Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
According to the answers, this is not true, but can anyone give me an explicit counterexample? Thanks. – josh Sep 22 '11 at 1:01
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 1:21
Oh, duh. I'm an idiot. Thank you! – josh Sep 22 '11 at 1:26

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.