Solving $AB - BA = C$ Suppose $C$ is an $n\times n$ matrix over complex numbers, with trace $0$.
Are there always $n\times n$ matrices $A,B$ such that $AB - BA = C$?
(Inspired by a recent question which asked for a trace free proof of non-existence of solutions for $C=I$).
 A: Yes, this is always true.  See a proof, for example, over here. The statement is proven by induction on $n$.
The key to the proof presented in the link is the following proposition:

Lemma 2: if $S \neq \lambda I$ for any scalar $\lambda \neq 0$, then $S$ is similar to a matrix with a $0$ in the $(1,1)$ entry.

There is a similarly useful extension of this statement in Horn and Johnson which says that every matrix is similar to some matrix whose diagonal entries are identical.
A: this is not an answer but a longer comment. let me try the case $n = 2.$ take the matrices $\pmatrix{a_1 & a_2 \cr a_3 & a4}, B = \pmatrix{b_1 & b_2 \cr b_3 & b_4}, \mbox{ and} \pmatrix{c_1 & c_2\cr c_3 & -c_1}.$ writing out $AB - BA = C$ as a string of linear equations i get an over determined system
$$\pmatrix{0 & 0 & a_2 & -a_3\cr -a_2 & a_1 - a_4 & 0 & a_2\cr a_3 & 0 & a_4 - a_1 & -a_3} \pmatrix{b_1 \cr b_2\cr b_3 \cr b_4} = \pmatrix{c_1 \cr c_2 \cr c_3} $$
of course, one has to worry about inconsistency.
A: In fact, there are many solutions in $(A,B)\in \mathbb{C}^{2n^2}$. Let $f:(A,B)\rightarrow AB-BA\in \{U;tr(U)=0\}\approx \mathbb{C}^{n^2-1}$. It can be shown that, for a generic $(A,B)$, $f$ is a submersion in a neighborhood of $(A,B)$ (that is $Df_{A,B}$ is surjective). Then, for a generic $C\in \mathbb{C}^{n^2}$, $f^{-1}(C)$ is an algebraic set of dimension $2n^2-(n^2-1)=n^2+1$ (degrees of freedom). 
Beware 1. it is not true for particular $C$; for example, $f^{-1}(0_n)$ has dimension $n^2+n$.
Beware 2. Although $dim(f^{-1}(C))>n^2$, we cannot randomly choose $A$ (for example). 
EDIT. Since $rank(Df_{A,B})\leq n^2-1$, the minimum of $dim(f^{-1}(C))$ is $n^2+1$ (generic case). Now, about the maximum, I think that it is $n^2+n$ but I am not sure...
