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I know I do it wrong but where is the mistake???
In E3* are given the points $A(1,0,0,0)$, $B(0,1,0,0)$, $D(0,0,1,0)$, $O(0,0,0,1)$ and $E(1,1,1,2)$.
The linear transformation $\Phi$ operates according to the rules:
$A\to A$; $B\to B$; $D\to D$; $O\to E$; and $E\to O$.

Find the matrix $C$ of the linear transformation (the analytical representment of $\Phi$). That means we have $5$ points in $3D$ space with homogeneous coordinates and we have to find the linear transformation. I start with this: $\Phi(A)=A \implies p_1(1,0,0,0) = C(1,0,0,0) \implies c_{11}=p_1, c_{21}=0, c_{31}=0, c_{41}=0$
(these are elements of matrix $C$);
$\Phi(B)=B \implies p_2(0,1,0,0) = C(0,1,0,0) \implies c_{12}=0, c_{22}=p_2, c_{32}=0, c_{42}=0$;
$\Phi(D)=D \implies p_3(0,0,1,0) = C(0,0,1,0) \implies c_{13}=0, c_{23}=0, c_{33}=p_3, c_{43}=0$;
$\Phi(O)=E \implies p_4(0,0,0,1) = C(1,1,1,2) \implies c_{14}=p_4, c_{24}=p_4, c_{34}=p_4, c_{44}=p_4??? $ ;
$\Phi(E)=O \implies p_5(0,0,0,1) = C(1,1,1,2)$

$c_{11} + c_{12} + c_{13} + c_{14} = 0$;
$c_{21} + c_{22} + c_{23} + c_{24} = 0$;
$c_{31} + c_{32} + c_{33} + c_{34} = 0$;
$c_{41} + c_{42} + c_{43} + c_{44} = 0$;

$\implies$
$p_1 + p_4 = 0$;
$p_2 + p_4 = 0$;
$p_3 + p_4 = 0$;
$p_4 = 1$

$\implies$ $p_1=p_2=p_3=-1$.

Where is the mistake?

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    $\begingroup$ Can five points in homogeneous 3-d space (which sits in 4-space) be linearly independent? If not it would seem the transform should be done via getting a spanning set from the original 5 points, and going from there. $\endgroup$
    – coffeemath
    Jun 29 '13 at 0:25
  • $\begingroup$ I now withdraw the above comment, since in homogeneous coordinates there are arbitrary constant multiples to think about. It can be done (see my answer...). $\endgroup$
    – coffeemath
    Jun 29 '13 at 7:35
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I see you are working with arbitrary multiples of vectors (as usual in homogeneous coordinates), so that I withdraw my comment above about dimensions. It has to be done more carefully, as you do it in your work shown, keeping track of possible multipliers.

From your steps:

$Φ(B)=B⟹p_2 (0,1,0,0)=C(0,1,0,0)⟹c_{12} =0,c_{22} =p 2 ,c_{32} =0,c_{42} =0$ ;

$Φ(D)=D⟹p_3 (0,0,1,0)=C(0,0,1,0)⟹c_{13} =0,c_{23} =0,c_{33} =p_ 3 ,c_{43} =0$ ;

$Φ(O)=E⟹p_4 (0,0,0,1)=C(1,1,1,2)⟹c_{14} =p_4 ,c_{24} =p_4 ,c_{34} =p_4 ,c_{44} =p_4$ ??? ;

Here the first two look OK, but since $p_4(0,0,0,1)=(0,0,0,p_4)$ it would seem that $c_{14}=c_{24}=c_{34}=0$ and only $c_{44}=p_4$.

Or maybe I have things backwards and the first three $c_{i4}$ are $p_4$, but in that case the last one should maybe be $2p_4$. Well, something must match the $(1,1,1,2)$ shape of vector $E$...

This correction may affect the solving of the rest of the question; I'll look at it again if you don't respond with something.

ADDED: By monkeying around with the above equations I found the matrix $M$ given by

\begin{bmatrix}2 & 0 & 0 & -1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 2 & -1 \\ 0 & 0 & 0 & -2 \end{bmatrix} which, viewing your four coordinate points as column vectors and multiplying on the right, send each point to a multiple of what it should be. It sends the first three vectors $A,B,D$ to twice themselves, and the point $(0,0,0,1)$ maps to $(-1,-1,-1,-2)=-1\cdot(1,1,1,2)$, while on the other hand the point $(1,1,1,2)$ maps to $(0,0,0,-4)=-4(0,0,0,1).$ I assume since you're working in homogeneous coordinates that this matrix (or scalar multiples of it) is what you want.

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A linear transformation in 3-space is determined by its images on 3 L.I. vectors. I think you have assumed too many (five) points.

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  • $\begingroup$ It's not usual 3-space, but instead it's the set of four-tuples with identification by multiplication by nonzero scalars. So effectively there is more room (I initially made the same "usual 3d" assumption, but OP makes it clear in the work shown that the scalar multiple identifications make a difference. $\endgroup$
    – coffeemath
    Jun 29 '13 at 7:56
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@coffeemath definitely this is the answer but I didn't understand how you solved it.. So again I begin with this: Φ(A)=A⟹p1(1,0,0,0)=C(1,0,0,0)⟹c11=p1,c21=0,c31=0,c41=0 (these are elements of matrix C); Φ(B)=B⟹p2(0,1,0,0)=C(0,1,0,0)⟹c12=0, c22=p2, c32=0, c42=0; Φ(D)=D⟹p3(0,0,1,0)=C(0,0,1,0)⟹c13=0, c23=0, c33=p3, c43=0; Φ(O)=E⟹p4(1,1,1,2)=C(0,0,0,1)⟹c14=p4, c24=p4, c34=p4, c44=2p4 ; and then what ?? Φ(E)=O⟹p5(0,0,0,1)=C(1,1,1,2);

p1+p4 = 0; p2+p4=0; p3+p4 = 0; 2p4=1 but these don't get us to your answer, which actually is the right answer..

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  • $\begingroup$ I think your use of $\Phi(E)=O$ should be re-thought. The equations on the $c_{ij}$ end up forcing the values $p_1,p_2,p_3,p_4$ to appear in the desired matrix at the positions you note (with a $2p_4$ in the lower right). This matrix already sends each of $A,B,D,O$ where you want them (or to scalar multiples, which is the same thing in homogeneous case). Now you just want to make that matrix carry $E$ back to $O$. So equations are $p1+2p_4=0$, $p_2+2p_4=0$, $p_3+2p_4=0$, and finally $2p4=c$ where $(0,0,0,c)$ is any multiple of $(0,0,0,1)$. In short you need $2p_4$ in there... $\endgroup$
    – coffeemath
    Jun 29 '13 at 10:21
  • $\begingroup$ @coffeemath, thank you very much! $\endgroup$
    – Nicholas
    Jun 29 '13 at 10:54
  • $\begingroup$ @coffeemath, can I ask you for help with one more example? this time in Ф:R^2 -> R^2; points are given: A(1,0,0)->A'(2,1,0); B(0,1,0)->B'(1,2,0); O(0,0,1)->O'(0,0,1); E(1,1,1)->E'(3,3,1); again in homogeneous coordinates, I have to find the matrix C of the linear transformation. $\endgroup$
    – Nicholas
    Jun 29 '13 at 11:27
  • $\begingroup$ This time the matrix ends up mapping everything even without scalar multiples: $M=[[2,1,0],[1,2,0],[0,0,1]]$ (those are the three rows) maps all four points (again regarded as column vectors to multiply $M$ on the right) into the required images. $\endgroup$
    – coffeemath
    Jun 29 '13 at 12:31
  • $\begingroup$ so in this case: p1(2,1,0) = C(1,0,0) => c11=???, c21=???, c31=0; p2(1,2,0)=C(0,1,0) => c12=???, c22=???, c32=0; p4(3,3,1)=E(1,1,1) => c13=???, c23=???, c33=??? p3(0,0,1)=C(0,0,1); and what are the eqations??? I can't catch it.. $\endgroup$
    – Nicholas
    Jun 29 '13 at 12:55

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