Let $A$ a spd (symmetric positive definite) matrix and $B$ a symmetric seminegative definite matrix. Is tr $AB \leq 0$ and more general is $AB$ seminegative definite?
I know that tr $AB \leq 0$ follows from $AB$ seminegative definite since the eigenvalues $\lambda$ of $AB$ are nonpositve and hence tr $AB=\sum_{\lambda \in spec\ A} \lambda \leq 0$. But I don't know how to find something out about the definitness of $AB$. I think in general there is nothing you can say about the eigenvalues of $AB$.
Thanks in advance!