Planar rotation ($R_{\theta }$)

A planar rotation about the origin is the linear map represented by the rotation matrix $R_{\theta }=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. It preserves inner products, lengths, oriented angles and area, with positive angles measured anticlockwise. Its inverse is $R_{-\theta }$ and its determinant is one.

For column coordinate vectors in the fixed Cartesian basis, an active planar rotation sends $(x,y{)}^T$ to $(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta {)}^T$. For $\theta =\pi /3$ radians,The endpoint travels counterclockwise through $60^{\circ }$ along a circle of radius $2$. Its Euclidean norm stays $\sqrt{1^2+(\sqrt{3}{)}^2}=2$. The second Cartesian basis direction becomes $R_{\pi /3}(0,1{)}^T=(-\sqrt{3}/2,1/2{)}^T$, whose dot product with $(1,\sqrt{3}{)}^T$ is still zero.

A quick sign check is to look at the first column: $R_{\theta }(1,0{)}^T=(\cos \theta ,\sin \theta {)}^T$ must point above the horizontal axis for $0<\theta <\pi$. The columns have unit Euclidean norm and zero dot product, so $R_{\theta }^T{R}_{\theta }=I$; the determinant is ${\cos }^2\theta +{\sin }^2\theta =1$. Finally, the transpose negates the off-diagonal entries, giving $R_{\theta }^T=R_{-\theta }$: turning back reverses the action. This is the same matrix used for passive rotation of coordinates, though that interpretation keeps the vector fixed.

A finite group of planar rotations about one point is a cyclic group. For a nontrivial group, choose its smallest positive rotation angle $\alpha$. Subtracting multiples of $\alpha$ from any other group angle leaves a group angle in $[0,\alpha )$, hence zero. Applying the same argument to $2\pi$ shows that $\alpha =2\pi /m$ for a positive integer $m$. The identity-only group is cyclic as well.