Let $X = \{Span(v)\ |\ 0\neq v\in\mathbb{F}^n\}$, and let $G = SL_n$. Define an action $$ A.Span(v) = Span(Av) $$ I need to show that this action is transitive. Here is my attempt:
Let $v,w\in\mathbb{F}^n$. If $w = \lambda v$, then let $(v_1,\ldots,v_n)$ be a basis to $\mathbb{F}^n$ with $v_1=v$. We therefore, in this basis representation, define $$ A_{ij} = \begin{cases} \lambda & i = j = 1 \\ \frac{1}{\lambda} & i = j = 2 \\ 1 & i = j\neq 1,2 \\ 0 & \text{otherwise} \end{cases} $$ Then, in this basis, $Av = w$, and therefore $A.Span(v) = Span(w)$, and $\det{A} = 1$.
If $v,w$ are linearly independent, then let $(v_1,\ldots,v_n)$ be a basis of $\mathbb{F}^n$ such that $v_1=v, v_2=w$. We define $$ A_{ij} = \begin{cases} 1 & i =1, j = 2 \\ -1 & i=2, j=1 \\ 1 & i = j\neq 2 \\ 0 & \text{otherwise} \end{cases} $$ Then, in this basis, $Av = w$, and therefore $A.Span(v) = Span(w)$ and $\det{A} =1$.
Is this OK?