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Suppose that $A$,$B$ are square $n\times n$ matrixes with real entries. $A$ is positive semidefinite and $B$ is negative semidefinite. Then we know that $AB$ is negative semidefinite. Suppose that $trace(AB)=0$.

Can we get some more information on the matrixes $A$ and $B$?

thank you

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The following is related… – copper.hat Oct 10 '12 at 19:34
up vote 0 down vote accepted

First, it is not true that $AB$ is negative semidefinite. Let $$ A=\begin{bmatrix}1&1\\1&1\end{bmatrix}, \ \ \ B=\begin{bmatrix}2&3\\3&1\end{bmatrix}. $$ Then $A$ is positive semidefinite, $B$ is negative semidefinite, but $AB$ is neither: $$ \begin{bmatrix}1\\1\end{bmatrix}^TAB\begin{bmatrix}1\\1\end{bmatrix}=18,\ \ \ \begin{bmatrix}11\\-12\end{bmatrix}^TAB\begin{bmatrix}11\\-12\end{bmatrix}=-7. $$

Regarding your concrete question, I don't think you can say anything about your matrices: let $A_0$ be any positive semidefinite matrix, $B_0$ any negative semidefinite matrix, and let $$ A=\begin{bmatrix}A_0&0\\0&0\end{bmatrix}, \ \ B=\begin{bmatrix}0&0\\0& B_0\end{bmatrix}. $$ Then $AB=0$, so $\mbox{tr}(AB)=0$.

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