Suppose $A$ and $P$ are symmetric, positive definite matrices and that we factor $P^{-1}=EE^\top.$ Is it true that the condition number of $PA$ is upper-bounded by the condition number of $E^{-1}AE^{-\top}$?


The two matrices clearly have the same eigenvalues, but $PA$ may not be symmetric. So, I wasn't sure if this still implies that they have the same singular values.


[Apologies if this is a goofy question! Just a potential hole in a proof of preconditioning for conjugate gradients...]


REVISION: The two matrices do not have the same singular values, but numerically it appears $\mathrm{cond}\ PA\geq\mathrm{cond}\ E^{-1}AE^{-\top}.$ I'm totally stuck how to prove this!



Condition number after preconditioning
Yes, $\kappa(E^{-1}AE^{-T}) \le \kappa(PA)$. For any matrix $X$, $|\lambda| \le \Vert X \Vert$ holds for any eigenvalue and any induced norm (since $\Vert \lambda u \Vert$ = $\Vert X u \Vert \le \Vert X \Vert \Vert u \Vert$ for the eigenvector $u$). Similarly, if $X$ is invertible, then $|\lambda^{-1}| \le \Vert X^{-1} \Vert$ for any eigenvalue. Now take $X = PA$ and apply these inequalities with the largest and smallest eigenvalues. Multiplying, we see that
$$ \kappa( E^{-1}AE^{-T} ) = |\lambda\_{\text{max}}|/|\lambda\_{\text{min}}| \le \Vert PA \Vert \Vert (PA)^{-1} \Vert = \kappa( PA ) . $$


