<s>[INST] Are there uncountably many $A\in M_3 (\mathbb {R})$ such that $A^8=I $? I'm working on the following problem:

Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then  
  
  
*
  
*the minimal polynomial of $A$ can only be of degree $2$.  
  
*the minimal polynomial of $A$ can only be of degree $3$.  
  
*either $A = I$ or $ A = -I$.  
  
*there are uncountably many such $A$.  

By taking $A=I $ we can eliminate options (1) & (2). For the minimal polynomial of $A$ in that case is of degree $1$. Now take 
$$A =
\begin{bmatrix}
     1 & 0 & 0 \\
     0 & -1 & 0 \\
     0 & 0 & -1
\end{bmatrix}$$
Then $A^8=I $ but $A$ is neither $I$ nor $-I$. So option (3) is eliminated. Now I don't know how to proceed about with option  (4). Can someone help me please? Thanks in advance.
 [/INST] Let $O(\theta)$ be the $2 \times 2$ orthogonal matrix
$O(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}; \tag 1$
set
$P = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}; \tag 2$
then
$O^{-1}(\theta) = O^T(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}; \tag 3$
we have
$O^T(\theta) P O(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$
$= \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2 \cos \theta \sin \theta \\ 2 \cos \theta \sin \theta & \sin^2 \theta - \cos^2 \theta \end{bmatrix}$
$= \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ \sin 2\theta & -\cos 2 \theta \end{bmatrix}, \tag 4$
which is clearly an uncountable family of $2 \times 2$ matrices; furthermore,
$(O^T(\theta) P O(\theta))^2 = O^T(\theta) P O(\theta)  O^T(\theta) P O(\theta) =  O^T(\theta) P I P O(\theta)$
$= O^T(\theta)P^2 O(\theta) = O^T(\theta)I O(\theta) = I, \tag 5$
whence
$(O^T(\theta) P O(\theta))^8 = I \tag 6$
as well; now set
$A(\theta) = \begin{bmatrix} O^T(\theta) P O(\theta) & 0 \\ 0 & -1 \end{bmatrix}; \tag 7$
then
$A^8(\theta) = \begin{bmatrix} (O^T(\theta) P O(\theta))^8 & 0 \\ 0 & (-1)^8 \end{bmatrix} = \begin{bmatrix} I & 0 \\ 0 & 1 \end{bmatrix} = I, \tag 8$
and the family $\{ A(\theta)\mid \theta \in \Bbb R \}$ is also uncountable.
</s>