Is the product of a positive definite matrix with it's inverse diagonal diagonalizable?

Question

For some implementation of Hamiltonian dynamics I'd like to diagonalize a specific matrix $F$. $F$ can be constructed in multiple ways. I have a positive definite matrix $A$. It is real and symmetric and therefore Hermitian, and diagonalizable.

Now for $F$ I have three options:

• $F = I A = A$
• $F = A^{-1} A = I$
• $F = (\text{maindiagonal}(A)\; )^{-1} A $.

The first two are always diagonalizable. From my implementation, it seems that the third option is no different, but providing a proof for this is a little harder. Is there a way to prove the diagonalizeability?

Answer 1

Yes, $F$ is diagonalizable with $>0$ eigenvalues because it's the product of $2$ positive symmetric matrices.