Given that $v \in V$ is an eigenvector of S with eigenvalue $\lambda$ , prove that T(v) is also an eigenvector of S with eigenvalue $\lambda$

Question

We have $S,T \in L(V,V)$ and can assume TS = ST How can we go from eigenvector v to eigenvector T(v)? I tried showing if $\lambda = 0$ then $v \in Ker(S)$ and if $\lambda \neq 0$ then $v \in Range(S)$ (not even sure if my argument for it being in the range is correct), but I'm stuck here. Any help?

Answer 1

Remark: We need to further impose the condition that $T(v)$ is non-zero as pointed out by Hagen.

let $v$ be an eigenvector of $S$, you want to verify $T(v)$ is an eigenvector as well.

so compute

$$S(T(v))=T(S(v))=T(\lambda v)$$

Do you see why $T(v)$ is an eigenvector now?

Answer 2

$v\in \mathrm{Null}(S-\lambda I)$ $$T(S-\lambda I)v=(S-\lambda I)Tv$$ as $TS=ST$.