Partition of $\Bbb R^2$ arising from matrix multiplication. Describe the partition of $\Bbb R^2$ arising from the action, by matrix multiplication, of the subgroup $H=\left\{\begin{pmatrix}
    1 & 0\\
    a & 1
\end{pmatrix}:a\in\Bbb R\right\}$ of $GL_2(\Bbb R)$.
First of all I don't think I really understand the question itself. Does it means I pick elements of $H$ and apply matrix multiplication to $\begin{pmatrix}
    x\\
    y
\end{pmatrix}$ where $x,y\in\Bbb R$? I don't know how to describe the partition of a cartesian in terms of matrix multiplication And don't know what partition to describe I am confused. And why is the answer like this:"The lines $x=c$ for $c\neq0$ and the points $(0,b)$"?

Let's first prove that indeed $H\le\operatorname{GL}_2(\mathbb{R})$, and that we really have a $H$-action on $\mathbb{R}^2$.

*

*Subgroup - closure:
$\space\begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix} 
1 & 0 \\
b & 1 \\
\end{pmatrix}=
\begin{pmatrix} 
1 & 0 \\
a+b & 1 \\
\end{pmatrix}\in H$;


*Subgroup - "closure by inverses":
$\space\begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}^{-1}=
\begin{pmatrix} 
1 & 0 \\
-a & 1 \\
\end{pmatrix}\in H$;


*Action - property #0:
$\space\begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix} 
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix} 
x \\
ax+y \\
\end{pmatrix} \in \mathbb{R}^2
$


*Action - property #1:
$\space\begin{pmatrix} 
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix} 
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix} 
x \\
y \\
\end{pmatrix}
,\space\space\forall \begin{pmatrix} 
x \\
y \\
\end{pmatrix}\in \mathbb{R}^2$


*Action - property #2:
$
\space\biggl(\begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix} 
1 & 0 \\
b & 1 \\
\end{pmatrix}
\biggr)
\begin{pmatrix} 
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix} 
1 & 0 \\
a+b & 1 \\
\end{pmatrix}
\begin{pmatrix} 
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix} 
x \\
(a+b)x+y \\
\end{pmatrix}
=
\begin{pmatrix} 
x \\
ax+bx+y \\
\end{pmatrix}
=
\begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix} 
x \\
bx+y
\end{pmatrix}
=
\begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}
\biggl(
\begin{pmatrix} 
1 & 0 \\
b & 1 \\
\end{pmatrix}
\begin{pmatrix} 
x \\
y \\
\end{pmatrix}
\biggr)
,\space\space
\forall \begin{pmatrix} 
1 & 0 \\
a & 1 \\
\end{pmatrix}, \begin{pmatrix} 
1 & 0 \\
b & 1 \\
\end{pmatrix}\in H, \space\space\forall \begin{pmatrix} 
x \\
y \\
\end{pmatrix}\in \mathbb{R}^2
$
From the general action theory, we know that the "acted upon" set, i.e. here $\mathbb{R}^2$, is partitioned into "orbits", given by (here $X$ is the generic point of $\mathbb{R}^2$):
$$O(X)=\{AX, A\in H\}=\biggl\{\begin{pmatrix} 
x \\
ax+y \\
\end{pmatrix}, \space \space a\in \mathbb{R}\biggr\}\tag 1$$
By $(1)$, the orbits by $x=0$ points "don't feel" the action of $H$ (via the parameter $a$): the points of the $y$-axis ($x=0$) constitue each a "degenerate" orbit:
$$O(X_{x=0})=O(y)=\biggl\{\begin{pmatrix} 
0 \\
y \\
\end{pmatrix}, \space \space a\in \mathbb{R}\biggr\}=\biggl\{\begin{pmatrix} 
0 \\
y \\
\end{pmatrix}\biggr\}$$
Off $y$-axis points ($x\ne 0$), instead, are moved all along the "vertical" direction by the point $(x,y)$. Summarizing, the partition of $\mathbb{R}^2$ induced by this action is made of single-point orbits (one for each point of the $y$-axis) and "vertical", straight lines for the rest of the real plane.
