Given two projective references (coordinate frames) in $\mathbb{RP}^2$: $R_1 = \{(-1:-2:-1), (2:2:2), (0:0:3); (2:3:3)\}$ $R_2 = \{(-6:-10:-5), (-1:-3:1), (3:4:7); (-4:-9:3)\}$ Determine the points of $\mathbb{RP}^2$ that have the same homogenous coordinates in both references.

My idea was calculating the matrix changing coordinates from one frame to the other and then looking for the vectors this matrix leaves invariant. Unfortuntely i already fail at calculating the matrix, i feel like i need to calculate an adapted basis for each of the reference systems and than calculate a change matrix for these but am unsure how to do that and even more feel like i have still not completely understood the idea of a projective reference. Thus appreciate any help both for solving the problem and regarding an explanation of references and their interplay with homogenous coordinates and the associated vector space very much. Thanks in advance!

  • $\begingroup$ What is a "projective reference system"? Google finds nothing. $\endgroup$ – Somos May 14 '18 at 13:01
  • $\begingroup$ @Somos i took the notation from Casas-Alvero, E. - Analytic Projective Geometry, A projective reference (or coordinate frame) of a projective space $\mathbb{P}_n$ is an ordered set of $n+2$ points $(p_0,...,p_n;A)$ satisfying any $n+1$ points extracted from $\{p_0,...,p_n,A\}$ are independent. (Compare chapter 2.1) $\endgroup$ – user526159 May 14 '18 at 13:31

The book by Hongyu Guo, Modern Mathematics and Applications in Computer Graphics and Vision on the bottom half of page 142 states:

2.4 Projective Frames

Any four points $A,B,C,D$ in a projective plane are said to be in general position, if no three points lie on the same line.

Theorem 3. (Projective frame) Given any four points $A,B,C,E$ in general position in a projective plane $P^2(V)$, we can choose a basis for $V$ in such a way that $A,B,C,E$ has homogeneous coordinates $(1,0,0), (0,1,0),(0,0,1)$ and $(1,1,1)$ respectively, and the projective coordinates of any other point is uniquely determined up to a scaling factor.

In a projective plane $P^2(V)$, we choose any four points $A,B,C$ and $E$ in general position and we choose a basis for $V$ determined in such a way by these four points $A,B,C$ and $E$, as described in Theorem 3. We say we have established a projective coordinate system, or projective frame, in the projective plane. $A,B$ and $C$ are called the fundamental points and $E$ is called the unit point. Hence in a projective plane, any four points in general position uniquely determine a projective coordinate frame.

The idea is that in Eucliden space any four points are linearly dependent and the dependency can be found using linear algebra. Applying this to your two sets of point, first let $\;p_0 = (2,3,3),\;$ $\;p_1 = (-1,-2,-1),\;$ $p_2 = (2,2,2),\;$ $p_3 = (0,0,3).\;$ Then $\;p_0 = -p_1 + \frac12 p_2 + \frac13 p_3.\;$ Second, let $\;q_0 \!=\! (-4,-9,3),\;$ $q_1 \!=\! (-6,-10,-5),\;$ $q_2 \!=\! (-1,-3,1),\;$ $q_3 \!=\! (3,4,7).\;$ Then $\;q_0 \!=\! q_1+q_2+q_3.$

In the first case, let ${\bar p_0} \!:=\! p_0,\;$ ${\bar p_1} \!:=\! -p_1 \!=\! (1,2,1),\;$ ${\bar p_2} \!:=\! \frac12 p_2 \!=\! (1,1,1),\;$ ${\bar p_3} \!:=\! \frac13 p_3 \!=\! (0,0,1).\;$ Now, $\;{\bar p_0} \!=\! {\bar p_1} \!+\! {\bar p_2} \!+\! {\bar p_3}.\;$ The 1st projective frame is $\;\{{\bar p_1},{\bar p_2},{\bar p_3},{\bar p_0}\}.\;$ The 2nd is $\;\{q_1,q_2,q_3,q_0\}.\;$

With each projective frame we can find homogeneous coordinates of any point. In the first frame we have, $\;(x,y,z) = (-x+y){\bar p_1} + (2x-y){\bar p_2} + (-x+z){\bar p_3}.\;$ In the second frame we have, $\;(x,y,z)25 = (-25x+10y+5z)q_1 + (50x-27y-6z)q_2 + (-25x+11y+8z)q_3.\;$

What points have the same homogeneous coordinates in both frames? Solving the equation $$\;(-x+y, 2x-y, -x+z)t = (-25x+10y+5z, 50x-27y-6z, -25x+11y+8z)\;$$ gives three solutions, namely, $\;(x,y,z,t) \in \{(1,0,0,25), (1,1,3,5), (7,12,6,-5)\}.\;$ For example, $\; (1,1,3) = 0{\bar p_1} + 1{\bar p_2} + 2{\bar p_3},\;$ and $\; (1,1,3)5 = 0q_1 + 1q_2 + 2q_3.\;$

  • $\begingroup$ Thank you very much! So i get it right that if $E$ is the standard reference a point $p = [x,y,z]_E = [(-x+y), (2x + y), (-x+z)]_{R_1} =\frac{1}{25} [(-25x+10y+5z),(50x-27y-6z),(-25x+11y+8z)]_{R_2}$ if i write coordinate vectors instead of parametric equations? $\endgroup$ – user526159 May 14 '18 at 17:37
  • 1
    $\begingroup$ Sure, you got it. No real difference. $\endgroup$ – Somos May 14 '18 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.