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.
Best Answer
This is in the same spirit to your proof, but presented in a different way. If $\lambda B = AB-BA$ for some $B\ne0$ and some $\lambda$ in the algebraic closure of $F$, then $(A-\lambda I)B=BA$ and $(A-\lambda I)^k B=BA^k$ for any $k\ge1$. In particular, $(A-\lambda I)^nB=0$. However, if $\lambda$ is nonzero, $A-\lambda I$ would be invertible and hence $B=0$, which is a contradiction.