Describe the set of cube roots of the identity matrix in $M_{n \times n}(\mathbb{R})$.
My attempt:
Let $I_n$ denote the identity matrix of dimension $n \times n$. To find the cube roots of $I_n$, one has to find matrices $A$ such that $A^3 = I_n$. A rotation matrix of dimension $n \times n$ has the property that its repeated application on a vector results in the vector being rotated by an angle which is a fraction multiple of $\frac{2\pi}{3}$. Thus, if we take the rotation matrix $R$ and raise it to the power $3$, we get the unit matrix $R^3 = I_n$. The set of cube roots of $I_n$ will consist of all possible combinations of rotation matrices raised to the power $k$, where $k\in \mathbb{Z}$: $$\{R^k \ | \ k \in \mathbb{Z}\}$$
Did I reason correctly and solve the problem? The set of cube roots of the identity matrix consists of all possible combinations of rotational matrices raised to degree $k$?
Notice that if $A$ satisfies $A^3 = I_n$, then so does any similar matrix $P A P^{-1}$, but the set of rotations is not closed under similarity. For example,$$A := \pmatrix{0&-1\\1&1}$$ satisfies $A^3 = I_2$, but since $A^\top A \neq I_2$, $A$ is not orthogonal and in particular not a rotation (though it is similar to a rotation by $\frac{2 \pi}{3}$).
Hint If $B$ satisfies $B^3 = I$, that is, if $p(B) = 0$, then $$p(x) := x^3 - 1,$$ hence the minimal polynomial of $B$ divides $x^3 - 1 = (x - 1) (x^2 + x + 1)$. (Notice that $A$ is the companion matrix of the quadratic factor, $x^2 + x + 1$.)