12
$\begingroup$
  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.

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

  2. 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.

$\endgroup$
0

2 Answers 2

9
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $n \ge 4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.