1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Randall Aug 16 '17 at 16:45
  • $\begingroup$ 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 $? $\endgroup$ – Peace - A Theme Aug 16 '17 at 17:25
  • $\begingroup$ Yes, you can use the matrix exponential for rotations and then multiply by any reflection for reflections. $\endgroup$ – Qiaochu Yuan Aug 16 '17 at 17:53
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.