# Introduction to Projective Geometry Solutions 5.3

## Collineations

A 22 minute read, posted on 21 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. Given $O:(1, 1, 2)$ and $l: 3x_1 - x_2 + x_3 = 0$. Find the equations of the transformation which assigns to a general point $P$ the intersection of $OP$ and $l$ as its image. Is this transformation singular or nonsingular? What points, if any, are invariant under this transformation? What points, if any, do not have a unique image under this transformation?

Let the general point $P$ be denoted by the coordinates $(p_1, p_2, p_3)$. Then we have

$$OP = (1, 1, 2) \times (p_1, p_2, p_3) = [p_3 - 2p_2, 2p_1 - p_3, p_2 - p_1]$$ $$P’: OP \cap l = [p_3 - 2p_2, 2p_1 - p_3, p_2 - p_1] \times [3, -1, 1] = \\ (p_1 + p_2 - p_3, -3p_1 + 5p_2 - p_3, -6p_1 + 2p_2 + 2p_3)$$

Hence the equations of this transformation can be represented by

$$P’ = AP$$ where

$$A = \begin{pmatrix} 1 & 1 & -1 \\ -3 & 5 & -1 \\ -6 & 2 & 2 \end{pmatrix}$$

As the determinant of this matrix is $0$, this transformation is singular.

Clearly if the point is on $l$, by the definition of the transformation, its image will be itself. Also, the image of the point $O$ will be the intersection of any line through $O$ and $l$. So it will not have a unique image. We can verify these assertions and the fact that there are no other invariant points or points with no unique images by calculating the characteristic values and characteristic vectors of the transformation matrix $A$.

The eigenvalues of this matrix are $4$ and $0$ (indicating once again that the matrix is singular) with multiplicities $2$ and $1$ respectively and the corresponding eigenvectors are $(1, 0, -3)$, $(0, 1, 1)$ for eigenvalue $4$ and $(1, 1, 2)$ for eigenvalue $0$.

As the eigenvalue $4$ is repeated, all linear combinations of its eigenvectors will also be invariant under this transformation. Hence the whole line determined by the eigenvectors of $4$ are the invariant points of this transformation. This is the line

$$(1, 0, -3) \times (0, 1, 1) = [3, -1, 1]$$

which is the line $l$. Hence the line $l$ is invariant.

The eigenvector of $0$, which is $O$, does not have a unique image as it is in the nullspace of $A$.

#### 2. (a) Given $\Gamma: x_1x_2 = x_3^2$ and $\lambda: x_1 + x_2 - x_3 = 0$. Find the equations of the transformation which maps a general point $P$ onto the intersection of $\lambda$ and the polar of $P$ with respect to $\Gamma$. Is this transformation singular or nonsingular? What points, if any, are invariant under this transformation? What points, if any, do not have a unique image under this transformation?(b) Work part (a), given $\Gamma:x_1^2 + x_1x_2 - x_2x_3 + x_3^2 = 0$ and $\lambda:-x_1 + x_2 + 7x_3 = 0$

Let the general point $P$ be denoted by the coordinates $(p_1, p_2, p_3)$. Expressing the conic in the form $X^TAX = 0$, we have $$A = \begin{pmatrix} 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$ And the polar of $P$ with respect to $\Gamma$ will be

$$P^TAX = (p_1, p_2, p_3)\begin{pmatrix} 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ which is $$p_2x_1 + p_1x_2 - 2p_3x_3 = 0$$

So the intersection of the polar of $P$ with $\lambda$ will be

$$P’: [p_2, p_1, -2p_3] \times [1, 1, -1] = (2p_3 - p_1, p_2 - 2p_3, p_2 - p_1)$$

Hence, the equations of transformation can be represented by the matrix

$$T = \begin{pmatrix} -1 & 0 & 2 \\ 0 & 1 & -2 \\ -1 & 1 & 0 \end{pmatrix}$$

Clearly, the determinant of $T$ is zero as adding rows one and two gives us row three. Hence this transformation is singular.

The eigenvalues of this transformation matrix are $-i\sqrt{3}$, $i\sqrt{3}$ and $0$ corresponding to eigenvectors $(1, \frac{-\sqrt{3}i - 1}{2}, \frac{-\sqrt{3}i + 1}{2})$, $(1, \frac{\sqrt{3}i - 1}{2}, \frac{\sqrt{3}i + 1}{2})$ and $(1, 1, \frac{1}{2})$ respectively. As we’re dealing with the $\Pi_2$, the classical projective plane, complex values are allowed. So the invariant points under this transformation are the two points $(1, \frac{-\sqrt{3}i - 1}{2}, \frac{-\sqrt{3}i + 1}{2})$ and $(1, \frac{\sqrt{3}i - 1}{2}, \frac{\sqrt{3}i + 1}{2})$

The point $(1, 1, \frac{1}{2})$ will not have a unique image under this transformation as it belongs to the nullspace of $T$. Looking at it another way, the polar of this point is $\lambda$ itself and hence each point on $\lambda$ is an image of $(1, 1, \frac{1}{2})$.

(b) In this case the matrix $A$ for the conic $X^TAX = 0$ will be

$$\begin{pmatrix} 1 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 1 \end{pmatrix}$$

And the polar of $P$ with respect to $\Gamma$ will be

$$P^TAX = (p_1, p_2, p_3)\begin{pmatrix} 1 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ which is $$(p_1 + \frac{1}{2}p_2)x_1 + (\frac{1}{2}p_1 - \frac{1}{2}p_3)x_2 + (p_3 - \frac{1}{2}p_2)x_3 = 0$$

So the intersection of the polar of $P$ with $\lambda$ will be

$$P’: [p_1 + \frac{1}{2}p_2, \frac{1}{2}p_1 - \frac{1}{2}p_3, p_3 - \frac{1}{2}p_2] \times [-1, 1, 1] = \\ (\frac{7}{2}p_1 + \frac{1}{2}p_2 - \frac{9}{2}p_3, -7p_1 - 3p_2 - p_3, \frac{3}{2}p_1 + \frac{1}{2}p_2 - \frac{1}{2}p_3)$$

Hence, the equations of transformation can be represented by the matrix

$$T = \begin{pmatrix} \frac{7}{2} & \frac{1}{2} & -\frac{9}{2} \\ -7 & -3 & -1 \\ \frac{3}{2} & \frac{1}{2} & -\frac{1}{2} \end{pmatrix}$$

As the determinant of $T$ is 0, the transformation is singular.

There is only one eigenvalue for this transformation matrix and that is $0$ with a multiplicity of 3. There is also only one eigenvector $(2, -5, 1)$ the polar of which is $\lambda$ and hence this point does not have a unique image. There are no invariant points under this transformation.

#### 3. (a) Given $l: x_1 + x_2 - 3x_3 = 0$ and $O:(0, 0, 1)$. Find the equations of the transformation which maps a general point $P$ onto the intersection of $l$ and the harmonic conjugate of $OP$ with respect to the lines $x_1 = 0$ and $x_2 = 0$. Is this transformation singular or nonsingular? What points, if any, are invariant under this transformation? What points, if any, do not have a unique image under this transformation? Work part (a), given $l: x_1 - 2x_2 = 0$.

Let the general point $P$ be denoted by the coordinates $(p_1, p_2, p_3)$. Then we have

$$OP = (0, 0, 1) \times (p_1, p_2, p_3) = [-p_2, p_1, 0]$$

Harmonic conjugate of $OP: [-p_2, p_1, 0]$ w.r.t. $[1, 0, 0]$ and $[0, 1, 0]$ is $[p_2, p_1, 0]$.

Intersection of the harmonic conjugate of $OP$ with $l$ is $$P’: [p_2, p_1, 0] \times [1, 1, -3] = (-3p_1, 3p_2, p_2 - p_1)$$

Hence the equations of transformation in matrix form will be $$A = \begin{pmatrix} -3 & 0 & 0 \\ 0 & 3 & 0 \\ -1 & 1 & 0 \end{pmatrix}$$

Clearly, the determinant of $A$ is zero as the last column of $A$ is 0. Hence this transformation is singular.

The eigenvalues of this transformation matrix are -3, 3, and 0 with eigenvectors $(1, 0, \frac{1}{3})$, $(0, 1, \frac{1}{3})$ and $(0, 0, 1)$ respectively. Thus the two invariant points are $(1, 0, \frac{1}{3})$, $(0, 1, \frac{1}{3})$. Note that lines formed by the invariant points with $O$ are $x_2 = 0$ and $x_1 = 0$. Hence they are their own harmonic conjugates.

The point $O: (0, 0, 1)$ does not have a unique image under this transformation (it maps to every point on $l$).

(b) Intersection of the harmonic conjugate of $OP$ with $l$ is $$P’: [p_2, p_1, 0] \times [1, -2, 0] = (0, 0, -2p_2 - p_1)$$

Hence the equations of transformation in matrix form will be $$A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ -1 & -2 & 0 \end{pmatrix}$$

Clearly, the determinant of $A$ is zero as the first two rows and the last column of $A$ are 0. Hence this transformation is singular.

This matrix has a single eigenvalue 0 and two eigenvectors $(2, -1, 0)$ and $(0, 0, 1)$. Points on the line determined by these two points, $[1, 2, 0]$, which is also the harmonic conjugate of $l$, does not have a unique image under this transformation. Each point on this line maps to all points on $l$.

There are no invariant points under this transformation.

#### 4. Is there a linear transformation which maps every point on $\Pi_2$ onto the point $(2, 1, -3)$?

This is an interesting question. While I can think of a transformation that maps most points to a given point (rank 1 with column space containing only multiples of vector $(2, 1, -3)$), there is no transformation that will map all points to the given point.

Consider a transformation that maps multiple points to one point. This transformation can not have an inverse as we can not really map one point back to its multiple preimages via a linear transformation. If the transformation can’t be inverted, the matrix of transformation must be singular. This means there is at least one vector in the nullspace of the matrix of transformation. This point either has no image or has multiple images and hence can’t be mapped to a single point $(2, 1, -3)$.

#### 5. Prove that the images of three noncollinear points under a linear transformation $X’ = AX$ will be collinear if and only if $A$ is a singular matrix.

Let’s denote three general noncollinear points by $X_1$, $X_2$, $X_3$ and their images by $X’_1$, $X’_2$, $X’_3$.

First, let us consider the necessity of $A$ being singular given that the images of three noncollinear points are collinear. As it is given that the three points $X’_1$, $X’_2$, $X’_3$ are collinear, from Theorem 1, Corollary 1, Sec 4.2, the following must hold $$\begin{vmatrix} X’_1 & X’_2 & X’_3 \end{vmatrix} = 0$$ Using the equations of transformation and the product rule of determinants, we can rewrite this as $$\begin{vmatrix} AX_1 & AX_2 & AX_3 \end{vmatrix} = 0$$ $$\implies|A|\begin{vmatrix} X_1 & X_2 & X_3 \end{vmatrix} = 0$$ As it is given that the three points $X_1$, $X_2$, and $X_3$ are not collinear, from Theorem 1, Corollary 1, Sec 4.2, the value of the second determinant can not be zero. Hence $|A|$ must be 0 and $A$ must be singular.

Now, let us discuss if $A$ being singular is sufficient for the images of three noncollinear points to be collinear. The value of the product of determinants below is 0 as it is given that $|A|$ is zero from virtue of it being singular. $$\implies|A|\begin{vmatrix} X_1 & X_2 & X_3 \end{vmatrix} = 0$$ Using the product rule of determinants and multiplying out the matrix A with each of the columns of the second matrix, we get $$\begin{vmatrix} AX_1 & AX_2 & AX_3 \end{vmatrix} = 0$$ Using the equations of transformations, we can further rewrite this as $$\begin{vmatrix} X’_1 & X’_2 & X’_3 \end{vmatrix} = 0$$ From Theorem 1, Corollary 1, Sec 4.2, this means that the images must be collinear. Hence the sufficiency of the condition is proved.

#### 6. (a) Given $\Gamma_1: x_1^2 + 2x_1x_2 + x_2^2 - x_3^2 = 0$ and $\Gamma_2: x_1^2 + x_1x_3 - x_2x_3 + x_3^2 = 0$. Find the equations of the transformation which maps a point $P$ onto the point $P’$ whose polar with respect to $\Gamma_2$ is the same as the polar of $P$ with respect to $\Gamma_1$. Is this transformation singular or nonsingular?(b) Find the equations of the transformation which maps $P$ onto the intersection of the polars of $P$ with respect to $\Gamma_1$ and $\Gamma_2$. Is this transformation a collineation?

Let’s represent both the conics using the matrices of their quadratic forms. $$A_1 = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$ $$A_2 = \begin{pmatrix} 1 & 0 & 0.5 \\ 0 & 0 & -0.5 \\ 0.5 & -0.5 & 1 \end{pmatrix}$$

Let us denote the general point $P$ by the coordinates $(p_1, p_2, p_3)$ and its image $P’$ by coordinates $(p’_1, p’_2, p’_3)$.

The polar of $P$ with respect to $\Gamma_1$ will be given by the following equation.

$$P^TA_1X = 0$$ $$\implies \begin{pmatrix} p_1 & p_2 & p_3 \end{pmatrix}\begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ $$\implies (p_1 + p_2)x_1 + (p_1 + p_2)x_2 - p_3x_3 = 0$$

The polar of $P’$ with respect to $\Gamma_2$ will be given by the following equation.

$$P’^TA_2X = 0$$ $$\implies \begin{pmatrix} p’_1 & p’_2 & p’_3 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0.5 \\ 0 & 0 & -0.5 \\ 0.5 & -0.5 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ $$\implies (p’_1 + 0.5p’_3)x_1 - 0.5p’_3x_2 + (0.5p’_1 - 0.5p’_2 + p’_3)x_3 = 0$$

If the two polars are to be the same, their line coordinates must be proportional. Hence,

$$p_1 + p_2 = k(p’_1 + 0.5p’_3)$$ $$p_1 + p_2 = k(-0.5p’_3)$$ $$-p_3 = k(0.5p’_1 - 0.5p’_2 + p’_3)$$

As these are homogeneous coordinates, it is no specialization to take $k$ as 1. So, the equations become

$$p_1 + p_2 = p’_1 + 0.5p’_3$$ $$p_1 + p_2 = -0.5p’_3$$ $$-p_3 = 0.5p’_1 - 0.5p’_2 + p’_3$$

Writing the image coordinates in terms of the object coordinates, we get $$p’_1 = 2(p_1 + p_2)$$ $$p’_2 = 2(p_3 - p_1 - p_2)$$ $$p’_3 = -2(p_1 + p_2)$$

Removing the scalar factor of 2 and writing this in matrix form, we get $$T = \begin{pmatrix} 1 & 1 & 0 \\ -1 & -1 & 1 \\ -1 & -1 & 0 \end{pmatrix}$$

As the first row is a multiple of the last row, it is clear that this matrix and hence this transformation is singular.

(b) The polar of $P$ with respect to $\Gamma_2$ will be given by the following equation.

$$P^TA_2X = 0$$ $$\implies \begin{pmatrix} p_1 & p_2 & p_3 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0.5 \\ 0 & 0 & -0.5 \\ 0.5 & -0.5 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ $$\implies (p_1 + 0.5p_3)x_1 - 0.5p_3x_2 + (0.5p_1 - 0.5p_2 + p_3)x_3 = 0$$

Intersection of the two polars will be given by $$[p_1 + p_2, p_1 + p_2, -p_3] \times [p_1 + 0.5p_3, -0.5p_3, 0.5p_1 - 0.5p_2 + p_3] = \\ (0.5p_1^2 + p_1p_3 - 0.5p_2^2 + p_2p_3 - 0.5p_3^2, \\ -0.5p_1^2 - 2p_1p_3 + 0.5p_2^2 - p_2p_3 - 0.5p_3^2, \\ -p_1^2 - p_1p_2 - p_1p_3 - p_2p_3)$$

This is clearly not a linear transformation. As a collineation must be a nonsingular linear transformation, this transformation does not qualify as a collineation.

#### 7. Given $\Gamma_1: X^TA_1X = 0$ and $\Gamma_2: X^TA_2X = 0$. Find the equation of the transformation which maps a point $P$ onto the point $P’$ whose polar with respect to $\Gamma_2$ is the same as the polar of $P$ with respect to $\Gamma_1$, and show that this transformation is a collineation if and only if $\Gamma_1$ and $\Gamma_2$ are nonsingular conics.

The equations of the two polars will be $$P^TA_1X = 0$$ $$P’^TA_2X = 0$$

If these two polars are to be the same, their line coordinates must be proportional. Hence, the following equation must hold within an arbitrary factor of proportionality.

$$P^TA_1 = P’^TA_2$$

Transposing both sides, we get

$$A_1^TP = A_2^TP’$$

If $A_2^T$ has a left inverse then we can multiply both sides by $(A_2^T)^{-1}$ on the left to get $$P’ = (A_2^T)^{-1}A_1^TP$$

This is the equation of the given transformation. Note that if $A_2^T$ does not have a left inverse then such a transformation is not possible.

For this transformation to be a collineation the matrix of transformation should be nonsingular. This means its determinant should be non-zero. Hence

$$|(A_2^T)^{-1}A_1^T| \ne 0$$ Using the product rule for determinants of matrices, we get $$|(A_2^T)^{-1}||A_1^T| \ne 0$$

For this product to be non-zero both the matrices $(A_2^T)^{-1}$ and $A_1^T$ must be nonsingular. This implies that the matrices $A_1$ and $A_2$ must be nonsingular. Hence, for the given transformation to be a collineation both the conics must be nonsingular. Applying the same argument in reverse, we can prove the sufficiency of the condition as well.

#### 8. What is the equation of the line transformation induced by the point transformation $X’ = AX$ if:

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

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

(c) $A = \begin{pmatrix} 1 & 0 & 3 \\ 2 & -1 & 0 \\ 1 & 1 & 7 \end{pmatrix}$

By Theorem 2, the equation of the line transformation will be $\Lambda’ = (A^{-1})^T\Lambda$, if $A$ is the matrix for the point transformation.

Denoting the inverse of the transpose by the symbol $A^{-T}$, the equations of the line transformation in each of these cases within an arbitrary factor of proportionality will be as follows.

(a) $\Lambda’ = \begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 1 \\ 2 & -1 & 0 \end{pmatrix}^{-T}\Lambda$ $\implies \Lambda’ = \begin{pmatrix} 1 & 2 & -2 \\ 1 & 2 & 5 \\ 3 & -1 & 1 \end{pmatrix}\Lambda$

(b) $\Lambda’ = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}^{-T}\Lambda$ $\implies \Lambda’ = \begin{pmatrix} -1 & -1 & 2 \\ 0 & -1 & 1 \\ 2 & 3 & -4 \end{pmatrix}\Lambda$

(c) $\Lambda’ = \begin{pmatrix} 1 & 0 & 3 \\ 2 & -1 & 0 \\ 1 & 1 & 7 \end{pmatrix}^{-T}\Lambda$ $\implies \Lambda’ = \begin{pmatrix} -7 & -14 & 3 \\ 3 & 4 & -1 \\ 3 & 6 & -1 \end{pmatrix}\Lambda$

#### 9. What is the relation between the characteristic roots of a nonsingular matrix, $A$, and the characteristic roots of its inverse, $A^{-1}$?

Let’s denote a general characteristic root of $A$ by $c$ and the corresponding characteristic vector as $x$. Then we have

$$Ax = cx$$

Multiplying on both sides on the left by $A^{-1}$, we get

$$x = A^{-1}cx$$ $$\implies x = c(A^{-1}x)$$ $$\implies A^{-1}x = \frac{1}{c}x$$

Hence, each characteristic vector of $A$ is also a characteristic vector of $A^{-1}$ and each characteristic value of $A^{-1}$ is a reciprocal of the corresponding characteristic value of $A$.

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

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

(b) $A = \begin{pmatrix} 3 & 0 & -1 \\ 10 & 4 & -8 \\ 6 & 1 & -3 \end{pmatrix}$

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

The answers are the eigenvectors of the non-zero eigenvalues of each of the given matrices. If there’s a repeated eigenvalue with distinct eigenvectors then there will be a line of fixed points.

(a) Point $(1, - 1, 1)$ and all points on the line $[-2, -1, 2]$ formed by points $(1, 0, 1)$ and $(0, 2, 1)$

(b) Points $(1, - 1, 1)$ and $(1, 2, 2)$

(c) Points $(1, \frac{3i - 7}{2}, \frac{5i - 3}{2})$, $(1, \frac{-3i - 7}{2}, \frac{-5i - 3}{2})$ and $(0, 1, 1)$

#### 11. If $A$ is a nonsingular matrix, the transformation $\Lambda = AX$ is called a correlation, because it associates, or correlates, a line $\Lambda$ with every point $P$, and conversely. Show that under any correlation the lines associated with the points of a range form a pencil.

Consider a range of points $B$. All points $X$ on this line must satisfy the incidence relation $B^TX = 0$. Using the equation of the correlation, we can rewrite this as $B^TA^{-1}\Lambda = 0$. Transposing this we get $\Lambda^T(A^{-1})^TB = 0$. This is the equation of the pencil of lines passing through the point $(A^{-1})^TB$. Hence under any correlation, the lines associated with the points of a range form a pencil.

#### 12. If $\Lambda = AX$ is a correlation, show that the set of points which lie on their associated lines is a conic locus, and find its equation. Show also that the set of lines which contain their associated points is a conic envelope, and find its equation. Show that in general the conic envelope is not the set of tangents to the conic locus.

The sets of points $X$ that lie on their associated lines must simultaneously satisfy the following two equations.

$$\Lambda = AX$$ $$\Lambda^TX = 0$$

Using the value of $\Lambda$ from the first equation in the second, we get $$(AX)^TX = 0$$ $$\implies X^TA^TX = 0$$

This is the equation of a conic locus with $A^T$ being the matrix of its quadratic form.

Similarly, the set of lines $\Lambda$ that contain their associated points must simultaneously satisfy the following two equations.

$$\Lambda = AX$$ $$\Lambda^TX = 0$$

Using the value of $X$ from the first equation in the second, we get $$\Lambda^TA^{-1}\Lambda = 0$$

This is the equation of the envelope of lines tangent to the conic $X^TAX = 0$.

The conic envelope will be the set of tangents to the conic locus if and only if $A^T = A$, i.e. when the matrix of transformation is symmetric.

#### 13. If $\Gamma: X^TAX = 0$ is an arbitrary nonsingular conic, show that the transformation which associates with an arbitrary point its polar with respect to $\Gamma$ is a correlation, and find its equation. What is the locus of points which lie on their image lines in this correlation? What is the envelope of the lines which contain their image points in this correlation?

The polar of a point $P$ with respect to the conic $\Gamma$ is $P^TAX = 0$. This can be written as $(A^TP)^TX = 0$. The line coordinates of this line will be $A^TP$. Hence the equation of correlation will be as follows.

$$\Lambda = A^TP$$

As $A$ is the matrix of the quadratic form of a conic, it must be symmetric. That means $A^T = A$. Hence we can rewrite the equation of the correlation as follows.

$$\Lambda = AP$$

Rewriting this in the standard form as $\Lambda = AX$, we see that this is the exact same correlation as in the solution to #12. Hence, the locus of points which lie on their image lines in this correlation will be $$X^TA^TX = 0$$ As $A$ is symmetric, we can rewrite this as $$X^TAX = 0$$

Similarly, the envelope of the lines which contain their image points in this correlation will be $$\Lambda^TA^{-1}\Lambda = 0$$ which is the envelope of lines tangent to the conic $X^TAX = 0$.

#### 14. Show that a correlation $\Lambda = AX$ is the polarity defined by some nonsingular conic if and only if $A$ is a symmetric matrix.

As the term polarity has not been defined before, I am going to assume its meaning from the surrounding context. According to my understanding, a polarity defined by a nonsingular conic is a correlation that associates a point with its polar with respect to the given conic.

The correlation $\Lambda = AX$ associates the line $(AP)^TX = 0$ or $P^TA^TX = 0$ with the point $P$. As the matrix $A$ is symmetric, this line is the equation of the polar of $P$ with respect to the conic $X^TAX = 0$. Hence the sufficiency of the condition holds.

If $\Lambda = AX$ is the polarity defined by some nonsingular conic then $(AP)^TX = 0$ or $P^TA^TX = 0$ must be the polar of the point $P$ with respect to that conic. This means $A^T$ must be the matrix of the quadratic form of the conic. We know that the matrix of the quadratic form of a conic is symmetric. Hence the necessity of the condition also holds.

#### 15. (a) Is the transformation which associates with each point its polar with respect to a given triangle a polarity? (b) Is the following transformation a correlation? Let $l_1$ and $l_2$ be two lines intersecting in a point $O$, let $A_1$ be a point on $l_1$ distinct from $O$, and let $A_2$ be a point on $l_2$ distinct from $O$. If $P$ is an arbitrary point, let $Q_1$ be the intersection of $PA_2$ and $l_1$, let $Q_2$ be the intersection of $PA_1$ and $l_2$, and, finally, let the line $Q_1Q_2$ be the image of the point $P$.

(a) This transformation is not even a correlation, so it can not be a polarity. This is because the vertices of the triangle do not map to unique lines under this transformation. See the solution to exercise 20 Sec 4.7 for details on how this transformation is defined.

(b) Similar to the solution to (a), the points $A_1$ and $A_2$ do not map to unique lines under this transformation. Hence this transformation can not be a correlation.

#### 16. Prove Theorem 2 by computing the coordinates of the image of a general line $P_1P_2$ from the coordinates of the images of $P_1$ and $P_2$.

Let $P_1$ and $P_2$ have coordinates $(x_1, y_1, w_1)$ and $(x_2, y_2, w_2)$ respectively. Let the matrix of transformation be $$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$$

So the coordinates of the images $P’_1$ and $P’_2$ will be $$P’_1 = AP_1 = (a_1a_{11} + b_1a_{12} + c_1a_{13}, \\ a_1a_{21} + b_1a_{22} + c_1a_{23}, \\ a_1a_{31} + b_1a_{32} + c_1a_{33})$$ $$P’_2 = AP_2 = (a_2a_{11} + b_2a_{12} + c_2a_{13}, \\ a_2a_{21} + b_2a_{22} + c_2a_{23}, \\ a_2a_{31} + b_2a_{32} + c_2a_{33})$$

Then, the line defined by $P’_1$ and $P’_2$ will be

$$P’_1 \times P’_2 = ((a_{22}a_{33} - a_{23}a_{32})(b_1c_2 - b_2c_1) + (a_{23}a_{31} - a_{21}a_{33})(c_1a_2 - c_2a_1) + (a_{21}a_{32} - a_{22}a_{31})(a_1b_2 - a_2b_1), \\ (a_{13}a_{32} - a_{12}a_{33})(b_1c_2 - b_2c_1) + (a_{11}a_{33} - a_{13}a_{31})(c_1a_2 - c_2a_1) + (a_{12}a_{31} - a_{11}a_{32})(a_1b_2 - a_2b_1), \\ (a_{12}a_{23} - a_{13}a_{22})(b_1c_2 - b_2c_1) + (a_{13}a_{21} - a_{11}a_{23})(c_1a_2 - c_2a_1) + (a_{11}a_{22} - a_{12}a_{21})(a_1b_2 - a_2b_1))$$

$$= \begin{pmatrix} a_{22}a_{33} - a_{23}a_{32} & a_{23}a_{31} - a_{21}a_{33} & a_{21}a_{32} - a_{22}a_{31} \\ a_{13}a_{32} - a_{12}a_{33} & a_{11}a_{33} - a_{13}a_{31} & a_{12}a_{31} - a_{11}a_{32} \\ a_{12}a_{23} - a_{13}a_{22} & a_{13}a_{21} - a_{11}a_{23} & a_{11}a_{22} - a_{12}a_{21} \end{pmatrix} \begin{pmatrix} b_1c_2 - b_2c_1 \\ c_1a_2 - c_2a_1 \\ a_1b_2 - a_2b_1 \end{pmatrix}$$

$$= (A^{-1})^T * (P_1 \times P_2)$$

Which is the same as what Theorem 2 states and yes that calculation was painful to type out.