Suppose $A,B$ are symmetric and positive definite (real) matrices such that $AB$ is not symmetric (equivalently $AB\neq BA$). My question is, what are sufficient and/or necessary conditions for $AB$ being positive definite in this case?
For clarification, I call a square matrix $C$ 'positive definite' if $x^TCx>0$ for any vector $x$, regardless of whether $C$ is symmetric or not.
Let $C=AB$ and $D=\frac{C+C^T}{2}$. Even though $C$ is not symmetric, $D$ is. Moreover, $x^TCx=x^TDx$. So a sufficient condition for $C$ to be positive definite is that $x^TDx>0$ for all nonzero $x$, i.e., that $D$ is positive definite. One sufficient condition for the latter is that all eigenvalues of $D$ are positive. (This answer does not use the fact that $A$ and $B$ are symmetric and positive definite.)