A 1-1 homomorphism from $\operatorname{Iso}(\mathbb{R}^2)$ to $GL(3,\mathbb{R})$ In class we saw A 1-1 homomorphism from $\operatorname{Iso}(\mathbb{R})$ to $GL(2,\mathbb{R})$
$$\operatorname{Iso}(\mathbb{R})\cong \left\{ \begin{pmatrix}\pm1 & x\\
0 & 1
\end{pmatrix}|x\in\mathbb{R}\right\}. $$
How can I get this result ? (It works, but it's probably not a guess, whats the idea ?)
Seeing this I think that there is a 1-1 homomorphism from $\operatorname{Iso}(\mathbb{R}^2)$ to $GL(3,\mathbb{R})$ , how can we find it ? (that's why I'm trying to gain a better understanding of $\operatorname{Iso}(\mathbb{R})$)
I could use some help with this.
 A: How about:
$$\operatorname{Iso}(\mathbb{R}^2)\cong \left\{ \begin{pmatrix}a & b & x\\ c & d & y\\ 0 & 0 & 1\end{pmatrix}\right\}. $$
where $$ \begin{pmatrix} a & b\\
c & d
\end{pmatrix}\in O_2(\mathbb{R})$$
and $(x,y)$ is a vector in $\mathbb{R}^2$. Represent the point $(x_0, y_0)$ by the vector $(x_0,y_0, 1)$.
A: The Mazur-Ulam theorem gives us that any isometry of $\mathbb R^n$ is of the form $f(x)=Ax+b$ where $A\in GL(n,\mathbb R)$ and $b\in \mathbb R^n$. Intuitively, this is just saying that every such isometry is a linear map composed with a translation. We can embed $GL(n,\mathbb R)$ in $GL(n+1,\mathbb R)$ by simply adding a $1$ in the bottom right corner and zeroes elsewhere on the last row and column. We can embed $\mathbb R^n$ in $GL(n+1,\mathbb R)$ by the map
$$\begin{pmatrix} x_1\\\vdots\\ x_n\end{pmatrix}\mapsto \begin{pmatrix} 1 & 0 & \cdots & x_1\\
& \ddots & \ddots & \vdots\\
0 & \cdots & 1 & x_n\\
0 &\cdots & 0 &1\\
\end{pmatrix}$$
which can be fairly easily verified by matrix multiplication. Putting these two together by multiplying the two matrices, we get an embedding $\mathrm{Iso}(\mathbb R^n)\to GL(n+1,\mathbb R)$. Note that combining two embeddings like this does not always work, but in this case it does because the images of the two embeddings intersect only at the identity $I\in GL(n+1,\mathbb R)$.
