Alternate Characterization of Linearity This question is prompted by this video on matrices and linear transformations, which I highly recommend as a pedagogical tool.  In it, the author characterizes linear transformations in the following way (I'm paraphrasing and formalizing)

Define a line in a vector space to be a set of the form $L = \{u + t\,v: t \in \Bbb R\}$
  for some vectors $u$ and $v$ (note: $L$ may consist of a single point).  That is, $S$ is an affine subspace of dimension at most $1$.
A function $T: \Bbb R^n \to \Bbb R^m$ is linear if:
  
  
*
  
*$T(0) = 0$
  
*For any line $L \subset \Bbb R^n$, the image $T(L)$ is a line in $\Bbb R^m$

I like this definition because of its geometric appeal and the fact that it manages to "put the line in linear".
Of course, the traditional definition of a linear map is one which preserves linear combinations.

My Question:

  
*
  
*How should one prove that a function satisfying this definition preserves linear combinations?
  
*Can this be proven in a beginner-friendly way?
  

I'll admit I haven't really banged my head against this one, but here are my thoughts: it is equivalent to prove that a function that satisfies only the second condition (i.e. maps lines to lines) is an affine transformation, i.e. that it preserves affine combinations.  From there, it would suffice to note that if $T$ is affine, then $x \mapsto T(x) - T(0)$ is linear.
That being said, I don't see a quick way to handle that proof off the top of my head.  Moreover, if this really is the quickest way to reach a proof, it seems that proving this in linear algebra 101 is a bit too ambitious (which is not to say that this fact fails to be pedagogically useful).  I'm guessing a little real-analysis might have to come in at some point.

A note for myself and future visitors:
If we are also given that  is bijective, then this is a consequence of the fundamental theorem of projective geometry.
 A: Only a partial answer, but at least something. I will prove the following statement:

Assume that $X$ and $Y$ are any real vector spaces and that $f:X\to Y$ is so that for any $u,v\in X$ there are $u',v'\in Y$ such that $f(\{u+tv\ |\ t\in\mathbb R\})=\{u'+sv'\ |\ s\in\mathbb R\}$ and $f(0)=0$. Then $f$ satisfies $f(u+qv)=f(u)+qf(v),\ u,v\in X,q\in\mathbb Q$.

The assumption can be rewritten as follows: There are functions $a,b:X\to Y$ and $g:\mathbb R\to\mathbb R$ such that
$$f(u+tv)=a(u)+g(t)b(v),\qquad u,v\in X,t\in\mathbb R.$$
We first obtain from
\begin{align}
0&=f(0+0\cdot v)=a(0)+g(0)b(v), \\
0&=f(0+t\cdot 0)=a(0)+g(t)b(0), \\
0&=f(tv-tv)=a(tv)+g(-t)b(v), \\
0&=f(tv+t(-v))=a(tv)+g(t)b(-v),
\end{align}
that
\begin{align*}
g(0)b(v)&=g(t)b(0), \\
a(tv)&=-g(-t)b(v)=-g(t)b(-v),\qquad v\in X,t\in\mathbb R.
\end{align*}
Case 1: $g(0)\neq 0:$ Then $b(v)=b(0)$. If $b(0)\neq0$, then $g(t)=g(0)$ and $a(v)=-g(1)b(-v)=-g(0)b(0)$ and $f\equiv0$. If $b(0)=0$, then $a(v)=-g(1)b(-v)=0$ and again $f\equiv0$.
Case 2: $g(0)=0$: Then $a(u)=f(u)$ for all $u\in X$. In particular,
\begin{align*}
f(u+tv)&=f(u)+g(t)b(v), \\
f(u+v)&=f(u)+g(1)b(v),\qquad u,v\in X,t\in\mathbb R.\tag{1}
\end{align*}
If $g(1)=0$, then $f(v)=f(0+v)=f(0)=0$ and $f\equiv0$. We therefore assume that $g(1)\neq0$. Then letting $u=0$ in (1) gives $f(v)=g(1)b(v)$. On the other hand, letting $v=-u$ in (1) gives $f(-v)=-g(1)b(v)$ and hence $f(-v)=-f(v)$ and 
\begin{align}
   f(u+tv)=f(u)+\tfrac{g(t)}{g(1)} f(v),\qquad u,v\in X,t\in\mathbb R.\tag{2}
\end{align}
Suppose that $f\not\equiv0$. Then there is some $x\in X$ such that $f(x)\neq0$. From (2) we obtain 
\begin{align}
f((s+t)x)&=\tfrac{g(s+t)}{g(1)}f(x), \\
f((s+t)x)&=f(sx+tx)=f(sx)+\tfrac{g(t)}{g(1)}f(x)=\left(\tfrac{g(s)}{g(1)}+\tfrac{g(t)}{g(1)}\right)f(x)
\end{align}
and consequently $g(s+t)=g(s)+g(t)$ for all $s,t\in\mathbb R$. Thus, $g(q)=qg(1)$ for all $q\in\mathbb Q$ and
$$
f(u+qv)=f(u)+qf(v),\qquad u,v\in X,q\in\mathbb Q.
$$
