Why is the stabilizer of a line segment a cylinder?

There is a passage in a paper I'm reading discussing the stabilizer of an edge.

For an edge (passing through the origin), its stabilizer (in $$\operatorname{GL}_2(\mathbb{R})$$) must fix the direction of the edge, i.e. it must have an eigenvector in that direction with an eigenvalue 1. The second eigenvector can be in any other direction, giving rise to a set isomorphic to $$\operatorname{SO}(2)$$, sans the direction of the first eigenvector, which in turn is isomorphic to the unit circle punctured at one point. Note that isomorphism here refers to topological isomorphism between sets. The second eigenvalue can be anything, but considering that the entire circle already accounts for every pair $$(\lambda,−\lambda)$$, the effective set is isomorphic to the positive half of the real-axis only. In summary, this stabilizer subgroup is: $$S_e = \operatorname{SO}(2)\times \mathbb{R}^+ \backslash \mathbb{R}^+$$. This space looks like a cylinder extended infinitely to one direction (Figure 3). More importantly, $$\operatorname{dim}(S_e) =2$$, and it is actually a non-compact set.

I don't understand their explanation here. I need help unraveling what they mean, mathematically.

This is a computer science paper, so naturally, there is no definition an "edge." One would think this means a line segment, but it may mean a line. We know the dimension will turn out to be $$2$$, so, let's find out.

To stabilize the line $$y=mx+b$$, we let $$(x^\prime,y^\prime)^\intercal={\bf A}(x,mx)^\intercal$$ and set $$y^\prime=mx^\prime$$. I get $$\begin{eqnarray*} a_{21} x + a_{22} m x&=& m (a_{11} x + a_{12} m x)\\ a_{21} x + a_{22} m x-(m (a_{11} x + a_{12} m x))&=&0\\ (a_{21} + a_{22} m -m a_{11} - a_{12} m^2)x&=&0\\ \end{eqnarray*}$$ implying that $${\bf A}$$ should have the form $$\left(\begin{array}{cc} r & s \\ -m(t-r-sm) & t\end{array}\right)$$, i.e. $$\operatorname{Stab}_{\operatorname{GL}_n(\mathbb{R})}\left(\{(x,y)\in\mathbb{R}^2:y=mx\}\right)=\operatorname{Span}\left\{\left(\begin{array}{cc} 1 & 0 \\ m & 0\end{array}\right),\left(\begin{array}{cc} 0 & 1 \\ m^2 & 0\end{array}\right),\left(\begin{array}{cc} 0 & 0 \\ -m & 1\end{array}\right)\right\}$$

So, ok, fine, that's the setwise stabilizer of a line, and it's 3-dimensional.

To stabilize a line segment, we don't just want the condition $$y^\prime=mx^\prime$$, we actually want $$(x,mx)^\intercal= {\bf A}(x,mx)^\intercal$$. So, using the earlier form of $${\bf A}$$, we get $$\left(\begin{array}{cc} r & s \\ -m(t-r-sm) & t\end{array}\right)\left(\begin{array}{c}x\\mx\end{array}\right)=\left(\begin{array}{c}rx+smx\\-m(t-r-sm)x+tmx\end{array}\right)=(r+sm)\left(\begin{array}{c}x\\mx\end{array}\right)$$ Since we would like for $$(x,mx)^\intercal= {\bf A}(x,mx)^\intercal=(r+sm)(x,mx)^\intercal$$, that implies that $$r+sm=1$$. Thus, continuing the never ending horror, $$\left(\begin{array}{cc} 1-sm & s \\ -m(t-(1-sm)-sm) & t\end{array}\right)=\left(\begin{array}{cc} 1-sm & s \\ -m(t-1) & t\end{array}\right)$$ That gives us matrices of the form $${\bf A} = \left(\begin{array}{cc} 1 & 0 \\ m & 0\end{array}\right)+s\left(\begin{array}{cc} -m & 1 \\ 0 & 0\end{array}\right)+t\left(\begin{array}{cc} 0 & 0 \\ -m & 1\end{array}\right)$$ and so now we will have to describe the stabilizer as an affine space. This is ok-- it is $$2$$-dimensional, as desired, so this must be what the author was talking about.

Except I don't see what the connection is here to the cylinder-shaped $$S_e$$ in the passage. How is that related? I understand you can express $$2$$-dimensional spaces in cylindrical coordinates, but what does that have to do with anything? It would be one thing if the line we wanted to be invariant were in $$\mathbb{R}^3$$ and we were rotating around it, but this is supposed to be describing an action on the plane.

How is it natural to view the stabilizer of a line segment as a cylinder?

For reference, this is what happens when I let $${\bf A}(s,t)$$ act on the plane. Red shows the (invariant) line segment from $$(-1,-1)$$ to $$(1,1)$$, then in blue we have the image of the unit circle along with the $$x$$- and $$y$$- axes from $$-1$$ to $$1$$. (I'm using $$m=1$$ here.)