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?
  • $\begingroup$ As for your first question: A complex number can be regarded as a vector with the real part/component along the "$x$-axis" and imaginary part along the "$y$-axis", so the $i$ is just a way of distinguishing between the two dimensions of the complex number/vector; you can instead distinguish these by using the unit vectors of $x$ and $y$. $\endgroup$ – Bobson Dugnutt Jun 16 '17 at 10:26
  • $\begingroup$ @Lovsovs Thank you for your comment. Perhaps I am just confused here and $e^\theta = sin(\theta) + cos(\theta)$ also applies without i included? I'm heaving hard time to get my head around why Real numbers can be rotated from a formula defined in the the domain of Complex numbers. $\endgroup$ – 01000001 Jun 16 '17 at 10:39
  • $\begingroup$ Use & to separate matrix elements that are on the same row so that they don’t run together. $\endgroup$ – amd Jun 17 '17 at 0:19

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

$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \end{pmatrix}$$

  • $\begingroup$ Thank you for your answer. I seem to understand why this matrix representation work, I just can't get my head around why $i$ can be emitted, or more precisely why can we apply a rotation matrix - derived from a formula with complex numbers - on real numbers. $\endgroup$ – 01000001 Jun 16 '17 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.