# The Existence of an Eigenvalue on Finite-Dimensional Complex Vector Spaces

We will now look at a crucially important theorem which tells us that a nonzero finite-dimensional vector space over the complex numbers $\mathbb{C}$ contains an eigenvalue.

Theorem 1: If $V$ is a finite-dimensional nonzero vector space over the complex numbers $\mathbb{C}$ then every operator $T \in \mathcal L (V)$ has an eigenvalue. |

**Proof:**Let $V$ be a finite-dimensional nonzero vector space over $\mathbb{C}$. Let $\mathrm{dim} (V) = n > 0$ (since $V \neq \{ 0 \}$). Let $T$ be a linear operator $T \in \mathcal L (V)$, and let $v \in V$ be such that $v \neq 0$. Consider the following set of vectors:

(1)

\begin{align} \quad \{ v, T(v), T^2(v), …, T^n(v) \} \end{align}

- This set contains $n + 1$ vectors and since $\mathrm{dim} (V) = n$ then this set cannot be linearly indepedent in $V$. Therefore, there exists complex numbers $a_0, a_1, …, a_n \in \mathbb{C}$ that are not all nonzero such that:

(2)

\begin{align} \quad 0 = a_0v + a_1T(v) + a_2T^2(v) + … + a_n T^n(v) \end{align}

- Let $m = 1, 2, …, n$ be the largest index such that $a_m \neq 0$ (noting that $a_0 \neq 0$ since not all $a_0, a_1, …, a_m \in \mathbb{C}$ are zero). Thus we have that $0 and:

(3)

\begin{align} \quad 0 = a_0v + a_1T(v) + a_2T^2 (v) + … + a_m T^m (v) \\ \quad 0 = (a_0v + a_1T(v) + a_2T^2(v) + … + a_m T^m)(v) \end{align}

- We can factor the polynomial above as:

(4)

\begin{align} \quad 0 = (c(T – \lambda_1I)(T – \lambda_2I)…(T – \lambda_mI))(v) \end{align}

- Since $v \neq 0$ (as assumed above) then we have that $(T – \lambda_jI)$ is not injective for some $j = 1, 2, …, m$ and so $T$ has an eigenvalue. $\blacksquare$

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