Let \(p\) be a prime and \(F = \mathbb{F}_p\) the field with \(p\) elements, and \(V = F^4\) the standard \(F\)-vector space of dimension \(4\). Count the number of \(F\)-linear subspaces of \(V\).

\(\textit{Solution.} \) Every \(F\)-linear subspace of \(V\) is of the form \(W = \text{span}(v_1,v_2)\) where \(v_1,v_2\in V\) are linearly independent. Thus, the set of \(F\)-linear subspaces is given by \[ X = \{\text{span}\{v_1,v_2\} \subset V : av_1+bv_2 = 0 \implies a = b = 0\}. \] Let \(G = GL_4(F)\). Then \(G\) acts transitively on \(X\) via left-translation. Consider the subspace \(W := \text{span}\left\{w_1,w_2\right\}\in X\) where \[ w_1:=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \ w_2 : =\begin{pmatrix}0\\1\\0\\0\end{pmatrix}. \] Denote the stabiliser of \(W\) by \(G_W\). If \[A = \begin{pmatrix}a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44}\end{pmatrix} \in G_W,\] then by definition of the stabiliser we must have \(A w_1,A w_2 \in W\). Thus, there exists \(x_1,x_2,x_,x_4 \in F\) such that \[ \begin{pmatrix}a_{11}\\a_{21}\\a_{31}\\a_{41}\end{pmatrix} = x_1w_1+ x_2 w_2 \ \ \text{and} \ \ \begin{pmatrix}a_{12}\\a_{22}\\a_{32}\\a_{42}\end{pmatrix} = x_3w_1 + x_4w_2. \] This implies that \(a_{31} = a_{41} = a_{32} = a_{42} = 0\) so that \[ A = \begin{pmatrix}a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ 0 & 0 & a_{33} & a_{34}\\ 0 & 0 & a_{43} & a_{44}\end{pmatrix} \] Further, \(A \in G\) so the remaining coefficients must be chosen so that \(A\) is invertible. There are \((p^2-1)(p^2-p)\) choices for the first two columns since there are independent if and only if we must have \(\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \in GL_2(F)\). Once the first two columns are chosen, we have \(p^4-p^2\) choices for column three and finally there are \(p^4 - p^3\) choices for the last column. Thus, \[ |G_W| = (p^2-1)(p^2-p)(p^4-p^2)(p^4-p^3). \] Since \(X\) is a transitive \(G\)-set, we conclude by the orbit equation that \[ |X| = [G:G_W] = \frac{(p^4-1)(p^4-p)(p^4-p^2)(p^4-p^3)}{(p^2-1)(p^2-p)(p^4-p^2)(p^4-p^3)} = (p+1)(p^2+p+1). \] \[ \hspace{17cm} \blacksquare\]