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?

I have deleted my previous post as it was stupid.One more question : do you consider diagonalisability over $\mathbb{R}$ or $\mathbb{C}$ ? — G. Fougeron, Dec 15 '17 at 17:38

score 1 · Accepted Answer · answered Dec 16 '17 at 11:12

1

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

answered Dec 16 '17 at 11:12

Proof here: https://math.stackexchange.com/questions/317334/why-is-this-matrix-product-diagonalizable – Hans Lundmark Dec 16 '17 at 11:24
The proof would have been nice in the original answer, but it checks out. – Lars Gebraad Dec 17 '17 at 01:08

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

1 Answers1