Eigenvalues of G-Representation are Roots of Unity

From ProofWiki
Jump to: navigation, search

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:

\(\displaystyle \vec v_\lambda\) \(=\) \(\displaystyle g^n \vec v_\lambda\) $\quad$ as $g^n = e$ and $e$ acts trivially on $V$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda g^{n-1} \vec v_\lambda\) $\quad$ as $G$ acts linearly on $V$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \vdots\) $\quad$ Continue exchanging $g^k \vec v_\lambda$ with $\lambda g^{k-1} \vec v_\lambda$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda^n \vec v_\lambda\) $\quad$ as $g^0 = e$ which acts trivially on $V$ $\quad$





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.

$\blacksquare$