is a character of a finite abelian group an eigenvector?

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at the moment I'm writing an essay for university and found a very interesting Lemma: lemma and proof

It basically says that characters of a finite abelian group are the eigenvectors and there is a pretty easy calculation to the corresponding eigenvalue. I tried for hours to understand it, but right now I'm more thinking it's wrong than correct. I tried it with the Cayley-Graph $G=(\mathbb{Z}_2^2,\{01,10\})$. It's basically a very small Hamming-Cube, which has an edge if it is exactly one bit different. The corresponding adjacency matrix is $\left(\begin{array}{rrr}0&1&1&0\\1&0&0&1\\1&0&0&1\\0&1&1&0\end{array}\right)$, the both characters and eigenvectors are for $\lambda=-2$ $\left(\begin{array}{rrr}1\\-1\\-1\\1\end{array}\right)$ and the trivial one $\left(\begin{array}{rrr}1\\1\\1\\1\end{array}\right)$ is the eigenvector to $\lambda=2$.

But for example $-2\neq\dfrac{1}{|S|}\sum_{s\in S}x(s)=\dfrac{1}{2}(-1+(-1))=-1$.

So either the Lemma is wrong (which would surprise me, as it's a renominated source) or I make some stupid mistakes.

I found this in a script of the Stanford University on the 2nd Page as Lemma 4: link

Thanks in advance!