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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.

asker
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    You weren't way out: Consider the trace instead of the determinant and remember that $\operatorname{tr}{AB} = \operatorname{tr}{BA}$. – t.b. Nov 07 '11 at 00:24
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    And if you don't remember that fact about the trace, try to prove it. – Gerry Myerson Nov 07 '11 at 00:37
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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. – Srivatsan Nov 07 '11 at 00:41
  • 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 $NM$. – Jonas Meyer Nov 07 '11 at 03:43

1 Answers1

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Thanks to @t.b.,

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

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

Contradiction.

asker
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  • Good. For future reference, this argument won't work when you come to learn about matrices with entries in finite fields (or other fields of finite characteristic). – Gerry Myerson Nov 07 '11 at 11:40
  • And at some point you can also accept this answer (and an answer to your other question) by clicking the tick on the left. – Rasmus Nov 07 '11 at 14:50
  • @GerryMyerson: Interesting, would you mind illustrating? – asker Nov 08 '11 at 11:59
  • @Rasmus: Thanks for the suggestion! I have tried doing that as you said but it keeps saying that there's an error... – asker Nov 08 '11 at 12:01
  • 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 $AB-BA=I$. – Gerry Myerson Nov 08 '11 at 12:16