# Learning and Proof

• 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

# 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.