Does there exist a positive integer $n$ such that $P^n = I$, where $P$ is a rotation matrix?

Question

Does there exist a positive integer $n$ such that $P^n = I$, where $P$ is a $2 \times 2$ rotation matrix for a rotation of the plane by an angle $2\pi q$ radians?

1. if $q$ is a rational number
2. if $q$ is an irrational number

How should I go about answering this question? Do I need to think geometrically?

Answer 1

For $q \in \Bbb R$, let $P_q$ denote the $2 \times 2$ matrix that represents rotation by $2\pi q$.

Hints:

1. Show that $P_q^n = P_{nq}$ for all integers $n \ge 1$.
2. Show that $P_q = P_{q'}$ iff $q - q' \in \Bbb Z$.
3. Note that $I = P_0$. Conclude that $P_q^n = I$ iff $nq \in \Bbb Z$.
4. Find an $n$ which works for rational $q$ (think about the denominator).
5. Show that no $n$ works for irrational $q$. (what if some $n$ worked? Get a contradiction.)

Answer 2

$$P^n$$ corresponds to a rotation by angle $2\pi qn$ radians, or $qn$ full turns. $qn$ can only be an integer when $q$ is rational.

Answer 3

For a rotation matrix that rotates 2πq, $Pˆn = I$ iff nq is an integer. Therefore, for any rational number, there is an n that makes nq an integer (a rational number can be represented as $a/b$, coprime and integers hence if n is a multiple of b it works.

An irrational number, by definition, can not be written as $a/b$ hence no integer n can make nq an intger

hope that helps