Find all $4 \times 4$ real matrices such that $A^3=I$ Find all $4 \times 4$ real matrices such that $A^3=I$.
The minimal polynomial must divide $x^3-1$.  Since the matrix is real, the minimal polynomial must be either $x-1$ or $x^3-1$ (i.e., if it contains one of the complex roots of unity, it must contain the conjugate).  Of course, we already know the identity matrix satisfies this equation.  If $A$ is any other non-identity real matrix satisfying the above properties, then its characteristic polynomial must be $(x-1)(x^3-1)$, by similar reasoning above.  This is as far as I could get.
edit: I missed a few cases, as pointed out in the comments.  The minimal polynomial can also be $x^2+x+1$ which yields additional possible characteristic polynomials.

Since $p(A) = 0$ where $p(x) = x^3 - 1 = (x-1) (x^2 + x + 1)$, and $x-1$ and $x^2 + x + 1$ are irreducible over $\mathbb{R}[x]$, then $\mathbb{R}^4$ can be decomposed as a direct sum of cyclic subspaces for $A$ corresponding to the polynomials $x-1$ and $x^2 + x + 1$.  Now, we know that a cyclic subspace with annihilator $x-1$ has matrix representation of the restricted operator equal to $\begin{bmatrix} 1 \end{bmatrix}$; and similarly, a cyclic subspace with annihilator $x^2 + x + 1$ has matrix representation of the restricted operator equal to $\begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix}$.  Therefore, any $4 \times 4$ matrices $A$ with $A^3 = I$ is similar to one of:
$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},
\begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},
\begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix}. $$
(And conversely, it's clear that any matrix similar to one of these matrices satisfies $A^3 = I$.)
