1

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 $V$?

Rohit Pandey
  • 7,547
  • 1
    Related questions (from same user, concerning same miniature): https://math.stackexchange.com/questions/4236378/linear-function-applied-to-sides-of-rectangle-and-its-tiles and https://math.stackexchange.com/questions/4241254/product-of-linear-function-applied-to-the-two-sides-of-a-rectangle-is-supposed-t and https://math.stackexchange.com/questions/4243544/linear-functions-over-vector-spaces-vs-those-from-analytic-geometry and https://math.stackexchange.com/questions/4242764/reals-as-a-vector-space-over-rationals-why-infinite-dimensional – Gerry Myerson Sep 11 '21 at 05:49
  • Another related question: https://math.stackexchange.com/questions/4247320/linear-function-from-mathbb-r-over-mathbb-q-to-mathbb-r-over-mathbb-q – Rohit Pandey Sep 11 '21 at 16:50

3 Answers3

5

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.

José Carlos Santos
  • 440,053
  • Then why did Matousek bother defining $f$ from $V \to \Bbb R$? His proof would have been shorter if he didn't even bother defining $V$ and said $f$ was from $\Bbb R \to \Bbb R$. Is it just to avoid assuming the axiom of choice? – Rohit Pandey Sep 11 '21 at 04:08
  • 4
    @RohitPandey: Almost certainly; note that the way it is given you can describe $f$ explicitly. Your $f$ is not explicit. For instance... what is its value at $\pi+e$? – Arturo Magidin Sep 11 '21 at 04:34
  • What does explicit mean here? Just that it can be defined for any real number? – Rohit Pandey Sep 11 '21 at 05:09
  • 1
    It means it can be evaluated for any $x$ in its domain. – Gerry Myerson Sep 11 '21 at 05:46
3

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.

Jyrki Lahtonen
  • 140,891
2

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.

Gerry Myerson
  • 185,413