8
$\begingroup$

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 $AB$.

Thanks in advance!

$\endgroup$
4
  • $\begingroup$ $AB$ is not necessarily symmetric, except if $A$ and $B$ commute. $\endgroup$ May 19 '12 at 16:01
  • $\begingroup$ Yes I know that. But the question is wheater $AB$ is seminegative definite or at least weather the trace is nonpositve. $\endgroup$
    – Julian
    May 19 '12 at 16:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – joriki
    May 19 '12 at 16:16
  • 1
    $\begingroup$ If you only cared about the trace, you could note that $\sqrt AB\sqrt A$ is negative semidefinite, and $\mathrm{tr}(\sqrt AB\sqrt A)=\mathrm{tr}(AB)$. $\endgroup$ May 25 '12 at 7:04
11
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I'm confused by something in your first answer. The product of two positive definite (PD) matrices is not PD, but your proof holds for two PD matrices regardless of whether they're symmetric. This would imply that the product of two PD matrices is PD. $\endgroup$ Mar 7 '20 at 19:32
  • $\begingroup$ Oh! I see now. You rely on the fact that $B=B'$. $\endgroup$ Mar 7 '20 at 19:35
  • $\begingroup$ 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? $\endgroup$ Mar 7 '20 at 19:55
  • $\begingroup$ 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. $\endgroup$ Mar 7 '20 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.