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I'm learning projective geometry and need help with the following problem :

Consider three distinct "points" of the projective space $\mathbb R P^1$, that is three distinct lines $l_1, l_2, l_3$ of $R^2 -\{0\}$ (the plane minus $0$). $(1)$ Show that we can choose homogeneous coordinates $[X_1 : X_2], [Y_1:Y_2]$ and $[Z_1:Z_2]$ for $l_i, i = 1, 2, 3$ such that

$$Z_1 = X_1 + Y_1, \quad Z_2 = X_2 + Y_2.$$

$(2)$ Find such coordinates for the lines $l_1 : -x+2y=0, \, l_2 = 2x+y = 0, \, l_3 : 5x-3y = 0.$

I'm sorry for the lack of effort on my part but I was not able to do much with this problem. I did go back several times through my notes but I don't know where to start here. Any help would be appreciated.

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A given line $l_1$ in $\mathbb{RP}^1$ can be specified by any homogeneous coordinate $[\lambda_1 X_1 : \lambda_1 X_2]$, with $\lambda_1$ nonzero. So the desired condition amounts to finding $\lambda_1,\lambda_2$, and $\lambda_3$ so that

$$ X_1 \lambda_1 + Y_1 \lambda_2 = Z_1 \lambda_3 \\ X_2 \lambda_1 + Y_2 \lambda_2 = Z_2 \lambda_3 $$

In other words, your problem amounts to solving a system of linear equations.


I'll take things a bit farther. If we let

$$ A= \begin{bmatrix} X_1 & Y_1\\ X_2 & Y_2 \end{bmatrix}, \quad x=\begin{bmatrix} \lambda_1 \\ \lambda_2 \end{bmatrix}, \quad b=\begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} $$

then the linear equation $Ax=b$ has a unique solution (why?) and yields a solution to your problem (why?).

Let me know if you'd like any further comments.

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  • $\begingroup$ Thank you very much for your answer. I'm still having difficulties to understand the theory of projective geometry and homogeneous coordinates but I get some intuition behind your answer. I think there is a small typo in the second row of the system of equations, it should be $Y_2 \lambda_2$. The homogeneous coordinates $[X_1 : X_2]$ correspond to a point on the line $\mathscr l_1$, right? So for the second question I could choose $[X_1 : X_2] = [2 : 1], [Y_1 : Y_2] = [1 : -2]$ and $[Z_1 : Z_2] = [3 : 5]$. Then set up the system of equations and solve for $\lambda_i$. Is that correct? $\endgroup$ – user347616 Jun 7 '17 at 17:31
  • $\begingroup$ Yep that's correct! Typo fixed. $\endgroup$ – Max Jun 7 '17 at 17:58
  • $\begingroup$ Ok, great! The only part which is not $100$% clear to me is for question $(1)$. Do I need to prove that the system you wrote always have a solution? Since we are not given any specific points in $(1)$ I don't know how to demonstrate this property. $\endgroup$ – user347616 Jun 7 '17 at 18:17
  • $\begingroup$ Sure, what do you know about when linear systems have solutions? $\endgroup$ – Max Jun 7 '17 at 18:24
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    $\begingroup$ That's the right idea. Try thinking about it in terms of linearly independent vectors. $\endgroup$ – Max Jun 7 '17 at 18:40
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OK, if you wish one more answer....

The homogeneous coordinates of a point $\mathscr l$ of the projective space $\Bbb RP_1$ are given by a pair $[T_1:T_2]$ of real numbers, such that not both $T_1$ and $T_2$ are zero. Moreover, if we multiply both $T_1$ and $T_2$ by any non-zero real number $\lambda$ then we obtain the homogeneous coordinates $[\lambda T_1: \lambda T_2]$ of the same point $\mathscr l$. This is all that you need to solve your problem, :-) so let’s start.

(1) Assume that we have three distinct points $\mathscr l_1$, $\mathscr l_2$, and $\mathscr l_3$ of $\Bbb RP_1$. Let the point $\mathscr l_1$ is given to us by its homogeneous coordinates $[x_1:x_2]$, the point $\mathscr l_2$ by $[y_1:y_2]$, and the point $\mathscr l_2$ by $[z_1:z_2]$. So we have to show that there exist non-zero real numbers $\lambda_1$, $\lambda_2$, and $\lambda_3$ such that

$$X_1=\lambda_1 x_1\mbox{, } X_2=\lambda_1 x_2,$$

$$Y_1=\lambda_2 y_1\mbox{, } Y_2=\lambda_2 y_2,$$

$$Z_1=\lambda_3 z_1\mbox{, } Z_2=\lambda_3 z_2,$$

$$Z_1=X_1+Y_1\mbox{, and } Z_2=X_2+Y_2.$$

Thus we have to show that this system of equations has a solution. This means that there exist non-zero real numbers $\lambda_1$, $\lambda_2$, and $\lambda_3$ such that

$$\lambda_1 x_1+\lambda_2 y_1-\lambda_3 z_1=0$$ $$\mbox{and}$$ $$\lambda_1 x_2+\lambda_2 y_2-\lambda_3 z_2=0.$$

We can tackle this problem as a question of solvability of a homogeneous system of linear equations, but there is a short-cut. This system means that the vector $(\lambda_1, \lambda_2,-\lambda_3)$ is orthogonal to both vectors $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. Such a property has their vector product

$$(x_1,y_1,z_1)\times (x_2,y_2,z_2)=\left|\begin{matrix}\bf{i} & \bf{j} & \bf{k}\\ x_1 & y_1 &z_1 \\ x_2 & y_2 &z_2 \end{matrix}\right|=(\lambda_1, \lambda_2, -\lambda_3).$$

It remains to check that $\lambda_i$’s are non-zero. This holds because the points $\mathscr l_i$ are distinct. For instance, $\lambda_1= \left|\begin{matrix} y_1 &z_1 \\ y_2 &z_2 \end{matrix}\right|=y_1z_2-y_2z_1$. If $\lambda_1$ is zero then $[y_1:y_2]z_2=[z_1:z_2]y_2$, which is impossible because points $\mathscr l_2$ and $\mathscr l_3$ are distinct.

(2) In this case we have $(x_1,y_1,z_1)=(2,-1,3)$ and $(x_2,y_2,z_2)=(1,2,5)$. Thus

$$(\lambda_1, \lambda_2, -\lambda_3)=(x_1,y_1,z_1)\times (x_2,y_2,z_2)=$$ $$ (2,-1,3)\times (1,2,5) =\left|\begin{matrix}\bf{i} & \bf{j} & \bf{k}\\ 2 & -1 & 3 \\ 1 & 2 & 5 \end{matrix}\right|=(-11,-7,6).$$

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