Spring2024

\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

The definition of $SO(2)$ is that

We are going to define a group of rotations by stating \(R2= \left\{ \left. \begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ \end{pmatrix} \right| \theta\in (-\pi,\pi] \right\}\)

Now we are going to show that $SO(2)=R2$. It is straight forward to show that if $x\in R2$ then $x\in SO(2)$. Therefore, here we focus on showing that if $x\in SO(2)$ then $x\in R2$. Let $x\in SO(2)$,

\[x= \begin{pmatrix} R_{11} & R_{12}\\ R_{21} & R_{22}\\ \end{pmatrix}.\]

We will then show that there exists a $\theta \in (-\pi, \pi]$ such that

\[\begin{pmatrix} R_{11} & R_{12}\\ R_{21} & R_{22}\\ \end{pmatrix}=\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ \end{pmatrix},\]

which is to say that we will find a $\theta \in (-\pi, \pi]$ such that \(\begin{align} R_{11}&=\cos(\theta)\label{eq:r1}\\ R_{21}&=\sin(\theta)\label{eq:r2}\\ R_{12}&=-\sin(\theta)\label{eq:r3}\\ R_{22}&=\cos(\theta)\label{eq:r4} \end{align}\)

It can be shown that there exists a $\theta$ (in fact a unique one) such that $\eqref{eq:r1}$ and $\eqref{eq:r2}$ are satisfied.

It follows from the conditions that $\mathsf{R}^{\mathsf{T}}\mathsf{R}=\mathsf{I}$ and $\text{Det}(\mathsf{R})=+1$ that

\[\begin{align} R_{22}&=R_{11}\label{eq:1}\\ R_{12}&=-R_{21}\label{eq:2} \end{align}\]

It now follows from $\eqref{eq:1}$ and $\eqref{eq:2}$ and the fact that $\eqref{eq:r1}$ and $\eqref{eq:r2}$ are satisfied that $\eqref{eq:r3}$ and $\eqref{eq:r4}$ are also satisfied.