Simple Trace Question I'm clearly lacking insight on the following problem...
So let $C$ be a $2\times2$ matrix with entries in some field.  I want to know when I can find $A, B$ (also $2\times2$) such that $C=AB-BA$.  In particular, you should be able to find them iff ${\rm Tr}(C)=0$ right?  I'm having trouble with the "given ${\rm Tr}(C)=0$, prove that $A, B$" exist direction.
Methinks the problem is intended to be done with minimal background...i.e. knowing what a determinant is is less than kosher (nor should I know that $AB-BA$ is $[A,B]$).  
 A: In addition to the question joriki linked to, there exists several more references on the subject. Notably this paper seems to be quite elementary, with the main focus of the proof being a clever lemma. Another reference is this, but it uses the rational canonical form which may be too much.
A: Let $$C=\pmatrix{r&s\cr t&-r\cr}$$ Let $$A=\pmatrix{a&b\cr c&d\cr}\qquad B=\pmatrix{x&y\cr z&w\cr}$$ In a deleted answer, NKS did the hard work of calculating $$
AB-BA = \left( \begin{array}{cc} bz - cy & (a-d)y + b(w-x)\\(d-a)z + c(w-x)&cy-bz \end{array}\right)
$$
Assuming that calculation is correct (I haven't checked it), let $b=1,z=r,c=y=0$, then all we need is $w-x=s,(d-a)r=t$. This is fine unless $r=0$ and $t\ne0$. If $r=0$, let $b=z=c=y=1$, and $a-d+w-x=s,d-a+w-x=t$. 
EDIT: Having checked the calculations, I think it should be $$
AB-BA = \left( \begin{array}{cc} bz - cy & (a-d)y + b(w-x)\\(d-a)z + c(x-w)&cy-bz \end{array}\right)
$$ If $r\ne0$ then we can take $b=1,z=r,c=y=0$ and take $a,d,w,x$ to satisfy $w-x=s,d-a=t/r$. If $r=0$, we can take $c=z=t,b=y=-s$ and take $a,d,w,x$ to satisfy $d+x-a-w=1$.  
A: In your particular case it is particularly simple:
1) Prove that $\,T_0:=\{A\in M_2(\Bbb F)\;\;|\;\;tr. A=0\}\,$ is a subspace of dimension 3 in $\,M_2(\Bbb F)\,$ (hint: $\,T_0=\ker tr.\,$, and since $\,trace=tr.\,$ is a non-zero linear functional its kernel is a hyperplane=a proper subspace of maximal dimension)
2) Show that $\,M':=\operatorname{Span}\,\{\,[A,B]:=AB-BA\;\;|\;\;A,B\in M_2(\Bbb F)\,\}\,$ is a vector subspace of dimension 3 (hint:
$$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\,,\,B=\begin{pmatrix}w&x\\y&z\end{pmatrix}\Longrightarrow AB-BA=\begin{pmatrix}k&**\\*&-k\end{pmatrix}$$ 
with $\,k\,,\,*\,,\,**\in\Bbb F\,$ , so using this show the following is a basis for $\,M'\,$:
$$\left\{\begin{pmatrix}0&1\\0&0\end{pmatrix}\,\,,\,\,\begin{pmatrix}0&0\\1&0\end{pmatrix}\,\,,\,\,\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right\}$$ 
3) Finally, it's easy to see that $\,M'\leq T_0\,$ , so...
A: This isn't the answer you are looking for, but it is stronger than the other answers so far. I.e. showing that given an nxn matrix, you can express it as the sum of $2n-1$ commutators.
You can show that given an nxn matrix $A$, it is the sum of 2 commutators. Let $A = D + N$, where $D$ is diagonal with trace zero and $N$ has zeros on its diagonal. You can then represent each of $D$, $N$ as a commutator. (This takes a little thought, but not too hard, I can give details if you need them)
