Permutation matrix/Cycle/Numbered/C/Eigenvalues/Fact/Proof

We have

${\displaystyle {}{\begin{aligned}M_{\rho }(v_{\zeta })&=M_{\rho }(\zeta ^{k-1}e_{1}+\zeta ^{k-2}e_{2}+\cdots +\zeta e_{k-1}+e_{k})\\&=\zeta ^{k-1}M_{\rho }(e_{1})+\zeta ^{k-2}M_{\rho }(e_{2})+\cdots +\zeta M_{\rho }(e_{k-1})+M_{\rho }(e_{k})\\&=\zeta ^{k-1}e_{2}+\zeta ^{k-2}e_{3}+\cdots +\zeta e_{k}+e_{1}\\&=e_{1}+\zeta ^{k-1}e_{2}+\zeta ^{k-2}e_{3}+\cdots +\zeta e_{k}\\&=\zeta (\zeta ^{k-1}e_{1}+\zeta ^{k-2}e_{2}+\cdots +\zeta e_{k-1}+e_{k})\\&=\zeta v_{\zeta }.\end{aligned}}}$

Since there are ${\displaystyle {}k}$ different ${\displaystyle {}k}$-th roots of unity in ${\displaystyle {}\mathbb {C} }$, these vectors are linearly independent due to fact, and they generate a ${\displaystyle {}k}$-dimensional linear subspace ${\displaystyle {}U}$ of ${\displaystyle {}K^{n}}$. In fact, we have

${\displaystyle {}U=\langle e_{i},\,i=1,\ldots ,k\rangle \,.}$

Since the vectors ${\displaystyle {}e_{i}}$, ${\displaystyle {}i\geq k+1}$, are fixed vectors, the ${\displaystyle {}v_{\zeta }}$ together with the ${\displaystyle {}e_{i}}$, ${\displaystyle {}i\geq k+1}$, form a basis consisting of eigenvectors of ${\displaystyle {}M_{\rho }}$. Hence, ${\displaystyle {}M_{\rho }}$ is diagonalizable.