Particular commutator matrix is strictly lower triangular, or at least annihilates last base vector Let $(e_1,e_2,\ldots,e_n)$ be the canonical basis of ${\mathbb C}^n$. Let $A$
be a $n\times n$ matrix such that $Ae_k=e_{k+1} (1\leq k \leq n-1$ (so everything in
$A$ is specified except for the last column). Let $B$ be another $n\times n$ matrix,
such that $C=AB-BA$ commutes with $A$. Can anyone prove or disprove that
(1) $Ce_n=0$.
(2) $C$ is a strictly lower triangular matrix.
Of course, (2) is much stronger than (1). I have checked (2) for $n\leq 3$. We know that $C$ is nilpotent (see $AB-BA$ is a nilpotent matrix if it commutes with $A$) ; not sure if that helps.

It's true when $charpoly(A)$ has only simple roots; otherwise it's false.
Assume $n=4$ and consider 
$A=\begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&2\\0&0&1&0\end{pmatrix},B=\begin{pmatrix}0&0&-1&0\\0&-1&0&-2\\-1&0&0&0\\0&0&0&1\end{pmatrix}$ where $charpoly(A)=charpoly(B)=(x-1)^2(x+1)^2$.
Then $C=\begin{pmatrix}0&1&0&1\\1&0&1&0\\0&-1&0&-1\\-1&0&-1&0\end{pmatrix}$ is not triangular and $Ce_4\not=0$.
EDIT. If $A$ has only simple eigenvalues $(\lambda_i)$, then we may assume that $A=diag(\lambda_i)$; thus $AB-BA$ is a nilpotent diagonal matrix, that is a zero matrix. Finally $AB=BA$ and $B$ is diagonal too.
