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Let $A\in\mathbb{R}^{n\times n}$ be a symmetric positive semi-definite matrix, and $B\in\mathbb{R}^{n\times n}$ is symmetric negative semi-definite.

How to prove the eigenvalue of $AB$ is either zero or negative?

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$-B$ is again nonnegative definite.

So it suffices to show the eigenvalues of $AB$ are nonnegative for $A, B$ nonnegative definite. We know the eigenvalues of $AB$ are the same as that of $A^{1/2}BA^{1/2}$, where $A^{1/2}$ is the unique square root of $A$. But $A^{1/2}BA^{1/2}$ is nonnegative definite....

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    Sorry, a minus sign is needed. So put $-B$ instead of $B$ in the above comment.2012-07-03