I'm trying to solve the following task:
Let $G:=\mathsf{GL}(2,\mathbb{R})$ acts on $\mathbb{R}^2$ as the matrix multiplication $(A,v):= Av$. Compute $G_{(0,0)}$, $G_{(1,0)}$ and find all the orbits. (Let $G\subset M_{2\times 2}(\mathbb{R})$ denote a group of $2\times 2$ matrices which are invertible, and let $X=\mathbb{R}^2$ ($G$ acts on $X$). Find all orbits of $X$ under the action of $G$.)
Additionally, I would like to calculate the stabilizer.
For the first task, i got $\begin{bmatrix}0\\0\end{bmatrix}$ and $\begin{bmatrix}a\\c\end{bmatrix}$.
However, I have no idea how to find the solution to the orbits of this actions. My best bet is that orbits would be of the form $\begin{bmatrix}a\\0\end{bmatrix}$,$\begin{bmatrix}0\\c\end{bmatrix}$ and
$\begin{bmatrix}a\\c\end{bmatrix}$, but i have to admit its mostly intuition.
How could I solve this task?
Best Answer
$\operatorname{GL}(2,\mathbb R)$ is the group of all invertible linear transformations on $\mathbb R^2$. So to find the stabilizer of a vector $v\in\mathbb R^2$, you have to find all invertible linear maps which fix $v$. You're supposed to do this for $(0,0)$ and $(1,0)$. So, which are the linear maps that fix the $0$-vector? All of them. And which are the linear maps that fix the first basis vector? The ones whose first column reads $\begin{pmatrix}1\\0\end{pmatrix}$.
As for the orbits: Remember that linear maps always map the $0$-vector to the $0$-vector, so one orbit is $[0]=\{0\}$. And then consider the fact that for $v,w\in\mathbb R^2$, the map whose matrix representation consists of the two columns $v$ and $w$ maps the first basis vector to $v$. Is there any $v\in\mathbb R^2$ you can't reach this way? The only restriction is that $v$ and $w$ be linearly independent, since that's a necessary and sufficient condition for invertibility of the matrix.