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I wonder why $\max\Vert R_1-R_2 \Vert =2\sqrt2$, where $R_1,R_2 \in SO(3)$, $<A, B>=\operatorname{tr}(A^{\operatorname{T}}B)$.

Rui Tachibana
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    A rotation matrix $R$ is a rotation around some axis passing through the origin in some angle $\theta$. Its eigenvalues are $1, e^{i\theta}, e^{-i\theta}$. So $\operatorname{tr}(R)=1+e^{i\theta}+e^{-i\theta}=1+2\cos(\theta)$ which is between $-1$ and $3$. – Chad K Dec 26 '23 at 07:16

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Hint: \begin{align*} \|R_1-R_2\|^2&=\operatorname{tr}((R_1-R_2)(R_1-R_2)^T)\\ &=\operatorname{tr}(2I-R_1R_2^T-R_2R_1^T)\\ &=2(3-\operatorname{tr}(R_1R_2^T)). \end{align*} Think about the form of the rotation matrices (in terms of the angle of rotation Rodrigues formula) and see if you can take if from here.

Anurag A
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