Considering reals as vectors over rationals, can I have a linear map that has uncountably infinite domain? In miniature 12 in Thirty-three Miniatures: Mathematical
and Algorithmic Applications of
Linear Algebra, Matousek considers the reals as an infinite dimensional vector space over rational numbers, $\Bbb Q$. He defines a linear map $f$ from $V \to \Bbb R$ where $V$ is a space spanned by a finite basis. Therefore, $V$ is countably infinite. But it makes me wonder, why did he bother? Why couldn't he just define $f$ from $\Bbb R \to \Bbb R$ and still satisfying the conditions: $f(1)=1$ and $f(\pi)=-1$ (he uses an irrational number $x$ which I've replaced with $\pi$). Is there something preventing this? And if not, why did he bother defining the sub-space $V$?
 A: Since $1$ and $\pi$ are linearly independent over $\Bbb Q$, yes, such a linear map exists. But the usual proof of this follows from the fact that $\{1,\pi\}$ can be extended to a basis of $\Bbb R$ over $\Bbb Q$. Proving this requires the axiom of choice. However, if we're working on a finite-dimensional subspace of $\Bbb R$, the axiom of choice doesn't have to be assumed.
A: Defining a map from all of $\Bbb{R}$ to $\Bbb{R}$ is not so easy unless you can give a formula. Compare.
It is easy to write such a formula for the $\Bbb{Q}$-span of $\{1,\pi\}$:
$$q_1+q_2\pi\mapsto q_1-q_2$$
for all rational numbers $q_1,q_2$.
But how do you describe something like this for all of $\Bbb{R}$ when you don't have a $\Bbb{Q}$-basis at hand? We can't, really! That is, we cannot do that all without calling the Axiom of Choice for help. If we don't assume the Axiom of Choice we don't know whether $\Bbb{R}$ has a basis at all!
The $\Bbb{Q}$-linear maps $\Bbb{R}\to\Bbb{R}$ that we can define without the Axiom of Choice end up being $\Bbb{R}$-linear, like $x\mapsto 2x$. These are not interesting here.

Answering the question in the title. Let $\mathcal{B}$ be a basis of $\Bbb{R}$ over $\Bbb{Q}$ such that $1,\pi\in\mathcal{B}$. Standard linear algebra tells us that the function $s:\mathcal{B}\to\Bbb{R}$, $1\mapsto1$,
$\pi\mapsto -1$, $s\mapsto s$ for all $s\in\mathcal{B}\setminus\{1,\pi\}$, can be uniquely extended to a $\Bbb{Q}$-linear transformation $s$ from $\Bbb{R}$ to itself. We then have $s(1-\pi)=0$ so $s$ has a non-trivial kernel. But $\mathcal{B}\setminus\{\pi\}$ is contained in the image of $s$, so the image is uncountable.
A: Matousek is trying to give an answer that can be understood and accepted by students who have never heard of the Axiom of Choice and have never studied its consequences. When you write a book, it's important to have an idea in your mind as to who, exactly, is your audience, and then write to that audience. That's what Matousek has done.
