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?
 A: I'm not sure why no one has yet posted
$$(1)^3=(1)$$
A: An easier approach is noticing that if $A^3 = A$, then $A^2 = I$, which should be much more tractable.  This also means that $A = A^{-1}$ and you can equate the generic abcd with its inverse.  Further, the determinant of A must be 1 or -1, since it squares to 1.  So that simplifies the inverse formula too.
Addendum: if $A^3 = A$, then $det(A)^3 = det(A)$ so $det(A)$ satisfies $x^3 = x$, which has solutions x = 1, 0, and -1. If x = 1 or -1, $A$ is invertible and the above description works.  If $A$ is not invertible, then it must have two rows or columns that are proportional, which is a further simplification for the search.
A: One approach is to notice that if $A^2 = A$, then $A^3 = A  A^2 = A^2 = A$. So it suffices to find a matrix with $A^2 = A$. Since $A$ is square these are matrices associated to projections (not neccesarily orthogonal projections). It's much easier to algebraicly solve $A^2 = A$ for the entires. As an example,
$$A = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$
will do what you want.
A: If $ad-bc=1,$ with $a,b,c,d$ non-zero, you can use:
$$\begin{pmatrix}-bc&-bd\\ac&ad\end{pmatrix}$$
For example, $a=2,b=1,c=1,d=1$ gives:
$$\begin{pmatrix}-1&-1\\2&2\end{pmatrix}$$
I got this by taking:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
Where $B=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ satisfies $B^2=B$ and hence $B^3=B.$
You can also choose $B=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ or $B=\begin{pmatrix}0&0\\0&-1\end{pmatrix}.$
A: One way to simplify the problem is to impose some further assumptions on $A$. For example, suppose we further assume that $A$ is singular. Then $A=uv^T$ for some entrywise vectors $u$ and $v$. The requirements that $A^3=A$ and $A\ne0$ entrywise are now equivalent to $(v^Tu)^2=1$ and $u,v\ne0$ entrywise, which are utterly easy to fulfil (when the characteristic of the underlying field is not $2$).
A: $$\begin{pmatrix}7&-6\\8&-7\end{pmatrix} $$
A: 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*}
A: $A^3=A$ can be written as $A(A-I)(A+I)=0$, which is satisfied by any matrix that is similar to a $2x2$ diagonal matrix with diagonals in the set $\{0,1,-1\}$.
A: $$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right) 
$$
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
$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right) 
\left(
\begin{array}{cc}
1+a & -b \\
c & 1+ a
\end{array}
\right) =
\left(
\begin{array}{cc}
1+a & -b \\
c & -1- a
\end{array}
\right) 
$$
