My question: suppose $A$ and $B$ are two Hermitian positive-definite matrices (they don't commute: $AB\neq BA$). Are all eigenvalues of $AB$ positive?

I am completely confused. I know because $A$ and $B$ don't commute, $AB$ is not positive-definite. But according to: https://en.wikipedia.org/wiki/Positive-definite_matrix having all positive eigenvalues is equivalant to being positive-definite. So, since $AB$ is not positive-definite, not all its eignevalues are positive.

But here: For real matrices, if $A$ and $B$ are both positive-definite, show that all of $AB$'s eigenvalues are positive. in the answer it is somehow shown that all eigenvalues of $AB$ are positive.

So, what is the difference between these two cases? and what is the correct answer to my question?



The key here is that for a symmetric matrix, all eigenvalues being positive is a necessary and sufficient condition for being positive definite.

While $A^T=A$ and $B^T=B$, we have that $(AB)^T=B^TA^T=BA \neq AB,$ so $AB$ cannot be a positive definite matrix because it isn't symmetric. It can still have only positive eigenvalues.

For example, we have $$\begin{pmatrix}1&0\\0&2\end{pmatrix}\begin{pmatrix}4&1\\1&4\end{pmatrix}=\begin{pmatrix}4&1\\2&8\end{pmatrix}$$.

The pairs of eigenvalues are $1,2$ for $A$, $3,5$ for $B$ and $6 \pm \sqrt{6}$ for $AB\neq BA.$.


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