---
title: Pre- and co-image
categories: lecture
---
# Lecture Notes
## Linear objects in 2D
+ The most important linear object is the line through the origin.
+ These are subspaces of dimension one.
+ The object is a set $\ell\subset\mathbb{R}^2$
+ Three descriptions
+ **functions**
$$\ell = \{ \vec{x}=(x,y) | y = a\cdot x, x\in\mathbb{R} \}$$
for some $a\in\mathbb{R}$
+ Exception: The vertical line would have $a=\infty$,
for infinitely steep
+ **equations**
$$\ell = \{ \vec{x}=(x,y) | \vec{x}\cdot\vec{x}^\bot \}$$
for some $\vec{x}^\bot\in\mathbb{R}^2$
+ Note that for $c\neq0$, $\vec{x}^\bot$ and $c\vec{x}^\bot$
define the same line.
+ **span**
$$\ell = \{ \vec{x}=(x,y) | a\cdot \vec{x}_0, a\in\mathbb{R} \}$$
for some $\vec{x}_0\in\mathbb{R}^2$
+ Exception: The vertical line would have $a=\infty$,
for infinitely steep
If we normalise $\vec{x}^\bot$, we can write $\vec{x}^\bot=(a,1)$
for $a\in\mathbb{R}$ unless we describe the vertical line, which has
$\vec{x}^\bot=(1,0)$, which we could imagine writing $(\infty,1)$.
+ We can normalise $\vec{x}_0$ in the same way.
+ The set of lines through origo is equivalent to $\mathbb{R}\cup\{\infty\}$,
which can be seen in either representation.
## Linear objects in 3D
We have the same situation in 3D, but we have more objects of interest.
+ In 2D, the line is defined by **one** function or **one** equation.
+ In 3D we have
+ the line $\ell= \{(x,y,z) | z = ax + by, (x,y)\in\mathbb{R}\}$
+ the plane $\mathcal{P}= \{(x,y,z) | z = ax, y = bx, x\in\mathbb{R}^2\}$
(two function)
+ Using equations to define it
+ The plane needs **one** equation
$$\mathcal{P}=\{\vec{x} | \vec{x}\cdot\vec{x}^\bot=0 \}$$
+ $\vec{x}^\bot$ is the dual space $\mathcal{P}$
+ The line needs **two** equation
$$\ell=\{\vec{x} | \vec{x}\cdot\vec{y}_1=0, \vec{x}\cdot\vec{y}_1=0\}$$
+ The space spanned by $\vec{y}_1$ and $\vec{y}_2$ is the
dual space $\ell^\bot$
+ What does it look like as **spans**?
+ An object needs
+ one function per dimension; or
+ Each adds one degree of freedom
+ one equation per *codimension*
+ Each equation removes one degree of freedom
## Projections from 3D to 2D
+ Recall that each point $x$ in the image plane is the image of any point on
a line through $O$
+ Correspondence between lines through $O$ and point in the image.
+ This line is called the **pre-image** of $x$.
Draw frontal model with image at $Z=1$. This gives projective
image co-ordinage $(x,y,1)$ embedded in 3D.
+ What about a line $l$ in the image plane? What is the pre-image?
+ Plane $P$ through the origin. The line $l$ is the intersection
of $P$ and the image plane
+ What is the image of a line $L$ in 3D?
+ if $O\in L$ we have a point, whose pre-image is $L$
+ if $P\not\in L$, we have a line $l$ whose pre-image is a plane $P\ni O$
+ $P$ is described by an orthogonal vector, the dual space $P^\bot$,
which we call the **co-image** of $l$
# Notes from the text book
*The following notes were made 2021 based on a textbook.
This exposition is not recommended because it is driven
by the definitions which only gain meaning later in the course.*
## Image and Image Plane
+ Image Plane is the universe where the image lives
$$ \text{image}\subset\text{image plane} $$
+ The Image Plane is a 2D World
+ The Image Plane exists in a 3D World
## Pre-image
+ Preimage is the set of points in 3D projecting onto the Image Plane
+ What is projection?
+ draw a line through the 3D point and origo (the pinhole)
+ the projection is the intersection with the image plane.
+ Thus
+ $\text{preimage} = \mathsf{span}(\text{image})$
+ $\text{image} = \text{preimage}\cap\text{image plane}$
+ The **span** of a set of points is the smallest linear subspace
containing all the points
## Points and Lines
| Image object | Pre-Image |
| :- | :- |
| Point (dimension 0) | Line through origo (dimension 1) |
| Line (dimension 1) | Plane through origo (dimension 2) |
+ Preimage is a linear subspaces, i.e. includes origo
+ A single point projects onto a point
+ any other point on the same line through origo projects onto the same
point
+ A line projects onto a line if it does not pass through origo
## Co-image
+ Coimage is the set of points (space) orthogonal on the preimage
$$\text{coimage} = \text{preimage}^\bot$$
$$\text{preimage} = \text{coimage}^\bot$$
## Points and Lines
| Image object | Pre-Image | Co-Image |
| :- | :- | :- |
| Point (dimension 0) | Line through origo (dimension 1) | Plane (co-dimension 1) |
| Line (dimension 1) | Plane through origo (dimension 2) | Line (co-dimension 2) |
+ Preimage and coimage are linear subspaces
+ origo is in both the pre- and co-image
# Notation
+ Recall $\hat u$ is a skew-symmetric matrix associated with $u$
+ $\mathsf{span}(\hat u) = u^\bot$
+ Associate an image point $x$ with either its pre-image or co-image
# Systems of Equations and Orthogonal Vectors
+ $\ell^Tx=0$ is an equation in three unknowns
+ This defines a plane (two unknowns)
+ e.g. $x_1+ax_2+bx_3$
+ If you have two points, say, $\ell^TL=0$, you have two equations
+ This defines a line (one unknowns)
+ e.g. $x_1+ax_2$
+ If you have two points $x_1$ and $x_2$ on a line
+ $x_1\times x_2$ is orthogonal on both of them
+ and on any other point on the line