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Note: There are other questions on this site similar to this one, but they are either unanswered or have a different definition of "lines remain lines".

3Blue1Brown uses the following intuitive definition of linearity in his video series on linear algebra:

A linear transformation is a transformation which fixes the origin and keeps lines straight.

I mainly care about this as an intuition for the plane, so I do not really want to generalise it to $\mathbb R^n$. So my question is, if $f\colon\mathbb R^2\to\mathbb R^2$ is a map such that

  1. $f(\boldsymbol 0) = \boldsymbol 0$, and
  2. for all $\boldsymbol a,\boldsymbol b\in\mathbb R^2$, there exist $\boldsymbol u, \boldsymbol v\in\mathbb R^2$ such that $f(\boldsymbol a+\mathbb R\boldsymbol b) = \boldsymbol u+\mathbb R\boldsymbol v$,

can I show that $f$ is a linear map on $\mathbb R^2$? (here $\boldsymbol a+\mathbb R\boldsymbol b$ denotes the obvious coset $\{\boldsymbol a+t\boldsymbol b:t\in\mathbb R\}$).

I've been playing around with these two properties a lot, but 2 doesn't seem to be strong enough to allow me to go from statements about lines to statements about individual vectors in an obvious way.

Luke Collins
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1 Answers1

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Let $k:\def\R{\mathbb R}\R\rightarrow\R$ be any non-linear bijection such that $k(0)=0$ (for instance, $k(x)=x^3$, or even something non-continuous). Define $f:\R^2\rightarrow\R^2$ by $a\boldsymbol x+b\boldsymbol y\mapsto k(a)\boldsymbol x$. Then every line (except vertical ones, which is mapped to a point) is mapped to the $x$-axis, but this map is obviously not linear.

UPDATE: If we assume bijectivity in addition (and do not assume $f(\boldsymbol0)=0$), the conditions give rise to a collineation, which I found in this linked question thanks to pregunton. According to Wikipedia, in the case of $\R^n(n\geq2)$:

In general, some collineations are not homographies, but the fundamental theorem of projective geometry asserts that is not so in the case of real projective spaces of dimension at least two.

So we arrive at the notion of a homography.

Tesla Daybreak
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  • what condition could be added in order to make the intuition correct? – Luke Collins Jan 14 '21 at 18:54
  • @LukeCollins Injectivity certainly does it. I'm not sure if that is the best condition though. –  Jan 14 '21 at 19:01
  • Yes, injectivity doesn't really appeal to the geometric intuition. – Luke Collins Jan 14 '21 at 19:03
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    @LukeCollins I feel like injectivity can be geometrically intuitive - every line is transformed into a distinct line and all the lines in the space are permuted/shuffled around. – rubikscube09 Jan 14 '21 at 19:33
  • You can go further, as we're only considering homographies mapping the affine plane to itself: If a collineation is injective, then parallel lines remain parallel, and you can use some configurations to show it's rational-ratio-preserving, and hence ratio-preserving by some weak form of continuity. Thus any affine ratio-preserving collineation is easily proved to be affine, and now you can add in $f(0)=0$ to get the desired linearity. –  Jan 15 '21 at 01:49