Prove that a 2 by 2 matrix C with tr(C)=0 can be written as C=AB-BA, where A and B are 2 by 2 matrices "Let C be a 2 by 2 matrix with tr(C) = 0. Prove that C can be written as C=AB-BA, where A and B are 2 by 2 matrices."
Although it doesn't say, I think the field is complex number field. Also, this question can be solved without using determinants or considering invertibility since it's from a introductory chapter that hasn't discussed invertibility and determinants.
Please provide a hint, and not the answer, to the solution.
 A: Personally, I would brute-force it: write $A$, $B$, and $C$ and try to solve for the variables of $C$ (where there is no redundancy) in terms of those of $A$ and $B$.
A: This would not be appropriate for the "beginning of the book", but anyway:
The condition of being a commutator is invariant under similarity, i.e. if
$C = AB-BA$ then $S C S^{-1} = (S A S^{-1}) (S B S^{-1}) - (S B S^{-1})(S A S^{-1})$.  Thus we can assume wlog that $C$ is in Jordan canonical form.
There are two possibilities:
$$ C = \pmatrix{a & 0\cr 0 & -a\cr}\ \text{and}\ C = \pmatrix{0 & 1\cr 0 & 0\cr}$$
In each case we can find an $A$ and a $B$ that each have only one nonzero entry.
A: Note : 
A=  \begin{bmatrix} a & b  \\ 
                    c & d 
  \end{bmatrix}
 B=
  \begin{bmatrix}
    a' & b' \\
    c' & d' 
  \end{bmatrix}
 C=
  \begin{bmatrix}
    x & y \\
    z & -x 
  \end{bmatrix}
C=AB-BA implies that :
$ x=bc'- b'c \\ y = ab' +bd'-b'd-a'b \\ z=a'c+c'd-ac'-cd' $ 
This System has 8 unknows and 3 equations we need to reduce the number of solutions to 1 just to proove the existence of A and B. By fixing $ a'=c=c'=d=d'=1 $
Now we have : 
$ x=b-b' \\ y= ab'-b' \\ z=1-a  $ 
And then you will find the value of $b, b', a$ , Now we have $ A , B $ such as $ C=AB-BA $ and $ Tr(C)=0 $ .
