# $AB=BA=0$ with zero trace

I have follow matrix equation ($A$ and $B$ are $N\times N$ matrices over some field)

$AB=BA=0$

with

$A\ne 0$

$\det{A}=0$

$\operatorname{Tr}{A}=0$

I suppose that

$\operatorname{Tr}{B}=0$

Is it so? And if it's so then what is the idea of the proof?

Thank you.

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Why? $A$ could be zero and then $B$ can be arbitrary. – t.b. Nov 14 '11 at 14:26
Since $A$ and $B$ are simultaneously triangularizable, we may assume they are upper triangular. Then you will readily find that the trace of $B$ need not vanish. In particular, just restrict our attention to the diagonal case. – Sangchul Lee Nov 14 '11 at 14:38
@Ximik: The question isn't stupid nor should it be closed. It is a well-posed and meaningful, even if the conclusion you were after isn't true. – Dimitrije Kostic Nov 14 '11 at 15:17
@t.b. why are you posting full, correct answers as comments? This way the operator cannot close the question and other people cannot answer it because we don't want to take the credits for your results. – Listing Nov 14 '11 at 15:19
@Listing: I don't really consider that as a full correct answer, but on your insistence I've posted it. – t.b. Nov 14 '11 at 15:48

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0\end{bmatrix}$ and let $B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$.