Let \(V\) be the space of \(n \times n\) matrices over a field \(F\), and let \(A\) be a fixed \(n \times n\) matrix over \(F\). Define a linear operator \(T\) on \(V\) by \(T(B) = AB - BA\). Prove that if \(A\) is a nilpotent matrix, then \(T\) is a nilpotent operator.
Solution
Let \[ L(B)=AB, \qquad R(B)=BA. \] Then \[ T=L-R. \]
1. The operators \(L\) and \(R\) commute and are nilpotent when \(A\) is nilpotent.
For every \(B\in V\), \[ LR(B)=A(BA)=(AB)A=RL(B), \] so \(LR=RL\). If \(A^m=0\), then \[ L^m(B)=A^mB=0, \qquad R^m(B)=BA^m=0 \] for every \(B\). Hence \(L^m=R^m=0\) [@HK71].
2. A sufficiently high power of \(T=L-R\) vanishes.
1.Therefore\[ T^{2m-1}=0. \]So \(T\) is nilpotent.