Today: eigenvalue sensitivity, eigenvalue algorithms Reminder: midterm starts today

Transcription

1 AM 205: lecture 22 Today: eigenvalue sensitivity, eigenvalue algorithms Reminder: midterm starts today Posted online at 5 PM on Thursday 13th Deadline at 5 PM on Friday 14th Covers material up to and including lecture 16 Open book. No collaboration allowed. Send questions as private messages on Piazza

2 Gershgorin s Theorem (from last time) Theorem: All eigenvalues of A C n n are contained within the union of the n Gershgorin disks of A

3 Sensitivity of Eigenvalue Problems We shall now consider the sensitivity of the eigenvalues to perturbations in the matrix A Suppose A is nondefective, and hence A = VDV 1 Let δa denote a perturbation of A, and let E V 1 δav, then V 1 (A + δa)v = V 1 AV + V 1 δav = D + E

4 Sensitivity of Eigenvalue Problems For a nonsingular matrix X, the map A X 1 AX is called a similarity transformation of A Theorem: A similarity transformation preserves eigenvalues Proof: We can equate the characteristic polynomials of A and X 1 AX (denoted p A (z) and p X 1 AX (z), respectively) as follows: p X 1 AX (z) = det(zi X 1 AX ) = det(x 1 (zi A)X ) = det(x 1 ) det(zi A) det(x ) = det(zi A) = p A (z), where we have used the identities det(ab) = det(a) det(b), and det(x 1 ) = 1/ det(x )

5 Sensitivity of Eigenvalue Problems The identity V 1 (A + δa)v = D + E is a similarity transformation Therefore A + δa and D + E have the same eigenvalues Let λ k, k = 1, 2,..., n denote the eigenvalues of A, and λ denote an eigenvalue of A + δa Then for some w C n, ( λ, w) is an eigenpair of (D + E), i.e. (D + E)w = λw

6 Sensitivity of Eigenvalue Problems This can be rewritten as w = ( λi D) 1 Ew This is a promising start because: we want to bound λ λ k for some k ( λi D) 1 is a diagonal matrix with entries 1/( λ λ k ) on the diagonal

7 Sensitivity of Eigenvalue Problems Taking norms yields w 2 ( λi D) 1 2 E 2 w 2, or ( λi D) E 2 Note that the norm of a diagonal matrix is given by its largest entry (in abs. val.) 1 max v 0 Dv v (D 11 v 1, D 22 v 2,..., D nn v n ) = max v 0 v { } max D v ii max i=1,2,...,n v 0 v = max D ii i=1,2,...,n 1 This holds for any induced matrix norm, not just the 2-norm

8 Sensitivity of Eigenvalue Problems Hence ( λi D) 1 2 = 1/ λ λ k, where λ k of A closest to λ is the eigenvalue Therefore it follows from ( λi D) E 2 that λ λ k = ( λi D) E 2 = V 1 δav 2 V 1 2 δa 2 V 2 = cond(v ) δa 2 This result is known as the Bauer Fike Theorem

9 Sensitivity of Eigenvalue Problems Hence suppose we compute the eigenvalues, λ i, of the perturbed matrix A + δa Then Bauer Fike tells us that each λ i must reside in a disk of radius cond(v ) δa 2 centered on some eigenvalue of A If V is poorly conditioned, then even for small perturbations δa, the disks can be large: sensitivity to perturbations If A is normal then cond(v ) = 1, in which case the Bauer Fike disk radius is just δa 2

10 Sensitivity of Eigenvalue Problems Note that a limitation of Bauer Fike is that it does not tell us which disk λ i will reside in Therefore, this doesn t rule out the possibility of, say, all λ i clustering in just one Bauer Fike disk In the case that A and A + δa are hermitian, we have a stronger result

11 Sensitivity of Eigenvalue Problems Weyl s Theorem: Let λ 1 λ 2 λ n and λ 1 λ 2 λ n be the eigenvalues of hermitian matrices A and A + δa, respectively. Then max λ i λ i δa 2. i=1,...,n Hence in the hermitian case, each perturbed eigenvalue must be in the disk 2 of its corresponding unperturbed eigenvalue! 2 In fact, eigenvalues of a hermitian matrix are real, so disk here is actually an interval in R

12 Sensitivity of Eigenvalue Problems The Bauer Fike Theorem relates to perturbations of the whole spectrum We can also consider perturbations of individual eigenvalues Suppose, for simplicity, that A C n n is symmetric, and consider the perturbed eigenvalue problem (A + E)(v + v) = (λ + λ)(v + v) Expanding this equation, dropping second order terms, and using Av = λv gives A v + Ev λv + λ v

13 Sensitivity of Eigenvalue Problems Premultiply A v + Ev λv + λ v by v to obtain v A v + v Ev λv v + λv v Noting that v A v = (v A v) = v Av = λ v v = λv v leads to v Ev λv v, or λ = v Ev v v

14 Sensitivity of Eigenvalue Problems Finally, we obtain λ v Ev v 2 2 v 2 Ev 2 v 2 2 = E 2, so that λ E 2 We observe that perturbation bound does not depend on cond(v ) when we consider only an individual eigenvalue this individual eigenvalue perturbation bound is asymptotic; it is rigorous only in the limit that the perturbations 0

15 Algorithms for Eigenvalue Problems

16 Power Method

17 Power Method The power method is perhaps the simplest eigenvalue algorithm It finds the eigenvalue of A C n n with largest modulus 1: choose x 0 C n arbitrarily 2: for k = 1, 2,... do 3: x k = Ax k 1 4: end for Question: How does this algorithm work?

18 Power Method Assuming A is nondefective, then the eigenvectors v 1, v 2,..., v n provide a basis for C n Therefore there exist coefficients α i such that x 0 = n j=1 α jv j Then, we have x k = Ax k 1 = A 2 x k 2 = = A k x 0 = n n A k α j v j = α j A k v j = j=1 n α j λ k j v j j=1 j=1 j=1 n 1 [ ] k = λ k λj n α n v n + α j v j λ n

19 Power Method Then if λ n > λ j, 1 j < n, we see that x k λ k nα n v n as k This algorithm converges linearly: the error terms are scaled by a factor at most λ n 1 / λ n at each iteration Also, we see that the method converges faster if λ n is well-separated from the rest of the spectrum

20 Power Method However, in practice the exponential factor λ k n could cause overflow or underflow after relatively few iterations Therefore the standard form of the power method is actually the normalized power method 1: choose x 0 C n arbitrarily 2: for k = 1, 2,... do 3: y k = Ax k 1 4: x k = y k / y k 5: end for

21 Power Method Convergence analysis of the normalized power method is essentially the same as the un-normalized case Only difference is we now get an extra scaling factor, c k R, due to the normalization at each step n 1 [ ] k x k = c k λ k λj n α n v n + α j v j λ n j=1

22 Power Method This algorithm directly produces the eigenvector v n One way to recover λ n is to note that y k = Ax k 1 λ n x k 1 Hence we can compare an entry of y k and x k 1 to approximate λ n We also note two potential issues: 1. We require x 0 to have a nonzero component of v n 2. There may be more than one eigenvalue with maximum modulus

23 Power Method Issue 1: In practice, very unlikely that x 0 will be orthogonal to v n Even if x 0 v n = 0, rounding error will introduce a component of v n during the power iterations Issue 2: We cannot ignore the possibility that there is more than one max. eigenvalue In this case x k would converge to a member of the corresponding eigenspace

24 Power Method An important idea in eigenvalue computations is to consider the shifted matrix A σi, for σ R We see that (A σi)v i = (λ i σ)v i and hence the spectrum of A σi is shifted by σ, and the eigenvectors are the same For example, if all the eigenvalues are real, a shift can be used with the power method to converge to λ 1 instead of λ n

25 Power Method Python example: Consider power method and shifted power method for [ ] 4 1 A =, 1 2 which has eigenvalues λ 1 = , λ 2 =

26 Inverse Iteration

27 Inverse Iteration The eigenvalues of A 1 are the reciprocals of the eigenvalues of A, since Av = λv A 1 v = 1 λ v Question: What happens if we apply the power method to A 1?

28 Inverse Iteration Answer: We converge to the largest (in modulus) eigenvalue of A 1, which is 1/λ 1 (recall that λ 1 is the smallest eigenvalue of A) This is called inverse iteration 1: choose x 0 C n arbitrarily 2: for k = 1, 2,... do 3: solve Ay k = x k 1 for y k 4: x k = y k / y k 5: end for

29 Inverse Iteration Hence inverse iteration gives λ 1 without requiring a shift This is helpful since it may be difficult to determine what shift is required to get λ 1 in the power method Question: What happens if we apply inverse iteration to the shifted matrix A σi?

30 Inverse Iteration The smallest eigenvalue of A σi is (λ i σ), where i = arg min λ i σ, i=1,2,...,n and hence... Answer: We converge to λ = 1/(λ i σ), then recover λ i via λ i = 1 λ + σ Inverse iteration with shift allows us to find the eigenvalue closest to σ

31 Rayleigh Quotient Iteration

32 Rayleigh Quotient For the remainder of this section (Rayleigh Quotient Iteration, QR Algorithm) we will assume that A R n n is real and symmetric 3 The Rayleigh quotient is defined as r(x) x T Ax x T x If (λ, v) R R n is an eigenpair, then r(v) = v T Av v T v = λv T v v T v = λ 3 Much of the material generalizes to complex non-hermitian matrices, but symmetric case is simpler

33 Rayleigh Quotient Theorem: Suppose A R n n is a symmetric matrix, then for any x R n we have λ 1 r(x) λ n Proof: We write x as a linear combination of (orthogonal) eigenvectors x = n j=1 α jv j, and the lower bound follows from r(x) = x T n Ax x T x = j=1 λ jαj 2 n j=1 α2 j n j=1 λ α2 j 1 n j=1 α2 j = λ 1 The proof of the upper bound r(x) λ n is analogous

34 Rayleigh Quotient Corollary: A symmetric matrix A R n n is positive definite if and only if all of its eigenvalues are positive Proof: ( ) Suppose A is symmetric positive definite (SPD), then for any nonzero x R n, we have x T Ax > 0 and hence λ 1 = r(v 1 ) = v T 1 Av 1 v T 1 v 1 > 0 ( ) Suppose A has positive eigenvalues, then for any nonzero x R n x T Ax = r(x)(x T x) λ 1 x 2 2 > 0

35 Rayleigh Quotient But also, if x is an approximate eigenvector, then r(x) gives us a good approximation to the eigenvalue This is because estimation of an eigenvalue from an approximate eigenvector is an n 1 linear least squares problem: xλ Ax x R n is our tall thin matrix and Ax R n is our right-hand side Hence the normal equation for xλ Ax yields the Rayleigh quotient, i.e. x T xλ = x T Ax

36 Rayleigh Quotient Question: How accurate is the Rayleigh quotient approximation to an eigenvalue? Let s consider r as a function of x, so r : R n R r(x) x j = x j (x T Ax) x T x (x T Ax) x j (x T x) (x T x) 2 = 2(Ax) j x T (x T Ax)2x j x (x T x) 2 2 = x T x (Ax r(x)x) j (Note that the second equation relies on the symmetry of A)

37 Rayleigh Quotient Therefore r(x) = 2 x T (Ax r(x)x) x For an eigenpair (λ, v) we have r(v) = λ and hence r(v) = 2 v T (Av λv) = 0 v This shows that eigenvectors of A are stationary points of r

38 Rayleigh Quotient Suppose (λ, v) is an eigenpair of A, and let us consider a Taylor expansion of r(x) about v: r(x) = r(v) + r(v) T (x v) (x v)t H r (v)(x v) + H.O.T. = r(v) (x v)t H r (v)(x v) + H.O.T. Hence as x v the error in a Rayleigh quotient approximation is r(x) λ = O( x v 2 2) That is, the Rayleigh quotient approx. to an eigenvalue squares the error in a corresponding eigenvector approx.

39 Rayleigh Quotient Iteration The Rayleigh quotient gives us an eigenvalue estimate from an eigenvector estimate Inverse iteration gives us an eigenvector estimate from an eigenvalue estimate It is natural to combine the two, this yields the Rayleigh quotient iteration 1: choose x 0 R n arbitrarily 2: for k = 1, 2,... do 3: σ k = x T k 1 Ax k 1/x T k 1 x k 1 4: solve (A σ k I)y k = x k 1 for y k 5: x k = y k / y k 6: end for

40 Rayleigh Quotient Iteration Suppose, at step k, we have x k 1 v ɛ Then, from the Rayleigh quotient in line 3 of the algorithm, we have σ k λ = O(ɛ 2 ) In lines 4 and 5 of the algorithm, we then perform an inverse iteration with shift σ k to get x k Recall the eigenvector error in one inverse iteration step is scaled by ratio of second largest to largest eigenvalues of (A σ k I) 1

41 Rayleigh Quotient Iteration Let λ be the closest eigenvalue of A to σ k, then the magnitude of largest eigenvalue of (A σ k I) 1 is 1/ σ k λ The second largest eigenvalue magnitude is 1/ σ k ˆλ, where ˆλ is the eigenvalue of A second closest to σ k Hence at each inverse iteration step, the error is reduced by a factor σ k λ σ k ˆλ = σ k λ (σ k λ) + (λ ˆλ) const. σ k λ as σ k λ Therefore, we obtain cubic convergence as k : x k v (const. σ k λ ) x k 1 v = O(ɛ 3 )

42 Rayleigh Quotient Iteration A drawback of Rayleigh iteration: we can t just LU factorize A σ k I once since the shift changes each step Also, it s harder to pick out specific parts of the spectrum with Rayleigh quotient iteration since σ k can change unpredictably Python demo: Rayleigh iteration to compute an eigenpair of A =

43 QR Algorithm

44 The QR Algorithm The QR algorithm for computing eigenvalues is one of the best known algorithms in Numerical Analysis 4 It was developed independently in the late 1950s by John G.F. Francis (England) and Vera N. Kublanovskaya (USSR) The QR algorithm efficiently provides approximations for all eigenvalues/eigenvectors of a matrix We will consider what happens when we apply the power method to a set of vectors this will then motivate the QR algorithm 4 Recall that here we focus on the case in which A R n n is symmetric

45 The QR Algorithm Let x (0) 1,..., x p (0) denote p linearly independent starting vectors, and suppose we store these vectors in the columns of X 0 We can apply the power method to these vectors to obtain the following algorithm: 1: choose an n p matrix X 0 arbitrarily 2: for k = 1, 2,... do 3: X k = AX k 1 4: end for

46 The QR Algorithm From our analysis of the power method, we see that for each i = 1, 2,..., p: ) x (k) i = (λ k nα i,n v n + λ k n 1α i,n 1 v n λ k 1α i,1 v 1 n ( ) k n p = λ k λj ( ) k λj n p α i,jv j + α i,jv j j=n p+1 λ n p j=1 λ n p Then, if λ n p+1 > λ n p, the sum in green will decay compared to the sum in blue as k Hence the columns of X k will converge to a basis for span{v n p+1,..., v n }

47 The QR Algorithm However, this method doesn t provide a good basis: each column of X k will be very close to v n Therefore the columns of X k become very close to being linearly dependent We can resolve this issue by enforcing linear independence at each step

48 The QR Algorithm We orthonormalize the vectors after each iteration via a (reduced) QR factorization, to obtain the simultaneous iteration: 1: choose n p matrix Q 0 with orthonormal columns 2: for k = 1, 2,... do 3: X k = A ˆQ k 1 4: ˆQ k ˆR k = X k 5: end for The column spaces of ˆQ k and X k in line 4 are the same Hence columns of ˆQ k converge to orthonormal basis for span{v n p+1,..., v n }

49 The QR Algorithm In fact, we don t just get a basis for span{v n p+1,..., v n }, we get the eigenvectors themselves! Theorem: The columns of ˆQ k converge to the p dominant eigenvectors of A We will not discuss the full proof, but we note that this result is not surprising since: the eigenvectors of a symmetric matrix are orthogonal columns of ˆQ k converge to an orthogonal basis for span{v n p+1,..., v n } Simultaneous iteration approximates eigenvectors, we obtain eigenvalues from the Rayleigh quotient ˆQ T A ˆQ diag(λ 1,..., λ n )

50 The QR Algorithm With p = n, the simultaneous iteration will approximate all eigenpairs of A We now show a more convenient reorganization of the simultaneous iteration algorithm We shall require some extra notation: the Q and R matrices arising in the simultaneous iteration will be underlined Q k, R k (As we will see shortly, this is to distinguish between the matrices arising in the two different formulations...)