[Math] Eigenvectors and Eigenvalues of Shift Matrix

Question

$$S:\mathbb{C}^n\rightarrow\mathbb{C}^n, $$
$$S(x_1,x_2,…,x_n)^T = (x_n,x_1,…,x_{n-1})^T.$$
How can the eigenvalues and eigenvectors of S be calculated?
I already have the standard matrix of S which is:
\begin{bmatrix}
0 & 0 & 0 & \dots & 0 & 1\\
1 & 0 & 0 & \dots & 0 & 0\\
0 & 1 & 0 & \dots & 0 & 0\\
0 & 0 & 1 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 1 & 0
\end{bmatrix}

Answer 1

As Rodrigo de Azevedo's edit to the question indicates, $\ S\ $ is a special case of a circulant matrix:$$ S=\pmatrix{c_0&c_{n-1}&\dots&c_2&c_1\\ c_1&c_0& c_{n-1}&\dots&c_2\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ c_{n-1}& c_{n-2}&\dots&c_1&c_0}\ , $$ where $\ c_1=1\ $ and $\ c_i=0\ $ for $\ i\ne1\ $ in this case.

The eigenvalues of $\ S\ $ are therefore the $\ n^\text{th}\ $ roots of unity, $\ w_j^{n-1}=e^{\frac{2\pi i j(n-1)}{n}}= w_{n-j}\ $, for $\ j=0,1,\dots, n-1\ $, where $\ w_j=e^{\frac{2\pi i j}{n}}\ $. An eigenvector $\ v_j\ $ corresponding to the eigenvalue $\ w_j\ $ is one whose $\ k^\text{th}\ $ entry is $\ w_j^{k-1}\ $: $$ v_j=\pmatrix{ 1& w_j& w_j^2& \dots & w_j^{n-1}}^T\ . $$ With $\ v_j\ $ thus defined, we have \begin{align} S v_j&= \pmatrix{ w_j^{n-1}& 1& w_j& \dots & w_j^{n-2}}^T\\ &= w_j^{n-1}v_j\ . \end{align}