Give an example of an element of infinite order in SO(2;R)

Question

I want a matrix in $SO_{2}(\mathbb{R})$ that has determinant equal to +1 and is orthogonal but has infinite order so: $\forall n \in \mathbb{N} $ when $A \in SO_{2}(\mathbb{R}): A ^{n} \neq id$ where

id = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} I can't find such a matrix can anyone give me an example?

Answer 1

Elements of $\operatorname{SO}_2(\mathbb{R})$ are rotations. Any rotation by an angle that is not a rational multiple of $\pi$ will work.

Answer 2

Take a prime number $p\ge5$. Then most likely the matrix below $$\pmatrix{\displaystyle\frac1p & -\displaystyle\frac{\sqrt{p^2-1}}p\cr \strut \displaystyle\frac{\sqrt{p^2-1}}p &\displaystyle \frac1p\cr}$$ for some prime should work. This is orthogonal and det = 1. I hope (no proof!) that $\cos\pi x$ for rational values of $x$ will miss out some $\frac1p$.