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I want to prove that all translations on an affine straight line are induced by projective transformations of the projective extension.

If we look at $\mathbb{R}$ as our affine straight line, then the extension is $\mathbb{P}^1(\mathbb{R})=\mathbb{P}(\mathbb{R}^2)$. Furthermore, I know $\mathbb{R}\subset\mathbb{P}(\mathbb{R}^2)$.
So a projective transformation $\mathbb{P}(\mathbb{R}^2)\rightarrow\mathbb{P}(\mathbb{R}^2)$ restricted to $\mathbb{R}$ should be a translation $f:\mathbb{R}\rightarrow\mathbb{R}$. But all projective transformations are induced by an injective linear map $\mathbb{R}\rightarrow\mathbb{R}^2$.

So, what I need to prove now is that for all translations on $\mathbb{R}$ there is a $2\times2$-matrix that induces this. How do I do that?
(In other words: how do these matrices look?)

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Yes, you can use homogeneous coordinates to represent all translations by $\tau$ on $\mathbb{R}$ as $$ T = \begin{pmatrix} 1 & \tau \\ 0 & 1 \end{pmatrix} $$ Your vectors use homogeneous coordinates as well $$ \begin{pmatrix} x \\ 1 \end{pmatrix} $$ It then works like $$ \begin{pmatrix} x' \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & \tau \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ 1 \end{pmatrix} = \begin{pmatrix} x + \tau \\ 1 \end{pmatrix} $$

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  • $\begingroup$ When you say translation by $\tau$ do you mean $f:\mathbb{R}\rightarrow\mathbb{R},f(x)=x+\tau$? $\endgroup$ – user316173 Mar 17 '16 at 14:28
  • $\begingroup$ Yes, indeed. This system is used heavily in 3D computer graphics and hardware, because translations can be specified as $4\times 4$ matrices and composition of transformations is then reduced to matrix multiplication. $\endgroup$ – mvw Mar 17 '16 at 14:30
  • $\begingroup$ Yes I see it now! Thank you! $\endgroup$ – user316173 Mar 17 '16 at 14:33
  • $\begingroup$ Oh. not unimportant: I do not know the definition of projective transformations, you must decide if this fits your need. $\endgroup$ – mvw Mar 17 '16 at 14:34

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