# Determining the matrix of a projection

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?

• $A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$? – amd Jan 11 at 21:07
• 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. – amd Jan 11 at 21:55
• 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. – amd Jan 11 at 21:57
• I’m guessing that when you write $IR$, you actually mean $\mathbb R$. Use \mathbb R to get this symbol. – amd Jan 11 at 21:58
• 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. – MvG Jan 14 at 23:01