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# Pre- and co-image

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--- title: Pre- and co-image categories: lecture title: Pre- and Co-Image categories: exercises --- # 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 + Exercise 3.9 are from Ma 2004 page 62ff. + Exercise 3.10 are from Ma 2004 page 62ff. ## Pre-image # First Exercise (3.9) + 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 Exercise 3.9 are from Ma 2004 page 62ff. ## Points and Lines ## Debrief Notes | Image object | Pre-Image | | :- | :- | | Point (dimension 0) | Line through origo (dimension 1) | | Line (dimension 1) | Plane through origo (dimension 2) | ### Part 1 + 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 You should first find the pre-image of the image of $L$. ## Co-image + What kind of object is the pre-image? + How did we describe such an object previously? + What is the relationship between this pre-image and a point $x\in L$? + What is the relationship between the pre-image and and the vector $\ell$? + Coimage is the set of points (space) orthogonal on the preimage ### Part 2 $$\text{coimage} = \text{preimage}^\bot$$ + If you read the points $x^1$ and $x^2$ as vectors in 3D, what do they look like? + Can you describe the pre-image in terms of $x^1$ and $x^2$? + maybe as a span? + What then is the relationship between $\ell$ and $x^1,x^2\in L$? $$\text{preimage} = \text{coimage}^\bot$$ How do you find a vector which is orthogonal on two known vectors in 3D? ## Points and Lines ### Part 3 | 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) | + Note that $x$ is an image point. + $\ell^1$ and $\ell^2$ are vectors in 3D, and co-images of two image lines + If you view $x$ as a 3D vector instead of a point, what does it look like? + What would be the relationship between this vector $x$ and $\ell^1$ and $\ell^2$? + How do we find vector $x$ with the right relationship with $\ell^1$ and $\ell^2$? + How do we make sure that the vector $x$ is an image point $x$? + Preimage and coimage are linear subspaces + origo is in both the pre- and co-image # Second Exercise (3.10) # Notation Exercise 3.10 are from Ma 2004 page 62ff. + 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 ## Debrief Notes # Systems of Equations and Orthogonal Vectors 1. Here, it is necessary to look at the pre-images of the two lines. + What does the pre-images look like? + What is the intersection of the pre-images? Could it be empty? + What is the intersection between the image plane and the pre-images? 2. Here, you need to look at the co-images. + What can you say about co-images of parallel lines? + What can you say about the relationship between the co-images and the images? Is there are relationship between one line and the co-image of the other line? + Now return to Part 3 of the previous exercise (3.9). 3. Because the two lines are parallel, they lie in the same plane (not necessarily through the origin). Consider the orientation of this plane. + Suppose first that it intersects the image plane close to the centre (image origin). Where is the vanishing point? + Suppose you turn the plane. Where does the vanishing point go? + At the extremity, the plane is parallel to the image plane. Where is the vanishing point now? + $\ell^Tx=0$ 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 # Debrief