Showing the norm' is non-negative

Question

Question: Prove that a norm is always non-negative.

Assume that the norm is negative.

Define: $f: \mathbb{R}^{n}\rightarrow \mathbb{R}^{-}_{0}$

$v \mapsto \left ( \vec{v} \right )f=\left \| \vec{v} \right \|$

$\left ( \vec{u}+\vec{v} \right )f=\left \| \vec{u}+\vec{v} \right \| \in \mathbb{R}^{-}_{0}$ and $\left ( \vec{u} \right )f=\left \| \vec{u} \right \|,\left ( \vec{v} \right )f=\left \| \vec{v} \right \|=\left \| \vec{v} \right \| \in \mathbb{R}^{-}_{0}$ At this point I am unable to progress further.

Any help is appreciated.

Answer 1

Usually, a norm is defined to be non-negative. But actually that is redundant.

Let $V$ be a vector space over a field $\mathbb K$. A norm is a function $\|\cdot\| \colon V \to \mathbb R$ with the following properties (norm axioms):

1. $\|\mathbf v\| = 0 \iff \mathbf v = \mathbf 0$
2. $\|a\mathbf v\| = |a|\|\mathbf v\|$ (scalability)
3. $\|\mathbf v + \mathbf w\| \leq \|\mathbf v\| + \|\mathbf w\|$ (triangle inequality)

for every $\mathbf v, \mathbf w \in V$ and $a \in \mathbb K$.

Now, using the fact that $\mathbf 0 = \mathbf v + (- \mathbf v)$, we have $$0 = \|\mathbf 0\| \leq \|\mathbf v\| + \|-\mathbf v\| = 2\|\mathbf v\|$$ where we used, in order, properties $1$, $3$ and $2$. It follows that $$\|\mathbf v\| \geq 0.$$ In other words, even if we start with codomain $\Bbb R$ for the norm, the range is contained in $\mathbb R_0^+$.