Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 8 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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.