# For rotation matrices in SO(3), solutions of AB = BA given A (or B)

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 $$SO(3)$$.

• See this thread :) – Paweł Czyż May 16 at 9:51
• thank you, nice to see another proof of the same fact stated in terms of the group. – 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 $$\pi$$.
• Shouldn't $ABa = Ba$ be $BAa = Ba$? – Paweł Czyż May 16 at 9:48
• 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. – user1151695 May 16 at 10:26