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

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.

  • 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
  • hi--I was wondering if you may be able to shed any light on this question mathoverflow.net/questions/277545/… that makes use of determinant inequalities? I would be much appreciative. thank you! – user2457324 Jul 30 '17 at 16:29

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.

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

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.

  • 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 at 16:13
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    Use directly the definition of PSD and the characterization A PSD <=> A=C C* for some C. – Manos Mar 19 at 14:42
  • Stupid question from my side. We can define $y=Bx$, else is straightforward. – Rajat Mar 19 at 16:09

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