# The eigenvalues of the product of a positive definite and a symmetric matrix.

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?

• "Obviously if B is either positive or negative definite..." Obviously? $A B$ is not even symmetric in general. I think you need to place more conditions. Oct 11, 2013 at 21:25
• @leonbloy What I said is true...see e.g. here or here. The eigenvalues of the product will be positive (or negative), even if the product is not symmetric. This follows since $\sqrt{A}B\sqrt{A}$ is symmetric and $\lambda(AB)=\lambda(\sqrt{A}B\sqrt{A})$ Oct 11, 2013 at 21:28
• Ah, it's ok then. But I doubt there is a simple rule, unless A and B have a common eigenvectors base. Oct 11, 2013 at 21:29
• I doubt you can say much or anything about the eigenvalues of $AB$, since you know so little about $B$. Note that $AB$ does not even need to be symmetric. I think trying all those matrix decompositions is a waste of time. Oct 12, 2013 at 1:00

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.