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Let $f: V \rightarrow V$ be a bijective map of a vector space to itself that preserves one-dimensional affine subspaces. Is $f$ already the composition of some invertible matrix and a translation? My intuition says yes, but writing down a matrix didn't work. Thanks!

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By "preserves" do you mean that the image of each one-dimensional affine subspace is contained in a one-dimensional affine subspace? Is the base field arbitrary? – Jonas Meyer May 24 '12 at 5:30
In one dimension this is wrong in general, as $x \mapsto x^3$ is bijective from $\mathbb R$ to $\mathbb R$. – martini May 24 '12 at 7:13
Jason Jeffers proves that this is indeed true in dimensions $n > 1$ in his article "Lost Theorems of Geometry", American Mathematical Monthly, Vol. 107, No. 9 (Nov., 2000) (pp. 800-812) – martini May 24 '12 at 9:52
@JonasMeyer: Sorry, I was a too imprecise: The image of a one-dimensional affine subspace is again a one-dimensional affine subspace. The ground field is arbitrary, but the dimension is indeed positive. Thanks martini, I will have a look into that. – Algol May 25 '12 at 18:15
I read Jeffers' article, and while it was very interesting, he works over the affine plane. I am not sure if this holds over arbitrary fields, or at least his proof doesn't seem to be applicable to the general case. Does anybody have any more information? I may assume that the ground field is finite, if this helps. – Algol Jun 13 '12 at 15:22

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