# Euclidian transformations, lengths and angles

## Linear metrics

### Metrics

We defined a norm as:

$$||v||=v^TMv$$

A metric is the distance between two vectors.

$$d(u,v)=||u-v||=(u-v)^TM(u-v)$$

#### Metric space

A set with a metric is a metric space.

### Inducing a topology

Metric spaces can be used to induce a topology.

### Translation symmetry

The distance between two vectors is:

$$(v-w)^TM(v-w)$$

So what operations can we do now?

As before, we can do the transformations which preserve $$u^TMv$$, such as the orthogonal group.

But we can also do other translations

$$(v-w)TM(v-w)$$

$$v^TMv+w^TMw-v^TMw-w^TMv$$

so symmetry is now $$O(3,1)$$ and affine translations

#### Translation matrix

$$[[1,x][0, 1]]$$ moves vector by $$x$$.

## Non-linear norms

### $$L_p$$ norms ($$p$$-norms)

#### $$L^P$$ norm

This generalises the Euclidian norm.

$$||x||_p=(\sum_{i=1}^{n}|x|^p_i)^{1/p}$$

This can defined for different values of $$p$$. Note that the absolute value of each element in the vector is used.

Note also that:

$$||x||_2$$

Is the Euclidian norm.

#### Taxicab norm

This is the $$L^1$$ norm. That is:

$$||x||_1=\sum_{i=1}^{n}|x|_i$$

## To linear forms

### Norms

We can use norms to denote the “length” of a single vector.

$$||v||=\sqrt {\langle v, v\rangle }$$

$$||v||=\sqrt {v^*Mv}$$

#### Euclidian norm

If $$M=I$$ we have the Euclidian norm.

$$||v||=\sqrt{v^*v}$$

If we are using the real field this is:

$$||v||=\sqrt{\sum_{i=1}^{n}v^2_i}$$

#### Pythagoras’ theorem

If $$n=2$$ we have in the real field we have:

$$||v||=\sqrt{v_1^2+v_2^2}$$

We call the two inputs $$x$$ and $$y$$, and the length $$z$$.

$$z=\sqrt {x^2+y^2}$$

$$z^2=x^2+y^2$$

### Angles

#### Recap: Cauchy-Schwarz inequality

This states that:

$$|\langle u,v\rangle |^2 \le \langle u, u\rangle \dot \langle v, v\rangle$$

Or:

$$\langle v,u\rangle\langle u,v\rangle \le \langle u, u\rangle \dot \langle v, v\rangle$$

#### Introduction

$$\langle v,u\rangle\langle u,v\rangle \le \langle u, u\rangle \dot \langle v, v\rangle$$

$$\dfrac{\langle v,u\rangle\langle u,v\rangle}{||u||.||v||} \le ||u||.||v||$$

$$\dfrac{||u||.||v||}{\langle v,u\rangle} \ge \dfrac{\langle u,v\rangle}{||u||.||v||}$$

$$\cos (\theta )=\dfrac{\langle u,v\rangle }{||u||.||v||}$$