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Show that any three points on a line cane be sent to any other three points on a line by projection.

Logically this makes sense. However, when trying to show that this works I am not sure where to begin. I thought about trying to use cross-ratio's but so far we have seen the use of cross ratio, but with four points and not three. If anyone has any suggestions that would be great. Thank you

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    $\begingroup$ What do you mean by "projection"? Do you mean a projective transformation? $\endgroup$ – Ted Shifrin Mar 8 '16 at 22:37
  • $\begingroup$ Probably "projectivity" was intended. $\endgroup$ – Michael Hardy Mar 8 '16 at 22:50
  • $\begingroup$ See page 37 of this document. $\qquad$ $\endgroup$ – Michael Hardy Mar 8 '16 at 22:56
  • $\begingroup$ @Michael Hardy I don't want to enter into polemics, but I don't find the document you have indicated very enlightning for somebody, like the author of the question, who enters into the subject. For example, it has almost no figures. I would advise this kind of document: morpheo.inrialpes.fr/people/Boyer/Teaching/M2R/geoProj.pdf $\endgroup$ – Jean Marie Mar 8 '16 at 23:18
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Consider a certain non degenerate homography (= projective transformation) of the projective plane:

$$x'=\dfrac{ax+by+ct}{gx+hy+it} \ , \ y'=\dfrac{dx+ey+ft}{gx+hy+it}$$

associated with matrix

$$H=\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}$$

Consider 3 points, $P_k$ with proj. coordinates $(x_k,y_k,t_k)^T$ (not necessarily aligned for the moment) and their image points $P'_k$ with coordinates $(x'_k,y'_k,t'_k)^T$ through homography $H$.

This can be given a compact presentation as the following matrix equality $HP=P'$:

$$\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix} \begin{bmatrix} x_1 & x_2 & x_3\\ y_1 & y_2 & y_3\\ t_1 & t_2 & t_3 \end{bmatrix} = \begin{bmatrix} x'_1 & x'_2 & x'_3\\ y'_1 & y'_2 & y'_3\\ t'_1 & t'_2 & t'_3 \end{bmatrix} $$

As a consequence, $det(HP) = det(P')$ i.e., $det(H) det(P) = det(P').$

But $det(H) \neq 0$ for a non degenerate homography.

Thus we have:

$$det(P)=0 \ \Leftrightarrow \ det(P')=0.$$

Otherwise said

$$\text{Points} \ P_k \ \text{aligned} \ \Leftrightarrow \ \text{Points} \ P'_k \ \text{aligned.}$$

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  • $\begingroup$ Is this answer convenient ? $\endgroup$ – Jean Marie Mar 9 '16 at 11:32

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