Eigenvalues of G-Representation are Roots of Unity

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:

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.