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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.

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    Look at the matrix form below. Perhaps it will answer some of your questions about why $Iso(\mathbb{R})$ is put together the way it is. I'd recommend testing out a few points, especially in the case where $(x,y)=0$ or the orthogonal submatrix in the upper left corner is the identity.2012-04-27

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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 $(x_0,y_0, 1)$.

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    Indeed, multiplying the above matrix with $(x_0,y_0,1)$ really makes me see this, thanks!2012-04-27
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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 $I\in GL(n+1,\mathbb R)$.

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    @DylanMoreland Isometries are in $O_n$ of course. However, since we just need an embedding and $O_n\subset GL_n$, we might as well use an embedding of $GL_n$.2012-04-28