<s>[INST] Orbit and stabaliser of $2\times2 * 2\times1$ matrices I have the group action of matrix multiplication, meaning:
$g((x,y))=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ \begin{pmatrix}x\\ y\end{pmatrix}$
$G=\left\{\begin{pmatrix}a&0\\0&b\end{pmatrix}:a,b\in\mathbb{R} - \{0\} \right\} g=\begin{pmatrix}a&0\\0&b\end{pmatrix}\in G$ and $X=\{(x,y):x,y\in\mathbb{R}\}$
I want to work out orbits and stabilizer.
I have only done this with permutation groups, so I have no idea how to do this:
Definition of orbit $orb((x,y))=\{g((x,y)):g\in G\}$
So I have $org((x,y))$ equalling the identity matrix?
$stab((x,y))=\{g\in G:g((x,y))=(x,y)\}$
That makes $stab((x,y))$ look like the identity though? Which am I misunderstanding?
 [/INST] If you write out the group action as $$g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax \\ by \end{pmatrix}$$
For the stabilizer, we look for elements $g \in G$ such that $$g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$$ This means that we are looking for $a$ and $b$ such that $ax = x$ and $bx = x$. If $x$ is nonzero, then we must have that $a = 1$, and similarly nonzero $y$ implies that our desired $b$ equals $1$. If $x =0$, then we can choose any nonzero $a$ (as $ax = a \cdot 0 = 0$ for each nonzero $a$), and if $y = 0$, we can choose any value of $b$. To summarize this, considering arbitrary nonzero $x$ and $y$, we have 
$$\mathrm{stab}_{G}\left( \begin{pmatrix} x \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\}, \: \: \mathrm{stab}_{G}\left( \begin{pmatrix} x \\ 0 \end{pmatrix} \right) = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & b \end{pmatrix} \bigg| b \in \mathbb{R} - \{0\} \right\}$$ $$\mathrm{stab}_{G}\left( \begin{pmatrix} 0 \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}\bigg| a \in \mathbb{R} - 0 \right\}, \:\mathrm{stab}_{G}\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right) = \left\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \bigg|\: a,b \in \mathbb{R} - 0 \right\}$$
Now, the orbit is the set of elements $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ such that there exists $g \in G$ for which $$ g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ Because $a$ and $b$ can be any nonzero real number, we then have that $$\mathrm{orb}_{G} \left( \begin{pmatrix} x \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} ax \\ by \end{pmatrix} \bigg| \: a,b \in \mathbb{R} - \{0\} \right\}$$  
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