let $A,B\in M_{n}(C)$ such that c is complex field and $AB^2-B^2A=B$ how prove $B^n=0$ Let $A,B\in M_{n}(C)$  such that $C$ is complex field and $AB^2-B^2A=B$. How prove $B^n=0$.
thanks in advance   
 A: First you should note that $AB^{n+2}-B^{n+2}A=(n+1)B^{n+1}$ because
$AB^{n+2} - B^{n+2}A=(AB^2-B^2A)B^{n}+B^2(AB^{n}-B^{n}A)=B^{n+1}+B^2(nB^{n-1})=(n+1)B^{n+1}$ (this is a proof by induction in $n$).
This implies that $tr(B^n)$ is $0$ for all $n$ (because the trace of a commutator is always $0$). It is a nice exercise to show that this implies that $B$ is nilpotent.
IDEA: You have to work with the eigenvalues of $B$. The "easy" case is when $B$ has simple eigenvalues.
EDIT: This is a particular case of Jacobson's lemma. A nice paper on the subject.
EDIT II: If $B$ has simple eigenvalues $\lambda_1,\dots\lambda_k$ then $0=tr(B^n)=\sum \lambda_i^n$. Let $M$ be a matrix such that $M_{ij}=\lambda_i^j$. This matrix is a non-singular Vandermonde matrix and the vector $(1,\cdots,1)$ is in the kernel, this is a contradiction which came from the assumption that $B$ has simple eigenvalues. You should adapt this proof to the case where $B$ has a non-zero eigenvalue (and reach the same contradiction showing that $B$ has to be $0$).
EDIT III: As noted by user1551 in his comment the first part of the "proof" above only works for even $n$. However the second part of the proof using the Vandermonde matrix still works (but needs some modifications).
A: The equality $AB^2-B^2A=B$ implies that $PAP^{-1}(PBP^{-1})^2-(PBP^{-1})^2PAP^{-1}=PBP^{-1}$ for any invertible matrix $P$. Therefore, we may assume WLOG that $B$ is already in its Jordan form. Let $B=J_1\oplus J_2\oplus\ldots\oplus J_k$, where $J_1,J_2,\ldots,J_k$ are Jordan blocks of possibly different sizes. Let those diagonal blocks of $A$ with conforming sizes be $A_1,A_2,\ldots,A_k$. Then $AB^2-B^2A=B$ implies that $A_iJ_i^2-J_i^2A_i=J_i$ for each $i$. Since the LHS has zero trace, the $J_i$ on the RHS must have a zero diagonal. Therefore $B$ is nilpotent and $B^n=0$.
