# Product of positve definite matrix and seminegative definite matrix

Let $A$ a spd (symmetric positive definite) matrix and $B$ a symmetric seminegative definite matrix. Is tr $AB \leq 0$ and more general is $AB$ seminegative definite?

I know that tr $AB \leq 0$ follows from $AB$ seminegative definite since the eigenvalues $\lambda$ of $AB$ are nonpositve and hence tr $AB=\sum_{\lambda \in spec\ A} \lambda \leq 0$. But I don't know how to find something out about the definitness of $AB$. I think in general there is nothing you can say about the eigenvalues of $AB$.

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$AB$ is not necessarily symmetric, except if $A$ and $B$ commute. –  Davide Giraudo May 19 '12 at 16:01
Yes I know that. But the question is wheater $AB$ is seminegative definite or at least weather the trace is nonpositve. –  Julian May 19 '12 at 16:05
I believe the term is "negative semidefinite". And I suspect that the reason Davide pointed out that $AB$ isn't necessarily symmetric is that sometimes symmetry is considered a prerequisite for positive/negative (semi)definiteness. –  joriki May 19 '12 at 16:16
If you only cared about the trace, you could note that $\sqrt AB\sqrt A$ is negative semidefinite, and $\mathrm{tr}(\sqrt AB\sqrt A)=\mathrm{tr}(AB)$. –  Jonas Meyer May 25 '12 at 7:04

First, take $A$, $B$ symmetric positive-definite. Suppose $\lambda$ is an eigenvalue of $AB$ with corresponding eigenvector $x\neq 0$, i.e. $ABx=\lambda x$.Then $BABx=\lambda Bx$ and so $x' BAB x = \lambda x' B x$. It is not hard to check that $BAB$ will also be positive definite. Since $x \neq 0$, $x'Bx \neq 0$, thus $\lambda = \frac{x' BAB x}{x'Bx}$. By the positive definiteness of $B$ we have $x' Bx >0$. By the positive definiteness of $BAB$ we will have $x' BAB x>0$. Thus $\lambda >0$, i.e. $AB$ has positive eigenvalues.
Now let $A>0$ and $B<0$. Apply the above result to $-(AB)=A(-B)$. Since $-B>0$, all eigenvalues of $A(-B)$ will be positive. But the eigenvalues of $A(-B)$ are the negative counterparts of the eigenvalues of $AB$. Thus $AB$ will have negative eigenvalues.
Note however, that $AB$ needs not be symmetric. The terminology "negative-definite" refers to hermitian (symmetric) matrices.