Normed Vector Space Requires Multiplicative Norm on Division Ring

Theorem

Let $R$ be a normed division ring with a submultiplicative norm $\norm {\, \cdot \,}_R$.

Let $V$ be a vector space that is not a trivial vector space.

Let $\norm {\, \cdot \,}: V \to \R_{\ge 0}$ be a mapping from $V$ to the positive real numbers satisfying the vector space norm axioms.

Then $\norm {\, \cdot \,}_R$ is a multiplicative norm.

That is:

$\forall r, s \in R: \norm {r s}_R = \norm r_R \norm s_R$

Proof

Since $V$ is not a trivial vector space:

$\exists \mathbf v \in V: \mathbf v \ne 0$
$\norm {\mathbf v} > 0$

Let $r, s \in R$:

 $\displaystyle \norm {r s}_R \norm {\mathbf v}$ $=$ $\displaystyle \norm {\paren {r s} \mathbf v}$ Norm axiom (N2) $\displaystyle$ $=$ $\displaystyle \norm {r \paren {s \mathbf v} }$ Vector Space Axiom $\text V 7$: Associativity with Scalar Multiplication $\displaystyle$ $=$ $\displaystyle \norm r_R \norm {s \mathbf v}$ Norm axiom (N2) $\displaystyle$ $=$ $\displaystyle \norm r_R \norm s_R \norm {\mathbf v}$ Norm axiom (N2)

By dividing both sides of the equation by $\norm {\mathbf v}$ then:

$\norm {r s}_R = \norm r_R \norm s_R$

The result follows.

$\blacksquare$