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Learning and Proof

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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.

  (\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)
$$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
  [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.