Is $v$ necessarily an eigenvector of $S\circ T$?

Question

Let $V$ be a vector space, $S,T:V\to V$ be linear s.t. $S\circ T=T\circ S$, $v\in V$ be an eigenvector of $T$ [with eigenvalue $\lambda$ and $\dim V_\lambda =1$]. Is $v$ necessarily an eigenvector of $S\circ T$?

Attempt.
$$S(T(v))=S(\lambda v)=\lambda S(v)=T(S(v))$$$S(v)$ is an eigenvector of $T$ with eigenvalue $\lambda\implies S(v)=v$. Therefore, $S\circ T(v)=\lambda v$.

Comment.
I don't see why $\dim V_\lambda=1$ is needed. In other words, how do the [square brackets] affect the problem?

Answer 1

First of all from $S(v)$ is an eigenvector of $T$ with eigenvalue $\lambda$ you can only conclude that $S(v) = \mu v$ for some $\mu \in \mathbb{K}$ where $\mathbb{K}$ is the scalar field of $V$. For this conclusion the condition $\dim V_\lambda =1$ is necessary.

In order to answer your question mentioned in your comment regard \begin{align*} T = \begin{pmatrix}1&0 \\ 0& 1\end{pmatrix} \quad \text{and}\quad S = \begin{pmatrix}1&0 \\ 1& 0\end{pmatrix} . \end{align*} Then $TS = ST$ and every vector $v \neq 0$ is an eigenvector of $T$ with eigenvalue $\lambda = 1$. However, the vector $v = \begin{pmatrix}1 \\ 0\end{pmatrix}$ is clearly not an eigenvector of $S$.

Answer 2

You conclude that because $S(v)$ is an eigenvector of $T$, it is equal to $v$. This is false actually, it has to be a multiple of $v$. But this is essentially where you need dimension 1. Else $S(v)$ could be any other eigenvector in $V_\lambda$, even one that is not a multiple of $v$.