existence of eigenvalue of operators — Jemoka Knowledge Base

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 > 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, … 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} … a_{n} can’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 + … 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’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 &= a_{0} v + a_{1} Tv + \dots a_{n} T^{n} v \\ &= (a_{0} I + a_{1} T + \dots a_{n} T^{n}) v \\ &= 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… 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