Suppose we have $\mathbf{AB} = \mathbf{BA}$, where $\mathbf{A},\mathbf{B} \in SO(3)$.

What facts does this imply about $\mathbf{A}$ and $\mathbf{B}$? Clearly $\mathbf{A} = \mathbf{I}_3$ and $\mathbf{B} = \mathbf{I}_3$ are always solutions.

Given $\mathbf{A} \neq \mathbf{I}_3$, what are the resulting constraints on $\mathbf{B}$? There are certain cases such as where both have rotation angle $\pi$ and perpendicular, coincident, or negative axes of rotation (Orthogonal axes of rotation of angle $\pi$ implies AB=BA), but what are the general constraints?

We may show for example that for any $m \times m \ \textbf{square}$ matrices $\mathbf{X,Y}$, the equation $\mathbf{XY} = \mathbf{YX}$ implies;

$$ (\mathbf{Y}^{\top}\otimes\mathbf{I}_m - \mathbf{I}_m\otimes\mathbf{Y})\mathbf{X}^s = \mathbf{0}$$

where $\otimes$ is the Kronecker product and $\{\cdot\}^s$ is the column stacking operator. This gives us a statement about $\mathbf{X}$ in terms of the nullspace of an $m^2 \times m^2$ matrix with coefficients dependent on $\mathbf{Y}$, but I am so far unable to constrain it further to $SO(3)$.

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    $\begingroup$ See this thread :) $\endgroup$ – Paweł Czyż May 16 at 9:51
  • $\begingroup$ thank you, nice to see another proof of the same fact stated in terms of the group. $\endgroup$ – user1151695 May 16 at 10:27

Let $\mathbf{a} \in \mathbb{R}^3$ be a vector colinear with the axis of rotation of $\mathbf{A}$. Then, $\mathbf{ABa} = \mathbf{Ba}$.

Then $\mathbf{Ba}$ must also be co-linear with the rotation axis of $\mathbf{A}$, $\mathbf{Ba} = k\mathbf{a}, k \in \mathbb{R}$.

Since $\mathbf{B}$ is a rotation matrix and hence scale preserving, $|k| = 1$. This has only two solutions, $k = \pm 1$. Hence $\mathbf{B}$ can correspond to identity, or rotation by $\pi$ about any axis orthogonal to $\mathbf{a}$. By symmetry, $\mathbf{A}$ must also rotate by $\pi$.

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  • $\begingroup$ Shouldn't $ABa = Ba$ be $BAa = Ba$? $\endgroup$ – Paweł Czyż May 16 at 9:48
  • $\begingroup$ Since $\mathbf{a}$ is colinear with rotation axis of $\mathbf{A}$, $\mathbf{Aa} = \mathbf{a}$ by definition. We left multiply both sides of $\mathbf{AB}=\mathbf{BA}$ by $\mathbf{a}$ to arrive at the result. $\endgroup$ – user1151695 May 16 at 10:26
  • $\begingroup$ sorry, right multiply $\endgroup$ – user1151695 May 16 at 10:38
  • $\begingroup$ Right, thank you! $\endgroup$ – Paweł Czyż May 16 at 11:54

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