Orbits of the Group Action from the Group of all $2 \times 2$ Diagonal non-zero Matrices on $\mathbb{R}^2$

Question

Consider the group $G = \left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} \;\middle|\; a, b \in \mathbb{R} \setminus \{0\}\right\}$ and the set $X = \{(x, y) \in \mathbb{R}^2\}$.

The group action $g$ is defined as, $g((x, y)) = (ax, by)$.

I need to find the orbits of this group, but I have no idea where to start showing them. I have a general idea of what the orbits are, i.e. If a point $(x, y)$ starts somewhere that is not on an axis, then the orbit is the whole Cartesian Plane except the axis. And if it starts on an axis, then the orbit is the entire axis except for the origin which stays at zero.

But I have no idea how to show this algebraically, or with a diagram of the Cartesian Plane.

Just pointer would be greatly appreciated.

Answer 1

Think of it as solving equations.

Supposing that $(x,y)$ is not on either axis, you want to prove that the orbit of $(x,y)$ is the set of all $(r,s)$ not on either axis. In other words, given $(x,y)$, $(r,s)$ not on either axis, you want to solve for $$\begin{bmatrix}a&0\\0&b\end{bmatrix} \in G $$ such that $$g((x,y)) = (r,s) $$ or, equivalently, $$(ax,by) = (r,s) $$ or, equivalently, $$ax=r \quad\text{and}\quad by=s $$ Since these are easily solved for $a,b$, you're done.

The other cases --- $(x,y)$ on the nonzero $x$-axis; and $(x,y)$ on the nonzero $y$-axis --- are similarly handled.

Answer 2

You action simply multiplies the $x$ and $y$ co-ordinates by two non-zero numbers. Take two vectors without any component being zero. By taking the raitos of corresponding co-ordinates as the entries of a diagonal matrix we can see that these two vectors are in each others orbit.

Now consider a vector with one component ($x$ or $y$) zero. In its orbits all the vectors will also have the same property. Conversely $(x,0)$ and $(x',0)$ are in each other's orbit if $x,x'\ne0$. And similarly for $y$ component. ANd origin is its own orbit.