TITLE: $T$ be linear operator on $V$ by $T(B)=AB-BA$. Prove that if A is a nilpotent matrix, then $T$ is a nilpotent operator.
QUESTION [12 upvotes]: Let $V$ be a vector space of $n\times n$ matrices over a field F, and let $A$ be a fixed  $n\times n$ matrix.  $T$ be linear operator on $V$ by $T(B)=AB-BA$. Prove that if A is a nilpotent matrix, then $T$ is a nilpotent operator.
I have done in a way that
$T^2(B)= T(AB-BA)= A^2B-2ABA+BA^2\Rightarrow T^3(B)=A^3B-3A^2BA+3ABA^2-BA^3$ 
Proceeding in this way since A is nilpotent $\exists m\in\Bbb N$ s.t   $T^m(B)=0$.
Hence T is also a nilpotent operator.
This a problem of Hoffman Kunze of Primary Decomposition Chapter. So can anyone give me any other solution because this solution depends on some sort of intution. Please don't use any Rational and Jordan Forms formula.

REPLY [2 votes]: Note that 
$$T=L-R
\qquad L=\textrm{Multiply $X$ at Left}\qquad  R=\textrm{Multiply $X$ at right}$$
Since 
$$L\circ R=R\circ L\qquad R^{n}=0\qquad L^{n}=0$$
if $A^n=0$, then $T^{2n}=0$ by binomial theorem.