What do eigenvalues have to do with modules

Question

Let $G$ be the cyclic group of order 4 , given by the presentation
$$G = \left\langle a: a^4 = 1\right\rangle.$$
Let $F$ be $\mathbb{R}$. Let $V$ be a 2-dimensional vector space over $F$ with basis $\mathcal{B} = \left\{v_1, v_2\right\}$. It turns out that $V$ can be made into a left $FG$-module in which the generator $a \in G$ acts as
$$a \cdot v_1 = v_2 \quad \text { and } \quad a \cdot v_2 = -v_1.$$
Let $\rho: FG \rightarrow GL_2(F)$ be the representation of $G$ corresponding to this module. Show that $V$ is an irreducible $FG$-module.

I am working on a homework problem to show that a certain FG-module is irreducible and my professor told us to "look at the eigenvectors." I wasn't sure why we should do that, but I went ahead and found that the two eigenvalues were $\lambda = \pm i$ with associated eigenvectors $<-i, 1>$ and $<i, 1>$. I was hoping to see the connection after finding them, but I can't see why finding these eigenvectors should show that an FG-module is irreducible.

My question is this: What do eigenvalues have to do with modules?

Answer 1

If anyone could write this up in an answer I would really appreciate it.

Sure

As David wrote in the comments:

If a 2-dimensional module is reducible, it must have a 1-dimensional submodule. That must be an eigenspace.

Now we are going to find the eigenspaces of our $\mathbb{R}G$-module $V$. You already did something like that:

found that the two eigenvalues were $\lambda=\pm i$ with associated eigenvectors $(−i,1)$ and $(i,1)$.

However, these are not real eigenvectors, as Mariano said:

What you called eigenvalues and eigenvectors are not eigenvalues and eigenvectors, because the first do not belong to the field you are working with, and the latter do not belong to the module you are working with

That is, our module action does not have eigenvectors when we restrict our numbers to $F=\mathbb{R}$; we have $(-i,1)\notin V$ and $\pm i\notin\mathbb{R}$.

Hence, we find that our action $\mathbb{R}G\to GL(V)$ does not have eigenvectors in $V$. Therefore, $V$ does not have a 1-dimensional submodule. Therefore, it is not reducible.