group oaction of $GL(n,\mathbb{R})$ on $\mathbb{R}^n$

Question

Let $A · v = Av\ $ be an action of group $\ GL(n, \mathbb{R})$ on $\ \mathbb{R}^n.$

I want to prove that there are only two orbits, the trivial one ${0}$ and ${\mathbb{R}^n}.$ It looks like a "surjectivity check", but I do not know how to choose $A$.

Answer 1

Take $v,w\in\mathbb{R}^n\setminus\{0\}$ then complete each to a basis (their singletons are independent sets): $v\in\mathcal{B},w\in\mathcal{C}$. Then there exists a unique $\varphi:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ such that $\varphi(\mathcal{B})=\mathcal{C}$ w.r.t. their order. Here you have $\varphi(v)=w$. Actually you are proving the transitivity of the action on the nonzero vectors, but the linear extension theorem states a sort of $n$-transitivity (given the linear indipendence).