Is there a trick to invert $I+ABA$ given a spectral decomposition of $A$ and $B$? If it matters, $A$ is positive definite, but $B$ may be indefinite. My goal is to avoid an additional factorization if possible.

I am aware of the very nice answer to a similar question here. And, certainly, we could accomplish this if we had a spectral decomposition of $ABA$ directly since if $VDV^T=ABA$, then $I+ABA=V(I+D)V^T$ and $(I+ABA)^{-1}=V(I+D)^{-1}V^T$. As such, I suppose an equivalent question would be whether we can find the spectral decomposition of $ABA$ given the spectral decomposition of $A$ and $B$.

Thanks in advance!

  • $\begingroup$ Are $A$ and $B$ square matrices or are they also normal matrices? $\endgroup$ – Andrew Shedlock Jul 8 '18 at 0:59
  • $\begingroup$ Both $A$ and $B$ are square, symmetric, and real. $A$ is positive definite, but $B$ may not be. Though, it's likely that it may if that helps. $\endgroup$ – wyer33 Jul 8 '18 at 1:15

Maybe to find an explicit formula for the spectral decomposition of $ABA$ from spectral decompositions of $A$ and $B$ could prove to be hard, but sure there is such a decomposition, since $A$ and $B$ being symmetric implies $ABA$ is symmetric too. So, your spectral decomposition $VDV^T = ABA$ sure exists.

  • $\begingroup$ Well, sure, it's symmetric, so it exists. Mostly, I'm trying to avoid a costly factorization. Really, what's happening is that I have the spectral decomposition of A and B in an explicit format without doing a computation and I don't want to break down and do one for ABA. $\endgroup$ – wyer33 Jul 8 '18 at 5:00

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