Let $v$ be an eigenvector of $B$ to $\lambda$ which might not be diagonalizable and $ABv=BAv$ with $A$ which might also not be diagonalizable, how does that imply $A^kBv=BA^kv$?
A trivial result that follows is that $Av$ is also an eigenvector of $B$ to the same eigenvalue $\lambda$, so the result would follow if one could show that the associated eigenspace of $B$ is one dimensional which would imply that $v$ is also an eigenvector of $A$
The reason I need this is for a theorem on simultaneous triangularization of matrices.
EDIT: I just realized that if one could show that $A(AB-BA)v=(AB-BA)Av$ then the statement would follow through induction. Can anyone show this?
EDIT2: Can anyone say anything if one assumes that the matrices A and B are diagonalizable?