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.
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.