It seems that the following two matrices have the same eigenvalues: $A^{-1/2} B A^{-1/2}$ and $L^{-1} B L^{-\top}$ where $L L^\top$ is the Cholesky decomposition of $A$. Both $A$ and $B$ are symmetric positive-definite matrices.
How can I prove this? I'm not even sure where to start from.
PS: There seems to be a common confusion between Cholesky factors $L$ and matrix square roots, see, e.g., this and this, and probably this property is one of the reasons. Yet I couldn't find this specific question asked elsewhere.
PPS: This is not a homework, I'm trying to understand why this yields the same eigenvalues since it seems to be a more computationally efficient way of finding geodesic distances between SPD matrices.
Best Answer
$AB$ and $BA$ have the same eigenvalues. That's why. So, if $\sigma(X)$ denotes the set of eigenvalues of $X$, then $$ \sigma(A^{-1/2}BA^{-1/2}) = \sigma(BA^{-1}) = \sigma(BL^{-T}L^{-1}) = \sigma(L^{-1}BL^{-T}). $$