Eigenvalues and Eigenvectors

Transcription

1 /88 Chia-Ping Chen Department of Computer Science and Engineering National Sun Yat-sen University Linear Algebra

2 Eigenvalue Problem /88

3 Eigenvalue Equation By definition, the eigenvalue equation for matrix A is Ax = λx Note that A must be square. This equation is satisfied by a vector x that, under the transform of A, is mapped to a vector in the subspace spanned by x. Solving an eigenvalue equation is an eigenvalue problem. 3/88

4 Consider the unknowns λ and x in the eigenvalue equation for A Ax = λx A solution for λ is an eigenvalue of A. The set of eigenvalues of A is called the spectrum of A. A solution for x is an eigenvector of A. Note that if x is an eigenvector of A, then cx is also an eigenvector of A for c 0. 4/88

5 Solving an Eigenvalue Equation An eigenvalue equation Ax = λx is non-linear, as it involves product of unknowns λx. How do we find a solution? The key is to turn the above non-linear equation to systems of linear equations. Find the eigenvalues by solving a non-linear equation in λ. For each eigenvalue, find the eigenvectors with that eigenvalue by solving a system of linear equations for x. 5/88

6 Finding Eigenvalues An eigenvalue λ for A must satisfy the characteristic equation of A as follows A λi = 0 Why? x 0 such that Ax = λx x 0 such that (A λi)x = 0 (A λi) is singular A λi = 0 6/88

7 Finding Eigenvectors For an eigenvalue λ of A, the eigenspace of A with λ is {x 0 Ax = λx} This set contains the eigenvectors of A with eigenvalue λ. Apart from a null vector, the eigenspace with eigenvalue λ is the same as the nullspace of (A λi). 7/88

8 Example Solve the eigenvalue equation for A = A λi = 0 λ λ = 0 λ =, 5 λ = s = λ = s = 8/88

9 Example: Projection Matrix Solve the eigenvalue equation for P = [ ] P λi = 0 λ λ = 0 λ =, 0 λ = s = λ = 0 s = 9/88

10 Example: Complex Eigenvalues Solve the eigenvalue equation for K = 0 0 K λi = 0 λ + = 0 λ = i, i λ = i s = i λ = i s = i 0/88

11 Number of Eigenvalues A matrix of order n n has at most n distinct eigenvalues. Why? The determinant a λ... a n A λi =..... a n... a nn λ is a polynomial of λ of order n, with no more than n roots. Thus, we can denote the spectrum of A by {λ,..., λ k }, k n /88

12 Eigenvalues vs. Distinct Eigenvalues Let the spectrum of an n n matrix A be {λ,..., λ k } Then the characteristic polynomial of A can be written as k A λi = c (λ λ i ) γ i i= where k γ i = n i= Sometimes we denote the (not-necessarily-distinct) eigenvalues by λ,..., λ n /88

13 Trace and Sum of Eigenvalues The sum of the diagonal elements of a square matrix is its trace. The sum of eigenvalues of a matrix equals its trace. a λ... a n A λi =..... = c(λ λ )... (λ λ n ) a n... a nn λ Equating the λ n and λ n terms on both sides ( λ) n = cλ n c = ( ) n ( λ) n (a + + a nn ) = cλ n (( λ ) + + ( λ n )) λ + + λ n = a + + a nn 3/88

14 Determinant and Product of Eigenvalues The product of the eigenvalues of a matrix equals its determinant. Why? n A λi = ( ) n (λ λ i ) i= n A = ( ) n ( λ i ) i= n = λ i i= 4/88

15 Diagonalization and Decomposition 5/88

16 Multiplicity of Eigenvalue Let A be a matrix of order n n with spectrum {λ,..., λ k } The characteristic polynomial of A can be written as f (λ) = A λi = ( ) n γ i is the algebraic multiplicity of λ i. k i= (λ λ i ) γ i For the eigenspace of A associated with λ i N (A λ i I) {0} g i = dim N (A λ i I) is the geometric multiplicity of λ i. 6/88

17 Sum of Multiplicities For any i γ i g i Since k γ i = n i= we have k g i n i= 7/88

18 Defective Matrix An n n matrix is defective if It is non-defective if k g i < n i= k g i = n i= For example [ 0 ] 0 0 is defective, while [ 0 ] is not. 8/88

19 Set of Independent Eigenvectors For a non-defective matrix of order n n, there is a set of n linearly independent eigenvectors (i.e. a basis for R n ) of the matrix. Why? Eigenvalue λ i corresponds to an eigenspace of dimension g i, contributing g i linearly independent eigenvectors to the basis. Eigenvectors associated with distinct eigenvalues are linearly independent. So n linearly independent eigenvectors can be found, forming a basis. 9/88

20 Eigenvalue Decomposition Let s,..., s n be linearly independent eigenvectors of a non-defective matrix A with eigenvalues λ,..., λ n. Then where A = SΛS S = [ s... s n ], Λ = diag(λ,..., λ n ) This is called eigenvalue decomposition of a (non-defective) matrix. As i = λ i s i AS = SΛ A = SΛS 0/88

21 Diagonalization It follows that Λ = S AS This is called diagonalization of a (non-defective) matrix. P = [ ] [ ] [ ] = K = = i i 0 i i i 0 0 i /88

22 Common Eigenvectors Non-defective matrices A and B have a common eigenvector matrix if they commute. Suppose AB = BA. Let x be an eigenvector of B with eigenvalue λ. Consider ABx and ABx. BAx = ABx = Aλx = λax So Ax is also an eigenvector of B with eigenvalue λ, which means Ax = λ x That is, x is an eigenvector of A for some eigenvalue λ. /88

23 Simultaneous Diagonalization Non-defective matrices A and B have a common eigenvector matrix only if they commute AB = BA Suppose A and B have a common eigenvector matrix S, with A = SΛ S, B = SΛ S Then AB = SΛ S SΛ S = SΛ Λ S = SΛ Λ S = SΛ S SΛ S = BA 3/88

24 Difference Equations 4/88

25 Power of a Matrix Suppose A is diagonalizable. The power of A can be simplifies by A k = (SΛS ) k = (SΛS )(SΛS )... (SΛS ) = SΛ(S S)Λ(S S)... (S S)ΛS = SΛ k S Note that SΛ k S is much easier to compute than A k. 5/88

26 Fibonacci Sequence Fibonacci numbers are defined recursively by F k+ = F k + F k We will look at the Fibonacci sequence which starts with F 0 = 0, F = The first few Fibonacci numbers are /88

27 Representation of Recursion with Matrix Combining the recurrence relation with a trivial equality F k+ = F k + F k F k = F k + 0 F k we have a system of matrix equations for the Fibonacci numbers Fk+ F k = Fk 0 F k with initial condition F F 0 = 0 7/88

28 Simplification by Matrix Define u k = Fk+ F k, A = 0 Let S be an eigenvector matrix of A and Λ be the associated eigenvalue matrix. We have u k = Au k = A(Au k ) = = A k u = A k u 0 = SΛ k S u 0 = [ s s ] [ λ k ] c λ k, c c c = S u 0 = c λ k s + c λ k s 8/88

29 Eigenvalue Problem of A Eigenvalues Eigenvectors A λi = 0 λ = + 5, λ = 5 [ λi λ i Fitting the initial condition c c = S u 0 = ] s i = 0 s i = [ λ λ ] [ 0 λi ] λ = λ λ λ 9/88

30 Fibonacci Numbers Thus u k = c λ k s + c λ k s = λ k λ λ λ So the Fibonacci numbers are λ λ λ k λ ( ) F k = λ k λ λ λ k = ( ) k ( ) k + 5 5, k = 0,, /88

31 A Markov Process Every year of the people outside Asia move in, and 0 people inside Asia move out. 0 of the Let y k be the number of people outside Asia and z k be the number of people inside Asia at the end of year k. Then yk+ z k yk = z k or simply x k+ = Ax k, k = 0,,... Matrix A is Markov, as the elements of A are non-negative the elements in every column of A sum to 3/88

32 Population as a Function of k Using the eigenvalue decomposition of A is we have A = SΛS = [ ] x k = Ax k = A(Ax k ) = = A k x 0 = SΛ k S x 0 That is yk z k = [ = (y 0 + z 0 )() k [ 3 3 ] k 0 y z 0 ] + (y 0 z 0 )(0.7) k [ 3 3 ] 3/88

33 Steady State We can see that x k converges to the first term x = y z = (y 0 + z 0 ) [ 3 3 ] Note that x is the eigenvector of A with eigenvalue Ax = x Thus x is called the steady state since it stays if entered. 33/88

34 Linear Differential Equations 34/88

35 Linear Differential Equation Consider an unknown function u(t) that satisfies where a is a constant. du(t) dt = au(t) The equation is a differential equation as it involves a derivative. What is the solution of such a linear differential equation? 35/88

36 Solution It is easy to verify that u(t) = u(0)e at where u(0) is the initial value. 36/88

37 System of Linear Differential Equations Next, consider two unknown functions, v(t) and w(t), that satisfy a system of linear differential equations where a, b, c, d are constants. dv = av + bw dt dw = cv + dw dt 37/88

38 Matrix Representation A system of linear differential equations can be represented by matrices. For the system of linear differential equations in the previous slide, we have du dt = Au where v a b u =, A = w c d 38/88

39 Sneak Preview It turns out the solution of can be expressed as du dt = Au u(t) = e At u(0) with an appropriate definition of matrix exponential e At 39/88

40 Working out an Example Consider the case with That is u(t) = v(t), A = w(t) dv = 4v 5w dt dw = v 3w dt /88

41 Looking for a Solution An exponential function, after differentiation, is still an exponential function. So we assume v(t) = e αt y, w(t) = e αt z Substitute into the differential equations, we have Eliminating e αt, we get αe αt y = 4e αt y 5e αt z αe αt z = e αt y 3e αt z αy = 4y 5z αz = y 3z which has the form of an eigenvalue equation Ax = αx 4/88

42 General Solution via A A solution of the differential equations consists of y, z and α, which are related to eigenvector and eigenvalue of A. If λ is an eigenvalue of A with associated eigenvectors x, then d ( e λt x ) = λe λt x = e λt Ax = A ( e λt x ) dt By linearity, any linear combination u = i c i e λ i t x i where λ i is an eigenvalue and x i is an eigenvector, satisfies the differential equation du dt = Au 4/88

43 The Solution Given initial condition u(0) = u 0, the solution of is u(t) = S du dt = Au where S is an eigenvector matrix of A. e λ t 0 0 e λ t S u 0 e u(t) = c e λt s +c e λt λ t 0 c c s = S 0 e λ t u(0) = u c 0 = S c c e = S λ t 0 u c 0 u(t) = S 0 e λ t S u 0 43/88

44 An Example of Initial Condition Solve du dt = Au, u = v(t), A = w(t) with initial condition v(0) = 8, w(0) = 5. A λi = 0 λ =, s = e t 0 u(t) = S 0 e t S u 0 = ] ] 4 5 3, s = [ 5 e t 0 0 e t ] [ 5 5 ] 8 5 = 3e t [ + e t [ 5 44/88

45 Matrix Exponential Let A be a square matrix. Matrix exponential is defined by e A k=0 A k k! = I + A + A! + A3 3! +... It has the same form as a scalar exponential e x = k=0 x k k! = + x + x! + x 3 3! /88

46 Diagonal Matrix For a diagonal matrix D = diag(d,..., d n ) e D = I + D + D! +... (( = diag + d + d! +... = diag ( ) e d,..., e dn ),..., ( + d n + d n! +... )) 46/88

47 Diagonalizable Matrices For a diagonalizable matrix A = SΛS e A = I + A + A! + = I + SΛS + (SΛS ) )! = S (I + Λ + Λ! +... S = Se Λ S /88

48 Solution of Differential Equation in Matrix The solution for is u(t) = i du dt = Au c i e λ i t s i = Se Λt S u(0) = e At u(0) since ( Se Λt S = S I + Λt + Λ t ) +...! S = I + (SΛS )t + (SΛS ) t +...! = I + At + A t +...! = e At 48/88

49 Higher-order Linear Differential Equations A high-order linear differential equation can be decomposed into a system of first-order linear differential equations. For example, a third-order linear differential equation can be converted to where y + by + cy = 0 u = Au y 0 0 u = v, A = 0 0 w 0 c b 49/88

50 Linear Partial Differential Equations A partial differential equation can be transformed to a system of linear differential equations if one variable is discretized. Consider a heat equation u t = u x When x is discretized, we have du dt = Au by defining u u =, A = u N 50/88

51 Complex Matrix 5/88

52 Vectors with Complex Numbers Inner product (x, y) = x y + + x n y n Length x = (x, x) = x x + + x n x n = x + + x n Orthogonality (x, y) = 0 x y 5/88

53 Example Decide the inner product, lengths, and orthogonality for x = 3 i + i and 5 y = i 53/88

54 Hermitian The Hermitian of matrix A is defined as A H = (A ) T Basically Hermitian = conjugation + transposition It can be shown that ( ) A H H = A (AB) H = B H A H 54/88

55 Hermitian and Inner Product The inner product of two vectors can be written as (x, y) = x H y It follows that (x, Ay) = x H Ay = (A H x) H y = (A H x, y) (Ax, y) = (Ax) H y = x H A H y = (x, A H y) 55/88

56 Hermitian Matrix By definition, a matrix A is Hermitian if A H = A For example is Hermitian, since [ ] 3 3i A = 3 + 3i 5 [ ] T [ ] A H = (A ) T 3 + 3i 3 3i = = = A 3 3i i 5 56/88

57 Properties of a Hermitian Matrix Let A be Hermitian. x H Ax is real for any x x H Ax = x H A H x = x H A H (x H ) H = (x H Ax) H = (x H Ax) The eigenvalues of A are real As i = λ i s i s H i As i = λ i s H i s i λ i = sh i As i s H i s i R Eigenvectors of A associated with distinct eigenvalues are orthogonal (As, s ) = (s, As ) (λ λ )(s, s ) = 0 (s, s ) = 0 57/88

58 Example [ ] 3 3i A = 3 + 3i 5 A λi = 0 λ = 8, (A 8I) s = 0 s = + i (A + I) s = 0 s = ( ) +i ( (s, s ) = + ( + i) + i ) = 0 58/88

59 Unitary Matrix By definition, a matrix U is unitary if U = U H i.e. UU H = U H U = I Vector length is invariant under transformation by U Ux = (Ux, Ux) = (x, U H Ux) = (x, x) = x The eigenvalues are of unit modulus Us i = λ i s i s i = Us i = λ i s i = λ i s i λ i = Eigenvectors associated with distinct eigenvalues are orthogonal (Us, Us ) = (s, U H Us ) = (s, s ) (Us, Us ) = (λ s, λ s ) = λ λ (s, s ) ( λ λ )(s, s ) = 0 (s, s ) = 0 59/88

60 Examples of Unitary Matrix Rotation matrix Fourier matrix Permutation matrix cos t sin t U = sin t cos t F = ω ω ω 3 ω ω 4 ω 6 ω 3 ω 6 ω P = /88

61 Skew-Hermitian By definition, a matrix K is skew-hermitian if K H = K Let [ ] [ 3 3i i K = ia = i = 3 + 3i i ] 3 + 3i 5i Indeed K H = (ia) H = ia H = ia = K [ K H i = 3 + 3i ] H 3 + 3i = 5i i 3 3i = K 3 3i 5i 6/88

62 Similarity Transformation 6/88

63 Similar Matrices By definition, matrix B is similar to matrix A if Similarity relation is denoted by M such that B = M AM B A Note that similarity is an equivalence relation. Thus, we also say A and B are similar. 63/88

64 Example A = 0, M 0 0 = M = b, M 0 = b, M = 0 A B = M AM = A B = M AM = [ ] b 0 0 [ ] 64/88

65 Eigenvalues of Similar Matrices If matrix A and matrix B are similar, then they have the same eigenvalues. Suppose A and B are similar, with B = M AM Let λ be an eigenvalue of B. Then B λi = 0 M AM λm M = 0 M (A λi)m = 0 A λi = 0 So λ is an eigenvalue of A. 65/88

66 Eigenvectors of Similar Matrices An eigenvector x of A corresponds to an eigenvector y = M x of B. Furthermore, x and y are associated with the same eigenvalue. Let λ be an eigenvalue of A with an associated eigenvector x. Then Ax = λx M Ax = λm x ( M AM ) ( M x ) = λ ( M x ) By = λy 66/88

67 Example A = 0, B 0 0 = b, B 0 0 = [ ] The eigenvalues of A are and 0, with eigenvectors 0, 0 B A. The eigenvectors of B are M = 0 0, M B A. The eigenvectors of B are ] M = 0 [, M 0 = 0 = b 67/88

68 Change of Basis The matrix representation for a linear transformation depends on the used basis. When there is a change of basis, the matrix changes to a similar matrix (with a similarity transformation). 68/88

69 Representation of a Vector Through a basis, a vector in a vector space can be represented by a column. The representation depends on the used basis. Specifically, suppose B = {v,..., v n } and then x = c v + + c n v n [x] B = {c i } 69/88

70 Representation of a Linear Transformation Through a basis for the domain D and a basis for the range R, a linear transformation from D to R can be represented by a matrix. The matrix representation of a linear transformation T : D R depends on the used bases. Here we use notation [T] UV for the matrix representation for T using basis U for D and basis V for R. 70/88

71 Construction of Matrix Representation Let U = {u,..., u n } be a basis for D and V = {v,..., v m } be a basis for R. Consider linear transformation Suppose T : D R m T(u j ) = a ij v i, i= j =,..., n Using basis U for D and basis V for R, we have [T] UV = {a ij } so that T(x) = y [y] V = [T] UV [x] U for x D and y R. 7/88

72 Proof Using basis U for D and basis V for R, suppose n m x = x j u j, T(x) = y = y i v i j= i= Using linearity of T n n m m n T(x) = x j T(u j ) = x j a ij v i = a ij x j v i j= j= i= i= j= Thus i.e. y i = a ij x j [y] V = [T] UV [x] U 7/88

73 Two Bases Consider a linear transformation within a vector space T : S S With basis G = {V,..., V n }, the matrix representation for T is b... b n n T(V j ) = b ij V i [T] GG =..... = B i= b n... b nn With basis H = {v,..., v n }, the matrix representation for T is a... a n n T(v j ) = a ij v i [T] HH =..... = A i= a n... a nn 73/88

74 Identity Transformation Consider the identity transformation I(x) = x Obviously I is linear, with [I] GG = I, [I] HH = I. How about [I] GH? Suppose the basis vectors are related by We have so n V j = m ij v i, j =,..., n i= n I(V j ) = V j = m ij v i, j =,..., n i= [I] GH = {m ij } = M 74/88

75 Change of Vector Representation How are [x] G = {x i } and [x] H = {x i } related? n n n n n x = x j V j = x j m ij v i = m ij x j v i j= j= i= i= j= n = x i v i i= Thus x i = j m ij x j in matrix form [x] H = [I] GH [x] G We also have [x] G = [I] GH [x] H = [I] HG [x] H [I] HG = [I] GH 75/88

76 Change of Matrix Representation How are [T] GG and [T] HH related? ( ) T(V j ) = T m ij v i = m ij (T(v i )) = i i k = b ij V i = b ij m ki v k, i i k i Thus a ki m ij = m ki b ij, j, k i i In matrix form [T] HH [I] GH = [I] GH [T] GG m ij a ki v k i 76/88

77 Relation to Similarity Transformation The identity [T] HH [I] GH = [I] GH [T] GG can be re-arranged as similarity transformation [T] HH = [I] GH [T] GG [I] GH [T] GG = [I] HG [T] HH [I] HG [T] GG = [I] HG [T] HH [I] GH (B = M A M) 77/88

78 Eigenvector of a Linear Transformation Consider a linear transformation T within a vector space. An eigenvector of T is defined by T(s) = λs The matrix representation of T using an eigenbasis G = {s,..., s n } consisting of eigenvectors of T is particular simple: [T] GG = diag(λ,..., λ n ) since Ts i = λ i s i 78/88

79 Eigenvalue Decomposition Let H = {v,..., v n } be another basis. Suppose s j = i s ij v i Then [I] GH = {s ij } = S The representation of T with H can be constructed by [T] HH = [I] GH [T] GG [I] GH 79/88

80 Example: Projection Let T be the projection to the line L at angle θ to x-axis. With an eigenbasis [T] GG = Λ = Let H be the standard basis. Then cos θ sin θ [I] GH = S = sin θ cos θ [ cos [T] HH = A = SΛS = ] θ cos θ sin θ cos θ sin θ sin θ 80/88

81 Normal Matrix and Orthonormal Eigenbasis 8/88

82 Schur Lemma For any matrix A, there exists a unitary matrix U such that T = U H AU is an upper-triangular matrix. Since A U H AU = T A is similar to an upper-triangular matrix. 8/88

83 Normal Matrix By definition, a matrix is normal if it commutes with its Hermitian, i.e. NN H = N H N 83/88

84 Diagonalizaility of a Normal Matrix A normal matrix can be diagonalized. Let A be a normal matrix. According to Schur lemma, there exists a unitary matrix U such that is upper-triangular. Furthermore T = U H AU TT H = U H AU(U H AU) H = U H AA H U = U H A H AU = U H A H UU H AU = T H T 84/88

85 Consider the first diagonal element Similarly (TT H ) = (T H T) t k tk = tkt k = t k k t k = 0, k > (TT H ) = (T H T) t k tk = tkt k = t k k t k = 0, k > The same argument applies to the remaining rows. Thus, T is diagonal, i.e., A is diagonalizable. 85/88

86 Complete Orthonormal Eigenvectors A normal matrix has sufficient orthonormal eigenvectors for a basis. Since T is diagonal U H AU = T AU = UT Au i = t ii u i so u i is an eigenvector of A associated with eigenvalue λ i = t ii. Since U is unitary U H U = I so the vectors u,..., u n are orthonormal. 86/88

87 Spectral Theorem A real symmetric matrix A can be factorized by A = QΛQ T where Λ is real and diagonal Q is real and orthogonal A is Hermitian, so the eigenvalues are real. The eigenvectors are also real, and orthonormal. 87/88

88 Example A = [ ] = [ ] [ ] [ ] = 3 [ ] + [ ] 88/88