Would I be correct in saying that the special orthogonal group SO(2) contains one matrix , namely;

$A=\begin{pmatrix}{} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}$

or would it have infinitely many matrices as $\theta $ can be any angle between 0 and 360, and different angles would produce different entries ?

  • $\begingroup$ The matrices are all of that form, but (assuming that the underlying field is $\mathbb R$ or something similar) there is an infinite number of them. $\endgroup$ – amd Mar 29 '18 at 1:12
  • 2
    $\begingroup$ Please: infinitely many matrices, not infinite matrices. $\endgroup$ – Robert Israel Mar 29 '18 at 1:21
  • $\begingroup$ @RobertIsrael Good point, I'll edit it now :) $\endgroup$ – excalibirr Mar 29 '18 at 1:22
  • $\begingroup$ As many matrices as there are options for $\theta$ $\endgroup$ – Clclstdnt Mar 29 '18 at 1:30
  • $\begingroup$ there are exactly as many matrices as there are numbers in $[0,2\pi)$, which is to say uncountably many $\endgroup$ – qbert Mar 29 '18 at 1:36

Infinitely many matrices: $$ {\rm SO}(2,\Bbb R) = \left\{A=\begin{pmatrix}{} \cos\theta& -\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}\mid \theta \in \Bbb R \right\}. $$For example, for $\theta = 0$ and $\theta = \pi/2$ we get $$\begin{pmatrix}{} 1 & 0\\ 0 & 1 \end{pmatrix}, \begin{pmatrix}{} 0 & -1\\ 1 & 0 \end{pmatrix} \in {\rm SO}(2,\Bbb R).$$


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