A fellow student posed the following question and I'd like to stop thinking about it so I can get back to work on my own research!
Suppose that $A>0$, i.e. $A$ is a real symmetric positive definite matrix, and $B$ is a real symmetric nonsingular matrix.
What can we say about the eigenvalues of $AB$? For instance, suppose $B$ has $n$ positive and $m$ negative eigenvalues. Will $AB$ have the same number of positive and negative eigenvalues?
Obviously if $B$ is either positive or negative definite, the result is straightforward, i.e. we have the `matrix sign rules'
$$
(+)\cdot (+)=(+)\qquad\text{and}\qquad (+)\cdot(-)=(-)
$$ Whereby we mean `a positive definite times a positive definite has positive eigenvalues' and 'a positive definite times a negative definite has negative eigenvalues'.
Playing around with matrix decompositions such as polar, spectral, etc, and identities such as $\{\lambda(AB)\}=\{\lambda(\sqrt{A}B\sqrt{A})\}$ (here $\lambda(\cdot)$ meaning `eigenvalues of') doesn't seem to lead to a quick result.
Any ideas?
Best Answer
Edit: Yes, the number of positive or negative eigenvalues in $AB$ and $B$ are the same. The spectrum of $AB$ is identical to the spectrum of $\sqrt{A}B\sqrt{A}$ and by Sylvester's law of inertia, $\sqrt{A}B\sqrt{A}$ and $B$ have the same number of positive/negative/zero eigenvalues.