TITLE: Find all matrices satisfying $X^3=I-X$
QUESTION [8 upvotes]: 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.

REPLY [11 votes]: 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
$$a+d=2\operatorname{Re}(\beta)\quad;\quad ad-cb=|\beta|^2$$