Example of matrix $A\neq I$ such that $A^3=I$ Let $A \neq I$ be a $3 \times 3$ matrix. I need to find an example of $A$ that satisfies $A^3 = I$.
Is there any "smart" way to do this? All I can think of is to either multiply $A$ for 3 times and then try to guess the factors or try to solve $A^2 = A^{-1}$. In either case I feel that it can be solved in a smarter way. Any ideas?
 A: Fix $3$ linearly independent vectors $e_1$, $e_2$, and $e_3$ of $\Bbb R^3$ (the vectors of the standard basis, say) and consider the map $f\colon\Bbb R^3\longrightarrow\Bbb R^3$ such that $f(e_1)=e_2$, $f(e_2)=e_3$ and that $f(e_3)=e_1$. Now, let $A$ be the matrix of $f$ with respect to some basis of $\Bbb R^3$. Can you check that such a matrix will work?
A: If you allow $A$ take complex values. Below is a solution.
By condition, $f(\lambda) = \lambda^3 - 1 = (\lambda - 1)(\lambda - \omega_1)(\lambda - \omega_2)$ is an annihilating polynomial of $A$, where
$\omega_1 = e^{i\frac{2}{3}\pi}, \omega_2 = e^{-i\frac{2}{3}\pi}$, but $g(\lambda) = \lambda - 1$ is not an annihilating polynomial. So there are some possibilities of how the minimal polynomial $d(\lambda)$ of $A$ can be. For example, if $d(\lambda) = \lambda - \omega_1$. Then $A$ has $3$ multiple eigenvalues $\omega_1$, in this case, $A = \mathrm{diag}(\omega_1, \omega_1, \omega_1)$.
You can check that $A \neq I$, but $A^3 = I$. You can also construct different $A$ by taking $d(\lambda) = \lambda - \omega_2$, $d(\lambda) = (\lambda - 1)(\lambda - \omega_1)$, etc.
If you require the entries to be real numbers, then it is easy to transform one complex solution such as $\mathrm{diag}(\omega_1, \omega_1, 1)$ to the real rotation matrix as below:
\begin{equation*}
A = \begin{pmatrix}
\cos\frac{2}{3}\pi & \sin\frac{2}{3}\pi & 0 \\
-\sin\frac{2}{3}\pi & \cos\frac{2}{3}\pi & 0 \\
0 & 0 & 1
\end{pmatrix} = 
\begin{pmatrix}
-\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
-\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{pmatrix}.
\end{equation*}
