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 $[A,B]$).

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 $t\
e0$. 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 $r\
e0$ 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 $d+x-a-w=1$.