--- title: Learning and Proof categories: lecture --- # Reading + Steinert, M., & Leifer, L. J. (2012). 'Finding One's Way': Re-Discovering a Hunter-Gatherer Model based on Wayfaring. International Journal of Engineering Education, 28(2), 251. + Ma 2004:Ch 5.1 # Learning and Study Technique ## Constructivist Learning Theory ## Learning is Design ## The Hunter-Gatherer Model # Relative Pose ## Repetition ### 3D Motion + Rotation by angle $\theta$ around the vector $\omega$ is given by $R=e^{\hat\omega\theta}$ assuming $\omega$ has unit length. **Rodrigues' formula** (2.16) $$e^{\hat\omega} = I + \frac{\hat\omega}{||\omega||}\sin(||\omega||) + \frac{\hat\omega^2}{||\omega||^2}(1-\cos(||\omega||))$$ See [Angular Motion]() for a more comprehensive summary. ### Basic Result on Relative Pose + Theorem 5.5 $$E = U\mathsf{diag}\{\sigma,\sigma,0\}V^T,$$ where $U,V\in\mathsf{SO}(3)$ + Tricky proof. Do not spend too much time on this. $$\begin{cases} (\hat T_1,R_1) &= (UR_Z(+\frac\pi2)\Sigma U^T, UR_Z(+\frac\pi2)V^T) \\ (\hat T_2,R_2) &= (UR_Z(-\frac\pi2)\Sigma U^T, UR_Z(-\frac\pi2)V^T) \end{cases}$$ where $$R_Z(+\frac\pi2) = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is a rotation by $\pi/2$ radians around the $z$-axis. + Note that there are two solutions from one $U\Sigma V^T$ decomposition. + Are there more solutions? > There exist exactly two relative poses $(R,T)$ with $R\in\mathsf{SO}(3)$ and $T\in\mathbb{R}^3$ corresponding to a nonzero essential matrix $E\in\mathcal{E}$ [Theorem 5.7] ## Proofs and Understanding ### Theorem 5.7 + **Demo** read the proof (debrief?) ### Lemma 5.6 > If $\hat T$ and $\hat TR$ are both skew-symmetric for $R\in\mathrm{SO}(3)$, then $R$ is a rotation by angle $\pi$ around $T$. + **Demo** read the proof (debrief?) + Skew-symmetry gives $(\hat TR)^T=-\hat TR$ + We also have $(\hat TR)^T=R^T\hat T^T=-R^T\hat T$ + Hence $\hat TR = R^T\hat T$, + and since $R^T=R^{-1}$, we have $$R\hat TR=\hat T$$ + Write $R=e^{\hat\omega\theta}$ for some $\omega$ of unit length and some $\theta$, to get $$e^{\hat\omega\theta}\hat Te^{\hat\omega\theta}=\hat T$$ + multiply by $\omega$ $$e^{\hat\omega\theta}\hat Te^{\hat\omega\theta}\omega=\hat T\omega$$ This represents a stationary rotation of the vector $\hat T\omega$. Note that $\omega$ is stationary under rotation by $R$, and hence it is an eigenvector associated with eigenvalue 1. Furthermore, it is the only such eigenvector, and $\hat T\omega$ cannot be such. Hence $\hat T\omega=T\times\omega=0$. This is only possible if $T\sim\omega$, and since $\omega$ has unit length, we get $$\omega = \pm\frac{T}{||T||}$$ We now know that $R$ has to be a rotation around $T$, and therefore $R$ and $T$ commute. This can be checked in Rodrigues' formula (Theorem 2.9). Hence $R^2\hat T = \hat T$. This looks like two half-round rotations to get back to start. If $\hat T$ had been a vector or a matrix of full rank, we would have been done. However, with the skew-symmetric $\hat T$ there is a little more fiddling.