Eigenvalues of G-Representation are Roots of Unity
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Theorem
Let $G$ be a finite group.
Let $\left({K, +, \cdot}\right)$ be a field.
Let $V$ be a $G$-module over $K$ (i.e. $V$ is a $K \left[{G}\right]$-module).
Then $\forall g \in G$, the eigenvalues of the action by the vector $g \in K \left[{G}\right]$ on $V$ are roots of unity.
Proof
Fix an arbitrary $g \in G$ and consider the corresponding vector $g \in K \left[{G}\right]$.
Let $\lambda$ be an eigenvalue of $g$, that is $\lambda$ is an eigenvalue of the map in $\operatorname{Aut} \left({V}\right): \vec v \mapsto g \vec v$.
Then by definition of an eigenvalue we have:
- $\exists \vec v_\lambda \in V : g \vec v_\lambda = \lambda \vec v_\lambda$
Let $n$ be the order of $g$ in $G$.
Then:
\(\ds \vec v_\lambda\) | \(=\) | \(\ds g^n \vec v_\lambda\) | as $g^n = e$ and $e$ acts trivially on $V$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda g^{n-1} \vec v_\lambda\) | as $G$ acts linearly on $V$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \vdots\) | Continue exchanging $g^k \vec v_\lambda$ with $\lambda g^{k-1} \vec v_\lambda$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda^n \vec v_\lambda\) | as $g^0 = e$ which acts trivially on $V$ |
Thus:
- $\vec v_\lambda = \lambda^n \vec v_\lambda$
which means:
- $\lambda^n = 1$
and thus by definition, $\lambda$ is an $n$th root of unity.
$\blacksquare$