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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 symmetric2012-02-16

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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. Tha$n$ks a lot!2012-02-17