Proof by Induction with different Eigenvalues

Question

$SA=AT$, where $S$ and $T$ have no eigenvalues in common. Prove by induction that $(T-\lambda I)^kv=0$ implies that $Av=0$.

Here is what I have so far:

$k=1$: $$ (T-\lambda I)v=0 $$$$ Tv=\lambda v $$$$ SAv=ATv=\lambda Av $$$$ SAv=\lambda Av $$ $Av$ must equal $0$ because $S$ and $T$ share no eigenvectors.

$k=n$: $$ (T-\lambda I)^nv=0 \textrm{ implies } Av=0 $$

$k=n+1$: now what?

Answer 1

Split the $k=n+1$ case into: $$ 0 = (T-\lambda I)^{n+1} v = (T - \lambda I)^n (T - \lambda I) v := (T - \lambda I)^n w $$ where $w = (T- \lambda I)v$. Now use the inductive step to conclude that $Aw = 0$.

This implies that $$ A(T - \lambda I)v = 0 \\ ATv = \lambda A v \\ S(Av) = \lambda (Av) $$ Hence $\lambda$ is an eigenvalue of $S$. But we know that $\lambda$ is an eigenvalue of $T$, and so $Av = 0$.