# Maths-->>Eigenvalues and eigenvectors

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Engineering

Published on April 26, 2014

Author: Jaydev_D_Kishnani

Source: slideshare.net

## Description

It is the topic of maths-2 in the second semester of engineering !!
I hope it is useful and satisfactory !!

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 ∈ ℜⁿ, then 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...λaa a...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)( n 2n 2 1n 1 nn ij 4 2 1 x ... x x

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 λ 1 ,..., λ 1 , λ 1 p n p 2 p 1 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...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-2 0 1 2 n n n-1 n-2 0 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) In-1 n-2 n-3 -1 0 1 2 n-1 n -1 n-1 n-2 n-3 0 1 2 n-1 n 0 =p A +p A +p A +...+p +p A 1 A =- [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

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 311 131 113 4 1 311 131 113 100 010 001 9 211 121 112 6 655 565 556 4 1 1 A A

16 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:-

17 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.AANow, (1)...5I)3A(A 3 1 A 2 21-

18 AAAAdj. A AAdj. Athat,knowWe 173 143 110 3 1 500 050 005 339 330 363 146 223 452 3 1 AFrom(1), 1 1 1

19 173 143 110 AAdj. 173 143 110 3 1 3)(AAdj. 3 113 110 121 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. 300 120 211 0 λ-300 1λ-20 21λ1- Example:-

24 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

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

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

27 Hence modal matrix is 2 1 00 11-0 2 1- 11 M MAdj. M 1-00 220 122- MAdj. 2M 200 21-0 311 M 1

28 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 1 00 11-0 2 1 11 AMM 1

29 Similarly, A3 = MD3M-1 = A3 = 2700 19-80 327-1 2 1 00 11-0 2 1 11 2700 080 001 200 21-0 311

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=PTAP 1. 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 204 060 402 66,2,λ 0λ)16(6λ)λ)(2λ)(6(2 0 λ204 0λ60 40λ2 Example :-

32 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 betheeigenvector 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

33 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 betheeigenvector 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

34 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

35 The normalised modal matrix is 1 1 0 2 2 N = 0 0 1 1 1 - 0 2 2 1 1 0 - 1 1 02 2 2 0 4 2 2 1 1 D =N'AN = 0 0 6 0 0 0 1 2 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 .

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. Quadratic Forms

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 1i jiijjiij bbwhere,xxb ).b(b 2 1 awherexxaxxb baandaawherebb2aSuppose jiijijji n 1j n 1i ijji n 1j n 1i ij iiiijiijijijij

38 Hence every quadratic form can be written as getweform,matrixin formsquadraticofexamplessaidabovethewritingNow .x,...,x,xXandaAwhere symmetric,alwaysisAmatrixthethatso AX,X'xxa n21ij ji n 1j n 1i ij y x bh ha y][xby2hxyax(i) 22

39 w z y x dnml ncgf mgbh lfha wzyx 2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii) z y x cgf gbh fha zyx2fzx2gyz2hxyczbyax(ii) 222 222

40 Two Theorems On Quadratic Form 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 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.

42 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 :-

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. 311 151 113 A

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. 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:-

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. 344 402 423 100 010 001 344 402 423 100 010 001 344 402 423

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 0 4 4 2 0 1 0 A 0 1 0 3 3 3 0 0 1 4 7 4 0 0 1 3 3 3 100 010 3 4 3 2 1 A 10 3 4 01 3 2 001 3 7 3 4 0 3 4 3 4 0 423 getweR.H.S),onfactorpostandL.H.S.onA(for 4/3)(C2/3),(COperating 3121

49 100 010 3 4 3 2 1 A 112 01 3 2 001 100 3 4 3 4 0 003 getwe(1),ROperating 32 APP'1, 3 4 3,Diagor 100 110 2 3 2 1 A 112 01 3 2 001 100 0 3 4 0 003 getwe(1),COperating 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 01 3 2 001 y y y z y x 2 3 2 2 2 1 3 2 1 321 yy 3 4 3y y y y 100 0 3 4 0 003 yyyAP)Y(P'Y'

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