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I've managed to prove that if $A$ and $B$ are positive definite then $AB$ has only positive eigenvalues. To prove $AB$ is positive definite, I also need to prove $(AB)^\ast = AB$ (so $AB$ is Hermitian). Is this statement true? If not, does anyone have a counterexample?

Thanks, Josh

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    According to the answers, this is not true, but can anyone give me an explicit counterexample? Thanks.2011-09-22
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    [Just try any two small Hermitian matrices; chances are, their product will not be Hermitian.](http://www.wolframalpha.com/input/?i=%7B%7B1,2%7D,%7B2,3%7D%7D*%7B%7B4,5%7D,%7B5,6%7D%7D)2011-09-22
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    Yes, but the matrices you gave are not positive definite. I'm having trouble finding positive definite matrices that act this way!2011-09-22
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    [Oh, it's easy to make them positive definite.](http://www.wolframalpha.com/input/?i=%7B%7B101,2%7D,%7B2,103%7D%7D%20times%20%7B%7B104,5%7D,%7B5,106%7D%7D)2011-09-22
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    Oh, duh. I'm an idiot. Thank you!2011-09-22
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    P.S. StackExchange ate the asterisk in my first comment's URL, so now it goes to WolframAlpha doing the Hadamard product of the matrices instead. It should be http://www.wolframalpha.com/input/?i=%7B%7B1,2%7D,%7B2,3%7D%7D%20times%20%7B%7B4,5%7D,%7B5,6%7D%7D.2011-09-22
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    For reference: there are two ways of producing some random symmetric positive (semi)definite matrix: generate some random matrix $\mathbf A$ and form $\mathbf A^\top\mathbf A$, or start with some symmetric matrix $\mathbf A$ and add a "sufficiently large" multiple of the identity.2011-09-22
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    Or generate $n$ orthonormal vectors $u_j$ and $n$ positive numbers $\lambda_j$ and take $\sum_j \lambda_j u_j u_j^T$.2011-09-22

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