Eigenvalues of G-Representation are Roots of Unity

Theorem
Suppose $G$ is a finite group.

Suppose $(K,+,\cdot)$ is an algebraically closed field of characteristic

Suppose $V$ is an $G$-module over $K$ (i.e. $V$ is a $K[G]$-module).

Then $\forall g \in G$, the eigenvalues of the action by the vector $g \in K[G]$ on $V$ are roots of unity.

Proof
Fix an arbitrary $g \in G$ and consider the corresponding vector $g \in K[G]$.

Suppose $\lambda$ is an eigenvalue of $g$, that is $\lambda$ is an eigenvalue of the map in $\operatorname{Aut}(V):\vec{v} \mapsto g\vec{v}$.

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 it follows,

Thus we have


 * $\vec{v}_{\lambda} = \lambda^{n} \vec{v}_{\lambda}$

Which means


 * $\lambda^{n}=1$

and thus by definition, $\lambda$ is an $n$th root of unity.