Are there uncountably many $A\in M_3 (\mathbb {R})$ such that $A^8=I $?

Question

I'm working on the following problem:

Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then

1. the minimal polynomial of $A$ can only be of degree $2$.

2. the minimal polynomial of $A$ can only be of degree $3$.

3. either $A = I$ or $ A = -I$.

4. 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.

Answer 1

Hint: What about $$\begin{bmatrix}1&-t&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&-1&0\\0&0&-1\end{bmatrix} \begin{bmatrix}1&t&0\\0&1&0\\0&0&1\end{bmatrix}$$

Answer 2

Hint The reflection $R$ about any $2$-dimensional subspace satisfies $R^2 = I$.

Answer 3

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.

Answer 4

In fact, we can show a lot more than the OP asks, without getting tired.

In particular, in other posts, it is shown that the general solution depends on, at least, $1$ real parameter.

$\textbf{Proposition}$. The general solution of the equation (in $X$) $X^2=I_3$ depends on exactly $4$ real parameters.

$\textbf{Proof}$. Note that any solution $X\not= \pm I_3$ is similar to $\pm A$, where $A=diag(1,1,-1)$; let $Z=\{P^{-1}AP;P\in GL_3(\mathbb{R})\}$ be the similarity class of $A$. The dimension of the algebraic set $Z$ is $3^2-dim(C(A))$, where $C(A)$ is the commutant of $A$; that is, $9-(2^2+1)=4$.