If $T \in L(\mathbb R^2)$ with $T^3 = I$, then is $T$ an isometry? I am aware that there are infinitely many $S \in L(\mathbb R^2)$ such that $S^2 = I$, and $S$ is not an isometry. Can I generalize the result for $T^n = I$, where $n >2$?
 A: An equation like $S^n = I$ will generally tell you something about the minimal polynomial of $S$. This in turn tells you something about the (complex) eigenvalues and to which matrices $S$ might be similar. However, being an isometry (i.e. orthogonal) is generally not preserved under similarity. For example, the matrix
$$
  \begin{pmatrix}0 && 1 \\ 1 && 0\end{pmatrix}
$$
is orthogonal whereas
$$
  \begin{pmatrix}1 && 1 \\ 0 && 1\end{pmatrix} \begin{pmatrix}0 && 1 \\ 1 && 0\end{pmatrix} \begin{pmatrix}1 && -1 \\ 0 && 1\end{pmatrix} = \begin{pmatrix}1 && 0 \\ 1 && -1\end{pmatrix}
$$
is not. (Note that the matrices I multiplied on the left and right are inverse to each other.) So you shouldn't expect an equation like that to tell you anything about the matrix being an isometry, except in very special cases (like $S^1 = I$). Because isometries are rare compared to all linear maps, I would expect to find infinitely many counter-examples.
For your specific problem, if $S$ is an endomorphism of $\mathbb{R}^2$ with $S^n = I$, then $S$ is a zero of the polynomial $t^n - 1$ which factors (over $\mathbb{R}$) into

*

*$t - 1$,

*$t + 1$ (if $n$ is even), and

*$(t - e^{2\pi ik/n})(t - e^{-2\pi ik/n}) = t^2 - 2 \cos(2\pi k/n) + 1$ for $1 \leq k < n/2$.

(This is because the complex roots are the $n$-th roots of unity, i.e. $e^{2\pi i k/n}$ for $0 \leq k < n$.)
The minimal polynomial of $S$ hence is a product of some of these polynomials. Its degree is at most $2$, so it can be

*

*$t - 1$, in which case $S = I$ and $S$ is orthogonal;

*$t + 1$, in which case $S = -I$ and $S$ is orthogonal;

*$(t - 1)(t + 1)$, (if $n$ is even), in which case $S$ is similar to the matrix considered above, (alternatively, in this case we have the stronger condition $S^2 = I$ and you know that infinitely many counter-examples exist); or

*one of the factors considered in 3. above and $S$ is similar to the companion matrix $$\begin{pmatrix}0 && -1 \\ 1 && 2 \cos (2\pi k/n)\end{pmatrix}$$ of that polynomial, which is not orthogonal (as $\cos(2 \pi k /n)$ is never $0$ for the values of $k$ under consideration).

For any $n \geq 3$, the last case can occur, and for $n = 2$, the third case can occur.
