Maths

J.Baskar Babujee Department of Mathematics

Anna University, Chennai-600 025.

MA6151MA6151

CONTENT

1.1 INTRODUCTION

1.2 TYPES OF MATRICES

1.3 CHARACTERISTIC EQUATION

1.4 EIGEN VECTORS

1.5 PROBLEMS

1.6 CAYLEY HAMILTON THEOREM

2

1.7 DIAGONALISATION OF A MATRIX

1.8 REDUCTION OF A MATRIX TO DIAGONAL

FORM

1.9 ORTHOGONAL TRANSFORMATION OF A

SYMMETRIC MATRIX TO DIAGONAL FORM

1.10 QUADRATIC FORMS

1.11 NATURE OF A QUADRATIC FORM

3

1.12 REDUCTION OF QUADRATIC FORM TO

CANONICAL FORM

1.13 INDEX AND SIGNATURE OF THE QUADRATIC

FORM

1.14 LINEAR TRANSFORMATION OF A

QUADRATIC FORM.

1.15 REDUCTION OF QUADRATIC FORM TO

CANANONICAL FORM BY ORTHOGONAL

TRANSFORMATION

4

1.1 INTRODUCTION:-A matrix is defined as a rectangular array (or

arrangement in rows or columns) of scalar subject to certain rules of operations.

If mn numbers (real or complex) or functions are arranged in the columns (vertical lines) then A is called an m n matrix. Each of the mn numbers is called an element of the matrix.

Unit 1 MATRICES

An m n matrix is usually written as

A matrix is usually denoted by a single capital letter A, B or C etc.

11 12 13 1

21 22 23 2

31 32 33 3

1 2 3

....

....

....

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

....

n

n

n

m m m mn

a a a a

a a a a

a a a a

a a a a

7

Thus, an m n matrix A may be written as

A = , where i = 1, 2, 3, … , m ;

j = 1, 2, 3, … , n

In Algebra, the matrices have their largest

application in the theory of simultaneous equations and linear transformations.

ija

E.g., The set of simultaneous equations

3333232131

2323222121

1313212111

bxaxaxa

bxaxaxa

bxaxaxa

may be symbolically represented by the equation

Where A = , X = , B =

333231

232221

131211

aaa

aaa

aaa

A X = B

3

2

1

x

x

x

3

2

1

b

b

b

720

135

1.2 TYPES OF MATRICES

(1) Real Matrix :- A matrix is said to be

real if all its elements are real numbers.

E.g., is a real matrix.

(2) Square Matrix:- A matrix in which the number of rows is equal to the number of columns is called a square matrix, otherwise, it is said to be rectangular matrix.

i.e., a matrix A = is a square matrix if m = n

a rectangular matrix if m nij m n

a

A square matrix having n rows and n columns is

called “ n – rowed square matrix”,

is a 3 – rowed square matrix

The elements of a square matrix are

called its diagonal elements and the diagonal

along with these elements are called principal or

leading diagonal.

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

33,2211 aa,a

The sum of the diagonal elements of a square matrix is called its trace or spur.

Thus, trace of the n rowed square matrix

A= is ija

n

1iijnn332211 aa....aaa

(3) Row Matrix :-

A matrix having only one row and any number of columns,

i.e., a matrix of order 1 n is called a row matrix.

[2 5 -3 0] is a row matrix.

Example:-

(4) Column Matrix:- A matrix having only one column

and any number of rows, ii.e., a matrix of order m 1 is called a column matrix.

is a column matrix.

1

0

2

Example:-

(5) Null Matrix:-

A matrix in which each element is zero

is called a null matrix or void matrix or zero matrix.

A null matrix of order m n is denoted by

=

16

nmO

Example :-

0000

000042O

(6) Sub - matrix :-

A matrix obtained from a given matrix A by deleting some of its rows or columns or both is called a sub – matrix of A.

is a sub – matrix of

41

03

2461

7053

5210

Example:-

(7) Diagonal Matrix :-

A square matrix in which all non – diagonal

elements are zero is called a diagonal matrix.

i.e., A = [a ] is a diagonal matrix if a = 0 for i j.

is a diagonal matrix.

ij nn ij

Example:-

000

010

002

(8) Scalar Matrix:-

A diagonal matrix in which all the diagonal elements are equal to a scalar, say k, is called a scalar matrix.

i.e., A = [a ] is a scalar matrix if

is a scalar matrix.

200

020

002

ij nn

jiwhenk

jiwhen0a ij

Example :-

(9) Unit Matrix or Identity Matrix:-

A scalar matrix in which each diagonal element is 1 is called a unit or identity matrix. It is denoted by .

i.e., A = [a ] is a unit matrix if

is a unit matrix.

nI

ij nn

jiwhen1

jiwhen0a ij

Example

10

01

(10) Upper Triangular Matrix.

A square matrix in which all the elements

below the principal diagonal are zero is called an

upper triangular matrix.

i.e., A = [a ] is an upper triangular matrix if a = 0

for i > j

is an upper triangular

matrix

300

510

432

ij nn ij

Example:-

(11) Lower Triangular Matrix.

A square matrix in which all the elements

above the principal diagonal are zero is called a

lower triangular matrix.

i.e., A = [a ] is a lower triangular matrix if a = 0

for i < j

is a lower triangular

matrix.

ij nn ij

Example:-

023

065

001

(12) Triangular Matrix:-

A triangular matrix is either upper triangular or lower triangular.

(13) Single Element Matrix:-

A matrix having only one element is called a single element matrix.

i.e., any matrix [3] is a single element matrix.

(14) Equal Matrices:-

Two matrices A and B are said to be equal iff they have the same order and their corresponding elements are equal.

i.e., if A = and B = , then A = B

iff a) m = p and n = q

b) a = b for all i and j.

qpij]b[ nmij]a[

ij ij

(15) Singular and Non – Singular Matrices:-

A square matrix A is said to be singular if |A| = 0 and non – singular if |A| 0.

A = is a singular

matrix since |A| = 0.

110

432

432

Example :-

1.3 CHARACTERISTIC EQUATION

If A is any square matrix of order n, we

can form the matrix , where is the nth

order unit matrix.

The determinant of this matrix equated to zero,

i.e.,

IλA I

0

λa...aa

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

a...λaa

a...aλa

λA

nnn2n1

2n2221

1n1211

I

is called the characteristic equation of A.

On expanding the determinant, we get

where k’s are expressible in terms of the elements a

The roots of this equation are called Characteristic

roots or latent roots or eigen values of the matrix A.

0k...λkλkλ1)( n2n

21n

1nn

ij

1.4 EIGEN VECTORS

Consider the linear transformation Y = AX ...(1)

which transforms the column vector X into the column

vector Y. We often required to find those vectors X

which transform into scalar multiples of themselves.

Let X be such a vector which transforms into

X by the transformation (1).

Then Y = X ... (2)

From (1) and (2), AX = X AX- X = 0

(A - )X = 0 ...(3)

This matrix equation gives n homogeneous linear equations

... (4)

I

0λ)x(a...xaxa

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

0xa...λ)x(axa

0xa...xaλ)x(a

nnn2n21n1

n2n222121

n1n212111

I

These equations will have a non-trivial solution only

if the co-efficient matrix A - is singular

i.e., if |A - | = 0 ... (5)

Corresponding to each root of (5), the

homogeneous system (3) has a non-zero solution

X = is called an eigen vector or latent vector

II

4

2

1

x

...

x

x

31

Properties of Eigen Values:-Properties of Eigen Values:-

1.The sum of the eigen values of a matrix is the sum

of the elements of the principal diagonal.

2.The product of the eigen values of a matrix A is

equal to its determinant.

3. If is an eigen value of a matrix A, then 1/ is

the eigen value of A-1 .

4. If is an eigen value of an orthogonal matrix, then

1/ is also its eigen value.

32

PROPERTY 1:-PROPERTY 1:- If λ1, λ2,…, λn are the eigen values of A, then

i. k λ1, k λ2,…,k λn are the eigen values of the matrix kA, where k is a non – zero scalar.

ii. are the eigen values of the inverse

matrix A-1.

iii. are the eigen values of Ap, where p is any positive integer.

n21 λ

1,...,

λ

1,

λ

1

pn

p2

p1 λ...,,λ,λ

33

Proof:-

i. Let λr be an eigen value of A and Xr the corresponding

eigen vector.

Then, by definition,

Multiplying both sides by k,

(kA)Xr = (kλr)Xr

Then k λr is an eigen value of kA and the corresponding

eigen vector is the same as that of λr, namely Xr.

AXr = λrXr

34

ii. Pre multiplying both sides of AXr = λrXr by A-1

A-1 (A Xr) = A-1 (λr Xr )

Xr = λr (A-1 Xr )

=> A-1 Xr = (Xr)

Hence is an eigen value of A-1 and the

corresponding eigen vector is the same as that of

λr , namely Xr.

rλ

1

rλ

1

35

iii. Pre multiplying both sides of AXr = λrXr by A

A (A Xr) = A (λr Xr )

A2 Xr = λr (A Xr )

= λr (λr xr)

= λr2 Xr.

Similarly, we can prove that A3Xr = λr3 Xr, …,

ApXr = λrp Xr, where p is any positive integer. Hence λr

p

is any eigen value of Ap and the corresponding eigen

vector is the same as that of λr, namely Xr.

36

THEOREM :-

A matrix A is singular if and only if 0 is an eigen value of A.

1.5 PROBLEMS1. Find the sum and product of the eigen values of

the matrix

without finding the eigen values.

021

612

322

A

37

Solution:-

Sum of the eigen values of A = sum of its diagonal elements.

= - 2 + 1 + 0

= -1.

Product of the eigen values of A = | A |

=

= 45.

021

612

322

38

2. Two eigen values of the matrix

are equal to 1 each. Find the third eigen value.

Solution:- Let a be the third eigen value of A.

Since sum of the eigen values = sum of the diagonal

elements,

1 + 1 + a = 2 + 3 + 2

a = 5

Therefore, the third eigen value of A is 5.

221

131

122

A

39

3. The product of two eigen values of the matrix

is 16. Find the third eigen value.

Solution:- Let a be the third eigen value of A. Since product of the eigen values = | A |

16a =

Therefore, a = 2.

312

132

226

A

312

132

226

40

4. Find the sum of the eigen values of the inverse of

Solution:-

The eigen values of the lower triangular matrix A is 1, -3, 2. Then the eigen values of A-1 are

Sum of the eigen values of A-1 =

=

250

032

001

A

.2

1,

3

1,1

.2

1

3

11

6

7

41

5. If , find the eigen values of 3A, A-1

and – 2A-1.

Solution:-

The eigen values of A are 2, -1, 4.

The eigen values of 3A are 3×2, 3×(-1), 3×4

i.e., 6, -3, 12

400

210

572

A

42

The eigen values of A-1 are

4

11,,

2

1i.e.,

4

1,

1

1,

2

1

2

12,1,i.e.,

4

121),2)((,

2

12

are2AofvalueseigenThe 1

43

6. Find the eigen values and eigen vectors of the matrix

A =

Solution:- The characteristic equation of the given matrix is

45

21

0|IA|

or

Thus,

Corresponding to =6, the eigen vectors are given by

(A – 6 ) X1 = 0

1,6

0652

045

21

I

0x

x

25

25or0

x

x

645

261or

2

1

2

1

the eigen values of A are 6, -1.

We get only one independent equation - 5x1 - 2x2 = 0

5

2kXarevectorseigenThe

5kx

2kx

(say)k5

x

2

x

11

12

11

121

Corresponding to = -1, the eigen vectos are given

by ( A + ) X2 = 0

I

1

1kXarevectorseigenThe

kx,kx

(say)k1

x

1

x

0xx

0

0

x

x

55

22

22

2221

221

21

2

1

7. Find the eigen values and eigen vectors of the

matrix A =

Solution:- The characteristic equation of the given

matrix is

021

612

322

0|IA|

0

21

612

322

Thus,

53,3,λ

05)3)(λ3)(λ(λ

015)2λ3)(λ(λ

it.satisfies3λtrial,By

04521λλλ

0λ)]1(143[6]2λ2[12]λ)λ(1λ)[2(

2

23

the eigen values of A are -3, -3, 5.

Corresponding to = - 3, the eigen vectors are

given by

0

1

2

k

1

0

3

k

k

k

k2k3

0x3x2

0

x

x

x

321

642

321

or0XI

21

1

2

21

32

3

2

1

1

1

2112213

1

X

bygivenarevectorseigenThe

2k3kxthenkx,kxLet

xequation tindependen oneonly get We

3A

Corresponding to λ = 5, the eigen vectors are given by (A – 5 I )X2 = 0.

052

032

0327

0

0

0

521

642

327

321

321

321

3

2

1

xxx

xxx

xxx

x

x

x

51

1

2

1

kX

kx2kx,kx

(say)k1

x

2

x

1

x22

x

53

x

6-10

x

32

333,231

3321

321

by given are vectors eigen the Hence

,equations. two first From


Solution:- The characteristic equation is

342

476

268

A

0|IA|

. are values eigen Hence

3,150,λ

045λ18λλ

0

λ342

4λ76

26λ8

23

15,3,0

53

Eigen vector X1 corresponding to λ1= 0 is given by

(3) ... 03xx42x

(2) ... 04x7x6x

(1) ... 02x6x8x

0

0

0

x

x

x

342

476

268

x

x

x

Xwhere0,I)Xλ(A

321

321

321

3

2

1

3

2

1

111

54

Solving equations (1) and (2) by cross-multiplication, we get

which satisfy equation (3) also. Required eigen vector corresponding to λ1 = 0 is

131211

11321

321

2kx,2kx,kx

0kwhere(say)k2

x

2

x

1

x3656

x

3212

x

1424

x

2

2

1

2

2 1

1

1

1

1 k

k

k

k

X

55


(6) ... 0x42x

(5) ... 04x4x6x

(4) ... 02x6x5x

0

0

0

x

x

x

042

446

265

x

x

x

Xwhere0,I)Xλ(A

21

321

321

3

2

1

3

2

1

222

56



232221

22321

321

2kx,kx,kx

0kwhere(say)k2-

x

1

x

2

x3620

x

2012

x

824

x

2

2

1

2

2

2

2

2

2

2

2 k

k

k

k

X

57


(9) ... 012xx42x

(8) ... 04x8x - 6x

(7) ... 02x6x7x-

0

0

0

x

x

x

12-42

48-6

267-

x

x

x

Xwhere0,I)Xλ(A

321

321

321

3

2

1

3

2

1

333

58



333231

33321

321

kx,kx,kx

0kwhere(say)k1

x

2-

x

2

x20

x

40-

x

40

x

22

1

2

2

2

2

3

3

3

3

3 k

k

k

k

X

59



312

132

226

A

0|IA|

8. 2, 2, are values eigen Hence

82,2,λ

03236λ12λλ

0

λ312

1λ32

22λ6

23

60

Eigen vector corresponding to is given by

221

0xx2x

0xx2x

02x2x4x

0

0

0

x

x

x

112

112

224

x

x

x

Xwhere0,I)Xλ(A

321

321

321

3

2

1

3

2

1

111

These equations are equivalent to a single equation … (1)

Let x3 = 2 k3 and x2 = 2 k2 then from (1)

2x1 – 2 k2 + 2 k3 = 0

=> x1 = k2 – k3

Required eigen vector corresponding to

is

02 321 xxx

21 2

2

0

1

0

2

1

2

2 32

3

2

32

1 kk

k

k

kk

X

Similarly, the eigen vector corresponding to = 8 is

given by,

3λ

....(4)05xx2x

....(3)0x5x2x

....(2)02x2x2x

0

0

0

x

x

x

512

152

222

0)X 8 -(A

x

x

x

X where0I)Xλ -(A

321

321

321

3

2

1

3

3

2

1

333



131211

1321

321

,,2

)(112

6612

kxkxkx

saykxxx

xxx

1

1

22

1

1

1

1

3 k

k

k

k

X

64

Show that if λ1,λ2, … λn are the latent roots of the

matrix A, then A3 has the latent roots

Solution:- Let λ be a latent root of the matrix A.

Then there exists a non – zero vector X such that

....,,, 332

31 n

Example 1:-

65

A X = λ X … (1)

=> A2 (AX) = A2 (λ X)

=> A3 X = λ (A2 X) [using (1)]

But A2 X = A ( A X) = A (λ X)

= λ (AX) = λ (λX) = λ2 X

Therefore, A3 X = λ (λ2 X) = λ3X

=> λ3 is a latent root of A3.

66

Therefore, If λ1,λ2, … λn are the latent roots of the matrix A, then are the latent roots of A3.

If λ1,λ2, … λn are eigen values of A then find eigen values of the matrix (A – λI)2 .

Solution:- (A – λI)2 = A2 – 2 λAI + λ2I2

= A2 – 2 λA + λ2I

332

31 ...,,, n

Example 2:-

67

Eigen values of A2 are

Eigen values of 2 λA are 2 λ λ1,2 λ λ2, …, 2 λ λn.

Eigen values of λ2I are λ2

Eigen values of ( A – λI)2 are

.λ...,λ,λ 2n

22

21

.λ)(λ,...,λ)(λ,λ)(λor

.λ2λλλ,...,λ2λλλ,λ2λλλ

2n

22

21

2n

2n

22

22

21

21

68

Find the eigen values and eigen vectors of the matrix


500

620

413

A

5. 2, 3, are values eigen Hence

52,3,λ

0λ)λ)(5λ)(2

0

λ500

6λ20

41λ3

3(

0|IA|

Example 3:-

69


0x0,x

02x

06xx

04xx

0

0

0

x

x

x

200

61-0

410

x

x

x

Xwhere0,)X3(A

23

3

32

32

3

2

1

3

2

1

11

I

The characteristic vector corresponding to eigen value λ1 = 3 is given by

When λ2 = 2, let X2 be the eigen vector then

(A – 2I) X2 = 0 where X2 = [x1 x2 x3]’

0k,kXwhere

0

0

k

x

x

x

X 111

1

3

2

1

1

0

0

0

300

600

411

3

2

1

x

x

x

71

0

1

1

k

0

k

k

XisvectoreigenRequired

0x

kx

kx

0kwhere(say)k0

x

1

x

1

x

xxor0xx

0x03x

06x

04xxx

22

2

2

3

22

21

22321

2121

33

3

321

72

When λ3 = 5, let X3 be the eigen vector then

(A – 5I) X3 = 0 where X3 = [x1 x2 x3]’

Solving above equations by cross – multiplication, we get

06x3x

04xx2x

0

0

0

x

x

x

000

630

412

32

321

3

2

1

73

Required eigen vector is

333231

33321

321

kx,2kx,3kx

0kwhere(say)k1

x

2

x

3

x6

x

12

x

126

x

1

2

3

k

k

2k

3k

X 3

3

3

3

3

1.6 CAYLEY HAMILTON

THEOREM

Every square matrix satisfies its own

characteristic equation.

Let A = [aij]n×n be a square matrix

then,

nnnn2n1n

n22221

n11211

a...aa

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

a...aa

a...aa

A

Let the characteristic polynomial of A be (λ)

Then,

The characteristic equation is

11 12 1n

21 22 2n

n1 n2 nn

φ(λ) = A - λI

a - λ a ... a

a a - λ ... a=

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

a a ... a - λ

| A - λI |= 0

Note 1:- Premultiplying equation (1) by A-1 , we

have

n n-1 n-20 1 2 n

n n-1 n-20 1 2 n

We are to prove that

p λ +p λ +p λ +...+p = 0

p A +p A +p A +...+p I= 0 ...(1)

I

n-1 n-2 n-3 -10 1 2 n-1 n

-1 n-1 n-2 n-30 1 2 n-1

n

0 = p A +p A +p A +...+p +p A

1A =- [p A +p A +p A +...+p I]

p

This result gives the inverse of A in terms of

(n-1) powers of A and is considered as a practical

method for the computation of the inverse of the

large matrices.

Note 2:- If m is a positive integer such that m > n

then any positive integral power Am of A is linearly

expressible in terms of those of lower degree.

Verify Cayley – Hamilton theorem for the matrix

A = . Hence compute A-1 .

Solution:- The characteristic equation of A is

211

121

112

tion)simplifica(on049λ6λλor

0

λ211

1λ21

11λ2

i.e.,0λIA

23

Example 1:-

To verify Cayley – Hamilton theorem, we have to

show that A3 – 6A2 +9A – 4I = 0 … (1)

Now,

222121

212221

212222

211

121

112

655

565

556

655

565

556

211

121

112

211

121

112

23

2

AAA

A

A3 -6A2 +9A – 4I = 0

= - 6 + 9

-4

=

This verifies Cayley – Hamilton theorem.

222121

212221

212222

655

565

556

211

121

112

100

010

001

0

000

000

000

81

Now, pre – multiplying both sides of (1) by A-1 , we have

A2 – 6A +9I – 4 A-1 = 0

=> 4 A-1 = A2 – 6 A +9I

311

131

113

4

1

311

131

113

100

010

001

9

211

121

112

6

655

565

556

4

1

1

A

A

82

Given find Adj A by using Cayley –

Hamilton theorem.

Solution:- The characteristic equation of the given matrix A is

113

110

121

A

tion)simplifica(on035λ3λλor

0

λ113

1λ10

1-2λ1

i.e.,0λIA

23

Example 2:-

83

By Cayley – Hamilton theorem, A should satisfy

A3 – 3A2 + 5A + 3I = 0

Pre – multiplying by A-1 , we get

A2 – 3A +5I +3A-1 = 0

339

330

363

3A

146

223

452

113

110

121

113

110

121

A.A ANow,

(1)...5I)3A(A3

1A

2

21-

84

AAAAdj.

A

AAdj.Athat,knowWe

173

143

110

3

1

500

050

005

339

330

363

146

223

452

3

1AFrom(1),

1

1

1

85

173

143

110

AAdj.

173

143

110

3

13)(AAdj.

3

113

110

121

ANow,

86

1.7 DIAGONALISATION OF A MATRIX

Diagonalisation of a matrix A is the

process of reduction A to a diagonal form.

If A is related to D by a similarity transformation,

such that D = M-1AM then A is reduced to the

diagonal matrix D through modal matrix M. D is

also called spectral matrix of A.

87

1.8 REDUCTION OF A MATRIX TO DIAGONAL FORM

If a square matrix A of order n has n linearly

independent eigen vectors then a matrix B can

be found such that B-1AB is a diagonal matrix.

Note:- The matrix B which diagonalises A is called

the modal matrix of A and is obtained by

grouping the eigen vectors of A into a square

matrix.

88

Similarity of matrices:-

A square matrix B of order n is said to be a similar to a square matrix A of order n if

B = M-1AM for some non singular matrix M.

This transformation of a matrix A by a non – singular matrix M to B is called a similarity transformation.

Note:- If the matrix B is similar to matrix A, then B

has the same eigen values as A.

89

Reduce the matrix A = to diagonal form by

similarity transformation. Hence find A3.

Solution:- Characteristic equation is

=> λ = 1, 2, 3 Hence eigen values of A are 1, 2, 3.

300

120

211

0

λ-300

1λ-20

21λ1-

Example:-

90

Corresponding to λ = 1, let X1 = be the eigen

vector then

3

2

1

x

x

x

0

0

1

kX

x0x,kx

02x

0xx

02xx

0

0

0

x

x

x

200

110

210

0X)I(A

11

3211

3

32

32

3

2

1

1

91


vector then,

3

2

1

x

x

x

0

1-

1

kX

x-kx,kx

0x

0x

02xxx

0

0

0

x

x

x

100

100

211-

0X)(A

22

32221

3

3

321

3

2

1

2

0,

I2

92


vector then,

3

2

1

x

x

x

2

2-

3

kX

xk-x,kx

0x

02xxx

0

0

0

x

x

x

000

11-0

212-

0X)(A

33

13332

3

321

3

2

1

3

3

2

2

3,

2

I3

k

x

93

Hence modal matrix is

2

100

11-02

1-11

M

MAdj.M

1-00

220

122-

MAdj.

2M

200

21-0

311

M

1

94

Now, since D = M-1AM

=> A = MDM-1

A2 = (MDM-1) (MDM-1)

= MD2M-1 [since M-1M =

I]

300

020

001

200

21-0

311

300

120

211

2

100

11-02

111

AMM 1

95

Similarly, A3 = MD3M-1

=

A3 =

2700

19-80

327-1

2

100

11-02

111

2700

080

001

200

21-0

311

96

1.9 ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM

A square matrix A with real elements is said

to be orthogonal if AA’ = I = A’A.

But AA-1 = I = A-1A, it follows that A is orthogonal if

A’ = A-1.

Diagonalisation by orthogonal transformation

is possible only for a real symmetric matrix.

97

If A is a real symmetric matrix then eigen

vectors of A will be not only linearly independent but

also pairwise orthogonal.

If we normalise each eigen vector and use

them to form the normalised modal matrix N then it

can be proved that N is an orthogonal matrix.

98

The similarity transformation M-1AM = D

takes the form N’AN = D since N-1 = N’ by a

property of orthogonal matrix.

Transforming A into D by means of the

transformation N’AN = D is called as orthogonal

reduction or othogonal transformation.

Note:- To normalise eigen vector Xr, divide each

element of Xr, by the square root of the sum of the

squares of all the elements of Xr.

99

Diagonalise the matrix A = by means of an

orthogonal transformation.

Solution:-

Characteristic equation of A is

204

060

402

66,2,λ

0λ)16(6λ)λ)(2λ)(6(2

0

λ204

0λ60

40λ2

Example :-

100

I

1

1 2

3

1

1

2

3

1 3

2

1 3

1 1 2 3 1

1 1

x

whenλ = -2,let X = x be theeigenvector

x

then (A +2 )X = 0

4 0 4 x 0

0 8 0 x = 0

4 0 4 x 0

4x + 4x = 0 ...(1)

8x = 0 ...(2)

4x + 4x = 0 ...(3)

x = k ,x = 0,x = -k

1

X = k 0

-1

101

2

2I

0

1

2

3

1

2

3

1 3

1 3

1 3 2

2 2 3

x

whenλ = 6,let X = x be theeigenvector

x

then (A - 6 )X = 0

-4 0 4 x 0

0 0 x = 0

4 0 -4 x 0

4x + 4x = 0

4x - 4x = 0

x = x and x isarbitrary

x must be so chosen that X and X are orthogonal among th

.1

emselves

and also each is orthogonal with X

102

2 3

3 1

3 2

3

1 α

Let X = 0 and let X = β

1 γ

Since X is orthogonal to X

α - γ = 0 ...(4)

X is orthogonal to X

α+ γ = 0 ...(5)

Solving (4)and(5), we get α = γ = 0 and β is arbitrary.

0

Taking β =1, X = 1

0

1 1 0

Modal matrix is M = 0 0 1

-1 1

0

103

The normalised modal matrix is

1 10

2 2N = 0 0 1

1 1- 0

2 2

1 10 - 1 1

02 2 2 0 4 2 21 1

D =N'AN = 0 0 6 0 0 0 12 2

4 0 2 1 1- 00 1 0

2 2

-2 0 0

D = 0 6 0 which is the required diagonal matrix

0 0 6

.

104

1.10 QUADRATIC FORMS

DEFINITION:- DEFINITION:-

A homogeneous polynomial of second degree in any number of variables is called a quadratic form.

For example,

ax2 + 2hxy +by2

ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and

ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw

are quadratic forms in two, three and four variables.

105

In n – variables x1,x2,…,xn, the general quadratic form

is

In the expansion, the co-efficient of xixj = (bij + bji).

n

1j

n

1ijiijjiij bbwhere,xxb

).b(b2

1awherexxaxxb

baandaawherebb2aSuppose

jiijijji

n

1j

n

1iijji

n

1j

n

1iij

iiiijiijijijij

106

Hence every quadratic form can be written as

getweform,matrixin

formsquadraticofexamplessaidabovethewritingNow

.x,...,x,xXandaAwhere

symmetric,alwaysisAmatrixthethatso

AX,X'xxa

n21ij

ji

n

1j

n

1iij

y

x

bh

hay][xby2hxy ax (i) 22

107

w

z

y

x

dnml

ncgf

mgbh

lfha

wzyx

2nzw 2myw 2lxw zx 2f 2gyz2hxy dw2 cz by ax (iii)

z

y

x

cgf

gbh

fha

zyx2fzx 2gyz 2hxy cz by ax (ii)

222

222

108

1.11 NATURE OF A QUADRATIC FORM

A real quadratic form X’AX in n variables is said to be

i. Positive definite if all the eigen values of A > 0.ii. Negative definite if all the eigen values of A < 0.iii. Positive semidefinite if all the eigen values of A 0

and at least one eigen value = 0.iv. Negative semidefinite if all the eigen values of A 0 and at least one eigen value = 0.v. Indefinite if some of the eigen values of A are + ve

and others – ve.

109

Find the nature of the following quadratic forms

i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx

ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy

Solution:-

i. The matrix of the quadratic form is

113

151

311

A

Example :-

110

The eigen values of A are -2, 3, 6.Two of these eigen values being positive and one being negative, the given quadratric form is indefinite.

ii. The matrix of the quadratic form is

The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form is positive definite.

311

151

113

A

111

1.12 REDUCTION OF QUADRATIC FORM TO CANONICAL FORM

A homogeneous expression of the second degree in any number of variables is called a quadratic form.

For instance, if

which is a quadratic form.

(i)....2hxy2gzx2fyzczbyaxAXX'then

],zyx[X'and

z

y

x

X,

cfg

fbh

gha

A

222

112

Let λ1, λ2, λ3 be the eigen values of the matrix A

and

be its corresponding eigen vectors in the

normalized form (i.e., each element is divided by

square root of sum of the squares of all the three

elements in the eigen vector).

3

3

3

3

2

2

2

2

1

1

1

1

z

y

x

X,

z

y

x

X,

z

y

x

X

113

Then B-1AB = D, a diagonal matrix.

Hence the quadratic form (i) is reduced to a sum of squares (i.e., canonical form).

λ1x2 + λ2y2 + λ3z2

And B is the matrix of transformation which is an orthogonal matrix.

Note:-

1. Here some of λi may be positive or negative or zero

2. If ρ(A) = r, then the quadratic form X’AX will contain only r terms.

114

1.13 INDEX AND SIGNATURE OF THE QUADRATIC FORM

The number p of positive terms in the

canonical form is called the index of the quadratic

form.

(The number of positive terms) – ( the number of

negative terms)

i.e., p – (r – p) = 2p – r is called signature of the

quadratic form, where ρ(A) = r.

115

1.14 LINEAR TRANSFORMATION OF A QUADRATIC FORM.

Let X’AX be a quadratic form in n- variables and let X = PY ….. (1) where P is a non – singular matrix, be the non – singular transformation.

From (1), X’ = (PY)’ = Y’P’ and hence

X’AX = Y’P’APY = Y’(P’AP)Y

= Y’BY …. (2)

where B = P’AP.

116

Therefore, Y’BY is also a quadratic form in n-

variables. Hence it is a linear transformation of

the quadratic form X’AX under the linear

transformation X = PY and B = P’AP.

Note. (i) Here B = (P’AP)’ = P’AP = B

(ii) ρ(B) = ρ(A)

Therefore, A and B are congruent matrices.

117

Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form.

Or

Diagonalise the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by linear transformations and write the linear transformation.

Or

Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the sum of squares.

Example:-

118

Solution:- The given quadratic form can be written as X’AX where X = [x, y, z]’ and the symmetric matrix

A =

Let us reduce A into diagonal matrix. We know tat A = I3AI3.

344

402

423

100

010

001

344

402

423

100

010

001

344

402

423

119

21 31OperatingR ( 2 / 3),R ( 4 / 3)

(for A onL.H.S.andpre factor on R.H.S.),weget

3 2 4 1 0 01 0 0

4 4 20 1 0 A 0 1 0

3 3 30 0 1

4 7 40 0 1

3 3 3

100

0103

4

3

21

A

103

4

013

2001

3

7

3

40

3

4

3

40

423

getweR.H.S),onfactorpostandL.H.S.onA(for

4/3)(C2/3),(C Operating 3121

120

100

0103

4

3

21

A

112

013

2001

1003

4

3

40

003

getwe(1),R Operating 32

APP'1,3

43,Diagor

100

110

23

21

A

112

013

2001

100

03

40

003

getwe(1),C Operating 32

121

The canonical form of the given quadratic form is

Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1.

Note:- In this problem the non-singular transformation which reduces the given quadratic form into the canonical form is X = PY.

i.e.,

3

2

1

112

013

2001

y

y

y

z

y

x

23

22

21

3

2

1

321

yy3

43y

y

y

y

100

03

40

003

yyyAP)Y(P'Y'

122

1.15 REDUCTION OF QUADRATIC FORM TO CANANONICAL FORM BY ORTHOGONAL TRANSFORMATION

Let X’AX be a given quadratic form. The modal

matrix B of A is the matrix whose columns are

characteristic vectors of A. If B represents the

orthogonal matrix of A,

123

then X = BY will reduce X’AX to Y’ diag(λ1, λ2,…, λn) Y,

where λ1, λ2,…, λn are characteristic values of A.

Note. This method works successfully if the

characteristic vectors of A are linearly independent

which are pairwise orthogonal.

124

Reduce 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz into canonical form by orthogonal reduction.

Solution:- The matrix of the quadratic form is

342

476

268

A

Example 1:-

125

The characteristic roots of A are given by 0|| IλA

153,0,λ

015)3)(λλ(λor

0

λ342

4λ76

26λ8

.,.ei

126

Characteristic vector for λ = 0 is given by

[A – (0)I] X1 = 0

'11

321

321

321

321

2)2,(1,kXvectoreigenthegiving

2

x

2

x

1

x

getwe,twofirstSolving

03x4x2x

04x7x6x

02x6x8xi.e.,

127

When λ = 3, the corresponding characteristic vector is given by [A – 3I] X2 = 0

i.e.,

Solving any two equations, we get X2 = k2 (2, 1, -2)’.

Similarly characteristic vector corresponding to

λ = 15 is X3 = k3 (2, -2, 1)’.

04x2x

04x4x6x

02x6x5x

21

321

321

128

Now, X1, X2, X3 are pairwise orthogonal

i.e., X1 . X2 = X2 . X3 = X3 . X1 = 0.

The normalised modal matrix is

3

1

3

2

3

23

2

3

2

3

13

2

3

1

3

2

B

129

Now B is orthogonal matrix and 1B

1500

000

003

3

1

3

2

3

23

2

3

2

3

13

2

3

1

3

2

A

3

1

3

2

3

23

2

3

2

3

13

2

3

1

3

2

ie.,

15}0,{3,diagDABBandBBi.e., 1T1

130

which is the required canonical form.

Note. Here the orthogonal transformation is X =BY, rank of the quadratic form = 2; index = 2, signature = 2. It is positive definite.

23

22

21

3

2

1

321

1

15y0.y3y

y

y

y

1500

000

003

yyy

DYY'AB)Y(BY'AXX'

131

TEST YOUR KNOWLEDGE

1. If then find the eigen value of

2. Write the matrix of the Quadratic form

3. Obtain the characteristic equation of the matrix whose

eigen values are 1,-2 and 3.

4. A = , then find the eigen values of

3A3+5A2-6A+2I.

1 2 3

0 3 2

0 0 2

A

23A I

1 2 3 1 2 32 2 2x x x x x x .

200

320

061

132

5. Write the quadratic form corresponding to the following symmetric matrix

6. Find the sum of the eigen values of the inverse of.

7. Obtain the latent roots of A4 where A =

8. If A is idempotent matrix then A2=A. What will be the eigen values of A.

342

411

210

1 0 0

2 3 0

0 5 2

5 4

1 2

133

9. If are the eigen values of the matrix A,

whose characteristic equation is

Obtain using the property.

10. (i) Using Cayley-Hamilton Theorem find the

inverse of the matrix

(ii) Find the Characteristic roots and

Characteristic vectors of the matrix

1 2 2 3 3 1

3 2 21 45 0 1 2 3, and

1 0 3

2 1 1

1 1 1

A

1 1 3

1 5 1

3 1 1

A

134

11. Reduce the quadratic form to the canonical form by orthogonal transformation. Also specify the matrix of transformation. Obtain its index, signature and nature of the quadratic form.

12. (i) Find the eigen value and eigen vector of the matrix

(ii) Using Cayley Hamiltonian find the inverse of

2 2 25 2 2 6x y z xy yz xz

6410

527

7411

111

112

301

135

13. Discuss the nature, index and signature of the quadratic form

14. Diagonalise the matrix by

orthogonal reduction and provide the normalized modal matrix.

15. Reduce the quadratic form 2x1x2+2x1x3-2x2x3 to the

canonical form by an Orthogonal transformation.

2 2 210 2 5 6 10 4x y z yz zx xy

8 6 2

6 7 4

2 4 3

A

136

16. (i) Find the eigen value and eigen vector of the matrix

(ii) For A = , compute the value of

, Using Cayley-

Hamilton theorem.

17. Reduce the quadratic form 3x12 + 5x2

2 + 3x32

-2x2x3+2x3x1-2x1x2 to a Canonical form by orthogonal reduction.

2 1 0

0 2 1

0 0 2

111

112

301

6 5 4 3 25 8 2 9 31 36A A A A A A I

THANK YOU

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