- Let $ S_n $ be the cyclic permutation operator on a tensor product space, defined by the following action $$ S_n (A_1 \otimes A_2 \otimes \cdots \otimes A_n ) = A_n \otimes A_1 \otimes A_2 \cdots \otimes A_{n-1} $$ where $A_i$ is an $n \times n $ matrix with elements in $ \mathbb{C}$
Is there a nice analytic formula for its eigenvectors?
2.Also suppose the tensor factors are instead vectors $ \psi_i $ from an n dimensional vector space again over $\mathbb{C}$. Same question, can the eigenvectors be written down simply? rather than $A_i$
Well I'm not sure if this is exactly what you are looking for, but here's one way to find the eigenvectors. Suppose $S$ is any linear map from a vector space to itself such that $S^n$ is the identity. If we fix $\zeta$, an $n$th root of unity then the maps
$v \to \frac{1}{n}(v+S(v)+S^2(v)+\dots + S^{n-1}(v))$
$v \to \frac{1}{n}(v+\zeta S(v)+\zeta^2S^2(v)+\dots + \zeta^{n-1}S^{n-1}(v))$
$v \to \frac{1}{n}(v+\zeta^2 S(v)+\zeta^4S^2(v)+\dots + \zeta^{2(n-1)}S^{n-1}(v))$
$\dots$
$v \to \frac{1}{n}(v+\zeta^{n-1} S(v)+\zeta^{2(n-1)}S^2(v)+\dots + \zeta^{(n-1)^2}S^{n-1}(v))$
are idempotents which project onto the eigenspaces of $S$. In particular, if you take an arbitrary vector and apply one of these maps it gives you an eigenvector, and every eigenvector is obtained this way.