MATH 537- Senstvty of Egenvalue Problems Prelmnares Let A be an n n matrx, and let λ be an egenvalue of A, correspondngly there are vectors x and y such that Ax = λx and y H A = λy H Then x s called A s (rght) egenvector; y s called A s left egenvector An egenvalue λ s called smple f A s characterstc polynomal can be factorzed as p A (λ) (λ λ ) q(λ), where q(λ) s a polynomal of degree n and q(λ ) Theorem Let A be an n n matrx, and let λ be A s egenvalue wth correspondng (rght) egenvector x and left egenvector y () Let µ be another egenvalue of A and w the correspondng (rght) egenvector (e, Aw = µw) If λ µ, then y H w = (2) If λ s smple, then y H x Proof: () s gven to the reader as an exercse Now we prove (2) By Schur Theorem, there s a untary matrx U such that U H AU = T ( n λ t H ) n T Snce λ s smple, λ s not T s egenvalue Ths means λi T s nonsngular Let z H = t H (λi T ), and set Z = ( n z H ), and then Z = n I ( n z H ) n I It can be verfed that ( Z U H z H AUZ = I ) ( λ t H T ) ( z H I ) ( λ λz = H + t H z H T T ) By the defnton of z, λz H + t H z H T = t H + z H (λi T ) = So we have λ AUZ = UZ and Z U H λ A = Z U H T T
Wrte X = UZ and Y H = Z U H We see that Xe s A s (rght) egenvector correspondng to λ and that Y e s A s left egenvector correspondng to λ Now snce λ s smple, A s (left and rght) egenspace correspondng to λ s -dmensonal (Why?) So where α = β; therefore αx = Xe = Ue and βy = Y e = U y H x = βα (U ( z ( z ) ) H (Ue ) = βα, as requred Example The case of a repeated egenvalue: A = whch has a repeated egenvalue wth correspondng (rght) egenvector x = and left egenvector y = It can be verfed that y H x = Example The dstnct egenvalue case: A = whose egenvalues and egenvectors 2 are λ =, x =, y =, and λ 2 = 2, x 2 =, y 2 = It can be verfed that ), y H x =, y H x 2 =, y H 2 x 2 =, y H 2 x = Theorem 2 (Gershgorn Dsk Theorem) Let A = (a j ) be an arbtrary matrx Then the egenvalues of A are located n the unon of the n dsks: λ : λ a kk a kj j k for k =, 2,, n Proof: Gven λ and x such that Ax = λx, let = x = x k for some k by scalng x f necessary (It also means that x j for j k) Then from the kth row of the equaton Ax = λx, we have N j= a kjx j = λx k = λ, so λ a kk = j k a kjx j Hence λ a kk j k a kj x j j k a kj Furthermore, t can also be shown that f th Gershgorn dsk s solated from the other dsks, then t contans precsely one of A s egenvalue (The proof depends on the contnuty of the egenvalues of a matrx as a functon of the elements of the matrx) 2
2 Egenvalue Senstvty Theorem 2 Let A be an n n matrx, and let λ be A s smple egenvalue wth correspondng (rght) egenvector x and left egenvector y Suppose A s perturbed to à A + δa, and consequently λ s perturbed to λ λ + δλ If δa 2 = ϵ s suffcently small, then Ths mples δλ = yh (δa)x y H x + O(ϵ 2 ) δλ y 2 x 2 y H δa 2 + O(ϵ 2 ) x Proof: Let x x + δx be à s egenvector correspondng to λ We have expandng whch leads to à x = λ x (A + δa)(x + δx) = (λ + δλ)(x + δx), Ax + A δx + δa x + δa δx = λx + λ δx + δλ x + δλ δx Ignorng second order terms δa δx and δλ δx and notcng that Ax = λx, we have A δx + δa x = λ δx + δλ x + O(ϵ 2 ), pre-multplyng the equaton by y H and notcng y H A = y H λ to get as requred Defnton 2 In Theorem 2, defne y H δa x = δλ y H x + O(ϵ 2 ), s λ def = yh x x 2 y 2 and cond(λ) def = s λ cond(λ) s called λ s ndvdual condton number Example Consder A = 2 3 4 5 4 s perturbed by δa = A s egenvalues are easly read and ts egenvectors can be computed: 8285 λ =, x =, y = 5523, 92 λ 2 = 4, x 2 = 5547 832, y 2 = 3 2,
λ 3 = 4, x 3 = 5547 832 2, y 3 = On the other hand, Ã s egenvalues computed by MATLAB s eg are λ =, λ2 = 39427, λ3 = 4582 The followng table compares λ λ wth ts upper bound cond(λ ) δa 2 s λ cond(λ ) cond(λ ) δa 2 λ λ 8285 27 2 2 7 4 6 3 6 57 3 7 4 6 3 6 57 Theorem 2 s useful only for suffcently small δa Such knd of error bound s called an asymptotc error bound On the other hand, We can remove the O(ϵ 2 ) term and get an error bound that s vald regardless of the sze δa of perturbaton s called a global error bound Theorem 22 (Bauer-Fke) Let A have all smple egenvalues (e, be dagonalzable) The egentrplets are denoted as (λ, x, y ) for =, 2,, n, normalzed so that x 2 = y 2 = Then egenvalues of A + δa le n dsks D, where D has center λ and radus n δa 2 y H x Proof: Snce A have all smple egenvalues, the egenvectors x, x 2,, x n form a bass Let X = (x, x 2,, x n ) It can be verfed that AX = XΛ, Λ = dag(λ,, λ n ) Agan snce A have all smple egenvalues, the left egenvectors y can be chosen so that y H x for all, and y H x j = for all j Let Y = (y, y 2,, y n ) and D = dag(y H x,, y H n x n ) It can be verfed that Y H X = D and so X = D Y H Now we prove every egenvalue of A + δa s n one of the dsks D To ths end, consder any egenvalue µ of A + δa If µ s also an egenvalue of A, then µ s n one of the dsks; no proof s necessary Suppose that µ s not an egenvalue of A We have A + δa µi = XΛX µi + δa = X[(Λ µi) + X δa X]X = X(Λ µi)[i + (Λ µi) X δa X]X whch s a sngular matrx whch n turn mples that I + (Λ µi) X δa X must be sngular Therefore (Λ µi) X δa X 2 (Λ µi) X 2 δa 2 X 2 (2) 4
We now estmate the rght-most quantty of (2) We have X 2 X F = n x 2 2 = n, Combne them wth (2) to get = (Λ µi) X 2 = (Λ µi) D Y H 2 (Λ µi) D 2 Y H 2 (Λ µi) D 2 Y H F = max λ µ y Hx n y 2 2 max = λ µ y Hx n max n λ µ y H x δa 2 λ k µ n δa 2 y H k x k, for some k n, e, µ s n D k, as was to be shown 3 Egenvector Senstvty Egenvectors may not be unquely determned n the case of a repeated egenvalue Example Consder + ϵ A = s perturbed to à = A + δa =, ϵ where ϵ and small Any vector s A s egenvector; whle à s egenvectors are e = (, ) T and e 2 = (, ) T (Why?) If you unfortunately choose an egenvector of A, say x = (, ) T, you would fnd t has nothng to do wth any of à s egenvector even though à and A could be made arbtrarly close Ths example alerts one to be careful when talkng about changes n egenvectors of a repeated egenvalue Example Consder ( ) + ϵ A = s perturbed to à = A + δa = + ϵ 3ϵ, ϵ 3ϵ ϵ where ϵ and small A egenvalues A egenvectors à egenvalues à egenvectors + ϵ, ϵ e, e 2 + 2ϵ, 2ϵ 3 2 e + 2 e 2, 3 2 e 2 e 2 What we see here s that the perturbaton δa makes small changes to A s egenvalues but ts egenvectors suffer enormous changes Ths s due to the closeness of A s egenvalues 5
as wll be clear from the next theorem whch roughly says the changes n an egenvector s proportonal to the recprocal of the dstance between the correspondng egenvalue and the rest of A s egenvalues Theorem 3 Suppose n n matrx A has n dstnct egenvalues wth correspondng (left and rght) egenvectors: egenvalues: λ, λ 2,, λ n, (rght) egenvectors: x, x 2,, x n, left egenvectors: y, y 2,, y n Normalze x and y j such that x 2 = = y j 2 for all and j Suppose A s perturbed to à = A + δa If δa 2 = ϵ s suffcently small, then for k =, 2,, n y H j (δa)x k x k = x k + (λ j k k λ j )yj Hx j + O(ϵ 2 ) These equatons mply that for k =, 2,, n x k x k 2 cond(λ j ) ϵ + O(ϵ 2 ) λ k λ j j k Proof: The condtons of ths theorem mply that each λ j s a smple egenvalue of A Thus by Theorem, y H j x k = f k j, and y H k x k (3) Notce also that A s egenvectors x, x 2,, x n consst of a bass of the n-dmensonal space snce A s egenvalues are parwse dstnct (Why?) Denote that under the perturbaton δa, λ k s changed to λ k λ k + δλ k and x k to x k x k + δx k Wrte n δx k = c x, = where c, c 2,, c n are small coeffcents Thus x k + δx k = ( + c k )x k + c x We may normalze x k + δx k by settng c k =, snce egenvectors are determned only up to a scalar multple and f c k we can always consder ( + c k )x k + c x + c k = x k + c k def x = x k + c k + c x k nstead Thus δx k = c x (32) The task s to fnd those c s To ths end, we expand (A + δa)(x k + δx k ) = (λ k + δλ k )(x k + δx k ) 6
as before to get Aδx k + δa x k = λ k δx k + δλ k x k + O(ϵ 2 ), substtutng (32) nto whch gves c λ x + δax k = λ k c x + δλ k x k + O(ϵ 2 ), snce Ax = λ x We have c (λ k λ )x = δax k δλ k x k + O(ϵ 2 ) Pre-multplyng the equaton by y H j (j k), together wth (3), yeld c j (λ k λ j )y H j x j = y H j δax k + O(ϵ 2 ), e, as requred c j = y H j δax k (λ k λ j )y H j x j + O(ϵ 2 ), 7