Orbit and stabaliser of $2\times2 * 2\times1$ matrices

Question

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?

Answer 1

The orbit of a vector $(x,y) \in X$ is the set of all $(a,b) \in X$ that can be "reached" by the group action. In other words, the orbit of $(x,y)$ is the set of all $(a,b)$ such that $(a,b) = g(x,y)$ for some $g\in G$.

The stabilizer of a vector $(x,y) \in X$ is the set (which turns out to be a subgroup) of all $g \in G$ such that $g(x,y) = (x,y)$.

Your problem can be naturally divided into four cases.

Case 1: $x = y = 0$.

Since $g(0,0) = (0,0)$ for every $g \in G$, the orbit of $(0,0)$ is just the singleton $\{(0,0)\}$.

The stabilizer of $(0,0)$ is the set of all $g\in G$ such that $g(0,0) = (0,0)$. Since this is true for any $g\in G$, the stabilizer of $(0,0)$ is the entire group $G$.

Case 2: $x = 0, y \neq 0$

Since $$\begin{pmatrix} a & 0 \\ 0 & b \\ \end{pmatrix} \begin{pmatrix} 0 \\ y \\ \end{pmatrix} = \begin{pmatrix} 0 \\ by \\ \end{pmatrix}$$ and $b$ is any nonzero real number, the orbit of any vector of the form $(0,y)$ (where $y \neq 0$) is the set of all vectors of the form $(0,z)$ where $z$ is any nonzero number.

The stabilizer of such an vector $(0,y)$ is the set of all $\begin{pmatrix} a & 0 \\ 0 & b \\ \end{pmatrix}$ such that $$\begin{pmatrix} a & 0 \\ 0 & b \\ \end{pmatrix} \begin{pmatrix} 0 \\ y \\ \end{pmatrix} = \begin{pmatrix} 0 \\ y \\ \end{pmatrix}$$ This forces $b=1$, but $a$ can be anything. So the stabilizer of $(0,y)$ is the set of all matrices in $G$ of the form $$\begin{pmatrix} a & 0 \\ 0 & 1 \\ \end{pmatrix}$$ You can easily check that this set is indeed a subgroup of $G$, as is any stabilizer. I'll let you work out the details of the other two cases, which are:

Case 3: $x \neq 0, y = 0$

and

Case 4: $x \neq 0, y \neq 0$

Two "sanity checks" you can use once you think you have a solution:

1. The orbits partition the set $X$. In other words, every element of $X$ belongs to exactly one orbit.

2. The stabilizer of each element of $X$ is a subgroup of $G$.

One more note: In general it's possible for different elements of $X$ to have the same stabilizer. Elements that are in the same orbit do not necessarily have the same stabilizer, and elements in different orbits can have the same stabilizer.

Answer 2

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 $$\mathrm{orb}_{G} \left( \begin{pmatrix} x \\ y \end{pmatrix} \right) = \left\{ \begin{pmatrix} ax \\ by \end{pmatrix} \bigg| \: a,b \in \mathbb{R} - \{0\} \right\}$$