R} = {ax3 + bx2 + cx − (a + b + c) | a, b, c ∈ R}. (c) First, we convert the polynomials in S into vectors in R4 : (x3 − x + 5) → [1, 0, −1, 5], (3x3 − 3x + 10) → [3, 0, −3, 10], (5x3 − 5x − 6) → [5, 0, −5, −6] and (6x − 6x3 − 13) → [−6, 0, 6, −13]. Now we use the Simplified Span Method on these vectors. To do this, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ 1 0 −1 0 1 0 −1 5 ⎢ 0 0 ⎢ 3 0 −3 0 1 ⎥ 10 ⎥ ⎢ ⎥. ⎥ A=⎢ ⎣ 5 0 −5 −6 ⎦ to obtain C = ⎣ 0 0 0 0 ⎦ 0 0 0 0 −6 0 6 −13 We must convert the nonzero rows of C to polynomial form: [1, 0, −1, 0] → (x3 − x), and [0, 0, 0, 1] → (1). Hence, span(S) is the set of linear combinations of these 2 polynomials. That is, span(S) = {a(x3 − x) + b(1) | a, b ∈ R} = {ax3 − ax + b | a, b ∈ R}. (3) (a) First, we convert the matrices in S into vectors in R4 : −1 1 0 0 −1 → [−1, 1, 0, 0], → [0, 0, 1, −1], and 0 0 1 −1 0

119

0 1

→ [−1, 0, 0, 1]. Now we use


the Simplified Span Method on these vectors. ⎡ ⎤ ⎡ −1 1 0 0 1 A = ⎣ 0 0 1 −1 ⎦ to obtain C = ⎣ 0 −1 0 0 1 0

Section 4.3

To do this, we row reduce the matrix ⎤ 0 0 −1 1 0 −1 ⎦. 0 1 −1

We must convert the nonzero rows of C to matrix form: 1 0 0 1 0 [1, 0, 0, −1] → , [0, 1, 0, −1] → , and [0, 0, 1, −1] → 0 −1 0 −1 1

0 −1

. Hence,

span(S) is the set of linear combinations of these 3 matrices. That is, span(S) = . . ) " " ) " a b 0 0 "" 1 0 0 1 " a, b, c ∈ R . a, b, c ∈ R = +c a +b c −a − b − c " 1 −1 " 0 −1 0 −1 (c) First, we convert the matrices in S into vectors in R4 : 1 −1 2 −1 −1 4 → [1, −1, 3, 0], → [2, −1, 8, −1], → [−1, 4, 4, −1], 3 0 8 −1 4 −1 3 −4 and → [3, −4, 5, 6]. Now we use the Simplified Span Method on these vectors. To do 5 6 this, we row ⎡ 1 ⎢ 2 A=⎢ ⎣ −1 3

reduce the matrix ⎡ ⎤ −1 3 0 ⎢ −1 8 −1 ⎥ ⎥ to obtain C = ⎢ ⎣ 4 4 −1 ⎦ −4 5 6

1 0 0 0

0 1 0 0

0 0 1 0

⎤ 0 0 ⎥ ⎥. 0 ⎦ 1

1 0

0 0

,

We must convert the nonzero rows of C to matrix form: [1, 0, 0, 0] → 0 1 0 0 0 0 [0, 1, 0, 0] → , [0, 0, 1, 0] → , and [0, 0, 0, 1] → . Hence, span(S) is the 0 0 1 0 0 1 set of linear combinations of these 4 matrices. ) 1 0 0 1 0 That is, span(S) = a +b +c 0 0 0 0 1 ) " . a b "" = a, b, c, d ∈ R = M22 . c d "

0 0

+d

0 0

0 1

" . " " a, b, c, d ∈ R "

(4) (a) W = {[a + b, a + c, b + c, c} | a, b, c ∈ R} = {a[1, 1, 0, 0] + b[1, 0, 1, 0] +c[0, 1, 1, 1] | a, b, c ∈ R} ⎤ ⎡ 1 1 0 0 = the row space of A, where A = ⎣ 1 0 1 0 ⎦. 0 1 1 1 ⎡ ⎤ 1 0 0 − 12 ⎢ ⎥ 1 ⎥ (b) A row reduces to B = ⎢ 2 ⎦. ⎣ 0 1 0 0

0

1

1 2

(c) From the Simplified Span Method, we know that W also equals the row space of B, which equals {a[1, 0, 0, − 12 ] + b[0, 1, 0, 12 ] + c[0, 0, 1, 12 ] | a, b, c ∈ R} = {[a, b, c, − 12 a + 12 b + 12 c] | a, b, c ∈ R}. (11) Suppose a(x3 − 2x2 + x − 3) + b(2x3 − 3x2 + 2x + 5) + c(4x2 + x − 3) + d(4x3 − 7x2 + 4x − 1) = 3x3 − 8x2 + 2x + 16. Equating the coefficients of the x3 term, the x2 term, etc., from each side of 120


Section 4.3

this equation produces the following linear system: ⎧ a + 2b + 4d = 3 ← x3 terms ⎪ ⎪ ⎨ −2a − 3b + 4c − 7d = −8 ← x2 terms . a + 2b + c + 4d = 2 ← x terms ⎪ ⎪ ⎩ −3a + 5b − 3c − d = 16 ← constant terms Using either Gaussian elimination or the Gauss-Jordan method on this system produces the solution set {[−2d − 1, −d + 2, −1, d] | d ∈ R}. Choosing d = 0 produces the particular solution [−1, 2, −1, 0]. This vector provides coefficients for the desired linear combination. That is, 3x3 − 8x2 + 2x + 16 = −1(x3 − 2x2 + x − 3) + 2(2x3 − 3x2 + 2x + 5) −1(4x2 + x − 3) + 0(4x3 − 7x2 + 4x − 1). Other particular solutions to the system will yield alternate linear combinations that produce 3x3 − 8x2 + 2x + 16. Finally, note that the augmented matrix that was row reduced in this solution has the coefficients of each polynomial as its columns. (14) (a) Theorem 1.13 shows that every 2 × 2 matrix A can be expressed as the sum of a symmetric matrix S and a skew-symmetric matrix V. Hence, every matrix in M22 is a linear combination of a matrix S in S1 and a matrix V in S2 . Therefore, M22 ⊆ span(S1 ∪ S2 ). But since span(S1 ∪ S2 ) ⊆ M22 , we have span(S1 ∪ S2 ) = M22 . " ⎡ ⎤ " ⎤ ⎡ 1 32 − 45 "" 0 10 15 −8 "" 0 ⎢ ⎥ " −8 "" 0 ⎦ to obtain ⎣ 0 0 0 " 0 ⎦. (16) (a) Row reduce [(1I3 − A)|0] = ⎣ 10 15 " 30 45 −24 " 0 0 0 0 " 0 The solution set to the associated linear system is {[− 32 b + 45 c, b, c] | b, c ∈ R}. Choosing b = 2 (to eliminate fractions) and c = 0 produces the fundamental eigenvector [−3, 2, 0]. Choosing b = 0 and c = 5 (to eliminate fractions) yields the fundamental eigenvector [4, 0, 5]. Hence, the desired set of two fundamental eigenvectors is {[−3, 2, 0], [4, 0, 5]}. (21) First, S1 ⊆ S2 ⊆ span(S2 ), by Theorem 4.5, part (1). Then, since span(S2 ) is a subspace of V containing S1 (by Theorem 4.5, part (2)), span(S1 ) ⊆ span(S2 ) by Theorem 4.5, part (3). (24) (b) Suppose S1 = {[1, 0, 0], [0, 1, 0]} and S2 = {[0, 1, 0], [0, 0, 1]}. Then S1 ∩ S2 = {[0, 1, 0]}, and so span(S1 ∩ S2 ) = {a[0, 1, 0] | a ∈ R} = {[0, a, 0] | a ∈ R}. Next, span(S1 ) = {b[1, 0, 0] + c[0, 1, 0] | b, c ∈ R} = {[b, c, 0] | b, c ∈ R}. Similarly, span(S2 ) = {d[0, 1, 0] + e[0, 0, 1] | d, e ∈ R} = {[0, d, e] | d, e ∈ R}. We want to show that span(S1 ∩ S2 ) = span(S1 ) ∩ span(S2 ). This is done by proving the two set inclusions span(S1 ∩ S2 ) ⊆ span(S1 ) ∩ span(S2 ) and span(S1 ∩ S2 ) ⊇ span(S1 ) ∩ span(S2 ). The first of these is already proven for any sets S1 and S2 in part (a). To prove the second, we must show that if x ∈ span(S1 ) ∩ span(S2 ), then x ∈ span(S1 ∩ S2 ). So, suppose x ∈ span(S1 ) ∩ span(S2 ). Thus, x is in both span(S1 ) and span(S2 ), and so must have the correct form for both sets. But x = [x1 , x2 , x3 ] ∈ span(S1 ) implies that x3 = 0. Similarly, x ∈ span(S2 ) implies that x1 = 0. Hence, x must have the form [0, x2 , 0], implying x ∈ span(S1 ∩ S2 ), since it has the proper form. (c) Suppose S1 = {[1, 0, 0], [0, 1, 0]} and S2 = {[1, 0, 0], [1, 1, 0]}. Then S1 ∩ S2 = {[1, 0, 0]}, and so span(S1 ∩ S2 ) = {a[1, 0, 0] | a ∈ R} = {[a, 0, 0] | a ∈ R}. Now x = [0, 1, 0] ∈ S1 , and so x ∈ span(S1 ), by part (1) of Theorem 4.5. Also, x = [0, 1, 0] = (−1)[1, 0, 0] + (1)[1, 1, 0] shows that x ∈ span(S2 ). Hence, x ∈ span(S1 )∩span(S2 ). But, since x has “1” as its second coordinate, it is not in span(S1 ∩ S2 ). Therefore, span(S1 ∩ S2 ) = span(S1 ) ∩ span(S2 ).

121


Section 4.3

(25) (c) Suppose S1 = {x5 } and S2 = {x4 }. Then S1 ∪ S2 = {x5 , x4 }. Now span(S1 ) = {ax5 | a ∈ R} and span(S2 ) = {bx4 | b ∈ R}. Consider p = x5 + x4 . Then p is clearly in neither span(S1 ) nor span(S2 ), since it does not fit the description of the polynomials in either set. Hence, p∈ / span(S1 ) ∪ span(S2 ). However, p ∈ span(S1 ∪ S2 ) because it is a linear combination of x5 and x4 . Therefore, span(S1 ∪ S2 ) = span(S1 ) ∪ span(S2 ). (28) We show that span(S) is closed under vector addition. Let u and v be two vectors in span(S). Then there are finite subsets {u1 , . . ., uk } and {v1 , . . ., vl } of S such that u = a1 u1 + · · · + ak uk and v = b1 v1 + · · · + bl vl for some real numbers a1 , . . ., ak , b1 , . . . , bl . The natural thing to do at this point is to combine the expressions for u and v by adding corresponding coefficients. However, each of the subsets {u1 , . . ., uk } or {v1 , . . ., vl } may contain elements not found in the other, so it is difficult to match up their coefficients. We need to create a larger set containing all the vectors in both subsets. Consider the finite subset X = {u1 , . . ., uk } ∪ {v1 , . . ., vl } of S. Renaming the elements of X, we can suppose X is the finite subset {w1 , . . ., wm } of S. Hence, there are real numbers c1 , . . ., cm and d1 , . . ., dm such that u = c1 w1 + · · · + cm wm and v = d1 w1 + · · · + dm wm . (Note that ci = aj if wi = uj and that ci = 0 if wi ∈ / {u1 , . . ., uk }. A similar formula gives the value of each di .) Then u+v =

m / i=1

ci wi +

m / i=1

di wi =

m /

(ci + di ) wi ,

i=1

a linear combination of the elements wi in the subset X of S. Thus, u + v ∈ span(S). (29) (a) False. The set S is permitted to be infinite. The restriction is that span(S) contains only the finite linear combinations from the set S. That is, in each separate linear combination, only a finite number of the vectors in S are used. (See Example 3 of Section 4.3 in the textbook for an example of the span of an infinite set.) (b) True. Span(S) is defined to be the set of all such finite linear combinations. (c) False. Theorem 4.5 shows that span(S) is the smallest subspace of V containing S, not the smallest set in V containing S. The smallest set containing S is clearly just S itself (no matter what S is). So, for example, if S = {[1, 0]} ⊆ R2 , then span(S) = {a[1, 0] | a ∈ R}. Hence, span(S) contains vectors such as [2, 0] and [−1, 0], and so is bigger than S itself. (d) False. Part (3) of Theorem 4.5 shows that the set inclusion given in this problem should be reversed. That is, span(S) ⊆ W. For a specific counterexample to the statement in the problem, consider V = W = R2 and S = {[1, 0]}. Then W contains S, but span(S) = {a[1, 0] | a ∈ R} clearly does not contain all of W, since [0, 1] ∈ W, but [0, 1] ∈ / span(S). (e) False. The row space of an m × n matrix is a subspace of Rn , since it consists of the set of linear combinations of the rows of the matrix, each of which has n entries. Therefore, the row space of a 4 × 5 matrix is a subspace of R5 , not R4 . (f) True. This statement summarizes the Simplified Span Method. (g) False. The eigenspace Eλ is the solution set of the homogeneous system (λIn − A)X = 0, which is generally different from the row space of (λIn − A). For a particular example in which these 1 0 two sets are different, consider the matrix A = and the eigenvalue λ = 1. Then 0 2 0 0 λI2 − A = I2 − A = . The row space of (1I2 − A) is {a[0, −1] | a ∈ R}, which is the 0 −1 set of all 2-vectors having a zero in the first coordinate. However, E1 is the solution set to the 122

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ) homogeneous system

0 −x2

= =

Section 4.4

0 , which is {a[1, 0] | a ∈ R}, the set of all 2-vectors having a 0

zero in the second coordinate. Hence, [1, 0] ∈ E1 , but [1, 0] is not in the row space of (1I2 − A). Therefore, E1 and the row space of 1I2 − A are different sets.

Section 4.4 (1) In each part, let S represent the given set. (a) Because S contains only 1 element, S is linearly independent since that element is not the zero vector. (See the discussion in the textbook directly after the definition of linear independence.) (b) Because S contains 2 elements, S is linearly dependent if and only if either vector in S is a linear combination of the other vector in S. But since neither vector in S is a scalar multiple of the other, S is linearly independent instead. (See the discussion in the textbook directly after Example 1.) (c) Because S contains 2 elements, S is linearly dependent if and only if either vector in S is a linear combination of the other vector in S. Since [−2, −4, 10] = (−2)[1, 2, −5], S is linearly dependent. (See the discussion in the textbook directly after Example 1.) (d) S is linearly dependent because S contains the zero vector [0, 0, 0], as explained in Example 4 in the textbook. (e) S is a subset of R3 that contains 4 distinct vectors. Hence, S is linearly dependent by Theorem 4.7. (2) In each part, let S represent the given set of vectors. (a) We follow the Independence Test Method. ⎡ ⎤ 1 3 −2 5 ⎦ , whose columns are the vectors in S. Step 1: Form the matrix A = ⎣ 9 4 −2 5 −7 ⎡ ⎤ 1 0 1 Step 2: A row reduces to B = ⎣ 0 1 −1 ⎦. 0 0 0 Step 3: There is no pivot in column 3, so S is linearly dependent. (b) We follow the Independence Test Method. ⎡ ⎤ 2 4 −2 0 ⎦, whose columns are the vectors in S. Step 1: Form the matrix A = ⎣ −1 −1 3 6 2 Step 2: A row reduces to B = I3 . Step 3: Since every column in I3 is a pivot column, S is linearly independent. (e) We follow the Independence Test Method. ⎡ 2 4 1 ⎢ 5 3 −1 Step 1: Form the matrix A = ⎢ ⎣ −1 1 1 6 4 −1

123

⎤ ⎥ ⎥ , whose columns are the vectors in S. ⎦


1

⎢ ⎢ 0 Step 2: A row reduces to B = ⎢ ⎣ 0 0

0 − 12 1 0 0

Section 4.4

⎤

⎥ ⎥ ⎥. 0 ⎦ 0

1 2

Step 3: There is no pivot in column 3, so S is linearly dependent. (3) In each part, let S represent the given set of polynomials. (a) First, we convert the polynomials in S into vectors in R3 : x2 + x + 1 → [1, 1, 1], x2 − 1 → [1, 0, −1], and x2 + 1 → [1, 0, 1]. Next, we use the Independence Test Method on this set of 3-vectors. ⎡ ⎤ 1 1 1 0 0 ⎦, whose columns are the vectors converted from the Step 1: Form the matrix A = ⎣ 1 1 −1 1 polynomials in S. Step 2: A row reduces to B = I3 . Step 3: Since every column in I3 is a pivot column, S is linearly independent. (c) First, we convert the polynomials in S into vectors in R2 : 2x − 6 → [2, −6], 7x + 2 → [7, 2], and 12x − 7 → [12, −7]. Next, we use the Independence Test Method on this set of 2-vectors. 2 7 12 Step 1: Form the matrix A = , whose columns are the polynomials in S converted −6 2 −7 to vectors. ( ' 1 0 73 46 . Step 2: A row reduces to B = 0 1 29 23 Step 3: There is no pivot in column 3, so S is linearly dependent. (Note that, in this problem, once we converted the polynomials to vectors in R2 , we could easily see from Theorem 4.7 that the vectors are linearly dependent, confirming the result we obtained using the Independence Test Method.) (4) In each part, let S represent the given set of polynomials. (a) First, we convert the polynomials in S into vectors in R3 : x2 −1 → [1, 0, −1], x2 +1 → [1, 0, 1], and x2 +x → [1, 1, 0]. We use the Independence Test Method on this set of 3-vectors. ⎡ ⎤ 1 1 1 Step 1: Form the matrix A = ⎣ 0 0 1 ⎦, whose columns are the polynomials in S converted −1 1 0 into vectors. Step 2: A row reduces to B = I3 . Step 3: Since every column in I3 is a pivot column, S is linearly independent. (c) First, we convert the polynomials in S into vectors in R3 : 4x2 + 2 → [4, 0, 2], x2 + x − 1 → [1, 1, −1], x → [0, 1, 0], and x2 − 5x − 3 → [1, −5, −3]. But this set of four vectors in R3 is linearly dependent by Theorem 4.7. Hence, S is linearly dependent. (e) No vector in S can be expressed as a finite linear combination of the others since each polynomial in S has a distinct degree, and no other polynomial in S has any term of that degree. By the remarks before Example 15 in the textbook, the given set S is linearly independent. 124


Section 4.4

(7) It is useful to have a simplified description of the vectors in span(S). Using the Simplified Span Method, ( ' 1 0 − 12 1 1 0 . Hence, span(S) = {[a, b, (− 12 a+ 12 b)] | a, b ∈ R}. we row reduce to obtain 1 −2 0 1 0 1 2 (b) We want to choose a vector v outside of span(S), because otherwise, S ∪ {v} would be linearly dependent, by the second “boxed” alternate characterization of linear dependence after Example 11 in the textbook. Let v = [0, 1, 0], which, using the simplified form of span(S), is easily seen not to be in span(S). Then S ∪ {v} = {[1, 1, 0], [−2, 0, 1], [0, 1, 0]}. We now use the Independence Test Method. ⎡ ⎤ 1 −2 0 0 1 ⎦, whose columns are the vectors in S ∪ {v}. Step 1: Form the matrix A = ⎣ 1 0 1 0 Step 2: A row reduces to B = I3 . Step 3: Since every column in I3 is a pivot column, S ∪ {v} is linearly independent. (c) Many other choices for v will work. In fact any vector in R3 that is outside of span(S) can be chosen here. In particular, consider v = [0, 0, 1], which does not have the proper form to be in span(S). You can verify that S ∪ {v} = {[1, 1, 0], [−2, 0, 1], [0, 0, 1]} is linearly independent using the Independence Test Method in a manner similar to that used in part (b). (d) Any nonzero vector u ∈ span(S) other than [1, 1, 0] or [−2, 0, 1] will provide the example we need. In particular, consider u = [1, 1, 0] + [−2, 0, 1] = [−1, 1, 1]. Then S ∪ {u} = {[1, 1, 0], [−2, 0, 1], [−1, 1, 1]}. You can verify that S ∪ {u} is linearly dependent by using the Independence Test Method. (Also, since u ∈ span(S), S ∪ {u} is linearly dependent by the alternate characterization of linear dependence mentioned in part (b).) (11) In each part, there are many different correct answers, but only one possibility is given here. (a) Let S = {e1 , e2 , e3 , e4 }. We will use the definition of linear independence to verify that S is linearly independent. Now a1 e1 + a2 e2 + a3 e3 + a4 e4 = [a1 , a2 , a3 , a4 ], which clearly equals the zero vector [0, 0, 0, 0] if and only if a1 = a2 = a3 = a4 = 0. Therefore, S is linearly independent. (c) Let S = {1, x, x2 , x3 }. We will use the definition of linear independence to verify that S is linearly independent. Now a1 (1) + a2 x + a3 x2 + a4 x3 clearly equals the zero polynomial if and only if a1 = a2 = a3 = a4 = 0. Therefore, S is linearly independent. ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 0 1 0 0 0 1 0 0 0 ⎬ ⎨ 1 0 0 (e) Let S = ⎣ 0 0 0 ⎦ , ⎣ 1 0 0 ⎦ , ⎣ 0 0 0 ⎦ , ⎣ 0 1 0 ⎦ . ⎩ ⎭ 0 0 0 0 0 0 1 0 0 0 0 0 Notice that each matrix in S is symmetric. We will use the verify that S is linearly independent. Now ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 1 0 0 0 1 0 0 0 1 a1 ⎣ 0 0 0 ⎦ + a2 ⎣ 1 0 0 ⎦ + a3 ⎣ 0 0 0 ⎦ + a4 ⎣ 0 0 0 0 0 0 1 0 0

definition of linear independence to 0 0 0

0 1 0

⎤ ⎡ 0 a1 0 ⎦ = ⎣ a2 a3 0

a2 a4 0

⎤ a3 0 ⎦, 0

which clearly equals the zero matrix O3 if and only if a1 = a2 = a3 = a4 = 0. Therefore, S is linearly independent. (13) (b) Let S be the given set of three vectors. We prove that v = [0, 0, −6, 0] is redundant. To do this, we need to show that span(S − {v}) = span(S). We will do this by applying the Simplified 125


Section 4.4

Span Method to both S − {v} and S. If the method leads to the same reduced row echelon form (except, perhaps, with an extra row of zeroes) with or without v as one of the rows, then the two spans are equal. For span(S − {v}): 1 1 0 0 1 row reduces to 1 1 1 0 0

1 0

For span(S): ⎡ ⎤ ⎡ 1 1 0 0 1 ⎣ 1 1 1 0 ⎦ row reduces to ⎣ 0 0 0 −6 0 0

0 1 1 0 0

0 0 0 1 0

. ⎤ 0 0 ⎦. 0

Since the reduced row echelon form matrices are the same, except for the extra row of zeroes, the two spans are equal, and v = [0, 0, −6, 0] is a redundant vector. For an alternate approach to proving that span(S − {v}) = span(S), first note that S − {v} ⊆ S, and so span(S − {v}) ⊆ span(S) by Corollary 4.6. Next, [1, 1, 0, 0] and [1, 1, 1, 0] are clearly in span(S − {v}) by part (1) of Theorem 4.5. But v ∈ span(S − {v}) as well, because [0, 0, −6, 0] = (6)[1, 1, 0, 0] + (−6)[1, 1, 1, 0]. Hence, S is a subset of the subspace span(S − {v}). Therefore, by part (3) of Theorem 4.5, span(S) ⊆ span(S − {v}). Thus, since we have proven subset inclusion in both directions, span(S − {v}) = span(S). (16) Let S = {v1 , . . . , vk } ⊆ Rn , with n < k, and let A be the n × k matrix having the vectors in S as columns. Then the system AX = O has nontrivial solutions by Corollary 2.6. But, if X = [x1 , . . . , xk ], then AX = x1 v1 + · · · + xk vk , and so, by the definition of linear dependence, the existence of a nontrivial solution to AX = 0 implies that S is linearly dependent. (19) (b) The converse to the statement is: If S = {v1 , . . . , vk } is a linearly independent subset of Rm , then T = {Av1 , . . . , Avk } is a linearly independent subset of Rn with Av1 , . . . , Avk distinct. We can construct specific counterexamples that contradict either (or both) of the conclusions of the converse. For a specific counterexample in which Av1 , . . . , Avk are distinct but T is linearly 1 1 dependent, let A = and S = {[1, 0], [1, 2]}. Then T = {[1, 1], [3, 3]}. Note that S is 1 1 linearly independent, but the vectors Av1 and Av2 are distinct and T is linearly dependent. For a specificcounterexample in which Av1 , . . . , Avk are not distinct but T is linearly independent, 1 1 use A = and S = {[1, 2], [2, 1]}. Then T = {[3, 3]}. Note that Av1 and Av2 both equal 1 1 [3, 3], and that both S and T are linearly independent. For a final counterexample in which both parts of the conclusion are false, let A = O22 and S = {e1 , e2 }. Then Av1 and Av2 both equal 0, S is linearly independent, but T = {0} is linearly dependent. (27) Suppose that S is linearly independent, and suppose v ∈ span(S) can be expressed both as v = a1 u1 + · · · + ak uk and v = b1 v1 + · · · + bl vl , for distinct u1 , . . ., uk ∈ S and distinct v1 , . . ., vl ∈ S, where these expressions differ in at least one nonzero term. Since the ui ’s might not be distinct from the vi ’s, we consider the set X = {u1 , . . . , uk } ∪ {v1 , . . . , vl } and label the distinct vectors in X as {w1 , . . ., wm }. Then we can express v = a1 u1 + · · · + ak uk as v = c1 w1 + · · · + cm wm , and also v = b1 v1 + · · · + bl vl as v = d1 w1 + · · · + dm wm , choosing the scalars ci , di , 1 ≤ i ≤ m with ci = aj if wi = uj , ci = 0 otherwise, and di = bj if wi = vj , di = 0 otherwise. Since the original linear combinations for v are distinct, we know that ci = di for some i. Now, v − v = 0 = (c1 − d1 )w1 + · · · + (cm − dm )wm . Since {w1 , . . ., wm } ⊆ S, a linearly independent set, each

126


Section 4.5

cj − dj = 0 for every j with 1 ≤ j ≤ m. But this is a contradiction since ci = di . Conversely, assume every vector in span(S) can be uniquely expressed as a linear combination of elements of S. Since 0 ∈ span(S), there is exactly one linear combination of elements of S that equals 0. Now, if {v1 , . . . , vn } is any finite subset of S, we have 0 = 0v1 + · · · + 0vn . Because this representation is unique, it means that in any linear combination of {v1 , . . . , vn } that equals 0, the only possible coefficients are zeroes. Thus, by definition, S is linearly independent. (28) (a) False. [−8, 12, −4] = (−4)[2, −3, 1]. Thus, one vector is a scalar multiple (hence a linear combination) of the other vector. Therefore, the set is linearly dependent by the remarks after Example 1 in the textbook. (b) True. This follows directly from Theorem 4.8. (c) True. See the discussion in the textbook directly after the definition of linear independence. (d) False. This statement directly contradicts the first “boxed” alternate characterization for linear independence following Example 11 in the textbook. (e) True. This is the definition of linear independence for a finite nonempty set. (f) True. This follows directly from Theorem 4.7. (g) False. In order to determine linear independence, we need to row reduce the matrix whose columns are the vectors in S, not the matrix whose rows are the vectors in S. (The matrix A whose rows are the vectors in S is used, instead, in the Simplified Span Method to find a simplified form for span(S).) For a specific counterexample to the given statement, consider S = {i, j, 0} in R2 . ⎡ ⎤ 1 0 Then A = ⎣ 0 1 ⎦, which row reduces to A itself. Note that there is a pivot in both columns 0 0 of A. However, S is linearly dependent since it contains the zero vector (see Example 4 in the textbook). (h) True. This follows directly from Theorem 4.9 (or from Theorem 4.10). (i) True. This follows directly from Theorem 4.8, since the given statement indicates that v3 is a linear combination of v1 and v2 . (Note: In the special cases in which either v3 = v1 or v3 = v2 , it follows that v1 = v2 , which is not allowed.)

Section 4.5 (4) (a) The set S is not a basis because S cannot span R4 . For if S spans R4 , then by part(1) of Theorem 4.13, 4 = dim(R4 ) ≤ |S| = 3, a contradiction. (c) The set S is a basis. To prove this, we will show that S spans R4 and then apply Theorem 4.13. ⎡ ⎤ 7 1 2 0 ⎢ 8 0 1 −1 ⎥ ⎥ We show that S spans R4 using the Simplified Span Method. The matrix A = ⎢ ⎣ 1 0 0 −2 ⎦ 3 0 1 −1 whose rows are the vectors in S reduces to I4 . Hence, span(S) = span({e1 , e2 , e3 , e4 }) = R4 . Now, |S| = 4 = dim(R4 ), and so part (1) of Theorem 4.13 implies that S is a basis for R4 . (e) The set S is not a basis since S is not linearly independent. For if S is linearly independent, then by part (2) of Theorem 4.13, 5 = |S| ≤ dim(R4 ) = 4, a contradiction.

127


Section 4.5

(5) (b) From part (a), the given set B is a maximal linearly independent subset of S. Theorem 4.14 then implies that B is a basis for span(S). Hence, dim(span(S)) = |B| = 2, by the definition of dimension. (c) No. Span(S) = R4 because dim(span(S)) = 2 = 4 = dim(R4 ). If span(S) and R4 were the same vector space, they would have the same dimension. (11) (b) Part (a) shows that the given set B is a basis for V. Since |B| = 5, the definition of dimension yields dim(V) = 5. (c) In part (a), we used the strategy of creating a basis for V by taking the polynomials 1, x, x2 , x3 , and x4 and multiplying them by (x − 2) so that the results would have x = 2 as a root. So, in this part we multiply 1, x, x2 , and x3 by (x − 2)(x − 3) so that both x = 2 and x = 3 are roots of the resulting polynomials. This produces the set D = {(x − 2)(x − 3), x(x − 2)(x − 3), x2 (x − 2)(x − 3), x3 (x − 2)(x − 3)}. We need to show that D is a basis for W. First, we show that D is linearly independent. We use the definition of linear independence. Suppose a1 (x − 2)(x − 3) + a2 x(x − 2)(x − 3) + a3 x2 (x − 2)(x − 3) + a4 x3 (x − 2)(x − 3) = 0. Then, factoring out (x − 2)(x − 3) yields (x − 2)(x − 3)(a1 + a2 x + a3 x2 + a4 x3 ) = 0. But since (x − 2)(x − 3) is zero only if x = 2 or x = 3, a1 + a2 x + a3 x2 + a4 x3 = 0 for all other values of x. Since this is a polynomial, it can equal zero for that many values of x only if a1 = a2 = a3 = a4 = 0. Hence, D is linearly independent. To show that D spans W, let p ∈ W. Then p(2) = p(3) = 0. Therefore, (x − 2) and (x − 3) are factors of p. Thus, since p has degree at most 5, dividing p by (x − 2)(x − 3) produces a polynomial of the form ax3 + bx2 + cx + d. That is, p(x) = (x − 2)(x − 3)(ax3 + bx2 + cx + d). Hence, p(x) = ax3 (x − 2)(x − 3) + bx2 (x − 2)(x − 3) + cx(x − 2)(x − 3) + d(x − 2)(x − 3), which shows that p ∈ span(D). Therefore, W ⊆ span(D). Since D is clearly a subset of W, part (3) of Theorem 4.5 shows that span(D) ⊆ W. Thus, span(D) = W. This completes the proof that D is a basis for W. (d) Part (c) shows that the set D defined there is a basis for W. Since |D| = 4, the definition of dimension yields dim(W) = 4. (12) (a) Let V = R3 , and let S = {[1, 0, 0], [2, 0, 0], [3, 0, 0]}. Now dim(V) = 3 = |S|. However, S does not span R3 . Every element of span(S) is of the form a[1, 0, 0] + b[2, 0, 0] + c[3, 0, 0], which must have zeroes in the second and third coordinates. Therefore, vectors in R3 such as [0, 1, 0] and [0, 0, 1] are not in span(S) because they are not of the correct form. (b) Let V = R3 and let T = {[1, 0, 0], [2, 0, 0], [3, 0, 0]}. Now dim(V) = 3 = |T |, but T is not linearly independent. This is clear, since [3, 0, 0] = [1, 0, 0] + [2, 0, 0], showing that one vector in T is a linear combination of the other vectors in T . (18) (b) First, span(B) is a subspace of V, and so by part (3) of Theorem 4.5, span(S) ⊆ span(B). Thus, since span(S) = V, V ⊆ span(B). Also, B ⊆ V, and so span(B) ⊆ V, again by part (3) of Theorem 4.5. Hence, span(B) = V. (c) Now B ⊂ C ⊆ S, but B = C, since w ∈ / B. Therefore, the definition for “B is a maximal linearly independent subset of S” given in the textbook implies that C is linearly dependent. (19) (b) Let S be a spanning set for a vector space V. If S is linearly dependent, then there is a subset C of S such that C = S and C spans V.

128


Section 4.5

(c) The statement in part (b) is essentially half of the “if and only if” statement in Exercise 12 in Section 4.4. We rewrite the appropriate half of the proof of that exercise. Since S is linearly dependent, then S is nonempty and a1 v1 + · · · + an vn = 0 for some {v1 , . . . , vn } ⊆ S and a1 , . . . , an ∈ R, with some ai = 0. Then ai+1 an vi = − aa1i v1 − · · · − ai−1 ai vi−1 − ai vi+1 − · · · − ai vn . Now, if w ∈ span(S), then w = bvi + c1 w1 + · · · + ck wk , for some {w1 , . . . , wk } ⊆ S − {vi } and b, c1 , . . . , ck ∈ R, where bai−1 bai+1 ban 1 b could be zero. Hence, w = − ba ai v1 − · · · − ai vi−1 − ai vi+1 − · · · − ai vn + c1 w1 + · · · + ck wk ∈ span(S − {vi }). So, span(S) ⊆ span(S − {vi }). Also, span(S − {vi }) ⊆ span(S) by Corollary 4.6. Thus, span(S − {vi }) = span(S) = V. Letting C = S − {vi } completes the proof. (22) (c) Suppose C ⊆ W with T ⊂ C and T = C. We must show that C is linearly dependent. We assume C is linearly independent and try to reach a contradiction. Then |C| ∈ A, by the definition of A. (Note that |C| is finite by Theorem 4.13 because it is a linearly independent subset of the finite dimensional vector space V.) But T ⊂ C and T = C implies that |C| > |T | = n. This contradicts the fact that n is the largest element of A. Therefore, C must be linearly dependent, proving that T is a maximal linearly independent subset of W. (d) In part (c) we proved that T is a maximal linearly independent subset of W. But W is a spanning set for itself since W is a subspace of V. Hence, Theorem 4.14 shows that T is a basis for W. (25) (a) True. This is the definition of a basis for a vector space. (b) False. As we saw in Example 8 in Section 4.5, dim(Pn ) = n + 1. Hence, dim(P4 ) = 5, implying that every basis for P4 has five elements, not four. (c) False. As we saw in Example 9 in Section 4.5, dim(Mmn ) = m · n. Hence, dim(M43 ) = 4 · 3 = 12, not 7. (d) False. Part (1) of Theorem 4.13 shows, instead, that |S| ≥ n. For a specific counterexample to the statement in the problem, consider S = {[1, 0], [0, 1], [1, 1]}, a subset of R2 . Now S spans R2 because every 2-vector can be expressed as a linear combination of vectors in S as follows: [a, b] = a[1, 0] + b[0, 1]. However, |S| = 3 > 2 = dim(R2 ). (e) False. Theorem 4.13 shows, instead, that |T | ≤ n. For a specific counterexample to the equation |T | = n in the given statement, consider T = {[1, 0]}, a subset of R2 . Then since [1, 0] = [0, 0], T is linearly independent by the comments in the textbook directly after the definition of linear independence in Section 4.4. But |T | = 1 < 2 = dim(R2 ). (f) True. This is proven using both parts of Theorem 4.13 as follows: |T | ≤ dim(W) (by part (2)) ≤ |S| (by part (1)). (g) False. Recall that V is a subspace of itself. Letting W = V gives a case in which dim(W) = dim(V), contradicting the strict inequality given in the problem. (h) False. Consider the infinite dimensional vector space P from Exercise 16 in Section 4.5. P5 is a finite dimensional subspace of P. (Note: dim(P5 ) = 6.) (i) False. We can only conclude that both S and T are bases for V (Theorems 4.14 and 4.15). If V is finite dimensional, that would imply that |S| = |T | but not necessarily that S = T . Any two different bases for the same vector space can be used to provide a counterexample. For a particular counterexample, let S = {[1, 0], [0, 1]} and let T = {[2, 0], [0, 2]}. Now S is a spanning set for R2 because [a, b] = a[1, 0] + b[0, 1]. Also, any proper subset of S has fewer than 2 elements and so by part (1) of Theorem 4.13 cannot span R2 . Hence, S is a minimal spanning set for R2 . Also, T is a linearly independent subset of R2 because a[2, 0] + b[0, 2] = [2a, 2b] = [0, 0] implies 2a = 2b = 0, yielding a = b = 0. But any larger set of vectors containing T will have more than 129


Section 4.6

2 elements, and so cannot be linearly independent, by part (2) of Theorem 4.13. Thus, T is a maximal linearly independent subset of R2 . Note that S = T . (j) True. Since A is nonsingular, A row reduces to I4 . The Simplified Span Method then shows that the span of the rows of A equals the span of the rows of I4 , which equals R4 . But, since A has exactly 4 rows, the size of this spanning set equals dim(R4 ) = 4. Part (1) of Theorem 4.13 then proves that the rows of A form a basis for R4 .

Section 4.6 (1) (a) In accordance with the Simplified Span Method, we row reduce the matrix whose rows are the vectors in the given set S. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 0 0 2 −2 1 2 3 −1 0 ⎢ ⎢ 3 0 1 ⎥ 6 8 −2 0 ⎥ ⎥. ⎢ ⎥ to obtain ⎢ 0 1 0 ⎣ 0 0 1 −1 ⎣ −1 −1 −3 0 ⎦ 1 1 ⎦ 0 0 0 0 0 −2 −3 −5 1 1 The set of nonzero rows of the row reduced matrix is a basis B for span(S). That is, B = {[1, 0, 0, 2, −2], [0, 1, 0, 0, 1], [0, 0, 1, −1, 0]}. (d) In accordance with the Simplified Span Method, we row reduce the matrix whose rows are the vectors in the given set S. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 0 0 −2 − 13 4 1 1 1 1 1 ⎢ 9 ⎥ ⎢ 1 2 3 4 ⎢ 0 1 0 ⎥ 3 5 ⎥ 2 ⎥ ⎢ ⎥ to obtain ⎢ . ⎢ ⎣ 0 1 2 3 ⎦ 1 4 0 −4 ⎥ ⎣ 0 0 1 ⎦ 0 0 4 0 −1 0 0 0 0 0 The set of nonzero rows of the row reduced matrix is basis B for span(S). That is, 9 1 B = {[1, 0, 0, −2, − 13 4 ], [0, 1, 0, 3, 2 ], [0, 0, 1, 0, − 4 ]}. (2) First, we convert the polynomials in S into vectors in R4 : (x3 − 3x2 + 2) → [1, −3, 0, 2], (2x3 − 7x2 + x − 3) → [2, −7, 1, −3], and (4x3 − 13x2 + x + 5) → [4, −13, 1, 5]. Next, using the Simplified Span Method, we row reduce the matrix whose rows are the converted vectors, above. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 −3 0 2 1 0 −3 0 ⎣ 2 −7 1 −3 ⎦ to obtain ⎣ 0 1 −1 0 ⎦. 4 −13 1 5 0 0 0 1 A basis B for span(S) is found by converting the nonzero rows of the reduced matrix into polynomials in P3 . This produces B = {x3 − 3x, x2 − x, 1}. (3) First, we convert the matrices in S into vectors in R6 . This is done by changing each 3 × 2 matrix A into its corresponding 6-vector [a11 , a12 , a21 , a22 , a31 , a32 ]. Doing this for each of the 4 matrices in S yields the vectors [1, 4, 0, −1, 2, 2], [2, 5, 1, −1, 4, 9], [1, 7, −1, −2, 2, −3], and [3, 6, 2, −1, 6, 12]. Next, using the Simplified Span Method, we row reduce the matrix whose rows are these vectors, above. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 4 1 0 2 0 3 3 1 4 0 −1 2 2 ⎢ ⎥ ⎢ 2 5 1 −1 4 9 ⎥ 0 1 − 13 − 13 0 0 ⎥ ⎢ ⎥ to obtain ⎢ ⎢ ⎥. ⎣ 1 7 −1 −2 2 −3 ⎦ ⎣ 0 0 0 0 0 1 ⎦ 3 6 2 −1 6 12 0 0 0 0 0 0 130


Section 4.6

A basis B for span(S) is found by converting the nonzero rows of the row reduced matrix back into ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 0 1 0 0 ⎬ ⎨ 1 0 3 × 2 matrices. Hence, B = ⎣ 43 31 ⎦ , ⎣ − 13 − 13 ⎦ , ⎣ 0 0 ⎦ . ⎭ ⎩ 0 1 2 0 0 0 (4) In parts (a), (c), and (e), we use the Independence Test Method, as described in Section 4.6. That is, we create a matrix whose columns are the vectors in the given set S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis for span(S), which is a subset of S. In part (h), we must use a different procedure because the set S is infinite. (a) Creating the matrix whose columns are the vectors in S, we get ⎤ ⎤ ⎡ ⎡ 3 0 6 1 0 0 ⎣ 1 0 2 ⎦, which row reduces to ⎣ 0 0 1 ⎦. −2 0 −3 0 0 0 Since columns 1 and 3 are pivot columns, we choose the 1st and 3rd columns of the original matrix for our basis for span(S). Hence, B = {[3, 1, −2], [6, 2, −3]}. (c) Creating the matrix whose columns are the vectors in S, we get ⎡ ⎤ ⎡ ⎤ 1 2 3 0 −2 1 0 −3 0 0 ⎣ 3 1 −6 1 1 ⎦, row reducing to ⎣ 0 1 3 0 −1 ⎦. −2 4 18 −1 −6 0 0 0 1 2 Since columns 1, 2, and 4 are pivot columns, we choose the 1st, 2nd, and 4th columns of the original matrix for our basis for span(S). Hence, B = {[1, 3, −2], [2, 1, 4], [0, 1, −1]}. (e) Creating the matrix whose columns are the vectors in S, ⎡ ⎤ ⎡ 3 1 3 −1 1 0 ⎣ −2 2 −2 −10 ⎦, which row reduces to ⎣ 0 1 2 −1 7 6 0 0

we get

⎤ 0 1 0 −4 ⎦. 1 0

Since columns 1, 2, and 3 are pivot columns, we choose the 1st, 2nd, and 3rd columns of the original matrix for our basis for span(S). Hence, B = {[3, −2, 2], [1, 2, −1], [3, −2, 7]}. (h) Because the given set S is infinite, we will use the Inspection Method. We notice that the 2nd coordinates of the vectors in S are described in terms of their 1st and 3rd coordinates, that is, x2 = −3x1 + x3 . This inspires us to consider the 1st and 3rd coordinates as “independent variables” and solve for the second coordinate, creating vectors in a manner similar to Step 3 of the Diagonalization Method of Section 3.4. Thus, we let x1 = 1 and x3 = 0 to obtain x2 = (−3)(1) + 0 = −3. This gives us the vector [1, −3, 0]. Next, letting x1 = 0 and x3 = 1 yields x2 = (−3)(0) + 1 = 1, producing [0, 1, 1]. Now the set B = {[1, −3, 0], [0, 1, 1]} is clearly linearly independent, since neither vector is a scalar multiple of the other. We claim that B is a maximal linearly independent subset of S, and hence is a basis for span(S) by Theorem 4.14. To do this, we need to show that every vector in S is a linear combination of the vectors in B, because then, adding another vector from S to B would make B linearly dependent. So, let v = [v1 , v2 , v3 ] ∈ S. Then, v2 = −3v1 + v3 . Hence, v = [v1 , −3v1 + v3 , v3 ] = v1 [1, −3, 0] + v3 [0, 1, 1], and so v is a linear combination of the vectors in B. (5) (a) First, we convert the polynomials in S into vectors in R4 : (x3 − 8x2 + 1) → [1, −8, 0, 1], (3x3 − 2x2 + x) → [3, −2, 1, 0], (4x3 + 2x − 10) → [4, 0, 2, −10],

131


Section 4.6

(x3 − 20x2 − x + 12) → [1, −20, −1, 12], (x3 + 24x2 + 2x − 13) → [1, 24, 2, −13], and (x3 + 14x2 − 7x + 18) → [1, 14, −7, 18]. Next, we use the Independence Test Method. That is, we create a matrix whose columns are the vectors above. We row reduce this matrix and note which columns are pivot columns. The polynomials for the vectors in the corresponding columns of the original matrix constitute a basis for span(S), which is a subset of S. Thus, we row reduce the matrix ⎡ ⎤ ⎡ ⎤ 1 0 0 0 −3 205 1 3 4 1 1 1 ⎢ 0 1 0 0 0 83 ⎥ ⎢ −8 −2 0 −20 24 14 ⎥ ⎥ ⎢ ⎥ to ⎢ ⎢ ⎥. 181 ⎣ 0 1 − 2 ⎦ 1 2 −1 2 −7 ⎦ ⎣ 0 0 1 0 1 0 −10 12 −13 18 0 0 0 1 0 −91 Since columns 1, 2, 3, and 4 are pivot columns, we choose the polynomials corresponding to the 1st, 2nd, 3rd, and 4th columns of the original matrix for our basis for span(S). Hence, B = {x3 − 8x2 + 1, 3x3 − 2x2 + x, 4x3 + 2x − 10, x3 − 20x2 − x + 12}. (c) Because the set S is infinite, we use the Inspection Method. We begin by choosing the nonzero polynomial x3 in S. Then we must choose more elements from S that are independent from those previously chosen. We choose x2 and x, each of which is clearly not in the span of the polynomials chosen before them. This gives us B = {x3 , x2 , x}. We must show that B is a maximal linearly independent subset of S, which will make it a basis for span(S), by Theorem 4.14. To do this, we must show that every polynomial in S can be expressed as a linear combination of the polynomials in B. For then, adding any polynomial from S to B would make B linearly dependent. Now every polynomial p in S is of the form p(x) = ax3 + bx2 + cx, since it is in P3 and has zero constant term. But this form clearly expresses p as a linear combination of the polynomials in B. (e) Because the set S is infinite, we use the Inspection Method. We begin by choosing the nonzero polynomial x3 + x2 in S. Then we must choose more elements from S that are independent from those previously chosen. We choose x and 1, each of which is clearly not in the span of the polynomials chosen before them. This gives us B = {x3 + x2 , x, 1}. We must show that B is a maximal linearly independent subset of S, which will make it a basis for span(S), by Theorem 4.14. To do this, we must show that every polynomial in S can be expressed as a linear combination of the polynomials in B. For then, adding any polynomial from S to B would make B linearly dependent. Now every polynomial p in S is of the form p(x) = ax3 + ax2 + bx + c, since it is in P3 and has the coefficients of its x2 and x3 terms equal. But then p(x) = a(x3 + x2 ) + bx + c(1), a linear combination of the polynomials in B. (6) (a) Recall that the standard basis for M33 is B = {Ψij | 1 ≤ i ≤ 3, 1 ≤ j ≤ 3}, where Ψij is the 3 × 3 matrix having 1 as its (i, j) entry and zeroes elsewhere. (See Example 3 in Section 4.5.) Note that B is a subset of the given set S. B is linearly independent and spans M33 . Now span(S) ⊇ span(B) = M33 , and so, since span(S) cannot be bigger than M33 , span(S) = M33 . Hence, B is a basis for span(S). (c) Because the given set S is infinite, we will use the method described in Section 4.6 for shrinking an infinite spanning set to a basis. The first step is to guess at such a basis. To help us make this guess, we first observe that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ a b c 1 0 0 0 1 0 0 0 1 0 0 0 ⎣ b d e ⎦ = a⎣ 0 0 0 ⎦ + b⎣ 1 0 0 ⎦ + c⎣ 0 0 0 ⎦ + d⎣ 0 1 0 ⎦ c e f 0 0 0 0 0 0 1 0 0 0 0 0

132

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎤ ⎡ ⎤ 0 0 0 0 0 0 + e ⎣ 0 0 1 ⎦ + f ⎣ 0 0 0 ⎦. Thus, we choose 0 1 0 0 0 1 ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 0 1 0 0 0 1 0 ⎨ 1 0 0 ⎣ 0 0 0 ⎦,⎣ 1 0 0 ⎦, ⎣ 0 0 0 ⎦, ⎣ 0 ⎩ 0 0 0 0 0 0 1 0 0 0

Section 4.6

⎡

B= 0 1 0

⎤ ⎡ 0 0 0 ⎦, ⎣ 0 0 0

0 0 1

⎤ ⎡ 0 0 1 ⎦,⎣ 0 0 0

0 0 0

⎤⎫ 0 ⎬ 0 ⎦ . ⎭ 1

Note that every matrix in B is symmetric. Also note that we have already shown that every matrix in S is in span(B), since we expressed every symmetric matrix as a linear combination of matrices in B. Finally, we need to check that B is linearly independent. But this is also clear, because none of the matrices in B can be expressed as a linear combination of the others, since each contains a 1 in at least one entry in which all of the other matrices in B have a 0. Now, no further matrices in S can be added to B and maintain linear independence since every matrix in S is already in span(B). Thus, B is a maximal linearly independent subset of S, and so is a basis for span(S). (7) (a) We use the Enlarging Method. Step 1: We let A = {e1 , e2 , e3 , e4 , e5 } be the standard basis for R5 . Step 2: We form the spanning set S = {[1, −3, 0, 1, 4], [2, 2, 1, −3, 1], e1 , e2 , e3 , e4 , e5 }. Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis B for span(S) = R5 . Since the vectors in T are listed first, this basis will contain T . Hence, we row reduce ⎡ 3 ⎤ 1 ⎤ ⎡ 1 0 0 0 0 13 13 1 2 1 0 0 0 0 ⎢ 0 1 0 0 0 −4 1 ⎥ ⎢ ⎢ −3 13 13 ⎥ 2 0 1 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ 7 5 ⎥ ⎢ ⎢ 0 1 0 0 1 0 0 ⎥ − 13 13 ⎥ to ⎢ 0 0 1 0 0 ⎢ ⎥. ⎢ ⎣ 1 −3 0 0 0 1 0 ⎦ 11 7 ⎥ ⎣ 0 0 0 1 0 13 13 ⎦ 4 1 0 0 0 0 1 4 1 0 0 0 0 1 − 13 13 Since there are pivots in the first 5 columns, the first 5 columns of the original matrix form the desired basis. That is, B = {[1, −3, 0, 1, 4], [2, 2, 1, −3, 1], [1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [0, 0, 1, 0, 0]}. (c) We use the Enlarging Method. Step 1: We let A = {e1 , e2 , e3 , e4 , e5 } be the standard basis for R5 . Step 2: We form the spanning set S = {[1, 0, −1, 0, 0], [0, 1, −1, 1, 0], [2, 3, −8, −1, 0], e1 , e2 , e3 , e4 , e5 }. Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis B for span(S) = R5 . Since the vectors in T are listed first, this basis will contain T . Hence, we row reduce ⎡ ⎤ 5 ⎤ ⎡ 1 0 0 0 − 94 −1 0 4 1 0 2 1 0 0 0 0 ⎢ 0 1 0 0 1 3 0 0 ⎥ ⎢ ⎢ 0 ⎥ 4 4 1 3 0 1 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ 1 1 ⎢ −1 −1 −8 0 0 1 0 0 ⎥ to obtain ⎢ 0 0 1 0 0 −4 0 ⎥ 4 ⎥ ⎢ ⎢ ⎥. ⎢ ⎥ ⎣ 0 7 3 1 −1 0 0 0 1 0 ⎦ 1 −4 0 ⎦ ⎣ 0 0 0 1 4 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 Since there are pivots in the first 4 columns and in column 8, the first 4 columns and column 8 of

133


Section 4.6

the original matrix form the desired basis. That is, B = {[1, 0, −1, 0, 0], [0, 1, −1, 1, 0], [2, 3, −8, −1, 0], [1, 0, 0, 0, 0], [0, 0, 0, 0, 1]}. (8) (a) First we convert the polynomials in T to vectors in R5 : (x3 − x2 ) → [0, 1, −1, 0, 0], and (x4 − 3x3 + 5x2 − x) → [1, −3, 5, −1, 0]. Now we use the Enlarging Method on this set of vectors to find a basis for R5 . Step 1: We let A = {e1 , e2 , e3 , e4 , e5 } be the standard basis for R5 . Step 2: We form the spanning set S = {[0, 1, −1, 0, 0], [1, −3, 5, −1, 0], e1 , e2 , e3 , e4 , e5 }. Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis B for span(S) = R5 . Since the vectors corresponding to the polynomials in T are listed first, this basis will contain T . Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 0 1 1 0 0 0 0 1 0 0 0 −1 −5 0 ⎢ 1 −3 0 1 0 0 0 ⎥ ⎢ 0 1 0 0 0 −1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1 ⎥ ⎢ 5 0 0 1 0 0 ⎥ to ⎢ 0 0 1 0 0 1 0 ⎥ ⎢ ⎥. ⎣ 0 −1 0 0 0 1 0 ⎦ ⎣ 0 0 0 1 1 2 0 ⎦ 0 0 0 0 0 0 1 0 0 0 0 0 0 1 Since there are pivots in columns 1, 2, 3, 4, and 7, columns 1, 2, 3, 4, and 7 of the original matrix form the basis {[0, 1, −1, 0, 0], [1, −3, 5, −1, 0], e1 , e2 , e5 } for R5 . Converting these 5-vectors back to polynomials in P4 yields the basis B = {x3 − x2 , x4 − 3x3 + 5x2 − x, x4 , x3 , 1}. (c) First we convert the polynomials in T to vectors in R5 : (x4 − x3 + x2 − x + 1) → [1, −1, 1, −1, 1], (x3 − x2 + x − 1) → [0, 1, −1, 1, −1], and (x2 − x + 1) → [0, 0, 1, −1, 1]. Now we use the Enlarging Method on this set of vectors to find a basis for R5 . Step 1: We let A = {e1 , e2 , e3 , e4 , e5 } be the standard basis for R5 . Step 2: We form the spanning set S = {[1, −1, 1, −1, 1], [0, 1, −1, 1, −1], [0, 0, 1, −1, 1], e1 , e2 , e3 , e4 , e5 }. Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis B for span(S) = R5 . Since the vectors corresponding to the polynomials in T are listed first, this basis will contain T . Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 ⎢ 0 1 0 1 1 0 0 ⎢ −1 0 ⎥ 1 0 0 1 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 1 −1 ⎥ 1 ⎥ 1 0 0 1 0 0 ⎥ to obtain ⎢ 0 0 1 0 1 0 0 ⎥. ⎢ ⎣ 0 0 0 0 0 1 0 −1 ⎦ ⎣ −1 1 −1 0 0 0 1 0 ⎦ 0 0 0 0 0 0 1 1 1 −1 1 0 0 0 0 1 Since there are pivots in columns 1, 2, 3, 6, and 7, columns 1, 2, 3, 6, and 7 of the original matrix form the basis {[1, −1, 1, −1, 1], [0, 1, −1, 1, −1], [0, 0, 1, −1, 1], e3 , e4 } for R5 . Converting these 5-vectors back to polynomials in P4 yields the basis B = {x4 − x3 + x2 − x + 1, x3 − x2 + x − 1, x2 − x + 1, x2 , x}. (9) (a) First, we convert the matrices in T to vectors in R6 . This is done by changing each 3 × 2 matrix A into the 6-vector [a11 , a12 , a21 , a22 , a31 , a32 ]. Doing this for both of the matrices in T yields the vectors [1, −1, −1, 1, 0, 0] and [0, 0, 1, −1, −1, 1]. Now we use the Enlarging Method on this set of vectors to find a basis for R6 . Step 1: We let A = {e1 , e2 , e3 , e4 , e5 , e6 } be the standard basis for R6 . 134


Section 4.6

Step 2: We form the spanning set S = {[1, −1, −1, 1, 0, 0], [0, 0, 1, −1, −1, 1], e1 , e2 , e3 , e4 , e5 , e6 }. Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis B for span(S) = R6 . Since the vectors corresponding to the matrices in T are listed first, this basis will contain T . Hence, we row reduce ⎤ ⎡ ⎤ ⎡ 1 0 0 0 0 1 0 1 1 0 1 0 0 0 0 0 ⎢ 0 1 0 0 0 ⎢ −1 0 0 1 ⎥ 0 0 1 0 0 0 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢ 0 0 1 0 0 −1 0 −1 ⎥ ⎥ ⎢ −1 1 0 0 1 0 0 0 ⎥. ⎥ to obtain ⎢ ⎢ ⎢ 0 0 0 1 0 ⎢ 1 −1 0 0 0 1 0 0 ⎥ 1 0 1 ⎥ ⎥ ⎢ ⎥ ⎢ ⎣ 0 0 0 0 1 ⎣ 0 −1 0 0 0 0 1 0 ⎦ 1 0 0 ⎦ 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 Since there are pivots in columns 1, 2, 3, 4, 5, and 7, columns 1, 2, 3, 4, 5, and 7 of the original matrix form the basis {[1, −1, −1, 1, 0, 0], [0, 0, 1, −1, −1, 1], e1 , e2 , e3 , e5 } for R6 . Converting these 6-vectors back to matrices in M32 yields the basis ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 1 −1 0 0 1 0 0 1 0 0 0 0 ⎬ ⎨ 1 ⎦ , ⎣ 1 −1 ⎦ , ⎣ 0 0 ⎦ , ⎣ 0 0 ⎦ , ⎣ 1 0 ⎦ , ⎣ 0 0 ⎦ . B = ⎣ −1 ⎭ ⎩ 0 0 −1 1 0 0 0 0 0 0 1 0 (c) First, we convert the matrices in T to vectors in R6 . This is done by changing each 3 × 2 matrix A into the 6-vector [a11 , a12 , a21 , a22 , a31 , a32 ]. Doing this for the four matrices in T yields the vectors [3, 0, −1, 7, 0, 1], [−1, 0, 1, 3, 0, −2], [2, 0, 3, 1, 0, −1], and [6, 0, 0, 1, 0, −1]. Now we use the Enlarging Method on this set of vectors to find a basis for R6 . Step 1: We let A = {e1 , e2 , e3 , e4 , e5 , e6 } be the standard basis for R6 . Step 2: We form the spanning set S = {[3, 0, −1, 7, 0, 1], [−1, 0, 1, 3, 0, −2], [2, 0, 3, 1, 0, −1], [6, 0, 0, 1, 0, −1], e1 , e2 , e3 , e4 , e5 , e6 }. Step 3: We apply the Independence Test Method. To do this, we create the matrix whose columns are the vectors in S. We row reduce this matrix and note which columns are pivot columns. The vectors in the corresponding columns of the original matrix constitute a basis B for span(S) = R6 . Since the vectors corresponding to the matrices in T are listed first, this basis will contain T . Hence, we row reduce ⎡ ⎤ 3 −1 2 6 1 0 0 0 0 0 ⎢ 0 0 0 0 0 1 0 0 0 0 ⎥ ⎢ ⎥ ⎢ −1 1 3 0 0 0 1 0 0 0 ⎥ ⎢ ⎥ ⎢ 7 3 1 1 0 0 0 1 0 0 ⎥ ⎢ ⎥ ⎣ 0 0 0 0 0 0 0 0 1 0 ⎦ 1 −2 −1 −1 0 0 0 0 0 1 ⎡ 35 1 4 53 ⎤ 1 0 0 0 0 0 101 303 303 303 ⎢ 0 1 0 0 − 8 ⎥ 23 32 0 − 303 0 − 121 ⎢ 101 303 303 ⎥ ⎢ ⎥ 4 3 113 58 ⎥ ⎢ 0 0 1 0 0 0 101 303 303 303 ⎢ ⎥. to obtain ⎢ 14 15 5 22 ⎥ ⎢ 0 0 0 1 0 − 101 − 101 0 − 101 ⎥ 101 ⎢ ⎥ ⎣ 0 0 0 0 0 1 0 0 0 0 ⎦ 0

0

0

0

0

0

0

0

1

0

Since there are pivots in columns 1, 2, 3, 4, 6, and 9, columns 1, 2, 3, 4, 6, and 9 of the original matrix form the basis {[3, 0, −1, 7, 0, 1], [−1, 0, 1, 3, 0, −2], [2, 0, 3, 1, 0, −1], [6, 0, 0, 1, 0, −1], e2 , e5 } 135


for R6 . Converting these 6-vectors back to matrices in M32 yields ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 3 0 −1 0 2 0 6 0 0 ⎨ 3 ⎦,⎣ 3 1 ⎦,⎣ 0 1 ⎦,⎣ 0 B = ⎣ −1 7 ⎦ , ⎣ 1 ⎩ 0 1 0 −2 0 −1 0 −1 0

the basis ⎤ ⎡ 1 0 0 ⎦,⎣ 0 0 1

Section 4.6

⎤⎫ 0 ⎬ 0 ⎦ . ⎭ 0

(10) To find a basis for U4 , we first guess at a basis B and then verify that B is, indeed, a basis for U4 . Now, recall that the standard basis for M44 is {Ψij | 1 ≤ i ≤ 4, 1 ≤ j ≤ 4}, where Ψij is the 4 × 4 matrix having 1 as its (i, j) entry and zeroes elsewhere. (See Example 3 in Section 4.5.) However, not all of these matrices are in U4 , since some have nonzero entries below the main diagonal. Let B be the set of these matrices that are upper triangular. That is, B = {Ψij | 1 ≤ i ≤ j ≤ 4}. There are 10 such matrices. We claim that B is a basis for U4 . First, ! we show that B is a linearly independent set. ! Suppose i≤j dij Ψij = O4 . Since the (i, j) entry of i≤j dij Ψij is dij , each dij = 0. Therefore, B is since each nonzero entry linearly independent. Next, we prove that B spans U4 . Let Q ∈ U4 . Then, ! of qij of Q is on or above the main diagonal of Q, we see that Q = i≤j qij Ψij , which shows that Q ∈ span(B). This completes the proof. (11) (b) To find a basis for V, we first guess at a basis B and then verify that B is, indeed, a basis for V. Now, recall that the standard basis for M33 is {Ψij | 1 ≤ i ≤ 3, 1 ≤ j ≤ 3}, where Ψij is the 3 × 3 matrix having 1 as its (i, j) entry and zeroes elsewhere. (See Example 3 in Section 4.5.) However, not all of these matrices are in V, since some have nonzero traces. However, the six matrices Ψij with i = j are all in V. But these 6 are not sufficient to span V, since all have zero entries on the main diagonal. Thus, to get our basis, we want to add more matrices to these, linearly independent of the first 6, and of each other. However, we cannot add more than two matrices, or else the basis would have 9 or more elements, contradicting Theorem 4.16, since V = M33 . Consider ⎡ ⎤ ⎡ ⎤ 1 0 0 0 0 0 0 ⎦ and C2 = ⎣ 0 1 0 ⎦, which are clearly in V. We claim that B = C1 = ⎣ 0 0 0 0 −1 0 0 −1 {Ψij | 1 ≤ i ≤ 3, 1 ≤ j ≤ 3, i = j} ∪ {C1 , C2 } is a basis for V.

! First, we show that B is a linearly independent set. Suppose i=j dij Ψij +c1 C1 +c2 C2 = O3 . ⎤ ⎡ ⎤ ⎡ d13 0 0 0 c1 d12 ⎦ = ⎣ 0 0 0 ⎦. Hence, each dij = 0 for i = j and c1 = c2 = 0. d23 Then ⎣ d21 c2 d31 d32 −c1 − c2 0 0 0

Therefore, B is linearly independent by definition. Next, we prove that B spans V. Let Q ∈ V. Then q33 = −q11 − q22 , and so Q has the ⎤ ⎡ q13 q11 q12 ! ⎦. Thus, Q can be expressed as Q = q23 form ⎣ q21 q22 i=j qij Ψij + q11 C1 + q22 C2 . q31 q32 −q11 − q22 Hence, Q ∈ span(B). This completes the proof that B is a basis. Therefore, dim(V) = |B| = 8. (d) To find a basis for V, we first guess at a basis B and then verify that B is, indeed, a basis for V. Consider that the definition of V involves three “independent variables” a, b, and c. Hence, use the polynomials obtained by first setting a = 1, b = 0, c = 0, then a = 0, b = 1, c = 0, and a = 0, b = 0, c = 1. This yields B = {x6 + x4 + x2 − x + 3, −x5 + x2 − 2, −x3 + x2 + x + 16}. Clearly, B spans V, since ax6 − bx5 + ax4 − cx3 + (a + b + c)x2 − (a − c)x + (3a − 2b + 16c) = a(x6 + x4 + x2 − x + 3) + b(−x5 + x2 − 2) + c(−x3 + x2 + x + 16). To show B is also linearly independent, set a(x6 + x4 + x2 − x + 3) + b(−x5 + x2 − 2) + c(−x3 + x2 + x + 16) = 0. Examining 136


Section 4.6

the coefficient of x6 yields a = 0. Similarly, looking at the coefficients of x5 and x3 produces b = 0 and c = 0. Hence, B is linearly independent by definition. This completes the proof that B is a basis. Therefore, dim(V) = |B| = 3. (12) (b) Let Cij = Ψij − ΨTij , where Ψij is the n × n matrix having 1 as its (i, j) entry and zeroes elsewhere. That is, Cij is the n × n matrix whose (i, j) entry is 1, (j, i) entry is −1, and all other entries are zero. A basis for the skew-symmetric n × n matrices is B = {Cij | 1 ≤ i < j ≤ n}. To show this is a basis, we first prove that B is linearly independent. But each Cij , for i < j, has its (i, j) entry equal to 1, while every other matrix in B has its (i, j) entry equal to zero. Hence, no Cij is a linear combination of the other matrices in B. Therefore, B is linearly independent. Also, if Q is skew-symmetric, then qji = −qij for all i < j. That is, entries qij above ! the main diagonal are paired Thus, ! with their additive inverse !qji below the main diagonal. ! (q Ψ + q Ψ ) = (q Ψ − q Ψ ) = q (Ψ − Ψ ) = q Q = ij ij ji ji ij ij ij ji ij ji i<j i<j i<j ij i<j ij Cij , which shows that B spans the skew-symmetric matrices. This completes the proof that B is a basis. Hence, dim(skew-symmetric matrices) = |B| = (n2 − n)/2. (The number (n2 − n)/2 is computed as follows: The number of nondiagonal entries in an n × n matrix is n2 − n. Half of these, (n2 − n)/2, are above the main diagonal. Since B contains one matrix corresponding to each of these entries, |B| = (n2 − n)/2.) (14) Let V and S be as given in the statement of the theorem. Let A = {k | a set T exists with T ⊆ S, |T | = k, and T linearly independent}. Step 1: The empty set is linearly independent and is a subset of S. Hence, 0 ∈ A, and so A is nonempty. Step 2: Suppose k ∈ A. Then every linearly independent subset T of S is also a linearly independent subset of V. Hence, using part (2) of Theorem 4.13 on T shows that k = |T | ≤ dim(V). Step 3: Suppose n is the largest number in A, which must exist, since A is nonempty and all its elements are ≤ dim(V). Let B be a linearly independent subset of S with |B| = n, which exists because n ∈ A. We want to show that B is a maximal linearly independent subset of S. Now, B is given as linearly independent. Suppose C ⊆ S with B ⊂ C and B = C. We assume C is linearly independent and try to reach a contradiction. Then |C| ∈ A, by the definition of A. (Note that |C| is finite, by an argument similar to that given in Step 2.) But B ⊂ C and B = C implies that |C| > |B| = n. This contradicts the fact that n is the largest element of A. Hence, C is linearly dependent, and so B is a maximal linearly independent subset of S. Step 4: The set B from Step 3 is a basis for V, by Theorem 4.14. This completes the proof. (15) (b) No; consider the subspace W of R3 given by W = {[a, 0, 0] | a ∈ R}. No subset of the basis B = {[1, 1, 0], [1, −1, 0], [0, 0, 1]} for R3 is a basis for W, since none of the vectors in B are in W. (To verify that B is a basis for R3 , note that no vector in B is a linear combination of the other vectors in B. Hence, B is a linearly independent subset of R3 with |B| = 3 = dim(R3 ). Thus, B is a basis for R3 by part (2) of Theorem 4.13.) (c) Yes; consider Y = span(B ). B is linearly independent because it is a subset of the linearly independent set B. Also, B clearly spans Y, so B is a basis for Y. Finally, B ⊆ B ⊆ V, and so Y = span(B ) ⊆ V, by part (3) of Theorem 4.5. Of course, Y is a subspace of V by part (2) of Theorem 4.5. (16) (b) In R3 , consider W = {[a, b, 0] | a, b ∈ R}. We could let W1 = {[0, 0, c] | c ∈ R} or W2 = {[0, c, c] | c ∈ R}. We need to show that W1 and W2 both satisfy the given conditions. For W1 : Suppose v = [v1 , v2 , v3 ] ∈ R3 . Let w = [v1 , v2 , 0] and w = [0, 0, v3 ]. Then w ∈ W and w ∈ W1 , and v = w+w , showing existence for the decomposition. To prove uniqueness, suppose u = [u1 , u2 , 0] ∈ W and u = [0, 0, u3 ] ∈ W1 , and v = u + u . Then [v1 , v2 , v3 ] = [u1 , u2 , u3 ] ⇒ 137


Section 4.7

v1 = u1 , v2 = u2 , and v3 = u3 ⇒ u = w, and u = w . For W2 : Suppose v = [v1 , v2 , v3 ] ∈ R3 . Let w = [v1 , v2 − v3 , 0] and w = [0, v3 , v3 ]. Then w ∈ W and w ∈ W2 , and v = w + w , showing existence. To prove uniqueness, suppose u = [u1 , u2 , 0] ∈ W and u = [0, u3 , u3 ] ∈ W2 , with v = u + u . Then [v1 , v2 , v3 ] = [u1 , u2 + u3 , u3 ]. Hence, v3 = u3 , so u = w . Now, w + w = v = u + u . Subtracting w = u from both sides of this equation yields u = w, finishing the proof. (20) (a) True. This is the statement of Theorem 4.17. (b) True. This is the statement of Theorem 4.18. (c) False. The Simplified Span Method produces a basis B for span(S), but B is not necessarily a subset of S. Instead, B contains vectors of a simpler form than those in S. For example, applying the Simplified Span Method to the spanning set S = {[1, 1], [1, −2]} for R2 produces the standard basis B = {i, j} for R2 . (Try it.) Notice that B is not a subset of S. (d) True. The Independence Test Method uses row reduction to select a maximal linearly independent subset of S, which, by Theorem 4.14, is a basis for span(S). (e) True. The Inspection Method, by repeatedly selecting vectors of S linearly independent from those already chosen, produces a maximal linearly independent subset of S, which, by Theorem 4.14, is a basis for span(S). (f) False. The basis B created by the Enlarging Method satisfies B ⊇ T rather than B ⊆ T . (This can be seen in Example 8 in Section 4.6.) (g) False. The statement has the two methods reversed. The Simplified Span Method actually places the vectors of the given spanning set S as rows in a matrix, while the Independence Test Method places the vectors in S as columns.

Section 4.7 (1) (a) We use the Coordinatization Method. Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors in the basis B, and then find its corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 7 1 5 0 "" 2 [A|v] = ⎣ −4 −7 −4 "" −1 ⎦ −→ ⎣ 0 1 0 "" −1 ⎦ 0 0 1 " −5 1 2 1 " 0 Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: [v]B = [7, −1, −5], the last column of the row reduced matrix. (c) We use the Coordinatization Method. Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors in the basis B, and then find its corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 7 1 0 0 "" −2 2 4 1 "" ⎢ 0 1 0 " 4 ⎥ ⎢ 3 3 2 "" −4 ⎥ ⎥ " ⎥ ⎢ ⎢ ⎥ " ⎥ " ⎢ 5 ⎥ −→ ⎢ 3 1 " [A|v] = ⎢ 1 ⎢ 0 0 1 " −5 ⎥. " " ⎦ ⎣ ⎣ −2 0 0 0 " 0 ⎦ 1 −1 " 13 " 0 0 0 " 0 2 −1 1 −13 Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. 138


Section 4.7

Step 4: We eliminate the last two rows of the row reduced matrix, since they contain all zeroes. Then [v]B = [−2, 4, −5], the remaining entries of the last column. (e) We convert the given polynomials to vectors in R3 by changing ax2 + bx + c to [a, b, c]. Then we use the Coordinatization Method. Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the polynomials in the basis B. Then find its corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 4 3 1 2 "" 13 2 3 "" −5 ⎦ −→ ⎣ 0 1 0 "" −5 ⎦. [A|v] = ⎣ −1 0 0 1 " 3 2 −3 −1 " 20 Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: [v]B = [4, −5, 3], the entries of the last column of the row reduced matrix. (Note that, even though we began the problem with polynomials for vectors, the answer to the problem, the coordinatization of the polynomial v, is a vector in R3 .) (g) We convert the given polynomials to vectors in R4 by changing ax3 + bx2 + cx + d to [a, b, c, d]. Then we use the Coordinatization Method. Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the polynomials in the basis B. Then find its corresponding reduced row echelon form: " " ⎤ ⎡ ⎤ ⎡ 1 0 0 "" −1 2 1 −3 "" 8 ⎥ " ⎢ ⎢ −1 2 −1 "" 11 ⎥ ⎥ −→ ⎢ 0 1 0 " 4 ⎥. [A|v] = ⎢ " " ⎣ ⎦ ⎣ 3 −1 0 0 1 " −2 ⎦ 1 " −9 " 0 0 0 " 0 11 −1 3 1 Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: We eliminate the last row of the row reduced matrix, since it contains all zeroes. Then [v]B = [−1, 4, −2], the remaining entries of the last column of the row reduced matrix. (Note that, even though we began the problem with polynomials for vectors, the answer to the problem, the coordinatization of the polynomial v, is a vector in R3 .) a b 4 (h) We convert the given matrices in M22 to vectors in R by changing to [a, b, c, d]. Then c d we use the Coordinatization Method. Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the matrices in the basis B. Then find its corresponding reduced row echelon form: " " ⎤ ⎡ ⎤ ⎡ 1 0 0 "" 2 1 2 1 "" −3 ⎥ " ⎢ ⎢ −2 −1 −1 " −2 ⎥ ⎥ −→ ⎢ 0 1 0 " −3 ⎥. " [A|v] = ⎢ " " ⎣ ⎦ ⎣ 0 0 0 1 " 1 ⎦ 1 3 " 0 " 0 0 0 " 0 3 1 0 1 Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: We eliminate the last row of the row reduced matrix, since it contains all zeroes. Then, [v]B = [2, −3, 1], the remaining entries of the last column of the row reduced matrix. (Note that, even though we began the problem with matrices for vectors, the answer to the problem, the coordinatization of the matrix v, is a vector in R3 .) 139

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition 6

(j) We convert the given matrices in M32 to vectors in R by changing

a b c d e f

Section 4.7 to [a, b, c, d, e, f ].

Then we use the Coordinatization Method. Steps 1 and 2: Form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the matrices in the basis B. Then find its corresponding reduced row echelon form: " " ⎤ ⎡ ⎤ ⎡ 1 0 "" 5 1 −3 "" 11 ⎢ 0 1 " −2 ⎥ ⎢ 3 1 "" 13 ⎥ ⎥ " ⎢ ⎥ ⎢ ⎢ 0 0 " 0 ⎥ ⎥ " ⎢ −1 7 " −19 ⎥ ⎥ " ⎢ −→ [A|v] = ⎢ ⎢ 0 0 " 0 ⎥. ⎢ 2 8 ⎥ 1 "" ⎥ " ⎢ ⎥ ⎢ ⎣ 0 0 " 0 ⎦ ⎣ 1 1 ⎦ 2 "" " 0 0 " 0 4 5 " 10 Step 3: The row reduced matrix contains no rows with all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: We eliminate the last four rows of the row reduced matrix, since they contain all zeroes. Then [v]B = [5, −2], the remaining entries of the last column of the row reduced matrix. (Note that, even though we began the problem with matrices for vectors, the answer to the problem, the coordinatization of the matrix v, is a vector in R2 .) (2) (a) We use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis B, and the columns of C are the vectors in the basis C. Then we find the reduced row echelon form for this matrix: " " ⎤ ⎤ ⎡ ⎡ 20 3 1 0 0 "" −102 1 1 1 "" 1 0 0 ⎣ 5 67 −13 −2 ⎦. 6 3 "" 0 1 0 ⎦ −→ ⎣ 0 1 0 "" " " 36 −7 −1 0 0 1 1 −6 14 0 0 1 The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ −102 20 3 67 −13 −2 ⎦. P=⎣ 36 −7 −1 (c) We convert the given polynomials to vectors in R3 by changing ax2 + bx + c to [a, b, c]. Then we use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors converted from the basis B, and the columns of C are the vectors converted from the basis C. Then we find the reduced row echelon form for this matrix: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 20 −30 −69 1 3 10 "" 2 8 1 ⎣ 3 4 17 " 3 1 0 ⎦ −→ ⎣ 0 1 0 " 24 −24 −80 ⎦. " " 11 31 0 0 1 " −9 1 1 5 " −1 1 6 The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 20 −30 −69 P = ⎣ 24 −24 −80 ⎦. −9 11 31 a b (d) We convert the given matrices in M22 to vectors in R4 by changing to [a, b, c, d]. Then c d we use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors converted from the basis B, and the columns of C are the vectors converted from the basis C. Then we find the reduced row echelon form for this matrix:

140

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition " −1 1 3 1 "" 1 2 3 ⎢ 1 0 −4 −1 " 3 1 1 ⎢ " ⎣ 3 0 −7 −2 " 5 0 1 " −1 1 4 1 " 1 4 0 ⎡

⎡ ⎤ 1 0 ⎢ 0 2 ⎥ ⎥ −→ ⎢ ⎣ 0 −4 ⎦ 0 1

0 1 0 0

0 0 1 0

0 0 0 1

Section 4.7

" " −1 −4 " " 4 5 " " 0 2 " " −4 −13

⎤ 2 −9 1 3 ⎥ ⎥. −3 1 ⎦ 13 −15

The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ −1 −4 2 −9 ⎢ 4 5 1 3 ⎥ ⎥. P=⎢ ⎣ 0 2 −3 1 ⎦ −4 −13 13 −15 (f) We convert the given polynomials to vectors in R5 by changing ax4 + bx3 + cx2 + dx + e to [a, b, c, d, e]. Then we use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors converted from the basis B, and the columns of C are the vectors converted from the basis C. Then we find the reduced row echelon form for this matrix: " " ⎤ ⎡ ⎤ ⎡ 1 2 1 0 0 "" 6 1 0 1 2 2 "" 6 ⎢ 0 1 0 " 1 ⎢ 3 7 1 2 ⎥ 5 5 ⎥ 5 "" 20 ⎥ " ⎢ ⎥ ⎢ " ⎢ ⎥ ⎢ 0 4 −3 " 7 7 17 ⎥ −→ ⎢ 0 0 1 " −1 −1 −3 ⎥ ⎥. " ⎢ ⎦ ⎣ 0 0 0 " 0 ⎣ 4 3 0 0 8 "" 19 −1 −10 ⎦ " " " 0 0 0 0 0 0 6 19 −2 1 −7 −4 Now, we eliminate the last two rows of the row reduced matrix, since it contains all zeroes. Then, the transition matrix consists of the remaining entries in the matrix on the right side of the row ⎡ ⎤ 6 1 2 1 2 ⎦. reduced matrix. That is, P = ⎣ 1 −1 −1 −3 (4) (a) In each subpart, we use the Transition Matrix Method. In order to find the transition matrix from basis X to basis Y , we form the augmented matrix [Y|X], in which the columns of X are the vectors in the basis X, and the columns of Y are the vectors in the basis Y . Then, in each case, we convert this matrix to reduced row echelon form. Since, in this problem, there are no rows of all zeroes in any of the row reduced matrices obtained, the matrices to the right of the augmentation bar are the desired transition matrices. (i) To find the transition matrix from B to C, we row reduce " " 31 13 31 1 0 "" 13 3 2 "" 3 7 . Hence, P = . , obtaining −18 −43 0 1 " −18 −43 7 5 " 1 2 (ii) To find the transition matrix from C to D, we row reduce " " 5 2 "" 3 2 1 0 "" −11 −8 −11 , obtaining . Hence, Q = 2 1 " 7 5 0 1 " 29 21 29 (iii) To find the transition matrix from B to D, we row reduce " " 3 1 1 0 "" 1 5 2 "" 3 7 . Hence, T = , obtaining −1 0 1 " −1 −4 2 1 " 1 2

−8 21

3 −4

.

. A straightforward

matrix multiplication verifies that T = QP. (c) First, we convert the polynomials in each of the given bases to vectors in R3 by changing ax2 +bx+c to [a, b, c]. Then, in each subpart, we use the Transition Matrix Method. Given bases of converted vectors X and Y , in order to find the transition matrix from X to Y , we form the augmented

141


Section 4.7

matrix [Y|X], in which the columns of X are the vectors in the basis X, and the columns of Y are the vectors in the basis Y . Then, in each case, we convert this matrix to reduced row echelon form. Since, in this problem, there are no rows of all zeroes in any of the row reduced matrices obtained, the matrices to the right of the augmentation bar are the desired transition matrices. (i) To find the transition matrix from B to C, we row reduce " " ⎤ ⎡ ⎤ ⎤ ⎡ ⎡ " 2 8 13 2 8 13 3 1 2 1 "" 1 3 " ⎣ 4 1 0 " 2 7 9 ⎦ to ⎣ I3 "" −6 −25 −43 ⎦. Hence, P = ⎣ −6 −25 −43 ⎦. " " 11 45 76 11 45 76 1 0 0 " 2 8 13 (ii) To find the transition " ⎡ 7 1 1 "" 1 2 ⎣ −3 7 −2 "" 4 1 2 −3 1 " 1 0

matrix from ⎤ ⎡ 1 0 ⎦ to ⎣ I3 0

C to D, we row reduce " ⎤ ⎡ " −24 −2 1 −24 " " 30 ⎦. Hence, Q = ⎣ 30 3 −1 " " 139 13 −5 139

−2 3 13

⎤ 1 −1 ⎦. −5

(iii) To find the transition matrix from B to D, we row reduce " " ⎤ ⎡ ⎤ ⎡ " −25 −97 −150 3 7 1 1 "" 1 3 " ⎣ −3 9 ⎦ to ⎣ I3 "" 31 120 185 ⎦ . 7 −2 "" 2 7 " 145 562 868 2 −3 1 " 2 8 13 ⎡ ⎤ −25 −97 −150 185 ⎦. A straightforward matrix multiplication verifies that Hence, T = ⎣ 31 120 145 562 868 T = QP. (5) (a) (i) In ⎡ 1 ⎣ 6 3

accordance with the Simplified ⎤ ⎡ −4 1 2 1 1 −4 −24 5 8 3 ⎦ to ⎣ 0 0 −12 3 6 2 0 0

Span Method, we row reduce ⎤ 0 −2 0 1 4 0 ⎦. 0 0 1

Hence, C = ([1, −4, 0, −2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]). (ii) We use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis B, and the columns of C are the vectors in the basis C. Then we convert this matrix to reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 6 3 1 0 0 "" 1 6 3 1 0 0 "" 1 ⎢ 0 1 0 " 1 5 3 ⎥ ⎢ −4 0 0 " −4 −24 −12 ⎥ ⎥ ⎥ ⎢ " ⎢ " ⎥ −→ ⎢ 0 0 1 " 1 3 2 ⎥. ⎢ 0 1 0 " 1 5 3 ⎥ ⎥ ⎢ " ⎢ " ⎣ −2 4 0 " 2 ⎣ 0 0 0 " 0 0 0 ⎦ 8 6 ⎦ " " 3 2 0 0 0 " 0 0 0 0 0 1 " 1 Now, we eliminate the last two rows of the row reduced matrix, since it contains all zeroes. Then the transition matrix consists of the remaining entries in the matrix on the right side of the row ⎡ ⎤ 1 6 3 reduced matrix. That is, P = ⎣ 1 5 3 ⎦. 1 3 2 ⎡ ⎤ 1 −3 3 0 ⎦, where we have computed P−1 (iii) According to Theorem 4.22, Q = P−1 = ⎣ 1 −1 −2 3 −1 by row reducing [P|I3 ] to [I3 |P−1 ]. (iv) We use the Coordinatization Method to find [v]B . To do this, we form the augmented matrix 142


[B|v], where the columns of B are the vectors reduced row echelon form of this matrix: " ⎡ ⎤ ⎡ 2 1 6 3 "" 1 ⎢ −4 −24 −12 " −8 ⎥ ⎢ 0 " ⎢ ⎥ ⎢ ⎢ 5 3 "" −2 ⎥ [B|v] = ⎢ ⎢ 1 ⎥ −→ ⎢ 0 ⎣ 2 ⎣ 0 8 6 "" −12 ⎦ 3 1 3 2 " 0

Section 4.7

in the basis B and then find the corresponding 0 1 0 0 0

0 0 1 0 0

" ⎤ " 17 " " 4 ⎥ ⎥ " " −13 ⎥ ⎥ " " 0 ⎦ " " 0

We eliminate the last two rows of the row reduced matrix, since they contain all zeroes. Then [v]B = [17, 4, −13], the remaining entries of the last column. Next, we compute [v]C . We form the augmented matrix [C|v], where the columns of C are the vectors in the basis C, and then find the corresponding reduced row echelon form of this matrix: " " ⎤ ⎡ ⎤ ⎡ 1 0 0 "" 2 2 1 0 0 "" ⎢ 0 1 0 " −2 ⎥ ⎢ −4 0 0 " −8 ⎥ ⎥ " ⎢ ⎥ " ⎢ ⎥ " ⎥ " ⎢ [C|v] = ⎢ 0 1 0 " −2 ⎥ −→ ⎢ ⎢ 0 0 1 " 3 ⎥ " " ⎦ ⎣ 0 0 0 ⎣ −2 4 0 −12 ⎦ " 0 " 0 0 0 " 0 3 0 0 1 " We eliminate the last two rows of the row reduced matrix, since they contain all zeroes. Then [v]C = [2, −2, 3], the remaining entries of the last column. (v) Since Q is the transition matrix from C to B, Theorem 4.20 implies that Q[v]C = [v]B . This can be verified for the answers computed above using matrix multiplication.. (c) (i) In accordance with the Simplified Span Method, we row reduce ⎡ ⎤ 3 −1 4 6 ⎢ 6 7 −3 −2 ⎥ ⎢ ⎥ to obtain I4 . ⎣ −4 −3 3 4 ⎦ −2 0 1 2 Hence, C = ([1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]), the standard basis for R4 . (ii) We use the Transition Matrix Method. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis B, and the columns of C are the vectors in in reduced row echelon⎤form. the basis C. However, since the matrix C equals I4 , [C|B] is already ⎡ 3 6 −4 −2 ⎢ −1 7 −3 0 ⎥ ⎥. Therefore, the transition matrix P actually equals B. That is, P = ⎢ ⎣ 4 −3 3 1 ⎦ 6 −2 4 2 ⎤ ⎡ 1 −4 −12 7 ⎥ ⎢ ⎥ ⎢ −2 9 27 − 31 2 ⎥ ⎢ −1 (iii) By Theorem 4.22, Q = P = ⎢ ⎥, where we have computed P−1 77 ⎥ ⎢ −5 22 67 − 2 ⎦ ⎣ 5 by row reducing [P|I3 ] to [I3 |P−1 ].

−23

−71

41

(iv) We use the Coordinatization Method to find [v]B . To do this, we form the augmented matrix [B|v], where the columns of B are the vectors in the basis B, and then find the corresponding reduced row echelon form of this matrix:

143


3 ⎢ −1 [B|v] = ⎢ ⎣ 4 6

" " ⎤ ⎡ " 2 6 −4 −2 "" 10 " ⎥ ⎢ " 1 " 7 −3 0 " 14 ⎥ " ⎢ −→ I 4 " −3 " ⎣ −3 3 1 " 3 ⎦ " " 7 " −2 4 2 12

Section 4.7

⎤ ⎥ ⎥. ⎦

Thus, [v]B = [2, 1, −3, 7], the last column of the row reduced matrix. Next we compute [v]C . But C is the standard basis for R4 , and so [v]C actually equals v. That is, [v]C = [10, 14, 3, 12]. (v) Since Q is the transition matrix from C to B, Theorem 4.20 implies that Q[v]C = [v]B . This can be verified for the answers computed above using matrix multiplication. (7) (a) To compute each of these transition matrices, we use the Transition Matrix Method. To do this, we form the augmented matrix [Ci |B], for 1 ≤ i ≤ 5, in which the columns of B are the vectors in the basis B, and the columns of Ci are the vectors in the basis Ci . Then, we convert each of these matrices to reduced row echelon form: " " ⎤ ⎡ ⎤ ⎡ " 0 1 0 3 2 3 2 −5 "" −5 " 9 "" 9 −9 −5 ⎦ → ⎣ I3 "" 0 0 1 ⎦ , [C1 |B] = ⎣ −9 −5 " 1 0 0 2 1 2 1 −1 " −1 " " ⎤ ⎤ ⎡ ⎡ " 0 0 1 3 2 2 −5 3 "" −5 " 9 −9 "" 9 −9 −5 ⎦ → ⎣ I3 "" 1 0 0 ⎦ , [C2 |B] = ⎣ −5 " 0 1 0 2 1 1 −1 2 " −1 " " ⎤ ⎡ ⎤ ⎡ " 1 0 0 3 2 −5 2 3 "" −5 " [C3 |B] = ⎣ 9 −5 −9 "" 9 −9 −5 ⎦ → ⎣ I3 "" 0 0 1 ⎦ , " 0 1 0 2 1 −1 1 2 " −1 " " ⎤ ⎡ ⎤ ⎡ " 0 1 0 3 2 3 −5 2 "" −5 " 9 −5 "" 9 −9 −5 ⎦ → ⎣ I3 "" 1 0 0 ⎦ , [C4 |B] = ⎣ −9 " 0 0 1 2 1 2 −1 1 " −1 " " ⎤ ⎤ ⎡ ⎡ " 0 0 1 3 2 2 3 −5 "" −5 " 9 "" 9 −9 −5 ⎦ → ⎣ I3 "" 0 1 0 ⎦. [C5 |B] = ⎣ −5 −9 " 1 0 0 2 1 1 2 −1 " −1 Thus, the transition ⎡ ⎤ ⎡ 0 1 0 0 ⎣ 0 0 1 ⎦, ⎣ 1 1 0 0 0

matrices from ⎤ ⎡ 0 1 1 0 0 ⎦, ⎣ 0 1 0 0

B to C1 , C2 , ⎤ ⎡ 0 0 0 0 1 ⎦, ⎣ 1 1 0 0

C3 , C4 , and ⎤ ⎡ 1 0 0 0 ⎦, ⎣ 0 1

C5 , respectively, are ⎤ 0 0 1 0 1 0 ⎦. 1 0 0

(10) According to Exercise 9, as suggested in the hint, if P is the matrix whose columns are the vectors in B, and Q is the matrix whose columns are the vectors in C, then Q−1 P is the transition matrix from B to C. Now suppose A is the transition matrix from B to C given in the problem. Then A = Q−1 P, implying QA = P. This yields Q = PA−1 . Hence, we first compute A−1 by row reducing [A|I3 ] to [I3 |A−1 ]. Then we use the result obtained from that computation to find Q, as follows: ⎡ ⎤⎡ ⎤ ⎡ ⎤ −2 1 −13 49 19 86 −142 −53 −246 5 ⎦ ⎣ −5 −2 −9 ⎦ = ⎣ 64 24 111 ⎦. Q = PA−1 = ⎣ 1 0 3 2 10 3 1 5 167 63 290 So, C = ([−142, 64, 167], [−53, 24, 63], [−246, 111, 290]), the columns of Q.

144


14 −15 (11) (b) First, Av = ⎣ 6 −7 3 −3

Section 4.7

⎤⎡ ⎤ ⎡ ⎤ −30 1 14 −12 ⎦ ⎣ 4 ⎦ = ⎣ 2 ⎦. −7 −2 5

We use the Coordinatization Method to find [Av]B . To do this, we form the augmented matrix [B|Av], where the columns of B are the vectors in the basis B, and then find its corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ " 2 5 1 2 "" 14 " [B|Av] = ⎣ 2 1 0 "" 2 ⎦ −→ ⎣ I3 "" −2 ⎦. " 3 1 0 1 " 5 Thus, [Av]B = [2, −2, 3], the last column of the row reduced matrix. Next, we use the Coordinatization Method to find [v]B . To do this, we form the augmented matrix [B|v] and find its corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ " 1 5 1 2 "" 1 " [B|v] = ⎣ 2 1 0 "" 4 ⎦ −→ ⎣ I3 "" 2 ⎦. " −3 1 0 1 " −2 Thus, [v]B = [1, 2, −3], the last column of the row reduced matrix. ⎡ ⎤⎡ ⎤ ⎡ ⎤ 2 0 0 1 2 0 ⎦ ⎣ 2 ⎦ = ⎣ −2 ⎦. Finally, we compute D[v]B = ⎣ 0 −1 0 0 −1 −3 3 And so, D[v]B = [Av]B = [2, −2, 3]. (13) Let V, B, and the ai ’s and w’s be as given in the statement of the theorem. Proof of Part (1): Suppose that [w1 ]B = [b1 , . . . , bn ] and [w2 ]B = [c1 , . . . , cn ]. Then, by definition, w1 = b1 v1 + · · · +bn vn and w2 = c1 v1 + · · · +cn vn . Hence, w1 + w2 = b1 v1 + · · · +bn vn + c1 v1 + · · · +cn vn = (b1 + c1 )v1 + · · · +(bn + cn )vn , implying [w1 + w2 ]B = [(b1 + c1 ), . . . , (bn + cn )], which equals [w1 ]B + [w2 ]B . Proof of Part (2): Suppose that [w1 ]B = [b1 , . . . , bn ]. Then, by definition, w1 = b1 v1 + · · · +bn vn . Hence, a1 w1 = a1 (b1 v1 + · · · +bn vn ) = a1 b1 v1 + · · · +a1 bn vn , implying [a1 w1 ]B = [a1 b1 , . . . , a1 bn ], which equals a1 [b1 , . . . , bn ]. Proof of Part (3): Use induction on k. Base Step (k = 1): This is just part (2). Inductive Step: Assume true for any linear combination of k vectors and prove true for a linear combination of k + 1 vectors. Now, [a1 w1 + · · · + ak wk + ak+1 wk+1 ]B = [a1 w1 + · · · + ak wk ]B + [ak+1 wk+1 ]B (by part (1)) = [a1 w1 + · · · + ak wk ]B + ak+1 [wk+1 ]B (by part (2)) = a1 [w1 ]B + · · · +ak [wk ]B + ak+1 [wk+1 ]B (by the inductive hypothesis). (14) Let v ∈ V. Then applying Theorem 4.20 twice yields (QP)[v]B = Q[v]C = [v]D . Therefore, (QP)[v]B = [v]D for all v ∈ V, and so Theorem 4.20 implies that QP is the (unique) transition matrix from B to D. (16) (a) False. Typically, changing the order that the vectors appear in an ordered basis also changes the order of the coordinates in the corresponding coordinatization vectors. For example, if v = [1, 2, 3] ∈ R3 , then since v = 1i + 2j + 3k, [v]B = [1, 2, 3]. However, since v = 2j + 3k + 1i, [v]C = [2, 3, 1]. (b) True. Now bi = 0b1 + · · · + 0bi−1 + 1bi + 0bi+1 + · · · + 0bn . Thus, [bi ]B is the n-vector having the coefficients 0, . . . , 0, 1, 0, . . . 0, respectively, as its entries. Hence, [bi ]B = ei . 145


Chap 4 Review

(c) True. This is seen by reversing the roles of B and C in the definition of the transition matrix from B to C. (d) False. This is close to the statement of Theorem 4.20, except that the positions of [v]B and [v]C are reversed. The correct conclusion is that P[v]B = [v]C . For a specific counterexample to the statement given in the problem, consider the ordered bases B = ([1, 1], [1, −1]) and C = 1 1 ([1, 0], [0, 1]) for R2 . The transition matrix from B to C is P = , and, for v = [2, 6], 1 −1 8 [v]B = [4, −2] and [v]C = [2, 6]. However, P[v]C = = [v]B . −4 (e) False. This is close to the statement of Theorem 4.21, except that the transition matrix T from B to D is QP, not PQ. For a specific counterexample to the statement given in the problem, consider the ordered bases B, C, and D for R2 given in Exercise 4(a) of this section, whose solution 13 31 −11 −8 appears above. Then, as computed in Exercise 4(a), P = , Q= , −18 −43 29 21 1 3 756 547 and T = . But PQ = = T. −1 −4 −1049 −759 (f) True. This is part of Theorem 4.22. (g) False. The matrix P−1 , not P, is the transition matrix from standard coordinates to a basis of fundamental eigenvectors for A, where P is the matrix from the Diagonalization Method whose columns are fundamental eigenvectors for A. This is explained in Example 11 of Section 4.7 in the textbook. Using A and P from that example gives a specific counterexample to the incorrect given statement. For, if P were the transition matrix from standard coordinates S to an ordered basis B = (b1 , b2 , b3 ) of eigenvectors for A, then P−1 would be the transition matrix from B to S. Hence, we would have b1 = P−1 [b1 ]B . But, [b1 ]B = e1 (see Exercise 6). So, b1 is the first column of P−1 , which is [1, −2, −1], as shown in Example 11. But [1, −2, −1] is not an eigenvector for A, since Av = [74, 32, 16], which is not a scalar multiple of [1, −2, −1].

Chapter 4 Review Exercises (2) By part (2) of Theorem 4.1, the zero vector 0 = 0 [x, y] = [0x + 4(0) − 4, 0y − 5(0) + 5] = [−4, 5]. Similarly, by part (3) of Theorem 4.1, the additive inverse of [x, y] = (−1) [x, y] = [(−1)x + 4(−1) − 4, (−1)y − 5(−1) + 5] = [−x − 8, − y + 10]. (3) (a) The set S = {[3a, 2a − 1, −4a] | a ∈ R} is not a subspace. The reason is that the zero vector, [0, 0, 0] is not an element of S. For [0, 0, 0] to be of the form [3a, 2a−1, −4a], we must have 3a = 0, implying a = 0. But then 2a − 1 = −1 = 0, which gives a contradiction. . ) " " 2a + b −4a − 5b " (c) The set V = " a, b ∈ R is a subspace of M22 . First, letting a = b = 0 0 a − 2b

0 0 shows that ∈ V, and so V is nonempty. By Theorem 4.2 it is enough to prove that V is 0 0 closed under addition and scalar multiplication. 2A + B −4A − 5B 2C + D −4C − 5D For addition, let , ∈ V. Then 0 A − 2B 0 C − 2D 2A + B −4A − 5B 2C + D −4C − 5D + 0 A − 2B 0 C − 2D 146


=

2(A + C) + (B + D) −4(A + C) − 5(B + D) 0 (A + C) − 2(B + D)

Chap 4 Review

, which is of the correct form to be in V,

with a = A + C and b = B + D. 2A + B −4A − 5B For scalar multiplication, let ∈ V and c ∈ R. Then 0 A − 2B 2A + B −4A − 5B c(2A + B) c(−4A − 5B) c = 0 A − 2B 0 c(A − 2B) 2(cA) + (cB) −4(cA) − 5(cB) = , which is of the correct form to be in V, with a = cA and 0 (cA) − 2(cB) b = cB. (d) The set S of all matrices in M22 that are both singular and symmetric is not a subspace of M22 1 0 0 0 because it is not closed under addition. For example, , ∈ S, 0 0 0 1 1 0 0 0 1 0 but + = ∈ / S. 0 0 0 1 0 1 (f) The set S of all polynomials in P4 whose highest order nonzero term has an even degree is not a subspace of P4 because it is not closed under addition. For example, x4 − x3 and −x4 + 2x3 are / S. both in S, but x4 − x3 + −x4 + 2x3 = x3 ∈ (4) (a) With S = {[3, 3, −2, 4], [3, 4, 0, 3], [5, 6, −1, 6], [4, 4, −3, 5]}, we follow the Simplified Span Method: ⎡ ⎤ 3 3 −2 4 ⎢ 3 4 0 3 ⎥ ⎥ Step 1: Form the matrix A = ⎢ ⎣ 5 6 −1 6 ⎦ . 4 4 −3 5 ⎡ ⎤ 1 0 0 5 ⎢ 0 1 0 −3 ⎥ ⎥. Step 2: A row reduces to C = ⎢ ⎣ 0 0 1 1 ⎦ 0 0 0 0 Step 3: span(S) = {a[1, 0, 0, 5] + b[0, 1, 0, −3] + c[0, 0, 1, 1] | a, b, c ∈ R} = {[a, b, c, 5a − 3b + c] | a, b, c ∈ R}. (b) The set B = {[1, 0, 0, 5], [0, 1, 0, −3], [0, 0, 1, 1]} is a basis for span(S). First, B clearly spans span(S), as can be seen from the description of span(S) in part (a). Also, B is linearly independent because each vector in B has a 1 in a coordinate in which every other vector has a zero. Since B has 3 elements, dim(span(S)) = 3. (7) (a) With S = {[3, 5, −3], [−2, −4, 3], [1, 2, −1]}, we follow the Independence Test Method. ⎡ ⎤ 3 −2 1 2 ⎦ , whose columns are the vectors in S. Step 1: Form the matrix A = ⎣ 5 −4 −3 3 −1 Step 2: A row reduces to B = I3 . Step 3: Since there is a pivot in every column, S is linearly independent. 147


Chap 4 Review

(b) Because the full set S is linearly independent, S itself is a maximal linearly independent subset of S. Also, S spans R3 by Theorem 4.14. (c) Because S is linearly independent, Theorem 4.9 guarantees that there is only one way to express v = [11, 20, −12] as a linear combination of the vectors in S. , (8) (a) First, we convert the polynomials in S -= −5x3 + 2x2 + 5x − 2, 2x3 − x2 − 2x + 1, x3 − 2x2 − x + 2, −2x3 + 2x2 + 3x − 5 into vectors in R4 : −5x3 + 2x2 + 5x − 2 → [−5, 2, 5, −2], 2x3 − x2 − 2x + 1 → [2, −1, −2, 1], x3 − 2x2 − x + 2 → [1, −2, −1, 2], and −2x3 + 2x2 + 3x − 5 → [−2, 2, 3, −5]. We follow the Independence Test Method. ⎡ ⎤ −5 2 1 −2 ⎢ 2 −1 −2 2 ⎥ ⎥ , whose columns are the n-vectors we found. Step 1: Form the matrix A = ⎢ ⎣ 5 −2 −1 3 ⎦ −2 1 2 −5 ⎡ ⎤ 1 0 3 0 ⎢ 0 1 8 0 ⎥ ⎥ Step 2: A row reduces to B = ⎢ ⎣ 0 0 0 1 ⎦. 0 0 0 0 Step 3: Since there is no pivot in column 3, S is linearly dependent. Using the entries in the third column of the matrix B, we see that the third vector equals 3 times the first vector plus 8times the second vector; that is, x3 − 2x2 − x + 2 = 3 −5x3 + 2x2 + 5x − 2 + 8 2x3 − x2 − 2x + 1 . (b) S does not span P3 since dim(span(S)) = 3. A maximal linearly independent subset of S must be a basis for span(S) by Theorem 4.14. Such a subset is {−5x3 + 2x2 + 5x − 2, 2x3 − x2 − 2x + 1, −2x3 + 2x2 + 3x − 5}. (c) Yes, there is an alternate linear combination 3 by2Theorem 4.9. One 3 such2alternate linear combi3 2 + 2x + 5x − 2 −5 2x − x − 2x + 1 + 1 x − 2x − x + 2 nation is −5 −5x − 1 −2x3 + 2x2 + 3x − 5 . This was found by row reducing the matrix " " ⎤ ⎡ ⎤ ⎡ 1 0 3 0 "" −2 −5 2 1 −2 "" 18 ⎥ " ⎢ ⎢ 2 −1 −2 2 "" −9 ⎥ ⎥, which yields ⎢ 0 1 8 0 " 3 ⎥ , and taking the particular ⎢ " " ⎦ ⎣ 0 0 0 1 ⎣ 5 −2 −1 3 " −19 ⎦ " −1 0 0 0 0 " 0 −2 1 2 −5 " 12 solution to the system that uses the value 1 for the independent variable corresponding to the third column. ⎡ ⎤ −2 3 −1 4 ⎢ 3 −3 2 −4 ⎥ ⎥ row reduces (12) (a) Using the Simplified Span Method, we find that the matrix ⎢ ⎣ −2 2 −1 3 ⎦ 3 −5 0 −7 to I4 , so the given set of 4 vectors spans R4 . By part (1) of Theorem 4.13, the given set is a basis for R4 . (13) (a) We will prove that W is a subspace of R4 using Theorem 4.2 by showing that W is nonempty and that it is closed under addition and scalar multiplication. First, we show that W is nonempty: 0 ∈ W because A0 = O. Closure under addition: If X1 , X2 ∈ W, then A(X1 + X2 ) = AX1 + AX2 = O + O = O. Closure under scalar multiplication: If X ∈ W, then A(cX) = cAX = cO = O.

148


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⎡

⎤ 1 −3 0 2 0 1 −1 ⎦. We use this to solve the homogeneous system AX = (b) A row reduces to ⎣ 0 0 0 0 0 O. Because there are no pivots in columns 2 and 4, those columns correspond to independent variables. Letting those variables equal 1 and 0, respectively, yields the solution [3, 1, 0, 0] for the homogeneous system. Setting these variables equal to 0 and 1, respectively, produces the solution [−2, 0, 1, 1]. Hence, we obtain the following basis for W: {[3, 1, 0, 0], [−2, 0, 1, 1]}. (c) Because the basis from part (b) has two elements, dim(W) = 2. Also, the reduced row echelon form for A from part (b) shows that rank(A) = 2. Therefore, dim(W)+rank(A) = 2+2 = 4 = dim(R4 ). (14) (a) First, we use direct computation to check that every polynomial in B is in V: If p(x) = x3 − 3x, then p (x) = 3x2 − 3, and so p (1) = 3 − 3 = 0. If p(x) = x2 − 2x, then p (x) = 2x − 2, and so p (1) = 2 − 2 = 0. If p(x) = 1, then p (x) = 0, and so p (1) = 0. Thus, every polynomial in B is actually in V. Next, we convert the polynomials in B to 4-vectors and use the Independence Test Method. Now, ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 ⎢ 0 1 0 ⎥ ⎢ 0 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −3 −2 0 ⎦ clearly row reduces to ⎣ 0 0 1 ⎦ , so B is linearly independent. Finally, since 0 0 0 0 0 1 the polynomial x ∈ / V, dim(V) < dim(P3 ) = 4 by Theorem 4.16. But then |B| ≥ dim(V), implying that |B| = dim(V) and B is a basis for V, by Theorem 4.13. Thus, dim(V) = 3. (b) If p(x) = ax3 + bx2 + cx + d, then p (x) = 3ax2 + 2bx + c, and p (x) = 6ax + 2b. Setting p (1) = 0 produces the equation 6a + 2b = 0, or b = −3a. Setting p (1) = 0 yields 3a + 2b + c = 0. Combining this with b = −3a gives us c = −3a − 2(−3a) = 3a. Note that d is arbitrary. Setting a = 1 and d = 0 produces the polynomial x3 − 3x2 + 3x. Setting a = 0 and d = 1 yields the constant polynomial 1. Note that C = {1, x3 − 3x2 + 3x} is a maximal linearly independent subset of W because dim(W) < dim(V) = 3 (since W is a proper subset of V). Thus, C is a basis for W by Theorem 4.14, and dim(W) = 2. (15) (a) We follow the Inspection Method: Choose v1 = [2, 3, 0, 1]. Skip [−6, 9, 0, −3], since it equals −3v1 . Choose v2 = [4, 3, 0, 4], since it is not a scalar multiple of v1 . Next, [8, −3, 0, 6] = 2v1 + v2 , so skip it. Finally, choose v3 = [1, 0, 2, 1], because it is not a linear combination of v1 and v2 , since it does not have a zero in its third coordinate. Thus, T = {[2, −3, 0, 1], [4, 3, 0, 4], [1, 0, 2, 1]} is a basis for V. (b) Yes. Since T is a basis for V, T is a maximal linearly independent subset of V. (See Exercise 20 in Section 4.5.) (17) To perform the Enlarging Method, we first add the standard basis for R4 to T to form a spanning set for R4 . We then use the Independence Test Method to reduce this spanning set to a basis for R4 . Thus, we row reduce ⎤ ⎡ 1 ⎡ ⎤ 1 0 25 0 0 10 2 1 1 0 0 0 ⎢ ⎥ ⎢ 0 1 15 0 0 − 15 ⎥ ⎢ 1 −2 0 1 0 0 ⎥ ⎥ ⎢ ⎥ to obtain ⎢ ⎢ 0 0 0 1 0 − 1 ⎥. ⎣ −1 2 0 0 1 0 ⎦ ⎣ 2 ⎦ 2 −4 0 0 0 1 1 0 0 0 0 1 2

149


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Since the 1st, 2nd, 4th, and 5th columns are pivot columns, we choose the vectors that correspond to those columns. Hence, the desired basis for R4 is {[2, 1, −1, 2], [1, −2, 2, −4], [0, 1, 0, 0], [0, 0, 1, 0]}. (20) (a) We use the Coordinatization Method. Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors in the basis B, and then find the corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" −3 2 5 −6 "" 1 2 "" −7 ⎦ −→ ⎣ 0 1 0 "" −1 ⎦ . [A|v] = ⎣ 1 0 0 0 1 " −2 2 1 1 " −9 Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: [v]B = [−3, −1, −2], the last column of the row reduced matrix. ⎡ ⎤ a b (c) We convert the given matrices in M23 to vectors in R6 by changing ⎣ c d ⎦ to [a, b, c, d, e, f ]. e f Then we use the Coordinatization Method. Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the matrices in the basis B. Then we find the corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" −3 −3 −10 2 "" −43 ⎢ 0 1 0 " 5 ⎥ ⎢ 3 3 11 "" −5 ⎥ ⎥ " ⎥ ⎢ ⎢ ⎥ ⎢ 0 0 1 " −1 ⎥ " ⎢ 11 28 10 " 97 ⎥ ⎥ ⎢ " −→ [A|v] = ⎢ ⎢ 0 0 0 " 0 ⎥. ⎢ 5 4 14 "" −9 ⎥ ⎥ ⎥ ⎢ " ⎢ ⎣ 0 0 0 " 0 ⎦ ⎣ −2 −6 −16 " −8 ⎦ " " 0 0 0 " 0 2 0 3 " −9 Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: We eliminate the last three rows of the row reduced matrix, since they contain all zeroes. Then [v]B = [−3, 5, −1], the remaining entries of the last column of the row reduced matrix. (Note that, even though v and the elements of B are matrices, the coordinatization of v is a vector in R3 .) (21) (a) First, to find [v]B , we use the Coordinatization Method. Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors in the basis B, and then find the corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 27 26 9 −3 "" 126 [A|v] = ⎣ −47 −16 10 "" −217 ⎦ −→ ⎣ 0 1 0 "" −62 ⎦ . 6 14 0 0 1 " −10 −1 37 " Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: [v]B = [27, −62, 6], the last column of the row reduced matrix. We use the Transition Matrix Method to find the transition matrix from B to C. To do this, we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis B, and the columns of C are the vectors in the basis C. Then we find the reduced row echelon form for this matrix:

150

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition " 2 −3 5 "" 26 9 ⎣ −3 5 −10 "" −47 −16 4 −1 −9 " −10 −1 ⎡

⎤ ⎡ −3 1 10 ⎦ −→ ⎣ 0 37 0

0 1 0

0 0 1

" " 4 " " −1 " " 3

Chap 4 Review

⎤ 2 5 0 1 ⎦. 1 −2

The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 4 2 5 1 ⎦. P = ⎣ −1 0 3 1 −2 Finally, we determine [v]C by computing P[v]B . Now, ⎡ ⎤⎡ ⎤ ⎡ ⎤ 4 2 5 27 14 1 ⎦ ⎣ −62 ⎦ = ⎣ −21 ⎦. Hence, [v]C = [14, −21, 7]. P[v]B = ⎣ −1 0 3 1 −2 6 7 (b) First, we solve for [v]B . We convert the given polynomials in B to vectors in R3 by changing ax2 + bx + c to [a, b, c]. Then we use the Coordinatization Method. Steps 1 and 2: We form the augmented matrix [A|v], where the columns of A are the vectors obtained by converting the polynomials in the basis B. Then we find the corresponding reduced row echelon form: " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" −4 1 3 −2 "" −13 4 "" −11 ⎦ −→ ⎣ 0 1 0 "" −1 ⎦. [A|v] = ⎣ 3 11 3 0 0 1 " 3 1 5 4 " Step 3: The row reduced matrix contains no rows having all zero entries on the left and a nonzero entry on the right, so we proceed to Step 4. Step 4: Then [v]B = [−4, −1, 3], the entries of the last column of the row reduced matrix. (Note that, even though v and the elements of B are polynomials, the coordinatization of v is a vector in R3 .) Next, we use the Transition Method to find the transition matrix from B to C. We convert the polynomials in C to vectors in R3 in the same manner as we did for the polynomials in B. Then we form the augmented matrix [C|B], in which the columns of B are the vectors in the basis B, and the columns of C are the vectors in the basis C. Then we find the reduced row echelon form for this matrix: " " ⎤ ⎤ ⎡ ⎡ 3 −2 1 0 0 "" 4 1 −2 −7 13 −16 "" 1 ⎣ 7 −7 0 ⎦. 4 ⎦ −→ ⎣ 0 1 0 "" 1 2 18 "" 3 11 " " −1 1 1 5 4 0 0 1 9 −13 22 1 The transition matrix is the matrix on the right side of the row reduced matrix. That is, ⎡ ⎤ 4 1 −2 0 ⎦. P=⎣ 1 2 −1 1 1 Finally, we determine [v]C by computing P[v]B . Now, ⎡ ⎤⎡ ⎤ ⎡ ⎤ 4 1 −2 −4 −23 0 ⎦ ⎣ −1 ⎦ = ⎣ −6 ⎦. Hence, [v]C = [−23, −6, 6]. P[v]B = ⎣ 1 2 −1 1 1 3 6 (23) (c) In part (a), your computations likely produced the following set of three fundamental eigenvectors: v1 = [4, 4, 13], v2 = [−3, 2, 0], and v3 = [3, 0, 4]. According to Example 11 in Section 4.7, the transition matrix from the ordered basis B = (v1 , v2 , v3 ) to standard coordinates is the matrix 151


Chap 4 Review

P from the Diagonalization Method; that is, it is the matrix whose columns are the vectors v1 , v2 , and v3 . (Also see part (b) of Exercise 8 in Section 4.7.) Hence, the transition matrix from B ⎡ ⎤ 4 −3 3 2 0 ⎦. to standard coordinates is ⎣ 4 13 0 4 (26) (a) True. To be a vector space, a set with given operations must satisfy all 10 properties in all cases. So, if we can find a single counterexample to any one of the properties, then the set cannot be a vector space. (b) True. We will use Theorem 4.2 to prove that V is a subspace of Rn by showing that V is nonempty and that it is closed under addition and scalar multiplication. First, we show that V is nonempty: 0 ∈ V because A0 = 0. Closure under addition: If X1 , X2 ∈ V, then A(X1 +X2 ) = AX1 +AX2 = 0 + 0 = 0. Closure under scalar multiplication: If X ∈ V, then A(cX) = cAX = c0 = 0. Therefore, since V is a subspace of Rn , V is itself a vector space. (Note: This is merely a generalized version of the proof for part (a) of Exercise 13 in the Chapter 4 Review Exercises.) (c) False. The integers are not closed under scalar multiplication. For example, using the scalar and the integer 1, 12 (1) = 12 , which is not in the set of integers.

1 2

(d) True. The reason is that every subspace is, itself, a vector space. Thus, any subspace must include the zero vector by property (3) from the definition of a vector space. (e) False. For example, let W1 = {[a, 0] | a ∈ R} and W2 = {[0, b] | b ∈ R}, both of which are subspaces of R2 . Then W1 ∪ W2 is the set of 2-vectors with a zero in at least one coordinate. However, this union is not closed under vector addition. For example, [1, 0] + [0, 1] = [1, 1] ∈ / W1 ∪ W2 . (f) True. By Theorem 4.5, span(S) is a subspace of V. If v is a nonzero vector in S, then all scalar multiples of v are in span(S), since span(S) is closed under scalar multiplication. But for each scalar c, cv is a different vector. (If cv = dv, then (c − d)v = 0, so part (4) of Theorem 4.1 shows that c − d = 0, or c = d.) Since the number of different scalars in R is infinite, the number of different vectors in span(S) is infinite. (g) True. See Exercise 27 in Section 4.3. (h) False. For a counterexample, let S1 = {[1, 0]} and S2 = {[2, 0]} in R2 . Then span(S1 ) = span(S2 ) = {[a, 0] | a ∈ R}. (i) False. The vectors found using the Simplified Span Method will usually be simpler in form than the original vectors. For example, applying the Simplified Span Method to the basis {[1, 1], [1, −1]} for R2 produces the standard basis {[1, 0], [0, 1]}. In fact, applying the Simplified Span Method to any basis for Rn produces the standard basis for Rn . (j) True. Since T is already a basis, it is linearly independent. The Independence Test Method only removes redundant vectors that are linear combinations of the other vectors in the set. Because T is linearly independent, there are no such redundant vectors in this case, and so no vectors are removed from T . (k) False. In Example 7 in Section 4.1, the “vector” 1 was shown to be the additive identity (that is, the zero vector) in this vector space. No set containing the additive identity is linearly independent. (l) False. The eigenvectors corresponding to a particular eigenvalue are linear combinations of the fundamental eigenvectors for that eigenvalue. In particular, the set consisting of any two distinct nonzero multiples of the same fundamental eigenvector is a linearly dependent set. Alternately, consider the matrix I2 . Then every nonzero vector in R2 is an eigenvector for I2 152


Chap 4 Review

corresponding to the eigenvalue 1 because I2 v = v for all v ∈ R2 . Hence, every subset of R2 of nonzero vectors is a set of distinct eigenvectors for I2 . In particular, {[1, 0], [0, 1], [1, 1]} is one such subset that is not linearly independent. (m) True. Exercise 10 in Section 4.4 proves this for 3 × 3 matrices. A similar proof works for n × n matrices. A different proof is suggested in part (d) of Exercise 15 in Section 4.5, in which you are asked to prove that the rows of a nonsingular n × n matrix form a basis for Rn and are therefore linearly independent. (n) True. See part (b) of Exercise 26 in Section 4.4. (o) False. For a specific counterexample, consider the linearly dependent set {[1, 0], [2, 0]} in R2 . Note that 2[1, 0] − 1[2, 0] = [0, 0]. (p) False. The problem here is that v1 could be the zero vector. For a specific counterexample, consider the linearly dependent set {[0, 0], [1, 1]} and note that there does not exist a scalar c such that [1, 1] = c[0, 0]. (q) True. For a set to be a basis for a vector space, it must both span the vector space and be linearly independent. We are given that T is linearly independent, and it spans span(T ) by the definition of span(T ). (r) False. The dimension of the trivial vector space is zero. The empty set, which contains no vectors, is a basis for the trivial vector space. This is discussed in Example 10 in Section 4.5. (s) True. Theorem 4.14 shows that T is a basis for V, and Theorem 4.15 shows that S is a basis for V. Theorem 4.12 then proves that |T | = |S|. (t) True. This is the contrapositive of the first two sentences of Theorem 4.16. (u) True. See part (a) of Exercise 12 in Section 4.6. (v) True. Since B is a basis for W, it is linearly independent. Theorem 4.18 then shows that there is a basis for V that contains B. (w) False. Theorem 4.22 states that P−1 , not PT , is the transition matrix from C to B. The vector space and bases discussed in Examples 9 and 10 in Section 4.7 provide a specific counterexample, since in that case, P−1 = PT . (x) True. If dim(V) = n, then P is an n × n matrix. To see why, notice that the columns of P are the vectors in B expressed in C-coordinates. Thus, the number of columns of P equals the number of vectors in B, which is n. Also, since each of these columns is the C-coordinatization of a vector in V, it has n entries. Therefore, P must have n rows as well. (y) True. See part (b) of Exercise 8 in Section 4.7. (z) False. After the row reduction, any rows consisting of all zeroes must first be removed. See Example 7 in Section 4.7.

153


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Chapter 5 Section 5.1 (1) (a) The function f is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation. Property (1): f ([x, y] + [w, z]) = f ([x + w, y + z]) = [3(x + w) − 4(y + z), −(x + w) + 2(y + z)] = [(3x − 4y) + (3w − 4z), (−x + 2y) + (−w + 2z)] = [3x − 4y, −x + 2y] + [3w − 4z, −w + 2z] = f ([x, y]) + f ([w, z]). Property (2): f (c[x, y]) = f ([cx, cy]) = [3(cx) − 4(cy), −(cx) + 2(cy)] = [c(3x − 4y), c(−x + 2y)] = c[3x − 4y, −x + 2y] = cf ([x, y]). Also, f is a linear operator because it is a linear transformation whose domain equals its codomain. (b) The function h is not a linear transformation. If it were, then part (1) of Theorem 5.1 would imply h(0) = 0. However, h([0, 0, 0, 0]) = [2, −1, 0, −3] = 0, so h cannot be a linear transformation. Because h is not a linear transformation, it cannot be a linear operator. (d) The function l is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation.

a b e f a+e b+f Property (1): l + =l c+g d+h c d g h (a + e) − 2(c + g) + (d + h) 3(b + f ) = −4(a + e) (b + f ) + (c + g) − 3(d + h) (a − 2c + d) + (e − 2g + h) 3b + 3f = −4a − 4e (b + c − 3d) + (f + g − 3h) a − 2c + d 3b e − 2g + h 3f = + −4a b + c − 3d −4e f + g − 3h

a b e f =l +l . c d g h

a b sa sb Property (2): l s =l c d sc sd (sa) − 2(sc) + (sd) 3(sb) = −4(sa) (sb) + (sc) − 3(sd)

a − 2c + d 3b a b s(a − 2c + d) s(3b) =s = sl . = s(b + c − 3d) −4a b + c − 3d c d s(−4a) Also, l is a linear operator because it is a linear transformation whose domain equals its codomain. (f) The function r is not a linear transformation since Property (2) of the definition fails. For example, r(8x3 ) = 2x2 , but 8 r(x3 ) = 8(x2 ). Note that Property (1) also fails, but once we have shown that one of the properties fails, it is not necessary to show that the other fails in order to prove that r is not a linear transformation. Also, r is not a linear operator because it is not a linear transformation. (h) The function t is a linear transformation. To prove this, we show that it satisfies the two properties in the definition of a linear transformation. Property (1): t((ax3 +bx2 +cx+d)+(ex3 +f x2 +gx+h)) = t((a+e)x3 +(b+f )x2 +(c+g)x+(d+h))

154


Section 5.1

= (a + e) + (b + f ) + (c + g) + (d + h) = (a + b + c + d) + (e + f + g + h) = t(ax3 + bx2 + cx + d) + t(ex3 + f x2 + gx + h). Property (2): t(s(ax3 + bx2 + cx + d)) = t((sa)x3 + (sb)x2 + (sc)x + (sd)) = (sa) + (sb) + (sc) + (sd) = s(a + b + c + d) = s t(ax3 + bx2 + cx + d). Finally, t is not a linear operator because its domain, P3 , does not equal its codomain, R. (j) The function v is not a linear transformation because Property (1) of the definition fails to hold. For example, v(x2 + (x + 1)) = v(x2 + x + 1) = 1, but v(x2 ) + v(x + 1) = 0 + 0 = 0. Note that Property (2) also fails, but once we have shown that one of the properties fails, it is not necessary to show that the other fails in order to prove that v is not a linear transformation. Also, v is not a linear operator because it is not a linear transformation. (l) The function e is not a linear transformation √ because Property (1) of the definition fails to hold. 2 2 For√example, e([3, √ 0] + [0, 4]) = e([3, 4]) = 3 + 4 = 5, but e([3, 0]) + e([0, 4]) 2 2 2 2 = 3 + 0 + 0 + 4 = 3 + 4 = 7. Note that Property (2) also fails (try a negative scalar), but once we have shown that one of the properties fails, it is not necessary to show that the other fails in order to prove that e is not a linear transformation. Also, e is not a linear operator because it is not a linear transformation. (10) (c) Since the positive z-axis rotates into the positive x-axis here, we insert the 2 × 2 matrix from Exercise 9 in the third and first rows and columns to perform the rotation in the xz-plane, and keep the second row and second column the same as the identity matrix to indicate that the y-axis ⎡ ⎤ sin θ 0 cos θ ⎦. 1 0 stays fixed. Thus, we get the matrix A = ⎣ 0 cos θ 0 − sin θ (26) We first need to express i and j as linear combinations of i + j and −2i + 3j. Hence, suppose ) a(i + j) + b(−2i + 3j) = i, and c(i + j) + d(−2i + 3j) = j. Setting the first and second coordinates of each side of these two equations equal to each other produces the linear systems ) ) c − 2d = 0 a − 2b = 1 . and c + 3d = 1 a + 3b = 0 1 3 1 Solving these systems yields a = 35 , b =− 15 , c = 25 , and d = 5 .1 Therefore,i = 5 (i + j) − 5 (−2i + 3j), 3 1 3 and so L(i) = L 5 (i + j) − 5 (−2i + 3j) = L 5 (i + j) + L − 5 (−2i + 3j) = 3 1 3 1 2 1 2j) =75 i − 11 5 L(i + j) − 5 L(−2i 5 j. Similarly, 5 (i + j) + 5 (−2i + 3j), j= 2 + 3j) = 15 (i − 3j) −5 (−4i + 2 1 2 1 and so L(j) = L 5 (i + j) + 5 (−2i + 3j) = L 5 (i + j) + L 5 (−2i + 3j) = 5 L(i + j) + 5 L(−2i + 3j) = 25 (i − 3j) + 15 (−4i + 2j) = − 25 i − 45 j.

(29) Suppose Part (3) of Theorem 5.1 is true for some n. We prove it true for n + 1. L(a1 v1 + · · · + an vn + an+1 vn+1 ) = L(a1 v1 + · · · + an vn ) + L(an+1 vn+1 ) (by property (1) of a linear transformation) = (a1 L(v1 ) + · · · + an L(vn )) + L(an+1 vn+1 ) (by the inductive hypothesis) = a1 L(v1 ) + · · · + an L(vn ) + an+1 L(vn+1 ) (by property (2) of a linear transformation), and we are done. (30) (b) The converse to the statement in part (a) is: If S = {v1 , . . . , vn } is a linearly independent set of vectors in V, then T = {L(v1 ), . . . , L(vn )} is a linearly independent set of vectors in W. For a counterexample, consider S = {v1 }, where v1 is any nonzero vector in V. (For instance, we could have V = W = R2 , with v1 = e1 .) Then S is linearly independent by definition. Let L 155


Section 5.2

be the zero linear transformation defined by L(v) = 0, for all v ∈ V. Then T = {0}, a linearly dependent subset of W. (31) First, we must show that L−1 (W ) is nonempty. Now, 0W ∈ W , so 0V ∈ L−1 ({0W }) ⊆ L−1 (W ). Hence, L−1 (W ) is nonempty. Next, if x, y ∈ L−1 (W ), then we must show that x + y ∈ L−1 (W ). But L(x), L(y) ∈ W by definition of L−1 (W ). Hence, L(x) + L(y) ∈ W , since W is closed under addition. Now L is a linear transformation, and so L(x) + L(y) = L(x + y). Therefore, L(x + y) ∈ W , which implies that x + y ∈ L−1 (W ). Finally, if x ∈ L−1 (W ), then we must show that ax ∈ L−1 (W ) for all a ∈ R. Now, L(x) ∈ W . Hence, aL(x) ∈ W (for any a ∈ R), since W is closed under scalar multiplication. This implies that L(ax) ∈ W . Therefore, ax ∈ L−1 (W ). Hence, L−1 (W ) is a subspace by Theorem 4.2. (36) (a) False. We are only given that property (2) of the definition of a linear transformation holds. Property (1) is also required to hold in order for L to be a linear transformation. For a specific example of a function between vector spaces for which property (2) holds but property (1) fails, consider f : R2 → R defined by f ([a, b]) = b whenever a = 0 and f ([0, b]) = 0. To show that property (2) holds for f , we consider 3 cases. First, if a = 0 and c = 0, f (c[a, b]) = f ([ca, cb]) = cb (since ca = 0) = cf ([a, b]). Next, if c = 0, then f (c[a, b]) = f ([0, 0]) = 0 = 0f ([a, b]). Finally, if a = 0, then f (c[0, b]) = f ([0, cb]) = 0 = c(0) = cf ([0, b]). Also, property (1) does not hold because f ([1, 1]) + f ([−1, 1]) = 1 + 1 = 2, but f ([1, 1] + [−1, 1]) = f ([0, 2]) = 0.1 (b) True. This is proven in Example 4 in Section 5.1 of the textbook. (c) False. L is not a linear transformation, and hence, not a linear operator. Since L(0) = [1, −2, 3] = 0, L does not satisfy part (1) of Theorem 5.1 and so cannot be a linear transformation. (We could also show, instead, that L satisfies neither of the two properties in the definition of a linear transformation.) (d) False. If A is a 4 × 3 matrix, then L(v) = Av is a linear transformation from R3 to R4 , not from R4 to R3 . (e) True. This is part (1) of Theorem 5.1. (f) False. In order for the composition of functions M1 ◦ M2 to be defined, the range of M2 must be a subset of the domain of M1 . However, in this case, the range of M2 is a subspace of X , while the domain of M1 is V. For a specific counterexample, consider M1 : M23 → M32 given by M1 (A) = AT , and M2 : M32 → R given by M2 (A) = a12 . Clearly here, M1 ◦ M2 is not defined. (Note that, in general, the composition M2 ◦ M1 is a well-defined linear transformation, since the codomain of M1 and the domain of M2 both equal W.) (g) True. This is part (1) of Theorem 5.3. (h) True. Since {0W } is a subspace of W, this statement follows directly from part (2) of Theorem 5.3.

Section 5.2 (2) In each part, we will use Theorem 5.5, which states that the ith column of ABC equals [L(vi )]C , where B is an ordered basis for the domain of L, C is an ordered basis for the codomain of L, and vi is the ith basis element in B. As directed, we will assume here that B and C are the standard bases in each case. 1 Note that property (1) alone also is insufficient to show that a function between real vector spaces is a linear transformation. However, describing an appropriate counterexample is quite complicated and beyond the level of this course. What makes constructing such a counterexample difficult is that property (1) for a function f does imply that f (cv) = cf (v) for every rational number c, but not for every real number.

156


Section 5.2

(a) Substituting each standard basis vector into the given formula for L shows that L(e1 ) = [−6, −2, 3], L(e2 ) = [4, 3, −1], and L(e3 ) = [−1, −5, 7]. Using each of these vectors as columns produces the ⎡ ⎤ −6 4 −1 3 −5 ⎦. desired matrix for L: ⎣ −2 3 −1 7 (c) Substituting each standard basis vector into the given formula for L shows that L(x3 ) = [4, 1, −2], L(x2 ) = [−1, 3, −7], L(x) = [3, −1, 5], and L(1) = [3, 5, −1]. Using each of these vectors as columns ⎡ ⎤ 4 −1 3 3 3 −1 5 ⎦. produces the desired matrix for L: ⎣ 1 −2 −7 5 −1 (3) In each part, we will use Theorem 5.5, which states that the ith column of ABC equals [L(vi )]C , where B is the given ordered basis for the domain of L, C is the given ordered basis for the codomain of L, and vi is the ith basis element in B. (a) Substituting each vector in B into the given formula for L shows that L([1, −3, 2]) = [4, −7], L([−4, 13, −3]) = [−1, 25], and L([2, −3, 20]) = [56, −24]. Next, we must express each of these answers in C-coordinates. We use the Coordinatization Method from Section 4.7. Thus, for each vector v that we need to express in C-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors in C, and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have three such systems to solve, we can solve all three simultaneously. Therefore, we row reduce " " 56 1 0 "" −47 128 −288 −2 5 "" 4 −1 . Hence, to 0 1 " −18 51 −104 −1 3 " −7 25 −24 [L([1, −3, 2])]C = [−47, −18], [L([−4, 13, −3])]C = [128, 51], and [L([2, −3, 20])]C = [−288, −104]. −47 128 −288 Using these coordinatizations as columns produces ABC = . (Note that −18 51 −104 ABC is just the matrix to the right of the bar in the row reduced matrix, above. This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.) (c) Substituting each vector in B into the given formula for L shows that L([5, 3]) = 10x2 + 12x + 6 and L([3, 2]) = 7x2 + 7x + 4. Next, we must express each of these answers in C-coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above two polynomials and the three polynomials in C into vectors in R3 by converting each polynomial of the form ax2 + bx + c to [a, b, c]. Then, for each converted vector v that we need to express in C-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors converted from C and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have two such systems to solve, we can solve both systems simultaneously. Therefore, we row reduce " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 22 14 3 −2 1 "" 10 7 ⎣ −2 2 −1 "" 12 7 ⎦ to ⎣ 0 1 0 "" 62 39 ⎦. Hence, [L([5, 3])]C = [22, 62, 68], and 0 0 1 " 68 43 0 −1 1 " 6 4 ⎡ ⎤ 22 14 [L([3, 2])]C = [14, 39, 43]. Using these coordinatizations as columns produces ABC = ⎣ 62 39 ⎦. 68 43

157


Section 5.2

(Note that ABC is just the matrix to the right of the bar in the row reduced matrix, above. This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.) (e) Substituting each vector in B into the given formula for L shows that L(−5x2 − x − 1) = 6 7 −19 0 5 −1 5 −3 −14 2 , and L(2x+1) = . Next, , L(−6x +3x+1) = −3 1 14 2 1 1 −6 −1 10 we must express each of these answers in C-coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above matrices and the six matrices in C into vectors in R6 by converting each matrix A to [a11 , a12 , a13 , a21 , a22 , a23 ]. Then, for each converted vector v that we need to express in C-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors converted from C, and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 3 such systems to solve, we can solve all 3 systems simultaneously. Therefore, we row reduce " " ⎤ ⎤ ⎡ ⎡ " 5 6 0 5 6 0 1 0 0 0 0 0 "" " " −11 −26 −6 ⎥ ⎢ ⎢ 0 −1 1 7 5 ⎥ 0 0 0 "" −3 " ⎥ ⎥ ⎢ ⎢ " ⎥ ⎥ ⎢ " ⎢ 0 0 1 0 0 0 " −14 −19 −1 ⎥ ⎢ I6 " −14 −19 −1 ⎥. ⎢ to obtain " ⎢ ⎥ " ⎢ 0 6 3 −2 ⎥ 2 ⎥ 0 0 −1 0 0 " −6 −3 " ⎥ ⎢ ⎢ " " ⎣ ⎦ ⎣ 0 1 1 ⎦ 1 1 0 0 0 1 0 " −1 " −1 " 11 13 0 14 1 0 0 0 0 1 1 " 10 Hence, [L(−5x2 − x − 1)]C = [5, −11, −14, 6, −1, 11], [L(−6x2 + 3x + 1)]C = [6, −26, −19, 3, 1, 13], and [L(2x+1)]C = [0, −6, −1, −2, 1, 0]. Using these coordinatizations as columns shows that ABC is just the 6 × 3 matrix to the right of the bar in the reduced matrix, above. (This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.) (4) (a) First, we find the matrix for L with respect to the standard basis B for R3 . Substituting each vector in B into the given formula for L shows that L(e1 ) = [−2, 0, 1], L(e2 ) = [1, −1, 0], and ⎡ ⎤ −2 1 0 L(e3 ) = [0, −1, 3]. Using each of these vectors as columns produces ABB = ⎣ 0 −1 −1 ⎦. 1 0 3 Next, we use Theorem 5.6 to compute ADE . To do this, we need the inverse of the transition matrix P from B to D and the transition matrix Q from B to E. Now P is the inverse of the matrix whose columns are the vectors in D, and Q is the inverse of the matrix whose columns ⎡ ⎤ 15 2 3 are the vectors in E (see Exercise 8(a) in Section 4.7). Hence, P−1 = ⎣ −6 0 −1 ⎦ and 4 1 1 ⎡ ⎤ ⎡ ⎤ 1 0 2 1 −2 −6 −1 3 −2 ⎦ . Using row reduction yields Q = Q−1 = ⎣ 1 −1 −4 ⎦ . Q−1 = ⎣ −3 1 −1 1 0 1 3 ⎡ ⎤ −202 −32 −43 Finally, using Theorem 5.6 produces ADE = QABB P−1 = ⎣ −146 −23 −31 ⎦. 83 14 18 (b) First, we find the matrix for L with respect to the standard basis B for M22 and the standard basis C for R2 . Recall that the standard basis B for M22 is {Ψij | 1 ≤ i ≤ 2, 1 ≤ j ≤ 2}, where Ψij is 158


Section 5.2

the 2 × 2 matrix having 1 as its (i, j) entry and zeroes elsewhere. (See Example 3 in Section 4.5.) Substituting each vector in B into the given formula for L shows that L(Ψ11 ) = [6, −2], L(Ψ12 ) = [−1, 3], L(Ψ21 ) = [3, −1], and L(Ψ22 ) = [−2, 4]. Using each of these vectors as columns produces 6 −1 3 −2 . Next, we use Theorem 5.6 to compute ADE . To do this, we need ABC = −2 3 −1 4 the inverse of the transition matrix P from B to D and the transition matrix Q from C to E. To compute P (or P−1 ), we need to convert the matrices in D into vectors in R4 . This is done by converting each 2×2 matrix A to [a11 , a12 , a21 , a22 ]. After we do this, P is the inverse of the matrix whose columns are the resulting vectors obtained from D, and Q is the inverse of the matrix whose ⎡ ⎤ 2 0 1 1 ⎢ 1 2 1 1 ⎥ ⎥ columns are the vectors in E (see Exercise 8(a) in Section 4.7). Hence, P−1 = ⎢ ⎣ 0 1 2 1 ⎦ 1 1 1 1 −1 −2 −1 2 1 and Q−1 = . Using Theorem 2.13 yields Q = Q−1 = . Finally, 5 2 −5 −2 21 7 21 16 −1 using Theorem 5.6 produces ADE = QABC P = . −51 −13 −51 −38 (6) (a) First, we use the method of Theorem 5.5. Substituting each vector in B into the given formula for L shows that L([4, −1]) = [9, 7] and L([−7, 2]) = [−16, −13]. Next, we must express each of these answers in B-coordinates. We use the Coordinatization Method from Section 4.7. Thus, for each vector v that we need to express in B-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors in B, and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 2 such systems to solve, we can solve both systems simultaneously. Therefore, we row reduce " " 1 0 "" 67 −123 4 −7 "" 9 −16 . yielding −68 0 1 " 37 −1 2 " 7 −13 Hence, [L([4, −1])]B = [67, 37] and [L([−7, 2])]B = [−123, −68]. Using these coordinatizations as 67 −123 . columns produces ABB = 37 −68 Next, we use the method of Theorem 5.6. We begin by finding the matrix for L with respect to the standard basis C for R2 . Substituting each vector in C into the given formula for L shows that L(i) = [2, 1] and L(j) = [−1, −3]. Using each of these vectors as columns produces 2 −1 . Next, we use Theorem 5.6 to compute ABB . To do this, we need the ACC = 1 −3 transition matrix P from C to B. Now P is the inverse of the matrix whose columns are the 4 −7 −1 . By Theorem 2.13, vectors in B (see Exercise 8(a) in Section 4.7). Hence, P = −1 2 −1 2 7 P = P−1 = . 1 4 67 −123 −1 Finally, using Theorem 5.6 produces ABB = PACC P = . 37 −68

159


Section 5.2

(b) First, we use the method of Theorem 5.5. Substituting each vector in B into the given formula for L shows that L(2x2 +2x−1) = 4x2 +3x+1, L(x) = x2 −1, and L(−3x2 −2x+1) = −4x2 −5x−2. Next, we must express each of these answers in B-coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above polynomials and the three polynomials in B into vectors in R3 by converting each polynomial of the form ax2 + bx + c to [a, b, c]. Then, for each converted vector v that we need to express in B-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors obtained from B and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 3 such systems to solve, we can solve all 3 systems simultaneously. Therefore, we row reduce " " ⎤ ⎤ ⎡ ⎡ 2 10 1 −4 1 0 0 "" −7 2 0 −3 "" 4 ⎣ 2 1 −2 " 3 0 −5 ⎦ to ⎣ 0 1 0 "" 5 −2 −9 ⎦. " 1 8 0 0 1 " −6 −1 0 1 " 1 −1 −2 Hence, [L(2x2 +2x−1)]B = [−7, 5, −6], [L(x)]B = [2, −2, 1], and [L(−3x2 −2x+1)]B = [10, −9, 8]. ⎡ ⎤ −7 2 10 Using these coordinatizations as columns produces ABB = ⎣ 5 −2 −9 ⎦. −6 1 8 Next, we use the method of Theorem 5.6. We begin by finding the matrix for L with respect to the standard basis C for P2 . Substituting each vector in C into the given formula for L shows to that L(x2 ) = 2x + 1, L(x) = x2 − 1, and L(1) = −2x2 + x − 1. We convert⎡ these results ⎤ 0 1 −2 0 1 ⎦. vectors in R3 . Then, using each of these vectors as columns produces ACC = ⎣ 2 1 −1 −1 Next, we use Theorem 5.6 to compute ABB . To do this, we need the transition matrix P from C to B. Now P is the inverse of the matrix whose columns are the polynomials in B, converted ⎡ ⎤ 2 0 −3 into vectors in R3 (see Exercise 8(a) in Section 4.7). Hence, P−1 = ⎣ 2 1 −2 ⎦ . Using −1 0 1 ⎡ ⎤ −1 0 −3 −1 2 ⎦. Finally, using Theorem 5.6 produces = ⎣ 0 1 row reduction yields P = P−1 −1 0 −2 ⎡ ⎤ −7 2 10 ABB = PACC P−1 = ⎣ 5 −2 −9 ⎦. −6 1 8 (7) (a) Computing the derivative of each polynomial in the standard basis B for P3 shows that L(x3 ) = 3x2 , L(x2 ) = 2x, L(x) = 1, and L(1) = 0. We convert these resulting polynomials in P2 to vectors in R3 . Then, using each of these vectors as columns produces ⎡ ⎤ 3 0 0 0 ABC = ⎣ 0 2 0 0 ⎦, where C is the standard basis for P2 . Now, by Theorem 5.5, 0 0 1 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 12 3 0 0 0 ⎢ ⎥ −5 ⎥ ⎣ ⎦ [L(4x3 − 5x2 + 6x − 7)]C = ABC [4x3 − 5x2 + 6x − 7]B = ⎣ 0 2 0 0 ⎦ ⎢ ⎣ 6 ⎦ = −10 . 6 0 0 1 0 −7 Converting back from C-coordinates to polynomials yields (4x3 − 5x2 + 6x − 7) = 12x2 − 10x + 6. 160


Section 5.2

(8) (a) To do this, we need to calculate L(i) and L(j). Rotating the vector i counterclockwise through an √ angle of π6 produces the vector [ 23 , 12 ]. Rotating j counterclockwise through an angle of π6 yields √ [− 12 , 23 ]. By Theorem 5.5, the matrix for L with respect to the standard basis is the matrix ( ' √ 3 − 12 2 √ . whose columns are L(i) and L(j); that is, 3 1 2

2

(9) (b) First, we need to compute L(bi ) for each matrix bi ∈ B. Since L is just the transpose transformation, the results of these computations are, respectively ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 1 0 0 0 0 0 0 −1 0 0 0 0 ⎣ 0 0 ⎦, ⎣ 1 0 ⎦, ⎣ 1 0 ⎦, ⎣ 0 0 ⎦, ⎣ 0 −1 ⎦, and ⎣ 0 0 ⎦. 0 0 −1 0 0 0 0 0 0 −1 0 1 Next, we must express each of these answers in C-coordinates. We use the Coordinatization Method from Section 4.7. To do this, we must first convert the above matrices and the 6 matrices in C into vectors in R6 by changing each 3 × 2 matrix A to the vector [a11 , a12 , a21 , a22 , a31 , a32 ]. Then, for each resulting vector v that we need to express in C-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors converted from C and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have 6 such systems to solve, we can solve all 6 systems simultaneously. Therefore, we row reduce " ⎤ ⎡ 0 0 0 0 0 1 1 0 0 0 0 "" 1 ⎢ 1 −1 0 0 0 −1 0 0 ⎥ 0 0 0 "" 0 ⎥ ⎢ " ⎢ 0 1 1 0 0 0 ⎥ 0 1 1 0 0 " 0 ⎥ ⎢ ⎢ 0 0 0 0 −1 0 ⎥ 0 1 −1 0 0 "" 0 ⎥ ⎢ ⎣ 0 0 0 0 ⎦ 0 0 0 1 1 "" 0 −1 0 0 0 0 −1 1 0 0 0 0 1 −1 " 0 " 1 ⎡ ⎤ " 0 0 − 12 0 0 " 2 ⎢ " 1 ⎥ 1 0 0 0 0 ⎥ ⎢ " 2 2 ⎢ " ⎥ 1 1 ⎢ " 0 0 − 12 0 ⎥ ⎢ " ⎥ 2 2 to obtain ⎢ I6 " ⎥. 1 1 1 ⎢ " 0 ⎥ 0 0 2 2 2 ⎢ " ⎥ ⎢ " 1 ⎥ 1 1 0 −2 " 0 −2 0 ⎣ ⎦ 2 " 1 1 " 0 −1 0 0 − 2 2 2 The columns of the matrix to the right of the bar give the C-coordinatizations of the images of the 6 matrices in B under L. Hence, that 6 × 6 matrix is ABC . (10) According to Theorem 5.7, the matrix ABD for the composition L2 ◦L1 is the matrix product ACD ABC . ⎡ ⎤ ⎡ ⎤ 4 −1 −12 12 −2 −2 3 −1 0 ⎦ 6 −2 ⎦. Hence, ABD = ⎣ 2 = ⎣ −4 4 0 −2 −1 −3 −10 −3 7 (13) In each part, we use the fact that, by Theorem 5.5, the ith column of ABB is the B-coordinatization of L(vi ). (a) For each i, L(vi ) = vi . But [vi ]B = ei . (See the comment just after Example 2 in Section 4.7.) Hence, the ith column of ABB is ei , and so ABB = In .

161


Section 5.2

(c) For each i, L(vi ) = cvi . But [cvi ]B = c[vi ]B (by part (2) of Theorem 4.19) = cei (see the comment just after Example 2 in Section 4.7). Hence, the ith column of ABB is cei , and so ABB = cIn . (e) Now L(v1 ) = vn , and for each i > 1, L(vi ) = vi−1 . But [vj ]B = ej . (See the comment just after Example 2 in Section 4.7.) Hence, the 1st column of ABB is en , and the ith column of ABB for i > 1 is ei−1 . Thus, ABB is the n × n matrix whose columns are en , e1 , e2 , . . ., en−1 , respectively. (15) First, note that [(L2 ◦L1 )(v)]D = [L2 (L1 (v))]D = ACD [L1 (v)]C = ACD (ABC [v]B ) = (ACD ABC )[v]B . Therefore, by the uniqueness condition in Theorem 5.5, ACD ABC is the matrix for L2 ◦L1 with respect to B and D. " " " x − 89 − 29 − 29 "" " " " 4 x − 59 (18) (a) pABB (x) = |xI3 −ABB | = " − 29 " = x(x−1)2 , where we have used basketweaving 9 " " 2 5 4 " − x− 9 " 9 9 and algebraic simplification. (b) For λ1 = 1, we row reduce [I3 − ABB |0] = " ⎤ ⎡ 1 ⎡ − 29 − 29 "" 0 9 1 −2 ⎥ ⎢ 2 4 4 " ⎣ 0 0 0 to obtain " ⎦ ⎣ −9 9 9 " 2 4 4 " 0 0 0 − 9

9

9

−2 0 0

" ⎤ " 0 " " 0 ⎦. " " 0

There are two independent variables, and hence Step 3 of the Diagonalization Method of Section 3.4 produces two fundamental eigenvectors for λ1 . Following that method, the basis for E1 = {[2, 1, 0], [2, 0, 1]}. For λ2 = 0, we row reduce [0I3 − ABB |0] = [−ABB |0] = " ⎤ ⎡ 8 ⎡ ⎤ 1 "" − 9 − 29 − 29 "" 0 1 0 2 " 0 ⎥ " ⎢ 2 4 ⎣ 0 1 −1 " 0 ⎦. ⎣ − 9 − 59 9 "" 0 ⎦ to obtain " 4 0 0 0 " 0 −2 −5 " 0 9

9

9

There is one independent variable. Setting it equal to 2 to eliminate fractions produces one fundamental eigenvector for λ2 . Following the Diagonalization Method, the basis for E0 = {[−1, 2, 2]}. Combining these bases for E1 and E0 yields C = ([2, 1, 0], [2, 0, 1], [−1, 2, 2]). (Note: The answer to part (c) will vary if the vectors in C are ordered differently.) (c) The transition matrix P from C to B is the matrix whose columns are the vectors in C (see ⎡ ⎤ 2 2 −1 2 ⎦. Exercise 8(b) in Section 4.7). Hence, P = ⎣ 1 0 0 1 2 (25) Let L : V −→ W be a linear transformation such that L(v1 ) = w1 , L(v2 ) = w2 , . . ., L(vn ) = wn , with {v1 , v2 , . . . , vn } a basis for V. If v ∈ V, then v = c1 v1 + c2 v2 + · · · + cn vn , for unique c1 , c2 , . . . , cn ∈ R (by Theorem 4.9). But then L(v) = L(c1 v1 + c2 v2 + · · · + cn vn ) = c1 L(v1 ) + c2 L(v2 ) + · · · + cn L(vn ) = c1 w1 + c2 w2 + · · · + cn wn . Hence, for each v ∈ V, L(v) is completely determined by the given values for w1 , . . . , wn and the unique values of c1 , . . . , cn . Therefore, L is uniquely determined. (26) (a) True. This is shown in Theorem 5.4. (b) True. This follows from Theorem 5.4.

162


Section 5.3

(c) False. Bases for the domain and codomain must be chosen. Different choices of bases typically produce different matrices for L. For instance, Examples 3 and 4 in Section 5.2 of the textbook show two different matrices for the same linear transformation. (d) False. This statement has the correct equation scrambled. The correct equation from Theorem 5.5 states that ABC [v]B = [L(v)]C . For a specific counterexample to the scrambled equation in the problem statement, consider L : R2 → R given by L([a, b]) = a and let B = (i, j), C = (1) and v = [1, 2]. Then ABC = [1, 0], a 1 × 2 matrix, [v]B = [1, 2], and [L(v)]C = [1]C = [1], a 1-vector. Note that ABC [L(v)]C is not even defined because the matrix and vector are of incompatible sizes. Hence, the scrambled equation cannot be correct. (e) True. This is part of the statement of Theorem 5.5. (f) False. The given matrix is the matrix (in standard coordinates) for the projection onto ⎡ 1 plane. The matrix for the projection L onto the xz-plane (in standard coordinates) is ⎣ 0 0 since L(i) = i, L(j) = 0, and L(k) = k.

the xy⎤ 0 0 ⎦, 1

0 0 0

(g) True. To see this, take the equation ADE = QABC P−1 from Theorem 5.6 and multiply both sides on the right by P. (h) True. This is shown just before Example 6 in Section 5.2. (i) True. This follows from Exercise 6 in Section 3.4. (j) False. According to Theorem 5.7, the matrix for L2 ◦ L1 with respect to the standard basis is 0 1 1 2 3 4 1 2 0 1 2 1 = , not = . 1 0 3 4 1 2 3 4 1 0 4 3

Section 5.3 (1) In each part, let A represent the given 3 × 3 matrix such that L(v) = Av. (a) Yes, because L([1, −2, 3]) = [0, 0, 0]. (c) No. If v = [2, −1, 4] were in range(L), then there would exist a vector x = [x1 , x2 , x3 ] such that Ax = v. This is equivalent to the system ⎧ 2 ⎨ 5x1 + x2 − x3 = + x3 = −1 , which has no solutions, since −3x1 ⎩ 4 x1 − x2 − x3 = " ⎡ ⎤ " ⎤ ⎡ 1 0 − 13 "" 13 5 1 −1 "" 2 ⎢ 2 " 1 ⎥ ⎣ −3 0 1 "" −1 ⎦ row reduces to ⎣ 0 1 3 "" 3 ⎦. 1 −1 −1 " 4 0 0 0 " 4 Note: In the above, we stopped the row reduction process at the augmentation bar. If you use a calculator to perform the row reduction, it may continue row reducing into the column beyond the bar, making that column equal e3 . (2) (a) No, since L(x3 − 5x2 + 3x − 6) = 6x3 + 4x − 9 = 0. (c) Yes. To show this, we need to find p(x) = ax3 + bx2 + cx + d such that L(p) = 8x3 − x − 1. Equating the coefficients of L(p) to those of 8x3 − x − 1 gives the system 163

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎧ ⎨ ⎩

2c −a − b

−c + d

= 8 = −1 = −1

Section 5.3

← x3 terms ← x terms ← constant terms

One solution is a = 1, b = 0, c = 4, d = 3. And so, L(x3 + 4x + 3) = 8x3 − x − 1, showing that 8x3 − x − 1 ∈ range(L). (3) In each part, let A represent the given matrix such that L(v) = Av. ⎡ ⎤ 1 0 2 (a) The reduced row echelon form of A is B = ⎣ 0 1 −3 ⎦. Then, by the Kernel Method, a basis 0 0 0 for ker(L) is found by considering the solution to the system BX = 0 computed by setting the one independent variable in the system equal to 1. This yields the basis {[−2, 3, 1]} for ker(L). Since ker(L) has a basis with 1 element, dim(ker(L)) = 1. Next, by the Range Method, a basis for range(L) consists of the first 2 columns of A, since those columns in B contain nonzero pivots. Thus, a basis for range(L) = {[1, −2, 3], [−1, 3, −3]} and dim(range(L)) = 2. Note that dim(ker(L)) + dim(range(L)) = 1 + 2 = 3 = dim(R3 ). ⎡ ⎤ 1 0 −1 1 ⎢ 0 1 3 −2 ⎥ ⎢ ⎥ ⎢ 0 0 ⎥ (d) The reduced row echelon form of A is B = ⎢ 0 0 ⎥. Then, by the Kernel Method, ⎣ 0 0 0 0 ⎦ 0 0 0 0 a basis for ker(L) is found by considering certain particular solutions to the system BX = 0. We note that x3 and x4 are independent variables in the system. Letting x3 = 1, x4 = 0 produces the vector [1, −3, 1, 0]. Using x3 = 0, x4 = 1 gives us [−1, 2, 0, 1]. This yields the basis {[1, −3, 1, 0], [−1, 2, 0, 1]} for ker(L). Since ker(L) has a basis with 2 elements, dim(ker(L)) = 2. Next, by the Range Method, a basis for range(L) consists of the first 2 columns of A, since those columns in B contain nonzero pivots. Thus a basis for range(L) = {[−14, −4, −6, 3, 4], [−8, −1, 2, −7, 2]} and dim(range(L)) = 2. Note that dim(ker(L)) + dim(range(L)) = 2 + 2 = 4 = dim(R4 ). (4) We will solve each part by inspection (mostly). (a) Now, ker(L) is the set of vectors sent to 0 by L. But, L([x1 , x2 , x3 ]) = [0, x2 ] = [0, 0] precisely when x2 = 0, and so ker(L) = {[x1 , 0, x3 ] | x1 , x3 ∈ R} = {[a, 0, c] | a, c ∈ R}, which has basis {[1, 0, 0], [0, 0, 1]}. Thus, dim(ker(L)) = 2. Also, range(L) = {[0, x2 ] | x2 ∈ R}, the set of images obtained using L, which can also be expressed as {[0, b] | b ∈ R}. Hence, a basis for range(L) = {[0, 1]} and dim(range(L)) = 1. Finally, note that dim(ker(L)) + dim(range(L)) = 2 + 1 = 3 = dim(R3 ). (d) Now, ker(L) is the set of vectors sent to 0 by L. But L(ax4 +bx3 +cx2 +dx+e) = cx2 +dx+e = 0 precisely when c = d = e = 0, and so ker(L) = {ax4 + bx3 | a, b ∈ R}, which has basis {x4 , x3 }. Thus, dim(ker(L)) = 2. Also, range(L) = {cx2 + dx + e | c, d, e ∈ R}, the set of images obtained using L. This is precisely P2 . Hence, a basis for range(L) = {x2 , x, 1} and dim(range(L)) = 3. Finally, note that dim(ker(L)) + dim(range(L)) = 2 + 3 = 5 = dim(P4 ). (f) Now, ker(L) is the set of vectors sent to 0 by L. But, L([x1 , x2 , x3 ]) = [x1 , 0, x1 −x2 +x3 ] = [0, 0, 0] precisely when x1 = 0 and x1 − x2 + x3 = 0. That is when x1 = 0 and x2 = x3 . Hence ker(L) = {[0, x2 , x2 ] | x2 ∈ R}, which can also be expressed as {[0, b, b] | b ∈ R}. Thus, ker(L) has basis {[0, 1, 1]}, and so dim(ker(L)) = 1. Also, range(L) = {[x1 , 0, x1 − x2 + x3 ] | x1 , x2 , x3 ∈ R}, the set of images obtained using L. 164


Section 5.3

Hence, range(L) = {x1 [1, 0, 1] + x2 [0, 0, −1] + x3 [0, 0, 1] | x1 , x2 , x3 ∈ R}, and so {[1, 0, 1], [0, 0, −1], [0, 0, 1]} spans range(L). However, this set is not linearly independent, since [0, 0, 1] = (−1)[0, 0, −1]. So, if we eliminate the last vector, we get the basis {[1, 0, 1], [0, 0, −1]} for range(L). (A simpler basis for range(L) can be found using the Simplified Span Method, yielding {[1, 0, 0], [0, 0, 1]}.) And so, dim(range(L)) = 2. Finally, note that dim(ker(L)) + dim(range(L)) = 1 + 2 = 3 = dim(R3 ). (g) Now, ker(L) is the set of vectors sent to O22 by L. But L(A) = AT = O22 precisely when A = O22 . Hence, ker(L) = {O22 }, having basis { } (the empty set), and dim(ker(L)) = 0. Also, every 2 × 2 matrix is the image of its transpose (since (AT )T = A), proving range(L) = M22 . Therefore, dim(range(L)) = 4, and a basis for range(L) = standard basis for M22 . Notice that dim(ker(L)) + dim(range(L)) = 0 + 4 = 4 = dim(M22 ). (i) Now, ker(L) is the set of vectors sent to 0 by L. Note that L(p) = [p(1), p (1)] = [0, 0] precisely when p(1) = 0 and p (1) = 0. But p(1) = 0 implies that (x − 1) is a factor of p. Thus, p = (x − 1)(ax + b). Next, p (1) = 0 implies (a(1) + b)(1) + ((1)− 1)(a) = 0 (using the product rule and substituting x = 1), which gives a = −b. Thus, p = (x−1)(ax−a) = a(x2 −2x+1). Therefore, a basis for ker(L) is {x2 − 2x + 1}, and dim(ker(L)) = 1. A spanning set for range(L) can be found by computing the images of a basis for P2 . Now L(x2 ) = [1, 2], L(x) = [1, 1], and L(1) = [1, 0]. The first 2 of these form a linearly independent set, and the fact that dim(R2 ) = 2 shows that the third vector is redundant. Hence, a basis for range(L) is {[1, 2], [1, 1]} and dim(range(L)) = 2. Since dim(range(L)) = dim(R2 ), range(L) = R2 , and so a simpler basis for range(L) = standard basis for R2 . Notice that dim(ker(L)) + dim(range(L)) = 1 + 2 = 3 = dim(P2 ). (6) First, we show that L is a linear transformation by verifying the two properties. !3 Property (1): L(A + B) = trace(A + B) = i=1 (aii + bii ) !3 !3 = i=1 aii + i=1 bii = trace(A) + trace(B) = L(A) + L(B). !3 !3 Property (2): L(cA) = trace(cA) = i=1 caii = c i=1 aii = c(trace(A)) = cL(A). Next, ker(L) is the set of all 3 × 3 matrices whose trace is 0. Thus, ⎫ ⎧⎡ ⎤" " c ⎬ ⎨ a b " ⎦ " a, b, c, d, e, f, g, h ∈ R . A detailed computation for a basis for ker(L) f ker(L) = ⎣ d e " ⎭ ⎩ g h −a − e " is given in the solution to Exercise 11(b) in Section 4.6, found earlier in this manual. The basis found there has eight elements, and so dim(ker(L)) = 8. Also, it is possible to obtain any real number a as the trace of some 3 × 3 matrix since trace( a3 I3 ) = a. Hence, range(L) = R, and so dim(range(L)) = 1. (8) First, ker(L) is the set of polynomials p such that x2 p(x) = 0 for all x. But since x2 is 0 only when x = 0, p(x) must be 0 for all other values of x. However, no nonzero polynomial can have an infinite number of roots. Hence, p = 0. Thus, ker(L) = {0}, and so dim(ker(L)) = 0. Next, the images under L are precisely those having x2 as a factor. Thus, range(L) = {ax4 + bx3 + cx2 | a, b, c ∈ R}. Clearly, a basis for range(L) is {x4 , x3 , x2 }, and so dim(range(L)) = 3. (10) First, let us consider the case k ≤ n. Taking the kth derivative of a polynomial reduces its degree by k, and so the polynomials of degree less than k are precisely those that have kth derivative equal to 0. Hence, when k ≤ n, ker(L) = {all polynomials of degree less than k} = Pk−1 . Therefore, dim(ker(L)) = k. Also, computing a kth antiderivative for a polynomial p of degree ≤ (n − k) will yield a pre-image under L for p. Hence, range(L) = Pn−k , and dim(range(L)) = n − k + 1. When k > n, L sends all of Pn to 0, and so ker(L) = Pn , dim(ker(L)) = n + 1, range(L) = {0}, and dim(range(L)) = 0. 165


Section 5.4

(12) First, ker(L) = {0} because L(X) = 0 ⇒ AX = 0 ⇒ A−1 AX = A−1 0 ⇒ X = 0. Similarly, every vector X is in the range of L since L(A−1 X) = A(A−1 X) = X. Thus, range(L) = Rn .

x 1 −1 x 1 −1 1 −1 (16) Consider L = . Since row reduces to , the Kernel y 1 −1 y 1 −1 0 0 Method and the Range Method show that ker(L) = range(L) = {[a, a] | a ∈ R}. (20) (a) False. By definition, the set {L(v) | v ∈ V} is range(L), not ker(L). These sets are usually unequal. For specific counterexamples, consider Examples 1, 2, and 3 in Section 5.3 of the textbook. (b) False. Range(L), in general, is a subspace of the codomain W, not of the domain V. A linear transformation such as that given in Example 4 in Section 5.3 provides a specific counterexample. (c) True. This is a direct consequence of the Dimension Theorem. (d) False. The Dimension Theorem relates the dimensions of ker(L) and the domain V of L with dim(range(L)), not with the dimension of the codomain W. For a specific counterexample, consider L : R5 → R3 given by L(v) = 0 for all v ∈ R5 . Then ker(L) = R5 , and so dim(ker(L)) = 5 = 2. (e) True. By part (2) of Theorem 5.9, dim(ker(L)) = n − rank(A). Now, rank(A) is the number of nonzero rows in the reduced row echelon form for A. Hence, it gives the number of pivots obtained from A, which is the number of pivot columns for A. But since A has n columns in all, it must have n − rank(A) nonpivot columns. (f) False. By part (1) of Theorem 5.9, dim(range(L)) = rank(A). See Examples 4 and 5 in Section 5.3 for a specific counterexample. In those examples, rank(A) = 3, and dim(range(L)) = 3 = 5 − rank(A). (g) False. By Corollary 5.11, rank(A) = rank(AT ). Hence, if rank(A) = 2, then rank(AT ) = 2 as well. (h) False. Even though Corollary 5.11 assures us that dim(row space of A) = dim(column space of A), the subspaces themselves are not necessarily the same. In fact, unless A is square, they are not even subspaces of the same vector space. See the comments after Example 7 in Section 5.3, which indicate that counterexamples to the statement are provided by Examples 4, 5, and 7 in Section 5.3.

Section 5.4 (1) In each part, we will check whether L is one-to-one using part (1) of Theorem 5.12, which states that L is one-to-one if and only if ker(L) = {0}. (a) L is not one-to-one because L([1, 0, 0]) = [0, 0, 0, 0]. Thus, [1, 0, 0] ∈ ker(L), and so ker(L) = {0}. Also, L is not onto because [0, 0, 0, 1] is not in range(L). The reason is that all vectors in range(L) have a zero in their fourth coordinate. (c) L is one-to-one because L([x, y, z]) = [0, 0, 0] implies that [2x, x + y + z, −y] = [0, 0, 0], which is equivalent to the system ⎧ = 0 ⎨ 2x x + y + z = 0 . A quick inspection shows that x = y = z = 0 is the only ⎩ −y = 0 solution, and hence ker(L) = {0}. Next, L is onto, because every vector [a, b, c] can be expressed as [2x, x + y + z, −y], where x = a2 , y = −c, and z = b − a2 + c. 166


Section 5.4

(e) L is one-to-one because L(ax2 + bx + c) = 0 implies that a + b = b + c = a + c = 0. Thus, a + b = a + c, and so b = c. Using this and b + c = 0 gives b = c = 0. Then a + b = 0 implies a = 0. Hence, ker(L) = {0}. Next, L is onto because every polynomial Ax2 + Bx + C can be expressed as (a + b)x2 + (b + c)x + (a + c), where a = (A − B + C)/2, b = (A + B − C)/2, and c = (−A + B + C)/2.

0 1 0 0 0 0 1 0 (g) L is not one-to-one because L = . Hence, ∈ ker(L), 1 0 −1 0 0 1 0 −1 A B and so ker(L) = {O23 }. However, L is onto because every 2 × 2 matrix can be C D a −c expressed as , where a = A, c = −B, e = C/2, d = D, and f = 0. 2e d + f 0 0 2 (h) L is one-to-one because L(ax + bx + c) = implies that a + c = b − c = −3a = 0, which 0 0 0 1 gives a = b = c = 0. Hence, ker(L) = {0}. Next, L is not onto, because is not in 0 0 range(L), since every matrix in range(L) has a zero for its (1,2) entry. (2) In each part, let A represent the given matrix such that L(v) = Av. (a) The matrix A row reduces to I2 . Hence, rank(A) = 2. Therefore, Theorem 5.9 implies that dim(ker(L)) = 0 and dim(range(L)) = 2. Since dim(ker(L)) = 0, L is one-to-one. Also, dim(range(L)) = 2 = dim(R2 ) implies that L is onto. ⎡ ⎤ 1 0 (b) The matrix A row reduces to ⎣ 0 1 ⎦. Hence, rank(A) = 2. Therefore, Theorem 5.9 implies 0 0 that dim(ker(L)) = 0 and dim(range(L)) = 2. Since dim(ker(L)) = 0, L is one-to-one. Also, dim(range(L)) = 2 < 3 = dim(R3 ) implies that L is not onto. ⎤ ⎡ 1 0 − 25 ⎥ ⎢ (c) The matrix A row reduces to ⎣ 0 1 − 65 ⎦. Hence, rank(A) = 2. Therefore, Theorem 5.9 0

0

0

implies that dim(ker(L)) = 1 and dim(range(L)) = 2. Since dim(ker(L)) = 0, L is not one-to-one. Also, dim(range(L)) = 2 < 3 = dim(R3 ) implies that L is not onto. (3) In each part, let A represent the given matrix for L. (a) The matrix A row reduces to I3 . Hence, rank(A) = 3. Therefore, Theorem 5.9 implies that dim(ker(L)) = 0 and dim(range(L)) = 3. Since dim(ker(L)) = 0, L is one-to-one. Also, dim(range(L)) = 3 = dim(P2 ) implies that L is onto. ⎤ ⎡ 19 1 0 − 10 11 11 ⎢ 3 9 ⎥ ⎥ ⎢ 0 1 − 11 11 ⎥. Hence, rank(A) = 2. Therefore, Theorem ⎢ (c) The matrix A row reduces to ⎢ ⎥ 0 0 0 0 ⎣ ⎦ 0 0 0 0 5.9 implies that dim(ker(L)) = 2 and dim(range(L)) = 2. Since dim(ker(L)) = 0, L is not one-to-one. Also, dim(range(L)) = 2 < 4 = dim(P3 ) implies that L is not onto. 167


Section 5.5

(9) (a) False. The second half of the given statement is the converse of what it means for a function to be one-to-one. That is, L is one-to-one if and only if L(v1 ) = L(v2 ) implies v1 = v2 . Any linear transformation that is not one-to-one, such as Example 2 in Section 5.4 of the textbook, provides a specific counterexample to the given statement. Note that the statement “v1 = v2 implies L(v1 ) = L(v2 )” is actually true for every function, not just one-to-one functions. (b) False. The correct definition of what it means for L : V → W to be onto is that for every w ∈ W there is a v ∈ V such that L(v) = w. The statement “for all v ∈ V, there is some w ∈ W such that L(v) = w” is true for every function L : V → W, not just those that are onto. Thus, any linear transformation that is not onto provides a counterexample. One such function is described just after Example 2 in Section 5.4 of the textbook. (c) True. This is one direction of the “if and only if” statement in part (1) of Theorem 5.12. (d) True. This follows from part (2) of Theorem 5.13. (See the discussion directly after the proof of Theorem 5.13 in the textbook.) Also, you are asked to prove this in part (b) of Exercise 8 in Section 5.4. (e) True. This is the contrapositive of one half of the if and only if statement in part (1) of Theorem 5.12. (f) True. This is the contrapositive of one half of the if and only if statement in part (2) of Theorem 5.12. (g) False. For a specific counterexample, consider the linear transformation L : R3 → R3 given by L(v) = 0 for all v ∈ R3 . The linearly independent set T = {i, j} has image L(T ) = {0}, which is linearly dependent. (h) False. For a specific counterexample, consider the linear transformation L : R3 → R2 given by L([a, b, c]) = [a, b]. Let S = {[1, 0, 0], [0, 1, 0]}, which does not span R3 . However, L(S) = {[1, 0], [0, 1]}, which does span R2 .

Section 5.5 (1) In each part, let A represent the given matrix for L1 and let B represent the given matrix for L2 . (a) (i) By Theorem 5.15, L1 and L2 are isomorphisms if and only if A and B are nonsingular. But cofactor expansions of each along the third column show that |A| = 1 and |B| = 3. Since neither of these is zero, both matrices are nonsingular. (ii) By the remark immediately following −1 v, where L−1 2 (v) = B ⎡ ⎤ ⎡ 1 0 0 1 A−1 = ⎣ 0 −1 0 ⎦ and B−1 = ⎣ 0 1 −2 0 2

−1 v and Theorem 5.15 in the textbook, L−1 1 (v) = A

0 0 1

0

⎤

− 13 ⎦ are obtained by row reduction. 0

(iii) By Theorem 5.7, (L2 ◦ L1 )(v) = (BA)v. Computing BA produces ⎤⎞ ⎡ ⎤⎡ ⎤ ⎛⎡ 0 −2 1 x1 x1 4 −2 ⎦ ⎣ x2 ⎦. (L2 ◦ L1 ) ⎝⎣ x2 ⎦⎠ = ⎣ 1 x3 x3 0 3 0 (iv) By the remark immediately following Theorem 5.15 in the textbook, (L2 ◦ L1 )−1 (v) = (BA)−1 (v). Using row reduction to compute the inverse of the matrix we found in part (iii) gives 168

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎤⎞ ⎡ 2 1 x1 ⎢ (L2 ◦ L1 )−1 ⎝⎣ x2 ⎦⎠ = ⎣ 0 0 x3 1 0 ⎛⎡

0 1 3 2 3

Section 5.5

⎤⎡

⎤ x1 ⎥⎣ ⎦ x2 ⎦. x3

−1 −1 −1 (v) By Theorem 5.7, (L−1 B )v. Using the matrices A−1 and B−1 we com1 ◦ L2 )(v) = (A puted in part (ii) yields ⎤ ⎡ 2 1 0 ⎥ ⎢ −1 −1 A−1 B−1 = ⎣ 0 0 13 ⎦. Hence, (L−1 , since 1 ◦ L2 ) is the same linear operator as (L2 ◦ L1 )

1

0

2 3

parts (iv) and (v) show that they have the same matrix with respect to the standard basis. (c) (i) By Theorem 5.15, L1 and L2 are isomorphisms if and only if A and B are nonsingular. But using basketweaving on each shows that |A| = −1 and |B| = 1. Since neither of these is zero, both matrices are nonsingular. −1 v and (ii) By the remark immediately following Theorem 5.15 in the textbook, L−1 1 (v) = A −1 −1 L2 (v) = B v, where ⎡ ⎤ ⎡ ⎤ 2 −4 −1 1 0 −1 1 −3 ⎦ are obtained by row reduction. A−1 = ⎣ 7 −13 −3 ⎦ and B−1 = ⎣ 3 5 −10 −3 −1 −2 2

(iii) By Theorem 5.7, (L2 ◦ L1 )(v) = (BA)v. Computing BA produces ⎤⎞ ⎡ ⎤⎡ ⎤ ⎛⎡ 29 −6 −4 x1 x1 (L2 ◦ L1 ) ⎝⎣ x2 ⎦⎠ = ⎣ 21 −5 −2 ⎦ ⎣ x2 ⎦. 38 −8 −5 x3 x3 (iv) By the remark immediately (BA)−1 (v). Using row reduction gives ⎛⎡ ⎤⎞ ⎡ x1 −9 (L2 ◦ L1 )−1 ⎝⎣ x2 ⎦⎠ = ⎣ −29 x3 −22

following Theorem 5.15 in the textbook, (L2 ◦ L1 )−1 (v) = to compute the inverse of the matrix we found in part (iii) −2 −7 −4

⎤⎡ ⎤ 8 x1 26 ⎦ ⎣ x2 ⎦. x3 19

−1 −1 −1 B )v. Using the matrices A−1 and B−1 we com(v) By Theorem 5.7, (L−1 1 ◦ L2 )(v) = (A puted in part (ii) yields ⎡ ⎤ −9 −2 8 −1 −1 , A−1 B−1 = ⎣ −29 −7 26 ⎦. Hence, (L−1 1 ◦ L2 ) is the same linear operator as (L2 ◦ L1 ) −22 −4 19

since parts (iv) and (v) show that they have the same matrix with respect to the standard basis. (5) (a) Now R(i) = j and R(j) = i. Since the columns of the matrix for R are the images of i and j, 0 1 respectively, the matrix is . 1 0 (8) Suppose dim(V) = n > 0 and dim(W) = k˙ > 0. Suppose L is an isomorphism. Let M be the matrix for L−1 with respect to C and B. Then −1 (L ◦ L)(v) = v for every v ∈ V =⇒ MABC [v]B = [v]B for every v ∈ V (by Theorem 5.7). Letting vi be the ith basis element in B, [vi ]B = ei , and so ei = [vi ]B = MABC [vi ]B = MABC ei = (ith column of MABC ). Therefore, MABC = In . Similarly, using the fact that (L ◦ L−1 )(w) = w for 169


Section 5.5

every w ∈ W implies that ABC M = Ik . Exercise 21 in Section 2.4 then implies that n = k, ABC is nonsingular, and M = A−1 BC . −1 Conversely, if ABC is nonsingular, then ABC is square, n = k, A−1 BC exists, and ABC ABC = −1 ABC ABC = In . Let K be the linear operator from W to V whose matrix with respect to C and −1 B is A−1 BC . Then, for all v ∈ V, [(K ◦ L)(v)]B = ABC ABC [v]B = In [v]B = [v]B . Therefore, (K ◦ L)(v) = v for all v ∈ V, since coordinatizations are unique. Similarly, for all w ∈ W, [(L ◦ K)(w)]C = ABC A−1 BC [w]C = In [w]C = [w]C . Hence, (L ◦ K)(w) = w for all w ∈ W. Therefore, K acts as an inverse for L. Hence, L is an isomorphism by Theorem 5.14. (13) (a) Because dim(R6 ) = dim(P5 ), Corollary 5.21 tells us that if L were one-to-one, L would also be onto. But since L is not onto, L is also not one-to-one. (b) Because dim(M22 ) = dim(P3 ), Corollary 5.21 tells us that if L were onto, L would also be one-to-one. But since L is not one-to-one, L is also not onto. (16) We want to use this exercise as part of the proof of the Dimension Theorem, so we will not use the Dimension Theorem in this proof. Because Y is a subspace of the finite dimensional vector space V, Y is also finite dimensional (Theorem 4.16). Suppose {y1 , . . . , yk } is a basis for Y. Because L is one-to-one, the vectors L(y1 ), . . . , L(yk ) are distinct. By part (1) of Theorem 5.13, {L(y1 ), . . . , L(yk )} = L({y1 , . . . , yk }) is linearly independent. Part (a) of Exercise 8 in Section 5.4 shows that span(L({y1 , . . . , yk })) = L(span({y1 , . . . , yk })) = L(Y). Therefore,{L(y1 ), . . . , L(yk )} is a k-element basis for L(Y). Hence, dim(L(Y)) = k = dim(Y). (17) (c) Let v ∈ T1 (ker(T ◦ T1 )). Thus, there is an x ∈ ker(T ◦ T1 ) such that T1 (x) = v. Then T (v) = T (T1 (x)) = (T ◦ T1 )(x) = 0W . Thus, v ∈ ker(T ). Therefore, T1 (ker(T ◦ T1 )) ⊆ ker(T ). Now suppose that v ∈ ker(T ). Then T (v) = 0W . Thus, (T ◦ T1 )(T1−1 (v)) = 0W as well. Hence, T1−1 (v) ∈ ker(T ◦T1 ). Therefore, v = T1 (T1−1 (v)) ∈ T1 (ker(T ◦T1 )), implying ker(T ) ⊆ T1 (ker(T ◦ T1 )), completing the proof. (e) Suppose y ∈ range(T2 ◦ T ). Hence, there is a v ∈ V such that (T2 ◦ T )(v) = y. Thus, y = T2 (T (v)). Clearly, T (v) ∈ range(T ), and so y ∈ T2 (range(T )). This proves that range(T2 ◦ T ) ⊆ T2 (range(T )). Next, suppose that y ∈ T2 (range(T )). Then there is some w ∈ range(T ) such that T2 (w) = y. Since w ∈ range(T ), there is a v ∈ V such that T (v) = w. Therefore, (T2 ◦ T )(v) = T2 (T (v)) = T2 (w) = y. This establishes that T2 (range(T )) ⊆ range(T2 ◦ T ), finishing the proof. (18) (a) Part (c) of Exercise 17 states (ker L2 ◦ L and that T1−1 (T ◦ T1 )) = ker (T ). Substituting T = −1 −1 ker L = ker (L yields L ◦ L ◦ L ◦ L). Now M = L ◦ L ◦ L T1 = L−1 2 2 2 1 1 1 1 . Hence, (ker (M )) = ker (L ◦ L). L−1 2 1 (b) Part (a) of Exercise 17 shows that ker (L2 ◦ L) = ker (L). This combined with part (a) of this exercise shows that L−1 1 (ker (M )) = ker (L). −1 (c) From part (b), dim(ker (L)) = dim(L−1 1 (ker (M ))). Apply Exercise 16 with L replaced by L1 −1 and Y = ker(M ) to show that dim(L1 (ker (M ))) = dim(ker(M )), completing the proof.

(23) (a) True. This is part of Theorem 5.14. (b) True. This is part of Theorem 5.14. (c) False. In fact, this statement contradicts Theorem 5.15, which implies that L is an isomorphism if and only if ABB is nonsingular.2 But ABB is nonsingular if and only if |ABB | = 0. Any 2 We

assume here that V is a nontrivial vector space, or else there is no matrix ABB , since if V is trivial, B is the empty set.

170


Section 5.6

linear operator on a nontrivial finite dimensional vector space provides a counterexample for the statement given in the problem. (d) False. While Corollary 5.21 shows this is true if V and W are finite dimensional vector spaces having the same dimension, the statement is not necessarily true otherwise. For example, the linear transformation of Example 2 in Section 5.4 of the textbook is onto, but is not one-to-one. Also, Exercise 2(b) in Section 5.4, whose solution appears in this manual, gives an example of a linear transformation that is one-to-one but not onto. (e) False. The problem is that ker(L) is a subspace of V while ker(L ◦ M ) is a subspace of X . For a particular counterexample, let L be the zero linear operator on R3 and let M : P2 → R3 be given by M (ax2 + bx + c) = [a, b, c]. Note that ker(L) = R3 while ker(L ◦ M ) = P2 . (Notice, however, that if the vector spaces are finite dimensional, then dim(ker(L)) = dim(ker(L ◦ M )).) (f) True. This is part (b) of Exercise 17. (g) True. By Theorem 5.14, L−1 is a linear transformation. The given statement then follows from part (3) of Theorem 5.1. (h) True. Since dim(R28 ) = dim(P27 ) = dim(M74 ) = 28, Corollary 5.19 shows that each of these vector spaces is isomorphic to the other two. (i) True. Because dim(R6 ) = dim(M32 ), the given statement follows from Corollary 5.21.

Section 5.6 (1) In all parts, we just need to perform Step 2 of the Generalized Diagonalization Method. However, this corresponds precisely with the Diagonalization Method from Section 3.4. Also, in each part, let A represent the given matrix. " " " x−2 −1 "" = (x − 2)2 . Hence, λ = 2 is the only eigenvalue for L, and the (a) First, pA (x) = "" 0 x−2 " algebraic multiplicity for λ is 2. To find a basis for E2 , we row reduce " " 0 −1 "" 0 0 1 "" 0 to obtain . [ 2I2 − A| 0] = 0 0 " 0 0 0 " 0 Setting the independent variable x1 of the associated system equal to 1 produces the basis {[1, 0]} for E2 . Since dim(E2 ) = 1, the geometric multiplicity of λ is 1. " " " x −1 −1 "" " 1 "" = x3 −2x2 −x+2 (using basketweaving) = (x−1)(x+1)(x−2). (c) First, pA (x) = "" 1 x − 4 " 1 −5 x+2 " Hence, the eigenvalues for L are λ1 = 1, λ2 = −1, and λ3 = 2, each having algebraic multiplicity 1. Next, we solve for a basis for each eigenspace. " " ⎤ ⎤ ⎡ ⎡ 1 0 −2 "" 0 1 −1 −1 "" 0 1 "" 0 ⎦ to obtain ⎣ 0 1 −1 "" 0 ⎦. For λ1 = 1, we row reduce [ 1I3 − A| 0] = ⎣ 1 −3 0 0 0 " 0 1 −5 3 " 0 Setting the independent variable x3 of the associated system equal to 1 produces the basis {[2, 1, 1]} for E1 . " " ⎤ ⎤ ⎡ ⎡ 1 0 1 "" 0 −1 −1 −1 "" 0 1 "" 0 ⎦ to obtain ⎣ 0 1 0 "" 0 ⎦. For λ2 = −1, we row reduce [ −1I3 − A| 0] = ⎣ 1 −5 0 0 0 " 0 1 −5 1 " 0

171


Section 5.6

Setting the independent variable x3 of the associated system equal to 1 {[−1, 0, 1]} for E−1 . " ⎤ ⎡ ⎡ 2 −1 −1 "" 0 1 "" 0 ⎦ to obtain ⎣ For λ3 = 2, we row reduce [ 2I3 − A| 0] = ⎣ 1 −2 1 −5 4 " 0

produces the basis " ⎤ 1 0 −1 "" 0 0 1 −1 "" 0 ⎦. 0 0 0 " 0

Setting the independent variable x3 of the associated system equal to 1 produces the basis {[1, 1, 1]} for E2 . Since each eigenspace is one-dimensional, each eigenvalue has geometric multiplicity 1. " " " x−2 0 0 ""

" x+3 6 "" = (x − 2) (x + 3)(x − 8) + 30 (using cofactor expansion along (d) pA (x) = "" −4 " 4 −5 x−8 " the 1st row) = (x − 2)2 (x − 3). Hence, the eigenvalues for L are λ1 = 2 and λ2 = 3. The algebraic multiplicities of these eigenvalues are 2 and 1, respectively. Next, we solve for a basis for each eigenspace. " ⎡ ⎤ " ⎤ ⎡ 1 − 54 − 32 "" 0 0 0 0 "" 0 ⎢ ⎥ " 5 6 "" 0 ⎦ to obtain ⎣ 0 0 0 " 0 ⎦. For λ1 = 2, we row reduce [ 2I3 − A| 0] = ⎣ −4 " " 4 −5 −6 0 0 0 0 " 0 Setting the independent variables x2 = 4 (to avoid fractions) and x3 = 0 produces the fundamental eigenvector [5, 4, 0]. Setting x2 = 0 and x3 = 2 yields [3, 0, 2]. Hence, {[5, 4, 0], [3, 0, 2]} is a basis for E2 , and the geometric multiplicity " " ⎤ ⎤ ⎡ ⎡ of λ1 is 2. 1 0 0 "" 0 1 0 0 "" 0 6 6 "" 0 ⎦ to obtain ⎣ 0 1 1 "" 0 ⎦. For λ2 = 3, we row reduce [ 3I3 − A| 0] = ⎣ −4 0 0 0 " 0 4 −5 −5 " 0 Setting the independent variable x3 equal to 1 produces the basis {[0, −1, 1]} for E3 , and the geometric multiplicity of λ2 is 1. (2) (b) Step 1: We let C = (x2 , x, 1), the standard basis for P2 . Then, since L(x2 ) = 2x2 −2x, L(x) = x−1, ⎡ ⎤ 2 0 0 1 0 ⎦ for L with respect to C. and L(1) = 0, we get the matrix A = ⎣ −2 0 −1 0 " " " x−2 0 0 "" " x − 1 0 "" = (x − 2)(x − 1)x (since the corresponding matrix is Step 2: First, pA (x) = "" 2 " 0 1 x " lower triangular). Hence, the eigenvalues for L for a basis for each eigenspace. ⎡ 0 For λ1 = 2, we row reduce [ 2I3 − A| 0] = ⎣ 2 0

are λ1 = 2, λ2 = 1, and " ⎤ ⎡ 0 0 "" 0 1 0 "" 0 ⎦ to obtain ⎣ 1 2 " 0

Setting the independent variable x3 of the associated {[1, −2, 1]} for E2 in R3 . " ⎡ −1 0 0 "" For λ2 = 1, we row reduce [ 1I3 − A| 0] = ⎣ 2 0 0 "" 0 1 1 "

λ3 = 0. Next, we solve " ⎤ 1 0 −1 "" 0 0 1 2 "" 0 ⎦. 0 0 0 " 0

system equal to 1 produces the basis ⎤ ⎡ 0 1 0 ⎦ to obtain ⎣ 0 0 0

0 1 0

0 1 0

" ⎤ " 0 " " 0 ⎦. " " 0

Setting the independent variable x3 of the associated system equal to 1 produces the basis 172


{[0, −1, 1]} for E1 in R3 . For λ3 = 0, ⎡

1 ⎣ 0 0

0 1 0

0 0 0

Section 5.6

" ⎤ −2 0 0 "" 0 we row reduce [ 0I3 − A| 0] = [ −A| 0] = ⎣ 2 −1 0 "" 0 ⎦ to obtain 0 1 0 " 0 " ⎤ " 0 " " 0 ⎦. Setting the independent variable x3 of the associated system equal to 1 " " 0 ⎡

produces the basis {[0, 0, 1]} for E0 in R3 . Since the union of the three bases is Z = {[1, −2, 1], [0, −1, 1], [0, 0, 1]}, which has 3 elements, L is diagonalizable. Step 3: Converting the vectors in Z into polynomials produces the basis B = (x2 −2x+1, −x+1, 1) for P2 . The matrix for L with respect to B is the diagonal matrix having the eigenvalues for L ⎡ ⎤ ⎡ ⎤ 2 0 0 1 0 0 on its main diagonal; that is, D = ⎣ 0 1 0 ⎦. Let P = ⎣ −2 −1 0 ⎦, the matrix whose 0 0 0 1 1 1 columns are the vectors in Z. Then row reduction yields P−1 = P. (This does not occur in general.) Matrix multiplication verifies that D = P−1 AP. (Try it!) (d) Step 1: We let C = (x2 , x, 1), the standard basis for P2 . Then, since L(x2 ) = −x2 − 12x + 18, L(x) = −4x, and L(1) = −5, we get the matrix ⎡ ⎤ −1 0 0 0 ⎦ for L with respect to C. A = ⎣ −12 −4 18 0 −5 " " " x+1 0 0 "" " 0 "" = (x + 1)(x + 4)(x + 5) (since the corresponding Step 2: First, pA (x) = "" 12 x + 4 " −18 0 x+5 " matrix is lower triangular). Hence, the eigenvalues for L are λ1 = −1, λ2 = Next, we solve for a basis for each eigenspace. ⎡ " ⎤ ⎡ 0 0 0 "" 0 ⎢ For λ1 = −1, we row reduce [ −1I3 − A| 0] = ⎣ 12 3 0 "" 0 ⎦ to obtain ⎣ −18 0 4 " 0 Setting the independent variable x3 of the associated produces the basis {[2, −8, 9]} for E−1 in R3 . ⎡ −3 For λ2 = −4, we row reduce [ −4I3 − A| 0] = ⎣ 12 −18

−4, and λ3 = −5. 1 0 0

" ⎤ − 29 "" 0 ⎥ 8 " 1 9 "" 0 ⎦. 0 0 " 0

0

system equal to 9 (to eliminate fractions) " ⎤ ⎡ 0 0 "" 0 1 0 0 "" 0 ⎦ to obtain ⎣ 0 0 1 " 0 0

0 0 0

0 1 0

" ⎤ " 0 " " 0 ⎦. " " 0

Setting the independent variable x2 of the associated system equal to 1 produces the basis {[0, 1, 0]} for E−4 in R3 . " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 0 −4 0 0 "" 0 For λ3 = −5, we row reduce [ −5I3 − A| 0] = ⎣ 12 −1 0 "" 0 ⎦ to obtain ⎣ 0 1 0 "" 0 ⎦. 0 0 0 " 0 −18 0 0 " 0 Setting the independent variable x3 of the associated system equal to 1 produces the basis {[0, 0, 1]} for E−5 in R3 . Since the union of the three bases is Z = {[2, −8, 9], [0, 1, 0], [0, 0, 1]}, which has 3 elements, L is diagonalizable. 173


Section 5.6

Step 3: Converting the vectors in Z into polynomials produces the basis B = (2x2 − 8x + 9, x, 1) for P2 . The matrix for L with respect to B is the diagonal matrix having the eigenvalues for ⎡ ⎤ ⎡ ⎤ −1 0 0 2 0 0 0 ⎦. Let P = ⎣ −8 1 0 ⎦ , the matrix L on its main diagonal; that is, D = ⎣ 0 −4 0 0 −5 9 0 1 ⎡ 1 ⎤ 0 0 2 ⎢ ⎥ whose columns are the vectors in Z. Then row reduction yields P−1 = ⎣ 4 1 0 ⎦. Matrix multiplication verifies that D = P

−1

− 92

AP. (Try it!)

(e) Step 1: We let C = (i, j), the standard basis for R2 . Then, since L(i) = [ 12 , ' √ ( 3 1 √ − 3 1 2 for L with respect to C. [− 2 , 2 ], we get the matrix A = √23 1 " " x− 1 " √2 Step 2: First, pA (x) = " " − 3 2

2 √ 3 2 x − 12

0 √ 3 2 ],

1 and L(j) =

2

" " " " = x2 − x + 1, which has no real roots. Hence, L has no "

eigenvalues, and so is not diagonalizable. (h) Step 1: Let C be the standard basis for M22 ; that is, C = {Ψij | 1 ≤ i ≤ 2, 1 ≤ j ≤ 2}, where Ψij is the 2 × 2 matrix having 1 in its (i, j) entry and zeroes elsewhere. (See Example 3 −4 0 0 −4 3 0 , L(Ψ12 ) = , L(Ψ21 ) = , and in Section 4.5.) Now L(Ψ11 ) = −10 0 0 −10 7 0 ⎡ ⎤ −4 0 3 0 ⎢ 0 3 0 −4 0 3 ⎥ ⎥ , and so we get the matrix A = ⎢ L(Ψ22 ) = ⎣ 0 7 −10 0 7 0 ⎦ 0 −10 0 7 for L with respect to C. " " " x+4 0 −3 0 "" " " 0 x+4 0 −3 "" . Using cofactor expansion along the 4th Step 2: First, pA (x) = "" 10 0 x − 7 0 "" " " 0 10 0 x−7 " row and cofactor 2nd row for each of expansion on the

the resulting 3 × 3 determinants yields pA (x) = 10 3 (x + 4)(x − 7) + 30 + (x − 7)(x + 4) (x + 4)(x − 7) + 30

2 = (x + 4)(x − 7) + 30 = (x − 1)2 (x − 2)2 . Hence, the eigenvalues for L are λ1 = 1 and λ2 = 2. Next, we solve for a basis for each eigenspace. For λ1 = 1, we row reduce [ 1I3 − A| 0] = ⎡ " ⎤ ⎡ 1 0 − 35 5 0 −3 0 "" 0 ⎢ ⎢ 0 5 0 −3 "" 0 ⎥ 0 1 0 ⎥ to obtain ⎢ ⎢ ⎢ ⎣ 10 0 −6 0 "" 0 ⎦ ⎣ 0 0 0 0 10 0 −6 " 0 0 0 0

" 0 "" " − 35 " " 0 "" 0 "

0

⎤

⎥ 0 ⎥ ⎥. 0 ⎦ 0

Setting the independent variables x3 = 5 (to avoid fractions) and x4 = 0 produces the fundamental

174


Section 5.6

eigenvector [3, 0, 5, 0]. Setting x3 = 0 and x4 = 5 yields [0, 3, 0, 5]. Hence, {[3, 0, 5, 0], [0, 3, 0, 5]} is a basis for E1 in R4 . For λ2 = 2, we row reduce [ 2I3 − A| 0] = " ⎡ ⎤ " ⎤ ⎡ 1 0 − 12 0 "" 0 6 0 −3 0 "" 0 ⎢ ⎥ " ⎢ 0 6 0 −3 "" 0 ⎥ 0 1 0 − 12 " 0 ⎥ ⎥ to obtain ⎢ ⎢ ⎢ ⎥. " ⎣ 10 0 −5 0 "" 0 ⎦ ⎣ 0 0 0 0 "" 0 ⎦ " 0 10 0 −5 0 0 0 0 0 " 0 Setting the independent variables x3 = 2 (to avoid fractions) and x4 = 0 produces the fundamental eigenvector [1, 0, 2, 0]. Setting x3 = 0 and x4 = 2 yields [0, 1, 0, 2]. Hence, {[1, 0, 2, 0], [0, 1, 0, 2]} is a basis for E2 in R4 . The union of the three bases is Z = {[3, 0, 5, 0], [0, 3, 0, 5], [1, 0, 2, 0], [0, 1, 0, 2]}. Since |Z| = 4, L is diagonalizable. Step 3: Converting the vectors in Z into matrices produces the basis

3 0 0 3 1 0 0 1 B= , , , for M22 . The matrix for L with respect to B 5 0 0 5 2 0 0 2 is the diagonal matrix having the eigenvalues for L on its main diagonal; that is, ⎤ ⎡ ⎤ ⎡ 3 0 1 0 1 0 0 0 ⎢ 0 3 0 1 ⎥ ⎢ 0 1 0 0 ⎥ ⎥ ⎢ ⎥ D=⎢ ⎣ 0 0 2 0 ⎦. Let P = ⎣ 5 0 2 0 ⎦ , the matrix whose columns are the vectors in 0 5 0 2 0 0 0 2 ⎡ ⎤ 2 0 −1 0 ⎢ 0 2 0 −1 ⎥ ⎥. Matrix multiplication verifies that Z. Then row reduction yields P−1 = ⎢ ⎣ −5 0 3 0 ⎦ 0 −5 0 3 D = P−1 AP. (Try it!) (4) (a) Consider C = (x2 , x, 1), the standard basis for P2 . Since a = 1 in this part, L(dx2 + ex + f ) = d(x + 1)2 + e(x + 1) + f . Therefore, L(x2 ) = (x + 1)2 = x2 + 2x + 1, L(x) = x + 1, and L(1) = 1. ⎡ ⎤ 1 0 0 Hence, we get the matrix A = ⎣ 2 1 0 ⎦ for L with respect to C. 1 1 1 " " " x−1 0 0 "" " x−1 0 "" = (x − 1)3 (since the corresponding matrix is lower Now, pA (x) = "" −2 " −1 −1 x−1 " triangular). Hence, the only eigenvalue for ⎡ 0 0 0 We row reduce [ 1I3 − A| 0] = ⎣ −2 −1 −1

L is " 0 "" 0 "" 0 "

λ = 1. Next, we solve for a basis for E1 . " ⎤ ⎡ ⎤ 0 1 0 0 "" 0 0 ⎦ to obtain ⎣ 0 1 0 "" 0 ⎦. 0 0 0 0 " 0

Setting the independent variable x3 of the associated system equal to 1 produces the basis {[0, 0, 1]} for E1 in R3 . Converting the vector in this basis back to polynomial form yields the basis E1 = {1} in P2 . (Note that our process has produced only one fundamental eigenvector for L, and so L is not diagonalizable.)

175


1 (7) (a) Let A = ⎣ 0 0

" ⎤ " x−1 1 −1 " 1 0 ⎦. Then, pA (x) = "" 0 " 0 0 1

−1 x−1 0

1 0 x−1

Section 5.6

" " " " = (x − 1)3 (since the cor" "

responding matrix is upper triangular). Hence, the only eigenvalue for A is λ = 1, which has algebraic multiplicity 3. Next, we solve for a basis for E1 . We row reduce " " ⎤ ⎤ ⎡ ⎡ 0 1 −1 "" 0 0 −1 1 "" 0 0 "" 0 ⎦. 0 0 "" 0 ⎦ to obtain ⎣ 0 0 [ 1I3 − A| 0] = ⎣ 0 0 0 0 " 0 0 0 0 " 0 Setting the independent variables x1 = 1 and x3 = 0 produces the fundamental eigenvector [1, 0, 0]. Setting x1 = 0 and x3 = 1 yields [0, 1, 1]. Hence, {[1, 0, 0], [0, 1, 1]} is a basis for E1 . Since this basis has 2 elements, λ = 1 has geometric multiplicity 2. " " ⎡ ⎤ " x−1 0 0 "" 1 0 0 " x − 1 0 "" = (x − 1)2 x (since the corresponding (b) Let A = ⎣ 0 1 0 ⎦. Then, pA (x) = "" 0 " 0 0 x " 0 0 0 matrix is upper triangular). Hence, λ = 1 is an eigenvalue for A which has algebraic multiplicity 2. (Zero is also an eigenvalue for A.) ⎡ Next, we "solve⎤for a basis for E1 . Performing the row 0 0 0 "" 0 operation 1 ↔ 3 puts [ 1I3 − A| 0] = ⎣ 0 0 0 "" 0 ⎦ into reduced row echelon form, yielding 0 0 1 " 0 " ⎤ ⎡ " 0 0 1 " 0 ⎣ 0 0 0 " 0 ⎦. Setting the independent variables x1 = 1 and x2 = 0 produces the fundamen" 0 0 0 " 0 tal eigenvector [1, 0, 0]. Setting x1 = 0 and x2 = 1 yields [0, 1, 0]. Hence, {[1, 0, 0], [0, 1, 0]} is a basis for E1 . Since this basis has 2 elements, λ = 1 has geometric multiplicity 2. (15) Suppose v ∈ B1 ∩ B2 . Then v ∈ B1 and v ∈ B2 . Now v ∈ B1 =⇒ v ∈ Eλ1 =⇒ L(v) = λ1 v. Similarly, v ∈ B2 =⇒ v ∈ Eλ2 =⇒ L(v) = λ2 v. Hence, λ1 v = L(v) = λ2 v, and so (λ1 −λ2 )v = 0. Since λ1 =λ2 , v = 0. But 0 cannot be an element of any basis. This contradiction shows that B1 ∩ B2 must be empty. (16) (a) Eλi is a subspace; hence, it is closed under linear combinations. Therefore, since ui is a linear combination of basis vectors for Eλi , it must also be in Eλi . !ki !n (b) First, substitute ui for j=1 aij vij in the given double sum equation. This proves i=1 ui = 0. are eigenvectors Now, the set of all nonzero ui ’s is linearly independent by Theorem 5.23, since !they n u corresponding to distinct eigenvalues. But then the nonzero terms in i=1 i would give a nontrivial linear combination (of vectors from a linearly independent set) equal to the zero vector. This contradiction shows that all of the ui ’s must equal 0. !ki aij vij , for each i. But {vi1 , . . . , viki } is linearly (c) Using part (b) and the definition of ui , 0 = j=1 independent, since it is a basis for Eλi . Hence, for each i, ai1 = · · · = aiki = 0. (d) We have shown that any linear combination of vectors in B that produces the zero vector must use all zero coefficients. Therefore, B is linearly independent by definition. (18) (a) False. The correct definition for the eigenspace Eλ is Eλ = {v ∈ V | L(v) = λv}. Note that the set S = {λL(v) | v ∈ V} is the set of images of L multiplied by λ. But since λL(v) = L(λv), it is not hard to show that S equals range(L). For a specific example in which Eλ = S, consider Example 7 in Section 5.6 of the textbook, using λ = 2. In that example, L is an isomorphism, so

176


Chap 5 Review

S = range(L) = R3 . However, the example shows that Eλ has dimension 1, since the algebraic multiplicity of λ = 2 is only 1. (b) True. This is the definition of the characteristic polynomial of a linear operator. (The characteristic polynomial for L is independent of the particular ordered basis chosen for V (because of Theorems 5.5 and 5.6). Therefore, the definition of a characteristic polynomial of a linear operator makes sense as given in the text.) (c) True. This follows directly from Theorem 5.22. (d) False. While one half of this “if and only if” statement is just Theorem 5.23, the converse of that statement (the other half of the “if and only if”) is false. For a specific counterexample, consider the zero linear operator L : R2 → R2 . Then λ = 0 is the only eigenvalue, and E0 = R2 . Thus, {i, j} is a set of linearly independent eigenvectors for the same eigenvalue λ = 0. (e) True. This is Theorem 5.25. (f) True. This follows from Theorem 5.28 (which indicates that a union of the bases for the distinct eigenspaces contains 6 vectors) and Theorem 5.25 (which indicates that the 6 vectors in the union of these bases are linearly independent, and hence form a basis for R6 ). (g) True. This is the purpose of the Method. (h) False. For a specific example in which the algebraic multiplicity of an eigenvalue is strictly larger than its geometric multiplicity, see Example 12 in Section 5.6 of the textbook. (Note that by Theorem 5.26, the algebraic multiplicity of an eigenvalue λ is always greater than or equal to its geometric multiplicity.) (i) False. According to Theorem 5.28, each geometric multiplicity must also equal its corresponding algebraic multiplicity in order for L to be diagonalizable. For a specific counterexample, consider L : R7 → R7 given by L(ei ) = ei+1 , for 1 ≤ i ≤ 6, and L(e7 ) = 0. Then it can be shown that pL (x) = x7 , and so λ = 0 is the only eigenvalue for L, having algebraic multiplicity 7. However, it can be shown that E0 has basis {e7 }, and so the geometric multiplicity of λ = 0 is 1. Hence, by Theorem 5.28, L is not diagonalizable. You should verify all of the claims made in this example. (j) True. Since pA (x) = (x−1)(x−4) = (1−x)(4−x), this follows directly from the Cayley-Hamilton Theorem.

Chapter 5 Review Exercises (1) (a) We present three different proofs that the function f is not a linear transformation. Solution 1: The function f is not a linear transformation because f ([0, 0, 0]) = [0, 1, 0]. The image of the zero vector under a linear transformation must be the zero vector in the codomain (part (1) of Theorem 5.1). Solution 2: Note that f ([1, 0, 0]) + f ([0, 1, 0]) = [0, 4, 5] + [−1, 1, 2] = [−1, 5, 7], but f ([1, 0, 0] + [0, 1, 0]) = f ([1, 1, 0]) = [−1, 4, 7], and so property (1) in the definition of a linear transformation is not satisfied. Solution 3: Note that f (2[1, 0, 0]) = f ([2, 0, 0]) = [0, 7, 10], while 2f ([1, 0, 0]) = 2[0, 4, 5] = [0, 8, 10], and so property (2) in the definition of a linear transformation is not satisfied. (3) f (A1 ) + f (A2 ) = CA1 B−1 + CA2 B−1 = C(A1 B−1 + A2 B−1 ) = C(A1 + A2 )B−1 = f (A1 + A2 ); f (kA) = C(kA)B−1 = kCAB−1 = kf (A).

177


Chap 5 Review

(4) Creating the matrix whose columns are the images of i, j, and k, we see that ⎡ ⎤⎡ ⎤ −3 5 −4 x 0 ⎦ ⎣ y ⎦ = [−3x + 5y − 4z, 2x − y, 4x + 3y − 2z]. Hence, L([x, y, z]) = ⎣ 2 −1 z 4 3 −2 ⎡ ⎤⎡ ⎤ ⎡ ⎤ −3 5 −4 20 6 0 ⎦ ⎣ 2 ⎦ = ⎣ 10 ⎦ . L([6, 2, −7]) = ⎣ 2 −1 −7 44 4 3 −2 (5) (b) By Theorem 5.2, L2 ◦ L1 : V → X is a linear transformation. Thus, by part (2) of Theorem 5.3, (L2 ◦ L1 )−1 (X ) is a subspace of V. (6) (a) We will use Theorem 5.5, which states that the ith column of ABC equals [L(vi )]C , where B is the given ordered basis for the domain of L, C is the given ordered basis for the codomain of L, and vi is the ith basis element in B. Substituting each vector in B into the given formula for L shows that L([−5, −3, −2]) = [−13, 1], L([3, 0, 1]) = [2, 12], and L([5, 2, 2]) = [10, 6]. Next, we must express each of these answers in C-coordinates. We use the Coordinatization Method from Section 4.7. Thus, for each vector v that we need to express in C-coordinates, we could row reduce the augmented matrix whose columns to the left of the bar are the vectors in C and whose column to the right of the bar is the vector v. But since this process essentially involves solving a linear system, and we have three such systems to solve, we can solve all three simultaneously. Therefore, we row reduce " " 2 10 1 0 "" 29 32 −2 4 −3 "" −13 . Hence, [L([−5, −3, −2])]C = [29, 43], to 1 12 6 0 1 " 43 42 −6 3 −2 " [L([3, 0, 1])]C = [32, 42], and [L([5, 2, 2])]C = [−2, −6]. Using these coordinatizations as columns 29 32 −2 . (Note that ABC is just the matrix to the right of the bar in produces ABC = 43 42 −6 the row reduced matrix, above. This is typical, but in general, any rows of all zeroes must be eliminated first, as in the Coordinatization Method.) C (7) (b) First, we find the matrix for L with respect to the standard basis B for P2 and the standard basis 6 0 for M22 . Substituting each vector in B into the given formula for L shows that L(x2 ) = , 2 1 −1 3 −1 2 L(x) = , and L(1) = . Converting these matrices to C-coordinates creates 0 −5 −4 1 ⎡ ⎤ 6 −1 −1 ⎢ 0 3 2 ⎥ ⎥. Next, we use Theorem 4-vectors, which we use as columns to produce ABC = ⎢ ⎣ 2 0 −4 ⎦ 1 −5 1 5.6 to compute ADE . To do this, we need the inverse of the transition matrix P from B to D and the transition matrix Q from C to E. To compute P (or P−1 ), we need to convert the polynomials in D into vectors in R3 . After we do this, P is the inverse of the matrix whose columns are the resulting vectors obtained from D, and Q is the inverse of the matrix whose ⎡ ⎤ ⎡ ⎤ 3 2 1 4 −5 3 −2 ⎢ 2 1 1 2 ⎥ ⎥ 1 ⎦ and Q−1 = ⎢ columns are the vectors in E. Hence, P−1 = ⎣ 2 −1 ⎣ 2 1 2 2 ⎦. 5 −1 3 5 4 4 7 178


Chap 5 Review

⎤ 0 −1 2 ⎥ ⎥. Finally, using Theorem 5.6 produces So, Q = Q−1 0 ⎦ −1 ⎤ 115 −45 59 ⎢ 374 −146 190 ⎥ ⎥. ADE = QABC P−1 = ⎢ ⎣ −46 15 −25 ⎦ −271 108 −137 " " " " x − 18 − 36 − 12 41 41 41 " " " " 31 24 x + (9) (a) pABB (x) = |xI3 − ABB | = " − 36 " = x3 − x2 − x + 1 = (x + 1)(x − 1)2 , where 41 41 41 " " 12 " − 24 x − 49 " −1 ⎢ −4 =⎢ ⎣ 0 3 ⎡

3 6 −1 −5

41

−1 −5 1 3

41

41

we have used basketweaving and algebraic simplification.

⎡

1 ⎢ 0 (10) (a) Let A represent the given matrix. The reduced row echelon form of A is B = ⎢ ⎣ 0 0

0 1 0 0

⎤ −2 3 3 −4 ⎥ ⎥. 0 0 ⎦ 0 0

Then, by the Kernel Method, a basis for ker(L) is found by considering the solution to the system BX = 0 computed by setting each independent variable in the system equal to 1, while the other is set to 0. This yields the basis {[2, −3, 1, 0], [−3, 4, 0, 1]} for ker(L). Since ker(L) has a basis with 2 elements, dim(ker(L)) = 2. Next, by the Range Method, a basis for range(L) consists of the first 2 columns of A, since those columns in B contain nonzero pivots. Thus, a basis for range(L) = {[3, 2, 2, 1], [1, 1, 3, 4]} and dim(range(L)) = 2. (12) (a) If v ∈ ker(L1 ), then L1 (v) = 0W . Thus, (L2 ◦ L1 )(v) = L2 (L1 (v)) = L2 (0W ) = 0X , and so v ∈ ker(L2 ◦ L1 ). Therefore, ker(L1 ) ⊆ ker(L2 ◦ L1 ). Hence, dim(ker(L1 )) ≤ dim(ker(L2 ◦ L1 )). (b) Let L1 be the projection onto the x-axis (L1 ([x, y]) = [x, 0]) and L2 be the projection onto the y-axis (L2 ([x, y]) = [0, y]). A basis for ker(L1 ) is {[0, 1]}, while ker(L2 ◦ L1 ) = R2 . Thus, dim(ker(L1 )) = 1 < 2 = dim(ker(L2 ◦ L1 )). (15) (a) Let A represent the given matrix. The reduced row echelon form of A is I3 . Then, by the Kernel Method, ker(L) = {0}, and so dim(ker(L)) = 0, and L is one-to-one. By Corollary 5.21, L is also onto. Hence, dim(range(L)) = 3. (18) (a) L1 and L2 are isomorphisms because their (given) matrices are nonsingular (by Theorem 5.15). Their inverses are given in part (b). (b) The matrix for L2 ◦ L1 with respect to the standard basis is found by multiplying the matrices for L2 and L1 with respect to the standard bases (by Theorem 5.7). In particular: ⎡ ⎤⎡ ⎤ 9 8 5 4 3 6 1 1 ⎢ 9 13 ⎢ 4 7 ⎥ 2 −2 1 ⎥ ⎥⎢ 5 ⎥ Matrix for L2 ◦ L1 = ⎢ ⎣ 5 9 2 5 ⎦⎣ 2 1 0 1 ⎦ −5 −2 −2 0 1 −1 −2 −1 ⎡ ⎤ 81 71 −15 18 ⎢ 107 77 −31 19 ⎥ ⎥. = ⎢ ⎣ 69 45 −23 11 ⎦ −29 −36 −1 −9 179


Chap 5 Review

−1 Also, computing inverses of the matrices for L1 and L2 gives the matrices for L−1 1 and L2 (by Theorem 5.15): ⎡ ⎤ 2 −10 19 11 1⎢ 5 −10 −5 ⎥ ⎢ 0 ⎥; = Matrix for L−1 1 ⎣ 3 −15 26 14 ⎦ 5 −4 15 −23 −17 ⎡ ⎤ −8 26 −30 2 1⎢ 41 −4 ⎥ ⎢ 10 −35 ⎥. Matrix for L−1 = 2 ⎣ 10 −30 34 −2 ⎦ 2 −14 49 −57 6

(21) (a) If p = ax4 + bx3 + cx2 , then L(p) = p + p = (4ax3 + 3bx2 + 2cx) + (12ax2 + 6bx + 2c) = 4ax3 + (12a + 3b)x2 + (6b + 2c)x + 2c. Hence, p ∈ ker(L) if and only if ⎧ 4a = 0 ⎪ ⎪ ⎨ 12a + 3b = 0 . 6b + 2c = 0 ⎪ ⎪ ⎩ 2c = 0 Since this system has only the trivial solution, ker(L) = {0}. Therefore, by part (1) of Theorem 5.12, L is one-to-one. (22) In all parts, we just need to perform Step 2 of the Generalized Diagonalization Method. However, this corresponds precisely with the Diagonalization Method from Section 3.4. Also, in each part, let A represent the given matrix. " " " x+9 −18 16 "" " −56 "" = x3 − 3x2 − x + 3 (using basketweaving) (a) (i) First, pA (x) = "" −32 x + 63 " −44 84 x − 75 " = (x − 1)(x + 1)(x − 3). Hence, the eigenvalues for L are λ1 = 1, λ2 = −1, and λ3 having algebraic multiplicity 1. Next, we solve for a basis for each eigenspace. ⎡ 1 " ⎤ ⎡ 1 0 4 " 10 −18 16 " 0 ⎢ ⎢ " ⎦ ⎣ −32 64 −56 " 0 to ⎣ 0 1 − 34 For λ1 = 1, we row reduce [ 1I3 − A| 0] = −44 84 −74 " 0 0 0 0

= 3, each " ⎤ " 0 " ⎥ " " 0 ⎥. " ⎦ " " 0

Setting the independent variable x3 of the associated system equal to 1 and then eliminating fractions produces the basis {[−1, 3, 4]} for E1 . " ⎤ ⎡ 1 " " ⎤ ⎡ 0 1 0 5 " " 8 −18 16 " 0 ⎥ ⎢ " 4 " ⎥ For λ2 = −1, we row reduce [ −1I3 − A| 0] = ⎣ −32 62 −56 "" 0 ⎦ to ⎢ ⎣ 0 1 − 5 " 0 ⎦. " " −44 84 −76 0 0 0 0 " 0 Setting the independent variable x3 of the associated system equal to 1 and then eliminating fractions produces the basis {[−1, 4, 5]} for E−1 . " ⎡ ⎤ 2 " " ⎤ ⎡ 1 0 9 " 0 " 12 −18 16 " 0 ⎢ " ⎥ 20 " ⎥ For λ3 = 3, we row reduce [ 3I3 − A| 0] = ⎣ −32 66 −56 "" 0 ⎦ to ⎢ ⎣ 0 1 − 27 " 0 ⎦. " " −44 84 −72 0 0 0 0 " 0 180


Chap 5 Review

Setting the independent variable x3 of the associated system equal to 1 and then eliminating fractions produces the basis {[−6, 20, 27]} for E3 . (ii) Since each eigenspace is one-dimensional, each eigenvalue has geometric multiplicity 1, which equals its algebraic multiplicity. Since the sum of the geometric multiplicities is 3, which equals the dimension of R3 , L is diagonalizable. (iii) The ordered basis B is the union of the bases of fundamental eigenvectors we found for each eigenspace. Thus, B = {[−1, 3, 4], [−1, 4, 5], [−6, 20, 27]}. The transition matrix P is the matrix whose columns are these vectors, and the diagonal matrix D = P−1 AP is the diagonal ⎡ ⎤ 1 0 0 matrix with the eigenvalues 1, −1, and 3 on the main diagonal. Hence, D = ⎣ 0 −1 0 ⎦ and 0 0 3 ⎡ ⎤ ⎡ ⎤ −1 −1 −6 −8 3 −4 4 20 ⎦. (Note that P−1 = ⎣ 1 3 −2 ⎦.) P=⎣ 3 1 −1 1 4 5 27 " " " x + 97 −20 −12 "" " x − 63 −36 "" = x3 − 5x2 + 3x + 9 (using basketweaving) = (c) (i) First, pA (x) = "" 300 " 300 −60 x − 39 " (x + 1)(x − 3)2 . Hence, the eigenvalues for L are λ1 = −1 and λ2 = 3, with braic multiplicity 1 and λ2 having algebraic multiplicity 2. Next, we solve for eigenspace. ⎡ " ⎤ ⎡ 1 96 −20 −12 "" 0 ⎢ " ⎦ ⎣ 300 −64 −36 " 0 to ⎣ 0 For λ1 = −1, we row reduce [ −1I3 − A| 0] = 300 −60 −40 " 0 0 Setting the independent variable x3 of the associated fractions produces the basis {[1, 3, 3]} for E−1 . ⎡ 100 −20 For λ2 = 3, we row reduce [ 3I3 − A| 0] = ⎣ 300 −60 300 −60

λ1 having algea basis for each " ⎤ 0 − 13 "" 0 ⎥ " 1 −1 "" 0 ⎦. 0 0 " 0

system equal to 1 and then eliminating −12 −36 −36

⎡ " ⎤ 1 − 15 " 0 " ⎢ " 0 ⎦ to ⎣ 0 0 " " 0 0 0

" ⎤ 3 " − 25 " 0 " ⎥ 0 "" 0 ⎦. 0 " 0

Setting the independent variable x2 of the associated system equal to 1 and x3 equal to 0 and then eliminating fractions produces the fundamental eigenvector [1, 5, 0] for E3 . Setting the independent variable x3 of the associated system equal to 1 and x2 equal to 0 and then eliminating fractions produces the fundamental eigenvector [3, 0, 25]. This gives us a basis {[1, 5, 0], [3, 0, 25]} for E3 . (ii) The eigenspace E−1 is one-dimensional, and so λ1 = −1 has geometric multiplicity 1, which equals its algebraic multiplicity. The eigenspace E3 is two-dimensional, and so λ2 = 3 has geometric multiplicity 2, which also equals its algebraic multiplicity. Since the sum of the geometric multiplicities is 3, which equals the dimension of R3 , L is diagonalizable. (iii) The ordered basis B is the union of the bases of fundamental eigenvectors we found for each eigenspace. Thus, B = {[1, 3, 3], [1, 5, 0], [3, 0, 25]}. The transition matrix P is the matrix whose columns are these vectors, and the diagonal matrix D = P−1 AP is the diagonal ⎡ ⎤ −1 0 0 matrix with the eigenvalues −1, 3, and 3 on the main diagonal. Hence, D = ⎣ 0 3 0 ⎦ and 0 0 3

181


1 P=⎣ 3 3

1 5 0

⎤ 3 0 ⎦. (Note that P−1 = 25

⎡

125 −25 1 ⎣ −75 16 5 −15 3

Chap 5 Review

⎤ −15 9 ⎦.) 2

(25) (a) True. This follows directly from Theorem 5.4. (b) False. There are many such linear transformations because the value of L(k) has not been specified. (Once the value of L(k) is known, the linear transformation is determined by Theorem 5.4.) Choosing either L(k) = i or L(k) = j would produce two different linear transformations having L(i) = j and L(j) = i. (c) True. See Example 11 in Section 5.1. Use θ = −45◦ . (d) True. The properties L(y1 + y2 ) = L(y1 ) + L(y2 ) and L(cy1 ) = cL(y1 ) hold for all vectors y1 , y2 in Y and scalars c because Y is a subset of V and L is a linear transformation on V. (e) True. By Theorem 5.5, the ith column of the matrix for the linear transformation is Bei . But this is the ith column of B. (f) False. Similar matrices must be square. The matrices ABC and ADE for the linear transformation are not square if dim(V) = dim(W). However, even if dim(V) = dim(W), the matrices ABC and ADE might not be similar. For example, consider L : R3 → P2 given by L([a, b, c]) = ax2 + bx + c. The matrix for L with respect to the standard bases is I3 . Note that I3 is the only matrix similar to I3 . However, if we continue to use the standard basis for R3 but use the basis {x2 + 1, x + 1, 1} ⎡ ⎤ 1 0 0 1 0 ⎦, which is not similar to I3 . for P2 , the matrix for L is ⎣ 0 −1 −1 1 (g) False. By the Dimension Theorem, dim(ker(L))+dim(range(L)) = 5. However, two equal integers added together cannot equal an odd integer. (h) True. Because dim(row space of A) = rank(A) (by the Simplified Span Method), the given statement is true by part (1) of Theorem 5.9. (i) True. See the discussion in the text just prior to the Range Method in Section 5.3. (j) False. Only range(L) is guaranteed to be finite dimensional. Counterexample: Define L : Pn → P by L(p) = p. The range of L is Pn , which is finite dimensional, but the codomain is P, which is infinite dimensional. (k) False. ker(L) is a subspace of the domain and is never empty, since it must contain the zero vector. Part (1) of Theorem 5.12 says that L is one-to-one if and only if ker(L) = {0}. The identity linear operator on R3 gives a counterexample to the given statement. (l) True. The Dimension Theorem states that dim(ker(L)) + dim(range(L)) = dim(V). Thus, dim(range(L)) = dim(V) if and only if dim(ker(L)) = 0. The given statement then follows by Part (1) of Theorem 5.12. (m) False. Counterexample: The zero linear operator on R3 is neither one-to-one nor onto. (n) True. The Dimension Theorem states that range(L) is finite dimensional. But, if L is onto, then range(L) = W, and so W is finite dimensional. (o) True. This is part (1) of Theorem 5.13. (p) False. f (x) = x3 is a one-to-one and onto function from R to R but is not a linear transformation. Also, translation functions on Rn are one-to-one and onto but are not linear transformations (see the discussion after Example 4 in Section 5.1).

182


Chap 5 Review

(q) False. Counterexample: The zero linear operator on R3 has O3 as its matrix with respect to any bases. O3 is square, but the zero operator is clearly not an isomorphism. (Note, however, that with V, W as given, if L is an isomorphism, then the matrix for L with respect to any bases for V and W is square.) (r) True. L ◦ L is the identity operator because every vector will reflect back to itself. (s) False. Only nontrivial finite dimensional vector spaces are isomorphic to Rn for some n. For example, P is not isomorphic to Rn for any value of n. (t) True. Both W1 and W2 are two-dimensional vector spaces, and so, by Corollary 5.19, they are isomorphic to each other. (u) True. This follows from part (a) of Exercise 17 in Section 5.5 with M in place of T2 and L in place of T . The given statement can be proven directly by showing that each set is a subset of the other. First, we show ker(L) ⊆ ker(M ◦ L). Let v ∈ ker(L). Then L(v) = 0. Thus, (M ◦L)(v) = M (L(v)) = M (0) = 0, and so v ∈ ker(M ◦L). Finally, we show ker(M ◦L) ⊆ ker(L). Let v ∈ ker(M ◦L). Then (M ◦L)(v) = 0. Thus, L(v) = ((M −1 ◦M )◦L)(v) = M −1 ((M ◦L)(v)) = M −1 (0) = 0, and so v ∈ ker(L). (v) False. Note that range(L) is a subspace of W, while range(M ◦ L) is a subspace of X . For a specific counterexample, let L be the identity linear operator on P3 (so range(L) = P3 ), and let M be the coordinatization isomorphism from P3 to R4 (so range(M ◦ L) = R4 ). (Note, however, that in general, dim(range(L)) = dim(range(M ◦ L)), since M is an isomorphism.) (w) True. This is merely a rewording of the definition of eigenspace in terms of the kernel of an operator. (x) True. Each algebraic multiplicity must be at least 1 by Theorem 5.26, and since there are n distinct eigenvalues, the sum of the algebraic multiplicities must be at least n. However, from the discussion just before Theorem 5.28, the sum of the algebraic multiplicities can be at most n. Therefore, each algebraic multiplicity must equal 1. (y) True. If x2 is a factor of the characteristic polynomial, then the eigenvalue 0 has algebraic multiplicity at least 2. But, if dim(E0 ) = 1, then the geometric multiplicity of the eigenvalue 0 is 1. Hence, by Theorem 5.28, L is not diagonalizable. (z) True. By Theorem 5.25, B1 ∪ B2 ∪ · · · ∪ Bk is a linearly independent subset of V. Therefore, it is a basis for the subspace W = span(B1 ∪ B2 ∪ · · · ∪ Bk ) of V.

183


Section 6.1

Chapter 6 Section 6.1 (1) (a) The vectors√are orthogonal because [3, −2] · [4, 6] = 0. The vectors are not orthonormal because [3, −2] = 13 = 1. 9 (c) The vectors are not orthogonal because √313 , − √213 · √110 , − √310 = √130 = 0. Since the vectors are not orthogonal, they are also not orthonormal. (f) The set of vectors is orthogonal because [2, −3, 1, 2]·[−1, 2, 8, 0] = 0, [2, −3, 1, 2]·[6, −1, 1, −8] = 0, and √ [−1, 2, 8, 0] · [6, −1, 1, −8] = 0. The vectors are not orthonormal because [2, −3, 1, 2] = 18 = 0. (2) In each part, let A represent the given matrix. (a) Straightforward computation shows that AAT = I2 . Therefore, A−1 = AT , and so A is an orthogonal matrix. ⎡ ⎤ 109 27 −81 19 −27 ⎦ = I3 . Therefore, A is not an orthogonal matrix. (c) AAT = ⎣ 27 −81 −27 91 (e) Straightforward computation shows that AAT = I4 . Therefore, A−1 = AT , and so A is an orthogonal matrix. √ √ (3) (a) First, B is an orthonormal basis for R2 because it has 2 elements and − 23 , 12 · 12 , 23 = 0, √ √ − 23 , 12 = 1, and 12 , 23 = 1. So, by Theorem 6.3, √ √ √ √ [v]B = [−2, 3] · − 23 , 12 , [−2, 3] · 12 , 23 = 2 23+3 , 3 23−2 . (c) First, B is an orthonormal basis for R4 because it has 4 elements and 1 1 1 1 3 1 1 1 √ √ √ √ 2 , − 2 , 2 , 2 · 2 3 , 2 3 , − 2 3 , − 2 3 = 0, 1 1 1 1 1 1 1 1 √2 , √1 , √1 √1 , √1 · 0, · 0, 0, − , − , , , − , , = 0, = 0, 2 2 2 2 2 2 2 2 6 6 6 2 2 3 1 1 1 3 1 1 1 √2 , √1 , √1 √1 , √1 √ √ √ √ √ √ √ √ , · 0, = 0, · 0, 0, − = 0, , − , − , , − , − 2 3 2 3 2 3 2 3 6 6 6 2 3 2 3 2 3 2 3 2 2 0, √26 , √16 , √16 · 0, 0, − √12 , √12 = 0, 12 , − 12 , 12 , 12 = 1, √ 1 1 1 √ √ √2 , √1 , √1 = 1, and 0, 0, − √1 , √1 = 1. So, by 0, = 1, , − , − 2 3 3 , 2√ 3 2 3 2 3 6 6 6 2 2

3 1 1 1 √ √ √ Theorem 6.3, [v]B = [8, 4, −3, 5] · 12 , − 12 , 12 , 12 , [8, 4, −3, 5] · 2√ , , − , − , 3 2 3 2 3 2 3 √ √ √ = 3, 133 3 , 5 3 6 , 4 2 . [8, 4, −3, 5] · 0, √26 , √16 , √16 , [8, 4, −3, 5] · 0, 0, − √12 , √12 (4) (a) First, let v1 = [5, −1, 2].

Next, v2 = [2, −1, −4] − [2,−1,−4]·[5,−1,2] [5, −1, 2] = [2, −1, −4] − [5,−1,2]·[5,−1,2]

3 30 [5, −1, 2]

=

3

9 21 2 , − 10 , − 5

.

We multiply v2 by 10 3 to simplify its form, yielding v2 = [5, −3, −14]. Hence, an orthogonal basis for the subspace is {[5, −1, 2], [5, −3, −14]}.

184


Section 6.1

(c) First, let v1 = [2, 1, 0, −1].

4 Next, v2 = [1, 1, 1, −1] − [1,1,1,−1]·[2,1,0,−1] [2,1,0,−1]·[2,1,0,−1] [2, 1, 0, −1] = [1, 1, 1, −1] − 6 [2, 1, 0, −1] = − 13 , 13 , 1, − 13 . Multiplying by 3 to eliminate fractions yields v2 = [−1, 1, 3, −1].

[1,−2,1,1]·[−1,1,3,−1] [2, 1, 0, −1] − Finally, v3 = [1, −2, 1, 1] − [1,−2,1,1]·[2,1,0,−1] [2,1,0,−1]·[2,1,0,−1] [−1,1,3,−1]·[−1,1,3,−1] [−1, 1, 3, −1] (−1) 5 7 5 3 = [1, −2, 1, 1] − (−1) 6 [2, 1, 0, −1] − 12 [−1, 1, 3, −1] = 4 , − 4 , 4 , 4 . Multiplying by 4 to eliminate fractions produces v3 = [5, −7, 5, 3]. Hence, an orthogonal basis for the subspace is {[2, 1, 0, −1], [−1, 1, 3, −1], [5, −7, 5, 3]}. (5) (a) First, we must find an ordinary basis for R3 containing [2, 2, −3]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the vector we have and use the Independence Test Method. Hence, we row reduce ⎤ ⎡ ⎡ ⎤ 1 0 0 − 13 2 1 0 0 2 ⎥ ⎣ 2 0 1 0 ⎦ to ⎢ ⎣ 0 1 0 3 ⎦, giving the basis {[2, 2, −3], [1, 0, 0], [0, 1, 0]}. We per2 −3 0 0 1 0 0 1 3

form the Gram-Schmidt Process on this set.

[1,0,0]·[2,2,−3] 2 Let v1 = [2, 2, −3]. Then v2 = [1, 0, 0] − [2,2,−3]·[2,2,−3] [2, 2, −3] = [2, 2, −3] = [1, 0, 0] − 17 13 4 6 by

17 to eliminate 17 , − 17 , 17 . Multiplying fractions yields v2 = [13, −4, 6]. Finally, [0,1,0]·[2,2,−3] [0,1,0]·[13,−4,6] [13, −4, 6] v3 = [0, 1, 0] − [2,2,−3]·[2,2,−3] [2, 2, −3] − [13,−4,6]·[13,−4,6] 2 9 6 13 = [0, 1, 0] − 17 [2, 2, −3] − (−4) 221 [13, −4, 6] = 0, 13 , 13 . Multiplying by 3 to simplify the form of the vector gives us v3 = [0, 3, 2]. Hence, an orthogonal basis for R3 containing [2, 2, −3] is {[2, 2, −3], [13, −4, 6], [0, 3, 2]}. (c) First, we must find an ordinary basis for R3 containing [1, −3, 1] and [2, 5, 13]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the given vector and use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ 1 0 0 − 13 5 ⎤ 44 44 1 2 1 0 0 ⎢ 3 ⎥ 1 ⎣ −3 5 0 1 0 ⎦ to ⎣ 0 1 0 44 44 ⎦, giving the basis {[1, −3, 1], [2, 5, 13], [1, 0, 0]}. 1 1 13 0 0 1 0 0 1 − 14 4 We perform the Gram-Schmidt Process on this set. Let v1 = [1, −3, 1]. Since we know that the given set of vectors is orthogonal, we may simply use v2 = [2, 5, 13]. (If the two given vectors were not already orthogonal, we would need to apply Gram-Schmidt to obtain v2 .) Finally,

[1,0,0]·[1,−3,1] [1,0,0]·[2,5,13] v3 = [1, 0, 0] − [1,−3,1]·[1,−3,1] [1, −3, 1] − [2,5,13]·[2,5,13] [2, 5, 13] 8 2 2 1 2 = [1, 0, 0] − 11 [1, −3, 1] − 198 [2, 5, 13] = 9 , 9 , − 9 . Multiplying by 92 to simplify the form of the vector gives v3 = [4, 1, −1]. Hence, an orthogonal basis for R3 containing the two given vectors is {[1, −3, 1], [2, 5, 13], [4, 1, −1]}. (e) First, we must find an ordinary basis for R4 containing [2, 1, −2, 1]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R4 to the given vector and use the Independence Test Method. Hence, we row reduce

185


2 ⎢ 1 ⎢ ⎣ −2 1

1 0 0 0

0 1 0 0

0 0 1 0

⎡ ⎤ 1 0 ⎢ 0 0 ⎥ ⎥ to ⎢ ⎣ 0 0 ⎦ 0 1

0 1 0 0

0 0 1 0

0 0 0 1

Section 6.1

⎤ 1 −2 ⎥ ⎥, giving the basis −1 ⎦ 2

{[2, 1, −2, 1], [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0]}. We perform the Gram-Schmidt Process on this set.

[1,0,0,0]·[2,1,−2,1] Let v1 = [2, 1, −2, 1]. Then, v2 = [1, 0, 0, 0] − [2,1,−2,1]·[2,1,−2,1] [2, 1, −2, 1] = [1, 0, 0, 0] − 2 3 1 2 , − 15 . Multiplying by 5 to eliminate fractions yields v2 = [3, −1, 2, −1]. 10 [2, 1, −2, 1] = 5 , − 5 , 5

[0,1,0,0]·[2,1,−2,1] [0,1,0,0]·[3,−1,2,−1] Next, v3 = [0, 1, 0, 0] − [2,1,−2,1]·[2,1,−2,1] [2, 1, −2, 1] − [3,−1,2,−1]·[3,−1,2,−1] [3, −1, 2, −1] = 1 [2, 1, −2, 1] − (−1) 0, 56 , 13 , − 16 . Multiplying by 6 to eliminate [0, 1, 0, 0] − 10 15 [3, −1, 2, −1] =

[0,0,1,0]·[2,1,−2,1] fractions gives us v3 = [0, 5, 2, −1]. Finally, v4 = [0, 0, 1, 0] − [2,1,−2,1]·[2,1,−2,1] [2, 1, −2, 1] −

[0,0,1,0]·[3,−1,2,−1] [0,0,1,0]·[0,5,2,−1] [3,−1,2,−1]·[3,−1,2,−1] [3, −1, 2, −1] − [0,5,2,−1]·[0,5,2,−1] [0, 5, 2, −1] 2 2 1 2 = [0, 0, 1, 0] − (−2) 10 [2, 1, −2, 1] − 15 [3, −1, 2, −1] − 30 [0, 5, 2, −1] = 0, 0, 5 , 5 . Multiplying by 5 to eliminate fractions produces v4 = [0, 0, 1, 2]. Hence, an orthogonal basis for R4 containing the given vector is {[2, 1, −2, 1], [3, −1, 2, −1], [0, 5, 2, −1], [0, 0, 1, 2]}. (7) In each part, let A represent the given matrix. Then, according to the directions for the problem, a vector in the direction of the axis of rotation is found by solving for an eigenvector for A corresponding to the eigenvalue λ = 1. (a) First, we must verify that A is an orthogonal matrix. Calculating AT and performing matrix multiplication easily shows that AAT = I3 . Therefore, A−1 = AT , and so A is orthogonal. Basketweaving quickly verifies that |A| = 1. Finally, we need to find a fundamental eigenvector for A corresponding to λ = 1. To do this, we row reduce " ⎤ ⎡ ⎤ ⎡ 9 " 9 6 1 "" − 11 1 0 11 11 " 0 3 " 0 ⎥ ⎢ 5 2 " 9 − 11 [1I3 − A|0] = ⎣ 11 " 0 ⎦ to ⎣ 0 1 −1 "" 0 ⎦. 11 " 5 " 0 0 0 " 0 0 −6 −7 11

11

11

Setting the independent variable x3 equal to 3 to eliminate fractions produces the fundamental eigenvector [−1, 3, 3]. This is a vector along the desired axis of rotation. (c) First, we must verify that A is an orthogonal matrix. Calculating AT and performing matrix multiplication easily shows that AAT = I3 . Therefore, A−1 = AT , and so A is orthogonal. Basketweaving quickly verifies that |A| = 1. Finally, we need to find a fundamental eigenvector for A corresponding to λ = 1. To do this, we row reduce " ⎤ ⎡ 1 " ⎤ ⎡ − 27 − 37 "" 0 7 1 0 −5 "" 0 ⎥ ⎢ 13 2 " ⎣ 0 1 −1 " 0 ⎦. [1I3 − A|0] = ⎣ − 37 7 7 "" 0 ⎦ to " 2 3 13 " 0 0 0 " 0 0 − − 7

7

7

Setting the independent variable x3 equal to 1 produces the fundamental eigenvector [5, 1, 1]. This is a vector along the desired axis of rotation. (8) (b) No. The vectors might not be unit vectors after scalar multiplication. For example, if we multiply every vector in the orthonormal set {[1, 0], [0, 1]} by 2, we get {[2, 0], [0, 2]}. This new set is orthogonal, but it is not orthonormal because each of its elements has length 2. 186


Section 6.1

(16) (b) By part (1) of Theorem 6.7, a 3×3 matrix is orthogonal if and only if its rows form an orthonormal basis for R3 . Thus, we must find an orthonormal basis containing √16 [1, 2, 1]. To do this, we will first find an orthogonal basis containing [1, 2, 1] and then normalize the vectors afterward. First, we must find an ordinary basis for R3 containing [1, 2, 1]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the vector we have and then use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 1 0 0 1 0 0 1 ⎣ 2 0 1 0 ⎦ to ⎣ 0 1 0 −1 ⎦, giving the basis {[1, 2, 1], [1, 0, 0], [0, 1, 0]}. Next, we per1 0 0 1 0 0 1 −2 form the Gram-Schmidt Process on this set.

5 1 1 1 Let v1 = [1, 2, 1]. Then v2 = [1, 0, 0] − [1,0,0]·[1,2,1] [1,2,1]·[1,2,1] [1, 2, 1] = [1, 0, 0] − 6 [1, 2, 1] = 6 , − 3 , − 6 . Multiplying by 6 to eliminate fractions yields v2 = [5, −2, −1].

Next, [0,1,0]·[1,2,1] [0,1,0]·[5,−2,−1] v3 = [0, 1, 0] − [1,2,1]·[1,2,1] [1, 2, 1] − [5,−2,−1]·[5,−2,−1] [5, −2, −1] = [0, 1, 0] − 26 [1, 2, 1] − 1 2 (−2) 30 [5, −2, −1] = 0, 5 , − 5 . Multiplying by 5 to eliminate fractions gives us v3 = [0, 1, −2]. basis Normalizing v1 ,v2 , and v3 produces 3 the orthonormal 4 1 2 1 5 2 1 1 2 √ , √ , √ , √30 , − √30 , − √30 , 0, √5 , − √5 for R3 . Using these vectors as rows for a 6 6 6 matrix, rationalizing denominators, and simplifying gives us the orthogonal matrix √ √ ⎤ ⎡ √6 6 6 ⎢ ⎢ ⎣

6 √ 30 6

0

3 √ 30 − 15 √ 5 5

6 √ 30 − 30 √ −255

⎥ ⎥, whose first row is the given vector ⎦

√1 [1, 2, 1]. 6

(17) Proof of the other half of part (1) of Theorem 6.7: Suppose that the rows of A form an orthonormal basis for Rn . Then the (i, j) entry of AAT = (ith row of A)·(jth column of AT ) ) 1 i=j = (ith row of A)·(jth row of A) = (since the rows of A form an orthonormal set). Hence, 0 i = j AAT = In , and A is orthogonal. Proof of part (2): A is orthogonal if and only if AT is orthogonal (by part (2) of Theorem 6.6) if and only if the rows of AT form an orthonormal basis for Rn (by part (1) of Theorem 6.7) if and only if the columns of A form an orthonormal basis for Rn . (22) (a) False. The remaining vectors in the set must also be perpendicular to each other. For example, the set {[0, 0], [1, 1], [2, 3]} contains the zero vector but is not orthogonal because [1, 1] · [2, 3] = 5 = 0. (b) True. The reason is that ei · ej = 0 for i = j, and ei = 1 for all i. (c) True. This is part of Theorem 6.3. (d) False. If the set of vectors {w1 , . . . , wk } is not orthogonal at the outset, adding more vectors will not make the original vectors perpendicular to each other. For example, the linearly independent set {[1, 1, 0], [1, 0, 1]} cannot be enlarged to an orthogonal basis for R3 because any such enlargement {[1, 1, 0], [1, 0, 1], v3 } will still have [1, 1, 0] · [1, 0, 1] = 1 = 0. (e) True. By Theorem 4.16, W has a finite basis B. Also, |B| > 0, since W is nontrivial. Then, by Theorem 6.4, the Gram-Schmidt Process can be used on the basis B to find an orthogonal basis for W. (f) True. Since AT A = In , Theorem 2.9 tells us that A−1 = AT . Hence, A is orthogonal by definition. 187


Section 6.2

(g) True. This follows directly from parts (3) and (1) of Theorem 6.6. (h) False. According to Theorem 6.7, the rows or columns of a matrix must be orthonormal, not just orthogonal, in order to guarantee that the matrix is orthogonal. So, for example, the rows of 1 2 A= form an orthogonal set, but A is not orthogonal, since AT A = 5I2 = I2 . 2 −1 (i) True. By part (1) of Theorem 6.7, the rows of A form an orthonormal basis for Rn . However, the set of rows of R(A) is the same as the set of rows of A, only listed in a different order. However, these vectors form an orthonormal basis for Rn regardless of the order in which they are listed. Hence, part (1) of Theorem 6.7 asserts that R(A) is also an orthogonal matrix. (j) True. This is the statement of Theorem 6.8.

Section 6.2 (1) If you need assistance combining the Gram-Schmidt Process with the Enlarging Method of Section 4.6, please review Example 5 in Section 6.1 of the textbook and Exercise 5 in Section 6.1. Solutions for the starred parts for Exercise 5 appear above in this manual. (a) The vector [2, 3], found by inspection, is clearly orthogonal to [3, −2]. Hence, [2, 3] ∈ W ⊥ . Now, / W ⊥ . Therefore, W ⊥ W ⊥ cannot be two-dimensional, or it would be all of R2 , and yet [3, −2] ∈ ⊥ must be one-dimensional, and so W = span({[2, 3]}). Note that dim(W) + dim(W ⊥ ) = 1 + 1 = 2 = dim(R2 ). (c) First, we find an orthogonal basis for W by performing the Gram-Schmidt Process on its basis {[1, 4, −2], [2, 1, −1]}. This results in the orthogonal basis B = {[1, 4, −2], [34, −11, −5]} for W. Next, we enlarge this set to a basis for R3 using the Enlarging Method of Section 4.6. This results in the basis C = {[1, 4, −2], [34, −11, −5], [1, 0, 0]} for R3 . Now we perform the Gram-Schmidt Process on the basis C to find an orthogonal basis D for R3 . Note that since the first two vectors in C are already perpendicular to each other, they will not be changed by the Gram-Schmidt Process. Completing the Gram-Schmidt Process produces the basis D = {[1, 4, −2], [34, −11, −5], [2, 3, 7]}. Since B is an orthogonal basis for W, Theorem 6.12 shows that {[2, 3, 7]} is an orthogonal basis for W ⊥ . Hence, W ⊥ = span({[2, 3, 7]}). Note that dim(W) + dim(W ⊥ ) = 2 + 1 = 3 = dim(R3 ). (e) Looking at the given equation for the plane W, we see that W is defined to be {[x, y, z] | [x, y, z] · [−2, 5, −1] = 0}. That is, W is the set of vectors orthogonal to [−2, 5, −1], and hence orthogonal to the subspace Y = span({[−2, 5, −1]}) of R3 . So, by definition of orthogonal ⊥ complement, W = Y ⊥ . Hence, by Corollary 6.14, W ⊥ = Y ⊥ = Y = span({[−2, 5, −1]}). Note that dim(W) + dim(W ⊥ ) = 2 + 1 = 3 = dim(R3 ). (f) First, we find an orthogonal basis for W by performing the Gram-Schmidt Process on its basis {[1, −1, 0, 2], [0, 1, 2, −1]}. This results in the orthogonal basis B = {[1, −1, 0, 2], [1, 1, 4, 0]} for W. Next, we enlarge this set to a basis for R4 using the Enlarging Method of Section 4.6. This results in the basis C = {[1, −1, 0, 2], [1, 1, 4, 0], [1, 0, 0, 0], [0, 1, 0, 0]} for R4 . Now we perform the Gram-Schmidt Process on the basis C to find an orthogonal basis D for R4 . Note that since the first two vectors in C are already perpendicular to each other, they will not be changed by the Gram-Schmidt Process. Completing the Gram-Schmidt Process produces the basis D = {[1, −1, 0, 2], [1, 1, 4, 0], [7, 1, −2, −3], [0, 4, −1, 2]}. Since B is an orthogonal basis for W, Theorem 6.12 shows that {[7, 1, −2, −3], [0, 4, −1, 2]} is an orthogonal basis for W ⊥ . Hence, W ⊥ = span({[7, 1, −2, −3], [0, 4, −1, 2]}). Note that dim(W) + dim(W ⊥ ) = 2 + 2 = 4 = dim(R4 ). 188


Section 6.2

(2) (a) First, we use the Gram-Schmidt Process on the basis {[1, −2, −1], [3, −1, 0]} for W to find the following orthogonal basis for W: {[1, this orthogonal basis for 4 3 −2, −1], [13, 4, 5]}. Normalizing 1 [13, 4, 5] . Hence, by definition, W produces the orthonormal basis √16 [1, −2, −1], √210

√1 [1, −2, −1] w1 = projW v = [−1, 3, 2] · √16 [1, −2, −1] 6

1 1 √ + [−1, 3, 2] · √210 [13, 4, 5] [13, 4, 5] 210 1 111 12 (9)[13, 4, 5] = − 33 = 16 (−9)[1, −2, −1] + 210 35 , 35 , 7 . 2 111 12 6 2 Finally, w2 = projW ⊥ v = v − projW v =[−1, 3, 2] − − 33 35 , 35 , 7 = − 35 , − 35 , 7 . (b) Looking at the given equation for the plane W, we see that W is defined to be {[x, y, z] | [x, y, z] · [2, −2, 1] = 0}. That is, W is the set of vectors orthogonal to [2, −2, 1], and hence orthogonal to the subspace Y = span({[2, −2, 1]}) of R3 . So, by definition of orthogonal ⊥ , - = Y = span 13 [2, −2, 1] , complement, W = Y ⊥ . Now, by Corollary 6.14, W ⊥ = Y ⊥ where we have normalized [2, −2, 1] to create an orthonormal 1 basis 1 for Y. Hence, by definition of v = [1, −4, 3] · [2, −2, 1] projY v, w2 = projW ⊥ v = proj 3 3 [2, −2, 1] Y 1 26 26 13 = 9 (13)[2, −2, 1] = 9 , − 9 , 9 . 17 10 14 26 13 = − 9 , − 9 , 9 . Note that Finally, w1 = projW v = v − projW ⊥ v = [1, −4, 3] − 26 9 ,− 9 , 9 in this problem we found it easier to compute projW ⊥ v first and use it to calculate projW v because it was easier to determine an orthonormal basis for W ⊥ than for W. (4) (a) Let v = [−2, 3, 1]. By Theorem 6.17, the minimum distance from P to W is v − projW v. Thus, we need to compute projW v. To do this, we need an orthonormal basis for W. We begin with the given basis {[−1, 4, 4], [2, −1, 0]} for W and apply the Gram-Schmidt Process to obtain3 {[−1, 4, 4], [20, −3, 8]}. Normalizing the vectors in this basis produces the orthonormal 4 √1 [−1, 4, 4], √ 1 [20, −3, 8] for W. Thus, by definition, 33 473

√1 [−1, 4, 4] projW v = [−2, 3, 1] · √133 [−1, 4, 4] 33

1 1 √ + [−2, 3, 1] · √473 [20, −3, 8] [20, −3, 8] 473

basis

=

1 33 (18)[−1, 4, 4]

+

1 473 (−41)[20, −3, 8]

105 64 = − 98 43 , 43 , 43 .

Hence, the minimum distance from P to W is v − projW v = [−2, 3, 1] − − 98 , 105 , 64 = 12 , 24 , − 21 = 43

43

43

43 43

43

√ 3 129 43 .

(d) Let v = [−1, 4, −2, 2]. By Theorem 6.17, the minimum distance from P to W is v − projW v = projW ⊥ v. To compute this, we will find an orthonormal basis for W ⊥ . Now W is defined to be {[x, y, z, w] | [x, y, z, w]·[2, 0, −3, 2] = 0}. That is, W is the set of vectors orthogonal to [2, 0, −3, 2], and hence orthogonal to the subspace Y = span({[2, 0, −3, 2]}). So, by definition of orthogonal ⊥ ⊥ complement, W = Y ⊥ . Thus, by Corollary 6.14, W ⊥ = Y ⊥ 3 = Y. Therefore, 4 a basis for W is {[2, 0, −3, 2]}. Normalizing, we get the orthonormal basis √117 [2, 0, −3, 2] for W ⊥ . So, by

√1 [2, 0, −3, 2] = 1 (8) [2, 0, −3, 2]. definition, projW ⊥ v = [−1, 4, −2, 2] · √117 [2, 0, −3, 2] 17 17 8√17 8 Hence, the minimum distance from P to W is projW ⊥ v = 17 [2, 0, −3, 2] = 17 .

(5) In each part, let A represent the given matrix. This problem utilizes the methods illustrated in Examples 8 and 9 and the surrounding discussion in Section 6.2 of the textbook. (a) The first step is to find the eigenvalues of A. Now, 189

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition " 2 " x − 11 " " 3 pA (x) = " 11 " 3 " 11

3 11

x− 1 11

10 11

3 11 1 11

x−

10 11

Section 6.2

" " " " " = x3 − 2x2 + x (by basketweaving) = x(x − 1)2 . Hence, A " "

has eigenvalues λ1 = 0 and λ2 = 1, as required for L to be an orthogonal projection. (L cannot be an orthogonal reflection since neither eigenvalue is −1.) Next, by row reducing [0I3 − A | 0] and [1I3 − A | 0], we find the following bases for the associated eigenspaces: basis for E0 = {[3, 1, 1]}, basis for E1 = {[−1, 3, 0], [−1, 0, 3]}. Since the basis vector [3, 1, 1] for E0 is perpendicular to both basis vectors for E1 , the operator L whose matrix is A with respect to the standard basis is an orthogonal projection onto the plane determined by E1 . But since E1 = E0⊥ , E1 is the plane consisting of the vectors orthogonal to [3, 1, 1]. Hence, L is the orthogonal projection onto the plane 3x + y + z = 0. (d) The first step is to find the eigenvalues of A. Now, " " 7 2 14 " " x − 15 15 15 " " " " 7 14 4 x − pA (x) = " " = x3 − x2 − x + 1 (by basketweaving) = (x − 1)2 (x + 1). 15 15 15 " " 1 4 " x + 25 " 5 5 Hence, A has eigenvalues λ1 = 1 and λ2 = −1, as required for L to be an orthogonal reflection. (L cannot be an orthogonal projection since neither eigenvalue is 0.) Next, by row reducing [1I3 − A | 0] and [−1I3 − A | 0], we find the following bases for the associated eigenspaces: basis for E1 = {[−1, 4, 0], [−7, 0, 4]}, basis for E−1 = {[2, 1, 3]}. However, [2, 1, 3] · [−1, 4, 0] = 2 = 0. Therefore, the eigenspaces are not orthogonal complements. Hence, the operator L is neither an orthogonal projection nor an orthogonal reflection. (6) In this problem, we utilize the ideas presented in Example 8 of Section 6.2 in the textbook. Let W = {[x, y, z] | 2x−y+2z = 0} = {[x, y, z] | [x, y, z]·[2, −1, 2] = 0}. Then, if Y = span({[2, −1, 2]}), W = Y ⊥ , and, by Corollary 6.14, W ⊥ = Y. So, {[2, −1, 2]} is a basis for W ⊥ , and by the Inspection Method of Section 4.6, {[1, 0, −1], [1, 2, 0]} is a basis for W. Now W = E1 , the eigenspace corresponding to the eigenvalue λ1 = 1, for L, since the plane W is left fixed by L. Also, W ⊥ = E0 , the eigenspace corresponding to the eigenvalue λ2 = 0 for L, since all vectors perpendicular to W are sent to 0 by L. So, we know by the Generalized Diagonalization Method of Section 5.6 that if A is the matrix for L with respect to the standard basis for R3 , then D = P−1 AP, where P is a matrix whose columns are a basis for R3 consisting of fundamental eigenvectors for L, and D is a diagonal matrix with the eigenvalues for L⎤on its main diagonal. Note that D = P−1 AP implies that A = ⎡PDP−1 . Setting ⎤ ⎡ 2 1 1 0 0 0 0 2 ⎦. D = ⎣ 0 1 0 ⎦ and, using the bases for W ⊥ = E0 and W = E1 from above, P = ⎣ −1 2 −1 0 0 0 1 ⎡ ⎤ ⎡ ⎤ 2 −1 2 5 2 −4 2 ⎦. Row reduction yields P−1 = 19 ⎣ 4 −2 −5 ⎦. Hence, A = PDP−1 = 19 ⎣ 2 8 1 4 1 −4 2 5 (9) (a) The following argument works for any orthogonal projection onto a plane through the origin in R3 . An orthogonal projection onto a plane in R3 has two eigenvalues, λ1 = 1 and λ2 = 0. The eigenspace E1 is the given plane, and thus is two-dimensional. Hence, the geometric multiplicity of λ1 is 2. The eigenspace E0 is the orthogonal complement of the plane, since the vectors perpendicular to the plane are sent to 0 by L. By Corollary 6.13, dim(E0 ) = 3−dim(E1 ) = 1. Hence, the 190


Section 6.2

geometric multiplicity of λ2 is 1. Since the geometric multiplicities of the eigenvalues add up to 3, L is diagonalizable, and each geometric multiplicity must equal its corresponding algebraic multiplicity. Therefore, the factor (x − 1) appears in pL (x) raised to the second power, and the factor of x appears in pL (x) raised to the first power. Hence, pL (x) = (x − 1)2 x = x3 − 2x2 + x. (c) The following argument works for any orthogonal reflection through a plane through the origin in R3 . An orthogonal reflection through a plane in R3 has two eigenvalues, λ1 = 1 and λ2 = −1. The eigenspace E1 is the given plane, and thus is two-dimensional. Hence, the geometric multiplicity of λ1 is 2. The eigenspace E−1 is the orthogonal complement of the plane, since the vectors perpendicular to the plane are reflected to their opposites by L. By Corollary 6.13, dim(E0 ) = 3 − dim(E1 ) = 1. Hence, the geometric multiplicity of λ2 is 1. Since the geometric multiplicities of the eigenvalues add up to 3, L is diagonalizable, and each geometric multiplicity must equal its corresponding algebraic multiplicity. Therefore, the factor (x − 1) appears in pL (x) raised to the second power, and the factor of (x + 1) appears in pL (x) raised to the first power. Hence, pL (x) = (x − 1)2 (x + 1) = x3 − x2 − x + 1. (10) (a) We will consider three different approaches to this problem. The third method is the easiest, but it involves using the cross product of vectors in R3 , and so can be used only in R3 and when dim(W) = 2. Method #1: First, we find an orthonormal basis for W. To do this, we need an orthogonal basis, which we get by applying the Gram-Schmidt Process to

the given basis {[2, −1, 1], [1, 0, −3]}. (−1) So, let v1 = [2, −1, 1]. Then v2 = [1, 0, −3]− [1,0,−3]·[2,−1,1] [2,−1,1]·[2,−1,1] [2, −1, 1] = [1, 0, −3]− 6 [2, −1, 1] = 4 1 17 3 , − 6 , − 6 . Multiplying by 6 to eliminate fractions yields v2 = [8, −1, −17]. We normalize these 3 4 1 vectors to find the following orthonormal basis for W: √16 [2, −1, 1] , √354 [8, −1, −17] . Now the matrix for L with respect to the standard basis is the 3 × 3 matrix whose columns are L(i), L(j), and L(k). operator, we get Using the definition

of the projection

1 1 √ √ [2, −1, 1] [2, −1, 1] L(i) = projW i = i · 6 6

1 √ 1 [8, −1, −17] = 1 [2, −1, 1] + 4 [8, −1, −17] = 1 [50, −21, −3], + i · √354 [8, −1, −17] 3 177 59 354

1 1 √ [2, −1, 1] L(j) = projW j = j · √6 [2, −1, 1] 6

(−1) (−1) 1 √ 1 [8, −1, −17] = + j · √354 [8, −1, −17] 6 [2, −1, 1] + 354 [8, −1, −17] 354

1 √1 [2, −1, 1] [−21, 10, −7], and, L(k) = projW k = k · √16 [2, −1, 1] = 59 6

(−17) 1 1 1 1 √ + k · √354 [8, −1, −17] [8, −1, −17] = 6 [2, −1, 1]+ 354 [8, −1, −17] = 59 [−3, −7, 58]. 354 ⎡ ⎤ 50 −21 −3 1 ⎣ −21 10 −7 ⎦. Therefore, the matrix for L with respect to the standard basis is A = 59 −3 −7 58 Method #2: The subspace W is a plane in R3 , and L is the orthogonal projection onto that plane. The plane W is also the eigenspace E1 for L, corresponding to the eigenvalue λ1 = 1. The eigenspace E0 , corresponding to the eigenvalue λ2 = 0 is W ⊥ . We are given the basis B = {[2, −1, 1], [1, 0, −3]} for W, but we also need a basis for W ⊥ . To find this basis for W ⊥ , we first expand the basis B for W to a basis C for R3 using the Enlarging Method, producing C = {[2, −1, 1], [1, 0, −3], [1, 0, 0]}. Next, we apply the Gram-Schmidt Process to C, obtaining the orthogonal basis D = {[2, −1, 1], [8, −1, −17], [3, 7, 1]} for R3 . Now {[2, −1, 1], [8, −1, −17]} is a 191


Section 6.2

basis for W = E1 , and so Theorem 6.12 shows that {[3, 7, 1]} is a basis for W ⊥ = E0 . Hence, each vector in D is an eigenvector for L. Thus, the matrix A for L with respect to the standard ⎡ ⎤ ⎡ ⎤ 1 0 0 2 8 3 −1 7 ⎦. Using row basis is given by A = PDP−1 , where D = ⎣ 0 1 0 ⎦ and P = ⎣ −1 0 0 0 1 −17 1 ⎡ ⎤ 118 −59 59 1 ⎣ 8 −1 −17 ⎦. Performing the matrix reduction to compute P−1 produces P−1 = 354 18 42 6 product PDP−1 gives the matrix A computed above using Method #1. Method #3: The subspace W is a plane is R3 , and L is the orthogonal projection onto that plane. The plane W is also the eigenspace E1 for L, corresponding to the eigenvalue λ1 = 1. The eigenspace E0 corresponding to the eigenvalue λ2 = 0 is W ⊥ . We are given the basis B = {[2, −1, 1], [1, 0, −3]} for W, but we also need a basis for W ⊥ . In Exercise 8 of Section 3.1 we defined the cross product of two vectors in R3 (v × w). In that exercise it is proven that v × w is orthogonal to both v and w. Thus, if we take the cross product of the two vectors in the basis B for W, we will get a vector orthogonal to both of these vectors, which places it in W ⊥ . Hence, [2, −1, 1] × [1, 0, −3] = [3, 7, 1] ∈ W ⊥ . But since dim(W ⊥ ) = 1 (by Corollary 6.13), {[3, 7, 1]} must be a basis for W ⊥ . So now we have a basis C = {[3, 7, 1], [2, −1, 1], [1, 0, −3]} for R3 consisting of eigenvectors. Thus, the matrix A for L with respect to the standard basis is given ⎡ ⎤ ⎡ ⎤ 0 0 0 3 2 1 0 ⎦. Using row reduction to by A = PDP−1 , where D = ⎣ 0 1 0 ⎦ and P = ⎣ 7 −1 1 1 −3 0 0 1 ⎡ ⎤ 3 7 1 1 ⎣ 21 −10 7 ⎦. Performing the matrix product PDP−1 compute P−1 produces P−1 = 59 8 −1 −17 gives the matrix A computed above using Method #1. (c) (Note: Three solutions using three different methods were presented for part (a) above. However, the third method used there will not work in part (c) because the cross product is not defined in R4 . Hence, we will solve this problem only two ways, corresponding to the first two methods presented in part (a).) Method #1: First, we find an orthonormal basis for W. To do this, we need an orthogonal basis, which we get by applying the Gram-Schmidt Process to the given basis {[1, 2, 1, 0], [−1, 0, −2, 1]}. So, let v1 = [1, 2, 1, 0]. Then v2 = [−1, 0, −2, 1] − [−1,0,−2,1]·[1,2,1,0] [1, 2, 1, 0] = [1,2,1,0]·[1,2,1,0] 1 3 [−1, 0, −2, 1] − (−3) 6 [1, 2, 1, 0] = − 2 , 1, − 2 , 1 . Multiplying by 2 to eliminate fractions yields v2 = [−1, 2, −3, 2]. We normalize 4 these vectors to find the following orthonormal basis for W: 3 1 √1 [1, 2, 1, 0], √ [−1, 2, −3, 2] 6 3 2

Now the matrix for L with respect to the standard basis is the 4 × 4 matrix whose columns are L(e1 ), L(e2 ), L(e3 ), and L(e4 ). Using the definition of the projection operator, we get

√1 [1, 2, 1, 0] L(e1 ) = projW e1 = e1 · √16 [1, 2, 1, 0] 6

1 1 1 √ √ + e1 · 3 2 [−1, 2, −3, 2] [−1, 2, −3, 2] = 16 [1, 2, 1, 0] + (−1) 18 [−1, 2, −3, 2] = 9 [2, 2, 3, −1], 3 2

√1 [1, 2, 1, 0] L(e2 ) = projW e2 = e2 · √16 [1, 2, 1, 0] 6 192


Section 6.2

1 1 √ + e2 · 3√ [−1, 2, −3, 2] [−1, 2, −3, 2] = 13 [1, 2, 1, 0] + 19 [−1, 2, −3, 2] = 19 [2, 8, 0, 2], 2 3 2

√1 [1, 2, 1, 0] L(e3 ) = projW e3 = e3 · √16 [1, 2, 1, 0] 6

(−1) 1 1 1 √ [−1, 2, −3, 2] = 1 [1, 2, 1, 0] + + e3 · 3√2 [−1, 2, −3, 2] 6 6 [−1, 2, −3, 2] = 3 [1, 0, 2, −1], 3 2

√1 [1, 2, 1, 0] and, L(e4 ) = projW e4 = e4 · √16 [1, 2, 1, 0] 6

1 1 √ √ [−1, 2, −3, 2] = (0)[1, 2, 1, 0] + 19 [−1, 2, −3, 2] = 19 [−1, 2, −3, 2]. + e4 · 3 2 [−1, 2, −3, 2] 3 2 ⎡ ⎤ 2 2 3 −1 ⎢ 2 8 0 2 ⎥ ⎥. Therefore, the matrix for L with respect to the standard basis is A = 19 ⎢ ⎣ 3 0 6 −3 ⎦ −1 2 −3 2 Method #2: The subspace W is a two-dimensional subspace of R4 , and L is the orthogonal projection onto that subspace. The subspace W is also the eigenspace E1 for L, corresponding to the eigenvalue λ1 = 1. The eigenspace E0 corresponding to the eigenvalue λ2 = 0 is W ⊥ . We are given the basis B = {[1, 2, 1, 0], [−1, 0, −2, 1]}} for W, but we also need a basis for W ⊥ . To find this basis for W ⊥ , we first expand the basis B for W to a basis C for R4 using the Enlarging Method, producing C = {[1, 2, 1, 0], [−1, 0, −2, 1], [1, 0, 0, 0], [0, 1, 0, 0]}. Next, we apply the Gram-Schmidt Process to C, obtaining the orthogonal basis D = {[1, 2, 1, 0], [−1, 2, −3, 2], [7, −2, −3, 1], [0, 1, −2, −4]} for R4 . Now {[1, 2, 1, 0], [−1, 2, −3, 2]} is a basis for W = E1 , and so Theorem 6.12 shows that {[7, −2, −3, 1], [0, 1, −2, −4]} is a basis for W ⊥ = E0 . Hence, each vector in D is an eigenvector for L. Thus, the matrix A for L with respect to the standard basis is given by A = PDP−1 , where ⎡ ⎡ ⎤ ⎤ 1 −1 7 0 1 0 0 0 ⎢ 2 ⎢ 0 1 0 0 ⎥ 2 −2 1 ⎥ −1 ⎢ ⎥ ⎥ D = ⎢ ⎣ 0 0 0 0 ⎦ and P = ⎣ 1 −3 −3 −2 ⎦. Using row reduction to compute P 0 0 0 0 0 2 1 −4 ⎡ ⎤ 21 42 21 0 ⎢ ⎥ −7 14 −21 14 1 ⎢ ⎥. Performing the matrix product PDP−1 gives produces P−1 = 126 ⎣ 14 −4 −6 2 ⎦ 0

6 −12

−24

the matrix A computed above using Method #1. (18) Let S be a spanning set for W, and let v ∈ W ⊥ . If u ∈ S, then u ∈ W, and so by definition of W ⊥ , v · u = 0. Hence, v is orthogonal to every vector in S. Conversely, suppose v · u = 0 for all u ∈ S. Let w ∈ W. Then w = a1 u1 + · · · + an un for some u1 , . . . , un ∈ S. Hence, v · w = a1 (v · u1 ) + · · · + an (v · un ) = 0 + · · · + 0 = 0. Thus, v ∈ W ⊥ . (20) We first prove that L is a linear operator. Suppose {u1 , . . . , uk } is an orthonormal basis for W. Then L(v1 + v2 ) = projW (v1 + v2 ) = ((v1 + v2 ) · u1 )u1 + · · · + ((v1 + v2 ) · uk )uk = ((v1 · u1 ) + (v2 · u1 ))u1 + · · · + ((v1 · uk ) + (v2 · uk ))uk = (v1 · u1 )u1 + · · · + (v1 · uk )uk + (v2 · u1 )u1 + · · · + (v2 · uk )uk = projW (v1 ) + projW (v2 ) = L(v1 ) + L(v2 ). Similarly, L(cv) = projW (cv) = ((cv) · u1 )u1 + · · · + ((cv) · uk )uk = c(v · u1 )u1 + · · · + c(v · uk )uk = c((v · u1 )u1 + · · · + (v · uk )uk ) = cprojW (v) = cL(v). Hence, L is a linear operator.

193


Section 6.2

To show W ⊥ = ker(L), we first prove that W ⊥ ⊆ ker(L). If v ∈ W ⊥ , then since v · w = 0 for every w ∈ W, L(v) = projW (cv) = (v · u1 )u1 + · · · + (v · uk )uk = 0u1 + · · · + 0uk = 0. Hence, v ∈ ker(L). Next, we prove that range(L) = W. First, note that if w ∈ W, then w = w + 0, where w ∈ W and 0 ∈ W ⊥ . Hence, by the statements in the text just before and just after Example 6 in Section 6.2, L(w) = projW w = w. Therefore, every w in W is in the range of L. Similarly, the same statements surrounding Example 6 imply projW v ∈ W for every v ∈ Rn . Hence, range(L) = W. Finally, by the Dimension Theorem, we have dim(ker(L)) = n − dim(range(L)) = n − dim(W) = dim(W ⊥ ). Since W ⊥ ⊆ ker(L), Theorem 4.16 implies that W ⊥ = ker(L). (23) Suppose that T is any point in W and w is the vector from the origin to T . We need to show that v − w ≥ v − projW v; that is, the distance from P to T is at least as large as the distance from P to the terminal point of projW v. Let a = v − projW v and b = (projW v) − w. First, we show that a ∈ W ⊥ . This follows from the statement just after Example 6 in Section 6.2 of the text. Next, we show that b ∈ W. Now, by the statement after Example 6, projW v ∈ W. Also, w ∈ W (since T is in W), and so b = (projW v) − w ∈ W. 2 2 2 Finally, v − w = v − projW v + (projW v) − w = a + b = (a + b) · (a + b) 2 2 2 2 = a · a + a · b + b · a + b · b = a · a + 0 + 0 + b · b = a + b ≥ a = v − projW v , which completes the proof. (26) (a) True. This is the definition of W ⊥ . (b) True. This follows from Theorem 6.10. (c) False. By Theorem 6.11, W ∩ W ⊥ = {0}, for every subspace W of Rn . (d) False. If the original basis {b1 , . . . , b7 } is an orthogonal basis, then this is true, but it is false in general. So, for example, if bi = ei for 1 ≤ i ≤ 6, and b7 = e1 + e7 , then {b1 , . . . , b7 } is a / W ⊥ because b7 · b1 = 1 = 0. In this case, basis for R7 . If W = span({b1 , . . . , b4 }), then b7 ∈ ⊥ {b5 , b6 , b7 } is not a basis for W . (e) True. This follows from Corollary 6.13. (f) False. The orthogonal complement of a subspace is not a setwise complement of the subspace. For example, if W is the subspace {[a, 0] | a ∈ R} of R2 , then W ⊥ = {[0, b] | b ∈ R}. However, the vector [1, 1] of R2 is in neither of these subspaces. (g) True. This is Corollary 6.14. (h) True. The orthogonal complement of a line through the origin was discussed in Example 5 and illustrated in Figure 6.2 in Section 6.2 of the textbook. Since this was found to be a plane through the origin, this discussion and Corollary 6.14, which states that (W ⊥ )⊥ = W, complete the justification of the given statement. (i) True. This is part of Theorem 6.16. (j) False. The eigenvalues for an orthogonal projection are 0 and 1, not −1 and 1. Vectors perpendicular to the plane are sent to 0 by the projection. No vector is sent to its opposite. The matrix for the projection diagonalizes to a matrix with 0, 1, and 1 on the main diagonal. See Example 8 in Section 6.2 of the textbook. (Keep in mind that orthogonal reflections have matrices that diagonalize to a diagonal matrix with −1, 1, 1 on the main diagonal.) (k) True. This follows from Theorem 6.17 and the “boxed” comment just after Example 6 in Section 6.2 of the textbook. (l) True. This follows from the “boxed” comment just after Example 6 in Section 6.2 of the textbook. 194


Section 6.3

(1) (a) The matrix for L with respect to the standard basis is A =

3 2

2 5

Section 6.3

. Since A is symmetric, L

is a symmetric operator by Theorem 6.18. [a,b,c] (d) Let v1 = [a,b,c] and let {v2 , v3 } be an orthonormal basis for the plane ax + by + cz = 0. Then B = {v1 , v2 , v3 } is an orthonormal basis for R3 , and the matrix for L with respect to B is the diagonal matrix with 0, 1, 1 on its main diagonal. Hence, L is an orthogonally diagonalizable linear operator. Therefore, L is symmetric by Theorem 6.20.

(e) Rotations in R3 about an axis through an angle of π3 have λ = 1 as their only eigenvalue. Also, the eigenspace E1 is one-dimensional, containing vectors parallel to the axis of rotation. Hence, the sum of the geometric multiplicities of the eigenvalues of L is 1. Since this is less than 3, L is not diagonalizable. Hence, L is not orthogonally diagonalizable. Therefore, Theorem 6.20 implies that L is not symmetric. (g) Direct computation shows that L(e1 ) = [4, 0, 3, 0], L(e2 ) = [0, 4, 0, 3], L(e3 ) = [3, 0, 9, 0], and L(e4 ) = [0, 3, 0, 9]. The matrix for L with respect to the standard basis is the 4 × 4 matrix whose columns are these four vectors in the order listed. Since this matrix is symmetric, L is a symmetric operator by Theorem 6.18. ( ' (2) (a) Let P =

3 5 4 5

4 5 − 35

, the matrix whose columns are the basis vectors for the eigenspaces E1

and E−1 . Then the desired matrix A equals PDP−1 , where D is the diagonal matrix with the eigenvalues 1 and −1 on its main diagonal. Theorem 2.13 shows that P−1 = P, and so ( ( ' ' 3 4 4 3 1 0 −7 24 5 5 5 5 1 = 25 A= 4 , which is symmetric. 4 0 −1 24 7 − 35 − 35 5 5 ⎡ ⎤ 12 12 −3 −2 ⎢ 3 −1 12 24 ⎥ ⎥, the matrix whose columns are the basis vectors for the (d) Let P = ⎢ ⎣ 4 7 0 −12 ⎦ 0 12 4 11 eigenspaces E−1 and E1 . Then the desired matrix A equals PDP−1 , where D is the diagonal matrix with the eigenvalues −1, −1, 1, and 1 on its main diagonal. Using row reduction to find ⎡ ⎤ 12 7 1 −12 ⎢ 0 −4 3 12 ⎥ 1 ⎢ ⎥. P−1 shows that P−1 = 169 ⎣ −11 12 24 −2 ⎦ 4 0 −12 3 ⎡ ⎤ −119 −72 −96 0 ⎢ 119 0 96 ⎥ 1 ⎢ −72 ⎥ , which is symmetric. Hence, A = PDP−1 = 169 ⎣ −96 0 119 −72 ⎦ 0 96 −72 −119 (3) In each part, we follow Steps 2 and 3 of the Orthogonal Diagonalization Method of Section 6.3. " " " x − 144 60 "" " = x2 − 169x = x(x − 169), giving us eigenvalues λ1 = 0 (a) Step 2(a): pA (x) = " 60 x − 25 " and λ2 = 169. Using row reduction to find particular solutions to the systems [0I2 − A|0] and 195


Section 6.3

[169I2 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[5, 12]} for E0 and {[−12, 5]} for E169 . Step 2(b): Since E0 and E169 are one-dimensional, the given bases are already orthogonal. To 1 [5, 12]} for E0 and create orthonormal bases, we need to normalize the vectors, yielding { 13 1 { 13 [−12, 5]} for E169 . 1 1 [5, 12], 13 [−12, 5] . Step 2(c): Z = 13 1 1 [5, 12], 13 [−12, 5] . Now P is the matrix Step 3: Since C is the standard basis for R2 , B = Z = 13 whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries 5 −12 0 0 1 and D = . are the eigenvalues 0 and 169. Hence, P = 13 12 5 0 169 Now P−1 = PT , and so D = PT AP, which is " 4 " x − 17 − 89 9 9 " " 4 x − 17 (c) Step 2(a): pA (x) = " − 89 9 9 " 4 4 11 " x − 9 9 9

easily verified by performing the matrix product. " " " " " = x3 − 5x2 + 7x − 3 = (x − 1)2 (x − 3), giving us " "

eigenvalues λ1 = 1 and λ2 = 3. Using row reduction to find particular solutions to the systems [1I3 − A|0] and [3I3 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[−1, 1, 0], [1, 0, 2]} for E1 and {[−2, −2, 1]} for E3 . orthogonal. To make it an Step 2(b): Since E3 is one-dimensional, the given basis is already , orthonormal basis, we need to normalize the vector, yielding 13 [−2, −2, 1] for E3 . Now E1 is two-dimensional, and so we must use the Gram-Schmidt Process to turn the basis we have into an [1,0,2]·[−1,1,0] [−1, 1, 0] = 12 , 12 , 2 . orthogonal basis. We let v1 = [−1, 1, 0]. Then v2 = [1, 0, 2]− [−1,1,0]·[−1,1,0] Multiplying by 2 to simplify produces v2 = [1, 1, 4]. Thus, we have 3 the orthogonal basis {v41 , v2 }

1 for E1 . Normalizing these vectors gives us the orthonormal basis √12 [−1, 1, 0], 3√ [1, 1, 4] . 2

1 [1, 1, 4], 13 [−2, −2, 1] . Step 2(c): Z = √12 [−1, 1, 0], 3√ 2

1 1 [1, 1, 4], [−2, −2, 1] . Step 3: Since C is the standard basis for R3 , B = Z = √12 [−1, 1, 0], 3√ 3 2 Now P is the matrix whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries are the eigenvalues 1, 1, and 3. Hence, ⎤ ⎡ 1 ⎡ ⎤ − √12 3√ − 23 2 1 0 0 ⎢ ⎥ 1 √1 √ ⎣ ⎦ − 23 ⎥ P=⎢ 2 3 2 ⎦ and D = 0 1 0 . ⎣ 0 0 3 1 4 0 3√2 3

Now P−1 = PT , and so D = PT AP, which is verified by performing the matrix product. (e) Step 2(a): The hint in the textbook indicates that pA (x) = (x−2)2 (x+3)(x−5), giving eigenvalues λ1 = 2, λ2 = −3, and λ3 = 5. Using row reduction to find particular solutions to the systems [2I4 −A|0], [−3I4 −A|0] and [5I4 −A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[3, 2, 1, 0], [−2, 3, 0, 1]} for E2 , {[1, 0, −3, 2]} for E−3 , and {[0, −1, 2, 3]} for E5 . bases are already Step 2(b): Since E−3 and E5 are one-dimensional, the given 3 4 orthogonal. To 1 make these bases orthonormal, we need to normalize, yielding √14 [1, 0, −3, 2] for E3 and 4 3 √1 [0, −1, 2, 3] for E5 . Now E2 is two-dimensional, and so typically, we would use the Gram14 Schmidt Process to turn the basis for E2 into an orthogonal basis. However, quick inspection shows us that the basis we have is already orthogonal! (This is not the typical situation.) Thus, 196


Section 6.3

3 4 we only need to normalize to get the orthonormal basis √114 [3, 2, 1, 0], √114 [−2, 3, 0, 1] for E2 .

Step 2(c): Z = √114 [3, 2, 1, 0], √114 [−2, 3, 0, 1], √114 [1, 0, −3, 2], √114 [0, −1, 2, 3] .

3 Step 3: Since C is the standard basis for R ,

1 1 1 B = Z = √14 [3, 2, 1, 0], √14 [−2, 3, 0, 1], √14 [1, 0, −3, 2], √114 [0, −1, 2, 3] . Now P is the matrix whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries are the eigenvalues 2, 2, −3, and 5. Hence, ⎡ ⎡ ⎤ ⎤ 3 −2 1 0 2 0 0 0 ⎢ 2 ⎢ 3 0 −1 ⎥ 0 0 ⎥ ⎥ and D = ⎢ 0 2 ⎥. P = √114 ⎢ ⎣ 1 ⎣ ⎦ 0 −3 2 0 0 −3 0 ⎦ 0 1 2 3 0 0 0 5

Now P−1 = PT , and so D = PT AP, which is " " x − 11 −2 10 " x − 14 −5 (g) Step 2(a): pA (x) = "" −2 " 10 −5 x + 10

easily verified by performing the matrix product. " " " " = x3 − 15x2 − 225x + 3375 = (x − 15)2 (x + 15), " "

giving us eigenvalues λ1 = 15 and λ2 = −15. Using row reduction to find particular solutions to the systems [15I3 − A|0] and [−15I3 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[1, 2, 0], [−5, 0, 2]} for E15 and {[2, −1, 5]} for E−15 . Step 2(b): Since E−15 is one-dimensional, the3given basis is4already orthogonal. To make the basis orthonormal, we need to normalize, yielding √130 [2, −1, 5] for E3 . Now E15 is two-dimensional, and so we must use the Gram-Schmidt Process to turn the basis

for E15 into an orthogonal [−5,0,2]·[1,2,0] basis. We let v1 = [1, 2, 0]. Then v2 = [−5, 0, 2] − [1,2,0]·[1,2,0] [1, 2, 0] = [−4, 2, 2]. Thus, we have the orthogonal basis 4 {v1 , v2 } for E1 . Normalizing these vectors gives the orthonormal basis 3 √1 [1, 2, 0], √1 [−2, 1, 1] . 5 6

Step 2(c): Z = √15 [1, 2, 0], √16 [−2, 1, 1], √130 [2, −1, 5] . 3

Step 3: Since C is the standard basis for R ,

B = Z = √15 [1, 2, 0], √16 [−2, 1, 1], √130 [2, −1, 5] . Now P is the matrix whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries are the eigenvalues 15, ⎤ ⎡ 1 √ √2 − √26 ⎡ ⎤ 5 30 15 0 0 ⎥ ⎢ ⎢ √1 − √130 ⎥ 0 ⎦. 15, and −15. Hence, P = ⎢ √25 ⎥ and D = ⎣ 0 15 6 ⎦ ⎣ 0 0 −15 √1 √5 0 6 30 Now P−1 = PT , and so D = PT AP, which is easily verified by performing the matrix product. (4) (a) We use the Orthogonal Diagonalization Method from Section 6.3: Step 1: The plane V is two-dimensional, and we are given the images under L of two vectors in V. Since the set {[−10, 15, 6], [15, 6, 10]} containing these two vectors is linearly independent, it forms a basis for V. A quick inspection shows this basis is already orthogonal. Thus, to find an orthonormal basis for V, we only need to normalize. Doing so produces the orthonormal basis 1 1 [−10, 15, 6], 19 [15, 6, 10] for V. C = 19 Next, we need to compute the matrix A for L with respect toC. Since V is two-dimensional, A will 1 1 [−10, 15, 6] = L ([−10, 15, 6]) = be a 2 × 2 matrix. Applying L to each vector in C yields L 19 19 1 1 1 1 [50, −18, 8] and L [15, 6, 10] = L ([15, 6, 10]) = [−5, 36, 22]. We need to express each of 19 19 19 19 197


Section 6.3

these results in C-coordinates. Hence, we row reduce ⎡ 10 15 " 50 " ⎤ ⎡ 5 ⎤ − 19 19 "" 19 − 19 1 0 "" −2 2 ⎢ 15 6 " 18 36 ⎥ ⎣ 0 1 " 2 1 ⎦. ⎣ 19 19 " − 19 19 ⎦ to obtain " " 10 " 22 6 8 0 0 " 0 0 19

19

19

19

(Note: In performing the above row reduction by hand, or if entering the matrix into a computer or calculator, it is easier to begin by first performing three type (I) operations to mentally multiply each row by 19.) Crossing out the last row of all zeroes in the above result and using the matrix −2 2 to the right of the bar gives the desired matrix A = . 2 1 " " " x+2 −2 "" Step 2(a): Now pA (x) = "" = x2 + x − 6 = (x − 2)(x + 3), giving eigenvalues λ1 = 2 −2 x−1 " and λ2 = −3. Using row reduction to find particular solutions to the systems [2I2 − A|0] and [(−3)I2 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[1, 2]} for E2 and {[−2, 1]} for E−3 . To Step 2(b): Since E2 and E−3 are one-dimensional, the given3bases are4 already orthogonal. 3 4 make them orthonormal bases, we need to normalize, yielding √15 [1, 2] for E2 and √15 [−2, 1] for E−3 .

Step 2(c): Z = √15 [1, 2], √15 [−2, 1] . 1 1 [−10, 15, 6], 19 [15, 6, 10] , we need to reverse the coordinatization isomorStep 3: Since C = 19 phism to calculate the vectors in a basis B for V. First, 1 1 √1 [1, 2] → √1 1 [−10, 15, 6] + 2 [15, 6, 10] = 191√5 [20, 27, 26]. Similarly, 19 19 5 5 1 1 √1 [−2, 1] → √1 (−2) [−10, 15, 6] + 1 [15, 6, 10] = 191√5 [35, −24, −2]. Hence, an or19 19 5 5

thonormal basis for V consisting of eigenvectors for L is B = 191√5 [20, 27, 26], 191√5 [35, −24, −2] . Now P is the matrix whose columns are the vectors in Z, and D, the matrix for L with respect to B, is the diagonal matrix whose main diagonal entries are the eigenvalues 2 and −3. Hence, 1 −2 2 0 1 and D = . P = √5 2 1 0 −3 Now P−1 = PT , and so D = PT AP, which is easily verified by performing the matrix product. (5) In each part, it is important to orthogonally diagonalize the matrix involved, not just diagonalize it. Otherwise, the matrix A obtained might not be orthogonally diagonalizable, and so will not correspond to a symmetric operator (by Theorem 6.20), implying that it will not be a symmetric matrix (by Theorem 6.18), as desired. 119 −108 1 and let C be the standard basis for R2 . First, we must orthogonally (a) Let B = 25 −108 56 diagonalize B. We use Steps 2 and 3 of the " 108 " x − 119 " 25 25 Step 2(a): Now pB (x) = " 108 56 " x − 25 25

Orthogonal Diagonalization Method of Section 6.3. " " " " = x2 − 7x − 8 = (x − 8)(x + 1), giving eigenvalues "

λ1 = 8 and λ2 = −1. Using row reduction to find particular solutions to the systems [8I2 − B|0] and [(−1)I2 − B|0] as in the Diagonalization Method of Section 3.4 produces the basis {[−4, 3]} for E8 and {[3, 4]} for E−1 . 198


Section 6.3

Step 2(b): Since E8 and E−1 are one-dimensional, the given ,bases are- already orthogonal. , - To make them orthonormal bases, we need to normalize, yielding 15 [−4, 3] for E8 and 15 [3, 4] for E−1 . Step 2(c): Z = 15 [−4, 3], 15 [3, 4] . Step 3: Since C is the standard basis for R2 , B = Z = 15 [−4, 3], 15 [3, 4] . Now P is the matrix whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries −4 3 8 0 1 and D = . are the eigenvalues 8 and −1. Hence, P = 5 3 4 0 −1 2 0 Now P−1 = PT , and so B = PDPT . Letting D2 = , whose diagonal entries are 0 −1 the cube roots of the eigenvalues 8 and −1 gives B = PD32 PT = PD2 (PT P)D2 (PT P)D2 PT = 23 −36 1 T 3 T PD2 P . Therefore, if we let A = PD2 P = 25 , then A3 = B, as desired. −36 2 ⎡ ⎤ 17 16 −16 41 −32 ⎦ and let C be the standard basis for R3 . First, we must orthogonally (c) Let B = ⎣ 16 −16 −32 41 diagonalize B. We use Steps 2 and 3 of the Orthogonal Diagonalization Method of Section 6.3. " " " x − 17 −16 16 "" " x − 41 32 "" = x3 −99x2 +1539x−6561 = (x−81)(x−9)2 , Step 2(a): Now pB (x) = "" −16 " 16 32 x − 41 " giving eigenvalues λ1 = 81 and λ2 = 9. Using row reduction to find particular solutions to the systems [81I3 − B|0] and [9I3 − B|0] as in the Diagonalization Method of Section 3.4 produces the basis {[−1, −2, 2]} for E81 and {[−2, 1, 0], [2, 0, 1]} for E9 . Step 2(b): Since E81 is one-dimensional, the given basis is -already orthogonal. To make the basis , orthonormal, we need to normalize, yielding 13 [−1, −2, 2] for E81 . Now E9 is two-dimensional, and so we must use the Gram-Schmidt Process an orthogonal basis. to turn the basis for E9 into 2 4 [2,0,1]·[−2,1,0] We let v1 = [−2, 1, 0]. Then v2 = [2, 0, 1] − [−2,1,0]·[−2,1,0] [−2, 1, 0] = 5 , 5 , 1 . Multiplying by 5 to eliminate fractions produces v2 = [2, 4, 5]. basis {v1 , v2 } for 3 Thus, we have an orthogonal 4

1 E9 . Normalizing gives the orthonormal basis √15 [−2, 1, 0], 3√ [2, 4, 5] for E9 . 5

1 [2, 4, 5] . Step 2(c): Z = 13 [−1, −2, 2], √15 [−2, 1, 0], 3√ 5

1 [2, 4, 5] . Step 3: Since C is the standard basis for R3 , B = Z = 13 [−1, −2, 2], √15 [−2, 1, 0], 3√ 5 Now P is the matrix whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries are the eigenvalues 81, 9, and 9. Hence, ⎤ ⎡ 1 2 ⎡ ⎤ − 3 − √25 3√ 5 81 0 0 ⎥ ⎢ 2 4 ⎥ and D = ⎣ 0 9 0 ⎦. √ √1 P=⎢ 5 3 5 ⎦ ⎣ −3 0 0 9 5 2 0 3√ 3 5 ⎡ ⎤ 9 0 0 Now P−1 = PT , and so B = PDPT . Letting D2 = ⎣ 0 3 0 ⎦, whose diagonal entries are the 0 0 3 2 square roots of the eigenvalues 81, 9, and 9 gives B = PD22 PT = PD2 (PT P)D2 PT = PD2 PT .

199


11 Therefore, if we let A = PD2 PT = 13 ⎣ 4 −4

4 17 −8

Chap 6 Review

⎤ −4 −8 ⎦, then A2 = B, as desired. 17

(6) For example, the matrix A in Example 7 of Section 5.6 is diagonalizable but not symmetric and hence not orthogonally diagonalizable. " " " x−a −b "" a b " = (x − a)(x − c) − b2 = x2 − (a + c)x + (ac − b2 ). Using (7) If A= , then pA (x) = " −b x−c " b c

the quadratic formula and simplifying yields the two eigenvalues λ1 = 12 a + c + (a − c)2 + 4b2

and λ2 = 12 a + c − (a − c)2 + 4b2 . Now, D is the diagonal matrix with these eigenvalues on its ' ( a + c + (a − c)2 + 4b2 0 1 main diagonal. Hence, D = 2 . 0 a + c − (a − c)2 + 4b2 (8) (b) Since L is symmetric, it is orthogonally diagonalizable by Theorem 6.20. But if 0 is the only eigenvalue of L, the geometric multiplicity of λ = 0 must be dim(V). Therefore, the eigenspace for 0 must be all of V, and so L(v) = 0v = 0 for all v ∈ V. Therefore, L must be the zero linear operator. (13) (a) True. The given condition on L is the definition of a symmetric operator. The statement then follows from Lemma 6.19. (b) False. A symmetric operator is guaranteed to have a symmetric matrix only with respect to an ordered orthonormal basis. For example, the symmetric operator on R2

basis, not every ordered x 1 0 x 1 −1 given by L = has matrix A = with respect to the basis y 0 2 y 0 2 {[1, 0], [1, 1]}. But A is not symmetric. (c) True. This is part of Theorem 6.18. (d) True. This follows from the definition of orthogonally diagonalizable and Theorem 6.20. (e) True. The Orthogonal Diagonalization Method states that D = P−1 AP, and that the matrix P is orthogonal. Thus, P−1 = PT . Hence, D = P−1 AP ⇒ PDP−1 = P(P−1 AP)P−1 ⇒ PDP−1 = In AIn ⇒ PDPT = A. (f) True. In the Orthogonal Diagonalization Method, P is the transition matrix from an ordered orthonormal basis of eigenvectors for A to standard coordinates. (We can assume we are using standard coordinates here because the problem refers only to a matrix rather than a linear transformation.)

Chapter 6 Review Exercises (1) (a) Let v1 = [1, 3, −2], v2 = [−1, 1,1], v3 = [5, 1, 4]. Then v1 · v2 = 0, v1 · v3 = 0, and v2 · v3 = 0, so (v·v1 ) (v·v2 ) (v·v3 ) 12 84 B is orthogonal. Then [v]B = ||v1 ||2 , ||v2 ||2 , ||v3 ||2 = −14 14 , 3 , 42 = [−1, 4, 2]. (2) (a) First, let v1 = [1, −1, −1, 1].

[5,1,1,5]·[1,−1,−1,1] Next, v2 = [5, 1, 1, 5]− [1,−1,−1,1]·[1,−1,−1,1] [1, −1, −1, 1] = [5, 1, 1, 5]− 84 [1, −1, −1, 1] = [3, 3, 3, 3]. We divide v2 by 3 to simplify its form, yielding v2 = [1, 1, 1, 1]. Hence, an orthogonal basis for the subspace is {[1, −1, −1, 1], [1, 1, 1, 1]}. 200


Chap 6 Review

(3) First, we must find an ordinary basis for R3 containing [6, 3, −6] and [3, 6, 6]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the given set of vectors, and use the Independence Test Method. Hence, we row reduce ⎡ ⎤ 1 ⎡ ⎤ 1 0 0 − 19 9 6 3 1 0 0 ⎢ ⎥ 1 ⎥, 1 ⎣ 3 6 0 1 0 ⎦ to ⎢ 0 1 0 ⎣ 9 18 ⎦ −6 6 0 0 1 1 0 0 1 −1 2 giving the basis {[6, 3, −6], [3, 6, 6], [1, 0, 0]}. We perform the Gram-Schmidt Process on this set. Let v1 = [6, 3, −6]. Since the given set of vectors is orthogonal, we let v2 = [3, 6, 6]. (If the two given vectors were not we would apply Gram-Schmidt to obtain v2 .) Finally,

need to

already orthogonal, [1,0,0]·[6,3,−6] [1,0,0]·[3,6,6] v3 = [1, 0, 0] − [6,3,−6]·[6,3,−6] [6, 3, −6] − [3,6,6]·[3,6,6] [3, 6, 6] 6 3 = [1, 0, 0] − 81 [6, 3, −6] − 81 [3, 6, 6] = 49 , − 49 , 29 . Multiplying by 92 to simplify the form of the vector gives v3 = [2, −2, 1]. Hence, an orthogonal basis for R3 containing the two given vectors is {[6, 3, −6], [3, 6, 6], [2, −2, 1]}. (6) (e) In Exercise 7 of Section 6.1, we saw that [−1, 3, 3] is an eigenvector for the matrix A corresponding to the eigenvalue λ = 1. Therefore, the axis of rotation is in the direction of the vector [−1, 3, 3]. First, we enlarge the set {[−1, 3, 3]} to an orthogonal basis for R3 . To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for R3 to the given vector and use the Independence Test Method. Hence, we row reduce ⎤ ⎡ 1 ⎡ ⎤ 1 0 0 3 −1 1 0 0 ⎥ ⎢ 1 ⎥, giving the basis {[−1, 3, 3], [1, 0, 0], [0, 1, 0]}. We per⎣ 3 0 1 0 ⎦ to ⎢ 0 1 0 ⎣ 3 ⎦ 3 0 0 1 0 0 1 −1 form the Gram-Schmidt Process on this set.

[1,0,0]·[−1,3,3] Let v1 = [−1, 3, 3]. Then v2 = [1, 0, 0] − [−1,3,3]·[−1,3,3] [−1, 3, 3] = [1, 0, 0] − (−1) 19 [−1, 3, 3] = 18 3 3 19 by 3 to eliminate fractions yields 19 , 19 , 19 . Multiplying

v2 = [6, 1, 1]. Finally, [0,1,0]·[−1,3,3] [0,1,0]·[6,1,1] v3 = [0, 1, 0] − [−1,3,3]·[−1,3,3] [−1, 3, 3] − [6,1,1]·[6,1,1] [6, 1, 1] 3 1 = [0, 1, 0] − 19 [−1, 3, 3] − 38 [6, 1, 1] = 0, 12 , − 12 . Multiplying by 2 to simplify the form of the vector gives us v3 = [0, 1, −1]. Hence, an orthogonal basis for R3 containing [−1, 3, 3] is {[−1, 3, 3], [6, 1, 1], [0, 1, −1]}. 3 Next, we normalize the vectors in this 4 orthogonal basis to get the orthonormal basis √1 [−1, 3, 3], √1 [6, 1, 1], √1 [0, 1, −1] . We create the orthogonal matrix 19⎡ 38 ⎤2 √6 √1 0 − 19 ⎢ 138 ⎥ 1 3 ⎢ ⎥, whose columns are the vectors in this orthonormal basis, using the √ √ √ Q = ⎣ 38 2 19 ⎦ √1 √3 − √12 38 19 T first ⎡ vector as the last column. ⎤Computing Q AQ yields ⎡ (approximately) ⎤ 0.1363636 0.9906589 0 cos θ −sin θ 0 ⎣ −0.9906589 0.1363636 0 ⎦, which is of the form ⎣ sin θ cos θ 0 ⎦. Since cos θ is posi0 0 1 0 0 1 tive and sin θ is negative, the angle θ is in the fourth quadrant. Hence, cos θ = 0.1363636 implies that θ ≈ 278◦ . This is a counterclockwise rotation about the vector [−1, 3, 3] as you look down from the point (−1, 3, 3) toward the plane through the origin spanned by [6, 1, 1] and [0, 1, −1]. 201


Chap 6 Review

(7) (a) First, we use the Gram-Schmidt Process on the basis {[8, 1, −4], [16, 11, −26]} for W to find the following orthogonal basis for W: {[8, 8, −14]}. Normalizing this orthogonal basis for , 1 1, −4], [−8, 1 [8, 1, −4], [−4, 4, −7] . Hence, by definition, W produces the orthonormal basis 9 9 1 w1 = projW v = [2, 7, 26] · 19 [8, 1, −4] [8, 1, −4] 9 1 [−4, 4, −7] + [2, 7, 26] · 19 [−4, 4, −7] 9 =

1 81 (−81)[8, 1, −4]

+

1 81 (−162)[−4, 4, −7]

= [0, −9, 18].

Finally, w2 = projW ⊥ v = v − projW v =[2, 7, 26] − [0, −9, 18] = [2, 16, 8]. (8) (a) Let v = [1, 29, −29, −2]. By Theorem 6.17, the minimum distance from P to W is v − projW v. Thus, we need to compute projW v. To do this, we need an orthonormal basis for W. We begin with the given basis {[2, 9, −6, 0], [2, 5, −12, 12]} for W and apply the Gram-Schmidt Process to obtain {[2,,9, −6, 0], [0, −4, −6, 12]}. Normalizing the vectors in this basis produces the orthonor1 [2, 9, −6, 0], 17 [0,−2, −3, 6] for W. Thus, by definition, mal basis 11 1 1 9, −6, 0] −6, 0] proj 11 [2, 9, W v = [1, 29, −29, 1 −2] · 11 [2, 1 + [1, 29, −29, −2] · 7 [0, −2, −3, 6] 7 [0, −2, −3, 6] =

1 121 (437)[2, 9, −6, 0]

+

1 49 (17)[0, −2, −3, 6]

≈ [7.223140, 31.810255, −22.710238, 2.081633].

Hence, the minimum distance from P to W is v − projW v ≈ [1, 29, −29, −2] − [7.223140, 31.810255, −22.710238, 2.081633] ≈ [−6.223140, −2.810255, −6.289762, −4.081633] ≈ 10.141294. (11) In this problem, we utilize the ideas presented in Example 8 of Section 6.2 in the textbook. Let W = {[x, y, z] | 2x−3y+z = 0} = {[x, y, z] | [x, y, z]·[2, −3, 1] = 0}. Then, if Y = span({[2, −3, 1]}), W = Y ⊥ , and, by Corollary 6.14, W ⊥ = Y. So, {[2, −3, 1]} is a basis for W ⊥ . Also, W can be expressed as {[x, y, −2x+3y]} = {x[1, 0, −2]+y[0, 1, 3]}, and so {[1, 0, −2], [0, 1, 3]} is a basis for W. Now W = E1 , the eigenspace corresponding to the eigenvalue λ1 = 1, for L, since the plane W is left fixed by L. Also, W ⊥ = E−1 , the eigenspace corresponding to the eigenvalue λ2 = −1 for L, since all vectors perpendicular to W are sent to their negatives by L because they are reflected through W. So, we know by the Generalized Diagonalization Method of Section 5.6 that if A is the matrix for L with respect to the standard basis for R3 , then D = P−1 AP, where P is a matrix whose columns are a basis for R3 consisting of eigenvectors for L, and D is a diagonal matrix with the eigenvalues for L on ⎡ ⎤ −1 0 0 its main diagonal. Note that D = P−1 AP implies that A = PDP−1 . Setting D = ⎣ 0 1 0 ⎦ 0 0 1 ⎡ ⎤ 2 1 0 0 1 ⎦ . Row reduction and, using the bases for W ⊥ = E−1 and W = E1 from above, P = ⎣ −3 1 −2 3 ⎡ ⎤ ⎡ ⎤ 10 −15 5 3 6 −2 1 ⎣ 14 0 −28 ⎦. Hence, A = PDP−1 = 17 ⎣ 6 −2 3 ⎦. yields P−1 = 70 6 5 3 −2 3 6 (12) (b) The following argument works for any orthogonal projection onto a line through the origin in R3 : An orthogonal projection onto a line in R3 has two eigenvalues, λ1 = 1 and λ2 = 0. The eigenspace E1 is the given line and thus is one-dimensional. Hence, the geometric multiplicity of λ1 is 1. The eigenspace E0 is the orthogonal complement of the line, since the vectors perpendicular to the line are sent to 0 by L. By Corollary 6.13, dim(E0 ) = 3 − dim(E1 ) = 2. Hence, the geometric multiplicity of λ2 is 2. Since the geometric multiplicities of the eigenvalues add up 202


Chap 6 Review

to 3, L is diagonalizable, and each geometric multiplicity must equal its corresponding algebraic multiplicity. Therefore, the factor (x−1) appears in pL (x) raised to the first power, and the factor of x appears in pL (x) raised to the second power. Hence, pL (x) = x2 (x − 1) = x3 − x2 . (13) (a) L([1, 0, 0, 0]) = [0, 3, 0, 0], L([0, 1, 0, 0]) = [0, 0, 2, 0], L([0, 0, 1, 0]) = [0, 0, 0, ⎡ 1], and L([0, 0, ⎤ 0, 1]) = 0 0 0 0 ⎢ 3 0 0 0 ⎥ ⎥ [0, 0, 0, 0]. Thus, the matrix for L with respect to the standard basis is ⎢ ⎣ 0 2 0 0 ⎦ , which 0 0 1 0 is not symmetric, and so L is not a symmetric operator. " " " x + 17 − 13 − 16 "" 30 15 " " " x − 11 − 13 " = x3 −2x2 −x+2 = (x−2)(x−1)(x+1), giving us (14) (a) Step 2(a): pA (x) = " − 13 15 15 " " " −1 −1 x − 11 " 6

3

6

eigenvalues λ1 = 1, λ2 = 2, and λ3 = −1. Using row reduction to find particular solutions to the systems [1I3 − A|0], [2I3 − A|0], and [(−1)I3 − A|0] as in the Diagonalization Method of Section 3.4 produces the bases {[−1, −2, 1]}, {[1, 2, 5]}, and {[−2, 1, 0]} for E1 , E2 , and E−1 , respectively. Step 2(b): Since all three bases are one-dimensional, they are already orthogonal. 3 To make each 4 an orthonormal basis, we need to normalize each vector, yielding the bases √16 [−1, −2, 1] , 4 3 4 3 √1 [1, 2, 5] , and √1 [−2, 1, 0] for E1 , E2 , and E−1 , respectively. 30 5

1 √ √1 [1, 2, 5], √1 [−2, 1, 0] . [−1, −2, 1], Step 2(c): Z = 6 30 5

3 Step 3: Since C is the standard basis for R , B = Z = √16 [−1, −2, 1], √130 [1, 2, 5], √15 [−2, 1, 0] . Now P is the matrix whose columns are the vectors in Z, and D is the diagonal matrix whose main diagonal entries are the eigenvalues 1, 2, and −1. Hence, ⎤ ⎡ ⎡ ⎤ − √16 √130 − √25 1 0 0 ⎢ ⎥ √2 √2 √1 ⎥ and D = ⎣ 0 2 0 ⎦. P=⎢ ⎣ − 6 30 5 ⎦ 0 0 −1 √1 √5 0 6 30 Now P−1 = PT , and so D = PT AP, which is verified by performing the matrix product. (17) (a) True. This is Theorem 6.1 and was proven as Result 7 in Section 1.3. (b) True. This is the step “Let v1 = w1 ” at the beginning of the formal Gram-Schmidt Process described in Section 6.1. You may end up with a scalar multiple of w1 if you decide to multiply by a scalar to put the vector in a simpler form. , . . . , wk−1 }) = span({v 1 , . . . , vk−1 }). Thus, Theorem (c) True. Note that w k ∈ span({w 1

6.3 shows wk ·vk−1 wk ·v1 wk ·v2 k ·v1 that wk = v1 ·v1 v1 + v2 ·v2 v2 + · · · + vk−1 ·vk−1 vk−1 , and so vk = wk − w v1 ·v1 v1 −

wk ·vk−1 wk ·v2 v2 ·v2 v2 − · · · − vk−1 ·vk−1 vk−1 is the zero vector. ( ' 3 4 ( ' 3 − 45 1 0 5 5 5 = as required by the definition of an orthog(d) True. Note that 3 4 0 1 − 45 53 5 5 onal matrix. 1 0 0 (e) False. A must also be square. The matrix A = is a counterexample. 0 1 0

203


Chap 6 Review

(f) False. The columns of an orthogonal matrix must be unit vectors. The matrix A =

2 0

0 3

is

a counterexample. (g) True. By definition, an orthogonal matrix has its transpose as its inverse. (h) True. v⊥w =⇒ v · w = 0. Theorem 6.9 then implies that 0 = v · w = Av · Aw = L(v) · L(w), and so L(v) ⊥ L(w). (i) True. Theorem 6.11 assures us that W ⊥ is a subspace of Rn for any subspace W of Rn . (j) True. Theorem 6.16 implies this since the statement is just one of the two properties of a linear transformation for the function L(v) = projW v. (k) True. First, the uniqueness assertion of the Projection Theorem guarantees that if w ∈ W, then projW w = w. Thus, W ⊆ range(L). But the definition of the projection shows that projW v ∈ W for all v ∈ Rn , and so range(L) ⊆ W. Hence, range(L) = W. (l) True. If w ∈ W, then projW ⊥ w = w − projW w = w − w = 0, and so w ∈ ker(L). This shows that W ⊆ ker(L). Next, if w ∈ ker(L), then 0 = projW ⊥ w = w − projW w. Therefore, w = projW w. But projW w ∈ W, and so w ∈ W. This shows that ker(L) ⊆ W. Hence, ker(L) = W. (m) False. The kernel of the projection operator is W ⊥ (see part (l), switching the roles of W and W ⊥ ). So, whenever W ⊥ is nontrivial, the projection operator has a nontrivial kernel, and so its matrix with respect to any basis is singular. Thus, the matrix cannot be orthogonal (see part (g)). For a specific counterexample, consider the subspace W = span({i}) in R2 . The matrix 1 0 for the orthogonal projection onto W is A = , which is not an orthogonal matrix, since 0 0 AAT = A = I2 . (n) True. Note that range(L) = W, and if w ∈ W, then projW w = w (see part (k)). Thus, for every v ∈ Rn , L(v) ∈ W and L(L(v)) = L(v). Hence, L ◦ L = L. (o) True. The eigenvalues are ±1. See Example 10 in Section 6.2 and the discussion that appears just before it. (p) True. Theorem 6.17 shows that the minimal distance is v − projW v. However, the comment appearing just after Example 6 in Section 6.2 shows that projW ⊥ v = v − projW v. (q) True. If v1 , v2 ∈ Rn , then L(v1 ) · v2 = (projW v1 ) · (projW v2 + projW ⊥ v2 ) = (projW v1 ) · (projW v2 ) + (projW v1 ) · (projW ⊥ v2 ) = (projW v1 ) · (projW v2 ) + 0 = (projW v1 ) · (projW v2 ) + (projW ⊥ v1 ) · (projW v2 ) = (projW v1 + projW ⊥ v1 ) · (projW v2 ) = v1 · L(v2 ). Hence, L is a symmetric operator by definition. As an illustration, note that the matrix for the orthogonal projection operator in Example 8 of Section 6.2 with respect to the standard basis is symmetric. Note: We could look at this problem from a different perspective. If B = {u1 , . . . , un } is an orthonormal basis for Rn , with {u1 , . . . , uk } being an orthonormal basis for W, then the matrix for L with respect to B is the diagonal matrix with k 1’s and (n − k) 0’s on its main diagonal. Therefore, L is a symmetric operator by Theorem 6.18. (r) False. By Exercise 9 of Section 6.3, the composition of symmetric operators is symmetric if and only if the operators commute. For a specific counterexample, consider L1 and L2 on the operators 1 −1 3 2 2 and . R whose respective matrices with respect to the standard basis are −1 2 2 −1 3 2 1 −1 1 1 = , The matrix for L2 ◦L1 with respect to the standard basis is 2 −1 −1 2 3 −4 which is not symmetric. Therefore, L2 ◦ L1 is not a symmetric operator. 204


Chap 6 Review

(s) False. The basis B must be an orthonormal basis of eigenvectors, not just any orthonormal basis. 1 2 2 For a specific counterexample, consider the symmetric operator on R with matrix A = 2 3 with respect to the standard basis. Of course, the standard basis is an orthonormal basis for R2 . The matrix for L with respect to this basis is just A, which is not diagonal. (t) True. By Lemma 6.19, the matrix has at least one eigenvalue. Since the matrix is orthogonally diagonalizable, every eigenvalue must have its geometric and algebraic multiplicities equal to each other. (u) True. This is the matrix version of part of Theorem 6.20. (v) False. A symmetric operator must be orthogonally diagonalizable. So, any orthogonal matrix that is not diagonalizable cannot be the matrix for a symmetric operator. For a specific counterexample, consider a counterclockwise rotation about the origin in R2 through an angle θ, where ◦ θ is not a multiple of 180 . Its matrix with respect to the standard basis is the orthogonal matrix cos θ −sin θ , which is not diagonalizable, and is not a symmetric matrix. Such rotations are sin θ cos θ not symmetric operators. (w) True. This is the matrix version of Exercise 11 in Section 6.3.

205


Section 7.1

Chapter 7 Section 7.1 (1) For extra help with adding and multiplying complex numbers, see Appendix C. (a) [2 + i, 3, −i] + [−1 + 3i, −2 + i, 6] = [2 + i − 1 + 3i, 3 − 2 + i, −i + 6] = [1 + 4i, 1 + i, 6 − i] (b) (−8 + 3i)[4i, 2 − 3i, −7 + i] = [(−8 + 3i)(4i), (−8 + 3i)(2 − 3i), (−8 + 3i)(−7 + i)] = [−12 − 32i, −7 + 30i, 53 − 29i] (d) (−4)[6 − 3i, 7 − 2i, −8i] = [−24 + 12i, −28 + 8i, 32i] = −24 + 12i, −28 + 8i, 32i = [−24 − 12i, −28 − 8i, −32i] (e) [−2 + i, 5 − 2i, 3 + 4i] · [1 + i, 4 − 3i, −6i] = (−2 + i) 1 + i + (5 − 2i) 4 − 3i + (3 + 4i) −6i = (−2 + i)(1 − i) + (5 − 2i)(4 + 3i) + (3 + 4i)(6i) = (−1 + 3i) + (26 + 7i) + (−24 + 18i) = 1 + 28i (3) For extra help with adding and multiplying complex numbers, see Appendix C. 2 + 5i −4 + i 9−i −3i (2 + 5i) + (9 − i) (−4 + i) + (−3i) (a) + = −3 − 6i 8 − 3i 5 + 2i 4 + 3i (−3 − 6i) + (5 + 2i) (8 − 3i) + (4 + 3i) 11 + 4i −4 − 2i = 2 − 4i 12 ⎤T ⎡ ⎡ ⎤T 1 + i −2i 6 + 4i 1−i 2i 6 − 4i T ⎦ =⎣ 0 ⎦ 3−i 5 3+i 5 (c) C∗ = C = ⎣ 0 10i 0 7 + 3i −10i 0 7 − 3i ⎡ ⎤ 1−i 0 10i ⎦ 3−i 0 = ⎣ 2i 6 − 4i 5 7 + 3i (−3i)(5 − i) (−3i)(−i) (−3i)(−3) −3 − 15i −3 9i (d) (−3i)D = = (−3i)(2 + 3i) (−3i)(0) (−3i)(−4 + i) 9 − 6i 0 3 + 12i 2 + 5i −4 + i 9−i −3i (f) = −3 − 6i 8 − 3i 5 + 2i 4 + 3i (2 + 5i)(9 − i) + (−4 + i)(5 + 2i) (2 + 5i)(−3i) + (−4 + i)(4 + 3i) (−3 − 6i)(9 − i) + (8 − 3i)(5 + 2i) (−3 − 6i)(−3i) + (8 − 3i)(4 + 3i) 1 + 40i −4 − 14i (23 + 43i) + (−22 − 3i) (15 − 6i) + (−19 − 8i) = = 13 − 50i 23 + 21i (−33 − 51i) + (46 + i) (−18 + 9i) + (41 + 12i) ⎤ ⎡ T 5 − i 2 + 3i T 5 − i −i −3 ⎦ 0 (i) CT D∗ = CT D = CT = CT ⎣ −i 2 + 3i 0 −4 + i −3 −4 + i ⎡ ⎤⎡ ⎤ 1+i 0 −10i 5 + i 2 − 3i ⎦⎣ i ⎦= 0 0 = ⎣ −2i 3 + i 6 + 4i 5 7 − 3i −3 −4 − i ⎡ ⎤ (1 + i)(5 + i) + 0(i) + (−10i)(−3) (1 + i)(2 − 3i) + 0(0) + (−10i)(−4 − i) ⎣ (−2i)(5 + i) + (3 + i)(i) + 0(−3) (−2i)(2 − 3i) + (3 + i)(0) + 0(−4 − i) ⎦ = (6 + 4i)(5 + i) + 5(i) + (7 − 3i)(−3) (6 + 4i)(2 − 3i) + 5(0) + (7 − 3i)(−4 − i) 206


(4 + 6i) + 0 + 30i ⎣ (2 − 10i) + (−1 + 3i) + 0 (26 + 26i) + (5i) + (−21 + 9i)

Section 7.1

⎤ ⎡ ⎤ (5 − i) + 0 + (−10 + 40i) 4 + 36i −5 + 39i ⎦ = ⎣ 1 − 7i −6 − 4i ⎦ (−6 − 4i) + 0 + 0 (24 − 10i) + 0 + (−31 + 5i) 5 + 40i −7 − 5i

(4) (b) Let Z be an m × n complex matrix and let W be an n × p matrix. Note that (ZW)∗ and W∗ Z∗ are both p × m matrices. We present two methods of proof that (ZW)∗ = W∗ Z∗ . First method: Notice that the rule (AB)T = BT AT holds for complex matrices, since matrix multiplication for complex matrices is defined in exactly the same manner as for real matrices, and hence the proof of Theorem 1.16 is valid for complex matrices as well. Thus, we have (ZW)∗ = (ZW)T = WT ZT = (WT )(ZT ) (by part (4) of Theorem 7.2) = W∗ Z∗ . Second method: We begin by computing the (i, j) entry of (ZW)∗ . Now, the (j, i) entry of ZW = zj1 w1i + · · · + zjn wni . Hence, the (i, j) entry of (ZW)∗ = ((j, i) entry of ZW) = (zj1 w1i + · · · + zjn wni ) = zj1 w1i + · · · + zjn wni . Next, we compute the (i, j) entry of W∗ Z∗ to show that it equals the (i, j) entry of (ZW)∗ . Let A = Z∗ , an n × m matrix, and B = W∗ , a p × n matrix. Then akl = zlk and bkl = wlk . Hence, the (i, j) entry of W∗ Z∗ = the (i, j) entry of BA = bi1 a1j + · · · + bin anj = w1i zj1 + · · · + wni zjn = zj1 w1i + · · · + zjn wni . Therefore, (ZW)∗ = W∗ Z∗ . (5) In each part, let A represent the given matrix. (a) The matrix A is skew-Hermitian because all of its main diagonal entries are pure imaginary (their real parts all equal zero) and, for i = j, aij = −aji . (b) The matrix A is not Hermitian because a11 is not a real number. Also, A is not skew-Hermitian because a11 is not a pure imaginary number. (c) The matrix A is Hermitian because all of its main diagonal entries are real and since all other entries are zero (and hence, for i = j, aij = aji ). (d) The matrix A is skew-Hermitian because all of its main diagonal entries are pure imaginary and since all other entries are zero (and hence, for i = j, aij = −aji ). (e) The matrix A is Hermitian because all of its main diagonal entries are real and, for i = j, aij = −aji . (9) Let Z be Hermitian. Then, since Z∗ = Z, ZZ∗ = ZZ = Z∗ Z. Similarly, if Z is skew-Hermitian, Z∗ = −Z. Therefore, ZZ∗ = Z(−Z) = −(ZZ) = (−Z)Z = Z∗ Z. (11) (a) False. Consider that [i, −1] · [1, i] = i 1 + (−1) i = i(1) + (−1)(−i) = 2i. (b) False. When computing the complex dot product, we find the complex conjugates of the entries in the second vector before multiplying corresponding coordinates. However, this is not done when computing each entry of the matrix product of complex matrices. For a specific example, consider the matrices Z and W in Example 2 in Section 7.1. In that example, the (1,1) entry of ZW is −5 − 8i. However, the complex dot product of the first row of Z with the first column of W is (1 − i) (−2i) + (2i) (−1 + 3i) + (−2 + i) (−2 + i) = (1 − i) (2i) + (2i) (−1 − 3i) + (−2 + i) (−2 − i) = (2 + 2i) + (6 − 2i) + 5 = 13. (c) True. Both are equal to the conjugate transpose of Z. This is stated in Section 7.1 just before Example 3. (d) False. By part (5) of Theorem 7.1, v1 · (kv2 ) = k(v1 · v2 ), which does not necessarily equal k(v1 · v2 ) when k is complex. For example, [1, 1] · (i[1, 1]) = [1, 1] · [i, i] = 1( i ) + 1( i ) = −2i, but i([1, 1] · [1, 1]) = i(1( 1 ) + 1( 1 )) = i(2) = 2i. 207


Section 7.2

(e) False. The matrix H in Example 4 in Section 7.1 is Hermitian but not symmetric. In fact, a Hermitian matrix is symmetric if and only if all of its entries are real. (f) True. In fact, the transpose of a skew-Hermitian matrix K is skew-Hermitian, since

T T ∗ T T K = (KT ) = (K∗ ) = (−K) = − KT . Hence, KT is normal by Theorem 7.4.

Section 7.2 (1) Remember: If z is a complex number, z1 = |z|z 2 . For Appendix C. 3 + i 5 + 5i (a) Start with the augmented matrix 1 + i 6 − 2i

further help with complex arithmetic, review " " 29 + 33i " " 30 − 12i .

Pivoting in the first column, we perform the following operations: 3 1 1 1 ← 3+i 1 , which is the same as 1 ← 10 − 10 i 1 2 ← −(1 + i) 1 + 2 . This results in the matrix " 1 2 + i "" 12 + 7i . Next, we pivot in column 2: 0 5 − 5i " 25 − 31i 1 1 1 2 ← 5−5i 2 , which is the same as 2 ← 10 + 10 i 2 " 1 2 + i "" 12 + 7i 3 This gives . Hence, z = 28 3 5 − 5 i. 0 1 " 28 − i 5 5 1 3 Back substitution yields w = (12 + 7i) − (2 + i) 28 5 − 5i = 5 + ⎡ 3i −6 + 3i 12 + 18i 25 − 2i (c) Begin with the augmented matrix ⎣ 3 + 2i 1 + 7i 1+i 2i 9+i ⎡ ⎤ 1 0 4 − 3i 2 + 5i −i 5 + 2i ⎦. we row reduce to ⎣ 0 1 0 0 0 0 Thus, the solution set is {( (2 + 5i) − (4 − 3i)c, (5 + 2i) + ic, (e) After we apply the row reduction method to the first column, " 3 − 2i 12 + 5i "" 3 + 11i 1 2 + 3i −1 + 3i to 5 + 4i −2 + 23i " −14 + 15i 0 0 3 + 4i

13 i. "5 ⎤ " −51 + 9i " " −13 + 56i ⎦ , " " −7 + 17i

c) | c ∈ C}. the matrix changes from .

The second row shows that the system has no solutions. (2) (b) Using a cofactor expansion on the row yields

first

|A| = i (1 − i)(2 − i) − (i)(−2) − (2) (1 + i)(2 − i) − (i)(4) + (5i) (1 + i)(−2) − (1 − i)(4) = i(1 − i) − (2)(3 − 3i) + (5i)(−6 + 2i) = −15 − 23i. Hence, |A| = −15 + 23i. " " " −i 1 − i 4 "" " 1 + i −2 "". Next, |A∗ | = "" 2 " −5i −i 2 + i " Again, using a cofactor expansion

on the first row, this determinant

equals

(−i) (1 + i)(2 + i) − (−2)(−i) − (1 − i) (2)(2 + i) − (−2)(−5i) + 4 (2)(−i) − (1 + i)(−5i) = (−i)(1 + i) − (1 − i)(4 − 8i) + 4(−5 + 3i) = −15 + 23i. 208


Section 7.2

(3) In each part, let A represent the given matrix. Your computations may produce complex scalar multiples of the vectors given here, although they might not be immediately recognized as such. " " " " x − (4 + 3i) 1 + 3i " = (x − (4 + 3i)) (x + (5 + 2i)) − (−8 + 2i)(1 + 3i) = " (a) pA (x) = " −8 + 2i x + (5 + 2i) " x2 + (1 − i)x − i = (x − i)(x + 1). Hence, the eigenvalues are λ1 = i and λ2 = −1. Now we solve for corresponding fundamental eigenvectors. " ' ( " 1 − 12 − 12 i "" 0 −4 − 2i 1 + 3i "" 0 to . For λ1 = i, we row reduce [iI2 − A|0] = " −8 + 2i 5 + 3i " 0 " 0 0 0 Setting the independent variable z2 in the associated system equal to 2 to eliminate fractions yields the fundamental eigenvector [1 + i, 2]. Hence, Ei = {c[1 + i, 2] | c ∈ C}. " ( " ' 7 6 1 − 17 − 17 i "" 0 −5 − 3i 1 + 3i "" 0 to For λ2 = −1, we row reduce [(−1)I2 −A|0] = . " −8 + 2i 4 + 2i " 0 " 0 0 0 Setting the independent variable z2 in the associated system equal to 17 to eliminate fractions yields the fundamental eigenvector [7 + 6i, 17]. Hence, E−1 = {c[7 + 6i, 17] | c ∈ C}. (c) Using cofactor expansion along the first row yields " " " x − (4 + 3i) 4 + 2i −4 − 7i "" " x − (−2 + 5i) −7 + 4i "" pA (x) = "" −2 + 4i " 4 + 2i −4 − 2i x + (4 + 6i) "

= (x − (4 + 3i)) (x − (−2 + 5i))(x + (4 + 6i)) − (−7 + 4i)(−4 − 2i)

−(4 + 2i) (−2 + 4i)(x + (4 + 6i)) − (−7 + 4i)(4 + 2i)

+(−4 − 7i) (−2 + 4i)(−4 − 2i) − (x − (−2 + 5i))(4 + 2i) = (x − (4 + 3i))(x2 + (6 + i)x + (2 − 6i)) −(4 + 2i)((−2 + 4i)x + (4 + 2i)) + (−4 − 7i)((−4 − 2i)x + (−2 + 4i)) = x3 + (2 − 2i)x2 + (−19 − 28i)x + (−26 + 18i) − ((−16 + 12i)x + (12 + 16i)) + ((2 + 36i)x + (36 − 2i)) = x3 +(2−2i)x2 +(−1−4i)x−2 = (x−i)2 (x+2). (A computer algebra system would help here.) Hence, the eigenvalues are λ1 = i and λ2 = −2. Now we solve for corresponding fundamental eigenvectors. For λ1 = i, we row reduce [iI3 − A|0] = " ⎡ ⎤ ⎡ 1 −1 −4 − 2i 4 + 2i −4 − 7i "" 0 ⎣ −2 + 4i 2 − 4i −7 + 4i " 0 ⎦ to ⎣ 0 0 " 4 + 2i −4 − 2i 4 + 7i " 0 0 0

3 2

" ⎤ +i " 0 " " 0 ⎦. 0 " " 0 0

Setting the independent variables z2 and z3 in the associated system equal to 1 and 0, respectively, yields the fundamental eigenvector [1, 1, 0]. Then setting the independent variables z2 and z3 in the associated system equal to 0 and 2 (to eliminate fractions), respectively, gives the fundamental eigenvector [−3 − 2i, 0, 2]. For λ2 = −2, we row reduce [(−2)I3 − A|0] = " ⎤ ⎡ ⎡ 1 0 −6 − 3i 4 + 2i −4 − 7i "" 0 ⎣ −2 + 4i −5i −7 + 4i "" 0 ⎦ to ⎣ 0 1 0 0 4 + 2i −4 − 2i 2 + 6i " 0 209

1 −i 0

" ⎤ " 0 " " 0 ⎦. " " 0


Section 7.3

Setting the independent variable z3 in the associated system equal to 1 yields the fundamental eigenvector [−1, i, 1]. Hence, Ei = {c[1, 1, 0] + d[(−3 − 2i), 0, 2] | c, d ∈ C} and E−2 = {c[−1, i, 1] | c ∈ C}. (4) (a) We use the solution to Exercise 3(a), above. Let A be the matrix given in Exercise 3(a). A is diagonalizable since two fundamental eigenvectors were found. The matrix P is the matrix whose columns are the fundamental eigenvectors we computed. The matrix D is the diagonal matrix whose main diagonal entries are the eigenvalues i and −1. Hence, 17 −7 − 6i 1 + i 7 + 6i i 0 1 −1 , by using P= and D = . Now P = 3+5i −2 1+i 2 17 0 −1 Theorem 2.13. Using

1 3+5i

=

1 34 (3

− 5i), we can verify that D = P−1 AP.

(6) (a) True. Performing the row operation 2 ← −i 1 + 2 on A produces a matrix B whose second row is zero. Hence, |B| = 0 (perform a cofactor expansion along its second row). But since B was obtained from A by performing a single type (II) row operation, |A| = |B|. Therefore, |A| = 0. (b) False. Example 2 in Section 7.2 of the textbook illustrates a 3 × 3 matrix A having eigenvalue 3 having algebraic multiplicity 1. What is true, however, is that the sum of all the algebraic multiplicities of all the eigenvalues of a complex n × n matrix must equal n. (c) False. If the number of fundamental eigenvectors for some eigenvalue does not equal its algebraic multiplicity, the matrix is still not diagonalizable when thought of as a complex matrix. For example, the real matrix given in Example 7 of Section 3.4 has −2 as an eigenvalue with algebraic multiplicity 2, but only one fundamental eigenvector is produced for this eigenvalue. Hence, this matrix is not diagonalizable, even when thought of as a complex matrix. (d) False. The Fundamental Theorem of Algebra guarantees only that an nth degree complex polynomial factors into n linear factors. However, if these factors are not different, the polynomial will not have n distinct roots. For example, x2 − 2ix − 1 = (x − i)2 has only one root, namely i.

Section 7.3 (2) In each part, let S represent the given set of vectors. (b) The set S is not linearly independent because [−3 + 6i, 3, 9i] = 3i[2 + i, −i, 3]. Thus, {[2 + i, −i, 3]} is a basis for span(S), and so dim(span(S)) = 1. ⎡ ⎤ 1 0 i (d) The matrix formed using the vectors in S as columns row reduces to ⎣ 0 1 −2i ⎦. Therefore, 0 0 0 by the Independence Test Method, {[3 − i, 1 + 2i, −i], [1 + i, −2, 4 + i]} is a basis for span(S). Hence, dim(span(S)) = 2. (3) In each part, let S represent the given set of vectors. (b) Since [−3 + 6i, 3, 9i] is not a real scalar multiple of [2 + i, −i, 3], S is linearly independent. (Consider the 3rd coordinate: 9i = 3c for any c ∈ R.) Hence, S is a basis for span(S), and dim(span(S)) = 2. (d) Separating the real and imaginary parts of each vector in S, we can think of them as vectors in R6 . That is, we rewrite each complex vector [a + bi, c + di, e + f i] as [a, b, c, d, e, f ]. We then create a matrix using each 6-vector as a column. This gives

210

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎤ ⎡ 3 1 3 1 ⎢ 0 −1 1 1 ⎥ ⎥ ⎢ ⎢ 1 −2 −2 ⎥ ⎥, which row reduces to ⎢ 0 ⎥ ⎢ 0 2 0 5 ⎥ ⎢ ⎣ 0 0 4 3 ⎦ −1 1 −8 0

0 1 0 0 0 0

0 0 1 0 0 0

Section 7.3

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

The Independence Test Method then shows that S is linearly independent. for span(S), and dim(span(S)) = 3. " ⎤ ⎡ ⎡ 1 2i 3 + i −3 + 5i "" 3 − i " −5 − 5i ⎦ row reduces to ⎣ 0 2i (4) (b) The matrix ⎣ −1 + 3i −2 " 0 4 1 − i −5 + 3i " 7 + i

Therefore, S is a basis 0 1 0

0 0 1

" ⎤ " i " " 1 + i ⎦. " " −1

Hence, [z]B = [i, 1 + i, −1]. (5) With C2 considered as a real vector space, we use the ordered basis B = ([1, 0], [i, 0], [0, 1], [0, i]). Now [L([1, 0])]B = [0, 1]B = [0, 0, 1, 0], [L([i, 0])]B = [0, −i]B = [0, 0, 0, −1], [L([0, 1])]B = [1, 0]B = [1, 0, 0, 0], ⎡ 0 0 ⎢ 0 0 ⎢ ⎣ 1 0 0 −1

and [L([0, i])]B = [−i, 0]B = [0, −1, 0, 0]. Therefore, the matrix for L with respect to B is ⎤ 1 0 0 −1 ⎥ ⎥. 0 0 ⎦ 0 0

(8) We present two methods for solving this problem: Method 1: Let B be the standard basis for C2 , C be the basis {[1 + i, −1 + 3i], [1 − i, 1 + 2i]} for C2 , ⎡ ⎤ 3−i 2+i 1 − 3i ⎦. and D be the standard basis for C3 . Then, by the given values for L on C, ACD = ⎣ 5 −i 3 Next, we compute the transition matrix P from B to C. Since B is the standard basis, we can do this 1 + 2i −1 + i 1+i 1−i −3+i . by computing the inverse of . Theorem 2.13 yields P = 10 1 − 3i 1 + i −1 + 3i 1 + 2i ⎡ ⎤ −3 + i − 25 − 11 5 i ⎢ 1 3 ⎥ ⎥. −i Finally, by Theorem 5.6, ABD = ACD P = ⎢ ⎣ 2 − 2i ⎦ −

1 2

+ 72 i

− 85 − 45 i

Method 2: We begin by expressing [1, 0] and [0, 1] as linear combinations of [1 + i, −1 + 3i] and " ' ( " 1 2 1 0 "" − 12 − 12 i 1 + i 1 − i "" 1 0 5 − 5i to obtain [1 − i, 1 + 2i] by row reducing . " −1 + 3i 1 + 2i " 0 1 0 1 " i − 25 − 15 i Therefore, [1, 0] = − 12 − 12 i [1 + i, −1 + 3i] + i[1 − i, 1 + 2i] and [0, 1] = 15 − 25 i [1 + i, −1 + 3i] + − 25 − 15 i [1 − i, 1 + 2i]. Hence, L([1, 0]) = − 12 − 12 i [3 − i, 5, −i] + i [2 + i, 1 − 3i, 3] = −3 + i, 12 − 32 i, − 12 + 72 i , 8 4 and L([0, 1]) = 15 − 25 i [3 − i, 5, −i] + − 25 − 15 i [2 + i, 1 − 3i, 3] = − 25 − 11 5 i, −i, − 5 − 5 i . Using L([1, 0]) and L([0, 1]) as columns produces the same 3 × 2 matrix ABD as in Method 1.

211


Section 7.4

(9) (a) True. This is the complex analog of Theorem 4.18. (b) False. If v = i, then L(iv) = = L(−1) = (−1) = −1. L(i(i)) However, iL(v) = iL(i) = i i = i(−i) = 1. (c) True. This follows from the complex analog of Theorem 4.19. (d) False. Complex vector spaces (and their complex subspaces) must have even dimension as real vector spaces, not necessarily as complex vector spaces. For example, the subspace W = span({[1, 0, 0]}) of C3 has complex dimension 1.

Section 7.4 (1) Remember to use the conjugates of the coordinates of the second vector when computing a complex dot product. (a) The given set is not orthogonal because [1 + 2i, −3 − i] · [4 − 2i, 3 + i] = −10 + 10i = 0. (c) The given set is orthogonal because [2i, −1, i] · [1, −i, −1] = 0, [2i, −1, i] · [0, 1, i] = 0, and [1, −i, −1] · [0, 1, i] = 0. (3) (a) First, we must find an ordinary basis for C3 containing [1+i, i, 1]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for C3 to the vector we have and use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1+i 1 0 0 1 0 0 1 ⎣ i 0 1 0 ⎦ to ⎣ 0 1 0 −1 − i ⎦, yielding the basis {[1 + i, i, 1], [1, 0, 0], [0, 1, 0]}. 1 0 0 1 0 0 1 −i We perform the Gram-Schmidt Process on this set. First, let v1 = [1 + i, i, 1].

1 1 1 [1,0,0]·[1+i,i,1] 1 1 [1+i, i, 1] = [1, 0, 0]− 1−i Next, v2 = [1, 0, 0]− [1+i,i,1]·[1+i,i,1] 4 [1+i, i, 1] = 2 , − 4 − 4 i, − 4 + 4 i . Multiplying by 4 to eliminate fractions yields v2 = [2, −1 − i, −1 + i].

[0,1,0]·[1+i,i,1] [0,1,0]·[2,−1−i,−1+i] Finally, v3 = [0, 1, 0] − [1+i,i,1]·[1+i,i,1] [1 + i, i, 1]− [2,−1−i,−1+i]·[2,−1−i,−1+i] [2, −1 − i, −1 + i] 1 1 (−i) (−1+i) = [0, 1, 0] − 4 [1 + i, i, 1] − 8 [2, −1 − i, −1 + i] = 0, 2 , 2 i . Multiplying by 2 to eliminate fractions produces v3 = [0, 1, i]. Hence, the desired orthogonal set is {[1 + i, i, 1], [2, −1 − i, −1 + i], [0, 1, i]}. (b) According to part (1) of Theorem 7.7, a 3 × 3 matrix is unitary if and only if its rows form an orthonormal basis for C3 . Normalizing the orthogonal basis we computed in part (a) yields an orthonormal basis whose first vector is a scalar multiple of [1 + i, i, 1]. Using the vectors from this ⎡ 1+i ⎤ i 1 ⎢ ⎢ orthonormal basis as the rows of a matrix produces the unitary matrix ⎢ ⎣

2

2

2

√2 8

−1−i √ 8

−1+i √ 8

0

√1 2

√i 2

⎥ ⎥ ⎥. ⎦

(7) (a) A is unitary if and only if A∗ = A−1 if and only if AT = A−1 if and only if (AT )T = (A−1 )T if and only if (AT )∗ = (A−1 )T if and only if (AT )∗ = (AT )−1 if and only if AT is unitary. 212

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition " " x − (1 − 6i) (10) (b) pA (x) = "" −2 + 10i

Section 7.4

" 10 + 2i "" = (x − (1 − 6i)) (x − 5) − (10 + 2i)(−2 + 10i) x−5 "

= x2 + (−6 + 6i)x + (45 − 126i). The quadratic formula produces the eigenvalues λ1 = 9 + 6i and λ2 = −3 − 12i. Now we solve for corresponding fundamental eigenvectors. For λ1 = 9 + 6i, we row reduce [(9 + 6i)I2 − A|0] = " ' ( " 1 12 − 12 i "" 0 8 + 12i 10 + 2i "" 0 to . " −2 + 10i 4 + 6i " 0 " 0 0 0 Setting the independent variable z2 in the associated system equal to 2 to eliminate fractions yields the fundamental eigenvector [−1 + i, 2]. For λ2 = −3 − 12i, we −4 − 6i 10 + 2i −2 + 10i −8 − 12i

row reduce [(−3 − 12i)I2 − A|0] = " " " 0 1 −1 + i "" 0 " " 0 to 0 " 0 . 0

Setting the independent variable z2 in the associated system equal to 1 yields the fundamental eigenvector [1 − i, 1]. Note that the two fundamental eigenvectors we have computed form an orthogonal basis for C2 . Normalizing each of these vectors and using them as columns for a matrix produces the unitary ' −1+i 1−i ( matrix P =

√ 6 √2 6

9 + 6i 0 0 −3 − 12i

√ 3 √1 3

. Direct computation shows that P−1 AP = P∗ AP =

, a diagonal matrix having the eigenvalues of A on its main diagonal.

" " x−1 −2 − i −1 + 2i " x+3 i (13) Let A represent the given matrix. Then, pA (x) = "" −2 + i " −1 − 2i −i x−2

" " " ". By basketweaving, " "

this determinant equals (x − 1)(x + 3)(x − 2) + (−2 − i)(i)(−1 − 2i) + (−1 + 2i)(−2 + i)(−i) −(x − 1)(i)(−i) − (−2 − i)(−2 + i)(x − 2) − (−1 + 2i)(x + 3)(−1 − 2i) = (x3 − 7x + 6) + (−5) + (−5) − (x − 1) − (5x − 10) − (5x + 15) = x3 − 18x − 8. Testing the divisors of (−8), the constant term, shows 2 that −4 is a root of pA (x). Then dividing pA (x) by (x + 4) yields √ pA (x) = (x √+ 4)(x − 4x − 2). Finally, 2 using the quadratic √ formula on√x − 4x − 2 gives the roots 2 + 6 and 2 − 6. Hence, the eigenvalues of A are −4, 2 + 6, and 2 − 6, which are all real. (Using a computer algebra system makes these computations much easier.) (15) (a) False. Instead, every Hermitian matrix is unitarily diagonalizable, by Corollary 7.10. But, for example, the Hermitian matrix given in Exercise 13 of Section 7.4 is not unitary, since its rows do not form an orthonormal basis for C3 . (b) True. Since the dot product of two vectors in Rn is the same as the complex dot product of those vectors (because a real number is its own complex conjugate), then any orthonormal set of n vectors in Rn is also an orthonormal set of n vectors in Cn . (c) True. The definition of unitarily diagonalizable is that A is unitarily diagonalizable if and only if there is a unitary matrix P such that P−1 AP is diagonal. But a matrix U is unitary if and only if U−1 = U∗ . Hence, A is unitarily diagonalizable if and only if there is a unitary matrix P such that P∗ AP is diagonal. Now notice that P is unitary if and only if P is unitary (by Exercise 6(a)) T if and only if P is unitary (by Exercise 7(a)) if and only if P∗ is unitary. Therefore, letting 213


Section 7.5

Q = P∗ then shows that A is unitarily diagonalizable if and only if there is a unitary matrix Q such that QAQ∗ is diagonal, which is what we needed to show. (d) True. If the columns of A form an orthonormal basis for Cn , then part (2) of Theorem 7.7 implies that A is unitary. Hence, part (1) of Theorem 7.7 shows that the rows of A form an orthonormal basis for Cn . (e) False. In general, A∗ , not AT , is the matrix for the adjoint of L with respect to the standard basis. This is explained in Section 7.4 of the textbook, just after Corollary 7.10.

Section 7.5 (1) (b) x, y = (Ax) · (Ay) = ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ 5 4 2 3 5 4 2 −2 ⎝⎣ −2 3 1 ⎦ ⎣ −2 ⎦⎠ · ⎝⎣ −2 3 1 ⎦ ⎣ 1 ⎦⎠ = [15, −8, 5] · [−8, 6, −3] = −183. 1 −1 0 4 1 −1 0 −1 Also, x = x, x . But x, x = (Ax) · (Ax) = ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ 5 4 2 3 5 4 2 3 ⎝⎣ −2 3 1 ⎦ ⎣ −2 ⎦⎠ · ⎝⎣ −2 3 1 ⎦ ⎣ −2 ⎦⎠ = [15, −8, 5] · [15, −8, 5] = 314. 1 −1 0 4 1 −1 0 4 √ Hence, x = 314. "π 5 5π π " (3) (b) f , g = 0 et sin t dt = (−et cos t) " − 0 (−et cos t) dt (using integration by parts) "π "π 5 0 π " " = (−et cos t) " + (et sin t) " − 0 et sin t dt (using integration by parts again) 0

0

"π "π " " = (−et cos t) " + (et sin t) " − f , g . Adding f , g to both ends of the equation produces 0 0 "π "π

" " t 2 f , g = (−e cos t) " + (et sin t) " = (−eπ cos π) − (−e0 cos 0) + (eπ sin π) − (e0 sin 0) 0

0

= eπ + 1. Dividing by 2 yields f , g = 12 (eπ + 1). 5π 5π f , f . But, f , f = 0 et et dt = 0 e2t dt = Next, f = f = 12 (e2π − 1).

"π "

1 2t " 2e "

0

=

1 2

e2π − 1 . Hence,

k < x, kx > (by property (5) of an inner product space) = kk < x, x > √ (by part (3) of Theorem 7.12) = |k|2 < x, x > = |k| < x, x > = |k| x. (9) (a) The distance between f and g is defined to be f − g = (f − g) , (f − g) . Now, 5π 5π (f − g) , (f − g) = 0 (t − sin t) (t − sin t) dt = 0 t2 − 2t sin t + sin2 t dt 5π 5π 5π = 0 t2 dt − 2 0 t sin t dt + 0 sin2 t dt. We compute each integral individually. "π " 5π 3 First, 0 t2 dt = 13 t3 "" = π3 . Next, using integration by parts, 0 " "π "π 5π "π 5 π " " t sin t dt = (−t cos t) − (− cos t) dt = (−t cos t) + sin t " " " = (π − 0) − (0 − 0) = π. Finally, 0 0 0 0 0 "π " 5 5 π π 1 2 2 1 using the identity sin t = 2 (1 − cos 2t), 0 sin t dt = 0 2 (1 − cos 2t) dt = 12 (t − 12 sin 2t)""

(6) kx =

√

< kx, kx > =

√

0

214


Section 7.5

3 3 − 0 −(0 − 0) = 12 π. Combining these produces (f − g) , (f − g) = π3 −2π+ 12 π = π3 − 3π 2 . 3 Hence, f − g = π3 − 3π 2 .

x·y (10) (b) The angle θ between x and y is defined to be arccos x y . First, we compute x, y . Now =

1

2π

⎛⎡

−2 x, y = (Ax) · (Ay) = ⎝⎣ 1 3

0 −1 −1

⎤⎡ ⎤⎞ ⎛⎡ 1 2 −2 2 ⎦ ⎣ −1 ⎦⎠ · ⎝⎣ 1 −1 3 3

0 −1 −1

⎤⎡ ⎤⎞ 1 5 2 ⎦ ⎣ −2 ⎦⎠ −1 2

2

= [−1, 9, 4] · [−8, 11, 15] = 167. Next, x = x, x = (Ax) · (Ax) = ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ −2 0 1 2 −2 0 1 2 ⎝⎣ 1 −1 2 ⎦ ⎣ −1 ⎦⎠ · ⎝⎣ 1 −1 2 ⎦ ⎣ −1 ⎦⎠ 3 −1 −1 3 3 −1 −1 3 2

= [−1, 9, 4] · [−1, 9, 4] = 98. Similarly, y = y, y = (Ay) · (Ay) = ⎛⎡ ⎤⎡ ⎤⎞ ⎛⎡ ⎤⎡ ⎤⎞ −2 0 1 5 −2 0 1 5 ⎝⎣ 1 −1 2 ⎦ ⎣ −2 ⎦⎠ · ⎝⎣ 1 −1 2 ⎦ ⎣ −2 ⎦⎠ 3 −1 −1 2 3 −1 −1 2 √ √ = [−8, 11, 15] · [−8, 11, 15] = 410. Therefore, x = 98 and y = 410, and so

√ θ = arccos √98167 ≈ arccos(0.83313) ≈ 0.586 radians, or, 33.6◦ . 410 (14) To prove that a set of vectors {v1 , . . . , vn } is orthogonal, we must show that vi , vj = 0 for i = j. 7 6 7 6 (a) Note that t2 , t + 1 = (1)(0) + (0)(1) + (0)(1) = 0, t2 , t − 1 = (1)(0) + (0)(1) + (0)(−1) = 0, and t + 1, t − 1 = (0)(0) + (1)(1) + (1)(−1) = 0. Hence, the given set is orthogonal. (c) Note that [5, −2], [3, 4] = (5)(3) − (5)(4) − (−2)(3) + (2)(−2)(4) = −15 = 0. Therefore, the given set is not orthogonal. (19) We begin with the basis {w1 , w2 , w3 } for P2 , with w1 = t2 − t + 1, w2 = 1, and w3 = t, and apply the Gram-Schmidt Process. (Note: The Enlarging Method is not needed here since it can be seen by inspection that {w1 , w2 , w3 } = {t2 − t + 1, 1, t} is linearly independent, and since dim(P2 ) = 3, 2 ,v1 {w1 , w2 , w3 } is a basis for P2 .) First, let v1 = w1 = t2 − t + 1. Next, v2 = w2 − w

v1 ,v1 v1 . But " 51 51 "1 w2 , v1 = −1 (1)(t2 − t + 1) dt = ( 13 t3 − 12 t2 + t)" = 83 , and v1 , v1 = −1 (t2 − t + 1)(t2 − t + 1) dt = −1 "1 51 4 " (8/3) 1 5 1 4 3 2 3 2 2 (t − 2t + 3t − 2t + 1) dt = ( t − t + t − t + t) " = 22 5 2 5 . Hence, v2 = (1) − (22/5) (t − t + 1). −1 −1

Multiplying by 33 to eliminate fractions yields v2 = −20t2 + 20t + 13. Finally, 51 51 3

w3 ,v2 3 ,v1 2 2 v3 = w3 − w

v1 ,v1 v1 − v2 ,v2 v2 . However, w3 , v1 = −1 (t)(t − t + 1) dt = −1 (t − t + t) dt = "1 51 51 " ( 14 t4 − 13 t3 + 12 t2 )" = − 23 , w3 , v2 = −1 (t)(−20t2 + 20t + 13) dt = −1 (−20t3 + 20t2 + 13t) dt = −1

(−5t4 +

20 3 3 t

"1

13 2 " 2 t )"

40 3 ,

51

(−20t2 + 20t + 13)(−20t2 + 20t + 13) dt ""1 51 = −1 (400t4 − 800t3 − 120t2 + 520t + 169) dt = 80t5 − 200t4 − 40t3 + 260t2 + 169t " = 418. Hence, v3 = (t) −

+

(−2/3) 2 (22/5) (t

−1

=

and v2 , v2 =

− t + 1) −

(40/3) 2 (418) (−20t

−1

−1

+ 20t + 13). Multiplying by 19 to eliminate fractions

215


Section 7.5

gives v3 = 15t2 + 4t − 5. Thus, we get the orthogonal basis {v1 , v2 , v3 }, with v1 = t2 − t + 1, v2 = −20t2 + 20t + 13, and v3 = 15t2 + 4t − 5. (22) (b) First, we prove part (4) of Theorem 7.19 as follows, using a proof similar to the proof of Theorem 6.12: Let {v1 , . . . , vn } be an orthogonal basis for V, with W = span({v1 , . . . , vk }). Let X = span({vk+1 , . . . , vn }). Since {vk+1 , . . . , vn } is linearly independent, it is a basis for W ⊥ if X = W ⊥ . We will show that X ⊆ W ⊥ and W ⊥ ⊆ X . To show X ⊆ W ⊥ , we must prove that any vector x of the form dk+1 vk+1 + · · · + dn vn (for some scalars dk+1 , . . . , dn ) is orthogonal to every vector w ∈ W. Now if w ∈ W, then w = c1 v1 + · · · + ck vk , for some scalars c1 , . . . , ck . Hence, x · w = (dk+1 vk+1 + · · · + dn vn ) · (c1 v1 + · · · + ck vk ), which equals zero when expanded because each vector in {vk+1 , . . . , vn } is orthogonal to every vector in {v1 , . . . , vk }. Hence, x ∈ W ⊥ , and so X ⊆ W ⊥ . To show W ⊥ ⊆ X , we must show that any vector x in W ⊥ is also in span({vk+1 , . . . , vn }). Let x ∈ W ⊥ . Since {v1 , . . . , vn } is an orthogonal basis for V, Theorem 7.16 tells us that x=

(x · v1 ) (x · vk ) (x · vk+1 ) (x · vn ) v1 + · · · + vk + vk+1 + · · · + vn . (v1 · v1 ) (vk · vk ) (vk+1 · vk+1 ) (vn · vn )

However, since each of v1 , . . . , vk is in W, we know that x · v1 = · · · = x · vk = 0. Hence, x=

(x · vk+1 ) (x · vn ) vk+1 + · · · + vn , (vk+1 · vk+1 ) (vn · vn )

and so x ∈ span({vk+1 , . . . , vn }). Thus, W ⊥ ⊆ X . Next, we prove part (5) of Theorem 7.19 as follows: Let W be a subspace of V of dimension k. By Theorem 7.17, W has an orthogonal basis {v1 , . . . , vk }. Expand this basis to an orthogonal basis for all of V. (That is, first expand to any basis for V by Theorem 4.18, then use the Gram-Schmidt Process. Since the first k vectors are already orthogonal, this expands {v1 , . . . , vk } to an orthogonal basis {v1 , . . . , vn } for V.) Then, by part (4) of Theorem 7.19, {vk+1 , . . . , vn } is a basis for W ⊥ , and so dim(W ⊥ ) = n − k. Hence, dim(W) + dim(W ⊥ ) = n. (d) By part (3) of Theorem 7.19, W ⊆ (W ⊥ )⊥ . Next, by part (5) of Theorem 7.19, dim(W) = n − dim(W ⊥ ) = n − (n − dim((W ⊥ )⊥ )) = dim((W ⊥ )⊥ ). Thus, by Theorem 4.16, or its complex analog, W = (W ⊥ )⊥ . (23) We begin with the basis B = {t3 + t2 , t − 1, 1, t2 } for P3 containing the polynomials t3 + t2 and t − 1. (Notice that the Enlarging Method is not needed here since it can be seen by inspection that B = {t3 + t2 , t − 1, 1, t2 } is linearly independent (since each element is a polynomial of a different degree), and since dim(P3 ) = 4, B is a basis for P3 .) We use the Gram-Schmidt Process on B. t−1,t3 +t2 Let v1 = t3 + t2 . Next, v2 = (t − 1) − t3 +t2 ,t3 +t2 (t3 + t2 ) = (t − 1) − 02 (t3 + t2 ) = t − 1. Also, 1,t3 +t2

1,t−1 1 1 v3 = (1) − t3 +t2 ,t3 +t2 (t3 + t2 ) − t−1,t−1 (t − 1) = (1) − 02 (t3 + t2 ) − (−1) 2 (t − 1) = 2 t + 2 . Multiplying by 2 to eliminate fractions yields v3 = t + 1. Finally, t2 ,t3 +t2 t2 ,t−1 t2 ,t+1 v4 = (t2 ) − t3 +t2 ,t3 +t2 (t3 + t2 ) − t−1,t−1 (t − 1) − t+1,t+1 (t + 1) 216


Section 7.5

= (t2 ) − 12 (t3 + t2 ) − 02 (t − 1) − 02 (t + 1) = − 12 t3 + 12 t2 . Multiplying by −2 to simplify the form of the result produces v4 = t3 − t2 . Hence, by part (4) of Theorem 7.19, W ⊥ = span({t + 1, t3 − t2 }). (25) As in the hint, let {v1 , . . . , vk } be an orthonormal basis for W. Now if v ∈ V, let w1 = v, v1 v1 + · · · + v, vk vk and w2 = v − w1 . Then, w1 ∈ W because w1 is a linear combination of basis vectors for W. We claim that w2 ∈ W ⊥ . To see this, let u ∈ W. Then u = a1 v1 + · · · + ak vk for some a1 , . . . , ak . Then u, w2 = u, v − w1 = a1 v1 + · · · + ak vk , v − ( v, v1 v1 + · · · + v, vk vk ) = a1 v1 + · · · + ak vk , v − a1 v1 + · · · + ak vk , v, v1 v1 + · · · + v, vk vk !k !k !k = i=1 ai vi , v − i=1 j=1 ai v, vj vi , vj . But vi , vj = 0 when i = j and vi , vj = 1 when !k !k i = j, since {v1 , . . . , vk } is an orthonormal set. Hence, u, w2 = i=1 ai vi , v − i=1 ai v, vi = !k !k ⊥ i=1 ai vi , v − i=1 ai vi , v = 0. Since this is true for every u ∈ W, we conclude that w2 ∈ W . Finally, we want to show uniqueness of decomposition. Suppose that v = w1 +w2 and v = w1 +w2 , where w1 , w1 ∈ W and w2 , w2 ∈ W ⊥ . We want to show that w1 = w1 and w2 = w2 . Now, w1 − w1 = w2 − w2 . Also, w1 − w1 ∈ W, but w2 − w2 ∈ W ⊥ . Thus, w1 − w1 = w2 − w2 ∈ W ∩ W ⊥ . By part (2) of Theorem 7.19, w1 − w1 = w2 − w2 = 0. Hence, w1 = w1 and w2 = w2 . 4 3 (26) From Example 8 in Section 7.5 of the textbook we know that √1π cos t, √1π sin t is an orthonor5π mal basis for W = span({cos t, sin t}). (We quickly verify this: cos t, sin t = −π (cos t sin t) dt = " 1 "π 2 " sin t = (0 − 0) = 0. Hence, {cos t, sin t} is an orthogonal basis for W. We normalize these 2 " −π

vectors to get an orthonormal basis. Now " "π 5π 5π cos t, cos t = −π cos2 t dt = −π 21 (1 + cos 2t) dt = 12 t + sin 2t "" = ( π2 + 0) − (− π2 + 0) = π, and −π

" 1 "π 5π 1 5π 2 sin t, sin t = −π sin t dt = −π 2 (1 − cos 2t) dt = 2 t − sin 2t "" = ( π2 − 0) − (− π2 − 0) = π. −π 3 4 √ √ Therefore, cos t = π and sin t = π, giving us the orthonormal basis √1π cos t, √1π sin t for W.) 8 9

8 9

√1 cos t + f , √1 sin t √1 sin t . So we have two Using f = k1 et , w1 = projW f = f , √1π cos t π π π 8 9 5π t 1 more inner products to compute. First, f , √1π cos t = k√ e cos t dt = π −π

""π 5π 1 1 √ et sin t "" − k√ (et sin t) dt, using integration by parts, k π π −π −π

=

1 √ et k π

8

""π sin t ""

−π

9

−

−1 √ et k π

""π cos t ""

−π

+

1 √ k π

5π −π

(−et cos t) dt, using integration by parts again.

8 9 1 t √1 cos t = Adding f , √1π cos t = k√ e cos t dt to both ends of this equation produces 2 f , π −π π

""π

""π π

−π 1 e e 1 π −π √ √ et cos t " et sin t "" − k−1 " = (0 − 0) − k√π − k√π = − √π , since k = e − e . Dividing k π π −π 9 8 9 8 −π 5π t 1 by 2 yields f , √1π cos t = − 2√1 π . Similarly, f , √1π sin t = k√ e sin t dt π −π 5π

217


Section 7.5

""π 5π 1 = cos t "" − k√ (−et cos t) dt, using integration by parts, π −π −π

""π

""π 5π t −1 1 1 t t " √ √ (e sin t) dt, using integration by parts again. = k π e cos t " + k π e sin t "" − k√ π −π

−1 √ et k π

8

−π

9

−π

5π

1 et sin t dt to both ends of this equation produces Adding f , √1π sin t = k√ π −π " 8

9

"π

""π π −π 1 −1 1 t t " 2 f , √π sin t = k√π e cos t " + k√π e sin t "" = ke√π − ke√π + (0 − 0) = −π

−π

√1 . π

9 8 Dividing by 2 yields f , √1π sin t = 2√1 π . Therefore, w1 = projW f

1 √1 cos t + √1 sin t = 1 (sin t − cos t). √ = − 2√1 π 2π π 2 π π 1 1 sin t + 2π cos t. Finally, we must verify that w1 , w2 = 0. Also, w2 = f − w1 = k1 et − 2π 1 t 5π 1 1 1 sin t + 2π cos t dt Now w1 , w2 = −π 2π (sin t − cos t) k e − 2π 5π t 5π 5π t 5π 5π 1 1 = 2kπ (e sin t) dt − 4π1 2 −π sin2 t dt − 2kπ (e cos t) dt − 4π1 2 −π cos2 t dt + 2π1 2 −π sin t cos t dt. −π −π 9 8

5π t 1 1 1 1 √1 sin t = √ √ √ We consider each integral separately. First, 2kπ f , (from (e sin t) dt = −π 2 π π 2 π 2 π 5 π 1 our earlier computations) = 4π . Next, 4π1 2 −π sin2 t dt = 4π1 2 sin t, sin t = 4π1 2 (π) (from our earlier 9 8

5π t 1 1 1 1 −1 1 √1 cos t √ √ √ computations) = 4π f , = − 4π . Similarly 2kπ (e cos t) dt = = and −π 2 π π 2 π 2 π 5 5 π π 1 1 1 1 1 1 2 4π 2 −π cos t dt = 4π 2 (π) = 4π . Finally, 2π 2 −π sin t cos t dt = 2π 2 sin t, cos t = 2π 2 cos t, sin t = 0

(from our earlier computations). Hence, w1 , w2 =

1 4π

1 1 1 − 4π −(− 4π )− 4π = 0, completing the problem.

(29) (b) First, we establish that range(L) = W. We start by showing that W ⊆ range(L). Let v ∈ W. Then we claim that L(v) = v. Note that v ∈ W implies that the unique decomposition of v into v = w1 + w2 with w1 ∈ W and w2 ∈ W ⊥ is w1 = v and w2 = 0. (The reason is that for these vectors w1 and w2 , we have w1 + w2 = v, and because by Theorem 7.20 this decomposition is unique.) Hence, since L(v) = projW v = w1 , we get L(v) = v. From this we can conclude that W ⊆ range(L), since every vector in W is the image of itself. Also, since the codomain of L is W, range(L) ⊆ W. Hence, range(L) = W. Next, we establish that ker(L) = W ⊥ . First, we show that W ⊥ ⊆ ker(L). Let v ∈ W ⊥ . We claim that L(v) = 0. Note that v ∈ W ⊥ implies that the unique decomposition of v into v = w1 + w2 with w1 ∈ W and w2 ∈ W ⊥ is w1 = 0 and w2 = v. (The reason is that for these vectors w1 and w2 , we have w1 + w2 = v, and because by Theorem 7.20 this decomposition is unique.) Hence, since L(v) = projW v = w1 , we get L(v) = 0. From this, we can conclude that W ⊥ ⊆ ker(L). We still need to prove that ker(L) ⊆ W ⊥ . So, let x ∈ ker(L). Then x = u1 + u2 with u1 ∈ W and u2 ∈ W ⊥ . Also, L(x) = u1 . But x ∈ ker(L) implies u1 = L(x) = 0. Therefore, x = u1 + u2 = 0 + u2 = u2 ∈ W ⊥ . This completes the proof. (30) (a) False. The correct equation is kx = |k| x. For example, in C2 with the usual complex dot [i, 0] · [i, 0] = product, consider x = [1, 0] and k = i. Then kx = i[1, 0] = [i, 0] = √ √ (i)( i ) + (0)(0) = 1 = 1. Also, x = [1, 0] · [1, 0] = 1 = 1. Hence, |k| x = |i|(1) = (1)(1) = 1 = kx, but k x = i(1) = (−i)(1) = −i = kx. (b) False. The distance between two vectors x and y is defined to be x − y = x − y, x − y , 218


Chap 7 Review

which is always real and positive (since x and y are distinct) by parts (1) and (2) of the definition of an inner product. Hence, the distance cannot be purely imaginary. (c) False. For a set S of vectors to be orthonormal, S must be an orthogonal set. This implies that S is linearly independent, but the converse is not true. There are3many sets independent 4 √ of√linearly 2 2 unit vectors that are not orthogonal sets. For example, S = [1, 0], 2 , 2 in R2 using the usual dot product is such a set. (d) True. For example, R2 has the usual dot product as one inner product, and the inner product of Example 2 in Section 7.5 in the textbook as a different inner product. (e) False. The uniqueness assertion is that, using a single, fixed inner product, and a single, fixed subspace W, the decomposition of a particular vector described in the Projection Theorem is unique. However, different inner products could result in different decompositions. For example, consider the subspace W = span({[1, 0]}) in R2 . Let v = [0, 1]. Using the dot product as the inner product on R2 , the decomposition of v is v = [0, 0] + [0, 1], with [0, 0] ∈ W and [0, 1] ∈ W ⊥ (since [0, 1] · [1, 0] = 0). However, consider the inner product from Example 2 in Section 7.5 in the textbook. It is easily verified that {[1, 0]} is an orthonormal basis for W with this inner product, and so, projW v = [0, 1], [1, 0] [1, 0] = (−1)[1, 0] = [−1, 0]. Therefore, the decomposition of v using this inner product is v = [−1, 0] + [1, 1].

Chapter 7 Review Exercises (1) (a) v · w = [i, 3 − i, 2 + 3i] · [−4 − 4i, 1 + 2i, 3 − i] = i(−4 − 4i) + (3 − i)(1 + 2i) + (2 + 3i)(3 − i) = i(−4 + 4i) + (3 − i)(1 − 2i) + (2 + 3i)(3 + i) = (−4 − 4i) + (1 − 7i) + (3 + 11i) = 0 (b) (1+2i)(v·z) = (1+2i)([i, 3−i, 2+3i]·[2+5i, 2−5i, −i]) = (1+2i)(i(2 + 5i)+(3−i)(2 − 5i)+(2+ 3i)(−i)) = (1+2i)(i(2−5i)+(3−i)(2+5i)+(2+3i)(i)) = (1+2i)((5+2i)+(11+13i)+(−3+2i)) = (1 + 2i)(13 + 17i) = −21 + 43i ((1+2i)v)·z =((1+2i)[i, 3−i, 2+3i])·[2+5i, 2−5i, −i] = [−2+i, 5+5i, −4+7i]·[2+5i, 2−5i, −i] = (−2+i)(2 + 5i)+(5+5i)(2 − 5i)+(−4+7i)(−i) = (−2+i)(2−5i)+(5+5i)(2+5i)+(−4+7i)(i) = (1 + 12i) + (−15 + 35i) + (−7 − 4i) = −21 + 43i v·((1+2i)z) = [i, 3−i, 2+3i]·((1+2i)[2+5i, 2−5i, −i]) = [i, 3−i, 2+3i]·[−8+9i, 12−i, 2−i] = (i)(−8 + 9i) + (3 − i)(12 − i) + (2 + 3i)(2 − i) = (i)(−8 − 9i) + (3 − i)(12 + i) + (2 + 3i)(2 + i) = (9 − 8i) + (37 − 9i) + (1 + 8i) = 47 − 9i (4) (a) Using the row operations ( 1 ← 3 2 − 13 i) 2 ), ( 1 3 ), ( 2 ← ( 13 " ⎡ i 1 + i "" −1 + 2i matrix ⎣ 1 + i 5 + 2i "" 5 − 3i 2 − i 2 − 5i " 1 − 2i

(−i) 1 ), ( 2 ← (−1 − i) 1 + 2 ), ( 3 ← (−2 + i) 1 + ← (−1 + i) 2 + 1 ), and ( 3 ← (−1 + 2i) 2 + 3 ), the " ⎤ ⎤ ⎡ 1 0 "" 4 + 3i ⎦ row reduces to ⎣ 0 1 " −2i ⎦. Thus, the system has " 0 0 0 "

the unique solution w = 4 + 3i, z = −2i. (d) Using the row operations ( 1 ← ( 12 − 12 i) 1 ), ( 2 ← (−4 − 3i) 1 + 2 ), ( 2 ← ( 12 − 12 i) 2 ), " 1+i 3−i 5 + 5i "" 23 − i row reduces and ( 1 ← (−1 + 2i) 2 + 1 ), the matrix 4 + 3i 11 − 4i 19 + 16i " 86 − 7i " 1 0 3 − i "" 2 + i . Hence, z is an independent variable. Setting the value of z equal to to 0 1 i " 7+i 219


Chap 7 Review

c yields the solution set {[(2 + i) − (3 − i)c, (7 + i) − ic, c] | c ∈ C} = {[(2 − 3c) + (1 + c)i, 7 + (1 − c)i, c] | c ∈ C}. " " " x+3 −5 10 "" " x+3 −8 "". Using basketweaving gives pA (x) = (6) (a) Step 1: pA (x) = |xI3 − A| = "" −2 " −2 3 x−7 " x3 − x2 + x − 1 = (x2 + 1)(x − 1) = (x − i)(x + i)(x − 1). Step 2: The eigenvalues of A are the roots of pA (x). Hence, there are three eigenvalues, λ1 = i, λ2 = −i, and λ3 = 1. Step 3: We solve for fundamental eigenvectors. For λ1 = i, we solve the homogeneous system whose augmented matrix is [(iI3 − A)|0] = " " ⎤ ⎤ ⎡ ⎡ 1 0 2 + i "" 0 3+i −5 10 "" 0 ⎣ −2 3 + i −8 "" 0 ⎦ , which reduces to ⎣ 0 1 −1 + i "" 0 ⎦. This yields the funda0 0 0 " 0 −2 3 −7 + i " 0 mental eigenvector [−2 − i, 1 − i, 1]. For λ2 = −i, we solve the homogeneous system whose " ⎤ ⎡ ⎡ 1 3−i −5 10 "" 0 ⎣ −2 3 − i −8 "" 0 ⎦ , which reduces to ⎣ 0 0 −2 3 −7 − i " 0

augmented matrix is [((−i)I3 − A)|0] = " ⎤ 0 2 − i "" 0 1 −1 − i "" 0 ⎦. This yields the funda0 0 " 0

mental eigenvector [−2 + i, 1 + i, 1]. For λ3 = 1, we solve the homogeneous system whose " ⎤ ⎡ ⎡ 1 0 0 4 −5 10 "" 0 ⎣ −2 4 −8 "" 0 ⎦ , which reduces to ⎣ 0 1 −2 0 0 0 −2 3 −6 " 0

augmented matrix is [(I3 − A)|0] = " ⎤ " 0 " " 0 ⎦. This yields the fundamental " " 0

eigenvector [0, 2, 1]. Step 4: Since n = 3, and we have found 3 fundamental eigenvectors, A is diagonalizable. Step 5: We form the 3 × 3 matrix whose columns are the 3 vectors we have found: ⎡ ⎤ −2 − i −2 + i 0 1 + i 2 ⎦. P=⎣ 1−i 1 1 1 Step 6: The matrix whose diagonal entries are the eigenvalues of A (in the same order) is ⎤ ⎡ ⎡ ⎤ i 0 0 −1 − i 1 + 2i −2 − 4i D = ⎣ 0 −i 0 ⎦. Also, by row reducing [P|I3 ], we get P−1 = 12 ⎣ −1 + i 1 − 2i −2 + 4i ⎦. 0 0 1 2 −2 6 It is easy to verify for these matrices that D = P−1 AP. (7) (a) One possibility: Consider L : C → C given by L(z) = z. Note that L(v+w) = (v + w) = v+w = L(v) + L(w). But L is not a linear operator on C because L(i) = −i, but iL(1) = i(1) = i, so the rule “L(cv) = cL(v)” is not satisfied. (b) The example given in part (a) is a linear operator on C, thought of as a real vector space. In that case we may use only real scalars, and so, if v = a + bi, then L(cv) = L(ca + cbi) = ca − cbi = c(a − bi) = cL(v). Note: For any function L from any vector space to itself (real or complex), the rule “L(v+w) = 220


Chap 7 Review

L(v) + L(w)” implies that L(cv) = cL(v) for any rational scalar c. Thus, any example for real vector spaces for which L(v + w) = L(v) + L(w) is satisfied but L(cv) = cL(v) is not satisfied for some c must involve an irrational value of c. Examples of such functions do exist, but the ideas behind their construction are slightly beyond the level of topics covered in this text. (8) (a) First, we must find an ordinary basis for C4 containing [1, i, 1, −i] and [1 + i, 2 − i, 0, 0]. To do this, we use the Enlarging Method from Section 4.6. We add the standard basis for C4 to the given vector and use the Independence Test Method. Hence, we row reduce ⎡ ⎤ ⎡ ⎤ 1 0 0 0 0 i 1 1+i 1 0 0 0 2 1 2 1 ⎢ ⎥ ⎢ i 2−i 0 1 0 0 ⎥ 0 5 + 5i 5 + 5 i ⎥, giving the basis ⎢ ⎥ to ⎢ 0 1 1 1 3 1 8 ⎣ 0 0 1 − − i 0 − − i ⎦ ⎣ 1 0 0 0 1 0 ⎦ 5 5 5 5 −i 0 0 0 0 1 0 0 0 0 1 −i {[1, i, 1, −i], [1 + i, 2 − i, 0, 0], [1, 0, 0, 0], [0, 0, 1, 0]}. We perform the Gram-Schmidt Process on this set. Let v1 = [1, i, 1, −i]. Then v2 = [1 + i, 2 − i, 0, 0]

− [1+i,2−i,0,0]·[1,i,1,−i] [1, i, 1, −i] = [1 + i, 2 − i, 0, 0] + 14 i[1, i, 1, −i] = 14 [4 + 5i, 7 − 4i, i, 1]. [1,i,1,−i]·[1,i,1,−i]

Multiplying by 4 to eliminate fractions yields v2 = [4 + 5i, 7 − 4i, i, 1]. Next,

[1,0,0,0]·[1,i,1,−i] v3 = [1, 0, 0, 0] − [1,i,1,−i]·[1,i,1,−i] [1, i, 1, −i]

7−4i, i, 1] − [4+5i,[1,0,0,0]·[4+5i, [4 + 5i, 7 − 4i, i, 1] 7−4i, i, 1]·[4+5i, 7−4i, i, 1] = [0, 1, 0, 0] − 14 [1, i, 1, −i] − =

1 27

(4−5i) 108

[4 + 5i, 7 − 4i, i, 1]

[10, −2 + 6i, −8 − i, −1 + 8i].

Multiplying by 27 to eliminate fractions gives us v3 =

[0,0,1,0]·[1,i,1,−i] [1, i, 1, −i] [10, −2 + 6i, −8 − i, −1 + 8i]. Finally, v4 = [0, 0, 1, 0] − [1,i,1,−i]·[1,i,1,−i]

[0,0,1,0]·[4+5i, 7−4i, i, 1]] − [4+5i, 7−4i, i, 1]·[4+5i, 7−4i, i, 1] [4 + 5i, 7 − 4i, i, 1]

[0,0,1,0]·[10, −2+6i, −8−i, −1+8i] [10, −2 + 6i, −8 − i, −1 + 8i] − [10, −2+6i, −8−i, −1+8i]·[10, −2+6i, −8−i, −1+8i]

= [0, 0, 1, 0] − 14 [1, i, 1, −i] − −

(−8+i) 270

(−i) 108

[4 + 5i, 7 − 4i, i, 1] [10, −2 + 6i, −8 − i, −1 + 8i] = 0, 0, 12 , 12 i . Multiplying by 2 to eliminate fractions

produces v4 = [0, 0, 1, i]. Hence, an orthogonal basis for C4 in which the first two vectors span the same subspace as {[1, i, 1, −i], [1 + i, 2 − i, 0, 0]} is {[1, i, 1, −i], [4 + 5i, 7 − 4i, i, 1] , [10, −2 + 6i, −8 − i, −1 + 8i] , [0, 0, 1, i]}. (9) (a) Note that A is Hermitian, "so A is unitarily diagonalizable. " " x − 3 −1 − i " " " = (x − 3)(x − 2) − (−1 − i)(−1 + i) = x2 − 5x + 4 = Now, pA (x) = |xI2 − A| = " −1 + i x − 2 " (x − 1)(x − 4). The eigenvalues of A are the roots of pA (x). Hence, there are two eigenvalues, λ1 = 1, and λ2 = 4. Next, we solve for fundamental eigenvectors. For λ1 = 1, we solve the homogeneous system whose augmented matrix is [(I2 − A)|0]. Now " ( ' " 1 12 + 12 i "" 0 −2 −1 − i "" 0 . Eliminating , which reduces to [(I2 − A)|0] = " −1 + i −1 " 0 " 0 0 0 221


Chap 7 Review

fractions, this yields the fundamental eigenvector v1 = [−1 − i, 2]. For λ2 = 4, we solve the homogeneous system whose augmented matrix " 1 −1 − i 1 −1 − i "" 0 [(4I2 − A)|0] = " 0 , which reduces to 0 0 −1 + i 2

is [(4I2 − A)|0]. Now " " 0 " " 0 . This yields the

fundamental eigenvector v2 = [1 + i, 1]. Now we normalize the two fundamental eigenvectors √ we have found√(which are orthogonal) to get an orthonormal basis B for C2 . Since v1 = 6 and v2 = 3, we get B = { √16 [−1 − i, 2],

√1 [1 3

⎡ P=⎣

+ i, 1]}. We form the 2 × 2 matrix whose columns are the 2 vectors in B:

√1 (−1 6 √2 6

− i)

√1 (1 3

+ i)

√1 3

⎤ ⎦.

The matrix whose diagonal entries are the eigenvalues of A (in the same order) is D =

1 0

0 4

.

Also, it is easy to verify for these matrices that D = P∗ AP. ⎡ ⎤ 80 105 + 45i 60 185 60 − 60i ⎦, and so A is normal. Hence, by (10) Both AA∗ and A∗ A equal ⎣ 105 − 45i 60 60 + 60i 95 Theorem 7.9, A is unitarily diagonalizable. 6 7 (13) The distance between x and x3 equals x − x3 = (x − x3 ), (x − x3 ) . But (x − x3 ), (x − x3 ) = 2 "1 51 51 35 42 15 8 x − x3 dx = 0 (x2 −2x4 +x6 ) dx = 13 x3 − 25 x5 + 17 x7 "0 = ( 13 − 25 + 17 )−0 = 105 − 105 + 105 = 105 . 0 8 ≈ 0.276. Thus, the distance is 105 (14) We are given w1 = [1, 0, 0], w2 = [0, 1, 0], and w3 = [0, 0, 1]. Following the steps in the Generalized Gram-Schmidt Process, we get v1 = w1 = [1, 0, 0]. Next, [0, 1, 0] − [0,1,0],[1,0,0]

[1,0,0],[1,0,0] [1, 0, 0] = [0, 1, 0] −

[−1,1,−3]·[1,−1,2] [1,−1,2]·[1,−1,2] [1, 0, 0]

= [0, 1, 0] −

(−8) 6 [1, 0, 0]

nate fractions to produce v2 = [4, 3, 0]. Finally, [0, 0, 1] −

= [ 43 , 1, 0]. We multiply by 3 to elimi [0,0,1],[1,0,0]

[0,0,1],[4,3,0]

[1,0,0],[1,0,0] [1, 0, 0] − [4,3,0],[4,3,0] [4, 3, 0] 5 [0, 0, 1] − 16 [1, 0, 0] − (−2) 3 [4, 3, 0] = [ 2 , 2, 1].

[2,3,1]·[1,−1,2] [2,3,1]·[1,−1,−1] [1, 0, 0] − [1,−1,−1]·[1,−1,−1] [4, 3, 0] = = [0, 0, 1] − [1,−1,2]·[1,−1,2] We multiply by 2 to eliminate fractions, producing v3 = [5, 4, 2]. Hence, the desired orthogonal basis is {[1, 0, 0], [4, 3, 0], [5, 4, 2]}.

(16) (a) True. Every real number is also a complex number, so an n-tuple of real numbers is also an n-tuple of complex numbers. (b) False. The angle between vectors in Cn is not defined. Since v · w can be a complex number, v·w could be complex, making the inverse of the cosine of this value a complex the expression vw number. (c) True. (A∗ Aw) · z = (A∗ (Aw)) · z = Aw · (A∗ )∗ z (by Theorem 7.3) = Aw · Az. (d) False. Example 5 in Section 7.1 provides a counterexample. (e) False. If A is skew-Hermitian, A∗ = −A, and so ajj = −ajj for each diagonal entry ajj of A. If ajj = c + di, then ajj = −ajj implies c − di = −c − di. This can be true only if c = 0. 222


Chap 7 Review

But, d can be any real number. Hence, the main diagonal entries of skew-Hermitian matrices are pure imaginarynumbers but not necessarily zeroes. A counterexample to the given statement is i 1 A= . −1 i (f) True. The Fundamental Theorem of Algebra shows that the characteristic polynomial for an n × n matrix must factor into n linear factors. Each linear factor contributes to the algebraic multiplicity of one of the eigenvalues of the matrix. (g) False. The system has a unique solution if and only if |A| = 0. However, systems with |A| = 0 can have infinitely many solutions. For example, the system Az = 0, with A being any square matrix with determinant zero (such as a 2 × 2 matrix with every entry equal to 1), has the trivial solution as well as other nontrivial solutions. (h) True. The vector [−1, − 1 + i, 1 + i] equals i[i, 1 + i, 1 − i]. Any nonzero scalar multiple of an eigenvector is also an eigenvector corresponding to the same eigenvalue because eigenspaces are closed under scalar multiplication. (i) False. Otherwise, every squarecomplex matrix would be diagonalizable. For a specific counterexi 1 ample, consider A = . This matrix has λ = i as its only eigenvalue, having algebraic 0 i multiplicity 2. However, Ei is one-dimensional, spanned by [1, 0]. " "2 " " (j) True. By parts (1) and (3) of Theorem 7.5, |AA∗ | = |A| |A∗ | = |A| (|A|) = "|A|" . (The last step uses the fact that, for any complex number z, zz = |z|2 . See Appendix C.) (k) True. If the vector space satisfies the 10 properties of a vector space using complex scalars, it will also satisfy them using real scalars, since the real numbers are a subset of the complex numbers. (l) True. This is stated in the text just after the definition of an orthogonal set of vectors in Cn . The proof for the complex case is almost identical to the proof for the real case. (m) False. Orthogonality in Cn is determined using the complex dot product, which involves using the complex conjugate. Thus, for the rows of a'complex matrix basis ' ( to form an orthonormal ( 3 4 7 24 − i − i − 5 5 25 25 , then AT A = , for Cn we need A∗ A = In . For example, if A = 3 24 7 − 45 i − i − 5 25 25 while A∗ A = I2 . Notice that the rows of A are orthogonal unit vectors in C2 . (n) True. This works because if all of the entries of A are real, AT = A∗ . (o) True. If A is Hermitian, then A is unitarily diagonalizable, by Corollary 7.10. But, by Theorem 7.11, all of the eigenvalues of A are real. Hence, A is unitarily similar to a diagonal matrix with its real eigenvalues on its main diagonal; that is, a matrix with all real entries. (p) True. If A is skew-Hermitian, then A is unitarily diagonalizable by Corollary 7.10. Thus, the geometric multiplicity of each eigenvalue must equal its algebraic multiplicity, just as in the case of a diagonalizable real matrix. (q) True. This warning is given in Section 7.4, just after Example 1. (r) True. This is discussed in Section 7.5 in Example 1. (s) True. Let {u1 , . . . , uk } be an orthonormal basis for W. Expand this to an orthonormal basis {u1 , . . . , un } for V. This will be a basis of fundamental eigenvectors for L because {u1 , . . . , uk } is a basis for E1 , and {uk+1 , . . . , un } is a basis for E0 . Hence, L is unitarily diagonalizable. (t) True. This is part (6) of Theorem 7.19. 223


Chap 7 Review

(u) True. As in Example 11 of Section 7.5, start with the basis {1, t, t2 , . . . , tn } for PnC . Performing the Gram-Schmidt Process on this basis and then normalizing will produce an ordered orthonormal basis in which the vectors have the correct degrees. (v) True. The distance between two vectors v and w is v − w = (v − w), (v − w) . 51 1 (w) False. For example, 1 is not orthogonal to cos(t) because −1 1 cos(t) dt = sin(t)|−1 = sin(1) − sin(−1) = 2 sin(1) = 0. However, if instead we had considered this set inside the inner product of continuous real-valued functions defined on [−π, π], with inner product given by f , g = 5space π f (t)g(t) dt, then the set would be an orthogonal set. (See the remarks after Example 8 in −π Section 7.5.)

224


Section 8.1

Chapter 8 Section 8.1 (1) In parts (a) through (d), use the fact the adjacency matrix for a graph is defined to be the matrix whose (i, j) entry is 1 if there is an edge between the vertices Pi and Pj and is 0 otherwise. In parts (e) through (h), use the fact the adjacency matrix for a digraph is defined to be the matrix whose (i, j) entry is 1 if there is an edge directed from the vertex Pi to the vertex Pj and is 0 otherwise. Inspection shows that the matrices in parts (a), (b), (c), (d), and (g) are symmetric, while those in parts (e), (f), and (h) are not. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 1 1 1 1 1 0 0 1 0 0 0 ⎢ 1 0 1 1 ⎥ ⎢ 1 0 0 1 0 ⎥ (c) G3 : ⎣ 0 0 0 ⎦ ⎥ ⎢ ⎥ (a) G1 : ⎢ ⎥ ⎣ 1 1 0 1 ⎦ 0 0 1 0 1 0 0 0 (b) G2 : ⎢ ⎢ ⎥ ⎣ 0 1 0 0 1 ⎦ 1 1 1 0 1 0 1 1 1 ⎤ ⎤ ⎤ ⎡ ⎡ ⎡ 0 1 0 1 0 0 0 1 0 0 0 1 1 0 ⎢ 1 0 1 1 0 0 ⎥ ⎢ 0 0 1 0 ⎥ ⎢ 0 1 1 1 ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ (f) D2 : ⎢ ⎢ 0 1 0 0 1 1 ⎥ (e) D1 : ⎣ 1 0 0 1 ⎦ ⎣ 0 1 0 1 ⎦ ⎥ (d) G4 : ⎢ ⎢ 1 1 0 0 1 0 ⎥ 1 1 0 0 0 0 0 1 ⎥ ⎢ ⎣ 0 0 1 1 0 1 ⎦ 0 0 1 0 1 0 ⎤ ⎡ ⎤ ⎡ 0 1 0 0 0 0 0 0 0 1 1 0 0 ⎢ 1 0 0 0 1 ⎥ ⎢ 0 0 1 0 0 0 0 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 1 0 0 0 0 ⎥ 1 0 0 1 0 (g) D3 : ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 0 0 0 0 1 0 0 0 ⎥ ⎣ 0 0 1 0 1 ⎦ ⎥ ⎢ (h) D4 : ⎢ ⎥ 0 1 0 1 0 ⎢ 0 0 0 0 0 1 0 0 ⎥ ⎢ 0 0 0 0 0 0 1 0 ⎥ ⎥ ⎢ ⎣ 0 0 0 0 0 0 0 1 ⎦ 1 0 0 0 0 0 0 0 (2) For a matrix to be the adjacency matrix for a simple graph or digraph, it must be a square matrix, all of whose entries are either 0 or 1. In addition, the adjacency matrix for a graph must be symmetric. A cannot be the adjacency matrix for either a graph or digraph because it is not square. B cannot be the adjacency matrix for either a graph or digraph because its (1,1) entry equals 2, which is not 0 or 1. C cannot be the adjacency matrix for either a graph or digraph because its (1,1) entry equals 6, which is not 0 or 1. D cannot be the adjacency matrix for either a graph or digraph because it is not square. E cannot be the adjacency matrix for either a graph or digraph because its (1,4) entry equals 6, which is not 0 or 1. F can be the adjacency matrix for either a graph or digraph (see Figure 5 on next page). G can be the adjacency matrix for a digraph only, since G is not symmetric (see Figure 6 on next page). H can be the adjacency matrix for a digraph only, since H is not symmetric (see Figure 7 on next page). I can be the adjacency matrix for a graph or digraph (see Figure 8 on next page). 225


226

Section 8.1


Section 8.1

J cannot be the adjacency matrix for either a graph or digraph because its (1,2) entry equals 2, which is not 0 or 1. K can be the adjacency matrix for a graph or digraph (see Figure 9 on previous page). L can be the adjacency matrix for a graph or digraph (see Figure 10 on previous page). M cannot be the adjacency matrix for either a graph or digraph because its (1,1) entry equals −2, which is not 0 or 1. (3) The digraph is shown in Figure 11 on the previous page, and the adjacency matrix is

A B C D E F

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

A B C 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0

D 0 1 0 0 1 1

E 0 1 1 0 0 0

F 0 0 0 1 0 0

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎦

where we have placed a 1 in the (i, j) entry of the matrix as well as a corresponding directed edge in the digraph, if author i influences author j. The transpose gives no new information. But it does suggest a different interpretation of the results: namely, the (i, j) entry of the transpose equals 1 if author j influences author i (see Figure 11 on previous page). (4) The adjacency matrix for the digraph in Figure ⎡ 0 ⎢ 1 ⎢ A=⎢ ⎢ 0 ⎣ 0 1

8.5 in Section 8.1 of the textbook is ⎤ 1 1 0 1 1 1 0 0 ⎥ ⎥ 1 0 1 0 ⎥ ⎥, 0 1 0 0 ⎦ 1 0 1 0

whose (i, j) entry is 1 if there is an edge directed from the vertex Pi to the vertex Pj and is 0 otherwise. ⎡ ⎤ ⎡ ⎤ 2 3 1 2 0 3 6 7 1 2 ⎢ 1 3 2 1 1 ⎥ ⎢ 4 7 5 3 1 ⎥ ⎢ ⎥ ⎢ ⎥ 2 3 ⎢ ⎥ ⎥ Direct computation shows that A = ⎢ 1 1 2 0 0 ⎥ and A = ⎢ ⎢ 1 4 2 2 1 ⎥. ⎣ 0 1 0 1 0 ⎦ ⎣ 1 1 2 0 0 ⎦ 1 2 3 0 1 3 7 3 4 1 (a) The number of paths of length 3 from P2 to P4 is the (2,4) entry of A3 . Hence, there are 3 such paths. (c) The number of paths of length 1, 2, and 3 from P3 to P2 is the (3,2) entry of A, A2 , and A3 , respectively. Adding these three matrix entries yields 1 + 1 + 4 = 6. Hence, the number of paths of length ≤ 3 from P3 to P2 is 6. (e) The number of paths of length k from P4 to P5 is the (4,5) entry of Ak . Thus, to find the shortest path from P4 to P5 , we look for the smallest positive integer power k for which the (4,5) entry of Ak is nonzero, indicating that a path of length k exists. Inspecting the matrices A, A2 , and A3 , above, we see that there is no path of length up to 3 from P4 to P5 , since each of these matrices has a 0 in its (4,5) entry. However, the (4,5) entry of A4 is the dot product of the 4th row of A with the 5th column of A3 , which is 1. Therefore, there is one path of length 4 from P4 to P5 , and the shortest path from P4 to P5 has length 4. 227


(5) The adjacency matrix for the digraph in Figure ⎡ 0 ⎢ 0 ⎢ A=⎢ ⎢ 1 ⎣ 0 1

Section 8.1

8.6 in Section 8.1 of the textbook is ⎤ 1 1 0 0 0 1 1 0 ⎥ ⎥ 1 0 1 0 ⎥ ⎥, 0 1 1 0 ⎦ 1 0 0 0

whose (i, j) entry is 1 if there is an edge directed from the vertex Pi to the vertex Pj and is 0 otherwise. ⎤ ⎤ ⎡ ⎡ 1 1 1 2 0 1 2 4 4 0 ⎢ 1 1 1 2 0 ⎥ ⎢ 1 2 4 4 0 ⎥ ⎥ ⎥ ⎢ ⎢ 2 3 ⎥ ⎥ ⎢ Direct computation shows that A = ⎢ 0 1 3 2 0 ⎥ and A = ⎢ ⎢ 3 3 3 6 0 ⎥. ⎣ 1 1 1 2 0 ⎦ ⎣ 1 2 4 4 0 ⎦ 0 1 2 1 0 2 2 2 4 0 (a) The number of paths of length 3 from P2 to P4 is the (2,4) entry of A3 . Hence, there are 4 such paths. (c) The number of paths of length 1, 2, and 3 from P3 to P2 is the (3,2) entry of A, A2 , and A3 , respectively. Adding these three matrix entries yields 1 + 1 + 3 = 5. Hence, the number of paths of length ≤ 3 from P3 to P2 is 5. (e) An inspection of the digraph shows that no edges go to P5 . Hence, there are no paths from any vertex of any length going to P5 . This can also be seen in the adjacency matrix. Since the 5th column of A has all zeroes, every power Ak of A (for integer k > 0) will also have all zeroes in its 5th column (because the 5th column of Ak equals Ak−1 multiplied by the 5th column of A). (6) In each part, we use the adjacency matrices for the digraphs and their powers as computed above for Exercises 4 and 5. (a) The number of cycles of length 3 connecting P2 to itself is the (2,2) entry of A3 . Using the matrix A3 from Exercise 4 yields 7 cycles of length 3 connecting P2 to itself in Figure 8.5. Using the matrix A3 from Exercise 5 yields 2 cycles of length 3 connecting P2 to itself in Figure 8.6. (c) The number of cycles of length ≤ 4 connecting P4 to itself is the sum of the (4,4) entries of A, A2 , A3 , and A4 . For Figure 8.5, since we did not compute A4 in Exercise 4, we still need the (4,4) entry of A4 . This is the dot product of the 4th row of A with the 4th column of A3 , resulting in 2. Using this and the (4,4) entries from the matrices from Exercise 4 yields 0 + 1 + 0 + 2 = 3 cycles of length ≤ 4 connecting P4 to itself in Figure 8.5. For Figure 8.6, since we did not compute A4 in Exercise 5, we still need the (4,4) entry of A4 . This is the dot product of the 4th row of A with the 4th column of A3 , resulting in 10. Using this and the (4,4) entries from the matrices from Exercise 5 yields 1 + 2 + 4 + 10 = 17 cycles of length ≤ 4 connecting P4 to itself in Figure 8.6. (7) (a) If the vertex is the ith vertex, then the ith row and ith column entries of the adjacency matrix all equal zero, except possibly for the (i, i) entry, which would equal 1 if there is a loop at the ith vertex. (b) If the vertex is the ith vertex, then the ith row entries of the adjacency matrix all equal zero, except possibly for the (i, i) entry. (Note: The ith column entries may be nonzero since there may be edges connecting other vertices to the ith vertex.) (8) (a) Since a 1 in the (i, i) entry indicates the existence of a loop at the ith vertex, the trace, which is the sum of the main diagonal entries, equals the total number of loops in the graph or digraph. 228


Section 8.1

(9) (a) The digraph in Figure 8.5 is strongly connected because there is a path of length ≤ 4 from any vertex to any other. In the solution to Exercise 4, above, we computed the adjacency matrix A for this digraph, and its powers A2 and A3 . Note that only the (4,5) entry is zero in all of the matrices A, A2 , and A3 simultaneously, showing that there is a path of length ≤ 3 from each vertex to every other vertex, except from P4 to P5 . However, in the solution to Exercise 4(e), above, we computed the (4,5) entry of A4 to be 1. Hence, there is one path of length 4 from P4 to P5 , which completes the verification that the digraph is strongly connected. The digraph in Figure 8.6 is not strongly connected since there is no path directed to P5 from any other vertex (see answer to Exercise 5(e), above). (10) (b) Yes, it is a dominance digraph, because no tie games are possible and because each team plays every other team. Thus, if Pi and Pj are two given teams, either Pi defeats Pj or vice versa. (11) We prove Theorem 8.1 by induction on k. For the base step, k = 1. This case is true by the definition of the adjacency matrix. For the inductive step, we assume the theorem is true for k, and prove that it is true for k + 1. Let B = Ak . Then Ak+1 = BA. Note that (i, j) entry of Ak+1

(ith row of B) · (jth column of A) n / = biq aqj =

=

q=1 n /

(number of paths of length k from Pi to Pq )

q=1

· (number of paths of length 1 from Pq to Pj ) =

number of paths of length k + 1 from Pi to Pj .

The last equation is true because any path of length k + 1 from Pi to Pj must be composed of a path of length k from Pi to Pq for some q followed by a path of length 1 from Pq to Pj . (12) (a) True. In a simple graph, no distinction is made regarding the direction of any edge. Thus, there is an edge connecting Pi to Pj if and only if there is an edge connecting Pj to Pi . The same edge is used in both cases. Hence, the (i, j) entry of the adjacency matrix must have the same value as the (j, i) entry of the matrix. (b) False. By definition, the adjacency matrix of a simple digraph can contain only the numbers 0 and 1 as entries. (c) True. The (1,2) entry of An gives the number of paths of length n from P1 to P2 . If this entry is zero for all n ≥ 1, then there are no paths of any length from P1 to P2 . (d) False. In a simple graph, a single edge connecting Pi and Pj for i = j places a 1 in both the (i, j) and the (j, i) entries of the adjacency matrix. Hence, summing 1’s in the adjacency matrix would count all such edges twice. For example, the graph G1 in Example 1 of Section 8.1 of the textbook has 4 edges, but its adjacency matrix contains eight 1’s. Note that G2 in that same example has 6 edges, and its adjacency matrix has eleven 1’s. (The loop connecting P2 to itself generates only one 1 in the adjacency matrix.) (e) True. In a simple digraph, since the edges are directed, each edge corresponds to exactly one nonzero entry in the associated adjacency matrix. That is, an edge from Pi to Pj places a 1 in the (i, j) entry of the matrix, and in no other entry.

229


Section 8.2

(f) True. To get from P1 to P3 in k + j steps, first follow the path of length k from P1 to P2 , and then follow the path of length j from P2 to P3 . (g) True. The (k, i) entry of the adjacency matrix gives the number of edges connecting Pk to Pi (either 0 or 1). Adding these numbers for 1 ≤ k ≤ n gives the total the number of edges connecting all vertices to Pi .

Section 8.2 (1) (a) The given circuit has two junctions and two loops. By Kirchhoff’s First Law, both junctions produce the same equation: I1 = I2 + I3 . The two loops starting and ending at the voltage source 36V are (1) (2)

I1 → I2 → I1 I1 → I3 → I1 .

Kirchhoff’s Second Law gives an Ohm’s Law equation for each of these two loops: 36V − I2 (4Ω) − I1 (2Ω) = 0 36V + 7V − I3 (9Ω) − I1 (2Ω) = 0 These relationships lead to ⎧ ⎨ I1 − I2 − I3 2I1 + 4I2 ⎩ + 9I3 2I1

(loop 1) (loop 2)

the following system of linear equations: = = =

0 36 . 43

Thus, we row reduce " ⎤ ⎡ ⎡ 1 1 −1 −1 "" 0 ⎣ 2 4 0 "" 36 ⎦ to obtain ⎣ 0 0 2 0 9 " 43

0 1 0

0 0 1

" ⎤ " 8 " " 5 ⎦. " " 3

This gives the solution I1 = 8, I2 = 5, and I3 = 3. (c) The given circuit has four junctions and four loops. By Kirchhoff’s First Law, the junctions produce the following equations: I1 I6 I6 I1

= = = =

I2 I3 I3 I2

+ + + +

I6 I4 I4 I6

+ I5 . + I5

The last two of these equations are redundant. Next, the four loops starting and ending at the voltage source 42V are (1) (2) (3) (4)

I1 I1 I1 I1

→ I6 → I6 → I6 → I2

→ I3 → I6 → I1 → I4 → I6 → I1 → I5 → I6 → I1 → I1

Kirchhoff’s Second Law gives an Ohm’s Law equation for each of these four loops: 42V 42V 42V 42V

− I3 (6Ω) − I1 (2Ω) = 0 − I4 (9Ω) − I1 (2Ω) = 0 − I5 (9Ω) − I1 (2Ω) = 0 + 7V − I2 (5Ω) − I1 (2Ω) = 0

(loop (loop (loop (loop

1) 2) 3) 4)

These relationships lead to the following system of linear equations: 230

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

I1

2I1 2I1 ⎪ ⎪ ⎪ ⎪ 2I1 ⎪ ⎪ ⎩ 2I1

−

I2 +

+

Thus, we ⎡ 1 −1 ⎢ 0 0 ⎢ ⎢ 2 0 ⎢ ⎢ 2 0 ⎢ ⎣ 2 0 2 5

I3 6I3

+

I4

+

9I4

+

I5

+

9I5

− I6 − I6

= = = = = =

5I2

0 0 42 42 42 49

row reduce 0 1 6 0 0 0

0 1 0 9 0 0

" ⎡ ⎤ 1 0 −1 "" 0 ⎢ 0 1 −1 "" 0 ⎥ ⎢ ⎥ ⎢ 0 0 "" 42 ⎥ ⎥ to obtain ⎢ 0 ⎢ 0 ⎥ " 0 0 " 42 ⎥ ⎢ " ⎣ 0 ⎦ 9 0 " 42 " 0 0 0 49

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

Section 8.3

0 0 0 0 1 0

0 0 0 0 0 1

" ⎤ " 12 " " 5 ⎥ ⎥ " " 3 ⎥ ⎥ " " 2 ⎥. ⎥ " " 2 ⎦ " " 7

This gives the solution I1 = 12, I2 = 5, I3 = 3, I4 = 2, I5 = 2, and I6 = 7. (2) (a) True. However, some of the equations produced by the junctions may be redundant, as in the solution to Exercise 1(c), above. (b) True. This is Ohm’s law: V = IR.

Section 8.3 (1) We use A and B to represent the matrices determined by the given points as described in Theorem 8.2. (a) Using the ⎡ 1 ⎢ 1 A=⎢ ⎣ 1 1

given data, the matrices A and B are ⎡ ⎤ ⎤ −8 3 ⎢ ⎥ 1 ⎥ ⎥ and B = ⎢ −5 ⎥. ⎦ ⎣ −4 ⎦ 0 −1 2 4 6 −18 T T Direct computation yields A A = and A B = . 6 14 −31

Row reducing the augmented matrix " " 4 6 "" −18 1 0 "" −3.3 gives , 6 14 " −31 0 1 " −0.8 and so the line of best fit is y = −0.8x − 3.3. Plugging in x = 5 produces y ≈ −7.3. (c) Using the ⎡ 1 ⎢ 1 ⎢ A=⎢ ⎢ 1 ⎣ 1 1

given data, the matrices A and B are ⎡ ⎤ ⎤ 10 −4 ⎢ 8 ⎥ −3 ⎥ ⎢ ⎥ ⎥ ⎥ ⎥ −2 ⎥ and B = ⎢ ⎢ 7 ⎥. ⎣ 5 ⎦ −1 ⎦ 4 0 5 −10 34 T T Direct computation yields A A = and A B = . −10 30 −83

Row reducing the augmented matrix

231


" 5 −10 "" 34 1 gives −10 30 " −83 0

Section 8.3

" 0 "" 3.8 , 1 " −1.5

and so the line of best fit is y = −1.5x + 3.8. Plugging in x = 5 produces y ≈ −3.7. (2) We use A and B to represent the matrices determined by the given points as described in Theorem 8.2. (a) Using the given data, the matrices A and B are ⎡ ⎤ ⎤ ⎡ 8 1 −4 16 ⎢ 5 ⎥ ⎢ 1 −2 4 ⎥ ⎥ and B = ⎢ ⎥ . A=⎢ ⎣ 3 ⎦ ⎣ 1 0 0 ⎦ 6 1 2 4 ⎡ ⎤ ⎡ ⎤ 4 −4 24 22 24 −64 ⎦ and AT B = ⎣ −30 ⎦. Direct computation yields AT A = ⎣ −4 24 −64 288 172 Row reducing the augmented matrix " ⎡ ⎤ ⎡ 4 −4 24 "" 22 1 ⎣ −4 24 −64 "" −30 ⎦ gives ⎣ 0 0 24 −64 288 " 172

0 1 0

0 0 1

" ⎤ " 3.600 " " 0.350 ⎦, " " 0.375

and so the least squares quadratic polynomial is y = 0.375x2 + 0.35x + 3.6. (c) Using the ⎡ 1 ⎢ 1 ⎢ A=⎢ ⎢ 1 ⎣ 1 1

given data, the matrices A ⎡ ⎤ −3 −4 16 ⎢ −2 −3 9 ⎥ ⎢ ⎥ ⎢ −2 4 ⎥ ⎥ and B = ⎢ −1 ⎣ 0 0 0 ⎦ 1 1 1 ⎡

and B are ⎤ ⎥ ⎥ ⎥. ⎥ ⎦

Row reducing the augmented matrix " ⎤ ⎡ ⎡ 1 5 −8 30 "" −5 ⎣ −8 30 −98 "" 21 ⎦ gives ⎣ 0 0 30 −98 354 " −69

⎤ ⎡ ⎤ 30 −5 −98 ⎦ and AT B = ⎣ 21 ⎦. 354 −69

−8 30 −98

5 Direct computation yields AT A = ⎣ −8 30 0 1 0

0 0 1

" ⎤ " 0.266 " " 0.633 ⎦, " " −0.042

and so the least-squares quadratic polynomial is y = −0.042x2 + 0.633x + 0.266. (3) We use A and B to represent the matrices determined by the given points as described in Theorem 8.2. (a) Using the ⎡ 1 ⎢ 1 ⎢ A=⎢ ⎢ 1 ⎣ 1 1

given data, the matrices A and B are ⎡ ⎤ ⎤ −3 −3 9 −27 ⎢ −1 ⎥ −2 4 −8 ⎥ ⎥ ⎢ ⎥ ⎥ ⎥ −1 1 −1 ⎥ and B = ⎢ ⎢ 0 ⎥. ⎣ ⎦ 1 ⎦ 0 0 0 4 1 1 1

232


5 ⎢ −5 Direct computation yields AT A = ⎢ ⎣ 15 −35

−5 15 −35 99

Row reducing the augmented matrix ⎡ " ⎡ ⎤ 1 5 −5 15 −35 "" 1 ⎢ ⎢ 0 ⎢ −5 15 −35 99 "" 15 ⎥ ⎥ gives ⎢ ⎢ ⎢ 0 ⎣ 15 −35 99 −275 "" −27 ⎦ ⎣ −35 99 −275 795 " 93 0 and so the least-squares cubic polynomial is y =

15 −35 99 −275

Section 8.3

⎡ ⎤ −35 1 ⎢ 99 ⎥ 15 ⎥ and AT B = ⎢ ⎣ −27 −275 ⎦ 795 93 " 0 "" " 0 " " 0 "" " 1 "

⎤ ⎥ ⎥. ⎦

⎤

37 35 25 ⎥ ⎥ 1 0 14 ⎥ , 25 ⎥ 0 1 28 ⎦ 1 0 0 4 1 3 25 2 25 37 4 x + 28 x + 14 x + 35 .

0

0

(4) We use A and B to represent the matrices determined by the computed points as described in Theorem 8.2. (a) First, we use the function y = x4 and the given x-values to generate the following data points: (−2, 16), (−1, 1), (0, 0), (1, 1), (2, 16). Using these data, the matrices A and B are ⎤ ⎡ ⎤ ⎡ 16 1 −2 4 ⎢ 1 ⎥ ⎢ 1 −1 1 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ 0 0 ⎥ and B = ⎢ A=⎢ 1 ⎢ 0 ⎥. ⎣ ⎦ ⎣ 1 1 ⎦ 1 1 16 1 2 4 ⎡ ⎤ ⎡ ⎤ 5 0 10 34 Direct computation yields AT A = ⎣ 0 10 0 ⎦ and AT B = ⎣ 0 ⎦. 10 0 34 130 Row reducing the augmented matrix " ⎤ ⎡ ⎡ 1 0 5 0 10 "" 34 ⎣ 0 10 0 ⎦ gives ⎣ 0 1 0 "" 0 0 10 0 34 " 130

0 0 1

" ⎤ " −2.0571 " " 0.0000 ⎦, " " 4.4286

and so the least-squares quadratic polynomial is y = 4.4286x2 − 2.0571. (c) First, we use the function y = ln x and the given x-values to generate the following data points: (1, 0), (2, 0.6931), (3, 1.0986), (4, 1.3863). Note that, while we list only 4 digits after the decimal point here, all computations were actually performed on a calculator displaying 12 significant digits of accuracy. Using these data, the matrices A and B are ⎡ ⎤ ⎤ ⎡ 0.0000 1 1 1 ⎢ ⎥ ⎢ 1 2 4 ⎥ ⎥ and B = ⎢ 0.6931 ⎥ . A=⎢ ⎣ 1.0986 ⎦ ⎣ 1 3 9 ⎦ 1.3863 1 4 16 ⎡ ⎤ ⎡ ⎤ 4 10 30 3.1781 Direct computation yields AT A = ⎣ 10 30 100 ⎦ and AT B = ⎣ 10.2273 ⎦. 30 100 354 34.8408 Row reducing the augmented matrix

233


4 ⎣ 10 30

10 30 100

30 100 354

" ⎤ ⎡ " 3.1781 1 " " 10.2273 ⎦ gives ⎣ 0 " " 34.8408 0

0 1 0

0 0 1

Section 8.3

" ⎤ " −0.8534 " " 0.9633 ⎦, " " −0.1014

and so the least-squares quadratic polynomial is y = −0.1014x2 + 0.9633x − 0.8534. (e) First, we use the function y = cos x and the given x-values to generate the following data points: (− π2 , 0), (− π4 ,

√ 2 2 ),

(0, 1), ( π4 ,

√ 2 2 ),

( π2 , 0). Making decimal approximations yields (−1.5708, 0),

(−0.7854, 0.7071), (0, 1), (0.7854, 0.7071), (1.5708, 0). Note that, while we list only 4 digits after the decimal point here, all computations were actually performed on a calculator displaying 12 significant digits of accuracy. Using these data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 0 1 −1.5708 2.4674 −3.8758 ⎢ 0.7071 ⎥ ⎢ 1 −0.7854 0.6169 −0.4845 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ and B = ⎢ 1 ⎥ 1 0 0 0 A = ⎢ ⎢ ⎥ . Direct computation yields ⎢ ⎥ ⎣ 0.7071 ⎦ ⎣ 1 0.7854 0.6169 0.4845 ⎦ 0 1 1.5708 2.4674 3.8758 ⎡ ⎤ ⎡ ⎤ 5 0 6.1685 0 2.4142 ⎢ ⎢ 0 6.1685 0 12.9371 ⎥ 0 ⎥ ⎥ and AT B = ⎢ ⎥ AT A = ⎢ ⎣ 6.1685 ⎣ 0.8724 ⎦. 0 12.9371 0 ⎦ 0 12.9371 0 30.5128 0 Row reducing the augmented matrix " ⎡ 5 0 6.1685 0 "" 2.4142 ⎢ 0 0 6.1685 0 12.9371 "" ⎢ ⎣ 6.1685 0 12.9371 0 "" 0.8724 0 0 12.9371 0 30.5128 "

⎤

⎡

⎢ ⎥ ⎥ gives ⎢ I4 ⎣ ⎦

" " 0.9706 " " 0 " " −0.3954 " " 0

⎤ ⎥ ⎥, ⎦

and so the least-squares cubic polynomial is y = 0x3 − 0.3954x2 + 0.9706. (5) We use A and B to represent the matrices determined by the given points as described in Theorem 8.2. This problem uses the following set of data points: (1, 3), (2, 3.3), (3, 3.7), (4, 4.1), (5, 4.6). ⎡ ⎤ ⎡ ⎤ 3.0 1 1 ⎢ 3.3 ⎥ ⎢ 1 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ and B = ⎢ 3.7 ⎥ . 1 3 (a) Using the given data, the matrices A and B are A = ⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 4.1 ⎦ ⎣ 1 4 ⎦ 4.6 1 5 5 15 18.7 Direct computation yields AT A = and AT B = . 15 55 60.1 Row reducing the augmented matrix " " 1 0 "" 2.54 5 15 "" 18.7 , gives 0 1 " 0.40 15 55 " 60.1 and so the line of best fit is y = 0.4x + 2.54. Next, to find out when the angle reaches 20◦ , set y = 20 in this equation and solve for x, producing x = 17.46 0.4 = 43.65. Hence, the angle reaches 20◦ in the 44th month. (b) Using the given data, the matrices A and B are

234

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ ⎢ ⎢ A=⎢ ⎢ ⎣

1 1 1 1 1

1 2 3 4 5

1 4 9 16 25

⎤

⎡

⎥ ⎢ ⎥ ⎢ ⎥ and B = ⎢ ⎥ ⎢ ⎦ ⎣

⎤ 3.0 3.3 ⎥ ⎥ 3.7 ⎥ ⎥. 4.1 ⎦ 4.6 ⎡

⎤ ⎡ ⎤ 15 55 18.7 55 225 ⎦ and AT B = ⎣ 60.1 ⎦. 225 979 230.1

5 Direct computation yields AT A = ⎣ 15 55 Row reducing the augmented matrix " ⎤ ⎡ ⎡ 1 5 15 55 "" 18.7 ⎣ 15 55 225 "" 60.1 ⎦ gives ⎣ 0 0 55 225 979 " 230.1

Section 8.3

0 1 0

0 0 1

" ⎤ " 2.74 " " 0.2286 ⎦, " " 0.02857

and so the least-squares quadratic polynomial is y = 0.02857x2 + 0.2286x + 2.74. Next, to find out when the angle reaches 20◦ , set y = 20 in this equation and solve for x. The Quadratic Formula produces the positive solution x ≈ 20.9. Hence, the angle reaches 20◦ in the 21st month. (7) We use A and B to represent the matrices determined by the given points as described in Theorem 8.2. Using the given data, the matrices A and B are ⎡ ⎤ ⎡ ⎤ 1 −2 4 6 0 0 ⎦ and B = ⎣ 2 ⎦ . A=⎣ 1 1 3 9 8 ⎡ ⎤ ⎡ ⎤ 3 1 13 16 Direct computation yields AT A = ⎣ 1 13 19 ⎦ and AT B = ⎣ 12 ⎦. 13 19 97 96 Row reducing the augmented matrix " ⎡ ⎤ " ⎤ ⎡ 1 0 0 "" 2 3 1 13 "" 16 " ⎥ ⎣ 1 13 19 " 12 ⎦ gives ⎢ ⎣ 0 1 0 " − 25 ⎦, and so the least-squares quadratic polynomial is " " " 13 19 97 96 0 0 1 " 45 y = 45 x2 − 25 x + 2. Since y = 6 when x = −2, y = 2 when x = 0, and y = 8 when x = 3, this polynomial is the exact quadratic through the given three points. ⎡ ⎤ ⎡ ⎤ 4 −3 12 5 ⎦ and B = ⎣ 32 ⎦ . (9) (a) The matrices A and B are A = ⎣ 2 3 1 21 29 1 175 and AT B = . Direct computation yields AT A = 1 35 145 Row reducing the augmented matrix " ' ( " 1 0 "" 230 29 1 "" 175 39 gives , " 1 35 " 145 0 1 " 155 39

235


230 39 ,

x2 =

− 3x2

=

11 23 ,

which is almost 12

+

=

31 23 ,

which is almost 32

=

21 23 ,

which is close to 21

giving the least-squares solutions x1 = ⎧ 4x1 ⎪ ⎪ ⎨ 2x1 ⎪ ⎪ ⎩ 3x1

5x2

+

x2

155 39 .

Section 8.4

Note that

.

Hence, the least-squares solution comes close to satisfying the original inconsistent system. (10) (a) True. Since the distance of each data point to the line is 0, the sum of the squares of these distances is also 0. That is the minimum possible value for the sum Sf , since all such sums are nonnegative. (b) False. The computed coefficient of x3 in the degree three least-squares polynomial might be zero. For example, if all of the data points lie on a straight line, that line will be the degree t least-squares polynomial for every t ≥ 1. (c) False. For example, it is easy to check that the least-squares line computed in Example 1 of Section 8.3 of the textbook does not pass through any of the data points used to compute the line. (d) False. Theorem 8.2 states that A is an n × (t + 1) matrix. Hence, AT is a (t + 1) × n matrix. Therefore, AT A is a (t + 1) × (t + 1) matrix.

Section 8.4 (1) A is not stochastic, since A is not square; A is not regular, since A is not stochastic. B is not stochastic, since the entries of column 2 do not sum to 1; B is not regular, since B is not stochastic. C is stochastic; C is regular, since C is stochastic and has all nonzero entries. D is stochastic; D is not regular, since every positive power of D is a matrix whose rows are the rows of D permuted in some order, and hence every such power contains zero entries. E is not stochastic, since the entries of column 1 do not sum to 1; E is not regular, since E is not stochastic. F is stochastic; F is not regular, since every positive power of F has all second row entries zero. G is not stochastic, since G is not square; G is not regular, since G is not stochastic. ⎡ 1 1 1 ⎤ ⎢ H is stochastic; H is regular, since H = ⎢ ⎣ 2

' (2) (a) First, p1 = Mp = 0.3102 ≈ . 0.6898

⎡

⎢ (c) First, p1 = Mp = ⎣

1 4 3 4

1 3 2 3

1 4 1 2 1 4

1 3 1 3 1 3

('

1 2 1 6 1 3

2 3 1 3

(

⎤⎡ ⎥⎢ ⎦⎣

2

4

4

1 4

1 2

1 4

1 4

1 4

1 2

' =

1 4 1 2 1 4

⎤

5 18 13 18

(

⎡

⎥ ⎢ ⎦=⎣

236

⎥ ⎥, which has no zero entries. ⎦ '

. Then p2 = Mp1 =

17 48 1 3 5 16

⎤ ⎥ ⎦.

1 4 3 4

1 3 2 3

('

5 18 13 18

(

' =

67 216 149 216

(

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ ⎢ Then p2 = Mp1 = ⎣

1 4 1 2 1 4

1 3 1 3 1 3

1 2 1 6 1 3

⎤⎡ ⎥⎢ ⎦⎣

17 48 1 3 5 16

⎤

⎡

⎥ ⎢ ⎦=⎣

(3) (a) We must solve the system (M − I2 )x = 0, " ⎤ ⎡ ⎡ 1 1 " 1 −2 3 " 0 ⎥ ⎢ ⎢ 1 1 " we row reduce ⎣ 2 − 3 " 0 ⎦ to ⎣ 0 " 0 1 1 " 1

205 576 49 144 175 576

⎤

Section 8.4

⎡

⎤ 0.3559 ⎥ ⎣ ⎦ ≈ 0.3403 ⎦. 0.3038

under the additional condition that x1 + x2 = 1. Thus, " ⎤ 0 "" 25 ⎥ " 1 " 35 ⎦, giving the steady state vector pf = 25 , 35 . " 0 " 0

(4) (a) We will round numbers to four places after 0.4167 0.3889 0.4028 M2 = , M3 = 0.5833 0.6111 0.5972 0.4001 0.3999 0.4000 M5 = , M6 = 0.5999 0.6001 0.6000

the decimal point. Then 0.3981 0.4005 0.3997 , M4 = , 0.6019 0.5995 0.6003 0.4000 0.4000 0.4000 7 ,M = . 0.6000 0.6000 0.6000

Since all entries of M6 and M7 agree to 4 places after the decimal point, we stop here. Hence, the steady state vector for this Markov chain is given by any of the columns of M7 , namely [0.4000, 0.6000], rounded to 4 significant digits. (5) (a) The initial probability vector p is [0.3, 0.15, 0.45, 0.1]. (Note that 10% of the citizens did not vote in the last election.) Then ⎤ ⎡ ⎤⎡ ⎤ ⎡ 0.340 0.7 0.2 0.2 0.1 0.30 ⎢ 0.1 0.6 0.1 0.1 ⎥ ⎢ 0.15 ⎥ ⎢ 0.175 ⎥ ⎥ ⎥⎢ ⎥ ⎢ p1 = Mp = ⎢ ⎣ 0.1 0.2 0.6 0.1 ⎦ ⎣ 0.45 ⎦ = ⎣ 0.340 ⎦ . 0.145 0.1 0.0 0.1 0.7 0.10 Thus, in the next election, 34% will vote for Party A, 17.5% will vote for Party C, and 14.5% will not vote at all. For the election after that, ⎡ ⎤⎡ ⎤ ⎡ 0.7 0.2 0.2 0.1 0.340 ⎢ 0.1 0.6 0.1 0.1 ⎥ ⎢ 0.175 ⎥ ⎢ ⎥⎢ ⎥ ⎢ p2 = Mp1 = ⎢ ⎣ 0.1 0.2 0.6 0.1 ⎦ ⎣ 0.340 ⎦ = ⎣ 0.1 0.0 0.1 0.7 0.145

for Party B, 34% will vote ⎤ 0.3555 0.1875 ⎥ ⎥. 0.2875 ⎦ 0.1695

Hence, in that election, 35.55% will vote for Party A, 18.75% will vote for Party B, 28.75% will vote for Party C, and 16.95% will not vote at all. (b) A lengthy computation shows that M100 p1 ≈ [0.36, 0.20, 0.24, 0.20]. Hence, in a century, the votes would be 36% for Party A and 24% for Party C. For an alternate approach, we assume that by the end of a century the system will have achieved its steady state. Hence, we can solve for the steady-state vector. Since the transition matrix M is regular (you can easily verify that M2 has no zero entries), part (5) of Theorem 8.5 shows that we can solve for pf by solving the system (M − I4 )x = 0, under the additional condition that x1 + x2 + x3 + x4 = 1. Hence, we row reduce " " ⎤ ⎤ ⎡ ⎡ 1 0 0 0 "" 0.36 −0.3 0.2 0.2 0.1 "" 0 ⎢ 0 1 0 0 " 0.20 ⎥ ⎢ 0.1 −0.4 0.1 0.1 "" 0 ⎥ ⎥ " ⎥ ⎢ ⎢ " 0 ⎥ to ⎢ 0 0 1 0 " 0.24 ⎥ . ⎢ 0.1 0.2 −0.4 0.1 ⎥ " ⎥ ⎢ " ⎢ ⎣ 0 0 0 1 " 0.20 ⎦ ⎣ 0.1 0.0 0.1 −0.3 "" 0 ⎦ " 0 0 0 0 " 0.00 1.0 1.0 1.0 1.0 " 1 237


Section 8.4

Therefore, the steady-state vector is [0.36, 0.20, 0.24, 0.20]. Again, we get that, in a century, the votes would be 36% for Party A and 24% for Party C. (6) (a) To compute the transition matrix M, we use the fact that the (i, j) entry mij of M is the probability that the rat moves from room j to room i, where we use the labels A, B, C, D, and E and 1, 2, 3, 4, and 5 for the rooms interchangeably. So, for example, mii = 12 for each i because the rat has a 50% chance of staying in the same room. To compute m21 , note that each of the 4 doorways leaving room A has the same probability of being used when a doorway is used, namely, 1 1 1 1 1 4 . But, since the rat uses a doorway only 2 of the time, m21 = 2 × 4 = 8 . Similarly, to compute 1 m41 , again note that each door out of room A has probability 4 of being used any time a door is used. Now combining the facts that a doorway is used onlyhalf the time and that there are two doors from room A to room D implies that m41 = 14 × 2 × 12 = 14 . Similar computations produce all of the other entries of M giving ⎤ ⎡ 1 1 1 1 0 2 6 6 5 ⎢ ⎥ ⎢ 1 1 0 0 1 ⎥ ⎢ 8 2 5 ⎥ ⎥ ⎢ ⎢ 1 1 1 ⎥ 1 M = ⎢ 8 0 2 10 10 ⎥ . ⎥ ⎢ ⎢ 1 1 ⎥ ⎥ ⎢ 4 0 16 21 5 ⎦ ⎣ 1 0 13 16 51 2 (b) If M is the stochastic matrix in part (a), then ⎡ 41 1 ⎢ ⎢ ⎢ ⎢ ⎢ 2 M =⎢ ⎢ ⎢ ⎢ ⎣

120

6

1 5

13 60

9 100

1 8

27 80

13 240

13 200

1 5

3 20

13 240

73 240

29 200

3 25

13 48

13 120

29 120

107 300

13 60

9 80

1 3

1 5

13 60

28 75

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎦

which has all entries nonzero. Thus, M is regular. (c) If the rat starts in room C, its initial state vector is p = [0, 0, 1, 0, 0]. By Theorem 8.4, the state vector p2 is given by M2 p. Using the matrix M2 computed in part (b) and the fact that p is the 13 73 29 1 vector e3 , we see that M2 p is just the third column of M2 , namely 15 , 240 , 240 , 120 , 5 . Hence, 29 the probability of the rat being in room D after two time intervals is 120 . (d) Since the transition matrix M is regular, part (5) of Theorem 8.5 shows that we can solve for pf by solving the system (M−I5 )x = 0, under the additional condition that x1 +x2 +x3 +x4 +x5 = 1. Hence, we row reduce " " ⎡ ⎤ ⎡ 1 ⎤ 1 1 1 1 0 0 0 0 "" 15 −2 0 "" 0 6 6 5 " 3 ⎥ ⎢ ⎥ ⎢ 1 1 " 0 0 ⎢ 0 1 0 0 0 " 20 ⎥ ⎢ 8 − 12 5 "" 0 ⎥ " ⎢ ⎥ ⎢ ⎥ 1 1 " 1 ⎢ 0 0 1 0 0 " 3 ⎥ ⎥ ⎢ 1 0 0 − " 20 ⎥ ⎢ ⎥ ⎢ 8 2 10 10 " " ⎥ to ⎢ ⎥. ⎢ 1 1 "" 1 1 ⎢ 0 0 0 1 0 " 14 ⎥ ⎥ ⎢ 4 0 0 − 6 2 5 " ⎢ ⎥ ⎢ " ⎥ " 1 ⎥ ⎢ ⎥ ⎢ " 1 1 1 − 12 " 0 ⎦ ⎣ 0 0 0 0 1 " 4 ⎦ ⎣ 0 3 6 5 " " 0 0 0 0 0 " 0 1 1 1 1 1 " 1

238


Section 8.5

3 3 1 1 Therefore, the steady-state vector is 15 , 20 , 20 , 4 , 4 . Over time, the rat frequents rooms B and C the least and rooms D and E the most. (9) We need to show that the product of any k stochastic matrices is stochastic. We prove this by induction on k. The Base Step is k = 2. Let A and B be n × n stochastic matrices. Then the entries of AB are nonnegative, since the entries ! of both A and B!are nonnegative. Furthermore, the n n sum of!the entries in the jth column of AB =! i=1 (AB)ij =! i=1 (ith row of A)·(jth !n column of n n n B) = i=1 (ai1 b1j + ai2 b2j + · · · + ain bnj ) = ( i=1 ai1 ) b1j + ( i=1 ai2 ) b2j + · · · + ( i=1 ain ) bnj = (1)b1j + (1)b2j + · · · + (1)bnj (since A is stochastic, and each of the summations is a sum of the entries of a column of A) = 1 (since B is stochastic). Hence, AB is stochastic. For the Inductive Step, we assume that the product of any k stochastic n × n matrices is stochastic, and we must show that the product of any k + 1 stochastic n × n matrices is stochastic. So, let A1 , . . . , Ak+1 be stochastic n × n matrices. Then A1 A2 · · · Ak Ak+1 = (A1 A2 · · · Ak )Ak+1 . Now A1 A2 · · · Ak is stochastic by the inductive hypothesis. Hence, (A1 A2 · · · Ak )Ak+1 is stochastic by the Base Step. (10) We prove Theorem 8.4 by induction on n. Suppose M is a k × k matrix. !k !k Base Step (n = 1): ith entry in Mp = (ith row of M)·p = j=1 mij pj = j=1 (probability of moving from state Sj to Si )(probability of being in state Sj ) = probability of being in state Si after 1 step of the process = ith entry of p1 . Inductive Step: Assume pk = Mk p is the probability vector after k steps of the process. Then, after an additional step of the process, the computation in the Base Step shows that probability vector pk+1 = Mpk = M(Mk p) = Mk+1 p. (11) (a) False. While the sum of the entries in each column of a stochastic matrix must equal 1, the sum of the entries in each row is not required to equal 1. For example, this is true of the stochastic matrix M in Example 1 of Section 8.4 of the textbook. Now the columns of the transpose of a matrix equal the rows of the original matrix. Thus, the transpose of a stochastic matrix might not have the sum of its entries along each column equal to 1, and so might not be stochastic. (b) True. If A is upper triangular, then Ak is also upper triangular for all k ≥ 1. Hence, Ak will have zeroes below the main diagonal. (c) True. By part (5) of Theorem 8.5, the given statement is true for the vector p = pf . (d) True. By part (3) of Theorem 8.5, if the Markov chain has a regular transition matrix, then there is a limit vector for every initial vector (and all of these limit vectors are equal). However, starting with the initial vector p, the given condition implies that Mk p = q when k is odd, and Mk p = p when k is even. Hence, lim Mk p cannot exist, since Mk p oscillates between p and q. Therefore, k→∞

M cannot be regular. (e) False. Each entry of the transition matrix gives the probability of changing from one specific state to another.

Section 8.5 (1) (a) First, we must convert the letters in the message “PROOF BY INDUCTION” into numbers by listing each letter’s position in the alphabet: P

R

O

O

F

B

Y

I

N

D

U

C

T

I

O

N

16

18

15

15

6

2

25

9

14

4

21

3

20

9

15

14

239


Next, we 3 A= 5 16 A 18 14 A 4

Section 8.5

take these numbers in pairs to form eight 2-vectors and multiply the matrix −4 times each of these 2-vectors in turn. This yields −7 −24 15 −15 6 10 25 = , A = , A = , A = −46 15 −30 2 16 9 26 21 51 20 24 15 = , A = , A = , A = 42 3 84 9 37 14

39 62

, .

−11 −23

Therefore, the encoded message is −24

−46

−15 −30

10

16

39

62

26

42

51

84

24

37

−11

(2) (a) If A is the given key matrix, by row reducing [A|I3 ] we find A−1 to be A−1

−23 . ⎡ −2 = ⎣ −6 9

−1 −4 4

⎤ 0 1 ⎦. 1

Next, we take the numbers representing the message to be decoded in triples to form eight 3-vectors and multiply the matrix A−1 times each of these 3-vectors in turn. This yields ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −62 8 −32 5 −142 18 A−1 ⎣ 59 ⎦ = ⎣ 23 ⎦, A−1 ⎣ 266 ⎦ = ⎣ 11 ⎦ , A−1 ⎣ 116 ⎦ = ⎣ 15 ⎦ , 107 13 67 15 223 9 ⎤ ⎤ ⎤ ⎡ ⎤ ⎤ ⎡ ⎤ ⎡ ⎡ ⎡ ⎡ 19 15 15 −160 −122 −122 A−1 ⎣ 301 ⎦ = ⎣ 7 ⎦, A−1 ⎣ 229 ⎦ = ⎣ 4 ⎦ , A−1 ⎣ 229 ⎦ = ⎣ 18 ⎦, 6 20 251 15 188 202 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −78 8 −111 15 A−1 ⎣ 148 ⎦ = ⎣ 5 ⎦, A−1 ⎣ 207 ⎦ = ⎣ 21 ⎦. 129 19 183 12 We take the results and associate with each number its corresponding letter in the alphabet: 8

15

13

5

23

15

18

11

9

19

7

15

H

O

M

E

W

O

R

K

I

S

G

O

15

4

6

15

18

20

8

5

19

15

21

12

O

D

F

O

R

T

H

E

S

O

U

L

Therefore, the message is “HOMEWORK IS GOOD FOR THE SOUL.” (3) (a) True. The reason is that the code used for a given letter is dependent on the letters that surround it in the message, not just upon the letter itself. Thus, the same letter may be replaced by several different numbers within the same message. (b) True. If the encoding matrix is singular, then there is no inverse matrix that can be used to decode the message. Decoding would involve solving systems having an infinite number of solutions. (c) False. The message is typically broken up into shorter strings of length k, where k is less than the length of the entire message (see Example 1 in Section 8.5 of the textbook). Of course, the higher the value of k, the more difficult it generally is to break the code. However, large values for k also make it more time consuming to invert the key matrix and to perform the necessary matrix products. 240


Section 8.6

Section 8.6 (1) In each part, let E represent the given matrix. (a) The operation (III): 2 ←→ 3 was performed on I3 to obtain E. The inverse operation is (III): 2 ←→ 3 . Since this row operation is its own inverse, the matrix E is its own inverse. (b) The operation (I): 2 ← −2 2 was performed on I3 to obtain E. The inverse operation is (I): ⎡ ⎤ 1 0 0 2 ←− − 12 2 . Hence, E−1 = ⎣ 0 − 12 0 ⎦, the matrix obtained by performing the inverse 0 0 1 row operation on I3 . (e) The operation (II): 3 ← −2 4 + 3 was performed ⎡ 1 0 0 ⎢ 0 1 0 −1 is (II): 3 ←− 2 4 + 3 . Hence, E = ⎢ ⎣ 0 0 1 0 0 0 the inverse row operation on I4 .

on I4 to obtain E. The inverse operation ⎤ 0 0 ⎥ ⎥, the matrix obtained by performing 2 ⎦ 1

(2) In each part, let A represent the given matrix. (a) We list the row operations needed to convert A to reduced row echelon form, the resulting matrix after each step, the inverse of the row operation used, and the elementary matrix corresponding to each inverse operation: Row Operation 1 ←

1 4

1

Resultant Matrix ' ( 1 94 '

2 ← −3 1 + 2 2 ← 4 2 1 ← − 94 2 + 1

3

7

1

9 4 1 4

0 1 0

9 4

1 0

0 1

Inverse Row Operation 1 ← 4 1

( 2 ← 3 1 + 2

1

2 ←

1 4

2

1 ←

9 4

2 + 1

Elementary Matrix for Inverse Operation 4 0 F4 = 0 1 1 0 F3 = 3 1 ' ( 1 0 F2 = 0 14 1 94 F1 = 0 1

Then A = F4 F3 F2 F1 I2 . (c) We list the row operations needed to convert A to reduced row echelon form, the resulting matrix after each step, the inverse of the row operation used, and the elementary matrix corresponding to each inverse operation:

241


Row Operation

1 ↔ 2

1 ← − 13 1

Resultant Matrix ⎡ ⎤ −3 0 0 −2 ⎢ 0 0 5 0 ⎥ ⎢ ⎥ ⎣ 0 6 −10 −1 ⎦ 3 0 0 3 ⎡ 2 ⎤ 1 0 0 3 ⎢ 0 0 5 0 ⎥ ⎢ ⎥ ⎣ 0 6 −10 −1 ⎦ ⎡

4 ← −3 1 + 4

2 ↔ 3

1 6

2

3 ←

1 5

3

2 ←

5 3

3 + 2

1 ← − 23 4 + 1

2 ←

1 6

4 + 2

1

⎢ 0 ⎢ ⎣ 0 0 ⎡ 1 ⎢ 0 ⎢ ⎣ 0 ⎡

2 ←

3

0

1 ⎢ ⎢ 0 ⎢ ⎣ 0 0 ⎡ 1 ⎢ ⎢ 0 ⎢ ⎣ 0 0 ⎡ 1 ⎢ ⎢ 0 ⎢ ⎣ 0 0 ⎡ 1 ⎢ ⎢ 0 ⎢ ⎣ 0 0 ⎡ 1 ⎢ 0 ⎢ ⎣ 0 0

0

0

3

0

0

2 3

0 6 0

5 −10 0

0

0

6 0 0

−10 5 0

0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0

0

1 ↔ 2

1 ← −3 1

⎤

0 ⎥ ⎥ −1 ⎦ 1 2 ⎤

4 ← 3 1 + 4

3

−1 ⎥ ⎥ 0 ⎦ 1 2 ⎤

2 ↔ 3

3

⎥ − 16 ⎥ ⎥ 5 0 ⎦ 0 1 2 ⎤ 0 3 ⎥ 5 − 3 − 16 ⎥ ⎥ 1 0 ⎦ 0 1 2 ⎤ 0 3 ⎥ 0 − 16 ⎥ ⎥ 1 0 ⎦ 0 1 ⎤ 0 0 ⎥ 0 − 16 ⎥ ⎥ 1 0 ⎦ 0 1 ⎤ 0 0 0 0 ⎥ ⎥ 1 0 ⎦ 0 1 − 53

Inverse Row Operation

2 ← 6 2

3 ← 5 3

2 ← − 53 3 + 2

1 ←

2 3

4 + 1

Section 8.6

Elementary Matrix for Inverse Operation ⎡ ⎤ 0 1 0 0 ⎢ 1 0 0 0 ⎥ ⎥ F9 = ⎢ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ −3 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ F8 = ⎢ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ F7 = ⎢ ⎣ 0 0 1 0 ⎦ 3 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 0 1 0 ⎥ ⎥ F6 = ⎢ ⎣ 0 1 0 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 6 0 0 ⎥ ⎥ F5 = ⎢ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ F4 = ⎢ ⎣ 0 0 5 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 0 ⎢ ⎥ ⎢ 0 1 − 53 0 ⎥ F3 = ⎢ ⎥ ⎣ 0 0 1 0 ⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 23 ⎢ 0 1 0 0 ⎥ ⎥ F2 = ⎢ ⎣ 0 0 1 0 ⎦ ⎡

2 ← − 16 4 + 2

0

0

0

1

0

0

⎢ ⎢ 0 1 0 F1 = ⎢ ⎣ 0 0 1 0 0 0

1 0

⎤

⎥ − 16 ⎥ ⎥ 0 ⎦ 1

Then A = F9 F8 F7 F6 F5 F4 F3 F2 F1 I4 . (7) A is nonsingular if and only if rank(A) = n (by Theorem 2.14) if and only if A is row equivalent to 242


Section 8.7

In (by the definition of rank) if and only if A = Ek · · · E1 In = Ek · · · E1 for some elementary matrices E1 , . . . , Ek (by Theorem 8.8). (10) (a) True. An elementary matrix is formed by performing a single row operation on In . Since In is square, the result of any row operation performed on it will also be square. (b) False. The equation B = EA is true only if B can be obtained from A using a single row operation – namely the row operation corresponding to the elementary matrix E. However, it may take several row operations to convert one matrix A to a row equivalent matrix B. For example, I2 and 5I2 are row equivalent, but it takes at least two row operations to convert I2 to 5I2 . −1 · · · E−1 (c) False. The inverse of E1 E2 · · · Ek is E−1 2 E1 , not Ek · · · E2 E1 . For example, if k 2 0 1 0 −1 and E2 = , then E1 E2 = 2I2 , and so (E1 E2 ) = 12 I2 . However, E1 = 0 1 0 2

E2 E1 = 2I2 . (d) True. If A is nonsingular, then A−1 is also nonsingular. The given statement then follows from Corollary 8.9. (e) True. The proof of Theorem 8.7 shows that the elementary matrix corresponding to the reverse of a row operation R is the inverse of the elementary matrix corresponding to R. The given statement follows from this fact and Theorem 8.6.

Section 8.7

√ (1) (c) For this parabola, the coefficients a, b, and c are a = 3, b = 1, and c = −2 3. Since, a = b, ' √3 ( 1

√ cos θ − sin θ 2 2 1 c 1 π = θ = 2 arctan a−b = 2 arctan − 3 = − 6 . Hence, P = . √ sin θ cos θ 3 − 12 2 √

√

Therefore, x = 23 u + 12 v and y = − 12 u + 23 v. Next, we substitute these expressions for x and y into the original equation for the conic section, producing *√ *√ +2 * +* √ +2 √ + √ 3 3 3 3 1 1 1 1 3 u+ v + − u+ v −2 3 u+ v − u+ v 2 2 2 2 2 2 2 2 √

− 1 + 12 3

*√

1 3 u+ v 2 2

Expanding the terms on the left √ 9 2 3 3 u + uv + 4 2 + *√

1 3 + 18 u − − 2 2

+

* √ + √

3 1 + 12 − 3 v + 26 = 0. − u+ 2 2

side of the equation yields √ √ 3 3 3 2 1 2 3 3 v + u − uv + v 2 + u2 − 3uv − v 2 4 4 2 4 2 2 * √ +

√ √ 3 3 + 6 3 v + −6 + u+ 6 3− v + 26 = 0. 2 2

Combining like terms, rearranging terms, and dividing by 2 yields the following equation in uvcoordinates: v = 2u2 − 12u + 13, which can be expressed, after completing the square on the right side, as (v + 5) = 2(u − 3)2 . From this form, we see that the conic sectionis a parabola and that 3 0.0981 its vertex in uv-coordinates is (3, −5). Its vertex in xy-coordinates is P ≈ ; −5 −5.830 243


Section 8.7

that is, the point (0.0981, −5.830). Figures 12 and 13 on the next page show the graph of the parabola in each of the two coordinate systems. (d) For this ellipse, the coefficients a, b, and c are a = 29, b = 36, and c = −24. Since, a = b, 1 c ≈ 0.6435 radians (about 36◦ 52 ). Let us compute exact θ = 2 arctan a−b = 12 arctan 24 7 values for sin θ and cos θ. Now 2θ = arctan( 24 7 ). So, using this right triangle,

25

24

2θ 7

7 √ 1+ 25 7 1+cos 2θ we see that cos 2θ = 25 , and so cos θ = = = 45 and sin θ = 1 − cos2 θ = 2 2 ' 4 ( − 35 5 cos θ − sin θ 3 1 − 16 = . Therefore, x = 45 u − 35 v and 25 = 5 . Hence, P = 4 3 sin θ cos θ 5

5

y = 35 u + 45 v. Next, we substitute these expressions for x and y into the original equation for the conic section, producing 2 2 29 45 u − 35 v +36 35 u + 45 v −24 45 u − 35 v 35 u + 45 v −118 45 u − 35 v +24 35 u + 45 v −55 = 0. Expanding the terms on the left side of the equation yields 261 2 324 2 864 576 2 288 2 168 288 2 464 2 696 u − uv + v + u + uv + v − u − uv + v 25 25 25 25 25 25 25 25 25 −

72 96 354 472 v + u + v − 55 = 0. u+ 5 5 5 5

Combining like terms, rearranging terms, and dividing by 5 yields the following equation in uvcoordinates: 4u2 − 16u + 9v 2 + 18v = 11, which can be expressed, after completing the squares 2 2 + (v+1) = 1. From this form, we can see that the conic section is an in both u and v, as (u−2) 9 4 ellipse and that its center in uv-coordinates is (2, −1). Its center in xy-coordinates is ' 11 ( 2 5 2 ; that is, the point 11 P = 5 , 5 . Figures 14 and 15 show the graph of the ellipse 2 −1 5

in each of the two coordinate systems. (f) (All answers are rounded to four significant digits.) For this hyperbola, a, b, and

the coefficients 16 1 c 1 c are a = 8, b = −5, and c = 16. Since, a = b, θ = 2 arctan a−b = 2 arctan 13 ≈ 0.4442 cos θ − sin θ 0.9029 −0.4298 radians (about 25◦ 27 ). Hence, P = = . sin θ cos θ 0.4298 0.9029

244


Section 8.7

Therefore, x = 0.9029u−0.4298v and y = 0.4298u+0.9029v. Next, we substitute these expressions for x and y into the original equation for the conic section, producing 2

2

8 (0.9029u − 0.4298v) − 5 (0.4298u + 0.9029v)

+16 (0.9029u − 0.4298v) (0.4298u + 0.9029v) − 37 = 0. Expanding the terms on the left side of this equation yields 245


Section 8.7

6.522u2 − 6.209uv + 1.478v 2 − 0.9236u2 − 3.881uv − 4.076v 2 +6.209u2 + 10.09uv − 6.209v 2 − 37 = 0. Combining and rearranging terms and dividing by 37 yields the following equation 2

2

u v From this form, we can see that the conic section in uv-coordinates: (1.770) 2 − (2.050)2 = 1. is a hyperbola and that its center in uv-coordinates is (0, 0). Its center in xy-coordinates is 0 0 P = ; that is, the origin. Figures 16 and 17 show the graph of the hyperbola in each 0 0

of the two coordinate systems.

(2) (a) True. The reason is that the coefficients of x2 and y 2 are equal. (b) False. Since the entire plane is rotated about the origin by the change in coordinates, the center of a hyperbola is rotated to a new position, unless the center is at the origin, in which case the √ 2 2 (u− 2) center stays fixed. For example. the hyperbola xy − x − y = 0, which changes to − v2 = 1 2 √ 2, 0 in after a 45◦ clockwise rotation, has its center move from (1, 1) in xy-coordinates to uv-coordinates. (c) True. This is explained in Section 8.7 of the textbook, just after Example 1. (d) False. For example, the conic section x2 − y = 0 is a parabola with vertex at the origin whose graph is symmetric with respect to the y-axis, but not with respect to the x-axis. In general, if there is no xy term, any axis of symmetry must be parallel to either the x- or y-axis, but it does not have to be one of these axes. For example, the ellipse (x − 1)2 + (y − 1)2 = 1 (which, when expanded, has no xy term) has the lines x = 1 and y = 1 as horizontal and vertical axes of symmetry.

246


Section 8.8

Section 8.8 (1) In each part we use the following list of vertices for the graphic in Figure 8.25(a) in Section 8.8 of the textbook: (5, 3), (5, 7), (8, 7), (10, 5), and (8, 3). (a) A translation in the direction of the vector [4, −2] is computed by merely adding this vector to each of the vertices in the graphic. This produces the new vertices (9, 1), (9, 5), (12, 5), (14, 3), and (12, 1). (c) A reflection through a line of slope m through the origin is computed by multiplying the vector 2m 1 − m2 1 . In this problem, m = 3, corresponding to each vertex by the matrix 1+m2 2m m2 − 1 −0.8 0.6 5 −2.2 5 0.2 and so the matrix used is A = . Now, A = ,A = , 0.6 0.8 3 5.4 7 8.6 8 −2.2 10 −5 8 −4.6 A = ,A = , and A = . Therefore, when we round 7 10.4 5 10 3 7.2 each of these results to the nearest whole number, the vertices for the reflected graphic are (−2, 5), (0, 9), (−2, 10), (−5, 10), and (−5, 7). (2) In each part we use the following list of vertices for the graphic in Figure 8.25(b) in Section 8.8 of the textbook: (6, 6), (8, 4), (10, 6), (14, 2), and (18, 6). (b) A rotation about the origin through an angle θ is computed by multiplying the vector correspond cos θ − sin θ ing to each vertex by . In this problem, θ = 120◦ , and so the matrix is sin θ cos θ ' √ ( − 12 − 23 −0.5 −0.866 6 −8.20 8 −7.46 √ A= ≈ . Now, A ≈ ,A ≈ , 3 0.866 −0.5 6 2.20 4 4.93 − 12 2 10 −10.20 14 −8.73 18 −14.20 A ≈ ,A ≈ , and A ≈ . 6 5.66 2 11.12 6 12.59 Therefore, when we round each of these results to the nearest whole number, the vertices for the rotated graphic are (−8, 2), (−7, 5), (−10, 6), (−9, 11), and (−14, 13). The new graphic is illustrated in Figure 18 on the next page. (d) A scaling about the origin with scale factors c in the x-direction and d in the y-direction is c 0 computed by multiplying the vector corresponding to each vertex by the matrix . In 0 d ' ( 1 0 6 3 2 1 . Now, A = , this problem, c = 2 and d = 3, and so the matrix is A = 6 18 0 3 8 4 10 5 14 7 18 9 A = ,A = ,A = , and A = . Therefore, 4 12 6 18 2 6 6 18 the vertices for the scaled graphic are (3, 18), (4, 12), (5, 18), (7, 6), and (9, 18). The new graphic is illustrated in Figure 19.

247


Figure 18

Figure 19

248

Section 8.8


Section 8.8

(3) In each part we use the following list of vertices for the graphic in Figure 8.25(c) in Section 8.8 of the textbook: (2, 4), (2, 10), (6.7), (8, 10), and (10, 4). In homogeneous coordinates, we express these points as the vectors [2, 4, 1], [2, 10, 1], [6, 7, 1], [8, 10, 1], and [10, 4, 1]. (a) When we use homogeneous coordinates, a rotation about the origin through an angle θ is computed ⎡ ⎤ cos θ −sin θ 0 cos θ 0 ⎦ . In this by multiplying the vector corresponding to each vertex by A = ⎣ sin θ 0 0 1 √ ⎤ ⎡ √2 ⎡ ⎤ − 22 0 0.707 −0.707 0 √2 √ ⎥ ⎢ 2 0.707 0 ⎦. Also, problem, θ = 45◦ , and so the matrix is A = ⎣ 22 0 ⎦ ≈ ⎣ 0.707 2 0 0 1 0 0 1 a reflection through a line of slope m through the origin is computed by multiplying the vector ⎡ ⎤ 2m 0 1 − m2 1 ⎣ 2m ⎦ . In this 0 m2 − 1 corresponding to each vertex by the matrix B = 1+m 2 2 0 0 1+m ⎡ ⎤ 0.6 0.8 0 problem, m = 12 , and so the matrix used is B = ⎣ 0.8 −0.6 0 ⎦. The composition of the 0 0 1 rotation followed by the reflection is therefore represented by the matrix ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0.990 0.141 0 2 2.545 2 3.394 BA ≈ ⎣ 0.141 −0.990 0 ⎦. Now BA ⎣ 4 ⎦ ≈ ⎣ −3.676 ⎦, BA ⎣ 10 ⎦ ≈ ⎣ −9.615 ⎦ , 0 0 1 1 1 1 1 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 8 9.332 10 10.464 6 6.930 BA ⎣ 7 ⎦ ≈ ⎣ −6.080 ⎦, BA ⎣ 10 ⎦ ≈ ⎣ −8.767 ⎦, and BA ⎣ 4 ⎦ ≈ ⎣ −2.545 ⎦. 1 1 1 1 1 1 Therefore, since each of the product vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number, we get the following vertices for the moved graphic: (3, −4), (3, −10), (7, −6), (9, −9), and (10, −3). (c) When we use homogeneous coordinates, a scaling about the origin with scale factors c in the x-direction and d in the y-direction is computed by multiplying the vector corresponding to each ⎤ ⎡ ⎡ ⎤ 3 0 0 c 0 0 ⎥ ⎢ vertex by A = ⎣ 0 d 0 ⎦ . In this problem, c = 3 and d = 12 , and so A = ⎣ 0 12 0 ⎦. Also, 0 0 1 0 0 1 a reflection through a line of slope m through the origin is computed by multiplying the vector ⎡ ⎤ 2m 0 1 − m2 1 ⎣ 2m ⎦. In this 0 m2 − 1 corresponding to each vertex by the matrix B = 1+m 2 2 0 0 1 + m ⎡ ⎤ −0.6 0.8 0 problem, m = 2, and so B = ⎣ 0.8 0.6 0 ⎦. The composition of the scaling followed by the 0 0 1

249


Section 8.8

⎡

⎤ −1.8 0.4 0 reflection is therefore represented by the matrix BA = ⎣ 2.4 0.3 0 ⎦ . Now, 0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 −2 2 0.4 6 −8 BA ⎣ 4 ⎦ = ⎣ 6 ⎦, BA ⎣ 10 ⎦ = ⎣ 7.8 ⎦ , BA ⎣ 7 ⎦ = ⎣ 16.5 ⎦, 1 1 1 1 1 1 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 10 −16.4 8 −10.4 BA ⎣ 10 ⎦ = ⎣ 22.2 ⎦, and BA ⎣ 4 ⎦ = ⎣ 25.2 ⎦. Therefore, since each of the product 1 1 1 1 vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number, we get the following vertices for the moved graphic: (−2, 6), (0, 8), (−8, 17), (−10, 22), and (−16, 25). (4) In each part we use the following list of vertices for the graphic in Figure 8.26(a) in Section 8.8 of the textbook: (5, 4), (5, 9), (8, 6), (8, 9), (8, 11), and (11, 4). In homogeneous coordinates, we express these points as the vectors [5, 4, 1], [5, 9, 1], [8, 6, 1], [8, 9, 1], [8, 11, 1], and [11, 4, 1]. (a) Since the given rotation is not about the origin, we use the Similarity Method. First, we must consider the translation along the vector [−8, −9] that moves the center of the rotation, (8, 9), to the ⎡ ⎤ 1 0 −8 origin. The matrix for this translation in homogeneous coordinates is T1 = ⎣ 0 1 −9 ⎦. Next, 0 0 1 we need to compute the matrix for a rotation of 120◦ about the origin. When we use homogeneous coordinates, a rotation about the origin through an angle θ is computed by multiplying the vector ⎡ ⎤ cos θ − sin θ 0 cos θ 0 ⎦ . In this problem, θ = 120◦ , corresponding to each vertex by the matrix A = ⎣ sin θ 0 0 1 √ ⎤ ⎡ ⎡ 1 ⎤ − 2 − 23 0 −0.5 −0.866 0 √ ⎥ ⎢ −0.5 0 ⎦. Finally, we need to and so the matrix used is A = ⎣ 23 − 12 0 ⎦ ≈ ⎣ 0.866 0 0 1 0 0 1 translate along the vector [8, 9] to bring the origin back to the point (8, 9). The matrix for this ⎡ ⎤ 1 0 8 translation is T2 = ⎣ 0 1 9 ⎦ . Hence, the rotation of 120◦ about (8, 9) is represented by the 0 0 1 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 13.830 −0.500 −0.866 19.794 5 matrix B = T2 AT1 ≈ ⎣ 0.866 −0.500 6.572 ⎦. Now B ⎣ 4 ⎦ ≈ ⎣ 8.902 ⎦ , 1 1 0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 5 9.500 8 10.598 8 8 8 6.268 B ⎣ 9 ⎦ ≈ ⎣ 6.402 ⎦, B ⎣ 6 ⎦ ≈ ⎣ 10.500 ⎦, B ⎣ 9 ⎦ = ⎣ 9 ⎦ , B ⎣ 11 ⎦ ≈ ⎣ 8.000 ⎦ 1 1 1 1 1 1 1 1

250


Section 8.8

⎡

⎤ ⎡ ⎤ 11 10.830 and B ⎣ 4 ⎦ ≈ ⎣ 14.098 ⎦. Therefore, since each of the product vectors has a 1 in the third 1 1 coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding these results to the nearest whole number, we get the following vertices for the moved graphic: (14, 9), (10, 6), (11, 11), (8, 9), (6, 8), and (11, 14). (c) Since the given scaling is not about the origin, we use the Similarity Method. First, we must consider the translation along the vector [−8, −4] that moves the center of the scaling, (8, 4), to the ⎡ ⎤ 1 0 −8 origin. The matrix for this translation in homogeneous coordinates is T1 = ⎣ 0 1 −4 ⎦. Next, 0 0 1 we need to find the matrix for a scaling about the origin with factors 2 in the x-direction and 13 in the y-direction. When we use homogeneous coordinates, a scaling about the origin with scale factors c in the x-direction and d in the y-direction is computed by multiplying the vector corre⎡ ⎤ c 0 0 sponding to each vertex by the matrix A = ⎣ 0 d 0 ⎦. In this problem, c = 2 and d = 13 , and 0 0 1 ⎤ ⎡ 2 0 0 ⎥ ⎢ so the matrix is A = ⎣ 0 13 0 ⎦. Finally, we need to translate along the vector [8, 4] to bring the 0

0 1

⎤ 1 0 8 origin back to the point (8, 4). The matrix for this translation is T2 = ⎣ 0 1 4 ⎦. Hence, the 0 0 1 scaling with factor 2 in the x-direction and 13 in the y-direction about the point (8, 4) is represented ⎡ ⎤ ⎤ ⎡ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 0 −8 2 5 2 5 ⎢ ⎥ 8 ⎥ ⎣ 4 ⎦ = ⎣ 4 ⎦, B ⎣ 9 ⎦ = ⎢ by the matrix B = T2 AT1 = ⎣ 0 13 ⎣ 17 3 ⎦ . Now, B 3 ⎦, 1 1 1 0 0 1 1 ⎤ ⎡ 8 ⎢ B⎣ 6 ⎦ = ⎣ 1 ⎡

⎤

⎤ ⎡ 8 14 ⎥ ⎣ 9 ⎦=⎢ ⎣ 3 ⎦, B 1 1 8

⎡

⎡

⎤

⎤ ⎡ 8 17 ⎥ ⎣ 11 ⎦ = ⎢ ⎣ 3 ⎦, B 1 1 8

⎡

⎤

⎤ ⎡ ⎤ 11 14 19 ⎥ ⎣ 4 ⎦ = ⎣ 4 ⎦. There3 ⎦, and B 1 1 1 8

⎡

fore, since each of the product vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number, we get the following vertices for the moved graphic: (2, 4), (2, 6), (8, 5), (8, 6), (8, 6), and (14, 4). (5) (b) We use the following list of vertices for the graphic in Figure 8.26(b) in Section 8.8 of the textbook: (7, 3), (7, 6), (10, 8), (13, 6), and (13, 3). In homogeneous coordinates, we express these points as the vectors [7, 3, 1], [7, 6, 1], [10, 8, 1], [13, 6, 1], and [13, 3, 1]. Since the given reflection is not about a line through the origin, we use the Similarity Method. First, we consider the translation along the vector [0, 10] that moves the y-intercept of the line of reflection, (0, −10), to the origin. The matrix for this translation in homogeneous coordinates

251


1 is T1 = ⎣ 0 0

0 1 0

Section 8.8

⎤ 0 10 ⎦. Next, we find the matrix for the reflection about the line y = 4x 1

(since we have shifted the y-intercept to the origin). A reflection through a line of slope m through the origin is computed by multiplying the vector corresponding to each vertex by A = ⎤ ⎡ ⎤ ⎡ 2m 0 1 − m2 −15 8 0 1 ⎣ 2m ⎦. In this problem, m = 4, and so A = 1 ⎣ 8 15 0 ⎦. 0 m2 − 1 1+m2 17 2 0 0 17 0 0 1+m Finally, we need to translate along the vector [0, −10] to bring the origin back to the point ⎡ ⎤ 1 0 0 (0, −10). The matrix for this translation is T2 = ⎣ 0 1 −10 ⎦. Hence, the reflection about 0 0 1 ⎡ ⎤ −15 8 80 1 ⎣ 8 15 −20 ⎦ . Now the line y = 4x − 10 is represented by the matrix B = T2 AT1 = 17 0 0 17 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 7 −0.059 7 1.353 10 −0.353 B ⎣ 3 ⎦ ≈ ⎣ 4.765 ⎦, B ⎣ 6 ⎦ ≈ ⎣ 7.412 ⎦, B ⎣ 8 ⎦ ≈ ⎣ 10.588 ⎦ , 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 13 −3.941 13 −5.353 B ⎣ 6 ⎦ ≈ ⎣ 10.235 ⎦, and B ⎣ 3 ⎦ ≈ ⎣ 7.588 ⎦. 1 1 1 1 Therefore, since each of the product vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number, we get the following vertices for the moved graphic: (0, 5), (1, 7), (0, 11), (−4, 10), and (−5, 8). (6) In each part we use the following list of vertices for the graphic in Figure 8.26(c) in Section 8.8 of the textbook: (5, 3), (6, 6), (8, 9), (9, 4), (9, 7), and (12, 8). In homogeneous coordinates, we express these points as the vectors [5, 3, 1], [6, 6, 1], [8, 9, 1], [9, 4, 1], [9, 7, 1], and [12, 8, 1]. (a) We begin with the rotation. Since the given rotation is not about the origin, we use the Similarity Method. First, we consider the translation along the vector [−12, −8] that moves the center of the rotation, (12, 8), to the origin. The matrix for this translation in homogeneous coordinates is ⎡ ⎤ 1 0 −12 T1 = ⎣ 0 1 −8 ⎦. Next, we need to compute the matrix for a rotation of 60◦ about the origin. 0 0 1 When we use homogeneous coordinates, a rotation about the origin through an angle θ is computed ⎡ ⎤ cos θ − sin θ 0 cos θ 0 ⎦ . by multiplying the vector corresponding to each vertex by the matrix A = ⎣ sin θ 0 0 1 √ ⎤ ⎡ 1 ⎡ ⎤ − 23 0 0.5 −0.866 0 2 √ ⎥ ⎢ 1 0.5 0 ⎦. To finish In this problem, θ = 60◦ , and so A = ⎣ 23 0 ⎦ ≈ ⎣ 0.866 2 0 0 1 0 0 1

252


Section 8.8

the rotation, we need to translate along the vector [12, 8] to bring the origin back to the point ⎡ ⎤ 1 0 12 (12, 8). The matrix for this is T2 = ⎣ 0 1 8 ⎦. Hence, the rotation of 60◦ about (12, 8) is 0 0 1 represented by B = T2 AT1 . Next, we compute the matrix for the reflection about y = 12 x + 6. Since the given reflection is not about a line through the origin, we use the Similarity Method. First, we must consider the translation along the vector [0, −6] that moves the y-intercept of the line of reflection, (0, 6),⎤to ⎡ 1 0 0 the origin. The matrix for this translation in homogeneous coordinates is T3 = ⎣ 0 1 −6 ⎦. 0 0 1 Next, we need to find the matrix for the reflection about the line y = 12 x (since we have shifted the y-intercept to the origin). A reflection through a line of slope m through the origin is computed⎤ ⎡ 1 − m2 2m 0 1 ⎣ 2m ⎦. 0 m2 − 1 by multiplying the vector corresponding to each vertex by C = 1+m 2 0 0 1 + m2 ⎡ ⎤ 0.6 0.8 0 In this problem, m = 12 , and so C = ⎣ 0.8 −0.6 0 ⎦. Finally, we need to translate along 0 0 1 the vector ⎡ 1 T4 = ⎣ 0 0

[0, 6] to bring the origin back to the point (0, 6). The matrix for this translation is ⎤ 0 0 1 6 ⎦. Hence, the reflection about the line y = 12 x + 6 is represented by the matrix 0 1

D = T4 CT3 . Finally, the composition of the rotation and the reflection is represented by the matrix ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0.993 −0.120 −2.157 5 2.448 DB = T4 CT3 T2 AT1 ≈ ⎣ −0.120 −0.993 23.778 ⎦ . Now DB ⎣ 3 ⎦ ≈ ⎣ 20.201 ⎦, 0 0 1 1 1 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 6 3.082 8 4.709 9 6.300 DB ⎣ 6 ⎦ ≈ ⎣ 17.103 ⎦ , DB ⎣ 9 ⎦ ≈ ⎣ 13.886 ⎦, DB ⎣ 4 ⎦ ≈ ⎣ 18.730 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 9 5.941 12 8.8 DB ⎣ 7 ⎦ ≈ ⎣ 15.752 ⎦, and DB ⎣ 8 ⎦ = ⎣ 14.4 ⎦. Therefore, since each of the product 1 1 1 1 vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number, we get the following vertices for the moved graphic: (2, 20), (3, 17), (5, 14), (6, 19), (6, 16), and (9, 14). The new graphic is illustrated in Figure 20.

253


Section 8.8

Figure 20 (c) We begin with the scaling. Since the given scaling is not about the origin, we use the Similarity Method. First, we must consider the translation along the vector [−9, −4] that moves the center of the scaling, (9, 4), to the origin. The matrix for this translation in homogeneous coordinates is ⎡ ⎤ 1 0 −9 T1 = ⎣ 0 1 −4 ⎦. Next, we need to find the matrix for a scaling about the origin with factors 0 0 1 1 in the x-direction and 2 in the y-direction. When we use homogeneous coordinates, a scaling 3 about the origin with scale factors c in the x-direction and d in the y-direction is computed by ⎡ ⎤ c 0 0 multiplying the vector corresponding to each vertex by A = ⎣ 0 d 0 ⎦. In this problem, c = 13 0 0 1 ⎡ 1 ⎤ 0 0 3 and d = 2, and so A = ⎣ 0 2 0 ⎦. Finally, we need to translate along the vector [9, 4] to bring 0 0 1 ⎡ ⎤ 1 0 9 the origin back to the point (9, 4). The matrix for this is T2 = ⎣ 0 1 4 ⎦. Hence, the scaling 0 0 1 1 with factor 3 in the x-direction and 2 in the y-direction about the point (9, 4) is represented by the matrix B = T2 AT1 . Next, we compute the matrix for the rotation. To do this, we first must consider the translation along the vector [−2, −9] that moves the center of the rotation, (2, 9), to the origin.

254


1 The matrix for this translation in homogeneous coordinates is T3 = ⎣ 0 0

Section 8.8

0 1 0

⎤ −2 −9 ⎦. Next, we 1

need to find the matrix for a rotation of 150◦ about the origin. When we use homogeneous coordinates, a rotation about the origin through an angle θ is computed by multiplying the vector ⎡ ⎤ cos θ −sin θ 0 cos θ 0 ⎦ . In this problem, θ = 150◦ , corresponding to each vertex by the matrix C = ⎣ sin θ 0 0 1 ⎤ ⎡ √3 ⎡ ⎤ − 12 0 − 2 −0.866 −0.5 0 √ ⎥ ⎢ 1 0.5 −0.866 0 ⎦. To finish the rotation, and so the matrix is C = ⎣ − 23 0 ⎦ ≈ ⎣ 2 0 0 1 0 0 1 we need to translate along the vector [2, 9] to bring the origin back to the point (2, 9). The matrix ⎡ ⎤ 1 0 2 for this translation is T4 = ⎣ 0 1 9 ⎦. Hence, the rotation of 150◦ about (2, 9) is represented 0 0 1 by the matrix D = T4 CT3 . Finally, the composition of the rotation and the reflection is represented by the matrix ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −0.289 −1 5.036 5 0.593 DB = T4 CT3 T2 AT1 ≈ ⎣ 0.167 −1.732 22.258 ⎦ . Now DB ⎣ 3 ⎦ ≈ ⎣ 17.896 ⎦, 0 0 1 1 1 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 6 −2.696 8 −6.274 9 −1.562 DB ⎣ 6 ⎦ ≈ ⎣ 12.866 ⎦ , DB ⎣ 9 ⎦ ≈ ⎣ 8.003 ⎦, DB ⎣ 4 ⎦ ≈ ⎣ 16.830 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 9 −4.562 12 −6.428 DB ⎣ 7 ⎦ ≈ ⎣ 11.634 ⎦, and DB ⎣ 8 ⎦ ≈ ⎣ 10.402 ⎦. Therefore, since each of the product 1 1 1 1 vectors obtained has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by dropping that 1. Doing this and rounding each of these results to the nearest whole number yields the following vertices for the moved graphic: (1, 18), (−3, 13), (−6, 8), (−2, 17), (−5, 12), and (−6, 10). The new graphic is illustrated in Figure 21.

255


Section 8.8

Figure 21 (9) (b) We use the following list of vertices for the graphic in Figure 8.26(b) in Section 8.8 of the textbook: (7, 3), (7, 6), (10, 8), (13, 6), and (13, 3). In homogeneous coordinates, we express these points as the vectors [7, 3, 1], [7, 6, 1], [10, 8, 1], [13, 6, 1] and [13, 3, 1]. Now according to part (b) of Exercise 7 in Section 8.8, a reflection about the line y = mx + b is ⎡ ⎤ 1 − m2 2m −2mb 1 ⎣ 2m ⎦ . In this problem, the line is 2b m2 − 1 represented by the matrix B = 1+m 2 0 0 1 + m2 ⎡ ⎤ −15 8 80 1 ⎣ 8 15 −20 ⎦ . Now, y = 4x − 10, and so m = 4 and b = −10. Hence, B = 17 0 0 17 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 7 −0.059 7 1.353 10 −0.353 B ⎣ 3 ⎦ ≈ ⎣ 4.765 ⎦, B ⎣ 6 ⎦ ≈ ⎣ 7.412 ⎦, B ⎣ 8 ⎦ ≈ ⎣ 10.588 ⎦, 1 1 1 1 1 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 13 −3.941 13 −5.353 B ⎣ 6 ⎦ ≈ ⎣ 10.235 ⎦, and B ⎣ 3 ⎦ ≈ ⎣ 7.588 ⎦. Therefore, since each of the product 1 1 1 1 vectors has a 1 in the third coordinate, each can be converted from homogeneous coordinates back to regular coordinates by merely dropping that 1. Doing this and rounding each of these results to the nearest whole number, we get the following vertices for the moved graphic: (0, 5), (1, 7), (0, 11), (−4, 10), and (−5, 8).

256


Section 8.9

(11) (b) Consider the reflection about the y-axis, whose matrix is given in Section 8.8 of the textbook as ⎡ ⎤ −1 0 0 A = ⎣ 0 1 0 ⎦ , and a counterclockwise rotation of 90◦ about the origin, whose matrix is 0 0 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 −1 0 0 1 0 0 −1 0 0 0 ⎦. Then AB = ⎣ 1 0 0 ⎦, but BA = ⎣ −1 0 0 ⎦, and so A and B B=⎣ 1 0 0 1 0 0 1 0 0 1 do not commute. In particular, starting from the point (1, 0), performing the rotation and then the reflection, yields (0, 1). However, performing the reflection followed by the rotation produces (0, −1). (13) (a) False. As explained in Section 8.8 of the textbook, the third coordinate of a vector in homogeneous coordinates must be nonzero. (b) False. Vectors in homogeneous coordinates that are nonzero scalar multiples of each other represent the same pixel. For example, [1, 1, 1] and [2, 2, 2], as homogeneous coordinates, both represent the pixel at the point (1, 1). (c) False. If A is a matrix for a rotation in homogeneous coordinates, then any nonzero scalar multiple of A also represents this same rotation. (d) True. As explained in Section 8.8 of the textbook, isometries are the composition of translations, rotations, and reflections only. However, any similarity can be expressed as a composition of translations, rotations, reflections, and scalings. But scalings that are not the identity or a reflection are not isometries. (e) True. The reason is that they do not send the origin to itself, violating part (1) of Theorem 5.1. (f) False. Rotations about the origin and reflections about lines through the origin are linear transformations. However, no other rotations or reflections are linear transformations since they do not send the origin to itself and therefore violate part (1) of Theorem 5.1.

Section 8.9 (1) (a) Now A =

13 −28 6 −13

, and so pA (x) =

" " x − 13 28 " " −6 x + 13

" " " = x2 − 1 = (x − 1)(x + 1). Hence, "

A has 2 eigenvalues, λ1 = 1 and λ2 = −1. Next, we solve for the corresponding fundamental eigenvectors. For λ1 = 1, we row reduce ( ' " " 1 − 73 " 0 −12 28 "" 0 " , [(1I2 − A) |0 ] = to obtain −6 14 " 0 0 0 " 0 producing the fundamental eigenvector [7, 3]. For λ2 = −1, we row reduce −14 28 [(−1I2 − A) |0 ] = −6 12

" " 0 1 " to obtain " 0 0

" −2 "" 0 , 0 " 0

producing the fundamental eigenvector [2, 1]. Therefore, by Theorem 8.11, the general form for the solution to F (t) = AF(t) is 7 2 t −t F(t) = b1 e + b2 e . 3 1 257


1 (c) Now A = ⎣ −1 1

4 2 1

" ⎤ " x−1 4 " 2 ⎦, and so pA (x) = "" 1 " −1 1

−4 x−2 −1

−4 −2 x−1

Section 8.9

" " " " = x3 − 4x2 + 3x " "

= x(x − 1)(x − 3). Hence, A has 3 eigenvalues, λ1 = 0, λ2 = 1, and λ3 = the corresponding fundamental eigenvectors. For λ1 = 0, we row reduce " ⎤ ⎡ ⎡ 1 −1 −4 −4 "" 0 [(0I3 − A) |0 ] = [−A |0 ] = ⎣ 1 −2 −2 "" 0 ⎦ to obtain ⎣ 0 0 −1 −1 −1 " 0

3. Next, we solve for

0 1 0

0 1 0

" ⎤ " 0 " " 0 ⎦, " " 0

producing the fundamental eigenvector [0, −1, 1]. For λ2 = 1, we row reduce " ⎡ ⎤ ⎡ 0 −4 −4 "" 0 1 [(1I3 − A) |0 ] = ⎣ 1 −1 −2 "" 0 ⎦ to obtain ⎣ 0 −1 −1 0 " 0 0

0 1 0

−1 1 0

" ⎤ " 0 " " 0 ⎦, " " 0

producing the fundamental eigenvector [1, −1, 1]. For λ3 = 3, we row reduce " ⎤ ⎡ ⎡ 1 2 −4 −4 "" 0 1 −2 "" 0 ⎦ to obtain ⎣ 0 [(3I3 − A) |0 ] = ⎣ 1 0 −1 −1 2 " 0

0 1 0

−2 0 0

" ⎤ " 0 " " 0 ⎦, " " 0

producing the fundamental eigenvector [2, 0, 1]. Therefore, by Theorem 8.11, the general form for the solution to F (t) = AF(t) is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 1 2 F(t) = b1 ⎣ −1 ⎦ + b2 et ⎣ −1 ⎦ + b3 e3t ⎣ 0 ⎦ . 1 1 1 ⎡

−5 (d) Now A = ⎣ −6 −6

−6 −5 −6

" ⎤ " x+5 6 −15 15 " ⎦ x+5 −15 15 , and so pA (x) = "" 6 " 6 6 x − 16 16

(x − 1)2 (x − 4). Hence, A has 2 eigenvalues, λ1 = 1 and λ2 = corresponding fundamental eigenvectors. For λ1 = 1, we row reduce ⎡ " ⎤ ⎡ 1 6 6 −15 "" 0 ⎢ " ⎦ ⎣ to obtain ⎣ 0 [(1I3 − A) |0 ] = 6 6 −15 " 0 6 6 −15 " 0 0 producing the fundamental eigenvectors [−1, 1, 0] and [5, 0, 2]. For λ2 = 4, we row reduce " ⎤ ⎡ ⎡ 1 9 6 −15 "" 0 [(4I3 − A) |0 ] = ⎣ 6 9 −15 "" 0 ⎦ to obtain ⎣ 0 0 6 6 −12 " 0 258

" " " " = x3 − 6x2 + 9x − 4 = " "

4. Next, we solve for the

0

" ⎤ − 52 "" 0 ⎥ " 0 " 0 ⎦, " 0 " 0

0 1 0

−1 −1 0

1 0

" ⎤ " 0 " " 0 ⎦, " " 0


producing the fundamental eigenvector [1, 1, 1]. Therefore, by Theorem 8.11, the general form for ⎡ ⎡ ⎤ −1 F(t) = b1 et ⎣ 1 ⎦ + b2 et ⎣ 0

Section 8.9

the solution to F (t) = AF(t) is ⎡ ⎤ ⎤ 1 5 0 ⎦ + b3 e4t ⎣ 1 ⎦ . 1 2

(There are other possible answers. For example, the first two vectors in the sum could be any basis for the two-dimensional eigenspace corresponding to the eigenvalue 1.) (2) In each part, we use the technique described in Section 8.9 of the textbook between Examples 4 and 5. (a) The characteristic equation is x2 + x − 6 = 0, found by using the coefficients of y , y , and y, in the given differential equation. Since x2 + x − 6 = (x − 2)(x + 3), the characteristic values (the roots of the characteristic equation) are λ1 = 2 and λ2 = −3. Hence, the general solution for the differential equation is y = b1 e2t + b2 e−3t . (c) The characteristic equation is x4 − 6x2 + 8 = 0, found by using the coefficients of y , y y , y , and y, in the given differential Since x4 − 6x2 + 8 = (x2 − 4)(x2 − 2) √ equation. √ = (x − 2)(x + 2) x − 2 x + 2 , the characteristic values (the roots of the characteristic √ √ equation) are λ1 = 2, λ2 = −2, λ3 = 2, and√λ4 = − √2. Hence, the general solution for the differential equation is y = b1 e2t + b2 e−2t + b3 e 2t + b4 e− 2t . " " " x + 11 6 −16 "" " 4 x+1 −4 "" = (4) (b) First, we find the general solution to F (t) = AF(t). Now pA (x) = "" " 12 6 x − 17 " x3 − 5x2 − x + 5 = (x − 5)(x2 − 1) = (x − 5)(x − 1)(x + 1). Hence, A has 3 eigenvalues, λ1 = 5, λ2 = 1, and λ3 = −1. Next, we solve for the corresponding fundamental eigenvectors. For λ1 = 5, we row reduce " " ⎤ ⎤ ⎡ ⎡ 1 0 −1 "" 0 16 6 −16 "" 0 0 "" 0 ⎦ , −4 "" 0 ⎦ to obtain ⎣ 0 1 [(5I3 − A) |0 ] = ⎣ 4 6 " 0 0 0 0 " 0 12 6 −12 producing the fundamental eigenvector For λ2 = 1, we row reduce ⎡ 12 6 [(1I3 − A) |0 ] = ⎣ 4 2 12 6

[1, 0, 1]. −16 −4 −16

" ⎤ ⎡ " 0 1 " " 0 ⎦ to obtain ⎣ 0 " " 0 0

producing the fundamental eigenvector [−1, 2, 0]. For λ3 = −1, we row reduce " ⎤ ⎡ ⎡ 1 10 6 −16 "" 0 −4 "" 0 ⎦ to obtain ⎣ 0 [((−1)I3 − A) |0 ] = ⎣ 4 0 0 12 6 −18 " 0

" ⎤ 0 " 0 " 0 1 "" 0 ⎦ , 0 0 " 0

1 2

0 1 0

−1 −1 0

" ⎤ " 0 " " 0 ⎦, " " 0

producing the fundamental eigenvector [1, 1, 1]. Therefore, by Theorem 8.11, the general form for the solution to F (t) = AF(t) is ⎡ ⎤ ⎡ ⎡ ⎤ ⎤ 1 −1 1 F(t) = b1 e5t ⎣ 0 ⎦ + b2 et ⎣ 2 ⎦ + b3 e−t ⎣ 1 ⎦ . 1 0 1 259


Section 8.9

Now we want to find the particular solution such that F(0) = [1, −4, 0]. Plugging 0 into the general solution yields F(0) = [b1 , 0, b1 ] + [−b2 , 2b2 , 0] + [b3 , b3 , b3 ] = [b1 − b2 + b3 , 2b2 + b3 , b1 + b3 ]. Setting this equal to [1, −4, 0] produces the system ⎧ 1 ⎨ b 1 − b2 + b3 = 2b2 + b3 = −4 ⎩ + b3 = 0 b1 " " ⎤ ⎤ ⎡ ⎡ 1 0 0 "" 2 1 −1 1 "" 1 2 1 "" −4 ⎦ to obtain ⎣ 0 1 0 "" −1 ⎦ shows that the unique solution Row reducing ⎣ 0 0 0 1 " −2 1 0 1 " 0 to the differential equation with the given initial condition is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 1 F(t) = 2e5t ⎣ 0 ⎦ − et ⎣ 2 ⎦ − 2e−t ⎣ 1 ⎦ . 1 0 1 (5) (b) Base Step (n = 1): A = [−a0 ], xI1 − A = [x + a0 ], and so pA (x) = x + a0 . Inductive Step: Assume true for k. Prove true for k + 1. For the case k + 1, ⎡ ⎤ 0 1 0 0 ··· 0 ⎢ 0 0 1 0 ··· 0 ⎥ ⎢ ⎥ ⎢ .. . . . .. ⎥ . . .. .. .. .. A=⎢ . . ⎥ ⎢ ⎥ ⎣ 0 0 0 0 ··· 1 ⎦ −a0 −a1 −a2 −a3 · · · −ak Then

⎡

x 0 .. .

⎢ ⎢ ⎢ (xIk+1 − A) = ⎢ ⎢ ⎣ 0 a0

−1 0 x −1 .. .. . . 0 0 a1 a2

··· ··· .. .

0 0 .. .

0 0 .. .

0 a3

··· ···

x ak−1

−1 x + ak

Using a cofactor expansion on the first column yields " " x −1 0 · · · " " .. .. .. . . " . . . |xIk+1 − A| = x " . " 0 0 0 ··· " " a1 a2 a3 · · · " " −1 " " x " (k+1)+1 + (−1) a0 " . " .. " " 0

260

⎤

0 0 .. .

0 .. . x ak−1

⎥ ⎥ ⎥ ⎥. ⎥ ⎦

" " " " " " −1 "" x + ak "

0 ··· −1 · · · .. .. . . 0 ···

0 .. .

" " " " " ". " " x −1 " 0 0 .. .

0 0 .. .


The first determinant equals |xIk − B|, where ⎡ 0 1 0 ⎢ 0 0 1 ⎢ ⎢ .. . .. .. B=⎢ . . ⎢ ⎣ 0 0 0 −a1 −a2 −a3

··· ··· .. .

0 0 .. .

0 0

··· ···

0 −ak−1

1 −ak

Section 8.10

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎦

Now, using the inductive hypothesis on the first determinant and Theorems 3.2 and 3.9 on the second determinant (since its corresponding matrix is lower triangular) yields pA (x) = x(xk + ak xk−1 + · · · + a2 x + a1 ) + (−1)k+2 a0 (−1)k = xk+1 + ak xk + · · · + a1 x + a0 . (7) (a) True. If F(t) = 0, then F (t) also equals 0. The equation 0 = A0 is clearly satisfied for every matrix A. (b) True. Example 6 in Section 4.1 of the textbook shows that the set of all real-valued functions on R forms a vector space. An analogous proof shows that the set V of all Rn -valued functions on R is also a vector space. It is easy to show that the subset of continuously differentiable functions that are solutions to F (t) = AF(t) is closed under addition and scalar multiplication, and so Theorem 4.2 shows that this set is a subspace of V, and is thus a vector space. 1 2 (c) True. Since A = is upper triangular, it is easy to see that the eigenvalues for A are 0 3 λ1 = 1 and λ2 = 3. It is also easy to verify (by direct multiplication) that [1, 0] is an eigenvector for . for λ2 . Hence, by Theorem 8.11, the solution set is ) λ1 andthat [1, 1] is"an eigenvector 1 1 "" t 3t + b2 e b , b ∈ R , which agrees with the given solution. b1 e 0 1 " 1 2 (d) False. The fact that the given matrix is not diagonalizable only means that the methods of Section 8.9 cannot be used to solve the problem. For this particular matrix, complex eigenvalues need to be used and will produce the general solution F(t) = [a sin t + b cos t, − b sin t + a cos t], which you can easily verify satisfies F (t) = AF(t). (Note for those who have some experience with complex functions: Using the method of Section 8.9 and complex eigenvalues to solve this problem will −i i a sin t + b cos t it −it +b2 e , which equals , where produce the solution F(t) = b1 e 1 1 −b sin t + a cos t a = b1 + b2 and b = i(−b1 + b2 ).)

Section 8.10

9 26 T (1) (a) Using the given matrix A, A A = and A b = . Thus, to solve 11 19 " ' ( " 1 0 "" 23 21 9 "" 26 30 T T , obtaining A Av = A b, we row reduce . " 9 11 " 19 0 1 " 11 10 11 Therefore, the unique least-squares solution is v = 23 30 , 10 . Now, ||Av − b|| = ⎡ 29 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 6 3 ' 23 ( 5 5 √ ⎢ ⎥ 30 1 ⎣ 1 −1 ⎦ − ⎣ 0 ⎦ ⎣ − 3 ⎦ − ⎣ 0 ⎦ = − 16 , − 13 , 16 = 66 ≈ 0.408. Note = 11 4 10 25 1 4 4 T

21 9

6

261


Section 8.10

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 5 3 5 5 1 − ⎣ 0 ⎦ = ⎣ 0 ⎦ − ⎣ 0 ⎦ that ||Az − b|| = ⎣ 1 −1 ⎦ = [0, 0, 1] = 1, which is 1 4 4 1 4 5 larger than ||Av − b||.

⎡

14 (c) When we use the given matrix A, AT A = ⎣ 8 6 AT Av = AT b, we row reduce ⎡

14 8 ⎣ 8 5 6 9

8 5 9

⎤ ⎡ ⎤ 6 18 9 ⎦ and AT b = ⎣ 7 ⎦. Thus, to solve 75 −35

⎡ " ⎤ 1 " 18 " ⎢ " ⎦ 7 , obtaining ⎣ 0 " " −35 0

6 9 75

0 1 0

" ⎤ −7 "" 17 3 " ⎥ 13 " − 23 3 ⎦. " 0 " 0

Therefore, there is an infinite number of least-squares solutions, all of the form 23 1 v = 7c + 17 3 , −13c − 3 , c . Using c = 0 and c = 3 produces the two particular least-squares 17 23 solutions 3 , − 3 , 0 and 8, −12, 13 . With v as the first of these vectors, ||Av − b|| = ⎡ 11 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 17 ⎤ ⎡ ⎤ 3 3 3 3 2 1 −1 √ 2 1 1 ⎢ ⎢ ⎣ ⎥ ⎥ , − , − = 6 ≈ 0.816. ⎣ 2 ⎦ = ⎣ 5 ⎦ − ⎣ 2 ⎦ 5 ⎦ ⎣ − 23 = − 3 2 ⎦ 3 3 3 3 3 3 6 6 17 1 0 −7 0 3 Using the other particular solution we computed for v, or any of the infinite number of least √

squares solutions instead, produces the same value of 36 for ||Av − b||. Note that ||Az − b|| = ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 1 −1 3 1 3 4 ⎣ 3 2 ⎦ ⎣ 1 ⎦ − ⎣ 2 ⎦ = ⎣ 0 ⎦ − ⎣ 2 ⎦ = [1, −2, 2] = 3, which is larger than 5 1 0 −7 6 −1 6 8 ||Av − b||.

⎡

⎤ −4 40 5 −8 ⎦ and AT b = −8 32 ⎡ " ⎤ 1 −4 40 "" 29 ⎢ 5 −8 "" −3 ⎦, obtaining ⎣ 0 −8 32 " 20 0

62 (2) (a) When we use the given matrix A, AT A = ⎣ −4 40 ⎡

62 solve AT Av = AT b, we row reduce ⎣ −4 40

⎡

⎤ 29 ⎣ −3 ⎦. Thus, 20 " 19 4 " 0 7 " 42 " 5 1 − 87 " − 21 " 0 0 " 0

Therefore, there is an infinite number of least-squares solutions, all of the form 8 5 4 19 19 v = − 47 c + 19 42 , 7 c − 21 , c . Now, setting − 7 c + 42 ≥ 0 yields c ≤ 24 . And setting produces , 4 −7c +

5 24 ≤ c, which 19 8 5 42 , 7 c − 21 , c

⎡

⎢ (3) (b) First, (λ I3 − C) = ⎣

8 7c

−

5 21

to ⎤ ⎥ ⎦.

≥0

also implies c ≥ 0. Hence, the desired solution set is " 5 " ≤ c ≤ 19 . 24

24

− 32

3

−11

12

5 − 72

2

⎤

⎥ −4 ⎦. We start with the homogeneous system 21 2

(λ I3 − C)x = 0 and add the equation x1 + x2 + x3 = 1. This produces the system Ax = b, with 262


⎤

Section 8.11

⎤ 0 ⎢ ⎥ ⎢ 0 ⎥ 5 − 72 −4 ⎥ ⎢ ⎥ A=⎢ ⎥ and b = ⎢ ⎣ 0 ⎦ . We find a least-squares solution for this system. ⎣ −11 12 21 ⎦ 2 1 1 1 1 ⎤ ⎡ 597 ⎡ ⎤ −153 − 275 4 2 1 ⎥ ⎢ 665 T ⎣ 1 ⎦. Thus, to solve AT Av = AT b, we 147 Now AT A = ⎣ −153 and A b = ⎦ 4 525 1 − 275 147 − 32

3

⎡

2

2

597 4

⎢ row reduce ⎣ −153 − 275 2

⎡

4

−153 665 4

147

" ⎤ ⎡ " 1 1 " ⎥ " 147 " 1 ⎦, obtaining ⎣ 0 " 525 " 0 1 4

− 275 2

0 1 0

0 0 1

" ⎤ " 0.46 " " −0.36 ⎦, " " 0.90

where we have rounded to two places after the decimal point. Hence, the approximate eigenvector is v ≈ [0.46, −0.36, 0.90]. Direct computation (λ I3 − C)v ≈ [0.03, −0.04, 0.07], which produces is close to the zero vector. (Note that (λ I3 − C)v ≈ 0.086, which is of the same order of magnitude as the difference between λ = 32 and the actual eigenvalue for C nearest to λ , which √ is 2 ≈ 1.414.) (6) Let b = [b1 , ..., bn ], let A be as in Theorem 8.2, and let W = {Ax | x ∈ Rn }. Since projW b ∈ W exists, there must be some v ∈ Rn such that Av = projW b. Hence, v satisfies part (1) of Theorem 8.12, so (AT A)v = AT b by part (3) of Theorem 8.12. This shows that the system (AT A)X = AT b is consistent, which proves part (2) of Theorem 8.2. Next, let f (x) = d0 + d1 x + · · · + dt xt and z = [d0 , d1 , . . . , dt ]. A short computation shows that ||Az − b||2 = Sf , the sum of the squares of the vertical distances illustrated in Figure 8.9, just before the definition of a least-squares polynomial in Section 8.3 of the textbook. Hence, minimizing ||Az−b|| over all possible (t + 1)-vectors z gives the coefficients of a degree t least-squares polynomial for the given points (a1 , b1 ), . . . , (an , bn ). However, parts (2) and (3) of Theorem 8.12 show that such a minimal solution is found by solving (AT A)v = AT b, thus proving part (1) of Theorem 8.2. Finally, when AT A is row equivalent to It+1 , the uniqueness condition holds by Theorems 2.14 and 2.15, which proves part(3) of Theorem 8.2. (7) (a) False. The least-squares solution of an inconsistent system Ax = b is, instead, any vector v such that the distance between Av and b is minimized. T T = (b) True. If A is an m×n matrix, then AT A is n×n, which is square. Also, AT A = AT AT AT A, and so AT A is symmetric. (c) True. If W is the subspace {Ax | x ∈ Rn }, then projW b is a well-defined vector in W. Hence, by the definition of W, there must exist a vector v in Rn with Av = projW b. Theorem 8.12 asserts that such a vector v is a least-squares solution to Ax = b. (d) True. By part (1) of Theorem 8.12, both Av1 and Av2 must equal projW b, and so they must equal each other. (e) True. This is illustrated in Example 3 in Section 8.10 of the textbook.

Section 8.11 (1) In each part, we use the fact that the entries for the upper triangular matrix C are given, for i ≤ j, by cij = (coefficient of xi xj ) in the formula for Q. Then, the entries of the symmetric matrix A are given, for i < j, by aij = aji = 12 cij , and, on the diagonal, aii = cii . 263


8 6 (a) Using the formulas given above produces C = and A = . 6 −9 ⎤ ⎡ ⎡ ⎤ 5 2 − 32 5 4 −3 ⎢ 5 ⎥ 5 ⎦ and A = ⎣ 2 −2 (c) Using the formulas given above yields C = ⎣ 0 −2 2 ⎦. 5 3 0 0 0 −2 0 2 8 0

12 −9

Section 8.11

(2) (a) The entries for the symmetric matrix A representing Q are given, for i = j, by aij = 12 (coefficient of xi xj ) in the expression for Q, and, on the diagonal, aii = (the coefficient of x2i ). Hence, 43 −24 A= . We need to orthogonally diagonalize A. −24 57 Now pA (x) = (x − 43)(x − 57) − 242 = x2 − 100x + 1875 = (x − 75)(x − 25). Thus, A has eigenvalues λ1 = 75 and λ2 = 25. Using row reduction to find particular solutions to the systems [75I2 − A|0] and [25I2 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[−3, 4]} for E75 and {[4, 3]} for E25 . Since E75 and E25 are one dimensional, the given bases are already orthogonal. To create orthonormal bases, we need to normalize the vectors, yielding { 15 [−3, 4]} for E75 and { 15 [4, 3]} for E25 . Thus, an orthonormal basis B for R2 is the set B = 15 [−3, 4], 15 [4, 3] . Now P is the matrix whose columns are the vectors in B, and D is the diagonal matrix whose main diagonal entries are the eigenvalues 75 and 25. Hence, −3 4 75 0 and D = . P = 15 4 3 0 25 Next, direct computation using the given formula for Q yields Q([1, −8]) = 4075. However, we can also calculate Q([1, −8]) as [1, −8]TB D[1, −8]B , where 1 1 −3 4 1 −35 −7 = PT = 15 = 15 = . [1, −8]B = P−1 −8 −8 4 3 −8 −20 −4 75 0 −525 −7 Then Q([1, −8]) = −7 −4 = −7 −4 = 4075. 0 25 −4 −100 (c) The entries for the symmetric matrix A representing Q are given, for i = j, by aij = 12 (coefficient of xi xj ) in the expression for Q, and, on the diagonal, aii = (the coefficient of x2i ). Hence, ⎡ ⎤ 18 48 −30 18 ⎦. We need to orthogonally diagonalize A. A = ⎣ 48 −68 −30 18 1 Now pA (x) = x3 + 49x2 − 4802x = x(x2 + 49x − 4802). Thus, λ1 = 0 is an eigenvalue for A. Using the quadratic formula on the second factor of pA (x) produces the eigenvalues λ2 = 49 and λ3 = −98. Then, using row reduction to find particular solutions to the systems [0I3 − A|0], [49I3 − A|0] and [−98I3 − A|0] as in the Diagonalization Method of Section 3.4 produces the basis {[2, 3, 6]} for E0 , {[−6, −2, 3]} for E49 , and {[3, −6, 2]} for E−98 . Since E0 , E49 , and E−98 are one-dimensional, the given bases are already orthogonal. To create orthonormal bases, we need , , , to normalize the vectors, yielding 17 [2, 3, 6] for E0 , 17 [−6, −2, 3] for E49 , and 17 [3, −6, 2] for E−98 . Thus, the orthonormal basis B for R3 is the set B = 17 [2, 3, 6], 17 [−6, −2, 3], 17 [3, −6, 2] . Now P is the matrix whose columns are the vectors in B, and D is the diagonal matrix whose

264


Section 8.11 ⎡

main diagonal ⎡

0 D=⎣ 0 0

0 49 0

2 entries are the eigenvalues 0, 49, and −98. Hence P = 17 ⎣ 3 6 ⎤ 0 0 ⎦. −98

−6 −2 3

⎤ 3 −6 ⎦ and 2

Next, direct computation using the given formula for Q yields Q([4, −3, 6]) = −3528. However, we can also calculate Q([4, −3, 6]) as [4, −3, 6]TB D[4, −3, 6]B , where ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 4 2 3 6 4 35 5 [4, −3, 6]B = P−1 ⎣ −3 ⎦ = PT ⎣ −3 ⎦ = 17 ⎣ −6 −2 3 ⎦ ⎣ −3 ⎦ = 17 ⎣ 0 ⎦ = ⎣ 0 ⎦ . 6 6 3 −6 2 6 42 6 ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 0 5 0 0 0 ⎦⎣ 0 ⎦ = 5 0 6 ⎣ 0 ⎦ = −3528. Then Q([4, −3, 6]) = 5 0 6 ⎣ 0 49 0 0 −98 6 −588 (4) Yes. If Q(x) = Σaij xi xj , 1 ≤ i ≤ j ≤ n, then xT C1 x and C1 upper triangular imply that the (i, j) entry for C1 is zero if i > j and aij if i ≤ j. A similar argument describes C2 . Thus, C1 = C2 . (6) (a) True. This is discussed in the proof of Theorem 8.13. (b) False. For this quadratic form, the coefficients of x2 and y 2 are zero, which are allowed in the definition of a quadratic form. (c) False. A quadratic form can be represented by many matrices. In particular, every quadratic form is represented by both an upper triangular matrix and a symmetric matrix (which are usually distinct). For example, the quadratic form Q([x, y]) = 2xy can be expressed both as 0 2 0 1 x x x y and as x y . 0 0 y 1 0 y (d) True. This is the conclusion of the Principal Axes Theorem (Theorem 8.15). (e) True. This can be inferred from the proof of the Principal Axes Theorem (Theorem 8.15) and the Orthogonal Diagonalization Method from Section 6.3 of the textbook. This fact is utilized in the Quadratic Form Method in Section 8.11 of the textbook.

265


Section 9.1

Chapter 9 Section 9.1 (1) (a) For the first system, row reduce 5 −2 5 −1.995

" " 10 1 " to obtain " 17.5 0

" 0 "" 602 . 1 " 1500

Hence, the solution to the first system is (602, 1500). For the second system, row reduce " " 1 0 "" 302 5 −2 "" 10 . to obtain 0 1 " 750 5 −1.99 " 17.5 Hence, the solution to the second system is (302, 750). These systems are ill-conditioned because a very small change in the coefficient of y leads to a very large change in the solution. (2) Answers to this problem may differ significantly from the following if you round at different places in the algorithm. We rounded the original numbers in the system and directly after performing each row operation. (a) Without partial pivoting, we perform the following sequence of row operations: 1 1 ← 0.00072 1 (new 1 is [1, − 5990, − 1370]) 2 ← −2.31 1 + 2 (new 2 is [0, 3960, 3030]) 2 ←

1 3960

2 (new 2 is [0, 1, 0.765])

1 ← 5990 2 + 1 (new 1 is [1, 0, 3210]) giving the solution (3210, 0.765). With partial pivoting, we perform the following sequence of row operations: 1 ↔ 2 1 ←

1 2.31

1 (new 1 is [1, − 4280, − 56.7])

2 ← −0.00072 1 + 2 (new 2 is [0, − 1.23, − 0.944]) 1 2 (new 2 is [0, 1, 0.767]) 2 ← − 1.23

1 ← 4280 2 + 1 (new 1 is [1, 0, 3230]) giving the solution (3230, 0.767). The actual solution is (3214, 0.765). (c) Without partial pivoting, we perform the following sequence of row operations: 1 1 (new 1 is [1, 722, 397, − 108]) 1 ← 0.00032 2 ← 241 1 + 2 (new 2 is [0, 174000, 95700, − 26600]) 3 ← −49 1 + 3 (new 3 is [0, − 35300, − 19500, 5580]) 2 ←

1 174000

2 (new 2 is [0, 1, 0.55, − 0.153])

1 ← −722 2 + 1 (new 1 is [1, 0, − 0.1, 2.47]) 3 ← 35300 2 + 3 (new 3 is [0, 0, − 85, 179]) 1 3 (new 3 is [0, 0, 1, − 2.11]) 3 ← − 85

1 ← 0.1 3 + 1 (new 1 is [1, 0, 0, 2.26]) 2 ← −0.55 3 + 2 (new 2 is [0, 1, 0, 1.01]) 266


Section 9.1

giving the solution (2.26, 1.01, −2.11). With partial pivoting, we perform the following sequence of row operations: 1 ↔ 2 1 1 (new 1 is [1, 0.9, 0.0332, 2.39]) 1 ← − 241

2 ← −0.00032 1 + 2 (new 2 is [0, 0.231, 0.127, − 0.0354]) 3 ← −49 1 + 3 (new 3 is [0, 0.9, 0.773, 166]) 2 ↔ 3 2 ←

1 0.9

2 (new 2 is [0, 1, 0.859, 184])

1 ← −0.9 2 + 1 (new 1 is [1, 0, − 0.740, − 163]) 3 ← −0.231 2 + 3 (new 3 is [0, 0, − 0.0714, − 42.5]) 1 3 (new 3 is [0, 0, 1, 595]) 3 ← − 0.0714

1 ← 0.74 3 + 1 (new 1 is [1, 0, 0, 277]) 2 ← −0.859 3 + 2 (new 2 is [0, 1, 0, − 327]) giving the solution (277, −327, 595). The actual solution is (267, −315, 573). (3) Answers to this problem may differ significantly from the following if you round at different places in the algorithm. We rounded directly after performing each row operation. (a) Without partial pivoting, we perform the following sequence of row operations: 1 1 (new 1 is [1, − 5989, − 1368]) 1 ← 0.00072 2 ← −2.31 1 + 2 (new 2 is [0, 3959, 3029]) 2 ←

1 3959

2 (new 2 is [0, 1, 0.7651])

1 ← 5989 2 + 1 (new 1 is [1, 0, 3214]) giving the solution (3214, 0.7651). With partial pivoting, we perform the following sequence of row operations: 1 ↔ 2 1 ←

1 2.31

1 (new 1 is [1, − 4275, − 56.62])

2 ← −0.00072 1 + 2 (new 2 is [0, − 1.234, − 0.9438]) 1 2 (new 2 is [0, 1, 0.7648]) 2 ← − 1.234

1 ← 4275 2 + 1 (new 1 is [1, 0, 3213]) giving the solution (3213, 0.7648). The actual solution is (3214, 0.765). (c) Without partial pivoting, we perform the following sequence of row operations: 1 1 (new 1 is [1, 723.1, 396.9, − 108]) 1 ← 0.00032 2 ← 241 1 + 2 (new 2 is [0, 174100, 95640, − 26600]) 3 ← −49 1 + 3 (new 3 is [0, − 35390, − 19450, 5575]) 2 ←

1 174100

2 (new 2 is [0, 1, 0.5493, − 0.1528])

1 ← −723.1 2 + 1 (new 1 is [1, 0, − 0.2988, 2.490]) 3 ← 35390 2 + 3 (new 3 is [0, 0, − 10.27, 167.4]) 1 3 (new 3 is [0, 0, 1, − 16.30]) 3 ← − 10.27

267


Section 9.1

1 ← 0.2988 3 + 1 (new 1 is [1, 0, 0, − 2.380]) 2 ← −0.5493 3 + 2 (new 2 is [0, 1, 0, 8.801]) giving the solution (−2.38, 8.801, −16.30). With partial pivoting, we perform the following sequence of row operations: 1 ↔ 2 1 1 (new 1 is [1, 0.9004, 0.0332, 2.390]) 1 ← − 241

2 ← −0.00032 1 + 2 (new 2 is [0, 0.2311, 0.1270, − 0.03532]) 3 ← −49 1 + 3 (new 3 is [0, 0.8804, 0.7732, 166.1]) 2 ↔ 3 2 ←

1 0.8804

2 (new 2 is [0, 1, 0.8782, 188.7])

1 ← −0.9004 2 + 1 (new 1 is [1, 0, − 0.7575, − 167.5]) 3 ← −0.2311 2 + 3 (new 3 is [0, 0, − 0.07595, − 43.64]) 1 3 (new 3 is [0, 0, 1, 574.6]) 3 ← − 0.07595

1 ← 0.7575 3 + 1 (new 1 is [1, 0, 0, 267.8]) 2 ← −0.8782 3 + 2 (new 2 is [0, 1, 0, − 315.9]) giving the solution (267.8, −315.9, 574.6). The actual solution is (267, −315, 573). (4) (a) First, we express the given system as :

x1 x2

= − 15 x2 = − 37 x1

+

26 5

−

6

.

Starting with the initial values x1 = 0 and x2 = 0, we plug these in the right side of these equations to obtain the new values x1 = 26 5 and x2 = −6. These are the values listed in the table, below, in the row labeled “After 1 Step.” These new values are again put in the right side to solve for x1 and x2 . After we round to 3 decimal places, this yields x1 = 6.400 and x2 = −8.229, the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps After 7 Steps After 8 Steps After 9 Steps

x1 0.000 5.200 6.400 6.846 6.949 6.987 6.996 6.999 7.000 7.000

x2 0.000 −6.000 −8.229 −8.743 −8.934 −8.978 −8.994 −8.998 −9.000 −9.000

After 9 steps, we get the same result as we had after 8 steps, and so we are done. The solution obtained for the system is (7, −9).

268


(c) First, we express the given system as ⎧ 1 ⎪ ⎨ x1 = − 7 x2 1 x2 = 6 x1 ⎪ ⎩ 1 x3 = 3 x1

+ − −

2 7 x3 1 6 x3 1 6 x2

62 7 9 2 13 3

− + −

Section 9.1

.

Starting with the initial values x1 = 0, x2 = 0, and x3 = 0, we plug these in the right side of these 9 13 equations to obtain the new values x1 = − 62 7 , x2 = 2 , and x3 = − 3 . These values, rounded to 3 places after the decimal point, are listed in the table, below, in the row labeled “After 1 Step.” These new values are again put in the right side to solve for x1 , x2 , and x3 . After we round to 3 decimal places, this yields x1 = −10.738, x2 = 3.746, and x3 = −8.036, the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps After 7 Steps After 8 Steps After 9 Steps After 10 Steps

x1 0.000 −8.857 −10.738 −11.688 −11.875 −11.969 −11.988 −11.997 −11.999 −12.000 −12.000

x2 0.000 4.500 3.746 4.050 3.975 4.005 3.998 4.001 4.000 4.000 4.000

x3 0.000 −4.333 −8.036 −8.537 −8.904 −8.954 −8.991 −8.996 −8.999 −9.000 −9.000

After 10 steps, we get the same result as we had after 9 steps, and so we are done. The solution obtained for the system is (−12, 4, −9). (5) (a) First, we express the given system as :

x1 x2

= − 15 x2 = − 37 x1

+

26 5

−

6

.

We start with the initial values x1 = 0 and x2 = 0. We plug x2 = 0 into the right side of the first 26 equation to obtain the new value x1 = 26 5 . Then we substitute x1 = 5 into the right side of the 288 second equation yielding x2 = − 35 . After we round to 3 places after the decimal point, we listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug x2 = −8.229 into the right side of the first equation to solve for x1 , which, after rounding to 3 decimal places, yields x1 = 6.846. Substituting x1 = 6.846 into the right side of the second equation produces x2 = −8.934. The two new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table:

269


Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps

x1 0.000 5.200 6.846 6.987 6.999 7.000 7.000

Section 9.1

x2 0.000 −8.229 −8.934 −8.994 −9.000 −9.000 −9.000

After 6 steps, we get the same result as we had after 5 steps, and so we are done. The solution obtained for the system is (7, −9). (c) First, we express the given system as ⎧ 1 ⎪ ⎨ x1 = − 7 x2 1 x2 = 6 x1 ⎪ ⎩ 1 x3 = 3 x1

+ − −

2 7 x3 1 6 x3 1 6 x2

− + −

62 7 9 2 13 3

.

We start with the initial values x1 = 0, x2 = 0, and x3 = 0. We plug x2 = x3 = 0 into the right side of the first equation to obtain the new value x1 = − 62 7 ≈ −8.857. Then we substitute x1 = −8.857 and x3 = 0 into the right side of the second equation, yielding x2 ≈ 3.024. Finally, we plug x1 = −8.857 and x2 = 3.024 into the right side of the third equation, yielding x3 ≈ −7.790. We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug x2 = 3.024 and x3 = −7.790 into the right side of the first equation to solve for x1 , which yields x1 ≈ −11.515. Substituting x1 = 11.515 and x3 = −7.790 into the right side of the second equation produces x2 ≈ 3.879. Finally, we plug x1 = −11.515 and x2 = 3.879 into the right side of the third equation, yielding x3 ≈ −8.818. The three new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps After 7 Steps

x1 0.000 −8.857 −11.515 −11.931 −11.990 −11.998 −12.000 −12.000

x2 0.000 3.024 3.879 3.981 3.997 4.000 4.000 4.000

x3 0.000 −7.790 −8.818 −8.974 −8.996 −8.999 −9.000 −9.000

After 7 steps, we get the same result as we had after 6 steps, and so we are done. The solution obtained for the system is (−12, 4, −9). (6) (a) A satisfies all of the criteria, so it is strictly diagonally dominant. (In row 1, | − 3| > |1|; that is, |a11 | > |a12 |. In row 2, |4| > | − 2|; that is, |a22 | > |a21 |.) (b) A is not strictly diagonally dominant because |a22 | < |a21 |. (Note that |a11 | = |a12 |, which also prevents A from being strictly diagonally dominant.) (c) A satisfies all of the criteria, so it is strictly diagonally dominant. (In row 1, |a11 | > |a12 | + |a13 |; in row 2, |a22 | > |a21 | + |a23 |; in row 3, |a33 | > |a31 | + |a32 |.) (d) A is not strictly diagonally dominant because |a22 | < |a21 | + |a23 |. 270


Section 9.1

(e) A is not strictly diagonally dominant because |a22 | = |a21 | + |a23 |. (7) (a) Put the third equation first and move the other two down. Then express the equations as follows: ⎧ 1 3 − + 25 ⎪ 8 x3 8 8 x2 ⎨ x1 = 2 1 x2 = − 13 x1 − 13 x3 . ⎪ ⎩ 1 2 26 x3 = − 15 x1 + 15 x2 + 15 We start with the initial values x1 = 0, x2 = 0, and x3 = 0. We plug x2 = x3 = 0 into the right side of the first equation to obtain the new value x1 = 25 8 = 3.125. Then we substitute x1 = 3.125 and x3 = 0 into the right side of the second equation, yielding x2 ≈ −0.481. Finally, we plug x1 = 3.125 and x2 = −0.481 into the right side of the third equation, yielding x3 ≈ 1.461. We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug x2 = −0.481 and x3 = 1.461 into the right side of the first equation to solve for x1 , which yields x1 ≈ 2.517. Substituting x1 = 2.517 and x3 = 1.461 into the right side of the second equation produces x2 ≈ −0.500. Finally, we plug x1 = 2.517 and x2 = −0.500 into the right side of the third equation, yielding x3 ≈ 1.499. The three new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps

x1 0.000 3.125 2.517 2.500 2.500

x2 0.000 −0.481 −0.500 −0.500 −0.500

x3 0.000 1.461 1.499 1.500 1.500

After 4 steps, we get the same result as we had after 3 steps, and so we are done. The solution obtained for the system is (2.5, −0.5, 1.5). (c) Put the second equation first, the fourth equation second, the first equation third, and the third equation fourth. Then express the equations as follows: ⎧ 1 1 49 x1 = − 29 x2 + − + ⎪ ⎪ 9 x3 9 x4 9 ⎪ ⎪ 86 ⎨ x2 = − 1 x1 − 3 x3 + 2 x4 − 17 17 17 17 . 1 1 2 120 ⎪ x = − x − x − x + ⎪ 3 13 1 13 2 13 4 13 ⎪ ⎪ ⎩ 3 1 55 x4 = − 17 x1 + 14 x2 − 14 x3 − 7 We start with the initial values x1 = 0, x2 = 0, x3 = 0, and x4 = 0. We plug x2 = x3 = x4 = 0 into the right side of the first equation to obtain the new value x1 = 49 9 ≈ 5.444. Then we substitute x1 = 5.444, x3 = 0, and x4 = 0 into the right side of the second equation, yielding x2 ≈ −5.379. Next, we substitute x1 = 5.444, x2 = −5.379, and x4 = 0 into the right side of the third equation, yielding x3 ≈ 9.226. Finally, we plug x1 = 5.444, x2 = −5.379, and x3 = 9.226 into the right side of the fourth equation, yielding x4 ≈ −10.447. We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug x2 = −5.379, x3 = 9.226, and x4 = −10.447 into the right side of the first equation to solve for x1 , which yields x1 ≈ 8.826. Substituting x1 = 8.826, x3 = 9.226, and x4 = −10.447 into the right side of the second equation produces x2 ≈ −8.435. Next, substituting x1 = 8.826, x2 = −8.435, and x4 = −10.447 into the right side of the third equation produces x3 ≈ 10.808. Finally, we plug x1 = 8.826, x2 = −8.435, and x3 = 10.808 into the right side of the fourth equation, yielding x4 ≈ −11.698. The four new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner, producing 271


Section 9.1

the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps After 7 Steps After 8 Steps

x1 0.000 5.444 8.826 9.820 9.973 9.995 9.999 10.000 10.000

x2 0.000 −5.379 −8.435 −8.920 −8.986 −8.998 −9.000 −9.000 −9.000

x3 0.000 9.226 10.808 10.961 10.994 10.999 11.000 11.000 11.000

x4 0.000 −10.447 −11.698 −11.954 −11.993 −11.999 −12.000 −12.000 −12.000

After 8 steps, we get the same result as we had after 7 steps, and so we are done. The solution obtained for the system is (10, −9, 11, −12). (8) First, we express the given system as ⎧ ⎨ x1 x2 ⎩ x3

= 5x2 = 6x1 = −7x1

+ x3 − 2x3 − x2

+ 16 − 13 . + 12

Using the Jacobi Method, we start with the initial values x1 = 0, x2 = 0, and x3 = 0, and plug these in the right side of these equations to obtain the new values x1 = 16, x2 = −13, and x3 = 12. These values are listed in the table, below, in the row labeled “After 1 Step.” These new values are put in the right side to again solve for x1 , x2 , and x3 , yielding x1 = −37, x2 = 59, and x3 = −87, the numbers in the row labeled “After 2 Steps.” We continue in this manner for 6 steps, producing the results given in the following table: Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps After 5 Steps After 6 Steps

x1 0.0 16.0 −37.0 224.0 −77.0 3056.0 12235.0

x2 0.0 −13.0 59.0 −61.0 907.0 2515.0 19035.0

x3 0.0 12.0 −87.0 212.0 −1495.0 −356.0 −23895.0

Note that the numbers are diverging rather than converging to a single solution. Next, we use the Gauss-Seidel Method. We start with the initial values x1 = 0, x2 = 0, and x3 = 0. We plug x2 = x3 = 0 into the right side of the first equation to obtain the new value x1 = 16. Then we substitute x1 = 16 and x3 = 0 into the right side of the second equation, yielding x2 = 83. Finally, we plug x1 = 16 and x2 = 83 into the right side of the third equation, yielding x3 = −183. We have listed these results in the table, below, in the row labeled “After 1 Step.” Next, we plug x2 = 83.0 and x3 = −183.0 into the right side of the first equation to solve for x1 , which yields x1 = 248. Substituting x1 = 248 and x3 = −183 into the right side of the second equation produces x2 = 1841. Finally, we plug x1 = 248 and x2 = 1841 into the right side of the third equation, yielding x3 = −3565. The three new results are the numbers in the row labeled “After 2 Steps.” We continue in this manner for four steps, producing the results given in the following table:

272


Initial Values After 1 Step After 2 Steps After 3 Steps After 4 Steps

x1 0.0 16.0 248.0 5656.0 124648.0

x2 0.0 83.0 1841.0 41053.0 909141.0

Section 9.2

x3 0.0 −183.0 −3565.0 −80633.0 −1781665.0

Again, the values are diverging rather than converging. To find the actual solution, we use Gaussian elimination. We reduce " ⎡ ⎤ " ⎤ ⎡ 1 −5 −1 "" 16 1 −5 −1 "" 16 ⎥ 4 " ⎣ 6 −1 −2 " 13 ⎦ to ⎢ 1 29 " − 83 ⎣ 0 29 ⎦ . " " 7 1 1 " 12 0 0 1 " 1 4 (1) − 83 The last row yields x3 = 1. The second row gives x2 = − 29 29 = −3, and the first row produces x1 = 5(−3) + 1(1) + 16 = 2. Hence, the actual solution for the system is (2, −3, 1).

(10) (a) True. This is the definition of roundoff error given in Section 9.1 of the textbook. (b) False. An ill-conditioned system is a system for which a very small change in the coefficients in the system leads to a very large change in the solution set. (c) False. The method of partial pivoting specifies that we should use row swaps to ensure that each pivot element is as large as possible in absolute value. (d) True. Roundoff errors do not accumulate in iterative methods since each iteration is using the previous iteration as a new starting point. Thus, in some sense, each iteration starts the solution process over again. (e) False. This is done in the Gauss-Seidel Method, not the Jacobi Method. (f) False. The values obtained should be x = 6 and y = 1 because the new value x = 6 is used in the computation of the new value of y.

Section 9.2 (1) (a) First, use the row operations 1 ← 12 1 , 2 ← 6 1 + 2 , and 2 ← 15 2 to obtain the upper 1 −2 . Using the formulas for the variables kij given in the proof of triangular matrix U = 0 1 Theorem 9.1 in Section 9.2 of the textbook, k11 = 2, k21 = −6, and k22 = 5. Then, using the ( ' 1 0 1 0 2 0 = and D = . formulas for L and D given in that same proof, L = −6 −3 1 0 5 1 2 (c) First, use the row operations 1 ← − 1 , 2 ← −2 1 + 2 , 3 ← −2 1 + 3 , 2 ← 12 2 , ⎡ ⎤ 1 −4 2 1 −4 ⎦. 3 ← −8 2 + 3 , and 3 ← 13 3 to obtain the upper triangular matrix U = ⎣ 0 0 0 1 Using the formulas for the variables kij given in the proof of Theorem 9.1 in Section 9.2 of the textbook, k11 = −1, k21 = 2, k31 = 2, k22 = 2, k32 = 8, and k33 = 3. Then, using the

273

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ ⎢ formulas for L and D given in that same proof, L = ⎣ ⎡

−1 D=⎣ 0 0

⎤ 0 0 ⎦. 3

0 2 0

1

0

2 −1 2 −1

8 2

1

0

Section 9.2 ⎤

⎡

1 ⎣ 0 ⎥ −2 = ⎦ −2 1

0 1 4

⎤ 0 0 ⎦ and 1

(e) First, use the row operations 1 ← − 13 1 , 2 ← −4 1 + 2 , 3 ← −6 1 + 3 , 4 ← 2 1 + 4 , 2 ← − 32 2 , 3 ← − 2 + 3 , 4 ← − 43 2 + 4 , and 3 ← 2 3 to obtain the upper triangular ⎤ ⎡ 1 1 − 13 − 13 3 ⎥ ⎢ 5 ⎢ 0 ⎥ 1 − 11 2 2 ⎥ ⎢ . Using the formulas for the variables kij given in the proof matrix U = ⎢ 0 1 3 ⎥ ⎣ 0 ⎦ 0 0 0 1 of Theorem 9.1 in Section 9.2 of the textbook, k11 = −3, k21 = 4, k31 = 6, k41 = −2, k22 = − 23 , k32 = 1, k42 = 43 , k33 = 12 , k43 = 0, and k44 ⎡ 1 0 0 ⎢ 4 1 0 ⎢ −3 that same proof, L = ⎢ 1 ⎢ 6 1 ⎣ −3 (−2/3) ⎡ ⎢ ⎢ D=⎢ ⎢ ⎣

−2

−3

0

0

− 23

0

0

0

0

0

0

−3 ⎤

(4/3) (−2/3)

0 (1/2)

= 1. Then, using the formulas for L and D given in ⎤ ⎤ ⎡ 0 1 0 0 0 ⎥ ⎢ 0 ⎥ ⎥ ⎢ − 43 1 0 0 ⎥ ⎥ and ⎥=⎢ ⎥ ⎢ 3 0 ⎥ ⎦ ⎣ −2 − 2 1 0 ⎦ 2 −2 0 1 1 3

⎥ 0 ⎥ ⎥. 1 0 ⎥ ⎦ 2 0 1 0

(3) (a) The coefficient matrix for the system AX = B is A =

−1 2

5 −13

−9 , with B = . We 21

need to find the KU decomposition for A. To do this, we use the row operations 1 ← − 1 , 1 −5 1 2 ← −2 1 + 2 , and 2 ← − 3 2 to obtain the upper triangular matrix U = . Using 0 1 the formulas for the variables kij given in the proof of Theorem 9.1 in Section 9.2 of the textbook, −1 0 . Now, k11 = −1, k21 = 2, and k22 = −3, yielding the lower triangular matrix K = 2 −3 AX = B is equivalent to K(UX) = B. Letting Y = UX, we must first solve KY = B. This is equivalent to the system

)

−y1 2y1

− 3y2

= −9 , = 21

whose solution is easily seen to be y1 = 9, y2 = −1; that is, Y = [9, −1]. Finally, we find X by solving the system UX = Y, which is ) x1 − 5x2 = 9 . x2 = −1 274


Section 9.3

Back substitution gives the solution x2 = −1, x1 = 4. Hence, the solution set for the original system is {(4, −1)}. ⎡ ⎤ ⎡ ⎤ −1 3 −2 −13 −7 ⎦, with B = ⎣ 28 ⎦. (c) The coefficient matrix for the system AX = B is A = ⎣ 4 −9 −2 11 −31 −68 We need to find the KU decomposition for A. To do this, we use the row operations 1 ← − 1 , 2 ← −4 1 + 2 , 3 ← 2 1 + 3 , 2 ← 13 2 , 3 ← −5 2 + 3 , and 3 ← − 12 3 to ⎡ ⎤ 1 −3 2 1 −5 ⎦. Using the formulas for the variables obtain the upper triangular matrix U = ⎣ 0 0 0 1 kij given in the proof of Theorem 9.1 in Section 9.2 of the textbook, k11 = −1, k21 = 4, k31 = −2, ⎡ ⎤ −1 0 0 0 ⎦. Now, k22 = 3, k32 = 5, and k33 = −2, yielding the lower triangular matrix K = ⎣ 4 3 −2 5 −2 AX = B is equivalent to K(UX) = B. Letting Y = UX, we must first solve KY = B. This is equivalent to the system ⎧ ⎨

−y1 4y1 ⎩ −2y1

+ +

3y2 5y2

− 2y3

= −13 = 28 , = −68

whose solution is easily seen to be y1 = 13, y2 = 13 (−4(13) + 28) = −8, y3 = − 12 (2(13) − 5(−8) − 68) = 1; that is, Y = [13, −8, 1]. Finally, we find X by solving the system UX = Y, which is ⎧ ⎨ x1 − 3x2 + 2x3 = 13 x2 − 5x3 = −8 . ⎩ 1 x3 = Back substitution gives the solution x3 = 1, x2 = 5(1) − 8 = −3, x1 = 3(−3) − 2(1) + 13 = 2. Hence, the solution set for the original system is {(2, −3, 1)}. (4) (a) False. Nonsingular matrices that require row swaps to reduce to In may not have LDU decom 0 1 positions. For example, has no LDU decomposition, as shown in Exercise 2 of Section 1 0 9.2. (b) True. The method for doing this is given in the proof of Theorem 9.1. (c) False. The given row operation adds a multiple of row 3 to row 2, which is above row 3. This is not allowed for lower type (II) operations. (d) False. First, solve for Y in KY = B; then solve for X in UX = Y.

Section 9.3 (1) In each part, we print only m digits after the decimal point for all computations, even though more digits are actually used and computed. We write all vectors as row vectors to conserve space, even when correct notation calls for a column vector. Also, let A represent the given matrix. 275


(a) We begin with u0 = [1, 0]. Then, w1 = Au0 = [2, 36], and so w1 =

√

Section 9.3

22 + 362 ≈ 36.06. Using

w1 1 ≈ 36.06 [2, 36] ≈ [0.06, 1.00]. These results are displayed in the first line of this gives u1 = w 1 the table, below. Continuing in this manner for a total of 9 iterations produces the following:

k 1 2 3 4 5 6 7 8 9

wk = Auk−1 [2, 36] [36.06, 24.96] [22.14, 42.69] [32.88, 36.99] [28.24, 41.11] [30.81, 39.34] [29.58, 40.30] [30.21, 39.84] [29.90, 40.08]

wk 36.06 43.85 48.09 49.49 49.87 49.97 49.99 50.00 50.00

wk uk = w k [0.06, 1.00] [0.82, 0.57] [0.46, 0.89] [0.66, 0.75] [0.57, 0.82] [0.62, 0.79] [0.59, 0.81] [0.60, 0.80] [0.60, 0.80]

Since u8 = u9 , we stop here. Hence, the Power Method produces the eigenvector [0.60, 0.80]. Since the vectors u8 and u9 are equal (rather than being the negations of each other), the sign of the corresponding eigenvalue is positive, and so the eigenvalue is w9 = 50. √ (c) We begin with u0 = [0, 1, 0]. Then, w1 = Au0 = [3, 0, 1], and so w1 = 32 + 02 + 12 ≈ 3.16. w1 1 Using this gives u1 = w ≈ 3.16 [3, 0, 1] ≈ [0.95, 0.00, 0.32]. These results are displayed in the 1 first line of the table, below. Continuing in this manner for a total of 7 iterations produces the following:

k 1 2 3 4 5 6 7

wk = Auk−1 [3, 0, 1] [1.58, 1.26, 3.16] [1.01, 1.26, 2.44] [1.15, 1.18, 2.30] [1.25, 1.22, 2.45] [1.23, 1.23, 2.47] [1.22, 1.23, 2.45]

wk 3.16 3.75 2.93 2.83 3.01 3.02 3.00

wk uk = w k [0.95, 0.00, 0.32] [0.42, 0.34, 0.84] [0.34, 0.43, 0.83] [0.41, 0.42, 0.81] [0.42, 0.41, 0.81] [0.41, 0.41, 0.82] [0.41, 0.41, 0.82]

Since u6 = u7 , we stop here. Hence, the Power Method produces the eigenvector [0.41, 0.41, 0.82]. Since the vectors u6 and u7 are equal (rather than being negations of each other), the sign of the corresponding eigenvalue is positive, and so the eigenvalue is w7 = 3.0. (e) We begin with u0 = [3, 8, 2, 3]. Then, w1 = Au0 = [17, 40, 14, 19], and so √ w1 1 w1 = 172 + 402 + 142 + 192 ≈ 49.457. Using this gives u1 = w ≈ 49.457 [17, 40, 14, 19] ≈ 1 [0.344, 0.809, 0.283, 0.384]. These results are displayed in the first line of the table, below. Continuing in this manner for a total of 15 iterations produces the following:

276


k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

wk = Auk−1 [17, 40, 14, 19] [2.123, 4.448, 2.143, 2.285] [1.825, 4.224, 1.392, 1.928] [1.935, 4.468, 1.440, 2.010] [1.878, 4.471, 1.232, 1.929] [1.893, 4.537, 1.190, 1.929] [1.878, 4.553, 1.116, 1.903] [1.879, 4.575, 1.084, 1.896] [1.874, 4.585, 1.053, 1.886] [1.873, 4.593, 1.036, 1.881] [1.871, 4.598, 1.022, 1.876] [1.870, 4.602, 1.013, 1.874] [1.869, 4.605, 1.007, 1.872] [1.869, 4.606, 1.003, 1.871] [1.868, 4.607, 1.000, 1.870]

wk 49.457 5.840 5.179 5.461 5.363 5.413 5.397 5.406 5.403 5.405 5.405 5.405 5.405 5.405 5.405

Section 9.3

wk uk = w k [0.344, 0.809, 0.283, 0.384] [0.364, 0.762, 0.367, 0.391] [0.352, 0.816, 0.269, 0.372] [0.354, 0.818, 0.264, 0.368] [0.350, 0.834, 0.230, 0.360] [0.350, 0.838, 0.220, 0.356] [0.348, 0.844, 0.207, 0.353] [0.347, 0.846, 0.201, 0.351] [0.347, 0.848, 0.195, 0.349] [0.346, 0.850, 0.192, 0.348] [0.346, 0.851, 0.189, 0.347] [0.346, 0.851, 0.187, 0.347] [0.346, 0.852, 0.186, 0.346] [0.346, 0.852, 0.185, 0.346] [0.346, 0.852, 0.185, 0.346]

Since u14 = u15 , we stop here. Hence, the Power Method produces the eigenvector [0.346, 0.852, 0.185, 0.346]. Since the vectors u14 and u15 are equal (rather than being negations of each other), the sign of the corresponding eigenvalue is positive, and so the eigenvalue is w15 = 5.405. (3) (b) Let λ1 , . . . λn be the eigenvalues of A with |λ1 | > |λj |, for 2 ≤ j ≤ n. Let {v1 , . . . , vn } be a basis of fundamental eigenvectors for Rn corresponding to λ1 , . . . , λn , respectively. Suppose the initial vector in the Power Method is u0 = a01 v1 + · · · + a0n vn and the ith iteration yields ui = ai1 v1 + · · · + ain vn . Thus, for 2 ≤ j ≤ n, λj = 0, and a0j = 0, we want to prove that " "i |ai1 | "" λ1 "" |a01 | = . |aij | " λj " |a0j | Proof: The Power Method states that each ui is derived from ui−1 as ui = ci Aui−1 , where ci is a normalizing constant. A proof by induction shows that ui = ki Ai u0 for some constant ki . (Base Step i = 0: u0 = In u0 = 1A0 u0 , with k0 = 1. Inductive Step: ui = ci Aui−1 = ci A(ki−1 Ai−1 u0 ) = ci ki−1 Ai u0 , with ki = ci ki−1 .) Now, if u0 = a01 v1 + · · · + a0n vn , then ui = ki Ai u0 = a01 ki Ai v1 +· · ·+a0n ki Ai vn . But because vj is an eigenvector for A corresponding to the eigenvalue λj , Ai vj = λij vj . Therefore, ui = a01 ki λi1 v1 + · · · + a0n ki λin vn . Since we also have ui = ai1 v1 + · · · + ain vn , the uniqueness of expression of a vector with respect to a basis shows that aij = a0j ki λij . Hence, " " "a01 ki λi1 " "" λ1 ""i |a01 | |ai1 | " =" " = " . "a0j ki λij " " λj " |a0j | |aij | (4) (a) False. The eigenvalue could be − Auk−1 . This occurs when uk = −uk−1 . For an example, −2 1 consider applying the Power Method to the matrix , starting with u0 = [1, 0]. Then 0 1 w1 = [−2, 0], and w1 = 2. Hence, u1 = [−1, 0] = −u0 , and the Power Method yields the eigenvector [−1, 0]. The eigenvalue corresponding to this eigenvector is −2. However, Au0 = w1 = 2. 277


Section 9.4

(b) True. Each iteration of the Power Method will just repeatedly produce the initial vector. (c) True. Starting with u0 = [1, 0, 0, 0], the Power Method will produce ui = 12 [1, 1, 1, 1] for all i ≥ 1 and wi = [2, 2, 2, 2] for all i ≥ 2. Note that [2, 2, 2, 2] = 4. (d) False. The dominant eigenvalue in this case is −3, not 2, because | − 3| > |2|. Hence, the Power Method will produce an eigenvector corresponding to the eigenvalue −3. (Note: The Power Method converges here since A is diagonalizable and there is a dominant eigenvalue.)

Section 9.4 (1) (a) Following the QR Factorization Method, we label the columns of A as w1 = [2, −2, 1], w2 = [6, 0, 6], and w3 = [−3, −9, −3]. Step 1: Beginning the Gram-Schmidt Process, we obtain = w1 = [2, −2, 1], and

2 ·v1 = w2 − w v1 ·v1 v1

v1 v2

=

[6, 0, 6] −

[6,0,6]·[2,−2,1] [2,−2,1]·[2,−2,1] [2, −2, 1]

=

[6, 0, 6] −

18 9 [2, −2, 1]

=

[2, 4, 4] .

Multiplying this vector by a factor of c2 = 12 to simplify, we let v2 = [1, 2, 2]. Finally,

w3 ·v2 3 ·v1 v − v3 = w3 − w 1 v1 ·v1 v2 ·v2 v2 [−3,−9,−3]·[2,−2,1] [2,−2,1]·[2,−2,1] [2, −2, 1]

=

[−3, −9, −3] −

=

[−3, −9, −3] − 99 [2, −2, 1] −

=

[−2, −1, 2] .

−

[−3,−9,−3]·[1,2,2] [1, 2, 2] [1,2,2]·[1,2,2]

−27 9 [1, 2, 2]

Thus, we obtain v3 = [−2, −1, 2]. Step 2: Normalizing v1 , v2 , v3 , we get 2 2 1 1 2 2 2 1 2 u1 = ,− , , u2 = , , , u3 = − , − , . 3 3 3 3 3 3 3 3 3 Step 3: From the QR Factorization Method, we Hence, ⎡ 2 1 Q = ⎣ −2 3 1

know that these vectors are the columns of Q. 1 2 2

⎤ −2 −1 ⎦ . 2

Step 4: ⎡

R = QT A

=

2 1 ⎣ 1 3 −2

−2 2 −1

278

⎤⎡ 1 2 2 ⎦ ⎣ −2 2 1

⎤ ⎡ 6 −3 3 0 −9 ⎦ = ⎣ 0 6 −3 0

6 6 0

⎤ 3 −9 ⎦. 3


Section 9.4

(c) Following the QR Factorization Method, we label the columns of A as w1 = [1, −2, 1], w2 = [5, −4, 5], and w3 = [−3, −2, −5]. Step 1: Beginning the Gram-Schmidt Process, we obtain = w1 = [1, −2, 1], and

2 ·v1 = w2 − w v1 ·v1 v1

v1 v2

=

[5, −4, 5] −

[5,−4,5]·[1,−2,1] [1,−2,1]·[1,−2,1] [1, −2, 1]

=

[5, −4, 5] −

18 6 [1, −2, 1]

=

[2, 2, 2] .

Multiplying this vector by a factor of c2 = 12 to simplify, we let v2 = [1, 1, 1]. Finally,

w3 ·v2 3 ·v1 v − v3 = w3 − w 1 v1 ·v1 v2 ·v2 v2 =

[−3, −2, −5] −

[−3,−2,−5]·[1,−2,1] [1,−2,1]·[1,−2,1] [1, −2, 1]

=

[−3, −2, −5] −

−4 6 [1, −2, 1]

=

[1, 0, −1] .

−

−

[−3,−2,−5]·[1,1,1] [1, 1, 1] [1,1,1]·[1,1,1]

−10 3 [1, 1, 1]

Thus, we obtain v3 = [1, 0, −1]. Step 2: Normalizing v1 , v2 , v3 , we get 1 1 1 2 1 1 1 1 u1 = √ , − √ , √ , u2 = √ , √ , √ , u3 = √ , 0, − √ . 6 6 6 3 3 3 2 2 Rationalizing the denominators produces '√ '√ √ √ ( '√ √ √ ( √ ( 6 2 6 6 3 3 3 2 2 u1 = , u2 = , u3 = . ,− , , , , 0, − 6 6 6 3 3 3 2 2 Step 3: From the QR Factorization Method, we know that these vectors are the columns of Q. Hence, √ √ √ ⎤ ⎡ 6 2 3 3 2 √ √ 1⎢ ⎥ Q = ⎣ −2 6 2 3 0 ⎦. 6 √ √ √ 6 2 3 −3 2 Step 4: ⎡ √ R = QT A

=

1 6

6

⎢ √ ⎣ 2 3 √ 3 2

√ −2 6 √ 2 3 0

√

6

⎤⎡

1 √ ⎥ 2 3 ⎦ ⎣ −2 √ 1 −3 2

5 −4 5

⎤ ⎡ √ 6 3√ 6 − 2 √ 6 ⎤ 3 −3 √ √ ⎥ ⎢ −2 ⎦ = ⎣ 0 2 3 − 10 3 ⎦. 3 √ −5 0 0 2

(2) (a) In this problem, we want to find a least-squares solution for the linear system AX = B, where ⎡ ⎤ ⎡ ⎤ 3 10 −8 A = ⎣ 4 −4 ⎦ and B = ⎣ 30 ⎦ 12 27 10 279


Section 9.4

using a QR factorization for A, as shown in Example 2 in Section 9.4. Following the QR Factorization Method, we label the columns of A as w1 = [3, 4, 12] and w2 = [10, −4, 27]. Step 1: Beginning the Gram-Schmidt Process, we obtain = w1 = [3, 4, 12], and

2 ·v1 = w2 − w v1 ·v1 v1

v1 v2

=

[10, −4, 27] −

[10,−4,27]·[3,4,12] [3,4,12]·[3,4,12] [3, 4, 12]

=

[10, −4, 27] −

338 169 [3, 4, 12]

=

[4, −12, 3] .

Hence, we let v2 = [4, −12, 3]. Step 2: Normalizing v1 , v2 , we get 3 4 12 4 12 3 u1 = , , , u2 = ,− , . 13 13 13 13 13 13 Step 3: From the QR Factorization Method, we know that these vectors are the columns of Q. Hence, ⎡ ⎤ 3 4 1 ⎣ 4 −12 ⎦ . Q= 13 12 3 Step 4:

R = QT A

Now,

=

1 13

3 4

4 −12

12 3

⎡

3 ⎣ 4 12

⎤ 10 13 −4 ⎦ = 0 27

26 13

.

1 216 , and hence, Q B= 13 −362 1 216 13 26 x RX = = . 0 13 y 13 −362 T

Since R is upper triangular, we can quickly find the solution by using back substitution. The last 362 equation asserts 13y = − 362 13 , which leads to y = − 169 ≈ −2.142. Finally, the first equation gives 216 362 13x + 26y = 13 . Substituting − 169 for y and solving for x leads to x = 940 169 ≈ 5.562. (c) In this problem, we want to find a least-squares solution for the linear system AX = B, where ⎡ ⎤ ⎡ ⎤ 7 1 15 1 ⎢ ⎥ ⎢ 4 −4 18 ⎥ ⎥ and B = ⎢ 11 ⎥ A=⎢ ⎣ ⎣ 0 ⎦ −5 ⎦ 8 −22 12 −8 10 −1 using a QR factorization of A, as shown in Example 2 in Section 9.4. Following the QR Factorization Method, we label the columns of A as w1 = [1, 4, 0, −8],

280


Section 9.4

w2 = [15, −4, 8, 10] and w3 = [1, 18, −22, −1]. Step 1: Beginning the Gram-Schmidt Process, we obtain v1 v2

= w1 = [1, 4, 0, −8], and

2 ·v1 = w2 − w v1 ·v1 v1 =

[15, −4, 8, 10] −

[15,−4,8,10]·[1,4,0,−8] [1,4,0,−8]·[1,4,0,−8] [1, 4, 0, −8]

=

[15, −4, 8, 10] −

−81 81 [1, 4, 0, −8]

=

[16, 0, 8, 2] .

Multiplying this vector by a factor of c2 = 12 to simplify, we let v2 = [8, 0, 4, 1]. Finally,

w3 ·v2 3 ·v1 v3 = w3 − w v1 ·v1 v1 − v2 ·v2 v2 =

[1, 18, −22, −1] −

[1,18,−22,−1]·[1,4,0,−8] [1,4,0,−8]·[1,4,0,−8] [1, 4, 0, −8]

=

[1, 18, −22, −1] −

81 81 [1, 4, 0, −8]

=

[8, 14, −18, 8].

Multiplying this vector by a factor of c2 =

1 2

−

−

[1,18,−22,−1]·[8,0,4,1] [8, 0, 4, 1] [8,0,4,1]·[8,0,4,1]

−81 81 [8, 0, 4, 1]

to simplify, we obtain v3 = [4, 7, −9, 4].

Step 2: Normalizing v1 , v2 , v3 , we get

' √ √ √ √ ( 1 4 8 2 2 2 2 2 7 2 8 4 1 . u1 = , , 0, − , u2 = , 0, , , u3 = , ,− , 9 9 9 9 9 9 9 18 2 9

Step 3: From the QR Factorization Method, we Hence, ⎡ 2 ⎢ ⎢ 8 1 ⎢ Q= 18 ⎢ 0 ⎣ −16

know that these vectors are the columns of Q. 16 0 8 2

√ 4 2 √ 7 2 √ −9 2 √ 4 2

⎤ ⎥ ⎥ ⎥. ⎥ ⎦

Step 4: ⎡

R = QT A

Now,

=

2 1 ⎣ 16 18 √ 4 2

8 0

√ 7 2

0 8

√ −9 2

⎡ 1 QT B = ⎣ 3 ⎡

9 ⎣ RX = 0 0

⎡ ⎤ 1 15 1 −16 ⎢ 4 −4 18 2 ⎦⎢ ⎣ 0 √ 8 −22 4 2 −8 10 −1

⎤

⎡ ⎤ 9 −9 9 ⎥ ⎥ = ⎣ 0 18 −9 ⎦. ⎦ √ 0 0 18 2

⎤ −15 ⎦ √16 , and hence, 33 2 ⎡ ⎤⎡ ⎤ ⎤ −9 9 −15 x 1 18 −9 ⎦ ⎣ y ⎦ = ⎣ √16 ⎦ . √ 3 z 33 2 0 18 2 281


Section 9.5

Since R is upper triangular, we √ can quickly find the solution by using back substitution. The last √ equation asserts 18 2z = 11 2, which leads to z = 11 18 ≈ 0.611. The middle equation states that 11 65 . Substituting z = into this equation yields y = 108 ≈ 0.602. Finally, the first 18y − 9z = 16 3 18 65 11 equation gives 9x − 9y + 9z = −5. Substituting 108 for y and 18 for z and solving for x leads to 61 ≈ −0.565. x = − 108 (5) (a) True. Suppose A is a nonsingular matrix having size n × n. Then the columns of A form a basis for Rn by part (c) of Exercise 15 in Section 4.5. Then by Theorem 9.2, Q is also an n × n matrix, and the columns of Q span Rn as well. Since the columns of Q are also an orthonormal set of vectors by Theorem 9.2, they form an orthonormal basis for Rn . Therefore, Q is an orthogonal matrix by Theorem 6.7. (b) True. Suppose A is a singular matrix having size n × n. Since A is singular, not all of the n columns of A are linearly independent, and so the more general QR Factorization Method (described at the end of Section 9.4) must be used here. Note that the matrix Q created by this more general method is also n × n and its columns form a linearly independent set. Thus, Q is a nonsingular matrix. Now suppose R has all of its main diagonal entries nonzero. Then |R| = 0 (by Theorem 3.2), and R is nonsingular. Then A = QR is nonsingular, a contradiction. (c) True. In the QR Factorization Method, the columns of Q are found by performing the GramSchmidt Process on the columns of A and then normalizing. Since A is an upper triangular matrix, the Gram-Schmidt Process produces vectors that are multiples of the standard basis vectors for Rn , and once these vectors are normalized, each resulting vector is either a standard basis vector in Rn or its opposite. Thus, the matrix Q that is created by the QR Factorization Method is a diagonal matrix. Note that the QR Factorization Method produces a matrix R with all main diagonal entries positive. Therefore, if A has all positive entries on the main diagonal, then Q equals In . However, if A has some negative entries on the diagonal, then the corresponding diagonal entries in Q would equal −1 instead of 1. (d) False. By Theorem 9.3, X = R−1 QT B, not RT Q−1 B. For a counterexample, consider A = 1 −1 3 and B = . Then the QR Factorization Method gives A = QR with Q = I2 and 0 1 2 R = A. The solution for the system is X = [5, 2], but RT Q−1 B = [3, −1]. (e) True. If QR is the QR factorization of A, then Q is a matrix having an orthonormal set of columns. Hence, QT Q = Ik . Therefore, R = Ik R = QT QR = QT A.

Section 9.5 (1) For each part, one possibility is given. 34 12 T , and pAT A (x) = (x − 34)(x − 16) − (−12)2 = x2 − 50x + 400 = (a) First, A A = 12 16 √ √ √ (x − 40)(x − 10). Hence, λ1 = 40 and λ2 = 10, implying σ 1 = 40 = 2 10 and σ 2 = 10. 1 −2 T , giving the fundamental eigenvector [2, 1]. Row reducing (40I2 − A A) produces 0 0 ( ' 1 12 Similarly, row reducing (10I2 − AT A) produces , giving the fundamental eigenvector 0 0 282


Section 9.5

3 4 [−1, 2]. Normalizing these vectors yields the set √15 [2, 1], √15 [−1, 2] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = σ11 Av1 and u2 = σ12 Av2 , where v1 and v2 are the right singular vectors. 3 4 2 10 1 1√ √1 √1 = = . Now, u1 = σ11 Av1 = 2√110 5 10 2 2 5 0 1 10 1 3 4 −1 5 1 1 1 1 1 1 √ = 5√2 = √2 . Similarly, u2 = σ2 Av2 = √10 5 5 0 2 −5 −1 Since we have two left singular vectors, and that is all we need for a basis for R2 , the set {u1 , u2 } of left singular vectors is complete. Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣVT √ 1 1 2 10 √0 2 −1 1 ,Σ= . A short computation with U = √2 , and V = √15 1 −1 1 2 0 10 √

1 1 2 2 1 3 4 10 0 1 1 √ √ = = A. verifies that √2 5 1 −1 −1 2 5 0 0 10 ⎡ ⎤ 130 140 −200 400 −340 ⎦, and pAT A (x) = x3 −900x2 +72900x = x(x−810)(x−90). (c) First, AT A = ⎣ 140 −200 −340 370 √ √ √ √ Hence, λ1 = 810, λ2 = 90, and λ3 = 0, implying σ 1 = 810 = 9 10, σ 2 = 90 = 3 10, ⎡ ⎤ 1 0 12 ⎢ ⎥ and σ 3 = 0. Row reducing (810I3 − AT A) produces ⎣ 0 1 1 ⎦, giving the fundamen0 0 0 ⎡ ⎤ 1 0 2 tal eigenvector [−1, −2, 2]. Similarly, row reducing (90I3 − AT A) produces ⎣ 0 1 −2 ⎦, 0 0 0 giving the fundamental eigenvector [−2, 2, 1]. Finally, row reducing (0I3 − AT A) produces ⎤ ⎡ 1 0 −1 ⎢ ⎥ ⎢ 0 1 − 1 ⎥, giving the fundamental eigenvector [2, 1, 2]. Normalizing these vectors yields 2 ⎦ ⎣ 0 0 0 ,1 the set 3 [−1, −2, 2], 13 [−2, 2, 1], 13 [2, 1, 2] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = σ11 Av1 and u2 = σ12 Av2 , where v1 and v2 are the first two right singular vectors. (Since Av3 = 0, we do not use v3 to compute a left singular vector.) ⎛ ⎡ ⎤⎞ −1 7 20 −17 ⎝ 1 ⎣ −81 −3 1 ⎦⎠ = √ √1 −2 Now, u1 = σ11 Av1 = 9√110 = . 3 27 10 10 −9 0 9 27 1 2 ⎤⎞ ⎛ ⎡ −2 7 20 −17 9 1 1 1 1 ⎣ 1 1 ⎦ ⎠ ⎝ √ √ √ 2 = 9 10 Similarly, u2 = σ2 Av2 = 3 10 = 10 . 3 −9 0 9 27 3 1

283


Section 9.5

Since we have two left singular vectors, and that is all we need for a basis for R2 , the set {u1 , u2 } of left singular vectors is complete. Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣVT with U = ⎡ ⎤ √ −1 −2 2 −3 1 9 10 √ 0 0 , and V = 1 ⎣ −2 √1 2 1 ⎦. A short computation ,Σ= 3 10 1 3 0 3 10 0 2 1 2 ⎛ ⎡ ⎤⎞

√ −1 −2 2 10 0 0 −3 1 9 7 20 −17 1 1 ⎣ ⎦ ⎠ ⎝ √ √ −2 2 1 verifies that = 3 10 1 3 −9 0 9 0 3 10 0 2 1 2 = A. 5 3 and pAT A (x) = x2 −10x+16 = (x−8)(x−2). Hence, λ1 = 8 and λ2 = 2, (f) First, AT A = 3 5 √ √ √ 1 −1 implying σ 1 = 8 = 2 2 and σ 2 = 2. Row reducing (8I2 − AT A) produces , giving 0 0 1 1 the fundamental eigenvector [1, 1]. Similarly, row reducing (2I2 −AT A) produces , giving 0 0 3 4 the fundamental eigenvector [−1, 1]. Normalizing these vectors yields the set √12 [1, 1], √12 [−1, 1] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = σ11 Av1 and u2 = σ12 Av2 , where v1 and v2 are the two right singular vectors. ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤

10 14 24 6 1 1 ⎝1 ⎣ 1 ⎣ ⎦⎠ √1 ⎦ = 1 ⎣ 3 ⎦. 12 0 12 Now, u1 = σ11 Av1 = 2√ = 7 28 7 2 2 1 1 7 8 2 ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤

10 14 4 2 −1 1 ⎣ −12 ⎦ = 17 ⎣ −6 ⎦. Similarly, u2 = σ12 Av2 = √12 ⎝ 17 ⎣ 12 0 ⎦⎠ √12 = 14 1 1 7 6 3 Since we have only two left singular vectors, we need one more for a basis for R3 . Thus, we need to expand the set {u1 , u2 } to an orthonormal basis. Row reducing the matrix whose columns are u1 , u2 , e1 , e2 , e3 shows, by the Independence Test Method, that the set {u1 , u2 , e1 } is linearly independent. Performing the Gram-Schmidt Process 1 [9, −6, −18], which on this set has no effect on u1 and u2 but does produce a new third vector 49 1 normalizes to u3 = 7 [3, −2, −6]. This completes the set {u1 , u2 , u3 } of left singular vectors. Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣVT ⎡ ⎤ ⎡ √ ⎤ 6 2 3 2 2 √0 1 −1 with U = 17 ⎣ 3 −6 −2 ⎦, Σ = ⎣ . A short computation 0 2 ⎦, and V = √12 1 1 2 3 −6 0 0 ⎤⎞ ⎡ √ ⎡ ⎤ ⎤ ⎛ ⎡ 6 2 3 2 2 √0 10 14 1 1 = 17 ⎣ 12 0 ⎦ = A. verifies that ⎝ 17 ⎣ 3 −6 −2 ⎦⎠ ⎣ 0 2 ⎦ √12 −1 1 2 3 −6 1 7 0 0 169 22 (2) (a) First, AT A = and pAT A (x) = x2 − 305x + 22500 = (x − 180)(x − 125). Hence, 22 136 284


Section 9.5

√ √ √ √ λ1 = 180 and λ2 = 125, implying σ 1 = 180 = 6 5 and σ 2 = 125 = 5 5. Row reducing 1 −2 T , giving the fundamental eigenvector [2, 1]. Similarly, row (180I2 − A A) produces 0 0 ' ( 1 12 T reducing (125I2 − A A) produces , giving the fundamental eigenvector [−1, 2]. Nor0 0 4 3 malizing these vectors yields the set √15 [2, 1], √15 [−1, 2] of right singular vectors for A.

Next, we find the left singular vectors for A by computing u1 = σ11 Av1 and u2 = σ12 Av2 , where v1 and v2 are the two right singular vectors. ⎤ ⎤ ⎛ ⎡ ⎤⎞ ⎡ ⎡

94 −128 60 2 2 1 ⎝ 1 ⎣ 1 ⎣ 1 ⎣ 95 110 ⎦⎠ √15 300 ⎦ = 15 10 ⎦. Now, u1 = σ11 Av1 = 6√ = 450 15 5 1 330 11 142 46 ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤

94 −128 −350 −14 −1 1 ⎝ 1 ⎣ 1 ⎣ 1 ⎣ 95 110 ⎦⎠ √15 125 ⎦ = 15 5 ⎦. Similarly, u2 = σ12 Av2 = 5√ = 375 15 5 2 142 46 −50 −2

Since we have only two left singular vectors, we need one more for a basis for R3 . Thus, we need to expand the set {u1 , u2 } to an orthonormal basis. Row reducing the matrix whose columns are u1 , u2 , e1 , e2 , e3 shows, by the Independence Test Method, that the set {u1 , u2 , e1 } is linearly independent. Performing the Gram-Schmidt Process on this set has no effect on u1 and u2 but does produce a new third vector 19 [1, 2, −2], which 1 [5, 10, −10]. This completes the set {u1 , u2 , u3 } of left singular normalizes to u3 = 13 [1, 2, −2] = 15 vectors. Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣVT ⎡ ⎤ ⎡ √ ⎤ 0 2 −14 5 6 5 √ 2 −1 1 ⎣ 1 ⎦ ⎣ ⎦ 5 10 , Σ = with U = 15 10 . The matrix 0 5 5 , and V = √5 1 2 11 −2 −10 0 0 ⎤ ⎡ 1 √ 0 0 6 5 ⎦ , and so A+ = VΣ+ UT Σ+ = ⎣ 1 0 5√ 0 5 ⎤⎞ 2 10 11 2 −1 104 70 122 ⎦ ⎝ 1 ⎣ −14 5 ⎦⎠ = 1 ⎣ −2 = √15 15 2250 1 1 2 −158 110 31 √ 0 5 5 0 5 10 −10 ⎡ 3 104 70 122 1 ⎣ 27 A least-squares solution to Ax = b is given by v = A+ b = 2250 −158 110 31 28 5618 2.4968 6823 454.86 1 1 = 2250 . Note that both AT Av and AT b equal 15 = = 3364 3874 1.4951 258.26 ⎡ ⎤ 229 −108 6 1 ⎣ −108 349 114 ⎦ and pAT A (x) = x3 − 14x2 + 49x − 36 (c) First, AT A = 49 6 114 108

⎡

1 √ 6 5

0

0

⎤⎛

⎡

285

. ⎤ ⎦ .


Section 9.5

√ = (x − 9)(x − 4)(x − 1). Hence, λ1 = 9, λ2 = 4, and λ3 = 1, implying σ 1 = 9 = ⎡ 3 1 0 2 √ √ ⎢ T σ 2 = 4 = 2, and σ 3 = 1 = 1. Row reducing (9I3 − A A) produces ⎣ 0 1 −3 0 0 0

3, ⎤ ⎥ ⎦,

giving the fundamental eigenvector [−3, 6, 2]. Similarly, row reducing (4I3 − AT A) produces ⎡ ⎤ 1 0 −2 ⎢ ⎥ ⎢ 0 1 − 2 ⎥, giving the fundamental eigenvector [6, 2, 3]. Finally, row reducing (1I3 − AT A) 3 ⎦ ⎣ 0 0 0 ⎤ ⎡ 1 0 13 ⎥ ⎢ 1 ⎥ produces ⎢ ⎣ 0 1 2 ⎦ , giving the fundamental eigenvector [−2, −3, 6]. Normalizing these vec0 0 0 , tors yields the set 17 [−3, 6, 2], 17 [6, 2, 3], 17 [−2, −3, 6] of right singular vectors for A. Next, we find the left singular vectors for A by computing u1 = σ11 Av1 , u2 = σ12 Av2 , and u3 = σ13 Av3 , where v1 , v2 , and v3 are the three right singular vectors. ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎛ ⎡ ⎤⎞ 23 −11 −6 −147 −1 −3 ⎜1 ⎢ 5 ⎟ ⎢ ⎥ ⎢ ⎥ 25 6 ⎥ ⎢ ⎥⎟ ⎝ 1 ⎣ 6 ⎦⎠ = 1 ⎢ 147 ⎥ = 1 ⎢ 1 ⎥. Now, u1 = σ11 Av1 = 13 ⎜ 14 7 294 2 ⎝ ⎣ 19 −17 ⎣ −147 ⎦ ⎣ −1 ⎦ 6 ⎦⎠ 2 1 19 18 147 1 ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎛ ⎡ ⎤⎞ 23 −11 −6 98 1 6 ⎜ ⎢ ⎟ 1 ⎢ 98 ⎥ ⎢ 1 ⎥ 25 6 ⎥ 1 1 ⎜ 1 ⎢ 5 1 1 ⎥⎟ ⎝ ⎣ 2 ⎦⎠ = ⎢ ⎥ ⎢ ⎥ Similarly, u2 = σ2 Av2 = 2 ⎝ 14 ⎣ 196 ⎣ 98 ⎦ = 2 ⎣ 1 ⎦. 19 −17 6 ⎦⎠ 7 3 1 19 18 98 1 ⎛ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎛ ⎡ ⎤⎞ 23 −11 −6 −49 −1 −2 ⎜1 ⎢ 5 ⎟ ⎢ ⎥ ⎢ ⎥ 25 6 ⎥ ⎢ ⎥⎟ ⎝ 1 ⎣ −3 ⎦⎠ = 1 ⎢ −49 ⎥ = 1 ⎢ −1 ⎥. Also, u3 = σ13 Av3 = 11 ⎜ 98 ⎣ 2 ⎣ ⎝ 14 ⎣ 19 −17 6 ⎦⎠ 7 49 ⎦ 1 ⎦ 6 1 19 18 49 1 Since we have only three left singular vectors, we need one more for a basis for R4 . Thus, we need to expand the set {u1 , u2 , u3 } to an orthonormal basis. Row reducing the matrix whose columns are u1 , u2 , u3 , e1 , e2 , e3 , e4 shows, by the Independence Test Method, that the set {u1 , u2 , u3 , e1 } is linearly independent. Performing the GramSchmidt Process on this set has no effect on u1 , u2 , and u3 but does produce a new fourth vector 1 1 4 [1, −1, −1, 1], which normalizes to u4 = 2 [1, −1, −1, 1]. This completes the set {u1 , u2 , u3 , u4 } of left singular vectors. Therefore, using the right singular vectors as columns for V, the left singular vectors as columns for U, and the singular values as diagonal entries for Σ, we can express A as UΣVT ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 3 0 0 −1 1 −1 1 −3 6 −2 ⎥ ⎢ ⎢ ⎥ 0 2 0 1 1 −1 −1 1 ⎣ ⎥ ⎥, Σ = ⎢ 6 2 −3 ⎦. The matrix with U = 12 ⎢ ⎣ 0 0 1 ⎦, and V = 7 ⎣ −1 1 1 −1 ⎦ 2 3 6 0 0 0 1 1 1 1

286


1 3

⎢ Σ+ = ⎢ ⎣ 0 0 ⎛ ⎡ −3 = ⎝ 17 ⎣ 6 2 ⎡ 36 1 ⎣ 12 = 84 −31

0

0

1 2

0

0

Section 9.5

⎤

⎥ + T + 0 ⎥ ⎦, and so A = VΣ U 0 1 0 ⎡ ⎤⎛ ⎡ ⎤⎞ 1 0 0 0 −1 3 6 −2 ⎢ ⎥ ⎜1 ⎢ 1 1 ⎢ ⎢ ⎥ ⎜ 2 −3 ⎦⎠ ⎣ 0 2 0 0 ⎦ ⎝ 2 ⎣ −1 3 6 1 0 0 1 0 ⎤ 24 12 0 36 −24 0 ⎦. −23 41 49

1 1 −1 −1

−1 1 1 −1

⎤⎞ 1 ⎟ 1 ⎥ ⎥⎟ ⎦ 1 ⎠ 1

⎡ ⎤ ⎤ 3 36 24 12 0 ⎢ ⎥ 2 ⎥ 1 ⎣ 12 36 −24 0 ⎦ ⎢ A least-squares solution to Ax = b is given by v = A+ b = 84 ⎣ 9 ⎦ −31 −23 41 49 4 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 264 44 127 1 ⎣ 1 ⎣ −108 ⎦ = 14 −18 ⎦. Note that both AT Av and AT b equal 17 ⎣ −30 ⎦. = 84 71 60 426 1 1 1 1 (3) (a) Using the results from the answers for part (a) of Exercise 1, u1 = √2 , u2 = √2 , −1 1 √ √ 2 −1 v1 = √15 , v2 = √15 , σ 1 = 2 10, and σ 2 = 10. Hence, A = σ 1 u1 v1T + σ 2 u2 v2T 1 2

√ √

1 1 1 1 1 √ √ √1 2 1 −1 2 = 2 10 √2 10 + . 5 2 5 1 −1 ⎡ ⎤ ⎡ ⎤ 6 2 (c) Using the results from the answers for part (f) of Exercise 1, u1 = 17 ⎣ 3 ⎦, u2 = 17 ⎣ −6 ⎦, 2 3 √ √ 1 −1 1 1 T v1 = √2 , v2 = √2 , σ 1 = 2 2, and σ 2 = 2. Hence, A = σ 1 u1 v1 + σ 2 u2 v2T 1 1 ⎤⎞ ⎛ ⎡ ⎤⎞ ⎛ ⎡ 6 2

√ √

+ 2 ⎝ 17 ⎣ −6 ⎦⎠ √12 −1 1 . = 2 2 ⎝ 17 ⎣ 3 ⎦⎠ √12 1 1 2 3 ⎡

(8) (a) The ith column of V is the right singular vector vi , which is a unit eigenvector corresponding to the eigenvalue λi of AT A. But −vi is also a unit eigenvector corresponding to the eigenvalue λi of AT A. Thus, if any vector vi is replaced with its opposite vector, the set {v1 , . . . , vn } is still an orthonormal basis for Rn consisting of eigenvectors for AT A. Since the vectors are kept in the same order, the λi do not increase, and thus {v1 , . . . , vn } fulfills all the necessary conditions to be a set of right singular vectors for A. For i ≤ k, the left singular vector ui = σ1i Avi , so when we change the sign of vi , we must adjust U by changing the sign of ui as well. For i > k, changing the sign of vi has no effect on U but still produces a valid singular value decomposition. (b) If the eigenspace Eλ for AT A has dimension higher than 1, then the corresponding right singular vectors can be replaced by any orthonormal basis for Eλ , for which there is an infinite number of choices. Then the associated left singular vectors u1 , . . . , uk must be adjusted accordingly.

287


Section 9.5

(9) (a) Each right singular vector vi , for 1 ≤ i ≤ k, must be an eigenvector for AT A. Performing the Gram-Schmidt Process on the rows of A, eliminating zero vectors, and normalizing will produce an orthonormal basis for the row space of A, but there is no guarantee that it will consist of 1 1 T , performing the Gram-Schmidt Process on eigenvectors for A A. For example, if A = 0 1 the rows of A produces the two vectors [1, 1] and − 12 , 12 , neither of which is an eigenvector for 1 1 . AT A = 1 2 (b) The right singular vectors vk+1 , . . . , vn are an orthonormal basis for the eigenspace E0 of AT A. Any orthonormal basis for E0 will do. By part (5) of Theorem 9.5, E0 equals the kernel of the linear transformation L whose matrix with respect to the standard bases is A. A basis for ker(L) can be found by using the Kernel Method. That basis can be turned into an orthonormal basis for ker(L) by applying the Gram-Schmidt process and normalizing. (10) If A = UΣVT , as given in the exercise, then A+ = VΣ+ UT , and so A+ A = VΣ+ UT UΣVT = VΣ+ ΣVT . Note that Σ+ Σ is an n × n diagonal matrix whose first k diagonal entries equal 1, with the remaining diagonal entries equal to 0. Note also that since the columns v1 , . . . , vn of V are orthonormal, VT vi = ei , for 1 ≤ i ≤ n. (a) If 1 ≤ i ≤ k, then A+ Avi = VΣ+ ΣVT vi = VΣ+ Σei = Vei = vi . (b) If i > k, then A+ Avi = VΣ+ ΣVT vi = VΣ+ Σei = V (0) = 0. (11) If A = UΣVT , as given in the exercise, then A+ = VΣ+ UT . Note that since the columns u1 , . . . , um of U are orthonormal, UT ui = ei , for 1 ≤ i ≤ m.

(a) If 1 ≤ i ≤ k, then A+ ui = VΣ+ UT ui = VΣ+ ei = V σ1i ei = σ1i vi . Thus, AA+ ui = A (A+ ui )

= A σ1i vi = σ1i Avi = ui . (b) If i > k, then A+ ui = VΣ+ UT ui = VΣ+ ei = V (0) = 0. Thus, AA+ ui = A (A+ ui ) = A (0) = 0. ⎛ ⎡ ⎤⎞ 1 ⎜ ⎢ 0 ⎥⎟ ⎜ 1 ⎢ ⎥⎟ 1 ⎢ ⎥⎟ (14) (a) A = σ 1 u1 v1T + σ 2 u2 v2T + σ 3 u3 v3T + σ 4 u4 v4T + σ 5 u5 v5T = 150 ⎜ ⎜ 3 ⎢ 2 ⎥⎟ 2 [1, 0, 1, −1, 0, −1] + ⎝ ⎣ 0 ⎦⎠ 2 ⎤⎞ ⎛ ⎡ ⎤⎞ ⎛ ⎡ 2 2 ⎜ ⎢ 0 ⎥⎟ ⎜ ⎢ 0 ⎥⎟ ⎥⎟ ⎜ ⎢ ⎜1 ⎢ ⎥⎟ ⎢ 1 ⎥⎟ 1 [1, 0, −1, 1, 0, −1] + 15 ⎜ 1 ⎢ −2 ⎥⎟ 1 [1, −1, 0, 0, 1, 1] + 30 ⎜ ⎜3 ⎢ ⎥⎟ 2 ⎜3 ⎢ ⎥⎟ 2 ⎝ ⎣ 0 ⎦⎠ ⎝ ⎣ 0 ⎦⎠ −2 1 ⎤⎞ ⎛ ⎡ ⎛ ⎡ ⎤⎞ 0 0 ⎜ ⎢ 3 ⎥⎟ ⎜ ⎢ 4 ⎥⎟ ⎥⎟ 1 ⎜ 1 ⎢ ⎥⎟ 1 ⎜1 ⎢ ⎢ ⎥⎟ ⎜ ⎢ ⎥⎟ 6⎜ ⎜ 5 ⎢ 0 ⎥⎟ 2 [1, 1, 0, 0, −1, 1] + 3 ⎜ 5 ⎢ 0 ⎥⎟ 2 [0, 1, 1, 1, 1, 0] ⎝ ⎣ −4 ⎦⎠ ⎝ ⎣ 3 ⎦⎠ 0 0

288


Section 9.5

⎤ ⎡ ⎤ 0 −25 10 0 −10 10 0 −10 ⎢ 0 0 ⎥ 0 0 0 0 0 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 0 −50 ⎥ 5 0 −5 5 0 −5 + ⎥ ⎢ ⎥ 0 0 ⎦ ⎣ 0 0 0 0 0 0 ⎦ 0 −50 −10 0 10 −10 0 10 ⎡ ⎤ ⎤ 10 −10 0 0 10 10 0 0 0 0 0 0 ⎢ ⎢ 0 0 0 0 0 0 ⎥ 9 9 0 0 −9 9 ⎥ ⎢ ⎢ ⎥ ⎥ 1 ⎢ 1 ⎢ ⎥ 10 0 0 −10 −10 ⎥ + 5 ⎢ 0 0 0 0 0 0 ⎥ + 2 ⎢ −10 ⎥ ⎣ ⎣ −12 −12 0 0 12 −12 ⎦ 0 0 0 0 0 0 ⎦ 5 −5 0 0 5 5 0 0 0 0 0 0 ⎤ ⎤ ⎡ ⎡ 40 −5 15 −15 5 −30 0 0 0 0 0 0 ⎢ 0 12 12 12 12 0 ⎥ ⎢ 1.8 3 1.2 1.2 −0.6 1.8 ⎥ ⎥ ⎥ ⎢ ⎢ 1 ⎢ ⎢ ⎥ 50 5 45 −45 −5 −60 ⎥ 0 0 0 0 ⎥=⎢ + 10 ⎢ 0 0 ⎥ ⎣ 0 9 0.9 3.3 −2.4 ⎦ 9 9 9 0 ⎦ ⎣ −2.4 −1.5 0.9 42.5 −2.5 60 −60 2.5 −37.5 0 0 0 0 0 0 ⎛ ⎡ ⎤⎞ ⎡ ⎤ 1 25 0 25 −25 0 −25 ⎜ ⎢ 0 ⎥⎟ ⎢ 0 0 0 ⎥ ⎜ 1 ⎢ ⎥⎟ 1 ⎥ ⎢ 0 0 0 T ⎢ ⎜ ⎢ ⎥ ⎟ (b) A1 = σ 1 u1 v1 = 150 ⎜ 3 ⎢ 2 ⎥⎟ 2 [1, 0, 1, −1, 0, −1] = ⎢ 50 0 50 −50 0 −50 ⎥ ⎥; ⎝ ⎣ 0 ⎦⎠ ⎣ 0 0 0 0 0 0 ⎦ 2 50 0 50 −50 0 −50 ⎤⎞ ⎛ ⎡ 2 ⎜ ⎢ 0 ⎥⎟ ⎥⎟ 1 ⎜1 ⎢ T T T ⎢ ⎥⎟ A2 = σ 1 u1 v1 + σ 2 u2 v2 = A1 + σ 2 u2 v2 = A1 + 30 ⎜ ⎜ 3 ⎢ 1 ⎥⎟ 2 [1, 0, −1, 1, 0, −1] ⎝ ⎣ 0 ⎦⎠ −2 ⎡ ⎤ ⎡ ⎤ 25 0 25 −25 0 −25 10 0 −10 10 0 −10 ⎢ 0 0 ⎢ 0 0 0 0 ⎥ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥+⎢ ⎥ 50 0 50 −50 0 −50 5 0 −5 5 0 −5 =⎢ ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 0 0 0 0 ⎦ ⎣ 0 0 0 0 0 0 ⎦ 50 0 50 −50 0 −50 −10 0 10 −10 0 10 ⎡ ⎤ 35 0 15 −15 0 −35 ⎢ 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎥; 55 0 45 −45 0 −55 =⎢ ⎢ ⎥ ⎣ 0 0 0 0 0 0 ⎦ 40 0 60 −60 0 −40 ⎤⎞ ⎛ ⎡ 2 ⎜ ⎢ 0 ⎥⎟ ⎥⎟ 1 ⎜1 ⎢ ⎢ ⎥⎟ A3 = σ 1 u1 v1T + σ 2 u2 v2T + σ 3 u3 v3T = A2 + σ 3 u3 v3T = A2 + 15 ⎜ ⎜ 3 ⎢ −2 ⎥⎟ 2 [1, −1, 0, 0, 1, 1] ⎝ ⎣ 0 ⎦⎠ 1 25 ⎢ 0 ⎢ =⎢ ⎢ 50 ⎣ 0 50 ⎡

0 25 −25 0 0 0 0 50 −50 0 0 0 0 50 −50

289

Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition ⎡ ⎢ ⎢ =⎢ ⎢ ⎣ ⎡ ⎢ ⎢ =⎢ ⎢ ⎣

35 0 55 0 40

0 15 −15 0 −35 0 0 0 0 0 0 45 −45 0 −55 0 0 0 0 0 0 60 −60 0 −40

⎡

⎤ ⎥ ⎥ ⎥+ ⎥ ⎦

1 2

⎢ ⎢ ⎢ ⎢ ⎣

40 −5 15 −15 5 −30 0 0 0 0 0 0 50 5 45 −45 −5 −60 0 0 0 0 0 0 42.5 −2.5 60 −60 2.5 −37.5

10 0 −10 0 5 ⎤

−10 0 0 10 0 0 0 0 10 0 0 −10 0 0 0 0 −5 0 0 5

10 0 −10 0 5

Section 9.5 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

⎥ ⎥ ⎥; ⎥ ⎦

A4 = σ 1 u1 v1T + σ 2 u2 v2T + σ 3 u3 v3T + σ 4 u4 v4T = A3 + σ 4 u4 v4T ⎤⎞ ⎛ ⎡ 0 ⎜ ⎢ 3 ⎥⎟ ⎥⎟ 1 ⎜1 ⎢ ⎢ ⎥⎟ = A3 + 6 ⎜ ⎜ 5 ⎢ 0 ⎥⎟ 2 [1, 1, 0, 0, −1, 1] ⎝ ⎣ −4 ⎦⎠ 0 ⎡ ⎡ ⎤ 0 0 40 −5 15 −15 5 −30 ⎢ ⎢ ⎥ 9 9 0 0 0 0 0 0 ⎢ ⎢ ⎥ ⎥+ 1⎢ 0 0 50 5 45 −45 −5 −60 =⎢ 5 ⎢ ⎢ ⎥ ⎣ −12 −12 ⎣ 0 0 0 0 0 0 ⎦ 0 0 42.5 −2.5 60 −60 2.5 −37.5 ⎡ ⎤ 40 −5 15 −15 5 −30 ⎢ 1.8 1.8 0 0 −1.8 1.8 ⎥ ⎥ ⎢ ⎥ 50 5 45 −45 −5 −60 =⎢ ⎥ ⎢ ⎣ −2.4 −2.4 0 0 2.4 −2.4 ⎦ 42.5 −2.5 60 −60 2.5 −37.5

0 0 0 0 0

0 0 0 0 0

0 −9 0 12 0

0 9 0 −12 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

(c) By Exercise 13, the singular values of Ai are σ 1 , . . . , σ i , and the singular values of A − Ai are σ i+1 , . . . , σ 5 . √ √ σ 21 + · · · + σ 25 = 1502 + 302 + 152 + 62 + 32 = 23670 Now, by Exercise 12, N (A) = √ √ ≈ 153.85. Similarly, N (A − A1 ) = σ 22 + · · · + σ 25 = 302 + 152 + 62 + 32 = 1170 ≈ 34.21, √ √ N (A − A2 ) = σ 23 + · · · + σ 25 = 152 + 62 + 32 = 270 ≈ 16.43, N (A − A3 ) = σ 24 + σ 25 √ √ √ √ = 62 + 32 = 45 ≈ 6.71, and N (A − A4 ) = σ 25 = 32 = 9 = 3. Hence, N (A − A1 )/N (A) ≈ 34.21/153.85 ≈ 0.2223, N (A − A2 )/N (A) ≈ 16.43/153.85 ≈ 0.1068, N (A − A3 )/N (A) ≈ 6.71/153.85 ≈ 0.0436, and N (A − A4 )/N (A) ≈ 3/153.85 ≈ 0.0195. Of course, instead of using Exercises 12 and 13 to help compute these values, we could have computed each N (A − Ai ) directly by finding each matrix A − Ai (using our answers from part (b)), summing the squares of its entries, and then taking the square root. We would, of course, have obtained the same results. (d) The method described in the text for the compression of digital images takes the matrix describing the image and alters it by zeroing out some of the lower singular values. This exercise illustrates how the matrices Ai that use only the first i singular values for a matrix A get closer to approximating A as i increases. The matrix for a digital image is, of course, much larger than the 5 × 6 matrix considered in this exercise. Also, you can frequently get a very good approximation of the image using a small enough number of singular values so that less data needs to be saved. When 290


Section 9.5

we use the outer product form of the singular value decomposition, only the singular values and the relevant singular vectors are needed to construct each Ai , so not all mn entries of Ai need to be kept in storage. (15) These are the steps to process a digital image in MATLAB, as described in the textbook: – Enter the command : edit – Use the text editor to enter the following MATLAB program: function totalmat = RevisedPic(U, S, V, k) T = V’ totalmat = 0; for i = 1:k totalmat = totalmat + U(:,i)*T(i,:)*S(i,i); end – Save this program under the name RevisedPic.m – Enter the command: A=imread(’picturefilename’); where picturefilename is the name of the file containing the picture, including its file extension. (preferably .tif or .jpg) – Enter the command: ndims(A) – If the response is “3”, then MATLAB is treating your picture as if it is in color, even if it appears to be black-and-white. If this is the case, enter the command: B=A(:,:,1); (to use only the first color of the three used for color pictures) followed by the command: C=double(B); (to convert the integer format to decimal) Otherwise, just enter: C=double(A); – Enter the command: [U,S,V]=svd(C); (This computes the singular value decomposition of C.) – Enter the command: W=RevisedPic(U,S,V,100); (The number “100” is the number of singular values you are using. You may change this to any value you like.) – Enter the command: R=uint8(round(W)); (This converts the decimal output to the correct integer format.) – Enter the command: imwrite(R,’Revised100.tif’,’tif’) (This will write the revised picture out to a file named “Revised100.tif”. Of course, you can use any file name you would like, or use the .jpg extension instead of .tif.) – You may repeat the steps: W=RevisedPic(U,S,V,k); R=uint8(round(W)); imwrite(R,’Revisedk.tif’,’tif’) with different values for k to see how the picture gets more refined as more singular values are used. 291


Section 9.5

– Output can be viewed using any appropriate software you may have on your computer for viewing such files. (16) (a) False. First, if A is an m × n matrix, AAT is m × m and AT A is n × n, so they cannot be equal T might not equal AT A even for square matrices. We can check that they if m = n. Second, AA 1 2 are not equal for A = . 3 4 (b) True. The singular values of A are the nonnegative square roots of the eigenvalues of AT A. These eigenvalues were proven to be nonnegative in the text at the beginning of this section. 1 2 (c) False. For a counterexample, consider A = , v = i, and w = j. Clearly, v · w = 0, 3 4 but (Av) · (Aw) = [1, 3] · [2, 4] = 14 = 0. You might get confused in this problem by part (2) of Lemma 9.4, which states that (Avi )⊥(Avj ), but that applies only to right singular vectors vi and vj (when i = j). (d) False. Only the first k left singular vectors are determined by the right singular vectors, where k = rank(A). The remaining left singular vectors can be chosen in any way so that {u1 , . . . , um } is an orthonormal basis for Rn . There could be an infinite number of ways to make the choices for uk+1 , . . . , um , depending on the values of k and m. (See Exercise 8.) (e) False. By part (5) of Theorem 9.5, {vk+1 , . . . , vn } forms an orthonormal basis for ker(L), not for (ker(L))⊥ . (f) True. We established this method for proving that matrices are equal at the beginning of the proof of Theorem 9.6. (g) False. See Exercises 4 and 8. However, the matrix Σ in the singular value decomposition of a matrix A is uniquely determined by A. (h) True. The product VΣT UT certainly equals AT and has the correct form for a singular value decomposition. Exercise 5 can then be used to show that this implies that it is a singular value decomposition for AT . (i) False. Because every matrix has a singular value decomposition, every matrix also has a pseudoinverse, which is computed using the singular value decomposition. (j) True. In part (c) of Exercise 10, we prove that A+ A is the matrix for the orthogonal projection onto (ker(L))⊥ with respect to the standard basis for Rn . But, for a nonsingular n × n matrix A, (ker(L))⊥ = Rn . Thus, A+ A = In , and so A+ = A−1 . A different approach to proving this can be made using Theorem 9.8 and the fact that for a nonsingular matrix A, Ax = b has the unique actual solution A−1 b (part (1) of Theorem 2.15), which must equal the least-squares solution. Hence, A+ b can be shown to equal A−1 b for every vector b, proving that A+ = A−1 . (k) True. Only the first k right singular vectors are used, corresponding to the k nonzero singular values. The remaining n − k right singular vectors are irrelevant, since they all correspond to σ k+1 = · · · = σ n = 0.

292


Appendix B

Appendices Appendix B (1) (a) f is not a function because it is not defined for every element of its stated domain. In particular, f is not defined for x < 1. (c) h is not a function because two values are assigned to each x = 1. (e) k is not a function because it is not defined for every element of its stated domain. In particular, k is not defined at θ = π2 n, where n is an odd integer. (f) l is a function. Its range is the set of all prime numbers. The range is contained in the set of prime numbers because the definition of l defines an image under l to be a prime number. The range contains every prime number because, if p is prime, l(p) = p, and so p is in the range. The image of 2 = l(2) = 2. The pre-image of 2 = {0, 1, 2}, the set of all natural numbers less than or equal to 2. (2) (a) The pre-image of {10, 20, 30} is the set of all integers whose absolute value is 5, 10, or 15, since 2 times the absolute value must be 10, 20, or 30. This is the set {−15, −10, −5, 5, 10, 15}. (c) The pre-image of the set of multiples of 4 is the set of all integers which, when doubled, have absolute values that are multiples of 4. This is the set of integers whose absolute value is a multiple of 2; that is, the set {. . . , −8, −6, −4, −2, 0, 2, 4, 6, 8, . . .}. √ 5x−1 2 3 1 2 2 = 3 + 2 = (3) (g ◦ f )(x) = g(f (x)) = g 5x−1 4 4 16 (25x − 10x + 1) + 2 = 4 75x − 30x + 35. √ √ (f ◦ g)(x) = f (g(x)) = f 3x2 + 2 = 14 (5 3x2 + 2 − 1).

x x 3 −2 x −4 4 3 −2 x (4) (g ◦ f ) = g f = g = = y y 1 4 y 0 2 1 4 y

−4 4 3 −2 x −8 24 x = . 0 2 1 4 y 2 8 y

x x −4 4 x 3 −2 −4 4 x (f ◦ g) = f g = f = = y y 0 2 y 1 4 0 2 y

3 −2 −4 4 x −12 8 x = . 1 4 0 2 y −4 12 y (8) The function f is not one-to-one because f (x + 1) = f (x + 3) = 1, which shows that two different domain elements have the same image. The function f is not onto because there is no pre-image for xn . For n ≥ 3, the pre-image of P2 is P3 . (10) The function f is one-to-one because f (A1 ) = f (A2 ) ⇒ B−1 A1 B = B−1 A2 B −1 ⇒ B(B A1 B)B−1 = B(B−1 A2 B)B−1 ⇒ (BB−1 )A1 (BB−1 ) = (BB−1 )A2 (BB−1 ) ⇒ In A 1 I n = In A 2 I n ⇒ A1 = A2 . 293


Appendix B

The function f is onto because for any C ∈ Mnn , f (BCB−1 ) = B−1 (BCB−1 )B = C. Also, f −1 (A) = BAB−1 . (12) (a) False. Functions may have two different elements of the domain mapped to the same element of the codomain. For example, f : R → R given by f (x) = x2 is such a function, since f maps both 2 and −2 to 4. (b) True. Since each element of X is assigned to exactly one element of Y , f is a function by definition. However, a function f having this property is not onto, since not every element of Y has a corresponding element of X. (c) False. Since f (5) = f (6), f is not one-to-one. Hence, f −1 does not exist. For example, the constant function f : R → R given by f (x) = 7 satisfies f (5) = f (6), but has no inverse, and so f −1 (5) is not defined. However, for any function f , f −1 ({5}) is defined and represents the set of all pre-images of 5. If f is the constant function f : R → R given by f (x) = 7, then f −1 ({5}) equals the empty set, while if f : R → R is given by f (x) = 5, then f −1 ({5}) = R. (d) False. For f to be onto, the codomain of f must equal the range of f , not the domain of f . For a specific counterexample, consider f : R →R given by f (x) = 0. Then the domain and codomain of f are both R; however, f is not onto because no nonzero element of the codomain R has a pre-image. (e) False. Instead, f is one-to-one if f (x1 ) = f (x2 ) implies x1 = x2 . Notice, by the way, that every function, by definition, has the property that x1 = x2 implies f (x1 ) = f (x2 ). Hence, any function that is not one-to-one (such as f : R → R given by f (x) = 0) provides a specific counterexample to the given statement. (f) False. Consider f : {0, 1} → {0, 1} given by f (x) = x, and g : {0, 1, 2} → {0, 1} given by g(0) = 0, and g(1) = g(2) = 1. Then simple computation shows that (g ◦ f )(x) actually equals f (x). Now, both f and g ◦ f are easily seen to be one-to-one, but g is not one-to-one because g(1) = g(2). (g) False. By Theorem B.2, in order for a function f to have an inverse, it must be both one-toone and onto. Hence, any function that is one-to-one but not onto (for example, f : R → R given by f (x) = arctan(x)) or that is onto but not one-to-one (for example, g : R → R given by g(x) = x3 − x) provides a counterexample. (h) False. Instead, by Theorem B.4, (g ◦ f )−1 = f −1 ◦ g −1 . For a specific counterexample to the statement in the problem, consider f : {0, 1, 2} → {0, 1, 2} given by f (0) = 1, f (1) = 2, and f (2) = 0, and g : {0, 1, 2} → {0, 1, 2} given by g(0) = 1, g(1) = 0, and g(2) = 2. Then, it is easily seen that f −1 : {0, 1, 2} → {0, 1, 2} is given by f −1 (0) = 2, f −1 (1) = 0, and f −1 (2) = 1, and g −1 : {0, 1, 2} → {0, 1, 2} is given by g −1 (0) = 1, g −1 (1) = 0, and g −1 (2) = 2. Also, g ◦ f : {0, 1, 2} → {0, 1, 2} is given by (g ◦ f )(0) = 0, (g ◦ f )(1) = 2, and (g ◦ f )(2) = 1. Hence, (g ◦ f )−1 : {0, 1, 2} → {0, 1, 2} is given by (g ◦ f )−1 (0) = 0, (g ◦ f )−1 (1) = 2, and (g ◦ f )−1 (2) = 1. However, g −1 ◦ f −1 : {0, 1, 2} → {0, 1, 2} is given by (g −1 ◦ f −1 )(0) = 2, (g −1 ◦ f −1 )(1) = 1, and (g −1 ◦ f −1 )(2) = 0. Therefore, (g ◦ f )−1 = g −1 ◦ f −1 .

294


Appendix C

Appendix C (1) (a) (6 − 3i) + (5 + 2i) = (6 + 5) + (−3 + 2)i = 11 − i

(c) 4 (8 − 2i) − (3 + i) = 4 (8 − 3) + (−2 − 1)i = 4(5 − 3i) = 20 − 12i

(e) (5 + 3i)(3 + 2i) = (5)(3) − (3)(2) + (5)(2) + (3)(3) i = 9 + 19i

(g) (7 − i)(−2 − 3i) = (7)(−2) − (−1)(−3) + (7)(−3) + (−1)(−2) i = −17 − 19i (i) 9 − 2i = 9 + 2i

(k) (6 + i)(2 − 4i) = (6)(2) − (1)(−4) + (6)(−4) + (1)(2) i = 16 − 22i = 16 + 22i √ (m) |−2 + 7i| = (−2)2 + (7)2 = 53 (2) In each part we use the formula

1 z

=

z |z|2

2

and the fact that, for z = a + bi, |z| = 2

(a) If z = 6 − 2i, then z = 6 + 2i. Also, |z| = 62 + (−2)2 = 40. Hence,

1 z

=

2

(c) If z = −4 + i, then z = −4 − i. Also, |z| = (−4)2 + (−1)2 = 17. Hence,

√

a2 + b2

6+2i 40

=

1 z

−4−i 17

=

3 20

2

+

= a2 + b2 .

1 20 i.

4 = − 17 −

1 17 i.

(3) In all parts, let z1 = a1 + b1 i and z2 = a2 + b2 i. (a) Part (1): z1 + z2 = (a1 + b1 i) + (a2 + b2 i) = (a1 + a2 ) + (b1 + b2 )i = (a1 + a2 ) − (b1 + b2 )i = (a1 − b1 i) + (a2 − b2 i) = z1 + z2 . Part (2): (z1 z2 ) = (a1 + b1 i)(a2 + b2 i) = (a1 a2 − b1 b2 ) + (a1 b2 + a2 b1 )i = (a1 a2 − b1 b2 ) − (a1 b2 + a2 b1 )i = (a1 a2 − (−b1 )(−b2 )) + (a1 (−b2 ) + a2 (−b1 ))i = (a1 − b1 i)(a2 − b2 i) = z1 z2 . (b) If z1 = 0, then

1 z1

exists. Hence,

1 z1 (z1 z2 )

=

1 z1 (0),

implying z2 = 0.

(c) Part (4): z1 = z1 ⇔ a1 + b1 i = a1 − b1 i ⇔ b1 i = −b1 i ⇔ 2b1 i = 0 ⇔ b1 = 0 ⇔ z1 is real. Part (5): z1 = −z1 ⇔ a1 + b1 i = −(a1 − b1 i) ⇔ a1 + b1 i = −a1 + b1 i ⇔ a1 = −a1 ⇔ 2a1 = 0 ⇔ a1 = 0 ⇔ z1 is pure imaginary. 2

(5) (a) False. Instead, zz = |z| . For example, if z = 1 + i, then z = 1 − i and zz = (1 + i)(1 − i) = 2, √ √ but |z| = 12 + 12 = 2. (b) False. Every real number z (such as z = 3) has the property that z = z. (c) True. A pure imaginary number is of the form z = 0+bi = bi, whose conjugate is z = 0−bi = −bi. (d) True. The additive inverse of z = a + bi is −z = −a − bi. (e) False. The complex number 0 has no multiplicative

inverse. However, every nonzero complex 1 1 number z has the multiplicative inverse z = |z|2 z.

295

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Elementary linear algebra with applications. Solutions

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