Find projective transformation Given four projective lines $L_1,L_2,L_3,L_4$ in projective plane, such that no three lines intersect in the same point and another four lines $M_1,M_2,M_3,M_4$ such that any three also do not intersect in the same point. Prove that there always exists a projective transformation $F: \mathbb{P}^2 \rightarrow \mathbb{P}^2$ such that $F(L_i)=M_i$, $i=1,2,3,4$.
My attempt was the following:
Since we are in projective space we know any two lines intersect at exactly one point, so we can define four points: 
$T_i = L_i \cap L_{i+1}$ for $i=1,2,3$ and $T_4 = L_4 \cap L_1$(and the same thing with lines $M_j$ to get points $P_j$) . Since line through two points is of the form $X = sA_1 + tA_2$, where $X = (x,y,z)$, $A_1$ and $A_2$ are points and $s$ and $t$ are parameters. Since $F$ is given by a matrix $X\cdot F = s X\cdot A_1 + tX\cdot A_2$ it is enough to find $F$ that maps $F(T_i)=P_i$.
Next I defined $A = [T_1, T_2, T_3]$ ($3 \times 3$ matrix) and $B = [aP_1, bP_2, cP_3]$ and we have to solve: $XA = B$. Since no three lines intersect at the same point $A$ is invertible we can write $X=BA^{-1}$ and by using the condition $XT_4 = \lambda P_4 $, where $\lambda = (a,b,c)$ we get our projective transformation.
I would like some insight if this is correct and if every thing is clear within my proof.
 A: Your proof is good.
Here is a cousin version.
Let us consider, the 6 points of intersection $A, B, C, D, F, G$ of lines $L_i$ plus the intersection of diagonals $E$. This kind of figure is called a complete quadrangle that we will denote by $(Q)$.
This quadrangle has in particular an important property of harmonic division, but we will not use it here.

Let us do the same thing on the second set of line $M_i$, with points $A', B', C', D', E', F',G'$ (complete quadrangle (Q')).
As shown here there is a projective mapping $f$ mapping (Q) onto a square [which amounts to say, if we prefer a "static" point of view,  that point $E$ has coordinates $(0,0)$, that  vertices $A,B,C,D$ have coordinates $(\pm 1,\pm 1)$ and that points $G$ and $F$ are sent to infinity].
Let us call $f'$ the similar projective mapping senting (Q') onto a square.
Taking  $\varphi := f'^{-1} \circ f$ gives the projective transformation mapping (Q) onto (Q'), therefore mapping all lines $L_i$ onto the corresponding lines $M_i$.
