TITLE: Suppose $A,B\in M_n(\mathbb{C})$ such that $AB-BA=A $ . Prove that $A$ is not invertible .
QUESTION [16 upvotes]: Suppose that $A,B\in M_n(\mathbb{C})$ such that $AB-BA=A $. Prove that $A$ is not invertible.

My work:
Suppose $A$ is invertible. Then $ABA^{-1}=I+B$ . So $B$ is similar to $I+B$  .Let $B$ have eigenvalues $c_1,c_2,\ldots,c_n \in \mathbb{C}$. So $B$ has basis such that $B$ is upper triangular with respect to it and  has $c_1,\ldots,c_n$ as diagonal entries . It is easy to see that $I+B$ is upper-triangular with respect to this basis and has entries $1+c_1,\ldots, 1+c_n$ . 
Hence $$c_1+c_2+\ldots +c_n=\operatorname{trace}(B)$$ $$=\operatorname{trace}(I+B)=(1+c_1)+\ldots+(1+c_n)=n+c_1+\ldots +c_n .$$ 
So $n=0$, contradiction. I'm not sure if my solution is correct one. It seems alright. I will be very thankful if you can confirm that the proof indeed is a correct one.
Any other possible solutions are welcomed.

REPLY [4 votes]: $\def\m{\mathfrak }$This result (and, in fact, orangeskid's observation) has the following far reaching generalization:

if $\m g$ is a finite dimensional (complex) Lie algebra and $\mathfrak r$ is its radical, then $[\m g,\m r]$ acts on any finite dimensional representation of $\m g$ nilpotently.

Indeed, in the question the matrices $A$ and $B$ span a Lie algebra $\m g$ of dimension (at most, really, but let us suppose) equal to $2$ which is solvable, so that the radical of $\m g$ is simply $\m g$ itself, and the theorem above tells us that $[\m g,\m g]$ acts nilpotently on finite dimensinal modules: since $[\m g,\m g]$ is spanned by $A$, the desired result follows.
Of course, this is immensely more general.
As an example: 

if $A$, $B$ and $C$ are square matrices such that $CA-AC = B$, $CB-BC = aA + B$, $AB-BA = 0$, then $B$ is nilpotent and,  more generally, al linear combinations of $A$ and $B$.

This comes from looking at a random 3-dimensional solvable Lie algebra.