2.8 Sparse Matrices in Matlab
3.1.5 Linear Perturbations of A
In this section we consider the family of matrices defined by
wheret belongs to the complex plane. We are interested in the behavior of the eigenelements ofA(t)whentvaries around the origin. Consider first the ‘param- eterized’ resolvent,
R(t, z) = (A+tH−zI)−1.
Noting thatR(t, z) =R(z)(I+tR(z)H)−1it is clear that if the spectral radius
oftR(z)H is less than one thenR(t, z)will be analytic with respect tot. More precisely,
Proposition 3.2 The resolventR(t, z)is analytic with respect totin the open disk
|t|< ρ−1(HR(z)).
We wish to show by integration over a Jordan curveΓthat a similar result holds for the spectral projectorP(t),i.e., thatP(t)is analytic fortsmall enough. The result would be true if the resolventR(t, z)were analytic with respect to t
for eachzonΓi. To ensure this we must require that |t|< inf
z∈Γρ
−1(R(z)H)).
The question that arises next is whether or not the disk of allt’s defined above is empty. The answer is no as the following proof shows. We have
ρ(R(z)H)≤ kR(z)Hk ≤ kR(z)kkHk.
The functionkR(z)kis continuous with respect to zforz ∈ Γand therefore it reaches its maximum at some pointz0of the closed curveΓand we obtain
ρ(R(z)H)≤ kR(z)Hk ≤ kR(z0)kkHk ≡κ . Hence, inf z∈Γρ −1(R(z)H)) ≥κ−1.
Theorem 3.4 LetΓbe a Jordan curve around one or a few eigenvalues ofAand let
ρa= inf
z∈Γ[ρ(R(z)H)]
−1.
Thenρa>0and the spectral projector
P(t) = −1 2πi
Z Γ
R(t, z)dz
is analytic in the disk|t|< ρa.
We have already proved thatρa>0. The rest of the proof is straightforward. As an immediate corollary of Theorem 3.4, we know that the rank ofP(t)will stay constant as long aststays in the disk|t|< ρa.
Corollary 3.1 The numbermof eigenvalues ofA(t), counted with their algebraic multiplicities, located inside the curveΓ, is constant provided that|t|< ρa.
In fact the condition ontis only a sufficient condition and it may be too restrictive since the real condition required is thatP(t)be continuous with respect tot.
While individual eigenvalues may not have an analytic behavior, their average is usually analytic. Consider the average
ˆ λ(t) = 1 m m X i=1 λi(t)
of the eigenvalues λ1(t), λ2(t), . . . , λm(t)of A(t)that are inside Γ where we assume that the eigenvalues are counted with their multiplicities. LetB(t)be a matrix representation of the restriction ofA(t)to the invariant subspaceM(t) = Ran(P(t)). Note that sinceM(t)is invariant underA(t)thenB(t)is the matrix representation of the rankmtransformation
A(t)|M(t)=A(t)P(t)|M(t)=P(t)A(t)|M(t)=P(t)A(t)P(t)|M(t) and we have ˆ λ(t)≡m1tr[B(t)] = 1 mtr[A(t)P(t)|M(t)] = 1 mtr[A(t)P(t)] (3.20)
The last equality in the above equation is due to the fact that for anyxnot inM(t)
we haveP(t)x= 0and therefore the extension ofA(t)P(t)to the whole space can only bring zero eigenvalues in addition to the eigenvaluesλi(t), i= 1, . . . , m. Theorem 3.5 The linear transformation A(t)P(t) and its weighted traceλˆ(t)are analytic in the disk|z|< ρa.
Proof. ThatA(t)P(t)is analytic is a consequence of the previous theorem. That
ˆ
λ(t)is analytic, comes from the equivalent expression (3.20) and the fact that the trace of an operatorX(t)that is analytic with respect totis analytic.
Therefore, a simple eigenvalueλ(t)ofA(t)not only stays simple around a neighborhood oft= 0but it is also analytic with respect tot. Moreover, the vector
ui(t) =Pi(t)uiis an eigenvector ofA(t)associated with this simple eigenvalue, withui = ui(0) being an eigenvector ofA associated with the eigenvalue λi. Clearly, the eigenvectorui(t)is analytic with respect to the variablet. However, the situation is more complex for the case of a multiple eigenvalue. If an eigen- value is of multiplicitymthen after a small perturbation, it will split into at most
mdistinct small branchesλi(t). These branches taken individually are not ana- lytic in general. On the other hand, their arithmetic average is analytic. For this reason it is critical, in practice, to try to recognize groups of eigenvalues that are likely to originate from the splitting of a perturbed multiple eigenvalue.
Example 3.1. That an individual branch of thembranches of eigenvaluesλi(t)
is not analytic can be easily illustrated by the example
A = 0 1 0 0 ! , H = 0 0 1 0 ! .
The matrixA(t)has the eigenvalues±√twhich degenerate into the double eigen- value0ast → 0. The individual eigenvalues are not analytic but their average remains constant and equal to zero.
In the above example each of the individual eigenvalues behaves like the square root oftaround the origin. One may wonder whether this type of behavior can be generalized. The answer is stated in the next proposition.
Proposition 3.3 Any eigenvalueλi(t)ofA(t)inside the Jordan curveΓsatisfies
|λi(t)−λi|=O(|t|1/li) whereliis the index ofλi.
Proof. Letf(z) = (z−λi)li. We have seen earlier (proof of Lemma 3.1) that
f(A)Pi= 0. For an eigenvectoru(t)of norm unity associated with the eigenvalue
λi(t)we have
f(A(t))P(t)u(t) = f(A(t))u(t) = (A(t)−λiI)liu(t)
= (λ(t)−λi)liu(t).
Taking the norms of both members of the above equation and using the fact that
f(A)Pi= 0we get
|λi(t)−λi|li = kf(A(t))P(t)u(t)k
≤ kf(A(t))P(t)k=kf(A(t))P(t)−f(A)Pik.
Sincef(A) =f(A(0)),Pi=P(0)andP(t), f(A(t))are analytic the right-hand- side in the above inequality isO(t)and therefore
|λi(t)−λi|li =O(|t|)
which shows the result.
Example 3.2. A standard illustration of the above result is provided by taking
Ato be a Jordan block andH to be the rank one matrixH =eneT 1: A= 0 1 0 1 0 1 0 1 0 H = 0 0 0 0 1 0 .
The matrix Ahas nonzero elements only in positions (i, i+ 1)where they are equal to one. The matrixH has its elements equal to zero except for the element in position(n,1)which is equal to one. Fort= 0the matrixA+tHadmits only the eigenvalueλ= 0. The characteristic polynomial ofA+tHis equal to
pt(z) = det(A+tH−zI) = (−1)n(zn−t)
and its roots areλj(t) = t1/ne2ijπn j = 1, . . . , n. Thus, ifn = 20 then for a
perturbation onAof the order of10−16,a reasonable number if double precision arithmetic is used, the eigenvalue will be perturbed by as much as0.158. .