If $A=AB-BA$, is $A$ nilpotent? Let matrix $A_{n\times n}$, be such that there exists a matrix $B$ for which
$$AB-BA=A$$
Prove or disprove that there exists $m\in \mathbb N^{+}$such 
$$A^m=0,$$
I know
$$tr(A)=tr(AB)-tr(BA)=0$$
then I can't.Thank you

For any $k$, you have  $A^{k+1}=AA^k=(AB-BA)A^k=ABA^k-BA^{k+1}$. So
$$
\text{tr}(A^{k+1})=0,\ \  k=0,1,2,\ldots
$$
We deduce that $\text{tr}(p(A))=0$ for every polynomial $p$ with $p(0)=0$. This implies that all eigenvalues of $A$  are zero (because otherwise we can get a polynomial $p$ that is 1 at all nonzero eigenvalues, and $p(A) $ would not have zero trace).  So $A$ is nilpotent,  i.e. there exists $m$ with $A^m=0$.
