TITLE: Is a finite order endomorphism a rotation?
QUESTION [5 upvotes]: Let $V$ be a real $2$-dimensional vector space, and $T\colon V\to V$ be an endomorphism such that
$$
T^q = Id \qquad \textrm{and} \qquad T^j\not= Id\quad\textrm{if}\ 0<j<q,
$$
where $T^0 = Id$ and $T^{j-1}\circ T = T^{j}$.
Moreover, let $T$ be orientation-preserving.
Can we say that there exist a basis $B$ for $V$ such that $T$ is the rotation of $\pm\frac{2\pi}{q}$, that is
$$
[T]_B = \begin{bmatrix}\cos(\frac{2\pi}{q}) & \pm\sin(\frac{2\pi}{q})\\
\mp\sin(\frac{2\pi}{q}) & \cos(\frac{2\pi}{q})\end{bmatrix}?
$$

REPLY [3 votes]: Note the following:

The determinant is real and satisfies $1=\det(\mathrm{Id})=\det(T^q)=\det(T)^q$, therefore $\det(T)=\pm1$. Because $T$ is orientation preserving we have $\det(T)=1$.
Something similar holds for the eigenvalues, i.e. $1=\lambda_i(\mathrm{Id})=\lambda_i(T^q)=\lambda_i(T)^q$, but they are allowed to be complex. Therefore the eigenvalues must be roots of unity, i.e. $\lambda=\exp(2\pi i \cdot p/q)$. Because $\det(T)=\lambda_1\lambda_2$, we have $\lambda_1=\lambda_2^{-1}=\bar\lambda_2$, i.e. they are conjugates of each other. We can write $\lambda_\pm=\exp(\pm2 \pi i\cdot p/q)$.
An example of a matrix with these eigenvalues is given by
$$\begin{pmatrix}\lambda_+ &0\\0& \lambda_-\end{pmatrix}=\begin{pmatrix}\lambda_+ &0\\0&\bar \lambda_+\end{pmatrix}$$
and it is well known that such a matrix is equivalent to the following real matrix, given $\lambda_+=\sigma+i\tau$:
$$\begin{pmatrix}\sigma&\tau\\-\tau&\sigma\end{pmatrix}.$$

The rest follows from the fact that
$$\lambda_\pm=\exp(\pm 2\pi \cdot p/q)=\cos( 2\pi \cdot p/q)\pm i\sin( 2\pi \cdot p/q).$$