Eigenvalues of a permutation operator in $\mathbb{C}^n$

Question

The problem that I'm trying to solve is the following:

Let $\pi$ be a permutation of the integers $\left\{1,2,...,n\right\}$. Find the eigenvalues of $A_{\pi}\in\mathcal{L}(\mathbb{C}^n)$ such that $\forall (x_1,...,x_n) \in \mathbb{C}^n: A_{\pi}(x_1,...,x_n)\mapsto(x_{\pi(1)},...,x_{\pi(n)})$.

Now that we're on the same page, I tried to solve it by using a theorem for permutations which says that any for any permutation $\pi$ of a finite number of elements there exists a least integer $m$ such that $\pi^n=e$ (nth composition). Where $e$ is the identity permutation. From there, $$A_{\pi}^m=I$$ With $I$ being the identity operator. By using $A_{\pi}x=\lambda x$ and the conclusion above I arrived at

$$\lambda x= \lambda^{m+1}x = ... = \lambda^{ml+1}x = ...$$ If I use only the two first terms in the equality above I can conclude that one of the eigenvalues is zero, which doesn't seem to make much sense because $A_{-\pi}$ exists, and also that the remaining eigenvalues are in the complex unit circle. In a case where $m=n!$ the number of eigenvalues is greater than the dimension of $\mathbb{C}^n$, which is also another contradiction. I would like to know what I did wrong or overlooking.

Many thanks.

Answer 1

Hints:

1. Consider the decomposition of $\pi$ into cycles. Then show that $\mathbb{C}^n$ decomposes into a direct sum of corresponding subspaces: for example, if one of the cycles is $(2 ~ 4 ~ 7 ~ 6)$ then $\mathbb{C} e_2 \oplus \mathbb{C} e_4 \oplus \mathbb{C} e_7 \oplus \mathbb{C} e_6$ is an invariant subspace for $A_\pi$.
2. Find the eigenvalues corresponding to each subspace. In this, the exact cycle won't matter much, just the length of the cycle: so as an example, try finding the eigenvalues corresponding to a cycle $(1 ~ 2 ~ 3 ~ 4 ~ 5)$.

Answer 2

Your matrix $A_\pi$ is a unitary matrix (each column has norm $1$ and every two distinct columns are orthogonal). Therefore, each of its eigenvalues has absolute value equal to $1$: Firthermore, of $n$ is the order of the permutation, then ${A_\pi}^n=\operatorname{Id}$. So, each eigenvalue is a root of unity.