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Let $L$, $G \subset \mathbb{P}^2$ be lines. Show that there exists a projective change of coordinates $T$, such that $T(L)=G$.

This is how we defined a projective change of coordinates in $\mathbb{P}^2$:

If $S$: $\mathbb{A}^3 \to\mathbb{A}^3$ is a linear affine change of coordiantes, then $S$ maps lines through the origin to lines through the origin again. So $S$ induces a map $S'$: $\mathbb{P}^2 \to\mathbb{P}^2$ which is called a projective change of coordinates.

I am not sure how to tackle this problem since I haven't got much more than the definition. I would appreciate any help.

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    $\begingroup$ Hint: If points in $\Bbb P^2$ correspond to lines through the origin in $\Bbb A^3$, what do lines in $\Bbb P^2$ correspond to? Can you figure out a way to map one of these objects on to another one of these objects? $\endgroup$ – KReiser Dec 12 '18 at 21:04
  • $\begingroup$ If I am not mistaken, then the lines of $\mathbb{P}^2$ are the planes of $\mathbb{A}^3$ that contain the $0$ (which are the $2$-dimensional subspaces of $\mathbb{A}^3$ ) $\endgroup$ – get rekt m8 Dec 12 '18 at 21:18
  • $\begingroup$ @KReiser Is there an easy argument why I can map (with an afiine change of coordinates) any plane in $\mathbb{A}^3$ going through $0$, to any other such plane? $\endgroup$ – get rekt m8 Dec 12 '18 at 21:20
  • $\begingroup$ Yes, you're correct that they're the planes. You ought to be able to do it with a linear map, right? $\endgroup$ – KReiser Dec 12 '18 at 21:27
  • $\begingroup$ @KReiser well, it seems at least intuitive that I can do it with a linear map, since all I would need to do is rotate the plane. Not sure if thats an argument for it being linear. Also: would a linear map be enough? after all I need a linear change of coordinates (is that the same in that case)? $\endgroup$ – get rekt m8 Dec 12 '18 at 21:33

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