Simple matrix product

Question

When $d,f\in\mathbb{R}$ and $N,m\in\mathbb{N}$ satisfy \begin{equation} \cos\theta = 1-\frac{d}{2f} ,\quad N\theta = m\pi, \end{equation} $\begin{pmatrix}1&d\\-\frac{1}{f}&1-\frac{d}{f}\end{pmatrix}^{2N}$ should be equal to the identity matrix. I confirmed this formula for some sets of $(d,f,N,m)$, but I cannot prove it. Could someone please help me?

Answer 1

The characteristic equation of $M:=\begin{pmatrix}1 & d\\ -\frac{1}{f} & 1-\frac{d}{f}\end{pmatrix}$ is $$ X^2-(2-\frac{d}{f})X +1=X^2-2\cos\theta X+1.$$

Provided $d\ne 0$, the eigenvalues are distinct $M$ is diagonalisable, and so similar to $\begin{pmatrix}e^{i\theta} & 0\\0 &e^{-i\theta}\end{pmatrix}$, whose $2N$-th power is clearly the identity.