# Product of positive definite and seminegative definite matrices

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 $AB$.

• $AB$ is not necessarily symmetric, except if $A$ and $B$ commute. 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. 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. 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 $\mathrm{tr}(\sqrt AB\sqrt A)=\mathrm{tr}(AB)$. 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.
• Oh! I see now. You rely on the fact that $B=B'$. Mar 7 '20 at 19:35
• I agree that $BAB$ is positive definite, but what prevents $x'BABx'$ from being zero for $x \neq 0$? More specifically, we could have $x \neq 0$ but $Bx = 0$, which would make $x'BABx$ zero and thus eigenvalue zero, meaning $AB$ is positive semi-definite. Why isn't this possible? Mar 7 '20 at 19:55
• Oh! I realized that because $B$ is positive definite, for $x \neq 0$, $Bx \neq 0$ because otherwise $x'Bx=0$, contradicting that $B$ is positive definite. Mar 7 '20 at 19:56