I have a symmetric positive-definite matrix $L$ and I know its eigendecomposition $L= V\Lambda V^T$. Is there a simple way to compute the eigendecomposition of $M = CL$, where $C$ is a diagonal non-singular matrix, by using my knowledge of $V$ and $\Lambda$? I know for instance that the eigenvalues of $M$ are the same as the eigenvalues of $N = C^{1/2}LC^{1/2}$ since $\det[CL - d I] = 0$ can be rewritten as $$\det[C^{1/2}(C^{1/2}LC^{1/2} - dI) C^{-1/2}] = \det[C^{1/2}]\det[C^{-1/2}]\det[C^{1/2}LC^{1/2} - dI] = 0.$$
Then if I have the eigendecomposition $C^{1/2}LC^{1/2} = QDQ^T$ I can multiply by $C^{1/2}$ from the left and by $C^{-1/2}$ from the right to get $CL = C^{1/2}Q D (C^{1/2}Q)^{-1}$ meaning that for the eigendecomposition $CL = PDP^{-1}$ I have $P = C^{1/2}Q$.
I don't see an easy way to relate the eigendecomposition of $L$ to the one of $C^{1/2}LC^{1/2}$ however. I am able to relate $L = V\Lambda V^T$ to $C^{1/2}LC^{-1/2} = C^{1/2}V\Lambda (C^{1/2}V)^{-1}$ but that is of course not the one I care about. The best I can do is $C^{1/2}LC^{1/2} = C^{1/2}V\Lambda (C^{-1/2}V)^{-1}$ which is not an eigendecomposition unless $C$ is the identity times a scalar. Is there anything that I am missing which would allow me to relate the eigendecompositions of $L$ and $CL$? It could be a purely numerical procedure too, since this arose in numerics where I can easily compute the eigendecomposition of $L$ and I want to use this to more cheaply estimate the eigendecomposition of $CL$.
