rotation_matrix

introduction

%%visits: 4 Let $A(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ where θ is the transformation in the anti-clockwise direction. ## intuition The inverse of A is A(−θ) since to invert the transformation you simply turn in the clockwise direction. The first component is simply the polar_coordinates form of a point, and the second point is it transposed. Properties of the rotation_matrix. Two reflections make a rotation, this is denoted as b(θ) = A(−θ)BA(θ), see changing_the_basis in the [[matrix]] page. ## rigour # The transpose is the inverse. $A = \begin{pmatrix} \cos(x) && \sin(x) \\ -\sin(x) && \cos(x) \end{pmatrix}$ $A^{T} = \begin{pmatrix} \cos(x) && -\sin(x) \\ \sin(x) && \cos(x) \end{pmatrix}$

$A A^{T} = \begin{pmatrix} \cos(x) && \sin(x) \\ -\sin(x) && \cos(x) \end{pmatrix} \begin{pmatrix} \cos(x) && -\sin(x) \\ \sin(x) && \cos(x) \end{pmatrix}$

$= \begin{pmatrix} \cos^2(x) + \sin^2 && - \cos(x)\sin(x) + \cos(x)\sin(x) \\ -\sin(x) \cos(x) + \cos(x)\sin(x) && \sin^2(x) + \cos^2(x) \end{pmatrix}$

$= \begin{pmatrix} 1 && 0 \\ 0 && 1 \end{pmatrix}$

$A^{T}A = \begin{pmatrix} \cos(x) && -\sin(x) \\ \sin(x) && \cos(x) \end{pmatrix} \begin{pmatrix} \cos(x) && \sin(x) \\ -\sin(x) && \cos(x) \end{pmatrix}$

$\begin{pmatrix} \cos^2(x) + \sin^2(x) && \cos(x) \sin(x) - \sin(x) \cos(x) \\ \sin(x) \cos(x) - \cos(x) \sin(x) && \sin^2(x) \cos^2(x)\end{pmatrix}$

$= \begin{pmatrix} 1 && 0 \\ 0 && 1 \end{pmatrix}$

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[[dihedral_group]] tags :math:linear_algebra_1:linear_algebra_2:groups_and_symmetries:

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