When we consider a vector space $V$ over some field $F$, I know that when the $F=\mathbb{R}$ or $ =\mathbb{C}$, by setting $\|x\|=\left\langle{x,x}\right\rangle^\frac{1}{2}$ we get a norm. However, since the inner product is a function with its image in $F$, what happens if we consider any $V$ over the rational numbers? For example, if we take $\mathbb{Q}^2$ over $\mathbb{Q}$ with the dot product, then $v=(1,1)$ has norm $\sqrt{2}$, which is not rational. How can one obtain a norm from a given inner product in such cases?

up vote 4 down vote accepted

As you noted, to construct an an inner-product space we require the basefield $F$ of the vector space $V$ to be a quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$ (i.e. every element of $F$ must have a square root in $F$).

However, a norm is a function $n:V\rightarrow [0,+ \infty)$, as opposed to an inner-product which is map $\langle \cdot ,\cdot \rangle :V\times V\to F$. In the case of the basefield $\mathbb{Q}$ therefore, we cannot construct an inner-product space from the dot product implied in your question, but we can construct a normed vector space from it.

An ordered field where $a^2+b^2$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each ordered field has a Pythagorean extension. If you really want $L^2$-norms you could always extend your ground field to a Pythagorean extension field.

  • 1
    Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean. – javierochomil Oct 15 at 2:30

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