I have a question about the correctness of my ideas regarding the following exercise.
Define
$A_0=[(1,0,0,0)]$
$B_0=[(0,1,0,0)]$
$A_1=[(0,0,1,0)]$
$B_1=[(0,0,0,1)] \in \mathbb RP^3$ where $\mathbb RP^3$ denotes real projective space.
Let $l_0$ be the line through $A_0$ and $B_0$ and $l_1$ be the line through $A_1$ and $B_1$. Let $l_2$ be an arbitrary line skew to both $l_0$ and $l_1$ and let $A_2=[a]$ and $B_2=[b]$ be two points of $l_2$ so that $P=[s \cdot a+ t \cdot b] $ is a general point of $l_2$. Let $\pi$ be the plane spanned by $l_0$ and P and define Q to be the intersection of $\pi$ with $l_1$. Let f be the map $f: l_2 \to l_1$ defined by $f(P)=Q.$ Find a matrix in $PGL(3,\mathbb R)$ corresponding to f.
My idea to find the matrix was this: We know that the representation vector $x$ of point $X$ in $\pi$ can be written as
$$x=s_0 \cdot (1,0,0,0)+t_0 \cdot (0,1,0,0)+k_0(sa+tb)\tag1$$
for $s_0,t_0,k_0 \in \mathbb R$.
The respresentation vector $y$ of a point $Y$ on $l_1$ can be written as
$$y=s_1 \cdot (0,0,1,0) + t_1 \cdot (0,0,0,1)\tag2$$
for $s_1, t_1 \in \mathbb R$.
Since the line $l_2$ is determined by the points $A_2$ and $B_2$ it is sufficient to find a matrix $A=(a_{ij})$ that maps the representation vectors $a$ and $b$ onto representation vectors of the points $(\pi A_2)\cap l_1$ and $(\pi B_2)\cap l_1$. The representation vectors of the image points must satisfy the equations $(1)$ and $(2)$. So we get the following equations
$Aa=s_0' \cdot (1,0,0,0)+t_0' \cdot (0,1,0,0)+a$
$Aa=s_1' \cdot (0,0,1,0) + t_1' \cdot (0,0,0,1)$
$Ab=s_0'' \cdot (1,0,0,0)+t_0'' \cdot (0,1,0,0)+b$
$Ab=s_1'' \cdot (0,0,1,0) + t_1'' \cdot (0,0,0,1)$
Since the matrix $A$ has 16 entries we get 16 equations in total. Solving this system with respect to the entries of $A$ should give me the matrix I need.
My question is:
Is there an easier way to solve this than solving a system of 16 equations?