I'm trying to understand how the two dimensional rotation matrix (i.e. $R \in \mathbb{R}^2$) can be derived from the Euler Formula ($e^{i\theta} = \cos \theta + i \sin \theta$). $R$ is given as:
$$ R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} $$
$$ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$ $$ x' = x \cos \theta - y \sin \theta $$ $$ y' = x \sin \theta + y \cos \theta $$
My questions are:
- Why can be $i$ omitted from the rotation matrix? (I tried to look for explanations 1,2 but none of these explanations goes beyond that $i$ is omitted)
- Why can we derive a rotation matrix for $\mathbb{R}^2$ from a form that is defined in $\mathbb{C}^2$? How comes we don't get complex numbers as a result after some rotations?
The complex number $a+bi$ can be represented by the matrix $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.
Note that $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \pm\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} a\pm c & -(b\pm d) \\ b\pm d & a\pm c \end{pmatrix}$$
Also we have $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \end{pmatrix}$$