I have have to find the number of orbits for the group action given by the set of all real $2\times 2$ invertible matrices acting on $\mathbb{R^2}$ by matrix multiplication.
I know that if a group $G$ acts on a set $X$ then for each $x\in X$,
$orbit(x) = \{gx: g\in G\}$, where $g$ runs over all the elements of the group $G$.
Thus here for any $x\in \mathbb{R^2}$ its orbit is given by set $\{Ax : A \in G\}$ and I have to keep in mind that orbits partition the set $X = \mathbb{R^2} $.
My confusion:
Here both group $G$ and set $X$ are of infinite order. So, how to find orbit of all the infinite elements of $\mathbb{R^2}$. Do I have to look at the basis elements of $\mathbb{R^2}$?
I am learning to find the orbits of a set. Any help and basic idea to solve such kind of problem would be very helpful to me.
Thank you very much for your kind consideration.
Best Answer
Take a point $p = (a,b) \in \mathbb{R}^2$. The claim is that taking any point $q =(x,y)\in\mathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$. The equation (1) corresponds to the system $\alpha a + \beta b = x \,\& \, \gamma a + \delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.