Linear Algebra
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Linear Jgebra C Y Hsiung Wuhan University
GYMao Wuhan Universi...
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Linear Algebra
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Linear Jgebra C Y Hsiung Wuhan University
GYMao Wuhan University of Technology
World Scientific Singapore'New Jersey'London 'Hong Kong
Published by World Scientific Publishing Co. Pie. Ltd. P O Box 128, Farrer Road, Singapore 912805 USA office: Suite IB, 1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE
British Library CataloguinginPublication Data A catalogue record for this book is available from the British Library.
LINEAR ALGEBRA Copyright © 1998 by World Scientific Publishing Co. Pie. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.
For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.
ISBN 9810230923
This book is printed on acidfree paper.
Printed in Singapore by U t o Print
PREFACE This book introduces the basic properties and operations of linear algebra as well as its basic theory and concepts. It is based on the first author's lecture notes on Linear Algebra during his teaching in the Mathematics Department of Wuhan University. Some amendments were made and many new ideas were added to the notes before the book was finalized. The book consists of eight chapters. The first six chapters introduce linear equations and matrices, and the last two chapters introduce linear spaces and linear transformations. They are arranged progressively from easy to difficult, simple to complex, and specific to general, which makes it easier for the readers to study on his own. A summary at the beginning of each chapter and each section gives the reader some idea of the topics and aims to be dealt with in the text. Both basic concepts and manipulation skills are equally emphasized, and enough examples are given for their illustration. There are also many exercises at the end of each section, with the answers atta.ched to the end of the book for reference. Basic concepts are specially stressed and great pains have been taken to explain the underlying thoughts and the approach. Furthermore, between chapters and between sections, a brief leader is given to preserve the coherence and continuity of the text. Chapter 1 was written by Professor JianKe Lu, who gave the definition of determinants in a very special way, quite different from those in other text books. This definition is easier for the reader to understand and master, and facilitates the proofs of some of their properties. The author would like to thank BangTeng Xu for his very helpful and meticulous work on the proofs of the other seven chapters.
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CONTENTS
Preface
v
1 Determinants 1.1. Concept of Determinants 1.2. Basic Properties of Determinants 1.3. Development of a Determinant 1.4. Cramer's Theorem
1 . 1 12 20 35
2 Systems of Linear Equations 2.1. Linear Relations between Vectors 2.2. Systems of Homogeneous Linear Equations . . . 2.3. Systems of Fundamental Solutions 2.4. Systems of Nonhomogeneous Linear Equations . 2.5. Elementary Operations
41 42 56 62 70 80
3 Matrix Operations 3.1. Matrix Addition and Matrix Multiplication . . 3.2. Diagonal, Symmetric, and Orthogonal Matrices . 3.3. Invertible Matrices
92 92 112 124
4 Quadratic Forms 4.1. Standard Forms of General Quadratic Forms . . 4.2. Classification of Real Quadratic Forms . . . . *4.3. Bilinear Forms
141 142 155 170
5 Matrices Similar to Diagonal Matrices 5.1. Eigenvalues and Eigenvectors 5.2. Diagonalization of Matrices
173 174 184
vii
viii
5.3. Diagonalization of Real Symmetric Matrices *5.4. Canonical Form of Orthogonal Matrices *5.5. CayleyHamilton Theorem and Minimum Polynomials 6 Jordan Canonical Form of Matrices 6.1. Necessary and Sufficient Condition for Two Matrices to be Similar 6.2. Canonical Form of AMatrices 6.3. Necessary and Sufficient Condition for Two AMatrices to be Equivalent 6.4. Jordan Canonical Forms
. . . .
198 212 217 230 230 236 242 252
7 Linear Spaces and Linear Transformations 273 7.1. Concept of Linear Spaces 273 7.2. Bases and Coordinates 286 7.3. Linear Transformations 304 7.4. Matrix Representation of Linear Transformations 321 *7.5. Linear Transformations from One Linear Space into Another . . 337 *7.6. Dual Spaces and Dualistic Transformations 341 8 Inner Product Spaces 349 8.1. Concept of Inner Product Spaces 349 8.2. Orthonormal Bases 362 8.3. Orthogonal Linear Transformations 373 8.4. Linear Spaces over Complex Numbers with Inner Products . . . 381 *8.5. Normal Operators 392 Answers to Selected Exercises
395
Index
440
CHAPTER 1 DETERMINANTS
In many practical problems, relations among variables may be simply expressed directly or approximately in terms of linear functions so that it is necessary to investigate such functions. Linear algebra is a branch of mathe matics dealing mainly with linear functions, in which systems of linear equations constitute its basic and also important part. In linear algebra, the notion of determinants is fundamental. Theory of determinants is established to satisfy the need for solving systems of linear equations. It has wide appli cations in mathematics, as well as in other scientific branches (for instance, physics, dynamics, etc.). The present chapter will mainly deal with the follow ing three problems: 1. Formulation of the concept of determinants, 2. Derivation of their basic properties and study of the relating calculations, 3. Solving systems of linear equations by using them as a tool.
1.1. C o n c e p t of D e t e r m i n a n t s In high school, we already learned how to solve a system of linear equations in 2 or 3 unknowns by using determinants of order 2 or 3 respectively. Is it possible, in general, to solve a system of linear equations in n unknowns in an analogous way? Determinants of an arbitrary order were introduced for such needs. The aim of the present section is to establish the concept of determinants of order n so as to answer the abovementioned problem 1. We shall first recall results familiar to us in high school. 1
2
Linear Algebra
First of all, let us solve the following system of linear equations in two unknowns x\, X2'. CLnXi +a\2X2
= b\ ,
(1) 021 Xl + 0.22 %2 — &2 j where b\, 62 are constant terms and a ^ ' s (i,j = 1,2) are called the coefficients of Xj. There are two subscripts in a^: the first one, i, and the second one, j , which signify t h a t it is the coefficient of Xj in the ith equation. For example, a\2 is the coefficient of X2 in the first equation. Eliminating X2, we obtain ( a n a 2 2  ai2 021 )xx = 61 022  012 b2 ■ Similarly, eliminating x\, we get ( a n a 2 2  a i 2 a 2 i ) x 2 = au b2  h a 2 i . Therefore, when D — an 022 — ai2 021 ^ 0, we have b\ 022 — 0,2 &2 Xl =
CLu i>2
All 022 — ^12^21
X2 =
au «22
b\ 021
a n 022 — 012021
T h a t is to say, if system (1) is solvable under solution must be (2). By direct verification, we solution of (1). Thus, (2) is the unique solution To facilitate memorization we introduce the an ^21
—
(2)
the proviso D ^ 0, then its assure t h a t (2) is actually the of (1) provided t h a t D ^ 0. notation
a i l «22  012 a 2 i ,
and call it a determinant of order 2. It consists of two (horizontal) rows a n d two (vertical) columns. T h e numbers appearing in a determinant are called its elements. From the above expression, we see t h a t a determinant of order 2 is the algebraic sum of two terms: one is the product of the two elements situated on the principal diagonal of the determinant, i.e., the diagonal from the left upper corner to the right lower corner, and the other is t h a t of t h e two elements situated on its subdiagonal, the other diagonal, with negative sign. For example, '
' = 1  5  (  2 )  3 = 11.
It is easily seen t h a t the two numerators in (2) may be respectively written as, by definition of determinants of order 2,
3
Determinants b\ 022 — ^12 &2
bi
a12
&2
022
anb2

an
ha2i
0.21
61 62
If we denote D = an
ai2
«21
Dx
^22
h
a12
62
a22
Da
an
61
a2i
b2
then the unique solution (2) of system (1) may be written as
Xi =
D
h
ai2
b2 an
«22
a2i
012
D2 X2 = D
a22
an
h
021
an
b2 ai2
^21
022
which may be easily memorized. E x a m p l e 1. Solve the system of linear equations f 2x + y = 7, \ x  Zy =  2 . Solution: D
2 1
In this case, 1 = 7, 3
£>i
7 2
1 3
19,
D2
11
and hence, the unique solution of the given system is £>i
19
D
7 '
y
Do
11
D
7
Let us then solve the following system of linear equations in 3 unknowns: a n %i + aw X2 + 01313 = 6 1 , «21 Z l + 022 X2 + 023 £3 = &2 ,
(3)
^31 X\ + 032 £2 + 033 £3 = 63 .
Analogous to the above, eliminating £3 both from the first two equations and from the last two equations gives rise to a new system of two linear equa tions in xi, X2, the solution of which may be obtained as before, and X3 is
4
Linear Algebra
then obtained by substitution in any of the given equations. By some rather lengthy calculations, we finally get Xl = jj(l>l «22 «33 + a i 2 «23 &3 + »13 &2 «32 — b\ 023 032 — Oi2 &2 «33 ~ a 1 3 ^22 fa) , X2 = 7 y ( a H ^2 «33 + h 023 031 + CI13 a 2 l 63
(4)
— ^11 023 &3  &1 021 133  O13 &2 fl3l) i xz = — ( a n a 2 2 &3 + ^12 &2 «3i + bi 021032 — a n 62 032 — a i 2 021 &3 — °i a 2 2 0 3 1 ) ,
provided that D
a n 022 033 + a i 2 023 031 + a i 3 021032  a n a23 a 3 2  a i 2 a 2 i a 3 3  a i 3 a 2 2 a 3 i ^ 0 .
(4) is actually a solution of system (3) which may be checked by direct substi tution. Therefore, (4) is the unique solution of (3) if D ^ 0. Analogous to the case n = 2, we define a determinant of order 3: an
ai2
an
021
022
023
031
032
A33
= a n 022 033 + ai2 023 031 + ai3 021 032 a n a23 032 — ai2 a2i 033 — ai3 022 031
(5)
which consists of three rows and three columns and is the algebraic sum of 6 terms. The summation may be memorized as follows: As in the following figure, we add up the products of the three elements situated on each solid line with a positive sign and those on each dotted line with a negative sign.
c ,x x
°13
= x. .K °21
°22
°31
°32
, 023 u
33
5
Determinants
For example 2 4
A
1 2 3 1 23 5
= 2  3  5 + l  l  2 + 2 (4) • 3  2 • 3 ■ 1 — 1 • (4) ■ 5 + 2 ■ 3 ■ 2 = 30 + 2  24  6 + 20  12 = 10 . In the expression (4), the common denominator of x\, X2, x% is D
an a2\
ai2
ai3
«22
2), Onl
the number of terms with positive sign and that with negative sign are equal to each other. Proof. Put a.ij = 1 in D, then
£> = ]T ±1 = m  n where m and n are numbers of terms with positive and negative signs respec tively. On the other hand, D = 0 by the abovementioned consequence 1. Hence m = n. Property 4. In a determinant, if we add k times of each element in a row to the corresponding element in another row, then the value of the determinant remains unchanged, i.e., an a,i + kdji «nl
("in T fvOjn
Oi„
Oil
Oln
=
Oil
■
O'in
Onl
•
onu
U * i) ■
This is readily proved by combining Property 2 with the previous conse quence 2. Note that in the above equality the j  t h row of the determinant on the lefthand side remains invariant.
15
Determinants For example, 2 1 2 4 3 1 2 3 5
2 4 + 22 2
2 1 2 0 5 5 2 3 5
1 3 + 12 3
2 1 + 22 5
2 1 2 O i l = 10. 2 3 5
Example 2. a d c d
c—a a— a c a c c—a a—
c 0 0 d b b d c 0 0
Example 3.
a+b a b
0 26 2a = 2 a b+c a 6 b c+ a
c c b+c a b c+ a
b a c 0 = 4a6c. 0 c
By the symmetry of row and column in the definition of a determinant, the above Properties 14 and consequences 1, 2 remain valid for "columns" in place of "rows". The following concept is fundamental. Given a determinant
D =
an
dl2
a^
021
«22
a2n
Onl
0,n2
if we change all its rows (columns) into columns (rows) with the orders pre served, then we get a new determinant an D' = au Oln
A21
CLnl
a22
an2
CL2n
called the transposed determinant of D. Obviously, the transpose of D' is D. So we may say that D and D' are transposes of each other. It is evident that
16
Linear Algebra
the element of D' at its ith row and j  t h column is a,ji, which is the element of D at its j  t h row and ith column (the elements of D and D' on their principal diagonals remain unchanged). Property 5. The value of a determinant D is the same as its transpose D' The same product ai P l • • • anPn will occur as a term of D and as a term of D'. An interchange between two rows (columns) of D is equivalent to one between two columns (rows) in £>'. Therefore, Property 5 follows immediately by the rule for determination of the sign of a term in a determinant. For example, 2 A' = 1 2
4 3 1
2 2 0 0 3 = 1 5 2 = 10 = A. 5 2 5 3
The above properties of determinants are important in both calculations and theoretical deliberations. We illustrate by some examples. Example 4. Prove that
b+c c+a q + r r +p y + z z+x
a+b a b c p+q = 2 p q r x+y x y z
Proof. By Properties 2 and 3, we see that b c+ a b+ c c+ a a+b q+r r+p p+q = q r+p y+z z +x x +y y z+x b c+a = q r+p y z+x b c = q r y Z
a p X
a+b p+q x+y a P
+
X
+
+
c r z
CT
a
r+p z+x
a+b p+q x+y
c a a +b r P p+q z X x+y
c x
a p
Z
X
b a b c q =2 p q r x y z y
The calculation of a determinant can be greatly simplified if it is first trans formed to a triangular one using the above properties.
17
Determinants
E x a m p l e 5. Calculate 1 D =
1
1
1
a,\
a
0,2
0,2
a
03
03
&3
0,3
a
a,2 a.1
Solution: Adding — oi, —02 and —03 times of the first row to the second, third and fourth rows respectively, we obtain 1 D
1
1
1
0 0
a — CLI 0
d2 — Qi a — a2
02 — 0,1 03—02
0
0
0
aa3
which is triangular. Then by Example 4 of the previous section D = (a — ai) (a  a 2 ) (a  a 3 ) . E x a m p l e 6. Calculate
D
3 1 1 1 1 3 1 1 1 1 3 1 1 1 1 3
Solution: Note that the sum of the elements of each row is 6. For each row, add to the element of the first column the elements of all the other columns. Then after taking out the factor 6, subtract, for each row, the element of the first column from the elements of all the other columns in turn. This results in a triangular determinant as follows:
6 1 1 1 6 3 11 =6 6 13 1 6 1 1 3
1 1 13 1 1 1 1 10 12 10 10
1 1 11 3 1 1 3 0 0 0 0 d 2 0 = 62 0 2
48.
18
Linear Algebra
Example 7. Calculate C a
b a a b 0 a a a 0
°  tc S i
Solution: 6 0 D = 6 0
0 b 0 b
6 a 0 a
a b a = 0
6 0 0 0
0 6 0 0
6+ a a+ 6 6 + 2a 2a+ 6
6 0 0 0
ii I) 2 2 + 6 + c 41) 0 0
a2 + 1 a2 = a2 c2 c2 2 c + 1
62 6 + l 62 2
62 1 0
62 6 +1 62 c2 + l 2
c2 c2 2 c +1
c2 0 = a2i b2 + c 1
19
Determinants
It is easy to verify that the equality holds in the case one or more of a, 6, c equal to zero. To end this section we introduce two important classes of determinants. If a,ij = a,ji (i, j = 1,2,..., n) in a determinant
D =
an
ai2
■ ■
0,1
A21
a22
■
■
a.2
ani
&n2
■
an
then D is called symmetric; if a.y = —a,ji (i,j = l,..., n, and so an = 0), then D is called skewsymmetric. For instance, the determinants in Examples 5 and 6 are symmetric while that in Section 1.3 below is skewsymmetric if a = 0. Example 9. Prove that any skewsymmetric determinant of odd order is equal to zero. Proof. Since a^ = — a,j, we have, by Properties 1 and 5,
D =
0
0,12
0ln
0
a21
Onl
a2l
0
02n
012
0
d„2
Onl
Qn2
0
Oln
= ("I)"
0
02l
Onl
ai2
0
an2
Oln
0,2n
' ■ •
^0,2n
= (!)"£>.
0
In the case n is odd, obviously D — 0. Exercises 1. Calculate the following determinants:
(1)
1 2 0 1 1 3 5 0 0 1 5 6 1 2 3 4
(3)
a — b— c 2a 26 bca 2c 2c
(2)
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2a 26 c— a — b
0
20
Linear Algebra
(4)
a b b b a b a b a a b a b b b a
1 3 3 3 2 3 3 3 3
(5)
3
3
3
n
2. Prove that
(1)
(2)
b d a+b a+b+c a+b+c+d 2 a + 6 3a + 26 + c 4a + 36 + 2c + d 3a + b 6a + 36 + c 10a + 66 + 3c + d by + az bx + ay bz + ax
bz + ax by + az bx + ay
bx + ay bz + ax by + az
x (a3 + 63) z y
y X
z
z y x
3. Prove that 2 3 4
n
3 4 5
n 1 2
1 2
*("i) n " + rin  l = (1)
n1
4. Prove Exercise 7, Sec. 1.1 using the properties of determinants. 5. Prove that an{t) an(t) d_ a2i(t) a 2(t) 2 dt a.3i{t) a32(t)
a13(i) a2z{t) a 33 (i)
au(t) «2lW
a
ai 2 (t) 22(0
031 (*)
032(*)
ai 3 (t) 23(0
a
a33(t)
a'uW a2i(t) a3i(t)
+
a
'i2(t) a22{t) a32(t)
an(t) a2\{t) a'3i(t)
a'13(t) a23(t) a33(t)
ai2(t) a22(t) a'32(t)
a13(t) a23(i) a'33(t)
and generalize it to the case of a determinant of an arbitrary order. 6. What is the relation between a determinant and one that is obtained by transposing around its subdiagonal?
1.3. Development of a Determinant We have given some basic properties of determinants, by which the calcu lations of many determinants may be simplified. We would now see a general
21
Determinants
method for such calculations. We know that the lower the order of a determi nant is, the simpler is its calculation. We therefore look for a general method to reduce the order of a determinant. To this aim, we first introduce some definitions. Taking certain k definite rows and k definite columns of a determinant D of order n(k < n), we construct a determinant of order k, called a minor of D of order k. The elements of the minor are those of D situated in the crossed area of the rows and columns taken with their relative orders preserved. In particular, by omitting all the elements of the ith row and j'th column at which the element a^ of D locates, the resultant determinant Mij, of order n — 1, is called the complement minor of a^ in D. For instance, for the determinant of order 4 an
0.12
ai3
an
«21
a22
a.23
a24
031 a4i
032
033
a34
a42
a.43
«44
a n and a23 are minors of order 1 and
Mn =
a22
0.23
«24
032
033
«34
042
a.43
»44
a
,
M23 =
n
031 an
ai2
ai4
«32
034 a44
Q42
are minors of order 3. It is easily seen that the complement of a^ in D is the transpose of its complement in the transpose D'. We expect to express a determinant D in terms of its complements. Let us consider a n M n first. Since any term of M n is of the form ± a 2 p 2 • ■ • anpn, any term ±ana2P2 ■ ■ ■ anPn in a n Mn is a term of D. Therefore, all the terms of a n M n form a part of the terms of D. Then, consider the general case a\k Mik, which we shall reduce to the above special case. Interchange the fcth column of D with its preceding adjacent column successively k — 1 times so that it is transferred to the first column; a\k is now situated at the left upper corner of the new determinant B. By Property 3, Sec. 1.2, we have
B=
(l)k~1D.
22
Linear Algebra
Note that each of the above interchanges does not change the relative order of the columns of Mm, in D and so the complement minor of a\k in B is still Mifc. Thus, all the terms in a\k Mifc are terms of B. Hence, all the terms of aife(l) fc_1 Mifc are terms of (  l ) *  1 ^ or D. Thus, we know that all the terms in the following products a n Mu, an M12,...
,ain(l)n_1Mn_i
are also terms of D, different from each other. But Mifc is a determinant of order n — 1 and so there are (n — 1)! terms in Mifc as well as in aifc(—1) Mifc. Therefore, we have got a total of n\ different terms of D, which means we have exhausted all the terms of D. Thus we can conclude D = a n M u  aw M12 + • ■ • + a i „ (  l ) n _ 1 M i „ , or, what is the same, £> = 5>ifc(l) 1 + f c Mifc. fc=l
As an example, we have 2 4 2
1 2 3 1 3 1 = 2 3 5 3 5
4 2
1 4 + 2 5 2
3 3
= 24 + 22  36 = 10. The above result is easy to extend to the ith row, instead of the first row. By interchanging the ith row with its preceding adjacent row successively, it becomes the first row of the new determinant B, with the relative orders of the remaining rows and columns unchanged, and so the complement minor of a,ij in B is exactly My in D. By the previous conclusion, we have B = an Mn  an Mi2 H
1 a i n (  l ) n _ 1 M i n ,
and consequently
D = {iylB
=
aniiy'Mn+aaiiyMa +  + ain(l)i1+n1Min,
or, what is the same, for fixed i,
3= 1
23
Determinants
For convenience, (l)* + J Mjj is called the algebraic complement or the (algebraic) cofactor of a^ in D, denoted by A^: Ail =
(l)i+JMij.
As an example, in the previous determinant of order 4 the algebraic com plements of a n , o23 are respectively An = (  1 ) 1 + 1 M „ = M a ,
A23 = (  l ) 2 + d M 2 3
"M 2 3
Thus, we have the following fundamental theorem: Theorem 1. The determinant of order n an
•••
air
D = Onl
a„
is equal to the sum of the products of all the elements of D in the ith row and their algebraic complements: D = an An + at2 Ai2 H
+ a, n Ai„ .
The theorem is frequently used and is usually formulated so as to develop a determinant about a row, say the ith row. It may also be developed about a column instead of a row: say the j'th column: D = aij Aij + a2j A2j \
+ anj Anj .
For instance, developing A about its second column and third column we have, respectively, 2 4 2
= 4
1 2 2 2 +3 3 5 2 5
2 2
1 3
=  4 + 1 8  4 = 10 and 2 4 2
1 2 4 3 2 1 2 3 1 = 2 2 3 — 2 3 + 5 4 3 5 =  3 6  4 + 50 = 10.
1 3
24
Linear Algebra
By developing a determinant about a chosen row or column, its calculation can be simplified by the use of complement minors. Which row or column should we take then? Usually, we choose one which contains most zero el ements since we can then avoid calculating the complement minors of such elements. Therefore, in practice, we often first simplify the determinant by us ing Property 4, Sec. 1.2 so that there occur as many zero elements as possible in a row or column, and then develop it. Let us illustrate by some examples. Example 1. Calculate the determinant of order 4
D =
3 1 5 1 2 0 1 5
1 2 3 4 1 1 3 3
Solution: There is already a zero element in row 3. By Property 4, Sec. 1.2, adding 2 times of column 4 to column 1 and then column 3 to column 4, we get a determinant with 3 zero elements in its row 3. Developing it about this column, we get
D
7 1  1 1 13 1 3 1 0 0 1 0  5  5 3 0
(1)
3+3
7 1 13 1 5  5
1 1 0
Then, adding row 1 of the determinant of order 3 on the righthand side to row 2, we obtain at length
D
7 6 5
1 2 5
1 6 0 = (1) 1+3 5 0
2 = 30 + 10 = 40 . 5
Example 2. Calculate
1+x D =
1 1 1
1 1x 1 1
1 1 1 1 1 1 + 2/ 1 12/
Solution: Assume xy ^ 0. By Theorem 1, the determinant may be modified by adding in a row and a column as follows:
25
Determinants
1
1+x D
1 1 1
1 1 1x 1 1
1 1 1 1 1 1 1 + 2/ 1 1 12/
1 1 1 1 1
0 0 0
1 0 x 0
1 0 0 y
0
0
By Property 4, Sec. 1.2, by adding  times of column 2, — times of column 3, i times of column 4, and — j times of column 5 altogether to column 1, it becomes a triangular determinant, and hence 1 0 D = 0 0 0
1 X
0 0 0
1 0 —x 0 0
1 0 0
l 0 0 0
y 0
2 2
= x'y*
y
Obviously, the above holds for x = 0 or y = 0 too, since D = 0 in such cases. It may also be proved using Property 2 of determinants. Example 3. Prove that a —b D = —c —dc
b a d
c —d a b
d c = (a 2 + b2 + c2 + d2)2 . —b a
P r o o f . Assume a ^ 0 for the time being. Then,
a
a2
a2 —ab —ac —ad
b c d a —d c d a —b — c b a
+ b2 + c2 + d2 a
a2
+ b2 + c2 + d2
a d c
a2
+
b2
+
a d a b
c
2
+
d2
1 0 0 0
b a d —c
c b a
(a 3 + bed  bed + ab2 + ac2 + ad2)
a