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 $(AB)^\ast = AB$, i.e., $AB$ is Hermitian. Is this statement true? If not, does anyone have a counterexample?
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According to the answers, this is not true, but can anyone give me an explicit counterexample? Thanks. – josh Sep 22 '11 at 01:01
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1Just try any two small Hermitian matrices; chances are, their product will not be Hermitian. – Sep 22 '11 at 01:11
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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! – josh Sep 22 '11 at 01:21
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1Oh, it's easy to make them positive definite. – Sep 22 '11 at 01:24
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Oh, duh. I'm an idiot. Thank you! – josh Sep 22 '11 at 01:26
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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. – Sep 22 '11 at 01:28
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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. – J. M. ain't a mathematician Sep 22 '11 at 02:56
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Or generate $n$ orthonormal vectors $u_j$ and $n$ positive numbers $\lambda_j$ and take $\sum_j \lambda_j u_j u_j^T$. – Robert Israel Sep 22 '11 at 03:52
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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.
suvrit
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1(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) – suvrit May 12 '17 at 14:40
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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! – user2457324 Jul 30 '17 at 16:29
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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.
Shaun Ault
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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.
Manos
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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? – Rajat Mar 18 '18 at 16:13
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1Use directly the definition of PSD and the characterization A PSD <=> A=C C* for some C. – Manos Mar 19 '18 at 14:42
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Stupid question from my side. We can define $y=Bx$, else is straightforward. – Rajat Mar 19 '18 at 16:09
