---
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
## Learning Outcomes.
1. The Learning Outcome is **not** a set of well-defined procedures
to be copied.
2. Textbook procedures work on textbook problems.
3. We need the theoretical understanding to adapt existing theory to
new situations.
Hence, *aim for understanding over working code*.
## Constructivist Learning Theory
1. Learning is **constructed by the learner**.
2. Knowledge has to be reinterpreted in your own personal context
- based on experience
- ... and on ambition
- ... and on background knowledge
Hence, seek to extend concepts and techniques with which
*you are familiar*.
## Learning is Design
1. Learning is no different from design and development.
2. There is always some problem or situation that we need to make sense
of.
3. In design, there is an ill-defined problem to comprehend.
4. In learning, there is a body of knowledge to make sense of.
## The Hunter-Gatherer Model
1. There are many good models to explain how we design, and how we
learn.
2. E.g. Simon, Gadamer, SchÃ¶n
3. The Hunter-Gatherer Model of Steinert and Leifer.
4. **Hunt** for the next big idea.
+ Look for clues, interpret the clues, guess, test your hunches,
discover new clues, and iterate.
6. Then **gather** the ideas
+ skin and partition the beast
+ assemble the ideas into a prototype/proof/presentation
to bring home.
7. Big ideas in learning are known as threshold concepts,
+ tricky to learn
+ but once learnt they explain and give meaning to a large body
of theory and knowledge
+ approach difficult concepts from different angles
+ test (intermediate) ideas, through implementation, argument,
prediction, or otherwise
# 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
+ $E = \hat TR$ where $(T,R)$ is the transformation between camera frames.
+ $E = U\mathsf{diag}\{\sigma,\sigma,0\}V^T,$
where $U,V\in\mathsf{SO}(3)$
+ $(\hat T,R) = (UR_Z(\pm\frac\pi2)\Sigma U^T, UR_Z(\pm\frac\pi2)V^T)$
where
$R_Z(\theta)$ is a rotation around the $z$-axis by an angle $\theta$
> 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
+ Proof of Theorem 5.7 depends on Lemma 5.6.
### 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.