From Vector Spaces to Periodic Functions

Vector Spaces

A fascinating result that appears in linear algebra is the fact that the set of real numbers $\mathbb{R}$ is a vector space over the set of rational numbers $\mathbb{Q}.$ This may appear surprising at first but it is easy to show that it is indeed so by checking that all eight axioms of vector spaces hold good:

1. Commutativity of vector addition:
   $x + y = y + x$ for all $x, y \in \mathbb{R}.$

2. Associativity of vector addition:
   $x + (y + z) = (x + y) + z$ for all $x, y, z \in \mathbb{R}.$

3. Existence of additive identity vector:
   We have $0 \in \mathbb{R}$ such that $x + 0 = x$ for all $x \in \mathbb{R}.$

4. Existence of additive inverse vectors:
   There exists $-x \in \mathbb{R}$ for every $x \in \mathbb{R}$ such that $x + (-x) = 0.$

5. Associativity of scalar multiplication:
   $a(bx) = (ab)x$ for all $a, b \in \mathbb{Q}$ and all $x \in \mathbb{R}.$

6. Distributivity of scalar multiplication over vector addition:
   $a(x + y) = ax + by$ for all $a \in \mathbb{Q}$ and all $x, y \in \mathbb{R}.$

7. Distributivity of scalar multiplication over scalar addition:
   $(a + b)x = ax + bx$ for all $a, b \in \mathbb{Q}$ and all $x \in \mathbb{R}.$

8. Existence of scalar multiplicative identity:
   We have $1 \in \mathbb{Q}$ such that $1 \cdot x = x$ for all $x \in \mathbb{R}.$

This shows that the set of real numbers $\mathbb{R}$ forms a vector space over the field of rational numbers $\mathbb{Q}.$ Another quick way to arrive at this fact is to observe that $\mathbb{Q} \subseteq \mathbb{R},$ that is, $\mathbb{Q}$ is a subfield of $\mathbb{R}.$ Any field is a vector space over any of its subfields, so $\mathbb{R}$ must be a vector space over $\mathbb{Q}.$

We can also show that $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{Q}.$ Let us assume the opposite, i.e., $\mathbb{R}$ is finite dimensional. Let $r_1, \dots, r_n$ be the basis for this vector space. Therefore for each $r \in \mathbb{R},$ we have unique $q_1, \dots, q_n \in \mathbb{Q}$ such that $r = q_1 r_1 + \dots + q_n r_n.$ Thus there is a bijection between $\mathbb{Q}^n$ and $\mathbb{R}.$ This is a contradiction because $\mathbb{Q}^n$ is countable whereas $\mathbb{R}$ is uncountable. Therefore $\mathbb{R}$ must be an infinite dimensional vector space over $\mathbb{Q}.$

Problem

Here is an interesting problem related to vector spaces that I came across recently:

If you want to think about this problem, this is a good time to pause and think about it. There are spoilers ahead.

Solution

The axiom of choice is equivalent to the statement that every vector space has a basis. Since the set of real numbers $\mathbb{R}$ is a vector space over the set of rational numbers $\mathbb{Q},$ there must be a basis $\mathcal{H} \subseteq \mathbb{R}$ such that every real number $x$ can be written uniquely as a finite linear combination of elements of $\mathcal{H}$ with rational coefficients, that is, $$x = \sum_{a \in \mathcal{H}} x_a a$$ where each $x_a \in \mathbb{Q}$ and $\{ a \in \mathcal{H} \mid x_a \ne 0 \}$ is finite. The set $\mathcal{H}$ is also known as the Hamel basis.

In the above expansion of $x,$ we use the notation $x_a$ to denote the rational number that appears as the coefficient of the basis vector $a.$ Therefore $(x + y)_{a} = x_a + y_a$ for all $x, y \in \mathbb{R}$ and all $a \in \mathcal{H}.$

We know that $b_a = 0$ for distinct $a, b \in \mathcal{H}$ because $a$ and $b$ are basis vectors. Thus $(x + b)_{a} = x_a + b_a = x_a + 0 = x_a$ for all $x \in \mathbb{R}$ and distinct $a, b \in \mathcal{H}.$ This shows that a function $f(x) = x_a$ is a periodic function with period $b$ for any $a \in \mathcal{H}$ and any $b \in \mathcal{H} \setminus \{ a \}.$

Let us define two functions: $$\begin{align*} g(x) & = \sum_{a \in \mathcal{H} \setminus \{ b \}} x_a a, & h(x) & = x_b b. \end{align*}$$ where $b \in \mathcal{H}$ and $x \in \mathbb{R}.$ Now $g(x)$ is a periodic function with period $b$ for any $b \in \mathcal{H}$ and $h(x)$ is a periodic function with period $c$ for any $c \in \mathcal{H} \setminus \{ b \}.$ Further, $$g(x) + h(x) = \left( \sum_{a \in \mathcal{H} \setminus \{ b \}} x_a a \right) + x_b b = \sum_{a \in \mathcal{H}} x_a a = x.$$ Thus $g(x)$ and $h(x)$ are two periodic functions such that their sum is the identity function.

References

• Vector Space by Eric W. Weisstein
• The Dimension of R over Q by Alex Youcis
• Sums of Periodic Functions by David Radcliffe