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So I was trying this exercise, but I have a lot of doubts: On a projective plane $P^2(\Bbb{R})$ I have these points: $$P_0=\{0,0,1\},\quad P_1=\{0,1,-1\},\quad P_2=\{1,-1,0\},\quad P_3=\{1,1,-3\}$$and $$Q_0=\{1,0,0\},\quad Q_1=\{1,0,-1\},\quad Q_2=\{0,1,-1\},\quad Q_3=\{0,1,1\}$$ I have to say if exist a projective transformation $f:P^2(\Bbb{R})\rightarrow P^2(\Bbb{R})$ such that

$f(P_j)=Q_j$, $\qquad j=0,1,2,3$

and, If $f$ exist, I have to show it.

So, I defined an arbitrary Matrix

$$A=\begin{bmatrix}a & b &c\\d & e &f\\g & h &i\end{bmatrix}$$

Now I calculate $A*P_0=Q_0$

$$\begin{bmatrix}a & b &c\\d & e &f\\g & h &i\end{bmatrix} \cdot \begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$$

And I find $ \left\{ \begin{array}{c} c=1 \\ f=0 \\ i=0 \end{array} \right. $

I do the same whit the others:

$A*P_1=Q_1 \Rightarrow$ $ \left\{ \begin{array}{c} b-c=1 \\ e-f=0 \\ h-i=-1 \end{array} \right. $

$A*P_2=Q_2 \Rightarrow$ $ \left\{ \begin{array}{c} a-b=0 \\ d-e=1 \\ g-h=-1 \end{array} \right. $

$A*P_3=Q_3 \Rightarrow$ $ \left\{ \begin{array}{c} a+b-3c=0 \\ d+e-3f=1 \\ g+h-3i=1 \end{array} \right. $

But I can't find any solution for this system... Does it mean that $f$ doesn't exist? And when it says "show $f$ explicitly" what does it mean exactly?

Arianna
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    You can’t use strict equality when you’re working with homogeneous matrices. By doing so you’re effectively limiting yourself to an affine transformation, and there is none that performs the desired mapping. So, for instance, your first constraint should be $AP_1=c_1 Q_1$ for some unknown scalar $c_1$. Since you’re working in $\mathbb P^2$ you can avoid introducing these extra unknowns by using cross products, e.g., $Q_1\times AP_1=0$. – amd May 24 '20 at 18:31
  • That said, this is basically a duplicate of https://math.stackexchange.com/q/296794/265466. See in particular this answer. – amd May 24 '20 at 18:34
  • I saw that question, but I don't know hot to apply it to my exercise – Arianna May 24 '20 at 21:25
  • For @amd I tried in that way and I found the matrix $A=\begin{bmatrix}3 & 3 &2\-1 & 0 &0\0 & -1 &0\end{bmatrix}$, is this possible? And I don't understand: the projective transformation is a matrix? – Arianna May 24 '20 at 21:28
  • Existence is assured by the fundamental theorem of projective geometry. A transformation is not a matrix, although it can be represented by one. Why were you trying to compute this matrix in the first place if not to find $f$ explicitly? – amd May 24 '20 at 22:40

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