Is a finite order endomorphism a rotation? 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
$$
[T]_B = \begin{bmatrix}\cos(\frac{2\pi}{q}) & \pm\sin(\frac{2\pi}{q})\\
\mp\sin(\frac{2\pi}{q}) & \cos(\frac{2\pi}{q})\end{bmatrix}?
$$
 A: No. Take $$(x, \, y) \mapsto (x, \, -y).$$
A: Notice that the minimal polynomial $m(x)$ of $T$ is a real polynomial of degree $2$ and divides $x^q - 1$. Using the condition that $\det(T) > 0$, we easily find that
$$ m(x) = (x - \omega)(x - \bar{\omega}) = x^2 - 2\cos (\alpha)x + 1 $$
for some complex zero $\omega = e^{i\alpha}$ of $\omega^q - 1$. So $T$ is diagonalizable over $\mathbb{C}$. Let $u$ be a complex eigenvector of $T$ corresponding to $\omega$ and write $u = v + iw$ for real vectors $v$ and $w$. Then both $v$ and $w$ are non-zero since $ T\bar{u} = \overline{Tu} = \overline{\omega u} = \bar{\omega}\bar{u} $ and hence $\{u, \bar{u}\}$ is $\mathbb{C}$-linearly independent.
$$ Tv = \operatorname{Re}(Tu) = \operatorname{Re}(\omega u) = \cos(\alpha)v - \sin(\alpha)w $$
and 
$$ Tw = \operatorname{Im}(Tu) = \operatorname{Im}(\omega u) = \sin(\alpha)v + \cos(\alpha)w. $$
Therefore, $B = \{v, w\}$ is a basis of $V$ such that
$$ [T]_B = \begin{pmatrix}
\cos\alpha & -\sin\alpha \\
\sin\alpha & \cos\alpha
\end{pmatrix}. $$
(Or simply invoke the real Jordan normal form of $T$.)
A: 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
$$\lambda_\pm=\exp(\pm 2\pi \cdot p/q)=\cos( 2\pi \cdot p/q)\pm i\sin( 2\pi \cdot p/q).$$
