I'm working on the orbit of $e_1$ in $\mathbb{R}^2$ under the action of matrix multiplication. For some matrix $A$ in $SL_2(\mathbb{R})$, I believe that the orbit is simply any vector $(a,b)$ in $\mathbb{R}^2$ that is not the zero vector. This seems obvious, but is it necessary to prove that any such vector is in the orbit? If so, how would I do that?
Then, what if I'm working with some matrix $A$ in the orthogonal group $O_2$ instead? Beyond writing down the definition of $O_2$, I'm unsure of how to proceed.
The case of $\operatorname{SL}(2, \Bbb R)$ is not obvious, inasmuch as the statement is not true for other common matrix groups, including $\operatorname{O}(2, \Bbb R)$.
Hint Recall that for any matrix $A \in M(2, \Bbb R)$, the image $A {\bf e}_1$ of the first standard basis vector ${\bf e}_1 := \pmatrix{1\\0}$ under the standard action of $A$ is just the first column of $A$. So, your claim is equivalent to the statement that for any nonzero vector $\pmatrix{a\\b} \in \Bbb R^2$ there is a matrix in $\operatorname{SL}(2, \Bbb R)$ of the form $\pmatrix{a&\ast\\b&\ast}$.