# 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.

• The minimal polynomial could also be $x^2 + x + 1$, and the characteristic polynomial could be $(x-1)^2 (x^2 + x + 1)$, $(x^2 + x + 1)^2$, etc. – Daniel Schepler Sep 26 '17 at 20:56
• Yes. I was just thinking about that now. I'll edit the question. – GillyB Sep 26 '17 at 20:58
• Can you tell whether or not $A$ must be diagonalizable? – Qiaochu Yuan Sep 26 '17 at 21:05
• So, a couple possible $2\times 2$ matrices with order 3 are: $\begin{bmatrix} 0 & -1 \\ 1 & -1\end{bmatrix}$, $\begin{bmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{bmatrix}$. Extend either of those by $I$ on the diagonal (or put two of them together on the diagonal) and you'll have nontrivial $4 \times 4$ examples. – Daniel Schepler Sep 26 '17 at 21:05
• And an example $3 \times 3$ permutation matrix: $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ (as before, add a 1 on the diagonal to get $4 \times 4$ examples). – Daniel Schepler Sep 26 '17 at 21:10

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 $A^3 = I$.)