Factorization of a matrix as a product of a symmetric and a skew-symmetric matrix

Question

1. When can an $n\times n$ matrix $M$ be written as a product $M=AB$, where $A^T=A$ and $B^T=-B$?

For example, a necessary condition is that the trace of $M$ vanishes. In this case, it is easy to check that for $n=2$ all traceless matrices $M$ admit such a factorization.

2. Is there a way to know all pairs of matrices $A$ and $B$ as above, for a given $M$?

3. What conditions should satisfy $M$ so that $A$ is non-singular, and what are all factorizations with $A$ non-singular?

For example, if $n=2$ and $M$ is non-singular, $A$ and $B$ are obtained easily, and are unique up to a scalar factor and non-singular. But it is possible for $A$ to be non-singular even when $M$ is singular, for example if $M=0_n$, $M=A0_n$ is a solution for any $A$.

I expect that there is literature about this kind of factorization, but I couldn't find anything.

Answer 1

This set of questions was considered in a recent paper by Stenzel: https://www.evernote.com/l/ABj_1ego5jZORbVzwCyxe_EL4EJVbk-4XQA

OK, it was 1920, but it seems like yesterday.

Answer 2

We have $M^T = B^T A^T = -BA$. Since $AB$ and $BA$ have the same characteristic polynomial, and so do $M$ and $M^T$, it follows that $M$ and $-M$ have the same characteristic polynomial. Equivalently, all of the odd coefficients (not just the trace) must vanish, and also equivalently, the eigenvalues of $M$ (counted by arithmetic multiplicity) come in pairs $\pm \lambda$. For orthogonally diagonalizable $M$ this necessary condition is also sufficient by reduction to the $2 \times 2$ case.

If either $A$ or $B$ is invertible it follows that $M$ and $-M$ are similar, so we get a stronger necessary condition involving the Jordan blocks for eigenvalues $\lambda$ and $-\lambda$ matching up. This condition only starts mattering when $n \ge 4$.