Compute the stabilizers and orbits of $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$ The natural action of $GL_2(\mathbb{R})$ on $\mathbb{R}^2$ is given by $A\cdot v=Av$, for $A\in GL_2(\mathbb{R})$ and $v\in\mathbb{R}^2$. Compute the stabilizers and orbits of $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$.
So I got that the orbit and stabilizer of $\begin{bmatrix}1\\0\end{bmatrix}$ is given by
\begin{align*}
Av=\lbrace\begin{bmatrix}a\\c\end{bmatrix}\rbrace
\qquad \textrm{Stab}(v)=\lbrace\begin{bmatrix}1&b\\0&d\end{bmatrix}\rbrace
\end{align*}
And the orbit and stabilizer of $\begin{bmatrix}1\\1\end{bmatrix}$ is given by
\begin{align*}
Av=\lbrace\begin{bmatrix}a+b\\c+d\end{bmatrix}\rbrace
\qquad \textrm{Stab}(v)=\lbrace\begin{bmatrix}1-b&b\\1-d&d\end{bmatrix}\rbrace
\end{align*}
I'm a little unsure about the stabilizers because the definitions just kind of seem to end at the second line with the equation. 
 A: If $G$ is acting on $X$, and $x \in X$, the stabilizer is
$$
G_x=\{ g \in G | gx=x \}
$$
Hence in our case, we are looking for all the matrices $A \in \mathrm{GL}_2 (\mathbb{R})$ such that $Av=v$, where $v$ is either $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ or $\begin{pmatrix} 1 \\ 1\end{pmatrix}$.
Now, if $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$, we have $\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix} 1 \\ 0\end{pmatrix}=\begin{pmatrix} a \\ c\end{pmatrix}$ and this is equal to $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ if and only if $a=1,c=0$, while $b$ can assume any value $\in \mathbb{R}$ and $d \in \mathbb{R} \setminus \{ 0 \}$ (otherwise the matrix is not invertible).
Similarly, $\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix} 1 \\ 1\end{pmatrix}=\begin{pmatrix} a+b \\ c+d\end{pmatrix}$ and this is equal to $\begin{pmatrix} 1 \\ 1\end{pmatrix}$ if and only if $a+b=c+d=1$ and thus $a=1-b,c=1-d$. Furthermore $b \neq d$ (in order to for the matrix to be invertible).
I don't know if this answer your question.
A: This is a rather unconventional way of writing these results. The orbits in both cases are simply $\mathbb R^2\setminus\{0\}$; there’s no need to parametrize them, and no reason to parametrize them differently in the two cases. The conventional way to write the stabilizers would be
$$
\operatorname{Stab}\left(\left[\matrix{1\\0}\right]\right)=\left\{\left[\matrix{1&b\\0&d}\right]\mid b,d\in\mathbb R,d\ne0\right\}
$$
and
$$
\operatorname{Stab}\left(\left[\matrix{1\\1}\right]\right)=\left\{\left[\matrix{1-b&b\\1-d&d}\right]\mid b,d\in\mathbb R,b\ne d\right\}\;,
$$
respectively. Note that you need to exclude the cases where the matrix is singular, as the general linear group consists only of the non-singular matrices.
