The Existence of an Eigenvalue on Fin.-Dim. Complex Vector Spaces
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:
\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:
\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 $0 < m ≤ n$ and:
\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:
\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. $\blacksquare$