# Introduction to Projective Geometry Solutions 5.7

## Collineations of types IV, V and VI

A 15 minute read, posted on 9 Sep 2019

Tags computer vision, projective geometry, problem solution

Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.

#### 1. Show that no collineation of type V can be periodic.

Multiplying the standard form of the matrix of a collineation of type V $n$ times with itself, we get $$\begin{pmatrix} k_1^n & 0 & nak^{n-1} \\ 0 & k_1^n & nbk^{n-1} \\ 0 & 0 & k_1^n \end{pmatrix}$$ As $a \ne 0$ and $b \ne 0$, this can never be a scalar multiple of the identity matrix and hence a collineation of type V can not be periodic.

#### 2. Under what conditions, if any, can a collineation of type IV be periodic?

Multiplying the standard form of the matrix of a collineation of type IV $n$ times with itself, we get $$\begin{pmatrix} k_1^n & 0 & 0 \\ na_{21}k^{n-1} & k_1^n & 0 \\ t & na_{32}k^{n-1} & k_1^n \end{pmatrix}$$ As $a_{21} \ne 0$ and $a_{32} \ne 0$, this can never be a scalar multiple of the identity matrix and hence a collineation of type IV can not be periodic under any conditions.

#### 3. Show that the composition of two collineations of type V with the same axis is always a collineation of type V.

Let $x_3 = 0$ be the common axis and the two centers be $(a, b, 0)$ and $(c, d, 0)$. Let the repeated characteristic roots of the collineation of type V be $k_1$ and $k_2$ respectively. Then the matrix of transformation of points for the composition of the two collineations of type V will be $$\begin{pmatrix} k_1 & 0 & a \\ 0 & k_1 & b \\ 0 & 0 & k_1 \end{pmatrix}\begin{pmatrix} k_2 & 0 & c \\ 0 & k_2 & d \\ 0 & 0 & k_2 \end{pmatrix} = \begin{pmatrix} k_1k_2 & 0 & k_1c + k_2a \\ 0 & k_1k_2 & k_1d + k_2b \\ 0 & 0 & k_1k_2 \end{pmatrix}$$

This is a collineation of type V with center $(k_1c + k_2a, k_1d + k_2b, 0)$, axis $x_3 = 0$ and repeated characteristic root $k_1k_2$.

#### 4. Show that the composition of two collineations of type V with distinct axes will also be a collineation of type V if and only if the center of each transformation is the point of intersection of the two axes.

This is a dual of the result in exercise #3 and as such needs no extra proof but we’ll do it anyway for fun and clarity.

Let $(0, 0, 1)$ be the common center and the two axes be $[a, b, 0]$ and $[c, d, 0]$. Let the repeated characteristic roots of the collineation of type V be $k_1$ and $k_2$ respectively. Then the matrix of transformation of lines for the composition of the two collineations of type V will be $$\begin{pmatrix} k_1 & 0 & a \\ 0 & k_1 & b \\ 0 & 0 & k_1 \end{pmatrix}\begin{pmatrix} k_2 & 0 & c \\ 0 & k_2 & d \\ 0 & 0 & k_2 \end{pmatrix} = \begin{pmatrix} k_1k_2 & 0 & k_1c + k_2a \\ 0 & k_1k_2 & k_1d + k_2b \\ 0 & 0 & k_1k_2 \end{pmatrix}$$ This is a collineation of type V with axes $[k_1c + k_2a, k_1d + k_2b, 0]$, center $(0, 0, 1)$ and repeated characteristic root $k_1k_2$.

#### 5. Find the equations of the most general collineation of type IV having the fixed elements (a) $F_1:(0, 1, 0), f_1: x_3 = 0$ (b) $F_1:(1, 0, 1), f_1: x_2 = 0$ (c) $F_1:(1, 0, 0), f_1: x_2 - x_3 = 0$

(a) Under this transformation the point $(0, 1, 0)$ will get transformed to $(0, k_1, 0)$. This implies that $a_{12} = a_{32} = 0$ and $a_{22} = k_1$. Further, as $x_3 = 0$ is the invariant line, any point with the last coordinate zero will be transformed to another point with the last coordinate zero. This implies $a_{31} = 0$. This means the matrix of transformation will look like $$A = \begin{pmatrix} a_{11} & 0 & a_{13} \\ a_{21} & k_1 & a_{23} \\ 0 & 0 & a_{33} \end{pmatrix}$$ For this matrix to have a single repeated characteristic root $k_1$, $a_{11} = a_{33} = k_1$ and for the characteristic matrix $A - k_1I$ to have rank two $a_{21} \ne 0$ and $a_{13} \ne 0$. Hence the equations of the most general collineation of type IV with the given fixed elements is $$x’_1 = k_1x_1 + a_{13}x_3$$ $$x’_2 = a_{21}x_1 + k_1x_2 + a_{23}x_3$$ $$x’_3 = k_1x_3$$ For this to be a collineation, i.e to keep the matrix nonsingular, $k_1 \ne 0$.

(b) Under this transformation the point $(1, 0, 1)$ will get transformed to $(k_1, 0, k_1)$. This implies $$k_1 = a_{11} + a_{13} \implies a_{11} = k_1 - a_{13}$$ $$0 = a_{21} + a_{23} \implies a_{21} = -a_{23}$$ $$k_1 = a_{31} + a_{33} \implies a_{33} = k_1 - a_{31}$$ As $x_2 = 0$ is the invariant line, any point with the second coordinate zero will be transformed to another point with the second coordinate zero. This implies $a_{21} = a_{23} = 0$. This means the matrix of transformation will look like $$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & 0 \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$$ The characteristic polynomial of this matrix is $$(a_{22} - k)((a_{11} - k)(a_{33} - k) - a_{13}a_{31})$$ For $k_1$ to be the single repeated root $a_{22} = k_1$. Substituting the values of $a_{11}$ and $a_{33}$ we derived earlier in the second component of the equation above, we get $$(k_1 - a_{13} - k)(k_1 - a_{31} - k) - a_{13}a_{31}$$ As $k_1$ is a repeated root of this quadratic polynomial, the discriminant must be zero and $$a_{13} + a_{31} = 0$$ This implies $$a_{13} = -a_{31}$$ $$a_{11} = k_1 + a_{31}$$ So the equations of transformation will be $$x’_1 = (k_1 + a_{31})x_1 + a_{12}x_2 - a_{31}x_3$$ $$x’_2 = k_1x_2$$ $$x’_3 = a_{31}x_1 + a_{32}x_2 + (k_1 - a_{31})x_3$$ For this to be a nonsingular transformation $k_1 \ne 0$ and for the matrix $A - k_1I$ to have rank 2, $a_{31} \ne 0$ and $a_{12} \ne a_{32}$.

(c) Under this transformation the point $(1, 0, 0)$ will get transformed to $(k_1, 0, 0)$. This implies $a_{11} = k_1$ and $a_{21} = a_{31} = 0$. Further, as $x_2 - x_3 = 0$ is the invariant line, any point that has the same last two coordinates must be transformed to another point which has identical second and third coordinates. This implies that $a_{22} + a_{23} = a_{32} + a_{33}$. The characteristic polynomial of this matrix is $$(a_{11} - k_1)((a_{22} - k)(a_{33} - k) - a_{32}a_{23})$$ For $k_1$ to be the single repeated root $a_{22} + a_{33} = 2k_1$ and $a_{23} = -a_{32}$. Expressing values in terms of $a_{22}$ we get $$a_{33} = 2k_1 - a_{22}$$ $$a_{23} = k_1 - a_{22}$$ $$a_{32} = -k_1 + a_{22}$$ Hence, the equations of transformation will be $$x’_1 = k_1x_1 + a_{12}x_2 + a_{13}x_3$$ $$x’_2 = a_{22}x_2 + (k_1 - a_{22})x_3$$ $$x’_3 = -(k_1 - a_{22})x_2 + (2k_1 - a_{22})x_3$$ For this to be a nonsingular transformation $k_1 \ne 0$ and for the matrix $A - k_1I$ to have rank 2, $a_{12} \ne -a_{13}$ and $a_{22} \ne k$.

#### 6. Find the equations of the most general collineation of type V having the fixed elements: (a) $F_1:(s, 0, t), f_1: x_2 = 0$ (b) $F_1:(1, 0, 1), f_1: x_1 - x_3 = 0$ (c) $F_1:(1, 1, 1), f_1: x_1 - x_2 = 0$

(a) Under the transformation, every point $(p_1, 0, p_3)$ will get transformed to $(k_1p_1, 0, k_1p_1)$. This implies that $a_{11} = a_{33} = k_1$ and $a_{21} = a_{23} = a_{13} = a_{31} = 0$. Furthermore as $k_1$ is a repeated triple root of the matrix of transformation, $a_{22} = k_1$. Hence the equations of transformation will be $$x’_1 = k_1x_1 + a_{12}x_2$$ $$x’_2 = k_1x_2$$ $$x’_3 = a_{32}x_2 + k_1x_3$$ Applying the collinearity condition of each point and its image with $F_1$, we get $a_{12}t - a_{32}s = 0$. This means that $a_{12} = \alpha s$ and $a_{32} = \alpha t$ for some $\alpha \ne 0$. So the final equations of transformation can also be written as $$x’_1 = k_1x_1 + \alpha sx_2$$ $$x’_2 = k_1x_2$$ $$x’_3 = \alpha tx_2 + k_1x_3$$

(b) Under the transformation, every point $(p_1, p_2, p_1)$ will get transformed to $(k_1p_1, k_1p_2, k_1p_1)$. This implies that $$a_{11} + a_{13} = k_1$$ $$a_{21} + a_{23} = 0$$ $$a_{31} + a_{33} = k_1$$ $$a_{22} = k_1$$ $$a_{12} = a_{32} = 0$$ The matrix of transformation look like $$A = \begin{pmatrix} k_1 - a_{13} & 0 & a_{13} \\ -a_{23} & k_1 & a_{23} \\ k_1 - a_{33} & 0 & a_{33} \end{pmatrix}$$ For $k_1$ to be a repeated root of this matrix, we must have $a_{33} - a_{13} = k_1$. So expressing the values in terms of $a_{13}$ we have $$a_{33} = k_1 + a_{13}$$ $$a_{31} = -a_{13}$$ So, the equations of transformation will be $$x’_1 = (k_1 - a_{13})x_1 + a_{13}x_3$$ $$x’_2 = -a_{23}x_1 + k_1x_2 + a_{23}x_3$$ $$x’_3 = -a_{13}x_1 + (k_1 + a_{13})x_3$$ Applying the collinearity condition of each point and its image with $F_1$, we get $a_{23} = 0$. Hence, the final equations of transformation can also be written as $$x’_1 = (k_1 - a_{13})x_1 + a_{13}x_3$$ $$x’_2 = k_1x_2$$ $$x’_3 = -a_{13}x_1 + (k_1 + a_{13})x_3$$

(c) Under the transformation, every point $(p_1, p_1, p_3)$ will get transformed to $(k_1p_1, k_1p_1, k_1p_3)$. This implies that $$a_{13} = a_{23} = 0$$ $$a_{11} + a_{12} = k_1$$ $$a_{21} + a_{22} = k_1$$ $$a_{31} + a_{32} = 0$$ $$a_{33} = k_1$$ The matrix of transformation look like $$A = \begin{pmatrix} k_1 - a_{12} & a_{12} & 0 \\ k_1 - a_{22} & a_{22} & 0 \\ -a_{32} & a_{32} & k_1 \end{pmatrix}$$ For $k_1$ to be a repeated root of this matrix, we must have $a_{22} - a_{12} = k_1$. So expressing the values in terms of $a_{12}$ we have $$a_{22} = k_1 + a_{12}$$ $$a_{21} = -a_{12}$$ So, the equations of transformation will be $$x’_1 = (k_1 - a_{12})x_1 + a_{12}x_2$$ $$x’_2 = -a_{12}x_1 + (k_1 + a_{12})x_2$$ $$x’_3 = -a_{32}x_1 + a_{32}x_2 + k_1x_3$$ Applying the collinearity condition of each point and its image with $F_1$, we get $a_{32} = a_{12}$. Hence, the final equations of transformation can also be written as $$x’_1 = (k_1 - a_{12})x_1 + a_{12}x_2$$ $$x’_2 = -a_{12}x_1 + (k_1 + a_{12})x_2$$ $$x’_3 = -a_{12}x_1 + a_{12}x_2 + k_1x_3$$

#### 7. Under what conditions, if any, will two collineations of type V commute with each other?

From the solutions to exercise #3 and its dual, #4, we can clearly see that two collineations of type V that either share the same axis or the same center commute.

Let’s assume that two commuting collineations, $T_A$ and $T_B$, have different axes.

Consider the transformation of a general point, $P$, on the axis of $T_A$, different from the point of intersection of the axes of the two collineations.

Under the transformation $T_BT_A$, $P$ will first be transformed by $T_A$ into itself, as it is invariant under $T_A$, and then transformed by $T_B$ into another point not equal to $P$, say $P’$, as $P$ is not invariant under $T_B$. So the image of $P$ under $T_BT_A$ will be $P’$.

Under the transformation $T_AT_B$, $P$ will be transformed by $T_B$ to $P’$ just as before, and $P’$ will be invariant under $T_A$ only if it lies on the axis of $T_A$. Now as the line defined by a point and its image must pass through the center of the transformation, $PP’$ must pass through the center of $T_B$. In other words, for a point on the axis of $T_A$ to have the same image under both $T_AT_B$ and $T_BT_A$, the axis of $T_A$ must pass through the center of $T_B$.

Applying the same logic for the transformation of a point on the axis of $T_B$ we will find that for a point on the axis of $T_B$ to have the same image under both $T_AT_B$ and $T_BT_A$, the axis of $T_B$ must pass through the center of $T_A$. Both conditions taken together imply that the two collineations can not commute if they do not share the same axis.

The proof for the shared center case is the dual of the shared axis case.

Therefore, two collineations of type V commute with each other if and only if they share the same axis or the same center.

#### 8. Show that in general the composition of two collineations of type V is a collineation of type I. Under what conditions, if any, will the composition be a collineation of type II? Of type III? Of type IV?

Consider two collineations of type V, $T_A$ and $T_B$. In general they will not share the same axis or the same center.

Let’s call the point of intersection of the two axes as $F_1$. This point will be invariant under the transformation defined by the composition of the two collineations as it lies on both their axes.

Let’s denote the line defined by the centers of the two collineations as $f_1$. This line will be invariant under the transformation defined by the composition of the two collineations as it passes through both their centers.

By Theorem 1 Sec 5.3, the mapping induced of $f_1$ onto itself by the composite collineation is a projectivity. As derived in Sec 4.8, this projectivity between cobasal ranges has a characteristic polynomial that, in general, has two distinct roots corresponding to the fixed points $F_2$ and $F_3$. This means that the composite collineation has a total of 3 fixed points, $F_1$, $F_2$ and $F_3$ which is characteristic of a collineation of type I.

Hence, in general, a composition of two collineations of type V is a collineation of type I.

Note that the line $f_1$ will be point-by-point invariant under the composite transformation of two collineations of type V only when they share the same axis. For a proof, consider the center of one of the collineations. Its image under the other collineation will be invariant only if the two share the same axis. Such a composite collineation will be of type V as we found out in #3.

When the projectivity induced on $f_1$ has only one fixed point, then we have a total of two fixed points including $F_1$ and two fixed lines. The line joining $F_1$ to the other fixed point, $f_2$, is not point-by-point invariant under the composite transformation. If it were point-by-point invariant then there must be another line $f’_2$ that it will be mapped to by one collineation with the second collineation mapping $f’_2$ back to $f_2$ to maintain its invariance under the composition. This would mean that the lines between the points on $f_2$ to their images on $f’_2$ were invariant under both the collineations of type V. However this is not possible as the invariant lines can’t belong to two different pencils (centers). Hence $f_2$ can’t be point-by-point invariant. Thus, this is a collineation of type II.

Using the same logic as above, we can deduce that no line other than a shared axis can be point-by-point invariant under a composite transformation of two collineations of type V. Hence, two collineations of type V can never compose to form a collineation of type III.

When the axis of one of the collineations of type V passes through the center of the other, the point of intersection is the center of the second collineation. No other point on the line will be invariant under the composite transformation. For a proof, take any point other than the point of intersection of the two axes. This point is invariant under the first transformation as it lies on its axis. The second collineation will then map this to some other point on the line as the point does not lie on the axis of the second collineation of type V. Hence this composite collineation with one fixed point and one fixed line is a collineation of type IV.

#### 9. What is the locus of a point with the property that it and its image under a collineation of type IV are collinear with a given point $O$? What is the locus of images of such points?

The answer to this question had me scratching my head. Axis of transformation and center for a collineation of type IV? Then I realized it must’ve been another typo in the book (it has quite a few). The question is actually asking about a collineation of type V.

Consider a general point $P: (p_1, p_2, p_3)$ and its image $P’:(p’_1, p’_2, p’_3)$.

Using the equations of transformation of a general collineation of type V, we get

$$p’_1 = k_1p_1 + ap_3, p_1 = \frac{k_1p’_1 - ap’_3}{k_1^2}$$ $$p’_2 = k_1p_2 + bp_3, p_2 = \frac{k_1p’_2 - bp’_3}{k_1^2}$$ $$p’_3 = k_1p_3, p_3 = \frac{p’_3}{k_1}$$

As per Sec 4.2, Theorem 1 if $P$ and its image are to be collinear with a given point $O: (o_1, o_2, o_3)$, we get

$$\begin{vmatrix} p_1 & p_2 & p_3 \\ k_1p_1 + ap_3 & k_1p_2 + bp_3 & k_1p_3 \\ o_1 & o_2 & o_3 \end{vmatrix} = 0$$

Expanding this we get the singular conic $$p_3(bo_3p_1 - a_3p_2 + (ao_2 - bo_1)p_3) = 0$$

This is the combination of two lines $p_3 = 0$ which is the axis of the collineation and $[bo_3, -ao_3, ao_2 - bo_1]$ which is $[o1, o2, o3] \times [a, b, 0]$ i.e. the line determined by the given point and the center of transformation.

Similarly, expressing in terms of image coordinates, we get $$\begin{vmatrix} \frac{k_1p’_1 - ap’_3}{k_1^2} & \frac{k_1p’_2 - bp’_3}{k_1^2} & \frac{p’_3}{k_1} \\ p’_1 & p’_2 & p’_3 \\ o_1 & o_2 & o_3 \end{vmatrix} = 0$$

Expanding this we get the same singular conic as before.

#### 10. Find the fixed points of the transformation $X’ = AX$ if:

(a) $A = \begin{pmatrix} 3 & -1 & 1 \\ 6 & -2 & 3 \\ 2 & -1 & 2 \end{pmatrix}$

(b) $A = \begin{pmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ -1 & -2 & 0 \end{pmatrix}$

(c) $A = \begin{pmatrix} 3 & 2 & -4 \\ 8 & 3 & -8 \\ 8 & 4 & -9 \end{pmatrix}$

(d) $A = \begin{pmatrix} -2 & -1 & 3 \\ 5 & 3 & -4 \\ -4 & -1 & 5 \end{pmatrix}$

I’m using maxima for this. You can use your favorite linear algebra software to find the eigenvectors.

(a) Type V, fixed line is $[2, -1, 1]$.
(b) Type IV, fixed point is $(1, -1, 1)$.
(c) Type V, fixed line is again $[-2, -1, 1]$.
(d) Type IV, fixed point is again $(1, -1, 1)$.