0
$\begingroup$

As the title states, the question is to prove that there's a unique Projective Transformation that maps four points of $\mathbb{R^2}$ to the projective plane. I tried defining the projective transformation as $\bf{x'} $$= M\cdot $$\bf x$ where $M$ is a $3 \times3$ matrix and defining four points $x_i = \pmatrix{a_i\cr b_i}$ but I don't think i'm getting anywhere. Any tips?

$\endgroup$
  • $\begingroup$ Count degrees of freedom. $\endgroup$ – amd Sep 10 '19 at 8:24
  • $\begingroup$ Alternatively, construct the transformation explicitly. One such construction is given here. You’ll find that you’ll need to impose some conditions on the points for your proposition to hold. $\endgroup$ – amd Sep 10 '19 at 10:50
0
$\begingroup$

To be clear, a unique determination of a collineation is only possible if you have been provided a mapping from one set of four points to another set of four points such that no three points in either set are collinear.

One way to do this, similar to your initial approach, is by showing that the determination of the required collineation is equivalent to solving a system of 12 homogeneous linear equations in 13 variables when the rank of the matrix of the coefficients is 12. I've written a detailed writeup of this approach in my blog in case you get stuck.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.