Projective transformation taking lines to lines

Question

Find a projective transformation of the complex projective plane $\mathbb{P}^2(\mathbb{C})$, which sends the lines $x + 2y + 3z = 0$, $x + z = 0$, $y = 0$ and $x − z = 0$ to $x = 0$, $x + y = 0$, $z = 0$ and $y + z = 0$ respectively.

Answer 1

This is the dual of the problem of finding the homography that maps one quadrilateral onto another. If you represent the lines as homogeneous coordinate vectors, the problem then becomes one of finding a projective transformation matrix $\mathcal H$ that makes the following mappings: $$\begin{align} [1:2:3] &\mapsto [1:0:0] \\ [1:0:1] &\mapsto [1:1:0] \\ [0:1:0] &\mapsto [0:0:1] \\ [1:0:-1] &\mapsto [0:1:1]. \end{align}$$ A simple method for doing this is described in this answer. This will give you a line transformation. To get the corresponding point transformation use the fact that, if points transform as $\mathbf p'=\mathcal M\mathbf p$, then lines transform as $\mathbf l' = \mathcal M^{-T}\mathbf l$, that is, the required point transformation is $\mathcal H^{-T}$. If you’re using the method in the linked answer, you can compute this matrix directly once you have the two intermediate mappings $A$ and $B$: $\mathcal H^{-T}=(B^{-1}A)^{-T}=B^{-T}A^T$