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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 weren't way out: Consider the trace instead of the determinant and remember that $\operatorname{tr}{AB} = \operatorname{tr}{BA}$. –  t.b. Nov 7 '11 at 0:24
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And if you don't remember that fact about the trace, try to prove it. –  Gerry Myerson Nov 7 '11 at 0:37
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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 7 '11 at 0:41
    
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 7 '11 at 3:43
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1 Answer

Thanks to @t.b.,

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

Contradiction.

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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 7 '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 7 '11 at 14:50
    
@GerryMyerson: Interesting, would you mind illustrating? –  asker Nov 8 '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 8 '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 8 '11 at 12:16
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