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Let $W$ be the space of $n × n$ matrices over the filed $F$, and let $W_0$ be the subspace spanned by the matrices $C$ of the form $AB - BA$. Prove that $W_0$ is exactly the subspace of matrices which have trace zero. (Hint: What is the dimension of the space of matrices of trace zero? Use the matrix 'units,' i.e. matrices with exactly one non-zero entry, to construct enough linearly independent mayrices of the form $AB-BA$. )

Let $K$ be the set of all $n × n$ matrices of the form $AB-BA$, amd let $U$ be the subspace of all $n × n$ matrices with trace zero.

First, I am not sure how $K$ is a subspace. Certainly it's closed under scalar multiplication, but how is it closed under addition?

We know that $K$ is a subset of $U$, but I am not sure how it's the other way around. We know that the $dim U = n^2-1$, and I could find $n^2-n$ 'unit' matrices (each of which has only one 1 on a non-diagonal entry) and express them as $AB-BA$. But I am not sure how to deal with the ones with at least one nonzero diagonal entry.

I am not even sure what the hint means. Any help?

José Carlos Santos
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    It isn't necessarily closed under addition. This is why $W_{0}$ is taken to be the subspace spanned by matrices of the form $AB - BA$ in the prompt. – Alex Wertheim Jun 24 '17 at 20:45
  • As for the problem/hint, your progress is correct so far. If you have a $1$ on the diagonal, how could you ''balance'' this by summing with another appropriate unit matrix (with the right scaling factor)? – Alex Wertheim Jun 24 '17 at 20:49
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    @Alex Wertheim: The fact that any matrix of trace $0$ is a commutator ( and not only a sum of commutators) is a bit more delicate. One can write any matrix with $0$ diagonal as a commutator in an explicit way. Then, more involved, one can show that any matrix of trace $0$ is similar to a matrix with a $0$ diagonal. I think this is discussed in a book by Horn and Johnson. – orangeskid Jun 24 '17 at 21:58
  • @orangeskid: neat, didn't realize! Thanks for sharing. – Alex Wertheim Jun 24 '17 at 23:09
  • [${AB-BA }$ where $A$ and $B$ are square matrices] is this set a subspace?....If it is not a subspace then how traceless matrices are commutators?@AlexWertheim – cmi Jan 09 '19 at 20:45

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The space of the matrices whose trace is $0$ has dimension $n^2-1$. Now, consider the matrices $E_{ij}$ such that the entry at the row $i$ and column $j$ is $1$ and all others are $0$. Then all matrices of the type $E_{ij}$ (with $i\neq j$) together with the matrices of the type $E_{ii}-E_{jj}$ (again, with $i\neq j$) form a set of $n^2-1$ linearly independent matrices whose trace is equal to $0$. It is not hard to prove that each of these matrices is of the form $AB-BA$ for two matrices $A$ and $B$ of the type $n\times n$. For instance, $E_{ii}-E_{jj}=E_{ij}E_{ji}-E_{ji}E_{ij}$.

José Carlos Santos
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