Let $V$ be a vector space over $\mathbb{C}$ and let $T: V \rightarrow V$ be a linear transformation such that $W \subset V$ is an invariant subspace. Let $V$ be a vector space over $\mathbb{C}$ and let $T: V \rightarrow V$ be a linear transformation such that $W \subset V$ is an invariant subspace.
Prove that there exists $\lambda$ an eigenvalues of $T$ and $v\notin W$ such that $(T- \lambda I)v \in W$.
I have been stuck with this exercise for hours, I thought about using minimal polynomials or to think about Jordans's form because it's what I'm learning but it didn't help.
I would be glad if you can give me a hint or a direction how to solve.
Thank you!
 A: Expansion of my comment/question above.
$W$ is invariant under $T$ implies, it induces a linear map $T':V/W\to V/W$. One is of course assuming that $W\neq V$, so $V/W\neq 0$ and thus $T'$ must have an eigen vector $v'\in V/W$ with $T'(v')=\lambda v'$. Lift everything to $V$, so if $v$ lifts $v'$, one has $v\not\in W$, $T(v)-\lambda v\in W$.
A: I assume $V$ is of finite dimension over $\Bbb C$:
$\dim_{\Bbb C} V < \infty. \tag 0$
Since $W$ is an invariant subspace of $T$, 
$T(W) \subseteq W, \tag 1$
the linear transformation $T$ factors through to a linear transformation $\tilde T$ defined on the quotient vector space $V/W$:
$\tilde T: V/W \to V/W \tag 2$
given by
$\tilde T(v + W) = T(v) + W; \tag 3$
$\tilde T$ is well-defined and depends only on the coset $v + W$, for 
$v_1 + W = v_2 + W \Longleftrightarrow v_1 - v_2 \in W, \tag 4$
which implies via (1)
$T(v_1) - T(v_2) = T(v_1 - v_2) \in W, \tag 5$
and thus
$T(v_1) + W = T(v_2) + W, \tag 6$
that is
$\tilde T(v_1 + W) = \tilde T(v_2 + W) \tag 7$
and thus we see that $\tilde T$ is well-defined.  In light of (0), 
$\dim_{\Bbb C} V/W < \infty, \tag 8$
and thus in the usual manner we infer that there exists some
$\lambda \in \Bbb C \tag 9$
and
$v \in V \setminus W \tag{10}$
with
$(\tilde T - \lambda I)(v + W) = 0 \in V/W; \tag{11}$
thus,
$(T - \lambda I)v + W = (\tilde T - \lambda I)(v + W) = 0 \in V/W, \tag{12}$
and therefore
$(T - \lambda I)v \in W; \tag{13}$
finally,
$v \notin W \tag{14}$
as indicated by (10).
