Normed Vector Space Requires Multiplicative Norm on Division Ring
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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:
- $\exists \mathbf v \in V: \mathbf v \ne 0$
By Norm Axiom $\text N 1$: Positive Definiteness:
- $\norm {\mathbf v} > 0$
Let $r, s \in R$:
\(\ds \norm {r s}_R \norm {\mathbf v}\) | \(=\) | \(\ds \norm {\paren {r s} \mathbf v}\) | Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {r \paren {s \mathbf v} }\) | Vector Space Axiom $\text V 7$: Associativity with Scalar Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm r_R \norm {s \mathbf v}\) | Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm r_R \norm s_R \norm {\mathbf v}\) | Norm Axiom $\text N 2$: Positive Homogeneity |
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.
$\blacksquare$