A 2 by 2 rotation matrix is a unitary matrix $$R(\theta)=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$$ whose eigenvalues are $\{e^{i\theta},e^{-i\theta}\}$.
Given two rotation matrices $R(\theta)$ and $R(\varphi)$ where $0\leq \theta,\varphi<2\pi$ are angles of rotation about the same axis. I am trying to show that $$\|R(\theta)-R(\varphi)\|\leq |\theta-\varphi|.$$
It's clear that $\|R(\theta)-R(\varphi)\|\leq |e^{i\theta}-e^{i\varphi}|$ since both matrices are diagonal in the same basis.
So I guess I'm wondering why$$|e^{i\theta}-e^{i\varphi}|\leq |\theta-\varphi|\,.$$
Am I missing something obvious here?
Best Answer
Note that $$ |e^{i \theta} - e^{i\varphi}| = |e^{i \varphi}(e^{i (\theta -\varphi)} - 1)| = |e^{i (\theta -\varphi)} - 1|, $$ and apply Euler's formula.
Alternatively, a nice approach is to think geometrically: the distance between two points on the unit circle is less than the arclength along the part of the circumference between them.
via Euler's formula:
$$ |e^{\theta} - 1|^2 = (\cos \theta - 1)^2 + \sin^2 \theta \\ = 2 (1 - \cos(\theta)) = 2 (1 - \cos(2[\theta/2])) \\ = 2(1 - (1 - 2\sin^2 (\theta/2)))\\ = 4 \sin^2(\theta/2) \leq 4(\theta/2)^2 = \theta^2 $$