How many \(2 \times 2\) matrices \(A \in \text{Mat}_{2\times 2}(\mathbb{F}_p)\) satisfy \(A^2 = A\)?

\(\textit{Solution.} \) Clearly \(A =0\) and \(A = 1\) satisfy \(A^2 = A\). Thus, the solution is \(|X|+2\) where \[X := \{ A \in \text{Mat}_{2\times 2}(\mathbb{F}_p) : A^2 = A, A \neq 0,1 \}.\] If \(A \in X\), then the minimal polynomial \(m_{A}(x)\) divides \(x^2 - x = x (x-1)\). Since \(A \neq 0\) and \(A \neq 1\), \(A\) is not a root of the polynomial \(x\) or the polynomial \(x-1\). Thus, \(m_A(x) = x(x-1)\). Since \(m_A(x)\) has no repeated roots, \(A\) is diagonalisable. In fact, since \(\text{deg}(m_A(x)) = 2\), we see that \(m_A\) is the characteristic polynomial for \(A\) and thus \(\text{spec}(A) = \{0,1\} \). Thus, \(A \) is similar to the matrix \(D =\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}\in X\), i.e., \(A = PDP^{-1}\) for some \(P \in GL_{2\times 2}(\mathbb{F}_p) \). Precisely, this means that \(X\) is a transitive \(G\)-set under the conjugation action of \(G =GL_2(\mathbb{F}_p)\) on \(X\). Let's compute the order of the stabiliser \(G_D\) of \(D\) under this action. If \(B = \begin{pmatrix} a & b\\ c & d\end{pmatrix} \in G_D\), then \[\begin{pmatrix} a & 0\\ c & 0\end{pmatrix} = \begin{pmatrix} a & b\\ 0 & 0\end{pmatrix}.\] Thus, \(b = c =0\) so that \(B = \begin{pmatrix} a & 0\\ 0 & d\end{pmatrix}.\) Since \(B \in G\), we can't have \(ad =0\) so there there are \(p-1\) choices for \(a,d\). Thus, \(|G_D| = (p-1)^2\). Finally, by the Orbit-Stabilizer theorem, we obtain \[ |X|+2 = [G:G_D] + 2 = \frac{(p^2-1)(p^2-p)}{(p-1)^2} + 2 = p^2 + p +2. \] \[ \hspace{17cm} \blacksquare\]