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I want to practice finding projective transformations but I'm not sure if I do it right.

For example:

$[1:1:0] \rightarrow [1:0:0]$

$[1:0:1] \rightarrow [0:1:0]$

$[1:1:1] \rightarrow [0:0:1]$

$[0:1:1] \rightarrow [1:1:1]$

Here is how I do it:

$A= \begin{bmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$

$A([x:y:z]) = [x':y':z']$, where $(x', y', z') = (x,y,z) \cdot A$

Now, $\alpha(1,0,0)=(1,1,0) \cdot A = (a_{11} + a_{21}, a_{12} + a_{22}, a_{13}+a_{23})$

$\alpha$ is a nonero scalar from the field over which we consider our projective plane.

Similarly, $\beta(0,1,0) = (1,0,1) \cdot A$, $\gamma(0,0,1) = (1,1,1)A$, $\delta(1,1,1) = (0,1,1)A$.

Then I get a system of linear equations, it turns out that $\delta = 2 \gamma$, $\alpha = -2 \gamma$, $\beta = -2 \gamma$, and the matrix x I eventually get is

$\begin{bmatrix} -2&-2&-1\\0&2&1\\2&0&1\end{bmatrix}$

However the solution here says it should be my matrix transposed and multiplied by $-1$. Could you tell me how to do that correctly?

I would really appreciate all your help because I can't find a thorough explanation anywhere.

Thank you.

MvG
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Bilbo
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    First, because projective transformations are defined only up to nonzero scalar, you shouldn't worry about the sign discrepancy. Second, I would guess that the convention in the solution is to consider matrices as acting on column vectors on the left, whereas your solution has matrices acting on row vectors on the right, so a simple transpose converts your answer to their answer (up to nonzero scalar). It looks like you're doing fine to me. – user43208 Oct 23 '13 at 17:33
  • All right. Thank you a lot :) – Bilbo Oct 23 '13 at 17:35
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    Bilbo's got a riddle; now that's rich! +1 for the question! – Robert Lewis Oct 23 '13 at 19:09
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    As for a thorough explanation: have you read this post of mine? Perhaps it can help in some way, although it, too, assumes the matrix acts on the left of a column vector, so it, too, would result in a transposed matrix. By the way, you seem to have a typo in the third of your four defining points. Will edit that, but please check my edit to ensure this is really what you meant. – MvG Oct 24 '13 at 21:20
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    @user43208: Do you want to turn this comment of yours into a proper answer, so we can get this question out of the “unanswered” queue? I doubt there is much more to say on the matter. – MvG Oct 24 '13 at 21:24

1 Answers1

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On request (to get the question off the unanswered list), I'm turning my comment into an answer.

First, remember that projective transformations are defined only up to nonzero scalar; to say it a bit more precisely, the group of projective transformations for $\mathbb{P}^2$ is $GL(3, F)/(F^\ast I_3)$, where invertible matrices are considered modulo nonzero scalar multiples of the identity. So the sign discrepancy is nothing to worry about.

Second, it appears that the convention of the linked solution is to have matrices act on the left (on column vectors), whereas the OP's convention is to use a right action (on row vectors). Hence by taking transpose (and ignoring sign differences), the OP's solution and the linked solution are reconciled.

user43208
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