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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 $dim(W)<n^2$?

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  • $\begingroup$ 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
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Every element of $W$ has $0$ trace.

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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.

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  • $\begingroup$ 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

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