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Fine the highest $n$ for which the following statements are true and give a counterexample for $n+1$:

$V$ is an 1-dimensional vector space

$A$ is an 1-dimensional affine space

$P$ is an 1-dimensional projective space

All over $\mathbb{R}$.

Statements:

1) For all $v_1,v_2,\cdots , v_n \in V\backslash \{0\}$ (all different) and all $v_1^*,v_2^*,\cdots , v_n^* \in V$ (all different) there is a linear mapping $f:V\rightarrow V$ with $f(v_1)=v_1^*, f(v_2)=v_2^*, \cdots $.

2) For all $p_1,p_2,\cdots , p_n \in A$ (all different) and $p_1^*,p_2^*,\cdots , p_n^*\in A$ (all different) there is an affinity $g:A\rightarrow A$ with $g(p_1)=p_1^*, g(p_2)=p_2^*, \cdots $.

3) For all $Q_1,Q_2,\cdots , Q_n \in P$ (all different) and all $Q_1^*,Q_2^*,\cdots , Q_n^*\in P$ (all different) there is a projective collineation with $h:P\rightarrow P$ with $h(Q_1)=Q_1^*, h(Q_2)=Q_2^*, \cdots $.

1) should be true for all $n$? What about the other two?

1 Answers 1

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Hints:

  1. The image of a single nonzero vector in a one dimensional vector space $V$ determines uniquely the map. (And there is always a map for any image point.)

  2. For affine line, an affine map is determined by the images of 2 distinct points (think, translation + stretch factor).

  3. (This one is the trickiest) The image of 3 non-pairwise colinear lines (3 lines in general position) in a two dimensional vector space determines (uniquely up to scaling) a linear map, and the map exists if the images are distinct lines - moving 2 lines you can do using linear algebra and the idea of basis, the third you move by rescaling the basis vectors independently from eachother (which doesn't change the lines they span, but will move other lines). Then check how this descends to projective space.

These have nothing to do with the reals, they are true for any field k.

(I'm thinking of projective space as $k^2 \setminus{0} / k^*$ (the "space" of lies in the plane). )