Find a $4\times 4$ matrix $A$ where $A\neq I$ and $A^2 \neq I$, but $A^3 = I$.

Solution 1:

Here is a $2\times2$ example $$ \begin{bmatrix} -\frac12&\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&-\frac12 \end{bmatrix}\tag{1} $$ This works because it is a matrix representation of $e^{2\pi i/3}=-\frac12+i\frac{\sqrt{3}}{2}$; that is, a rotation by $\frac{2\pi}{3}$. Thus, squaring it gives another $2\times2$ example $$ \begin{bmatrix} -\frac12&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&-\frac12 \end{bmatrix}\tag{2} $$ $(1)$ and $(2)$ can easily be extended to $4\times4$ examples in many ways. Here is one using $(1)$: $$ \begin{bmatrix} -\frac12&\frac{\sqrt{3}}{2}&0&0\\ -\frac{\sqrt{3}}{2}&-\frac12&0&0\\ 0&0&\hspace{7pt}1\hspace{7pt}\vphantom{-\frac12}&0\\ 0&0&0&\hspace{5pt}1\hspace{5pt}\vphantom{-\frac12} \end{bmatrix}\tag{3} $$

Yet another $2\times2$ matrix:

As Hagen von Eitzen points out, we can consider the unit vectors $u,v,w$, which are separated by $\frac{2\pi}{3}$

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and note that $u+v+w=0$ to get that $w=-u-v$. Let $\begin{bmatrix}a\\b\end{bmatrix}$ be a coordinate vector using the basis vectors $\{u,v\}$. Then rotation by $\frac{2\pi}{3}$ has the following action: $$ \begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\mapsto\begin{bmatrix}v&w\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}0&-1\\1&-1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\tag{4} $$ Thus, the matrix for rotating by $\frac{2\pi}{3}$ under the basis $\{u,v\}$ is $$ \begin{bmatrix}0&-1\\1&-1\end{bmatrix}\tag{5} $$ which, as Gerry Myerson points out, is the companion matrix for $x^2+x+1$. A companion matrix is annihilated by its polynomial, so $(5)$ is annihilated by $x^3-1=(x-1)(x^2+x+1)$.

The square of $(5)$ also satisfies the specified conditions: $$ \begin{bmatrix}-1&1\\-1&0\end{bmatrix}\tag{6} $$ The answer given by Marc van Leeuwen uses $(6)$.

Solution 2:

Here's a $3\times3$: $$\pmatrix{0&1&0\cr0&0&1\cr1&0&0\cr}$$ You should be able to get a $4\times4$ out of this.

Solution 3:

$$\begin{pmatrix}-1&1&0&0\\-1&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$

Solution 4:

Since powers of diagonal matrices just raise the elements to those powers individually, we can simply pick a diagonal matrix such that all the diagonal elements are cube roots of $1$. If you're allowed complex matrices, the solution is easy.