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I have two quadrilaterals in a projective plane ($\mathbb{R}^{3\times 1}$) with standard basis $E=(e_0,e_1,e_2)$:

$A=\mathbb{R}e_0$

$B=\mathbb{R}e_1$

$C=\mathbb{R}e_2$

$D=\mathbb{R}(e_0+e_1+e_2)$

How to find the collineation mapping this quadrilateral to the following one:

$A=\mathbb{R}(-1,1,2)^T$

$B=\mathbb{R}(2,-5,1)^T$

$C=\mathbb{R}(1,2,-3)^T$

$D=\mathbb{R}(0,0,1)^T$

I found theorems which prove that there is always one (and only one) such collineation. But I wonder how I can calculate a specific on if I have such an numerical example?

1 Answers 1

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Send $e_{0} \mapsto a\cdot(-1,1,2)^{T}$, $e_{1} \mapsto b\cdot (2,-5,1)^{T}$, $e_{2} \mapsto c \cdot (1,2,-3)^{T}$.

This gives you a matrix $A = \begin{bmatrix} -a&2b&c\\a&-5b&2c\\2a&b&-3c \end{bmatrix}$

Now you need to have that $$A(e_{0}+e_{1}+e_{2}) = A \begin{bmatrix} 1\\1\\1\end{bmatrix} = \begin{bmatrix} -a+2b+c\\a-5b+2c\\2a+b-3c \end{bmatrix}$$ is a scalar multiple of $(0,0,1)^{T}$. So you can set up equations to find $a$, $b$, and $c$ (you can either assume WLOG that $a=1$, or that $A(e_{0}+e_{1}+e_{2}) = (0,0,1)^{T}$, whichever is more convenient).

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    Thx. Two questions: is it not -a in the first row/column in A. And what is the difference in solving such a problem in affine plane and in projective plane? – 2017-01-23
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    Ah yes, thanks I missed the negative sign. In a projective plane we are working with homogeneous coordinates (notice that we only need $Ae_{0}$ to be a *multiple* of $(-1,1,2)$. In an affine plane we would need to map exactly to this vector; on the other hand we would also be able to work with *translations* in an affine plane. – 2017-01-23