I'm stuck with this exercise: I have 4 points in the projective plane $\Bbb{P}^2(\Bbb{R})$
$$P_0=\{0,0,1\},\quad P_1=\{0,1,-1\},\quad P_2=\{1,-1,0\},\quad P_3=\{1,1,-3\}$$
and I have to compute their cross ratio, so I tried to see if they are collinear. I used the theorem that say:
4 points $(P,Q,T,R)$ are collinear $\Leftarrow \Rightarrow$ $rg \begin{pmatrix}p_0&p_1&p_2\\t_0&t_1&t_2\\q_0&q_1&q_2\\r_0&r_1&r_2 \end{pmatrix}\le2$
but the rg of my matrix is $=3$. Am I doing something wrong? And, is the collinear's condition necessary?
You didn't need to check if they were collinear. They were in general position (and the other 4 points also were in general position), so (after multiplying the coordinates by some constants) you had two projective frames. The fundamental points of these two projective frames are the projectivization of two bases of $\Bbb{R}^3$. We have a theorem that states that $f$ exists and its associated isomorphism is the one you find when you change the coordinates between the two bases.
(Plus the cross ratio (as far as we know) only works in $\Bbb{P}^1$)