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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.2011-11-14
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    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.2011-11-14
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    @t.b., sorry, forgot about this. I've edited the post.2011-11-14
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    Yes, I see. So the question is stupid and can be closed. Thank you.2011-11-14
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    @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.2011-11-14
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    @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.2011-11-14
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    @Listing: I don't really consider that as a full correct answer, but on your insistence I've posted it.2011-11-14

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