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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

4 Answers 4

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EDIT: Changed example to use strictly positive definite $A$ and $B$.

To complement the nice answers above, here is a simple explicit counterexample:

$$A=\begin{bmatrix}2 & -1\\\\ -1 & 2\end{bmatrix},\qquad B = \begin{bmatrix}10 & 3\\\\ 3 & 1\end{bmatrix}. $$ Matrix $A$ has eigenvalues (1,3), while $B$ has eigenvalues (0.09, 10).

Then, we have $$AB = \begin{bmatrix} 17 & 5\\\\ -4 & -1\end{bmatrix}$$

Now, pick vector $x=[0\ \ 1]^T$, which shows that $x^T(AB)x = -1$, so $AB$ is not positive definite.

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    (some authors use more general definition of positive definite; this matrix is **not** symmetric positive definite immediately due to lack of symmetry, but it is not positive definite even by the more general definition)2017-05-12
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    hi--I was wondering if you may be able to shed any light on this question https://mathoverflow.net/questions/277545/satisfying-the-following-determinant-inequality that makes use of determinant inequalities? I would be much appreciative. thank you!2017-07-30
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In general no, because for Hermitian $A$ and $B$, $(AB)^* = AB$ if and only if $A$ and $B$ commute. On the other hand, $ABA$ and $BAB$ can be proven to be positive definite.

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$AB$ is not necessarily Hermitian (or symmetric).

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As already noted, $AB$ is not necessarily Hermitian. However, the eigenvalues of $AB$ are all real and in fact positive. Let $\lambda$ be eigenvalue with associated eigenvector $\xi$. Then $AB\xi = \lambda \xi$ and multiplying from the left by $\xi^*B^*$ yields $\xi^*B^*AB\xi=\lambda \xi^*B^*\xi$ and so $\lambda = \frac{\xi^*B^*AB\xi}{\xi^*B^*\xi}$ which is positive since $B^*AB$ is positive-definite.

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    I was trying to prove that $B^{\star}AB$ is PSD, but was not successful. Could anyone put some light how to prove it?2018-03-18
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    Use directly the definition of PSD and the characterization A PSD <=> A=C C* for some C.2018-03-19
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    Stupid question from my side. We can define $y=Bx$, else is straightforward.2018-03-19