Let $F$ be a field.

Let $W$ be the subspace of $M_n(F)$ generated by elements of the form $AB-BA$.

How do I prove that $dim(W)<n^2$?

  • $\begingroup$ Do you mean $W=\{AB-BA\colon A,B\in M_n(\mathbb F)\}$ or $W=\langle\{AB-BA\colon A,B\in M_n(\mathbb F)\}\rangle$? $\endgroup$ – Git Gud Apr 21 '15 at 21:47
  • $\begingroup$ @GitGud I meant the subspace not just that set, but if $W$ is the ideal generated by that, then is $W=M_n(F)$? $\endgroup$ – Rubertos Apr 21 '15 at 21:54
  • $\begingroup$ To me it's not even obvious that $\{AB-BA\colon A,B\in M_n(\mathbb F)\}$ is a vector space. How do you prove that the sum is closed? Answering your question,$\langle X \rangle$ here denotes the smallest subspace containing $X$, so if $\{AB-BA\colon A,B\in M_n(\mathbb F)\}$ is indeed a vector space, then $\langle \{AB-BA\colon A,B\in M_n(\mathbb F)\}\rangle=\{AB-BA\colon A,B\in M_n(\mathbb F)\}$. $\endgroup$ – Git Gud Apr 21 '15 at 21:58
  • $\begingroup$ @GitGud I don't know that set is indeed a subspace or not. However, as I wrote in my post, I wrote $W$ to denote the subspace generated by that set. And even I think it's not going to be easy to prove or disprove that the set $\{AB-BA:A,B\in M_n(F)\}$ is indeed a subspace, but that wasn't my question. $\endgroup$ – Rubertos Apr 21 '15 at 22:25
  • $\begingroup$ Again, to me it's not obvious how the answers help. Why should it be the case that all the matrices in $\langle\{AB-BA\colon A,B\in M_n(\mathbb F)\}\rangle$ are of the form $AB-BA$? $\endgroup$ – Git Gud Apr 21 '15 at 22:53

Every element of $W$ has $0$ trace.


Look at the trace. $$ tr(AB-BA)= tr(AB)-tr(BA)=tr(AB)-tr(AB)=0 $$ Since (presumably) you are looking at linear combinations and products, you can't make the trace nonzero. There are elements with nonzero trace in $M_n$, which you can't get to.

  • $\begingroup$ Is it seems to me that you're assuming that every matrix in $\langle\{AB-BA\colon A,B\in M_n(\mathbb F)\}\rangle$ is of the form $AB-BA$. Is this true? $\endgroup$ – Git Gud Apr 21 '15 at 23:01
  • $\begingroup$ @GitGud Since the trace is linear and we are looking at finite linear combinations, I don't see that there's anything that needs mentioning separately. $\endgroup$ – Chappers Apr 21 '15 at 23:26
  • $\begingroup$ You want to prove that $\forall M\in \langle\{AB-BA\colon A,B\in M_n(\mathbb F)\}\rangle(\text{tr}(M)=0)$, correct? To this effect you take an arbitrary $M\in \langle\{AB-BA\colon A,B\in M_n(\mathbb F)\}\rangle$, correct? Now what I get from your answer is that you assume without explanation that $M=AB-BA$ for some matrices $A,B$. And why this is the case, I have no idea. $\endgroup$ – Git Gud Apr 21 '15 at 23:31
  • $\begingroup$ It's a subspace spanned by elements of the form $AB-BA$, according to the OP. The trace is linear, so the trace and finite sums can be interchanged. Hence each element of the subspace is of the form $\sum_i \lambda_i C_i$, where $C_i$ is of the form $AB-BA$. Then $tr(\sum_i \lambda_i C_i) = \sum \lambda_i tr(C_i)$, and $tr(C_i) = 0$. $\endgroup$ – Chappers Apr 22 '15 at 0:06
  • $\begingroup$ "Hence each element of the subspace is of the form (...)" How does that follow from what's before? $\endgroup$ – Git Gud Apr 22 '15 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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