Finding $\text{ker}(Tr)$ Defining the trace in the usual way as a function $Tr: F^{n\times n} \rightarrow F$, where $F$ is some field.  I want to show that $\text{ker}(Tr)=\text{span}_F(\{AB-BA|A,B\in F^{n\times n}\})$.
So far I've certainly deduced that $\text{im}(Tr)=F$, and so $\text{dim}(\text{ker}(Tr))=n^2-1$. Now I need only find a basis for $S$ (which I'll use to denote the spanning set above), since clearly $S\subseteq \text{ker}(Tr)$.
I've fooled around writing $A$ and $B$ as linear combinations of the standard $E_{ij}$, and messing around with the indices to find redundancies etc, but I can't seem to make any headway.  Am I missing some clever trick.  Perhaps I need to rewrite the $E_{ij}$ in some more useful way?
 A: Choosing $A=E_{ij}$ and $B=E_{jk}$ with $i\not=k$ gives $E_{ik}$ in $S$.
The same choice with $i=k$ and $j=i+1$ gives $E_{ii}-E_{i+1,i+1}$.
Those matrices clearly form an independent set in the kernel of trace of the appropriate size, so they form a base.
A: There are a few approaches. First the commutators over $F$ are in fact a subspace. Every commutator is a traceless matrix and conversely, every traceless matrix can be expressed as a commutator. This immediately shows that the span of commutators is the span of traceless matrices i.e. the subspace of all traceless matrices. Admittedly this is not constructive and not quite satisfying.
Alternatively, the following holds
$$\left[E_{i1},\ E_{1j}\right] = \begin{cases} E_{ij} & i\neq j \\ E_{ii}-E_{11} & i=j\end{cases}$$
which clearly provides $(n^2 - n) + (n-1) = n^2 - 1$ linearly independent commutators which span the space of traceless matrices.
Letting $S$ be the space of traceless matrices, the above shows 
$$S\subseteq\rm{span}\{AB-BA|A,B\in F^{n\times n}\}$$
Since commutators are traceless, every linear combination of commutators remains traceless, so we naturally have $$\rm{span}\{AB-BA|A,B\in F^{n\times n}\} \subseteq S$$
This shows that the two sets coincide.
