Suppose $X \in S^{n}_{>0}$ $i$.$e$. positive definite, $V \in S^{n}$.i.e. symmetric. I'm curious if $X^{-1/2}VX^{-1/2}$ is diagonalizable or not?
This arises from $X+tV = X^{1/2}(I+tX^{-1/2}VX^{-1/2})X^{1/2}$. I think the matrix is not normal, but I suspect it is diagonal. However, I cannot prove this, and probably neglected something very simple.
Any hints?
Edit: The matrix is normal by definition. So this is of course true. I made a mistake before.
All your matrices are symmetric.
Because $X$ is symmetric and positive definite, so is $X^{-1/2}$. Then you can use that if $A$ is symmetric and $B$ is any real matrix, $$ B^TAB\qquad\text{ is symmetric.} $$ In your case, $$ X^{-1/2}VX^{-1/2}=(X^{-1/2})^TVX^{-1/2}. $$ And real symmetric matrices are orthogonaly diagonalizable.