---
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.