I've been trying to wrap my head around this for a while now. Apparently, a map is a linear map if it preserves scalar multiplication and addition. So let's say I have the mapping:
$$f(x) = (x,x)$$
This is not a mapping to a lower or equal dimension, but to a higher one. Yet it seems to preserve scalar multiplication and addition:
$$f(ax) = (ax,ax) = a(x,x) = af(x)$$
$$f(x+y) = (x+y,x+y) = (x,x) + (y,y) = f(x) + f(y)$$
I must have made an error in my logic somewhere, but I can't seem to find it. Or are linear maps simply defined this way? I would really appreciate to know this.
Best Answer
You can indeed have a linear map from a "low-dimensional" space to a "high-dimensional" one - you've given an example of such a map, and there are others (e.g. $x\mapsto (x, 0)$).
However, such a map will "miss" most of the target space. Specifically, given a linear map $f: V\rightarrow W$, the range or image of $f$ is the set of vectors in $W$ that are actually hit by something in $V$: $$im(f)=\{w\in W: \exists v\in V(f(v)=w)\}.$$ This is in contrast to the codomain, which is just $W$. (The distinction betwee range/image and codomain can feel slippery at first; see here.)
The point is that $im(f)$ is a subspace of $W$, and always has dimension $\le$ that of $V$. (Proof hint: show that if $I\subseteq im(f)$ is linearly independent in $W$, then $f^{-1}(I)$ is linearly independent in $V$.) So in this sense, linear maps can't "increase dimension".