Problems of Eigenvalues/Eigenvectors

67 Problems of Eigenvalues/Eigenvectors Reveiw of Eigenvalues and Eigenvectors Gerschgorin s Disk Theorem Power and Inverse Power Methods Jacobi Transform for Symmetric Matrices Spectrum Decomposition Theorem Singular Value Decomposition with Applications QR Iterations for Computing Eigenvalues A Markov Process e A and Differential Equations Other Topics with Applications

68 Definition and Examples Let A R n n. If v 0 such that Av = λv, λ is called an eigenvalue of matrix A, and v is called an eigenvector corresponding to (or belonging to) the eigenvalue λ. Note that v is an eigenvector implies that αv is also an eigenvector for all α 0. We define the Eigenspace(λ) as the vector space spanned by all of the eigenvectors corresponding to the eigenvalue λ. Examples: 2 0. A = 0 2 2. A = 0 3 3. A = 3 0 4. A = 0 3 0 5. B = 8 3 6. C = 3, λ =2,u =, λ =2,u =, λ =4,u =, λ = j, u = 0 0, thenλ =3,u =, λ 2 =,u 2 =, λ 2 =,u 2 =, λ 2 =2,u 2 = j 0, λ 2 = j, u 2 =, thenτ =4,v = 5 2 5 2 2... j ; λ 2 =, u 2 = ; τ 2 =2,v 2 =, j =. 0. 2 2. Ax = λx (λi A)x = 0, x 0 det(λi A) =P (λ) =0.

69 Gershgorin s Disk Theorem Note that u i 2 =and v i 2 =fori =, 2. Denote U =[u, u 2 ]andv =[v, v 2 ], then 3 0 4 0 U BU =, V CV = 0 0 2 Note that V t = V but U t U. Let A R n n,thendet(λi A) is called the characteristic polynomial of matrix A. Fundamental Theorem of Algebra A real polynomial P (λ) =λ n + a n λ n + + a 0 of degree n has n roots {λ i } such that ( n ) ( P (λ) =(λ λ )(λ λ 2 ) (λ λ n )=λ n n ) λ i λ n + +( ) n λ i i= i= n i= λ i = n i= a ii = tr(a) n i= λ i = det(a) Gershgorin s Disk/Circle Theorem Every eigenvalue of matrix A R n n lies in at least one of the disks D i = {x x a ii a ij }, i n j i 3 Example: B = 0 4, λ,λ 2,λ 3 S S 2 D 3, where S = {z z 3 2}, S 2 = 2 2 5 {z z 4 }, S 3 = {z z 5 4}. Notethatλ =6.566, λ 2 =3.0000, λ 3 =2.4383. Amatrixissaidtobediagonally dominant if j i a ij < a ii, i n. A diagonally dominant matrix is invertible.

70 Theorem: Let A, P R n n, with P nonsingular, then λ is an eigenvalue of A with eigenvector x iff λ is an eigenvalue of P AP with eigenvector P x. Theorem: Let A R n n and let λ be an eigenvalue of A with eigenvector x. Then (a) αλ is an eigenvalue of matrix αa with eigenvector x (b) λ μ is an eigenvalue of matrix A μi with eigenvector x (c) If A is nonsingular, then λ 0andλ is an eigenvalue of A with eigenvector x Definition: A matrix A is similar to B, denotebya B, iff there exists an invertible matrix U such that U AU = B. Furthermore, a matrix A is orthogonally similar to B, iff there exists an orthogonal matrix Q such that Q t AQ = B. Theorem: Two similar matrices have the same eigenvalues, i.e., A B λ(a) = λ(b).

7 Diagonalization of Matrices Theorem: Suppose A R n n has n linearly independent eigenvectors v, v 2,..., v n corresponding to eigenvalues λ, λ 2,..., λ n. Let V = [v, v 2,..., v n ], then V AV = diag[λ,λ 2,...,λ n ]. If A R n n has n distinct eigenvalues, then their corresponding eigenvectors are linearly independent. Thus, any matrix with distinct eigenvalues can be diagonalized. Not all matrices have distinct eigenvalues, therefore not all matrices are diagonalizable. Nondiagonalizable Matrices A = 2 0 0 2 0 0 2, B = 0 0 2 0 3 5 2 Diagonalizable Matrices C =, D = 2 0, E = 0 2 0 0 2 2 0 3 0, K = 0 Spectrum Decomposition Theorem* Every real symmetric matrix can be diagonalized.

72 Similarity transformation and triangularization Schur s Theorem: A R n n, an orthogonal matrix U such that U t AU = T is upper- Δ. The eigenvlues must be shared by the similarity matrix T and appear along its main diagonal. Hint: By induction, suppose that the theorem has been proved for all matrices of order n, and consider A R n n with Ax = λx and x 2 =,then ahouseholder matrix H such that H x = βe, e.g., β = x 2, hence H AH t e = H A(H e )=H A(β x)=h β Ax = β λ(h x)=β λ(βe )=λe Thus, λ H AH t = O A () Spectrum Decomposition Theorem: Every real symmetric matrix can be diagonalized by an orthogonal matrix. Q t AQ =ΛorA = QΛQ t = n i= λ i q i q t i Definition: A symmetric matrix A R n n is nonnegative definite if x t Ax 0 x R n, x 0. Definition: A symmetric matrix A R n n is positive definite if x t Ax > 0 x R n, x 0. Singular Value Decomposition Theorem: Each matrix A R m n can be decomposed as A = UΣV t, where both U R m m and V R n n are orthogonal. Moreover, Σ R m n = diag[σ,σ 2,...,σ k, 0,...,0] is essentially diagonal with the singular values satisfying σ σ 2... σ k > 0. A = UΣV t = k i= σ i u i v t i Example: 2 A =, B = 2 0 0 0 0 0 0, C = 0 0

73 A Jacobi Transform (Givens Rotation) J(i, k; θ) = 0 0.... 0 0 c s 0..... 0 s c 0 0.... 0 0 0 J hh =ifh i or h k, wherei<k J ii = J kk = c =cosθ J ki = s = sin θ, J ik = s =sinθ Let x, y R n,theny = J(i, k; θ)x implies that y i = cx i + sx k y k = sx i + cx k c = x i x 2 i +x 2 k, s = x k x 2 i +x 2 k, x = 2 3 4, cos θ sin θ = / 5 2/ 5, then J(2, 4; θ)x = 20 3 0

74 Jacobi Transforms (Givens Rotations) The Jacobi method consists of a sequence of orthogonal similarity transformations such that JKJ t K t J2J t AJ t J 2 J K J K =Λ where each J i is orthogonal, so is Q = J J 2 J K J K. Each Jacobi transform (Given rotation) is just a plane rotation designed to annihilate one of the off-diagonal matrix elements. Let A =(a ij ) be symmetric, then B = J t (p, q, θ)aj(p, q, θ), where b rp = ca rp sa rq b rq = sa rp + ca rq for r p, r q for r p, r q b pp = c 2 a pp + s 2 a qq 2sca pq b qq = s 2 a pp + c 2 a qq +2sca pq b pq =(c 2 s 2 )a pq + sc(a pp a qq ) To set b pq =0,wechoosec, s such that α = cot(2θ) = c2 s 2 2sc = a qq a pp 2a pq () For computational convenience, let t = s c,thent2 +2αt = 0 whose smaller root (in absolute sense) can be computed by t = sgn(α) α2 ++ α, and c =, s = ct, τ = s +t 2 +c (2) Remark b pp = a pp ta pq b qq = a qq + ta pq b rp = a rp s(a rq + τa rp ) b rq = a rq + s(a rp τa rq )

75 Algorithm of Jacobi Transforms to Diagonalize A A (0) A for k = 0,,, until convergence Let a (k) pq = Max i<j{ a (k) ij } Compute α k = a(k) t = qq a (k) pp 2a (k) pq sgn(α) α 2 ++ α, solve cot(2θ k )=α k for θ k. c = +t 2,, s = ct τ = s +c A (k+) J t k A(k) J k,wherej k = J(p, q, θ k ) endfor

76 Convergence of Jacobi Algorithm to Diagonalize A Proof: Since a (k) pq a (k) ij for i j, p q, then a pq (k) 2 off(a (k) /2N, wheren = n(n ),and 2 off(a (k) )= ( ) n (k) 2, i j a ij the sum of square off-diagonal elements of A (k) Furthermore, off(a (k+) ) = off(a (k) ) 2 ( a (k) pq ) 2 ( ) 2 +2 a (k+) pq = off(a (k) ) 2 ( ) 2 a pq (k), since a (k+) pq =0 off(a (k) ) ( N ), since a (k) pq 2 off(a (k) /2N Thus off(a (k+) ) ( N ) k+ off(a (0) ) 0 as k Example: 4 2 0 c s 0 A = 2 3, J(, 2; θ) = s c 0 0 2 0 0 Then 4c 2 4cs +3s 2 2c 2 + cs 2s 2 s A () = J t (, 2; θ)aj(, 2; θ) = 2c 2 + cs 2s 2 3c 2 +4cs +4s 2 c s c Note that off(a () )=2< 0 = off(a (0) )=off(a)

77 Example for Convergence of Jacobi Algorithm A (0) =.0000 0.5000 0.2500 0.250 0.5000.0000 0.5000 0.2500 0.2500 0.5000.0000 0.5000 0.250 0.2500 0.5000.0000, A () =.5000 0.0000 0.5303 0.2652 0.0000 0.5000 0.768 0.0884 0.5303 0.768.0000 0.5000 0.2652 0.0884 0.5000.0000 A (2) =.8363 0.0947 0.0000 0.497 0.0947 0.5000 0.493 0.0884 0.0000 0.493 0.6637 0.2803 0.497 0.0884 0.2803.0000, A (3) = 2.0636 0.230 0.76 0.0000 0.230 0.5000 0.493 0.0405 0.76 0.493 0.6637 0.2544 0.0000 0.0405 0.2544 0.7727 A (4) = 2.0636 0.230 0.095 0.0739 0.230 0.5000 0.0906 0.254 0.095 0.0906 0.4580 0.0000 0.0739 0.254 0.0000 0.9783, A (5) = 2.0636 0.08 0.095 0.02 0.08 0.469 0.0880 0.0000 0.095 0.0880 0.4580 0.027 0.02 0.0000 0.027.0092 A (6) = 2.070 0.0000 0.0969 0.00 0.0000 0.4627 0.0820 0.0064 0.0969 0.0820 0.4580 0.027 0.00 0.0064 0.027.0092, A (5) = 2.0856 0.0000 0.0000 0.0000 0.0000 0.5394 0.0000 0.0000 0.0000 0.0000 0.3750 0.0000 0.0000 0.0000 0.0000.0000

78 Power of A Matrix and Its Eigenvalues are eigen- Theorem: Let λ,λ 2,,λ n be eigenvalues of A R n n. Then λ k,λk 2,,λk n values of A k R n n with the same corresponding eigenvectors of A. Thatis, Av i = λ i v i A k v i = λ k i v i i n Suppose that the matrix A R n n has n linearly independent eigenvectors v, v,, v n corresponding to eigenvalues λ,λ 2,,λ n.thenanyx R n can be written as x = c v + c 2 v 2 + + c n v n Then A k x = λ k c v + λ k 2c 2 v 2 + + λ k nc n v n In particular, if λ > λ j for 2 j n and c 0,thenA k x will tend to lie in the direction v when k is large enough.

79 Power Method for Computing the Largest Eigenvalues Suppose that the matrix A R n n is diagonalizable and that U AU = diag(λ,λ 2,,λ n ) with U =[v, v 2,, v n ]and λ > λ 2 λ 3 λ n. Given u (0) R n,then power method produces a sequence of vectors u (k) as follows. for k =, 2, z (k) = Au (k ) r (k) = z (k) m = z (k),forsome m n. u (k) = z (k) /r (k) endfor λ must be real since the complex eigenvalues must appear in a relatively conjugate pair. Let u (0) = 2 A = 2 [ 0 ] [,thenu (5) = λ =3 λ 2 =.0 0.998 ], v = 2 [,andr (5) =2.9756. ], v 2 = [ 2 ]

80 QR Iterations for Computing Eigenvalues % % Script File: shiftqr.m % Solving Eigenvalues by shift-qr factorization % Nrun=5; fin=fopen( datamatrix.txt ); fgetl(fin); % read off the header line n=fscanf(fin, %d,); A=fscanf(fin, %f,[n n]); A=A ; SaveA=A; for k=:nrun, s=a(n,n); A=A-s*eye(n); [Q R]=qr(A); A=R*Q+s*eye(n); end eig(savea) % % datamatrix.txt % Matrices for computing eigenvalues by QR factorization or shift-qr 5.0 0.5 0.25 0.25 0.0625 0.5.0 0.5 0.25 0.25 0.25 0.5.0 0.5 0.25 0.25 0.25 0.5.0 0.5 0.0625 0.25 0.25 0.5.0 4 for shift-qr studies 2.9766 0.3945 0.498.59 0.3945 2.7328-0.3097 0.29 0.498-0.3097 2.5675 0.6079.59 0.29 0.6097.723

8 Norms of Vectors and Matrices Definition: A vector norm on R n is a function that satisfies τ : R n R + = {x 0 x R} () τ(x) > 0 x 0, τ(0) =0 (2) τ(cx) = c τ(x) c R, x R n (3) τ(x + y) τ(x)+τ(y) x, y R n Hölder norm (p-norm) x p =( n i= x i p ) /p for p. (p=) x = n i= x i (Mahattan or City-block distance) (p=2) x 2 =( n i= x i 2 ) /2 (Euclidean distance) (p= ) x = max i n { x i } ( -norm)

82 Definition: A matrix norm on R m n is a function τ : R m n R + = {x 0 x R} that satisfies () τ(a) > 0 A O, τ(o) =0 (2) τ(ca) = c τ(a) c R, A R m n (3) τ(a + B) τ(a)+τ(b) A, B R m n Consistency Property: τ(ab) τ(a)τ(b) A, B (a) τ(a) =max{ a ij i m, j n} (b) A F = [ mi= nj= a 2 ij] /2 (Fröbenius norm) Subordinate Matrix Norm: A = max x 0 { Ax / x } () If A R m n,then A = max j n ( m i= a ij ) (2) If A R m n,then A = max i m ( nj= a ij ) (3) Let A R n n be real symmetric, then A 2 = max i n λ i,whereλ i λ(a)

83 Theorem: Let x R n and let A =(a ij ) R n n. Define A = Sup u ={ Au } Proof: For u =, A = Sup{ Au } = n n n n n n a ij u j a ij u j = u j a ij i= j= j= i= j= i= Then n n n A Max j n { a ij } u j = Max j n { a ij } i= j= i= On the other hand, let n i= a ik = Max j n { n i= a ij } and choose u = e k,which completes the proof. Theorem: Let A =[a ij ] R m n, and define A = Max u ={ Au }. n Show that A = Max i m a ij j= n n Proof: Let a Kj = Max i m a ij, for any x Rn with x =,wehave j= j= Ax = Max i m { nj= a ij x j } Max i m { nj= a ij x j } Max i m { nj= a ij x } Max i m { nj= a ij } = n j= a Kj In particular, if we pick up y R n such that y j = sign(a Kj ), j n, then y =,and Ay = n j= a Kj, which completes the proof.

84 Theorem: Let A =[a ij ] R n n, and define A 2 = Max x 2 ={ Ax 2 }. Show that A 2 = ρ(a t A)= maximum eigenvalue of A t A (spectral radius) (Proof) Letλ, λ 2,, λ n be eigenvalues and their corresponding unit eigenvectors u, u 2,, u n of matrix A t A,thatis, (A t A)u i = λ i u i and u i 2 = i n. Since u, u 2,, u n must be an orthonormal basis based on spectrum decomposition n theorem, for any x R n,wehavex = c i u i.then i= A 2 = Max x 2 ={ Ax 2 } = Max x 2 ={ Ax 2 2 } = Max x 2 ={x t A t Ax} = = n Max x 2 = λ i c 2 i i= Max j n { λ j }

85 AMarkovProcess Suppose that 0% of the people outside Taiwan move in, and 20% of the people indside Taiwan move out in each year. Let y k and z k be the population at the end of the k th year, outside Taiwan and inside Taiwan, respectively. Then we have y k z k = 0.9 0.2 0. 0.8 y k λ =.0, λ 2 =0.7 z k y k z k = k 0.9 0.2 y 0 0. 0.8 z 0 = 3 2 k 0 0 (0.7) k 2 y 0 z 0 A Markov matrix A is nonnegative with each colume adding to. (a) λ = is an eigenvalue with a nonnegative eigenvector x. (b) The other eigenvalues satisfy λ i. (c) If any power of A has all positive entries, and the other λ i <. Then A k u 0 approaches the steady state of u which is a multiple of x as long as the projection of u 0 in x is not zero. Check Perron-Fröbenius theorem in Strang s book.

86 e A and Differential Equations e A = I + A! + A2 2! + + Am m! + du dt = λu u(t) =e λt u(0) du dt = Au = 2 u u(t) =e ta u(0) 2 A = UΛU t for an orthogonal matrix U, then e A = Ue Λ U = Udiag[e λ,e λ 2,...,e λn ]U t Solve x 3x +2x =0. Let y = x, z = y = x,andletu =[x, y, z] t. The problem is reduced to solving 0 0 u = Au = 0 0 u 0 2 3 Then u(t) =e ta u(0) = 3 2 2 3 2 0 4 3 2 0 e 2t 0 0 0 e t 0 0 0 0 2.293 2.293 0 3.464.732.5000 0.5000 u(0)

87 Problems Solved by Matlab A = LetA,B,H,x, y, u, b be matrices and vectors defined below, and H = I 2uu t 2 4 6 0 2 7 2, B = 3 0 3 0 0 0 3, u = /2 /2 /2 /2, b = 6 2 5, x =, y =. Let A=LU=QR, find L, U; Q, R. 2. Find determinants and inverses of matrices A, B, and H. 3. Solve Ax = b, how to find the number of floating-point operations are required? 4. Find the ranks of matrices A, B, and H. 5. Find the characteristic polynomials of matrices A and B. 6. Find -norm, 2-norm, and -norm of matrices A, B, and H. 7. Find the eigenvalues/eigenvectors of matrices A and B. 8. Find matrices U and V such that U AU and V BV are diagonal matrices. 9. Find the singular values and singular vectors of matrices A and B. 0. Randomly generate a 4 4 matrix C with 0 C(i, j) 9.