Are there infinitely many $A\in \mathbb{C}^{2 \times 2}$ satisfying $A^3 = A$? Let $A$ be a $2 × 2$-matrix with complex entries. The number of $2 × 2$-matrices
$A$ with complex entries satisfying the equation $A^3 = A$ is infinite.
Is the above statement true? I know that $0$ and I are two solutions. But are there any more solutions?
 A: More generally, $A^3=A$ if and only if $A(A^2-I)=0$. Hence the minimal polynomial of $A$, $m_A(x)$ divides $x(x-1)(x+1)$, so the characteristic polynomial is $p_A(x)=x^i(x-1)^j(x+1)^k$ such that $i+j+k=2$. Take any matrix with all of it's eigenvalues $0,\pm1$ (some of those) and you will have a matrix satisfying your equation. There are some 'simple' matrices satisfying this equation:
$$\begin{pmatrix}0&0\\0&0\end{pmatrix},\begin{pmatrix}1&0\\0&1\end{pmatrix},\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\\
\begin{pmatrix}0&0\\0&1\end{pmatrix},\begin{pmatrix}0&0\\0&-1\end{pmatrix},\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
And those are all the 'different' matrices satisfying your equation. Any other solution to your equation is similar to one of those, i.e. take any invertible matrix $P$, take any matrix $E$ from the list I just mentioned and $A=PEP^{-1}$ is a solution.
In short - there are infinitely many solutions, each is of the form $A=PEP^{-1}$ for an invertible $P$ and $E$ from the six above.
A: The matrices $A_r=\begin{pmatrix} 1 & 0 \\ r & 0 \end{pmatrix}, \ r \in \mathbb{R}$ satisfy $A_r^2=A_r$, thus $A_r^3=A_r$. 
More generally for $B=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, it is easy to verify that $B^3=B$ and therefore if $A$ is any matrix that is similar to $B$, let's say $A=P^{-1}BP$, then
$$
A^3=P^{-1}B^3P=P^{-1}BP=A.
$$
In our case the matrices $A_r$ are similar to the matrix $B$ for any $r\in\mathbb R$.
A: It is indeed infinite. For example all the matrices
$$
\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix},\ \ t\in[0,1]
$$
satisfy your equality (they actually satisfy $A^2=A$, which of course implies $A^3=A$).
