---
title: "NUS-MATH530 5.A Problem 35/36"
source: https://www.jemoka.com/posts/kbhnus_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?
&ldquo;Not invertible&rdquo; =&gt; not injective&mdash;
\(T\) 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.
&ldquo;Not invertible&rdquo; =&gt; not surjective&mdash;
\(T\) 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.
