Symmetries

Suppose $A,B\in\Hom(E)$ satisfy $[A,B]=\b B$. Prove by induction on $n$ that for all $n\in\N$ and all $\l\in\C$: $(A-\l-\b)^nB=B(A-\l)^n$.

Since $[A-\l,B]=[A,B]$, we may assume w.l.o.g. that $\l=0$. For $n=1$ we have: $AB=\b B+BA$, i.e. $(A-\b)B=BA$. Now assume $(A-\b)^nB=BA^n$, then \begin{eqnarray*} (A-\b)^{n+1}B &=&(A-\b)(A-\b)^nB =(A-\b)BA^n\\ &=&ABA^n-\b BA^n =(\b B+BA)A^n-\b BA^n =BA^{n+1} \end{eqnarray*}