How many solutions $X^{10} - I=0$ has in $M_2(\mathbb{R})$? How many solutions $X^{10} - I=0$ has in $M_2(\mathbb{R})$? Where $M_2(\mathbb{R})$ denotes the set of $2 \times 2$ real matrices.
I absolutely have no idea of where I should start from. $I$ and $-I$ solve the equation, but are there any other non-trivial solutions? Are there infinitely many solutions?
 A: Answer : infinitely many. Any matrix whose eigenvalues are $1$ and
$-1$ does the job, for example ; you can take
$$
A=\begin{pmatrix} 2y+1 & -2 \\ 2y(1+y) & -(2y+1) \end{pmatrix}
$$
All those $A$ satisfy $A^2=I$. There are other solutions ...
A: There's a way to write all those matrices:
we know that a matrix $X$ satisfying those conditions must be diagonalizable in $\mathbb{C}$, and his minimal polynomial $q_X(t)$ must be real.
If $q_X(t)$ has degree one, then $X=cI$, where $c$ is either $1$ or $-1$, since they are the only solutions to $x^{10}=1$ in the real numbers.
If $q_X(t)$ has degree $2$, it must be 
$$q_X(t)=(x-1)(x+1) \quad or \quad q_X(t)=(x-z_{10})(x-\overline{z_{10}})$$
with $z_{10}$ a $10$-th complex root of $1$.
$X$ solves $X^{10}=I$, if and only if for every $M$ invertible matrix $MXM^{-1}$ solves it. This means that all the solutions are the real matrices diagonalizable with eigenvalues
$$
(1,1)\quad (1,-1)\quad (-1,-1)\quad (\overline{z_{10}},z_{10}) 
$$
You can also write these as the matrices similar to 
$$
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\quad or \quad
\begin{pmatrix} \cos\big(n\frac{2\pi}{10}\big) & \sin\big(n\frac{2\pi}{10}\big) \\ -\sin\big(n\frac{2\pi}{10}\big) & \cos\big(n\frac{2\pi}{10}\big) \end{pmatrix}
$$
for some integer $n$.
For example, the matrices
$$
\begin{pmatrix} 1 & s \\ 0 & -1 \end{pmatrix}
$$
are similar to the first one, for all $s$ real, so you have an infinite set of matricial solutions to your equation.
A: Let $A \in M_2(\mathbb{R})$ be such that $A^{2}=I$. Then of course $A^{10}-I=0$.
One such matrix is 
$
D=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
$
Then all matrices of the form $A=PDP^{-1}$ satisfy $A^{2}=I$.
So, there are infinitely many solutions for $X^{10}-I=0$.
A: In a non-commutative setting any polynomial equation is likely to be infinitely many solutions. Indeed, if $A$ is, say, an $\Bbb R$-algebra and $a\in A$ satisfies the polinomial relation
$$
P(a)=0
$$
for some polynomial $P(x)\in\Bbb R[x]$ then for any invertible $g\in A$ the element $a^\prime=gag^{-1}$ will also satisfy
$$
P(a^\prime)=0.
$$
This is certainly the case for a ring of matrices, unless $a$ is a scalar matrix (which satisfies a degree 1 polynomial relation), i.e. a central element.
A: If we choose a $X$ such that $X^{10}=I_2$ then it is a solution.
So let $X=\begin{pmatrix}0&a\\\frac{1}{a}&0\end{pmatrix}$ for any arbitrary $a\in\mathbb{R}$. There are infinite solutions.
