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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.

  • 13
    You weren't way out: Consider the trace instead of the determinant and remember that $\operatorname{tr}{AB} = \operatorname{tr}{BA}$.2011-11-07
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    And if you don't remember that fact about the trace, try to prove it.2011-11-07
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    asker: Since you are new to Math.SE, let me point out that you are explicitly permitted and encouraged in this site to answer your own question; so if you understood @t.b.'s comment, please consider posting your answer. That way, this post will appear as answered in the question, and other users will get a chance to upvote your solution.2011-11-07
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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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