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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!

  • 0
    $AB$ is not necessarily symmetric, except if $A$ and $B$ commute.2012-05-19
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    Yes I know that. But the question is wheater $AB$ is seminegative definite or at least weather the trace is nonpositve.2012-05-19
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    I believe the term is "negative semidefinite". And I suspect that the reason Davide pointed out that $AB$ isn't necessarily symmetric is that sometimes symmetry is considered a prerequisite for positive/negative (semi)definiteness.2012-05-19
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    If you only cared about the trace, you could note that $\sqrt AB\sqrt A$ is negative semidefinite, and $\mathrm{tr}(\sqrt AB\sqrt A)=\mathrm{tr}(AB)$.2012-05-25

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