How to prove that $B$ is nilpotent. 
Let $A$ and $B$ be complex matrices with $AB^2-B^2A=B$. Prove that $B$ is nilpotent.

By the way: This problem is from American Mathematical Monthly, Problem 10339,and this solution post 1996 American Mathematical Monthly, page 907.
My question: This problem has other nice solutions? Thank you.


Without loss of generality, assume that $B$ is already in its Jordan normal form. Partition $A$ according to the Jordan block structure of $B$. Denote the $(i,j)$-th submatrix of $A$ by $A_{ij}$ and similarly for the subblocks of $B$. Now the given condition implies that $A_{ii}B_{ii}^2-B_{ii}^2A_{ii}=B_{ii}$. Since the LHS is traceless, the Jordan block on the RHS must be nilpotent. Thus all Jordan blocks of $B$ are nilpotent and $B$ is nilpotent as well.
