Let $K$ be a field equipped with an absolute value $\vert-\vert:K\rightarrow \mathbb{R}_{\geq 0}$ satisfying
- $\vert k \vert = 0$ iff $k=0$
- $\vert x+y \vert\leq\vert x\vert+\vert y\vert$
- $\vert x\cdot y \vert = \vert x \vert\cdot\vert y\vert$
- $K$ is complete with respect to the topology induced by the absolute value
(I think of $\mathbb{R,C}$ but also $\mathbb{Q}_p$ and $\mathbb{C}_p$)
Let $V$ be a $K$-vector space equipped with a norm $\Vert - \Vert: V\rightarrow \mathbb{R}_{\geq 0}$ satisfying
- $\Vert v \Vert = 0$ iff $v=0$
- $\Vert k\cdot v \Vert = \vert k \vert\cdot\Vert v\Vert$
- $\Vert x+y \Vert \leq \Vert x \Vert + \Vert y \Vert$
Under which circumstances can we normalize any nonzero vector? More precisely, given $0\neq v \in V$ with norm $\Vert v \Vert = r \neq 1$, when does there exist some $k\in K$ such that $\vert k \vert = \frac{1}{r}$, ie. such that $\Vert kv \Vert = 1$?
One sufficient criterion is of cause that the absolute value $\vert-\vert:K\rightarrow\mathbb{R}_{\geq 0}$ is a surjective function (as is the case for $\mathbb{R}$ and $\mathbb{C}$).
In fact it is sufficient and necessary that the norm function $\Vert-\Vert:V\rightarrow\mathbb{R}_{\geq 0}$ factors through the image of the absolute value function. A priori I see no reason why this should always be the case. I am especially interested in the case, where $K$ is a local field in the sense of Wikipedia.
Thank you for your time.
If $\Vert \cdot\Vert_1$ is any norm, then so is $\Vert \cdot \Vert_2 :=c \cdot \Vert \cdot \Vert_1$ for any $c \in (0,\infty)$, and these two norms are equivalent (cf. e.g. 1). So as soon as $\lvert K \rvert \subsetneq [0,\infty)$, which is always the case for a non-archimedean local field $(K, \lvert \cdot \rvert)$, you can enforce that some vectors cannot be scaled to $1$.
However, e.g. on finite dimensional vector spaces, all norms are equivalent if $K$ is complete as assumed. (Some standard proofs state this only for $\mathbb R$ and $\mathbb C$, and use local compactness, see e.g. 2, 3; for the general statement, look e.g. at Keith Conrad's Equivalence of Norms. In fact, just read that, it says everything I say here and more, better.). So one can always find a norm, equivalent to the given one, where every vector can be scaled to $1$. In fact, you can just choose any basis $b_1,...,b_n$, and with respect to this basis define the maximum norm $\Vert \kappa_1 b_1 + ... + \kappa_n b_n\Vert:=\max|\kappa_i|$. This way, any basis can play the role of a normal (unit) basis, without changing the topology.