# Perturbation theory for the generalized eigenvalue problem

Is there a standard reference for the perturbation theory of the generalized eigenvalue problem?

More specifically, I would like to get a systematic expansion for the problem

$(A_0 + \epsilon A_1)v = \lambda B v$

in terms of the solutions to

$A_0 v = \lambda B v$

where $A_0$, $A_1$ and $B$ are known, $n\times n$ Hermitian matrices, $\epsilon >0$ is a "small" parameter, and $\lambda\in \mathbb{C}$ and $v\in\mathbb{C}^n$ are the unknowns. When $B$ is positive-definite, the usual (Rayleigh-Schroedinger) perturbation series can easily be generalized by using $B$ to define an alternative inner product. However, in my problem, $B$ is not positive-definite (although it is still nonsingular).

-
Well, if $B$ is invertible, why don't you multiply both sides by $B^{-1}$? Then at least your system is a little simpler-looking. – MTS Jun 4 '12 at 3:55
Various nice properties (e.g., orthogonality) of the eigenvectors follow easily from the Hermiticity of $A$ and $B$, and I was hoping that the perturbation theory would be better-behaved in this form. The general theory of perturbations of eigenvalues (of a possibly non-normal matrix, which $B^{-1}A$ may be) seemed a little scary, but maybe it will turn out to be ok, and I may end up trying that. The case of positive-definite $B$ was an almost trivial generalization of the usual perturbation theory for a Hermitian matrix with all the usual niceties, so I was hoping to have something similar. – user142 Jun 4 '12 at 4:07
For one example of what can go wrong if $B$ is not positive-definite, consider the case $A = \pmatrix{1 & 1\cr 1 & 1\cr}$, $B = \pmatrix{1 & 0\cr 0 & -1\cr}$, where $Av = \lambda B v$ is "missing" a generalized eigenvector: the Jordan form of $B^{-1} A$ is $\pmatrix{0 & 1\cr 0 & 0\cr}$. – Robert Israel Jun 4 '12 at 6:06
Thanks, Robert, that was an eye-opener. – user142 Jun 4 '12 at 6:37

In any case, here is a nifty trick that can help you get a first-order expansion, and that can perhaps be generalized. If the eigenvalue is simple, we know that the perturbed eigenvalue and eigenvector is an analytic function of the perturbation. Therefore, we may take derivatives in $$Av=\lambda Bv$$ to get $$(\Delta A)v+A\dot{v}=\dot{\lambda}Bv+\lambda (\Delta{B})v+\lambda B\dot{v}.$$ Now multiply by the left eigenvector $u^*$ to get $$u^* (\Delta A)v = \dot{\lambda} u^* Bv+\lambda u^*(\Delta B)v.$$ Solve for $\dot{\lambda}$ and you get a first-order expression for the perturbed eigenvalue. (Yes, I know it sounds fishy, but I think everything can be made rigorous with a little work).