Prove that $MA \neq A^T M$, where M is positive definite, and A doesn’t have distinct roots

Question

\Prove or refute the following statement

Let $A \in \mathbb{R}^{n×n}$ be a matrix such that $MA \neq A^TM$ for all positive definite matrices $M \in \mathbb{R}^{n×n}$, then the characteristic polynomial for $A$ must have a complex (non-real) root or a root with algebraic multiplicity greater than 1.

I’ve attempted this question from a mock paper for hours and I just can’t get anywhere. I’ve tried using JNF, by contradiction assuming A is diagonizable but I’ve got nowhere and would like a little push in the right direction.

Answer 1

Hint. Suppose the contrary. Then $A$ is diagonalisable over $\mathbb R$. In this case, can you construct a positive definite matrix $M$ such that $MA$ is real symmetric?

A further hint:

Let $A=PDP^{-1}$ be a diagonalisation over $\mathbb R$. To make $MA=(MP)DP^{-1}$ a symmetric matrix, can you relate $MP$ to $P^{-1}$?