Let $k \geq 1$ and let $S=\{A_1,A_2,\dots,A_k \}$ be a subset of $\mathcal{M}_{n}(\mathbb{C})$, such that $I_n \in S$. Also, if $X \in S$, then $X^{-1} \in S$ and if $X,Y \in S$, then $XY \in S$. Let $$M=A_1+A_2+\dots+A_k$$ Prove that the only eigenvalues of $M$ are from the set $\{0,k \}$.
I am not sure if the statement is entirely correct. It seems like for $k=3$ and $S=\{I_n,A,A^{-1} \}$ the conclusion doesn't hold. I'm wondering if there is a correction that could be made, or maybe I'm just missing something.
Hint. Since $S$ is a finite multiplicative subgroup of $GL_n(\mathbb C)$, we must have $A_jS=S$ for every $j$. Now, what is $M^2$?
Related: Finite multiplicative group of matrices with sum of traces equal to zero.