Permutation matrix/Cycle/Characteristic polynomial/Fact/Proof

Proof

We may assume that the cycle has the form ${}1\mapsto 2\mapsto 3\mapsto \ldots \mapsto k\mapsto 1$. The corresponding permutation matrix ${}M_{\rho }$ looks with respect to ${}e_{k+1},\ldots ,e_{n}$ like the identity matrix and has, with respect to the first ${}k$ standard vectors, the form

${\begin{pmatrix}0&0&\ldots &0&1\\1&0&0&\ldots &0\\0&1&0&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&1&0\end{pmatrix}}.$

The determinant of ${}XE_{n}-M_{\rho }$ is ${}(X-1)^{n-k}$ multiplied with the determinant of

${\begin{pmatrix}X&0&\ldots &0&-1\\-1&X&0&\ldots &0\\0&-1&X&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&-1&X\end{pmatrix}}.$

The expansion with respect to the first row yields

${}X^{k}+(-1)^{k+1}(-1)(-1)^{k-1}=X^{k}-1\,.$

To fact