<s>[INST] Find a matrix $A$ with no zero entries such that $A^3=A$ I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?
 [/INST] It is useful to think about a problem like that in geometric terms.
We want to have $ A^3=A $ which could be thought as some kind of rotation of the plane.
Along these lines what we want is if we rotate a vector by $\theta $ under $A$ to be equal to a rotation by $3 \theta$ corresponding to $A^3$.
A rotation by $\theta$ is given by $A_\theta= 
\left (\begin{array}{cc}
\cos \theta & -\sin \theta\\
\sin \theta  &  \cos \theta
\end{array} \right )$.
So if we choose $\theta = \pi = 3\theta  ~\text{mod} 2 \pi $
We get
$$A_{\pi/2}= 
\left (\begin{array}{cc}
-1 & 0\\
0  &  1
\end{array} \right ).$$
Which should satisfy $$ A_{\pi}^3 = A_{\pi} $$
We can also change the coordinates by $\pi/3$ so to eliminate the zeros.
So we define
\begin{align*}
A &= A_{\pi/3}  A_{\pi} A_{-\pi/3} \\
&=  \left (\begin{array}{cc}
\frac{1}{2} & -\frac{\sqrt{3}}{{2}}\\
\frac{\sqrt{3}}{{2}} &  \frac{1}{2}
\end{array} \right ) \cdot  \left (\begin{array}{cc}
-1 & 0\\
0  &  1
\end{array} \right ) \cdot \left (\begin{array}{cc}
\frac{1}{2} & \frac{\sqrt{3}}{{2}}\\
-\frac{\sqrt{3}}{{2}} &  \frac{1}{2}
\end{array} \right ) 
\end{align*}
</s>