NUS-MATH530 5.A Problem 35/36

Warmup: 35
Suppose V is finite dimensional, T \in \mathcal{L}(V) and U is invariant under T. Prove each eigenvalue of T / U is an eigenvalue of T.
Now, \lambda is an eigenvalue of T / U. That is:

\begin{equation} Tv + U = \lambda v + U \end{equation}

Meaning:

\begin{equation} (T-\lambda I) v \in U, \forall v \in V \end{equation}

Suppose for the sake of contradiction \lambda is not an eigenvalue of T. This means no \lambda such that Tv = \lambda v; specifically, that means also no \lambda such that T|_{u} u = \lambda u. Now, that means T|_{u} - \lambda I is invertible given finite dimensional V.
The previous statement means that (T|_{u} - \lambda I) is subjective across u:

\begin{equation} \forall v, \exists u: (T-\lambda I)v = (T|_{u}-\lambda I) u \end{equation}

And so:

\begin{equation} Tv - \lambda v = Tu - \lambda u \end{equation}

Finally, then:

\begin{equation} T(v-u) = \lambda (v-u) \end{equation}

Now, v + U being an eigenvector of T / U requires that v + U \neq 0, which means v \not \in U. And so, v \neq u meaning v-u \neq 0. Hence, the above expression demonstrates \lambda to be an eigenvalue of T, reaching contradiction. \blacksquare
Now: 36
Removing finite-dimensional from the requirements above, demonstrate the result above breaks.
Let V = \mathcal{P}(\mathbb{F}) and let T be differentiation. Now, let U be P_{2}(\mathbb{F}). Now:

\begin{equation} T / U (v + U) = \lambda v + U \end{equation}

let v \in \mathcal{P}_{3}(\mathbb{F}). Now, then, T / U (v + U) = Tv + U, with Tv \in \mathcal{P}_{2}(\mathbb{F}). Hence, T / U (v + U) = Tv + U = 0 + U. This makes 0 an eigenvalue and u \in \mathcal{P}_{2}(\mathbb{F}) eigenvectors.
Of course this does not hold for T in general as all Tv \in \mathcal{P}_{2}(\mathbb{F}) are not identically 0.
Having shown a counter-example, \blacksquare
Do we have finite-dimensions?
“Not invertible” => not injective—
T being not injective means that null\ T has more than just the zero vector.
Hence:

\begin{equation} \exists v: Tv = 0 = 0 v \end{equation}

That would make all nonzero v \in null\ T eigenvectors and 0 an eigenvalue.
“Not invertible” => not surjective—
T being not surjective means that range\ T \subset V strictly. So then T|_{range\ T} is an operator so range\ T is an invariant subspace under T.
Either way, we have that an eigenvalue exist.