TITLE: Find a $4\times 4$ matrix $A$ where $A\neq I$ and $A^2 \neq I$, but $A^3 = I$.
QUESTION [11 upvotes]: Give an example of a $4 \times 4$ matrix where $A \neq I$, $A^2 \neq I$, and $A^3 = I$.
I found a $2 \times 2$ matrix where $A \neq I$ and $A^2 = I$, but this problem is more complex and has me completely stumped.

REPLY [4 votes]: Nobody has really highlighted the geometric interpretation yet, so I will. You're looking for a linear mapping (in 4 dimensions) that will take a vector $x$ to itself if applied three times in a row, yet will yield distinct intermediate vectors $Ax$, $A^2x$.
Let's start with the 2-dimensional case. There, it's easy to see that the mapping you want is a rotation.  If $A$ represents the rotation by an $n$-th of a full circle (i.e, $360/n$ degree, or $2\pi/n$ red), then $A^n$ rotates by a full circle or equivalently not at all.
The rotation by $\varphi$ rad is given by $\begin{pmatrix} \cos\varphi & \sin\varphi \\ -\sin\varphi & \cos \varphi\end{pmatrix}$. Thus, $$A = \begin{pmatrix} \cos\frac{2\pi}{n} & \sin\frac{2\pi}{n} \\ -\sin\frac{2\pi}{n} & \cos\frac{2\pi}{n}\end{pmatrix}$$ is a mapping with $A^0 \ldots A^{n-1} \neq I$ and $A^n = I$.
To extend this to more than two dimensions, you can simply use the identity mapping for the other dimensions. Or, in the 4-dimensional case, you could also combine two rotation matrixes. The resulting matrix then has block-diagonal form with two $2x2$ blocks.