<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] $$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right) 
$$
with eigenvectors as columns of
$$
\left(
\begin{array}{cc}
\cos \frac{t}{2} & -\sin  \frac{t}{2} \\
\sin  \frac{t}{2} &  \cos \frac{t}{2}
\end{array}
\right) \; \; , \; \;
$$
this confirmed by the identity
$$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right) 
\left(
\begin{array}{cc}
\cos \frac{t}{2} & -\sin  \frac{t}{2} \\
\sin  \frac{t}{2} &  \cos \frac{t}{2}
\end{array}
\right) =
\left(
\begin{array}{cc}
\cos \frac{t}{2} & \sin  \frac{t}{2} \\
\sin  \frac{t}{2} & - \cos \frac{t}{2}
\end{array}
\right)
$$
It does not need to be symmetric, although the reflection is what first came to mind for $M^2 = I,$  which happens when $M^3 = M$ and $M$ is nonsingular. For real matrices, we need just trace zero and determinant $-1$
As  long as $$ a^2 + bc = 1 $$  we can use
$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right) 
$$
for example
$$
\left(
\begin{array}{cc}
7 & 8 \\
-6 & - 7
\end{array}
\right) 
$$
We see the eigenvectors as columns
$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right) 
\left(
\begin{array}{cc}
1+a & -b \\
c & 1+ a
\end{array}
\right) =
\left(
\begin{array}{cc}
1+a & -b \\
c & -1- a
\end{array}
\right) 
$$
</s>