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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?

1 Answers 1

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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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    $\lambda I-AB=\lambda A^{1/2}A^{-1/2}-A^{1/2}A^{1/2}B=A^{1/2}(\lambda I-A^{1/2}BA^{1/2})A^{-1/2}$, so $|\lambda I-AB|=|A^{1/2}(\lambda I-A^{1/2}BA^{1/2})A^{-1/2}|=|(\lambda I-A^{1/2}BA^{1/2})|$2012-04-04
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    I'm sorry, the above proof is only correct for invertible matrix $A$, but for semi-definite (singular) $A$, we would not have $A^{1/2}A^{-1/2}=I$. Then can we still say $AB$ and $A^{1/2}BA^{1/2}$ have the same eigenvalues? How to prove then?2012-04-04
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    For any square matrices $XY$, its spectrum coincides with that of $YX$. You may first assume $X$ is invertible, otherwise use an $\epsilon$ argument.2012-04-04
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    what is $\epsilon$ argument? can you give me a hint?2012-04-05
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    See e.g., page 57 of Fuzhen Zhang, Matrix Theory: Basic Results and Techniques Second Edition2012-04-05
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    $A^{1/2}BA^{1/2}=(A^{1/2}B^{1/2})(B^{1/2}A^{1/2})=HH^T$, where $H=A^{1/2}B^{1/2}$ and we used that $(CD)^T=D^TC^T$ and $(A^{1/2})^T=A^{1/2},\,(B^{1/2})^T=B^{1/2}$.2012-07-03
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    Sorry, a minus sign is needed. So put $-B$ instead of $B$ in the above comment.2012-07-03