The number of the solutions of $‎ x^{10}=‎ ‎ ‎\begin{bmatrix}1&0\\‎ ‎0&1‎ ‎\end{bmatrix}‎$ How many solutions does the following equation have in $ M_{2}(\mathbb R)$ and why?
$$‎  x^{10}=‎ ‎\begin{bmatrix}1&0\\‎ ‎0&1‎ ‎\end{bmatrix}‎$$
Every hint is appreciated.
 A: You are asking about all linear transformations of $\mathbb R^2$ that, when applied 10 consecutive times, yield the identity.
This immediately gives the answer to your question: There are infinitely many solutions to the equation, as every reflection matrix
$$\begin{bmatrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta
\end{bmatrix}$$
for an angle $\theta \in [0,2\pi]$ fulfills the condition as pointed out by Marc (an even number of reflections over the same line gives the identity).
There are other solution types, notably rotations. Any rotation
$$\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}$$
about an angle $\theta$ with the property $10\theta = 2\pi k$ for some $k \in \mathbb N$ satisfies the equation.
If you were to expand the condition $x\in M_2(\mathbb R)$ to $x\in M_2(\mathbb C)$ you would also get "complex scaling" solutions of the form
$$\begin{bmatrix}
z & 0 \\
0 & z
\end{bmatrix}$$
where $z$ is any $10$th root of unity, i.e. $e^{(\pi i k)/5}$ for $k \in 0,...,9$. Of course, these solutions are just rotations in disguise.
A: Any solution of the matrix equation has the form
$$
C \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix} C^{-1}, C \in Mat_2(\mathbb{C}), \det(C) \neq 0, \alpha^{10}=\beta^{10}=1.
$$
A: hint: Every such matrix is diagonalizable over the field $C$. 
