# 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

# 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

## 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.
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.
5. Then gather the ideas
• skin and partition the beast
• assemble the ideas into a prototype/proof/presentation to bring home.
6. 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

• 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.