If $A$ and $B$ are positive-definite matrices, is $AB$ positive-definite? 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.
