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Jun 4, 2012 at 18:18 answer added Salvo Tringali timeline score: 6
Jun 4, 2012 at 18:06 vote accept user142
Jun 4, 2012 at 17:10 answer added Federico Poloni timeline score: 9
Jun 4, 2012 at 6:06 comment added Robert Israel 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 $\pmatrix{0 & 1\cr 0 & 0\cr}$.
Jun 4, 2012 at 4:07 comment added user142 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.
Jun 4, 2012 at 3:55 comment added MTS 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.
Jun 4, 2012 at 1:51 history asked user142 CC BY-SA 3.0