---
title: "existence of eigenvalue of operators"
type: concept
related: [Complex Number, Vector, Polynomial Operator, Polynomial, List]
source: https://www.jemoka.com/posts/kbhoperators_on_complex_vector_spaces_have_an_eigenvalue/
confidence: high
status: active
---

A result so important it gets a page.
Every operator on a finite-dimensional, non-zero, complex vector space has an eigenvalue.
Proof:
Suppose \(V\) is a complex vector space with dimension \(n &gt; 0\), and \(T \in \mathcal{L}(V)\). Choose \(v \in V, v\neq 0\) (possible as \(V\) is non-zero):
Construct a list of \(n+1\) vectors:
\begin{equation} v, Tv, \dots T^{n} v \end{equation}
because we managed to cram \(n+1\) vectors into a list for a vector space with dimension \(n\), that list is linearly dependent.
And thus, by definition of linearly dependence, exists a set of \(a_0, &hellip; a_{n} \in \mathbb{C}\), which not all are \(0\), such that:
\begin{equation} 0 = a_0 v + a_1 T v + \dots + a_{n} T^{n} v \end{equation}
Note that, because \(v \neq 0\), \(a_{1} &hellip; a_{n}\) can&rsquo;t all be \(0\) either because otherwise \(a_0 = 0\) making all \(a_{j}=0\).
Now, this polynomial can be completely factored because of the fundamental theorem of algebra into linear factors, \(a_{0} + a_{1}z + &hellip; a_{n}z^{n} = c(z-\lambda_{1}) \dots (z- \lambda_{m})\). We have to invoke the fundamental theorem of algebra with complex factors \(z\) because we haven&rsquo;t shown it holds for polynomial operators yet.
However, the existence of such a complete factoring over the complex numbers means that, with possibly complex number \(\lambda_{j}\) values:
\begin{align} 0 &amp;= a_{0} v + a_{1} Tv + \dots a_{n} T^{n} v \\ &amp;= (a_{0} I + a_{1} T + \dots a_{n} T^{n}) v \\ &amp;= c(T - \lambda_{1} I) \dots (T- \lambda_{m} I)v \end{align}
note that \(m\) is not necessarily \(n\) because different multiplicities.
Now, \(c\) cannot be \(0\) because \(a_0 \neq 0\), and multiplying everything out out&hellip; makes the ending not zero?
Given \(c \neq 0\), \(v \neq 0\), and yet the map maps \(v\) to \(0\), at least one of the maps has to be non-injective. And because the properties of eigenvalues, some \((T- \lambda_{j} I)\) being non-injective for a finite-dimensional vector space means that \(\lambda_{j}\) is an eigenvalue of \(T\). \(\blacksquare\)
