Find all matrices satisfying $X^3=I-X$ We are studying for a qualifying exam and have come across the following problem in a previous exam.

Determine the solutions (if any) of the matrix equation $X^3=I-X$ in the $2 \times 2$-matrices over $\mathbb{R}$.

Any hints in the right direction would be much appreciate for this.
 A: Hint: The matrix satisfies $X^3+X-I=0$. What can you say about the minimal polynomial?
A: The equation is $x^3+x-1=0$ now a two by two matrix has a minimal polynomial of degree $\leq 2$. So which quadratic or linear polynomials divide tbe above cubic ?
A: Hint
You can start looking at the eigenvalues of $X$. What kind of relationship there is? Can you find the eigenvalues of $X$?
A: Note that this polynomial has one real root and two complex conjugates.  For a real matrix, if one of the complex solutions is an eigenvalue then so is the other.  A real $2 \times 2$ matrix with eigenvalues $a \pm bi$ is 
$$ \pmatrix{ s & t \cr -((a-s)^2 + b^2)/t & 2a - s }$$
for arbitrary $s$ and $t \ne 0$.
On the other hand, if the real root $r$ is an eigenvalue, it is the only one, and
has geometric multiplicity $2$, so the matrix is $r I$.
A: The polynomial $x^3+x-1$ has one real root $\alpha$ and two conjugate complex non real roots $\beta$ and $\overline \beta$. Since this polynomial annihilates the matrix $X$ then their eigenvalues belong to the set of roots of the polynomial and since $X$ is a real matrix then there's two cases:


*

*$\alpha$ is the only eigenvalue of $X$ and if $X$ isn't diagonalizable then the minimal polynomial is $(x-\alpha)^2$ divides the given polynomial which isn't true so $X$ is diagonalizable and then $X$ is similar to $\alpha I_2$.

*$\beta$ and $\overline\beta$ are the eigenvalues of $X$ so the minimal polynomial of $X$ is $(x-\beta)(x-\overline\beta)=x^2-2\operatorname{Re}(\beta) x+|\beta|^2$ so let
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
and since
$$\chi_X(x)=\det(X-xI_2)=(a-x)(d-x)-cb=x^2-(a+d)x+ad-cb$$
hence we choose $a,b,c$ and $d$ such that
$$a+d=2\operatorname{Re}(\beta)\quad;\quad ad-cb=|\beta|^2$$

