Symmetrizing matrix properties A symmetrizer $P$ is a $n\times n$ symmetric matrix such that for a $n\times n$ matrix $A$ it holds that $AP=PA^T$. There exists a symmetrizer for any square matrix, and in general it is not unique. Furthermore, if $A$ has complex eigenvalues, then there does not exist a positive definite $P$ that symmetrizes $A$.
I am looking for more information (or literature) on existence and properties of symmetrizers. Particularly:
Are there conditions on $A$ that ensure that a positive (semi-)definite $P$ exists? I am particularly interested in the case $A\in\mathbb{R}^{n\times n}$.
Are there conditions on the existence of symmetrizer, that symmetrizes two distinct matrices $A_1, A_2 \in \mathbb{R}^{n\times n}$?
What can be said about the spectrum of $PA$?
 A: Two classical papers on this topic are:

O. Taussky and H. Zassenhaus, On the similarity transformation between a matrix and its transpose, Pacific J. Math., 9 (1959), 893-896.
Olga Taussky, Positive-definite matrices and their role in the study of the characteristic roots of general matrices, Advances in Mathematics, Volume 2, Issue 2, June 1968, 175-186.

In the first paper, the authors proved that for every square matrix $A$ over any field $F$, there exists a nonsingular symmetric matrix $S$ such that $S^{-1}AS=A^\top$. The matrix $S$ is unique if and only if the minimal polynomial of $A$ is equal to the characteristic polynomial of $A$, i.e. if and only if $A$ is similar to a companion matrix.
In the second paper, Olga Taussky proved that if $A$ is a real square matrix, then $AS=SA^\top$ for some symmetric positive definite matrix $S$ if and only if $A$ is diagonalisable over $\mathbb R$. And when such a positive definite matrix $S$ exists, there is a one-to-one correspondence between the signs of the eigenvalues of $AS$ and the signs of the eigenvalues of $A$. (The similar conclusion holds for $SA$, because $SA$ is similar to $AS$.)
The proof isn't difficult. If $A=XDX^{-1}$ is a diagonalisation over $\mathbb R$, then $AS=SA^\top$ where $S=XX^\top\succ0$. Conversely, suppose $H=AS=SA^\top$ for some $S\succ0$. Then $A$ is similar to the real symmetric matrix $S^{-1/2}AS^{1/2}=S^{-1/2}HS^{-1/2}$ (hence $A$ is diagonalisable over $\mathbb R$) and its eigenvalues have the same signs as those of $H$, because $S^{-1/2}HS^{-1/2}$ is congruent to $H$.
