Projective Transformations: “If all the points lie on a plane, then the linear mapping reduces to …”

Page 7 of my computer vision textbook, Multiple View Geometry in Computer Vision, says the following:

In applying projective geometry to the imaging process, it is customary to model the world as a $$3$$D projective space, equal to $$\mathbb{R}^3$$ along with points at infinity. Similarly the model for the image is the $$2$$D projective plane $$\mathbb{P}^2$$. Central projection is simply a map from $$\mathbb{P}^3$$ to $$\mathbb{P}^2$$. If we consider points in $$\mathbb{P}^3$$ written in terms of homogeneous coordinates $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}, \mathrm{T})^T$$ and let the centre of projection be the origin $$(0, 0, 0, 1)^T$$, then we see that the set of all points $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}, \mathrm{T})^T$$ for fixed $$\mathrm{X}$$, $$\mathrm{Y}$$, and $$\mathrm{Z}$$, but varying $$\mathrm{T}$$ form a single ray passing through the point centre of projection, and hence all mapping to the same point. Thus, the final coordinates of $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}, \mathrm{T})$$ is irrelevant to where the point is imaged. In fact, the image point is the point in $$\mathbb{P}^2$$ with homogeneous coordinates $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z})^T$$. Thus, the mapping may be represented by a mapping of $$3$$D homogeneous coordinates, represented by a $$3 \times 4$$ matrix $$\mathrm{P}$$ with the block structure $$P = [I_{3 \times 3} | \mathbf{0}_3]$$, where $$I_{3 \times 3}$$ is the $$3 \times 3$$ identity matrix and $$\mathbf{0}_3$$ a zero 3-vector. Making allowance for a different centre of projection, and a different projective coordinate frame in the image, it turns out that the most general imaging projection is represented by an arbitrary $$3 \times 4$$ matrix of rank $$3$$, acting on the homogeneous coordinates of the point in $$\mathbb{P}^3$$ mapping it to the imaged point in $$\mathbb{P}^2$$. This matrix $$\mathrm{P}$$ is known as the camera matrix.

In summary, the action of a projective camera on a point in space may be expressed in terms of a linear mapping of homogeneous coordinates as

$$\begin{bmatrix} x \\ y \\ w \end{bmatrix} = \mathrm{P}_{3 \times 4} \begin{bmatrix} \mathrm{X} \\ \mathrm{Y} \\ \mathrm{Z} \\ \mathrm{T} \\ \end{bmatrix}$$

Furthermore, if all the points lie on a plane (we may choose this as the plane $$\mathrm{Z} = 0$$) then the linear mapping reduces to

$$\begin{bmatrix} x \\ y \\ w \end{bmatrix} = \mathrm{H}_{3 \times 3} \begin{bmatrix} \mathrm{X} \\ \mathrm{Y} \\ \mathrm{T} \\ \end{bmatrix}$$

which is a projective transformation.

The aforementioned section of the textbook is available freely here.

My questions are as follows:

1. Where it says

Thus, the final coordinates of $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}, \mathrm{T})$$ is irrelevant to where the point is imaged.

shouldn't the vector be $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}, \mathrm{T})^T$$?

1. What is $$\mathrm{H}_{3 \times 3}$$ supposed to be?

I would greatly appreciate it if people could please take the time to clarify these.

1) It really doesn't make a difference whether you think of it as a row or column vector. I suppose if you're being consistent, yes, it's still the column vector $$(X,Y,Z,T)^\top$$, but clearly the final component of $$(X,Y,Z,T)^\top$$ is the same as the final coordinate of $$(X,Y,Z,T)$$.
2) "the most general imaging projection is represented by an arbitrary 3×4 matrix of rank 3, acting on the homogeneous coordinates of the point in $$\mathbb{P}^3$$ mapping it to the imaged point in $$\mathbb{P}^2$$. This matrix $$P$$ is known as the camera matrix."
So if $$P$$ is arbitrary of rank 3, and we are imposing the linear condition $$Z = 0$$, $$P$$ will reduce to a matrix $$H$$ that is simply $$P$$ with the 3rd column removed. This may have rank 2 or 3. Try it out for yourself, constructing examples of matrices $$P$$.
• I don’t understand why the linear condition $\mathrm{Z} = 0$ will remove the 3rd column? Please elaborate. – The Pointer Nov 27 '18 at 19:55
• @amd But why does the transformation become $$\begin{bmatrix} x \\ y \\ w \end{bmatrix} = \mathrm{H}_{3 \times 3} \begin{bmatrix} \mathrm{X} \\ \mathrm{Y} \\ \mathrm{T} \\ \end{bmatrix}$$ instead of $$\begin{bmatrix} x \\ y \\ w \end{bmatrix} = \mathrm{P}_{3 \times 4} \begin{bmatrix} \mathrm{X} \\ \mathrm{Y} \\ 0 \\ \mathrm{T} \\ \end{bmatrix}$$? – The Pointer Nov 28 '18 at 2:26
• That's what I mean when I say I don't see why having $\mathrm{Z} = 0$ necessitates getting rid of the entire 3rd column. – The Pointer Nov 28 '18 at 2:27
• @ThePointer $H$ is a map (homography) between two planes. Homogeneous coordinate vectors on the planes only have three components. You could just as well have chosen any plane in the ambient projective space and a coordinate system for that plane, as is explained elsewhere in the book. – amd Nov 28 '18 at 2:53