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 $C=I$).
On
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.
On
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...
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.