Norm

In mathematics, a norm is a function in vector spaces with certain special properties.

Conventional definition in inner product spaces

By convention, in inner product space $V$, the norm of a vector $v \in V$, denoted $\left\| v \right\|$, is defined:

$$ \| v \| = \sqrt{\langle v, v \rangle} $$

Formal definition in all vector spaces

Let $V$ be a vector space over $\mathbb{F}$.

Map

$$ \begin{aligned} \| \cdot \|: V &\to \mathbb{R}\\ v &\mapsto \| v \| \end{aligned} $$

is a norm on $V$ if 3 conditions are satisfied:

• Definiteness: $\forall v\in V,\; \left\| v \right\|=0 \iff v=\vec 0$;
• Homogeneity: $\forall a\in\mathbb{F},\;\forall v\in V,\; \left\| av \right\|=|a|\left\| v \right\|$;
• Triangle inequality: $\forall v,w\in V,\; \left\| v+w \right\|\le \left\| v \right\|+\left\| w \right\|$

Notice the conventional definition $\| v \| = \sqrt{\langle v, v \rangle}$ is a special case of the mathematical definition.

Derived properties

Positive definiteness

$$\forall v \in V \| v \|\ge 0$$

Proof:

$$\begin{gathered} \| v \| = \frac{1}{2}(2\| v \|)= \frac{1}{2}(\| v \| + \| -v \|) \\ \stackrel{\text{Triangle inequality}}{\ge} \frac{1}{2}\| v-v \| = 0 \end{gathered}$$