Matrices on cycles

Question

Find all matrices $A \in M_{2x2}(\mathbb{R})$ such that $A^3=I$ and $A \not = I$ Or prove they don’t exist.

In the case of replacing the cube with a square there are many examples such as:
$$-I$$ $$\left[\begin{matrix} 0 & 1 \\ 1 & 0 \\\end{matrix}\right]$$ And (probably) many more.
The equation can be rephrased as:
$$A^3 - I = 0$$ It can be factorized into:
$$(A^2+A+1)(A-1)=0$$ But, in the ring of matrices there are 0 divisors so it isn’t much help.
Can any of you help me?
things I do know: $\det(A)=1$
$A$ is not diagnalizeable. Proof:
Assume $A= P^{-1}MP$ $$A^3 = P^{-1} M^3 P= I$$ $$M^3 P = P$$ $$M^3 = I$$ Since $I$ is the only diagonal matrix satisfying that equation: $$M= I$$ $$P^{-1}IP= A$$ $$A=I$$ Contradiction! Because it is non-diagnalizeable we can say that it has one eigenvector with eigenvalue 1.

Answer 1

There is an infinity of real solution.

1. $A$ is diagonalizable over $\mathbb{C}$. Indeed $(x^2+x+1)(x-1)$ has only simple roots.

2. $A^2+A+I_2=0_2$. Indeed, the eigenvalues of $A$ are conjugate or real. Thus, $\chi_A$, the characteristic polynomial of $A$, is $(x-1)^2$ or $x^2+x+1$; it's not $(x-1)^2$ because $A\not= I$ and $A$ is diagonalizable. By Cayley Hamilton, $A^2+A+I_2=0_2$.

3. The set of solutions is $\{PCP^{-1};P\in GL(\mathbb{R}^2)\}$, where $C=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$. Indeed, $spectrum(A)=spectrum(C)=\{j,j^2\}$, where $j=e^{2i\pi/3}$. Then $A,C$ are both similar over $\mathbb{C}$ to $diag(j,j^2)$, ; thus $A$ is similar to $C$ over $\mathbb{R}$. Conversely, if $A$ is similar to $C$, then $A^3=I_2$.

4. In other words, the set of solutions is $\{\begin{pmatrix}a&b\\c&d\end{pmatrix};a,b,c,d\in\mathbb{R},a+d=-1,ad-bc=1\}$. This set depends on $2$ real parameters.

Answer 2

Rotation by $\pm {2 \over 3}\pi$. i.e. $A=\begin{pmatrix} \cos({2 \over 3}\pi) & \mp \sin({2 \over 3}\pi)\\ \pm\sin({2 \over 3}\pi) & \cos({2 \over 3}\pi) \end{pmatrix}=\begin{pmatrix} -{1 \over 2} & \mp {{\sqrt{3}}\over 2}\\ \pm {{\sqrt{3}}\over 2}& -{1 \over 2} \end{pmatrix}$

I think these are the only two such matrices (but I don't know how to prove it).