Two conflicting answers: Problem in linear algebra involving quotient spaces and T-invariant subspaces I was presented this scary looking problem in my linear algebra class involving quotient spaces: I am given finite dimensional vector space V over the complex numbers C and linear operator $ T:V \rightarrow V $, and a T-invariant proper subspace U of V. I am asked to show that there exists an eigenvalue $ \lambda $ of T and a vector $ w \notin U  $ such that $ (T- \lambda I)w \in U $, as a hint I am asked to look at a quotient space. The only thing I can figure out here is that due to T being over a finite dimensional complex vector space there is necessarily a complex eigenvalue. I have no idea how to do it, so really need the help.
 A: Suppose that $T$ has a proper invariant subspace $U \subsetneq V$.  An element of $V/U$ is an equivalence class $[x]=\{ x + m : m \in U\}$. Note that $V/U$ is a non-trivial vector space (contains something other than the $[0]$ vector) because $U \ne V$.
$T$ induces a linear map $\dot{T} : V/U \rightarrow V/U$ defined by
$$
                      \dot{T}[x] = [Tx]
$$
This is well-defined as $[x]=[y]$ implies $[Tx]=[Ty]$ because $T(x-y) \in TU\subseteq U$. Therefore $\dot{T}$ is a linear map on a non-trivial linear space $V/U$ over $\mathbb{C}$ (i.e., there are non-zero vectors in $V/U$) and, hence, $\dot{T}$ has an eigenvalue $\lambda$. So, there exists $[x] \ne [0]$ (equivalently, $x \notin U$) such that
$$
                    \dot{T}[x]=\lambda [x].
$$
In other words $[(T-\lambda I)x]=[0]$ with $x \notin U$. But that means $(T-\lambda I)x \in U$ and $x\notin U$.
The only part that remains to be checked is that $\lambda$ is an eigenvalue of $T$. If $m$ is the minimal polynomial for $T$, then $m(T)=0$ gives $m(\dot{T})=0$. Hence, the minimal polynomial $\dot{m}$ of $\dot{T}$ must divide $m$, which means that any eigenvalue of $\dot{T}$ is also an eigenvalue of $T$.
