Action of $GL(\Bbb F_2^3)$ on the sub-spaces of $\Bbb F_2^3$ of dimension $2$ Let $V=\Bbb F_2^3$ and let $G=GL(V)$ act naturally on the set $X=\{W\subset V:\text{sub-vector space,}\dim =2\}$
If $W\in X$ how do you determine the $Stab_G(W)$? and why shoud the cardinality of $Stab_G(W)$ be the same for different $W$'s?
If we want to prove that the action is transitive why is it enough to verify that there exists a matrix that sends $e_1$ to $u$, $e_2$ to $v$ and $e_3$ to $e_3$ for arbitrary linearly independent vectors $u,v$?
 A: To prove transitivity it would be enough to verify that there exists a matrix that sends $e_1$ to $u$, $e_2$ to $v$ and $e_3$ to $e_3$ for arbitrary linearly independent vectors $u,v$, but unfortunately such a matrix does not always exist. After all, the vectors $u$, $v$ and $e_3$ may be linearly dependent. 
Fortunately, it is already enough to show that for every pair of linearly independent vectors $u,v\in V$ there exists a matrix in $G$ that maps $e_1$ to $u$ and $e_2$ to $v$. This would imply that the hyperplanes $\operatorname{span}(e_1,e_2)$ and $\operatorname{span}(u,v)$ are in the same orbit under the action of $G$, and hence that all hyperplanes are in the same orbit, i.e. that the action is transitive.
It is a basic fact (or simple exercise) on group actions that elements in the same $G$-orbit have conjugate stabilizers; in particular their stabilizers have the same cardinality.
A: Your question on transitivity has been answered in the comments, but with regards to your question on $\operatorname{Stab}_{G}(W)$ for $W$ a two dimensional subspace of $V$. Note that because we have transitivity, $\operatorname{Stab}_{G}(W) \cong \operatorname{Stab}_{G}(W')$ for any pairs $W,W' \in X$. Indeed, suppose that $W, W' \in X$. Then by transitivity, there exists $g \in G$ such that $gW = W'$. Now suppose that $h \in \operatorname{Stab}_{G}(W')$, then $h\cdot w' \in w'$ for every $w' \in W'$, and so $(hg) \cdot w \in W'$ for every $w \in W$, and so finally $(g^{-1}hg)\cdot w \in W$ for every $w \in W$. This should make it clear that the map $\operatorname{Stab}_{G}(W') \rightarrow \operatorname{Stab}_{G}(W)$ such that $h \mapsto g^{-1}hg$ is an isomorphism. Thus, the stabalisers of pairs of elements of $X$ have the same cardinality, and so the stabiliser of all elements of $X$ must have the same cardinality.
Note that the above also allows us to calculate $\operatorname{Stab}_{G}(W)$ for any $W$ by calculating $\operatorname{Stab}_{G}(W)$ for our favourite subspace $W = \left< e_1, e_2 \right> \subseteq V$. It should be easy to see that for this $W$ we have
$$
\operatorname{Stab}_{G}(W) = \left\{ \begin{pmatrix} a & b & p \\ c & d & q \\ 0 & 0 & r  \end{pmatrix} : ad - bc \neq 0, r \neq 0 \right\} 
$$ 
and so 
$$
\left|\operatorname{Stab}_{G}(W) \right| = \left| \operatorname{GL}_{2}\left(\mathbb{F}_2 \right)\right|\left|\mathbb{F}_2^{\times} \right|\left| \mathbb{F}_2 \right|^{2}   = 4\left| \operatorname{GL}_{2}\left(\mathbb{F}_2 \right)\right| = 4\cdot 6 = 24
$$
