Invertible matrix/Finite order/C/Diagonalizable/Fact/Proof

Proof

The matrix is trigonalizable and can be brought, due to fact, into Jordan normal form. We show that the Jordan blocks

${\begin{pmatrix}\lambda &1&0&\cdots &\cdots &0\\0&\lambda &1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda &1&0\\0&\cdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &\cdots &0&\lambda \end{pmatrix}}$

are trivial. Because of the finite order, ${}\lambda$ is a root of unity. By multiplying with ${}\lambda ^{-1}E_{n}$, we can assume that we have a matrix of the form

${\begin{pmatrix}1&a&0&\cdots &\cdots &0\\0&1&a&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&a&0\\0&\cdots &\cdots &0&1&a\\0&\cdots &\cdots &\cdots &0&1\end{pmatrix}}$

(with ${}a\neq 0$). If this is not an ${}1\times 1$-matrix, then there exist two vectors ${}u,v$, where ${}u$ is an eigenvector and where ${}v$ is sent to ${}v+au$. The ${}k$-th iteration of the matrix sends ${}v$ to ${}v+kau$, and this is never ${}v$, contradicting the property of being of finite order.

To fact