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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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1 Answer 1

up vote 4 down vote accepted

$-B$ 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 $|\lambda I-AB|=|A^{1/2}(\lambda I-A^{1/2}BA^{1/2})A^{-1/2}|=|(\lambda I-A^{1/2}BA^{1/2})|$ –  Shiyu Apr 4 '12 at 16:11
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? –  Shiyu Apr 4 '12 at 18:02
For any square matrices $XY$, its spectrum coincides with that of $YX$. You may first assume $X$ is invertible, otherwise use an $\epsilon$ argument. –  Sunni Apr 4 '12 at 19:08
what is $\epsilon$ argument? can you give me a hint? –  Shiyu Apr 5 '12 at 2:05
See e.g., page 57 of Fuzhen Zhang, Matrix Theory: Basic Results and Techniques Second Edition –  Sunni Apr 5 '12 at 13:45

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