Let $V$ be a vector space, $T:V\to V$ be linear, $\lambda$ be an eigenvector of $T$, $U$ be an invariant subspace of $V$ s.t. $V=U\oplus V_\lambda$, $v\in V$ be an eigenvector of $T$ (not necessarily belonging to $\lambda$).
Is it necessary that $v\in U\cup V_\lambda$?
Attempt:
Trying to prove it resulted in:
Let $v=u+v'$ where $u\in U$ and $v'\in V_\lambda$ and let $\lambda^*$ be an eigenvalue of $T$. We apply $T$ to get:
$$\lambda^*v=T(v)=T(u+v')=T(u)+T(v')=T(u)+\lambda v'$$Thus we get
$$v=\frac{1}{\lambda^*}T(u)+\frac{\lambda}{\lambda^*}v'\notin U\cup V_\lambda$$It looks like there might be a counterexample or something that I'm missing.
If $v \in U$ then $v'=0$.
So suppose $ v \notin U$, so $v' \ne 0$. Then since $T(v') = \lambda v'$, in order to make $T(u+v')$ a multiple of $u+v'$, we must have $T(u)=\lambda u$ and so $T(v) = \lambda v$. But if $T(v) = \lambda v$ then $v \in V_{\lambda}$.
So either $v \in U$ or $v \in V_{\lambda}$.