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Eigenval ues and Eigenvec tors

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### Transcript of Maths-->>Eigenvalues and eigenvectors

Eigenvalues and

Eigenvectors

Eigenvalues and Eigenvectors

• If A is an n x n matrix and λ is a scalar for which Ax = λx has a nontrivial solution x ∈ ⁿ, then ℜ λ is an eigenvalue of A and x is a corresponding eigenvector of A.

Ax=λx=λIx (A-λI)x=0

• The matrix (A-λI ) is called the characteristic matrix of a where I is the Unit matrix.

• The equation det (A-λI )= 0 is called characteristic equationof A and the roots of this equation are called the eigenvalues of the matrix A. The set of all eigenvectors is called the eigenspace of A corresponding to λ. The set of all eigenvalues of a is called spectrum of A.

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.,

is called the characteristic equation of A.

IλA

0

λa...aa............a...λaaa...aλa

λA

nnn2n1

2n2221

1n1211

I

• 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.

•X = is called an eigen vector or latent vector

0k...λkλkλ1)( n2n

21n

1nn

ij

4

2

1

x...xx

5

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.

6

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 λ1,...,

λ1,

λ1

pn

p2

p1 λ...,,λ,λ

Algebraic & Geometric Multiplicity

• If the eigenvalue λ of the equation det(A-λI)=0 is repeated n times then n is called the algebraic multiplicity of λ. The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding matrix A- λI and is known as geometric multiplicity of eigenvalue λ.

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...aaa...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 - λIa - 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 A1A =- [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

211121

112

tion)simplifica(on049λ6λλor

0λ211

1λ2111λ2

i.e.,0λIA

23

Example 1:-

To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1)

Now,

222121212221212222

211121

112

655565

556

655565

556

211121

112

211121

112

23

2

AAA

A

A3 -6A2 +9A – 4I = 0 = - 6 + 9

-4

=

This verifies Cayley – Hamilton theorem.

222121212221212222

655565

556

211121

112

100010001

0000000000

15

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

311131113

41

311131113

100010001

9211121

1126

655565

5564

1

1

A

A

16

Given find Adj A by using Cayley –

Hamilton theorem.

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

113110121

A

tion)simplifica(on035λ3λλor

0λ1131λ101-2λ1

i.e.,0λIA

23

Example 2:-

17

By Cayley – Hamilton theorem, A should satisfyA3 – 3A2 + 5A + 3I = 0

Pre – multiplying by A-1 , we get A2 – 3A +5I +3A-1 = 0

339330363

3A

146223452

113110121

113110121

A.A ANow,

(1)...5I)3A(A31A

2

21-

18

173143110

31

500050005

339330363

146223452

31AFrom(1),

1

1

1

19

173143110

173143110

3113110121

ANow,

Similarity of Matrix

• If A & B are two square matrices of order n then B is said to be similar to A, if there exists a non-singular matrix P such that,B= P-1AP

1. Similarity matrices is an equivalence relation.2. Similarity matrices have the same determinant.3. Similar matrices have the same characteristic polynomial

and hence the same eigenvalues. If x is an eigenvector corresponding to the eigenvalue λ, then P-1x is an eigenvector of B corresponding to the eigenvalue λ where B= P-1AP.

Diagonalization

• A matrix A is said to be diagonalizable if it is similar to diagonal matrix.

• A matrix A is diagonalizable if there exists an invertible matrix P such that P-1AP=D where D is a diagonal matrix, also known as spectral matrix. The matrix P is then said to diagonalize A of transform A to diagonal form and is known as modal matrix.

22

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.

23

Reduce the matrix A = to diagonal form by

similarity transformation. Hence find A3.

Solution:- Characteristic equation is

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

300120

211

0

λ-3001λ-20

21λ1-

Example:-

24

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

vector then

3

2

1

xxx

001

kX

x0x,kx02x0xx02xx

000

xxx

200110

2100X)I(A

11

3211

3

32

32

3

2

1

1

25

Corresponding to λ = 2, let X2 = be the eigen

vector then,

3

2

1

xxx

01-1

kX

x-kx,kx0x

0x02xxx

000

xxx

100100

211-

0X)(A

22

32221

3

3

321

3

2

1

2

0,

I2

26

Corresponding to λ = 3, let X3 = be the eigen

vector then,

3

2

1

xxx

22-3

kX

xk-x,kx

0x02xxx

000

xxx

00011-0

212-

0X)(A

33

13332

3

321

3

2

1

3

3

2

23,

2

I3

k

x

27

Hence modal matrix is

210011-0

21-11

1-00220122-

2M

20021-0

311M

1

28

Now, since D = M-1AM

=> A = MDM-1

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

= MD2M-1 [since M-1M = I]

300020001

20021-0

311

300120

211

210011-0

2111

AMM 1

29

Similarly, A3 = MD3M-1

=

A3 =

270019-80327-1

210011-0

2111

2700080001

20021-0

311

Orthogonally Similar Matrices

• If A & B are two square matrices of order n then B is said to be orthogonally similar to A, if there exists orthogonal matrix P such that B= P-1AP

Since P is orthogonal,P-1=PT

B= P-1AP=PTAP1. A real symmetric of order n has n mutually orthogonal real

eigenvectors.2. Any two eigenvectors corresponding to two distinct

eigenvalues of a real symmetric matrix are orthogonal.

31

Diagonalise the matrix A = by means of an

orthogonal transformation.Solution:- Characteristic equation of A is

204060402

66,2,λ0λ)16(6λ)λ)(2λ)(6(2

0λ204

0λ6040λ2

Example :-

32

I

1

1 2

3

1

1

2

3

1 3

2

1 3

1 1 2 3 1

1 1

xwhenλ = -2,let X = x be theeigenvector

x

then (A +2 )X = 0

4 0 4 x 00 8 0 x = 04 0 4 x 0

4x + 4x = 0 ...(1)8x = 0 ...(2)

4x + 4x = 0 ...(3)x = k ,x = 0,x = -k

1X = k 0

-1

33

2

2I

0

1

2

3

1

2

3

1 3

1 3

1 3 2

2 2 3

xwhenλ = 6,let X = x be theeigenvector

x

then (A - 6 )X = 0

-4 0 4 x 00 0 x = 04 0 -4 x 0

4x + 4x = 04x - 4x = 0

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

.1

emselvesand also each is orthogonal with X

34

2 3

3 1

3 2

3

1 αLet X = 0 and let X = β

1 γ

Since X is orthogonal to Xα - γ = 0 ...(4)

X is orthogonal to Xα+ γ = 0 ...(5)

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

Taking β =1, X = 10

1 1 0Modal matrix is M = 0 0 1

-1 1

0

35

The normalised modal matrix is1 1 02 2

N = 0 0 11 1- 02 21 10 - 1 1 02 2 2 0 4 2 21 1D =N'AN = 0 0 6 0 0 0 12 2

4 0 2 1 1- 00 1 02 2

-2 0 0D = 0 6 0 which is the required diagonal matrix

0 0 6.

36

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.

37

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(b21awherexxaxxb

baandaawherebb2aSuppose

jiijijji

n

1j

n

1iijji

n

1j

n

1iij

iiiijiijijijij

38

Hence every quadratic form can be written as

.x,...,x,xXandaAwheresymmetric,alwaysisAmatrixthethatso

AX,X'xxa

n21ij

ji

n

1j

n

1iij

)(by2hxy ax (i) 22

ii

yx

bhha

y][xby2hxy ax (i) 22

39

wzyx

dnmlncgfmgbhlfha

wzyx

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

zyx

cgfgbhfha

zyx2fzx 2gyz 2hxy cz by ax (ii)

222

222

40

Theorem(1): A quadratic form can always be expressed with respect to a given coordinate system as

where A is a unique symmetric matrix.Theorem2: Two symmetric matrices A and B represent

the same quadratic form if and only if B=PTAP where P is a non-singular matrix.

AxxY T

Nature of Quadratic FormA 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.

42

Find the nature of the following quadratic formsi. x2 + 5y2 + z2 + 2xy + 2yz + 6zxii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xySolution:- iii. The matrix of the quadratic form is

113151311

A

Example :-

43

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.

311151

113A

Linear Transformation of a Quadratic Form

44

• 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.

45

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.

46

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

OrDiagonalise the quadratic form 3x2 + 3z2 + 4xy +

8xz + 8yz by linear transformations and write the linear transformation.

OrReduce the quadratic form 3x2 + 3z2 + 4xy + 8xz +

8yz into the sum of squares.

Example:-

47

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.

344402423

100010001

344402423

100010001

344402423

48

21 31OperatingR ( 2 / 3),R ( 4 / 3)(for A onL.H.S.andpre factor on R.H.S.),weget

3 2 4 1 0 0 1 0 04 4 20 1 0 A 0 1 03 3 3

0 0 14 7 40 0 13 3 3

10001034

321

A

1034

0132

001

37

340

34

340

423

getweR.H.S),onfactorpostandL.H.S.onA(for4/3)(C2/3),(C Operating 3121

49

10001034

321

A

112

0132

001

10034

340

003

getwe(1),R Operating 32

APP'1,343,Diagor

100110

2321

A

112

0132

001

100

0340

003

getwe(1),C Operating 32

50

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

0132

001

yyy

zyx

23

22

21

3

2

1

321

yy343y

yyy

100

0340

003yyyAP)Y(P'Y'

THANK YOU