TITLE: let $A,B\in M_{n}(C)$ such that c is complex field and $AB^2-B^2A=B$ how prove $B^n=0$
QUESTION [6 upvotes]: 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

REPLY [2 votes]: 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 $B^n=0$.