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Suppose we have two matrices $A$ and $B$ in which $A \preceq 0$ and $B \succeq 0$. If $\operatorname{tr}(AB) = 0$, can we conclude $AB = 0$ ? Say, $A$ and $B$ are all symmetric.

Thanks a lot!

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There is really no need to say A B are all symmetric – yaoxiao Feb 16 '12 at 23:50
up vote 3 down vote accepted

Yes. WLOG we can assume $A$ is diagonal. Then $\text{tr}(AB) = \sum_j a_{jj} b_{jj}$. Since $a_{jj} \le 0$ and $b_{jj} \ge 0$, the only way this can be $0$ is all $a_{jj} b_{jj} = 0$. But if $B$ is positive semidefinite and $b_{jj} = 0$, we must have $b_{jk} = 0$ for all $k$, so $(AB)_{jk} = a_{jj} b_{jk} = 0$.

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Got you. Thanks a lot! – mining Feb 17 '12 at 0:07

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