# If All the Points Lie On a Plane, Then Why Does the Linear Mapping Reduce to …?

I previously asked a question with regards to what the matrix $$\mathrm{H}_{3 \times 3}$$ is/represents in the following textbook excerpt:

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.

It is now clear to me that I didn't understand this section properly. Specifically, it is not clear to me why choosing the plane $$\mathrm{Z} = 0$$ means that the linear mapping

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

reduces to

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

More specifically, I don't understand why setting $$\mathrm{Z} = 0$$ necessitates getting rid of the entirety of the 3rd column of $$\mathrm{P}_{3 \times 4}$$, which gets us $$\mathrm{H}_{3 \times 3}$$ (as per bounceback's answer in the aforementioned question)?

And I'm also wondering whether choosing the plane $$\mathrm{Z} = 5$$ (or any other plane) instead of $$\mathrm{Z} = 0$$ would still reduce the transformation from

$$\begin{bmatrix} x \\ y \\ w \end{bmatrix} = \mathrm{P}_{3 \times 4} \begin{bmatrix} \mathrm{X} \\ \mathrm{Y} \\ 0 \\ \mathrm{T} \\ \end{bmatrix}$$

to

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

where

$$\mathrm{H}_{3 \times 3} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$$

?

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

Lets write $$P_{3\times4}=\begin{bmatrix}r_{11}&r_{12}&r_{13}&\tau_1\\ r_{21}&r_{22}&r_{23}&\tau_2\\r_{31}&r_{32}&r_{33}&\tau_3\\\end{bmatrix}$$ and call the $$r$$ terms $$R_{3\times3}$$ and the $$\tau$$ terms $$\tau_{3\times1}$$ thus $$P_{3\times 4}=[R,\tau]$$ (I dropped the dimension indices for brevity, but $$\tau$$ is a column vector). Now $$P_{3\times 4} U_4$$ becomes $$R \begin{bmatrix} X\\Y\\Z \end{bmatrix} +T\tau=R U_3 +T\tau$$. where I defined $$U_3 =\begin{bmatrix} X\\Y\\Z \end{bmatrix}$$.

Before dealing with the $$Z=0$$ case, lets look at the case where the plane does not pass through the origin: all points $$U_3$$ are on a plane, hence they obey $$N^T U_3 =d$$ where $$N$$ is a unit vector normal to the plane and $$-d\neq0$$ is the distance of the origin from the plane. For this case we just have $$1= {1\over d} N^T U_3$$ and therefor $$R U_3 +T\tau= R U_3 +(T\tau){1\over d}(N^T U_3)$$ so that finally $$P_{3\times 4} U_4$$ can be written as $$[R +{T\over d}(\tau N^T)]U_3$$ and $$H_{3\times3}=[R+ {T\over d}(\tau N^T)]$$ is called a "Homography".

The degenerate case is where $$d=0$$ and which also contains the case $$Z=0$$ by setting $$N^T=[0,0,1]$$. In this case instead of pushing $$1= {1\over d} N^T U_3$$ we push $$0= N^T U_3$$ . So $$R U_3 +T\tau = (R (I-N N^T) +\tau N^T )(U_3+T N)$$ , In the specific case $$N^T=[0,0,1]$$ we have $$\left[R \begin{bmatrix} 1&0&0\\0&1&0\\0&0&0\end{bmatrix} +\begin{bmatrix} 0&0&\tau_x\\0&0&\tau_y\\0&0&\tau_z\end{bmatrix}\right]\begin{bmatrix} X\\Y\\T\end{bmatrix}$$

• Thanks for the answer. Can you please explain what happened in the first line? I have many questions. (1) What does it mean to "split" $U_4$ into $U_3=\begin{bmatrix} X \\ Y \\ Z\end{bmatrix}$ and $T$? (2) What does the notation $[R,\tau]$ mean? (3) What are $R$ and $\tau$ supposed to be? (4) And so how does $P_{3\times 4} U_4$ become $R U_3 +T\tau$? My apologies for all the questions. – The Pointer Dec 5 '18 at 15:39
• I just wrote $P_{3\times3}=\begin{bmatrix}r_{11}&r_{12}&r_{13}&\tau_1\\ r_{21}&r_{22}&r_{23}&\tau_2\\r_{31}&r_{32}&r_{33}&\tau_3\\\end{bmatrix}$ and called the $r$ terms $R$ and the $\tau$ terms as $\tau$. Then I wrote the product $P_{3\times4} \begin{bmatrix} X\\Y\\Z\\T \end{bmatrix}$ in terms of this. – user617446 Dec 5 '18 at 15:48
• Thanks for the clarification. After reading your answer further, it is not clear that some parts make any sense. For instance, you state that $H_{3\times3}=[R+ {T\over d}(\tau N^t)]$, but it is not clear how $[R+ {T\over d}(\tau N^t)]$ is a $3 \times 3$ matrix? You specified that $N^t$ is a vector, and we have that $\tau$ is also a vector. And based on your previous response, we can assume that $R$ is a $3 \times 3$ matrix. Given this, I'm not sure how it is that $[R+ {T\over d}(\tau N^t)]$ can be considered a $3 \times 3$ matrix? And [...] – The Pointer Dec 8 '18 at 5:28
• [...] the addition doesn't make any sense either, since we're adding the $3 \times 3$ matrix $R$ to ${T\over d}(\tau N^t)$, which, whatever it is (this is not clear either), it is not a $3 \times 3$ matrix. – The Pointer Dec 8 '18 at 5:33
• Despite this, I will award you the bounty for effort. – The Pointer Dec 8 '18 at 5:34

If 2D is initial affine space we compose 1P - projective space by $$\left( \begin{matrix} \tilde{x} \\ \tilde{y} \end{matrix} \right)= \left( \begin{matrix} a & b \\ c& d \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right)$$.

Left hand are homogeneous coordinates of 1P - point. Right hand coordinates represent a vector in 2D.

In a similar way we compose 2P from 3D: $$\left( \begin{matrix} \tilde{x} \\ \tilde{y} \\ \tilde{z} \end{matrix} \right) = \left( \begin{matrix} a_{11}& a_{12} & a_{13} \\ a_{21}& a_{22}& a_{23} \\ a_{31}& a_{32}& a_{33} \\ \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \\ \end{matrix} \right)$$.

$$\frac{x}{\omega}=\frac{p_{11}X+p_{12}Y+p_{13}Z+p_{14}T}{p_{31}X+p_{32}Y+p_{33}Z+p_{34}T} \\ \frac{y}{\omega}=\frac{p_{21}X+p_{22}Y+p_{23}Z+p_{24}T}{p_{31}X+p_{32}Y+p_{33}Z+p_{34}T}$$.

If Z=0 then
$$\frac{x}{\omega}=\frac{p_{11}X+p_{12}Y+p_{14}T}{p_{31}X+p_{32}Y+p_{34}T} \\ \frac{y}{\omega}=\frac{p_{21}X+p_{22}Y+p_{24}T}{p_{31}X+p_{32}Y+p_{34}T}$$.

If Z=5 then
$$\frac{x}{\omega}=\frac{p_{11}X+p_{12}Y+p_{13}.5T+p_{14}T}{p_{31}X+p_{32}Y+p_{33}.5T+p_{34}T} \\ \frac{y}{\omega}=\frac{p_{21}X+p_{22}Y+p_{23}.5T+p_{24}T}{p_{31}X+p_{32}Y+p_{33}.5T+p_{34}T}$$.