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.
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Essentially a duplicate of https://math.stackexchange.com/q/2487881/265466, and ultimately, of https://math.stackexchange.com/a/339033/265466. The lines are different, and there are more of them, but the method is the same. Hint: this is the dual problem to finding a projectivity between two quadrilaterals. – amd Oct 26 '17 at 01:57
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I've actually seen both of those, and I've tried the method of plucking two random points from each line but it doesn't seem to work here. I think we no longer have that freedom because now we have 4 lines now instead of 2 so there are more constraints.. – SSF Oct 26 '17 at 10:12
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That’s right. You need to work with the four pairs of lines directly. – amd Oct 26 '17 at 17:56
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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 $B$: $\mathcal H^{-T}=(B^{-1}A)^{-T}=B^{-T}A^T$
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