Solving $AB - BA = C$
Solution 1:
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.