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Let $M,N$ be $n\times n$ matrices. Then why is it that $MN-NM=I_n$ cannot be true, where $I_n$ is the $n\times n$ identity matrix?

I am thinking of perhaps there is an argument using determinants? (Of course I am probably way out.)

Thanks.

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    You could note that $MN$ and $NM$ have the same eigenvalues, but $MN=NM+I_n$ would imply that the eigenvalues of $MN$ are all shifted by $1$ from those of $NM$.2011-11-07

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Thanks to @t.b.,

$\operatorname{tr}(AB-BA)=\operatorname{tr}(I) \implies \operatorname{tr}(AB)-\operatorname{tr}(AB)=n$

Contradiction.

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    Do you know about modular arithmetic? Working modulo 2 (for example), $1+1=0$, so the $2\times2$ identity matrix with entries taken from the integers mod 2 has trace zero. So in that setting, trace of $AB$ equals trace of $BA$ doesn't contradict $AB-BA=I$.2011-11-08