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$\textbf{Lemma:}$ An isometry $f$ that has the form $m=t_a \rho_{\theta} $, with $\theta \neq 0$, is a rotation through the angle $\theta$ about a point in the plane.

$\forall x \in M_{2,1}(\mathbb{R}) , t_a(x)=x+a$ (translation)

$\forall x= \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in M_{2,1}(\mathbb{R}), \rho_{\theta}(x)=\begin{pmatrix} \cos\theta & -\sin\theta \\ \cos\theta & \sin\theta \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} $ (rotation)

What does that mean ? Could someone rephrase it differently ?

amir
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1 Answers1

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Can you check your source again - maybe it should be $m = t_a \rho_\theta t_{-a}$?

Anyway, the map $\rho_\theta$ will rotate every point around the origin; the origin will not move, every other point will rotate around it.

What if you wanted to rotate around a different point though, so that point does not move, and all other points rotate relative to it? Say the point you want to rotate around is $a$.

Writing out the form of this map in one step would be difficult, but we can more easily write it as a composition of simpler maps.

In particular, if we first translate the whole plane so that $a$ is at the origin ($t_{-a}$), then rotate around the origin ($\rho_\theta$), and then move the plane back to its original position ($t_a$), we will have accomplished the desired movement.

I suggest you graph the behavior of this operator (with, for example, $\theta = \pi/4$, $a = (1,0)^T$) to get a feel for it.

BaronVT
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