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I am asked to find a projective transformation which maps the lines $L_1, L_2$ to $L^{'}_1, L^{'}_2$ respectively, where

$$L_1 : X - 2Y = 0 \:, \quad L_2 : Y+Z=0 \:, \quad L^{'}_1 : 2X+Y+Z=0 \:, \quad L^{'}_2 : X+3Y =0$$

I've been trying to think of a way to do this but I'm struggling. I had thought if I perhaps take 2 points from each line and then try to find a transformation which maps the two from each of $L_1$ and $L_2$ to the two from $L^{'}_1$ and $L^{'}_2$ respectively but I just can't seem to get it to work out. Any help is much appreciated!!

EDIT:

To clarify, the method I had tried to use consisted of taking 2 points from each line. So as an example say I take the following points;

$$P_1 = \left[0,0,1\right] \quad,\quad Q_1 = \left[2,1,0\right] $$ $$P_2 = \left[1,0,0\right] \quad,\quad Q_2 = \left[0,1,-1\right] $$ $$P^{'}_1 = \left[1,-1,-1\right] \quad,\quad Q^{'}_1 = \left[0,1,-1\right] $$ $$P^{'}_2 = \left[3,-1,0\right] \quad,\quad Q^{'}_2 = \left[0,0,1\right] $$

from their respective lines. Now I am trying to find the transformation that maps the points to the $'$ equivalent. In this case I have tried to use a method I have in my notes but this method only ever takes me through to the standard basis.

Killuminati
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  • What are $X$, $Y$ and $Z$ here? On the face of it, these are equations of planes – amd Oct 24 '17 at 18:25
  • I'm not sure how this helps? – Killuminati Oct 24 '17 at 18:58
  • I’m asking you to help us by clarifying the terms that you’re using in your question. – amd Oct 24 '17 at 19:02
  • Oh dear I'm sorry! I guess $X, Y$ and $Z$ are points within the projective plane $\mathbb{P}^{2}$. So these are lines with the projective plane. Sadly I am completely lost within this module. – Killuminati Oct 24 '17 at 19:11
  • Ah, so $[X:Y:Z]$ are the homogeneous coordinates of points in $\mathbb{RP}^2$. That helps. I’d approach this by representing those lines as homogeneous vectors and then using the fact that if a point transformation is given by the matrix multiplication $M\mathbf p$, then lines transform as $M^{-T}\mathbf l$. – amd Oct 24 '17 at 19:15
  • Okie dokie. However I would add that it's $\mathbb{C}\mathbb{P}^2$ not the reals. Not that I think this will make much difference – Killuminati Oct 24 '17 at 19:17
  • Yeah, that’s not going to make a difference. Another way to go is to find a projective transformation that preserves the intersection points and maps the direction vectors—the points at infinity of the lines—appropriately. You should in fact be able to find an affine transformation that works. – amd Oct 24 '17 at 19:21
  • Let us continue this discussion in chat. – Killuminati Oct 24 '17 at 19:29

1 Answers1

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Your basic idea is sound: Projective transformations preserve incidence relations, so a projectivity that maps one pair of points to another will map the line through the first pair to the line through the second pair. A convenient choice for two sets of corresponding points is the points at infinity of corresponding lines. If you then map the intersection of the two lines to itself, you’ll end up with an affine transformation that maps one line to the other.

The line intersections are easily found via cross products of homogeneous vectors. The rest of the construction is also pretty simple if you remember that the columns of a transformation matrix are the images of the basis vectors. Using that fact you can construct a matrix $M$ that maps the origin and coordinate axes to the first set of lines and their intersection, and a second matrix $M'$ for the other pair of line. The desired projective map is then $M'M^{-1}$.

Based on your update, it looks like you’re trying to build the homography that maps the quadrilateral formed from pairs of points on $L_1$ and $L_2$ to a quad formed from pairs of points on $L_1'$ and $L_2'$. This method should also work, again because projective transformations preserve incidence relationships. You can find a nice description of one way to construct this map in this answer. Following that method, solve the equation $$\begin{bmatrix} P_1 &P_2 & Q_1\end{bmatrix}\begin{bmatrix}\lambda \\ \mu \\ \tau \end{bmatrix} = Q_2 $$ and form the matrix $M=\begin{bmatrix}\lambda P_1 & \mu P_2 & \tau Q_1 \end{bmatrix}$. Similarly, form the matrix $M'=\begin{bmatrix} \lambda'P_1' & \mu'P_2' & \tau'Q_1' \end{bmatrix}$. The transformation that maps one of these quads to the other is, as above, $H=M'M^{-1}$. You should end up with $$H = \begin{bmatrix} -\frac35 & \frac35 & 0 \\ \frac35 & -\frac15 & -\frac25 \\ \frac35 & 0 & \frac25 \end{bmatrix} \begin{bmatrix}0&-2&2 \\ 0&0&1 \\ -1&0&0 \end{bmatrix}^{-1} = \begin{bmatrix} -\frac3{10} & \frac35 & \frac35 \\ \frac1{10} & -\frac35 & -\frac35 \\ 0 & \frac25 & -\frac35 \end{bmatrix}.$$ The intersection of $L_1$ and $L_2$ is $R = [1:-2:0]\times[0:1:1]=[-2:-1:1]$, the intersection of $L_1'$ and $L_2'$ is $R' = [2:1:1]\times[1:3:0] = [-3:1:5]$ and $HR=\left[\frac35:-\frac15:-1\right] = R'$, so that checks out. To check that the lines are mapped appropriately by $H$, you could compute $H^{-T}L_1$ and $H^{-T}L_2$, but it’s easier to check what $H$ does to the points at infinity on those lines. Their images should lie on the corresponding image lines, which amounts to checking that $[2:1:1]\,H\,[2:1:0]^T=0$ and $[1:3:0]\,H\,[1:0:0]^T=0$.

amd
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