# Introduction to Projective Geometry Solutions 5.5

## Collineations of Type I

A 13 minute read, posted on 26 Aug 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. Find the fixed points of the collineation $X’ = AX$ if:

(a) $A = \begin{pmatrix} 0 & 6 & 2 \\ 1 & -1 & -1 \\ 2 & 6 & 0 \end{pmatrix}$ (b) $A = \begin{pmatrix} 1 & -4 & -4 \\ -1 & 1 & -1 \\ 1 & -2 & 0 \end{pmatrix}$

(c) $A = \begin{pmatrix} 1 & 2 & 2 \\ -2 & 6 & 3 \\ 2 & -4 & -1 \end{pmatrix}$ (d) $A = \begin{pmatrix} 0 & -5 & -3 \\ 0 & 3 & 2 \\ -1 & -5 & -2 \end{pmatrix}$

This one is pretty straightforward and you can use any software to find the eigenvectors of the given matrices.

#### 2. Prove that no collineation of type I can be of period 2.

In terms of a coordinate system whose triangle of reference is the triangle whose vertices are the invariant points, the equations of collineation of type I in matrix form are

$$X’ = AX$$ $$A = \begin{pmatrix} k_1 & 0 & 0 \\ 0 & k_2 & 0 \\ 0 & 0 & k_3 \end{pmatrix}$$

In order for the transformation to be of period 2, $A^2$ must be a scalar multiple of $I$ and it is no specialization to assume the scalar to be 1. In this case,

$$A^2 = \begin{pmatrix} k_1^2 & 0 & 0 \\ 0 & k_2^2 & 0 \\ 0 & 0 & k_3^2 \end{pmatrix}$$

For this to be $I$, $k_1^2 = k_2^2 = k_3^2 = 1$. This means at least two of the characteristic values must be equal (as there are only 2 distinct square roots of 1). However, a collineation of type I has three distinct characteristic values by definition. Hence no collineation of type I can be of period 2.

#### 3. Under what conditions, if any, can a collineation of type I be of period 3? of period 4? of period n?.

Building on our proof to the previous exercise the conditions will be

$k_1^3 = k_2^3 = k_3^3 = 1$, with all $k_i$s being distinct for type 3

$k_1^4 = k_2^4 = k_3^4 = 1$, with all $k_i$s being distinct for type 4

$k_1^n = k_2^n = k_3^n = 1$, with all $k_i$s being distinct for type n

This is equivalent to saying that for the collineation of type I to be of period $n$, the characteristic roots must be distinct $p^{th}$ roots of 1, where $p \le n$, with at least one root being an $n^{th}$ root of unity (else the period will be the highest $p$).

#### 4. Is a rotation of the euclidean plane about an arbitrary point the specialization of a collineation of type I?

If we change coordinates such that the arbitrary point is the origin in the new coordinate system, then Theorem 2 guarantees that the clockwise rotation of the euclidean plane about this point is the specialization of a collineation of type I. We can change coordinates without affecting the classification of the collineation because from section 5.4 we know that the structure of the configuration of invariant points of a collineation is independent of the particular coordinate system in terms of which its equation is expressed. Hence, rotation of the euclidean plane about an arbitrary point is the specialization of a collineation of type I.

#### 5. Determine the nature of the euclidean specialization of a collineation of type I whose characteristic values are $1$, $e^{i\theta}$, $e^{-i\theta}$ if the corresponding fixed points are $(0, 0, 1)$, $(1, -i, 0)$, $(1, i, 0)$

The requirement that $(0, 0, 1) \rightarrow (0, 0, 1)$ implies that $a_{13} = a_{23} = 0$. Similarly, the requirements that $(1, -i, 0) \rightarrow (e^{i\theta}, -e^{i\theta}i, 0)$ and $(1, i, 0) \rightarrow (e^{-i\theta}, e^{-i\theta}i, 0)$ imply that

$$a_{11} - a_{12}i = e^{i\theta}$$ $$a_{11} + a_{12}i = e^{-i\theta}$$

$$a_{21} - a_{22}i = -e^{i\theta}i$$ $$a_{21} + a_{22}i = e^{-i\theta}i$$

$$a_{31} - a_{32}i = 0$$ $$a_{31} + a_{32}i = 0$$

Solving these equations we get $$a_{11} = \frac{e^{i\theta} + e^{-i\theta}}{2} = cos(\theta)$$ $$a_{12} = \frac{-e^{i\theta} + e^{-i\theta}}{2i} = -sin(\theta)$$ $$a_{21} = \frac{e^{i\theta} - e^{-i\theta}}{2i} = sin(\theta)$$ $$a_{22} = \frac{e^{i\theta} + e^{-i\theta}}{2} = cos(\theta)$$ $$a_{31} = a_{32} = 0$$

Hence, the equations of the collineation become $$x’_1 = cos(\theta)x_1 - sin(\theta)x_2$$ $$x’_2 = sin(\theta)x_1 + cos(\theta)x_2$$ $$x’_3 = x_3$$

The equations of the euclidean specialization of this transformation will be $$x = cos(\theta)x - sin(\theta)y$$ $$y = sin(\theta)x + cos(\theta)y$$

We can also write these as $$x = cos(-\theta)x + sin(-\theta)y$$ $$y = -sin(-\theta)x + cos(-\theta)y$$

Hence these are the equations of rotation of the euclidean plane about the origin in the anticlockwise direction through an angle whose measure is $\theta$.

#### 6. Is the composition of two collineations of type I always, sometimes, or never a collineation of type I?

Consider two collineations of type I represented by the matrices $A$ and $B$.

The composition $AB$ will surely be a collineation of type I if both $A$ and $B$ share the same characteristic vectors. Any characteristic vector of $A$ or $B$ will be a fixed point of $AB$. So we have at least eliminated “never” from the list of possible answers.

Consider a collineation of type I and its inverse (which is also of type I). The composition of these two will be the identity transformation that has one repeated characteristic root and hence is not a collineation of type I. Hence, the composition of two collineations of type I will not always be a collineation of type I.

#### 7. Find the equations of a collineation of type I whose characteristic values are $k_1$, $k_2$, $k_3$ if the corresponding fixed points are

(a) $(1, 0, 0), (0, 1, 0), (1, 1, 1)$

(b) $(1, 0, 0), (0, 1, 1), (0, 1, -1)$

(c) $(0, 1, 1), (1, 0, 1), (1, 1, 0)$

Because I already know about eigendecomposition of matrices1, I’m just going to directly use the formula (yay!).

(a) $$A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} k_1 & 0 & 0 \\ 0 & k_2 & 0 \\ 0 & 0 & k_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$ $$= \begin{pmatrix} k_1 & 0 & k_3 - k_1 \\ 0 & k_2 & k_3 - k_2 \\ 0 & 0 & k_3 \end{pmatrix}$$

(b) $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{pmatrix}^{-1} \begin{pmatrix} k_1 & 0 & 0 \\ 0 & k_2 & 0 \\ 0 & 0 & k_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{pmatrix}$$ $$= \begin{pmatrix} k_1 & 0 & 0 \\ 0 & \frac{k_3 + k_2}{2} & \frac{k_2 - k_3}{2} \\ 0 & \frac{k_2 - k_3}{2} & \frac{k_3 + k_2}{2} \end{pmatrix}$$

(c) $$A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} k_1 & 0 & 0 \\ 0 & k_2 & 0 \\ 0 & 0 & k_3 \end{pmatrix} \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$$ $$= \frac{1}{2}\begin{pmatrix} k_3 + k_2 & k_3 - k_1 & k_2 - k_1 \\ k_3 - k_2 & k_3 + k_1 & k_1 - k_2 \\ k_2 - k_3 & k_1 - k_3 & k_2 + k_1 \end{pmatrix}$$

The answer for (c) in the book is incorrect, it should be transposed.

#### 8. Show that there are no nonsingular conics which are transformed into themselves by a collineation of type I.

Consider a collineation of type I, $T$ such that $X’ = TX$. In coordinate system with the triangle of reference formed by the invariant points of $T$, the equations of transformation will be as follows. $$T = \begin{pmatrix} k_1 & 0 & 0 \\ 0 & k_2 & 0 \\ 0 & 0 & k_3 \end{pmatrix}$$ $$T^{-1} = \begin{pmatrix} \frac{1}{k_1} & 0 & 0 \\ 0 & \frac{1}{k_2} & 0 \\ 0 & 0 & \frac{1}{k_3} \end{pmatrix}$$

Let’s consider a general conic $X^TAX = 0$, where $$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix}$$

So, the equation of the transformed conic will be $(T^{-1}X)^TA(T^{-1}X) = X^T((T^{-1})^TAT^{-1})X$. Now it is clear that $T^T = T$ and the same is true for the inverse. So the equation of the image of the conic will become $X^T(T^{-1}AT^{-1})X$. Expanding the matrix of the quadratic form of this conic we get

$$T^{-1}AT^{-1} = \begin{pmatrix} \frac{1}{k_1} & 0 & 0 \\ 0 & \frac{1}{k_2} & 0 \\ 0 & 0 & \frac{1}{k_3} \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \begin{pmatrix} \frac{1}{k_1} & 0 & 0 \\ 0 & \frac{1}{k_2} & 0 \\ 0 & 0 & \frac{1}{k_3} \end{pmatrix}$$

$$= \begin{pmatrix} \frac{a_{11}}{k_1^2} & \frac{a_{12}}{k_1k_2} & \frac{a_{13}}{k_1k_3} \\ \frac{a_{12}}{k_1k_2} & \frac{a_{22}}{k_2^2} & \frac{a_{23}}{k_2k_3} \\ \frac{a_{13}}{k_1k_3} & \frac{a_{23}}{k_2k_3} & \frac{a_{33}}{k_3^2} \end{pmatrix}$$

For this to be the same conic as $A$, all of the following must hold.

$$(\frac{1}{k_1^2} - 1)a_{11} = 0$$ $$(\frac{1}{k_2^2} - 1)a_{22} = 0$$ $$(\frac{1}{k_3^2} - 1)a_{33} = 0$$ $$(\frac{1}{k_1k_2} - 1)a_{12} = 0$$ $$(\frac{1}{k_1k_3} - 1)a_{13} = 0$$ $$(\frac{1}{k_2k_3} - 1)a_{23} = 0$$

As all of the $k_i$s must be distinct and non-zero, atleast one of the $a_{jj}$ terms must be zero and two of the $k_i$s must be 1 and -1 for the first three equations to hold.

If this is the case, all three $a_{jk}$s must be 0 for the last three equations to hold. This however results in a singular matrix.

Hence, there are no nonsingular conics which are transformed into themselves by a collineation of type I.

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

Let the coordinates of $P$, $P’$ and $O$ be $(p_1, p_2, p_3)$, $(p’_1, p’_2, p’_3)$, $(o_1, o_2, o_3)$ respectively. Let the characteristic roots of the transformation be $k_1, k_2, k_3$.

From the equations of transformation in the coordinate system where the fixed points of the collineation are the vertices of the triangle of reference, we know that $$p’_1 = k_1p_1, p_1 = \frac{1}{k_1}p’_1$$ $$p’_2 = k_2p_2, p_2 = \frac{1}{k_2}p’_2$$ $$p’_3 = k_3p_3, p_3 = \frac{1}{k_3}p’_3$$

As per Theorem 1 section 4.2, we have $$\begin{vmatrix} p_1 & p_2 & p_3 \\ p’_1 & p’_2 & p’_3 \\ o_1 & o_2 & o_3 \end{vmatrix} = 0$$

Expressing the image coordinates in terms of the object coordinates we get $$\begin{vmatrix} p_1 & p_2 & p_3 \\ k_1p_1 & k_2p_2 & k_3p_3 \\ o_1 & o_2 & o_3 \end{vmatrix} = 0$$

$$\implies (k_3 - k_2)o_1p_2p_3 + (k_1 - k_3)o_2p_1p_3 + (k_2 - k_1)o_3p_1p_2 = 0$$ This is the locus of points $P$ that are collinear with their image and a given point $O$.

Expressing the object coordinates in terms of the image coordinates we get $$\begin{vmatrix} \frac{p’_1}{k_1} & \frac{p’_2}{k_2} & \frac{p’_3}{k_3} \\ p’_1 & p’_2 & p’_3 \\ o_1 & o_2 & o_3 \end{vmatrix} = 0$$

$$\implies (k_3 - k_2)k_1o_1p’_2p’_3 + (k_1 - k_3)k_2o_2p’_1p’_3 + (k_2 - k_1)k_3o_3p’_1p’_2 = 0$$ This is the locus of the images of points $P$ that are collinear with their image and a given point $O$.

#### 10. Let $\Gamma_1$ and $\Gamma_2$ be two conics having the triangle of reference as a common self-polar triangle, and let $T_1$ and $T_2$ be the polarities determined by $\Gamma_1$ and $\Gamma_2$, respectively. Under what conditions, if any, is the transformation $T_2^{-1}T_1$ a collineation of type I?

Let the vertices of the given triangle be denoted by $V_1$, $V_2$, $V_3$.

From exercise 4, Sec 4.9, Page 175, we know that a triangle is said to be self-polar with respect to conic $\Gamma$ if each side of the triangle is the polar of the opposite vertex with respect to $\Gamma$.

This means $T_1V_i = b_i(V_j \times V_k)$ and $T_2V_i = c_i(V_j \times V_k)$. Combining this with the fact that correlations are invertible, we get

$$T_1V_i = k_iT_2V_i, k_i = \frac{b_i}{c_i}$$ $$\implies T_2^{-1}T_1V_i = k_iV_i$$

Hence, $T_2^{-1}T_1$ will be a collineation of type I with the vertices of the self-polar triangle being the fixed points of the transformation if the value of each $k_i$ is distinct.

#### 11. Let $T$ be a collineation of type I with the characteristic values $k_1$, $k_2$, $k_3$ and corresponding fixed points $F_1:(1, 0, 0)$, $F_2:(0, 1, 0)$, $F_3:(0, 0, 1)$. Let $\Gamma$ be the conic $ax_2x_3 + bx_1x_3 + cx_1x_2 = 0$, and let $\Gamma_1$ be the image of $\Gamma$ under $T$. Clearly, $\Gamma$ and $\Gamma_1$ intersect in the fixed points of $T$ and in some fourth point, $P$. Find the coordinates of $P$. If this process be iterated indefinitely, show that the intersection of $\Gamma_n$ and $\Gamma_{n+1}$ which is distinct from $F_1$, $F_2$, $F_3$ is $T^nP$.

From the equations of transformation in the coordinate system where the fixed points of the collineation are the vertices of the triangle of reference, we know that $$x’_1 = k_1x_1, x_1 = \frac{1}{k_1}x’_1$$ $$x’_2 = k_2x_2, x_2 = \frac{1}{k_2}x’_2$$ $$x’_3 = k_3x_3, x_3 = \frac{1}{k_3}x’_3$$ Hence the image $\Gamma_1$ of $\Gamma$ will be $$\frac{ax_2x_3}{k_2k_3} + \frac{bx_1x_3}{k_1k_3} + \frac{cx_1x_2}{k_1k_2} = 0$$ or, $$k_1ax_2x_3 + k_2bx_1x_3 + k_3cx_1x_2 = 0$$

Clearly the fixed points solve the equations of both conics and hence lie on both of them. To find the fourth point of intersection, let us substitute the value of $x_1$ from the equation of $\Gamma$ into the equation of $\Gamma_1$.

$$x_1 = -\frac{ax_2x_3}{bx_3 + cx_2}$$

$$k_1ax_2x_3 - \frac{(k_2bx_3 + k_3cx_2)(ax_2x_3)}{bx_3 + cx_2} = 0$$ $$(ax_2x_3)(k_1bx_3 + k_1cx_2 - k_2bx_3 - k_3cx_2) = 0$$ As $a \ne 0$ and $x_2 = 0$ and $x_3 = 0$ will give us $F_3$ and $F_2$ respectively, to find $P$ we must solve the second component of the equation above. $$(k_1 - k_2)bx_3 + (k_1 - k_3)cx_2 = 0$$ $$\implies \frac{x_2}{x_3} = \frac{b(k_2 - k_1)}{c(k_1 - k_3)}$$

Using this we can find the ratio $\frac{x_1}{x_3}$ as $$\frac{x_1}{x_3} = -\frac{\frac{ab(k_2 - k_1)}{c(k_1 - k_3)}}{b + \frac{cb(k_2 - k_1)}{c(k_1 - k_3)}}$$

As $k_1 \ne k_3$, and $c \ne 0$ we get $$\frac{x_1}{x_3} = -\frac{ab(k_2 - k_1)}{bc(k_1 - k_3) + cb(k_2 - k_1)}$$

As $b \ne 0$, we get $$\frac{x_1}{x_3} = \frac{a(k_2 - k_1)}{c(k_3 - k_2)}$$

Hence, the fourth point of intersection is $(\frac{a(k_2 - k_1)}{c(k_3 - k_2)}, \frac{b(k_2 - k_1)}{c(k_1 - k_3)}, 1)$. As we’re dealing with homogeneous coordinates, we can multiply the whole thing by $\frac{c}{k_2 - k_1}$ to get $P:(\frac{a}{k_3 - k_2}, \frac{b}{k_1 - k_3}, \frac{c}{k_2 - k_1})$

Now $\Gamma_n$ and $\Gamma_{n+1}$ will respectively be $$k_1^nax_2x_3 + k_2^nbx_1x_3 + k_3^ncx_1x_2 = 0$$ $$k_1^{n+1}ax_2x_3 + k_2^{n+1}bx_1x_3 + k_3^{n+1}cx_1x_2 = 0$$ Clearly the fixed points solve the equations of both conics and hence lie on both of them.

Solving these two equations, we will get the fourth point of intersection as $(\frac{ak_1^n}{k_3 - k_2}, \frac{bk_2^n}{k_1 - k_3}, \frac{ck_3^n}{k_2 - k_1})$ which is indeed $T^nP$.

#### 12. In the proof of Theorem 2, why is it necessary to assume that $\theta \ne n\pi$?

If $\theta = n\pi$, then as $cos(n\pi) = (-1)^n$ and $sin(n\pi) = 0$, the matrix of transformation becomes

$$\begin{pmatrix} (-1)^n & 0 & 0 \\ 0 & (-1)^n & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

which is clearly not a collineation of type I as it has between two and three identical characteristic roots depending on whether $n$ is odd or even. Hence it is necessary to assume that $\theta \ne n\pi$ in the proof of Theorem 2.