1

It can be shown that every matrix in $ \mathrm{O}_{2}(\mathbb{R}) $ is of the form

$$ \begin{pmatrix} \cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \varphi \end{pmatrix} $$

or

$$ \begin{pmatrix} \cos \varphi & \sin \varphi \\ \sin \varphi & -\cos \varphi \end{pmatrix} $$

for some angle $ \varphi $ where the former is a counter-clockwise rotation by $ \varphi $ and the latter is a reflection about the line which passes through the origin and $ (\cos \varphi/2, \sin \varphi/2) $.

My question is: can a similar "general" form can be found for matrices in $ \mathrm{O}_{3}(\mathbb{R}) $?

So far, I attempted this by showing that the first column corresponds to a point on the unit sphere in $ \mathbb{R}^3 $. Therefore, it must be of the form $ (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta) $ for some choice of angles $ \theta $ and $ \varphi $. However, I am unable to determine the other two columns from this.

shoteyes
  • 1,425
  • 1
    The nicest form I know gives three matrices, each dependent on a choice of angle (rotations about the three axes) that can be shown to generate $SO(3)$. I'm sure someone here knows more. – Randall Aug 16 '17 at 16:45
  • That could be useful. After all, if $ H = \mathrm{SO}{3}(\mathbb{R}) $ and $ K = \Bigl{ \Bigl( \begin{smallmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{smallmatrix} \Bigr), \Bigl( \begin{smallmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{smallmatrix} \Bigr) \Bigr} $, is it not true that $ \mathrm{O}{3}(\mathbb{R}) = HK $? – shoteyes Aug 16 '17 at 17:25
  • Yes, you can use the matrix exponential for rotations and then multiply by any reflection for reflections. – Qiaochu Yuan Aug 16 '17 at 17:53

1 Answers1

1

I'm a little rusty, but here goes. Given a rotation angle $\theta$, we have rotations by $\theta$ radians $$ R_z(\theta) = \begin{pmatrix} \cos{\theta} & -\sin{\theta} & 0 \\ \sin{\theta} & \cos{\theta} & 0 \\ 0 & 0 & 1 \\\end{pmatrix} $$

$$ R_y(\theta) = \begin{pmatrix} \cos{\theta} & 0 & -\sin{\theta} \\ 0 & 1 & 0 \\ \sin{\theta} & 0 & \cos{\theta} \\\end{pmatrix} $$

$$ R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos{\theta} & -\sin{\theta} \\ 0 & \sin{\theta} & \cos{\theta} \\\end{pmatrix} $$ in the $xy$-, $xz$-, and $yz$-planes respectively (subscript denotes the axis of rotation). You can always use these to generate any rotation in $SO(3)$ by moving the axis first (then moving it back).

Let $A$ be a non-identity element of $SO(3)$. Then $A$ has a 1-diml eigenspace (the axis of rotation), about which we rotate by an angle $\theta$. This eigenspace intersects the unit sphere somewhere, so write this point of intersection as $$ \mathbf{n} = \langle \cos{\alpha}\sin{\beta},\sin{\alpha}\sin{\beta},\cos{\beta}\rangle^T $$ with spherical coordinates (and as a column vector).

Note that $R_y(\beta)R_z(\alpha)^{T}\mathbf{n} = \langle 0,0,1\rangle^T$ so we have moved the axis so that this rotation occurs in the $xy$-plane. We now rotate by our original angle $\theta$ and then rotate our axis back to its original position. You can do the math and check that $$ A = R_z(\alpha)R_y(\beta)^TR_z(\theta)R_y(\beta)R_z(\alpha)^{T} $$ accomplishes this.

Randall
  • 18,542
  • 2
  • 24
  • 47