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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?

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  • $\begingroup$ $A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$? $\endgroup$ – amd Jan 11 at 21:07
  • $\begingroup$ The wording of this problem is troublesome. “Let $f$ be the map $f:l_2\to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,\mathbb R)$), but asks for a matrix in $PGL(3,\mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space. $\endgroup$ – amd Jan 11 at 21:55
  • $\begingroup$ I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map. $\endgroup$ – amd Jan 11 at 21:57
  • $\begingroup$ I’m guessing that when you write $IR$, you actually mean $\mathbb R$. Use \mathbb R to get this symbol. $\endgroup$ – amd Jan 11 at 21:58
  • $\begingroup$ I can imagine two ways to read the question. Either you want $Q=[s'\cdot a'+t'\cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2\times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4\times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $\textrm{PGL}(3,\mathbb R)$ playing a role, since it has equivalence classes of invertible $3\times3$ matrices. Please double-check the question statement. $\endgroup$ – MvG Jan 14 at 23:01

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