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The question is from Artin's Algebra. If $A$ and $B$ are two square matrices with real entries, show that $AB-BA=I$ has no solutions. I have no idea on how to tackle this question. I tried block multiplication, but it didn't appear to work.

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What happens when you take the trace of both sides? – user38268 Jan 23 '13 at 9:22
Thanks, I got it. – Ishan Banerjee Jan 23 '13 at 9:23
If you have, please answer your own question so it does not keep getting bumped to the main page. – user38268 Jan 23 '13 at 9:48
or accept Panu's answer. – robjohn Jan 23 '13 at 10:33
up vote 12 down vote accepted

As you should have known by now, for real matrices, the equation $AB-BA=I$ has no solution because the LHS has zero trace but the RHS is not traceless. The same conclusion holds for complex matrices. For other fields, an in-depth discussion can be found in the answers to a related question. In particular, it has been proven that a square matrix $M$ is a commutator (i.e. $M=AB-BA$ for some square matrices $A$ and $B$) if and only if $M$ is traceless:

A.A. Albert and Benjamin Muckenhoupt (1957), "On matrices of trace zeros". Michigan Math. J., 4(1):1-3.

It follows that $AB-BA=I_n$ has a solution if and only if $I_n$ is traceless over the underlying field.

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Let a $n\times n$ matrix. Take trace of both sides $$\operatorname{trace}(AB-BA)= \operatorname{trace}(I)\Rightarrow \operatorname{trace}(AB)- \operatorname{trace}(BA) =n\Rightarrow0=n$$

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