Euclidean Space is Normed Vector Space

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Theorem

The Euclidean space $\R^n$ is a normed vector space.


Proof

Let $\norm {\, \cdot \,}$ denote the Euclidean norm on $\R^n$.

We prove that $\norm {\, \cdot \,}$ is indeed a norm on $\R^n$ by proving it fulfils the norm axioms.


Norm Axiom $\text N 1$: Positive Definiteness

As $\mathbf 0 = \tuple {0, \ldots, 0}$, it follows that:

$\norm {\mathbf 0} = \sqrt {0^2 + \ldots + 0^2} = 0$


Suppose instead that $\norm {\mathbf v} = 0$ for some $\mathbf v = \tuple {v_1, \ldots, v_n}$ with $v_1, \ldots, v_n \in \R$.

As $\ds \sqrt {\sum_{i \mathop = 1}^n v_n^2 } = 0$, it follows from squaring both sides that:

$\ds \sum_{i \mathop = 1}^n v_n^2 = 0$

From Even Power is Non-Negative, it follows that for all $i$: $v_i = 0$.

Hence $\mathbf v = \mathbf 0$.

Thus Norm Axiom $\text N 1$: Positive Definiteness holds

$\Box$


Norm Axiom $\text N 2$: Positive Homogeneity

Let $\mathbf v = \tuple {v_1, \ldots, v_n} \in \R^n$ with $v_1, \ldots, v_n \in \R$.

Let $r \in \R$.

Then:

\(\ds \norm { r \mathbf v }\) \(=\) \(\ds \sqrt {\sum_{i \mathop = 1}^n \paren { r v_i }^2 }\) Definition of Euclidean Norm
\(\ds \) \(=\) \(\ds \sqrt {\sum_{i \mathop = 1}^n \paren { r^2 v_i^2 } }\) Power of Product
\(\ds \) \(=\) \(\ds \sqrt{ r^2 \sum_{i \mathop = 1}^n v_i^2 }\)
\(\ds \) \(=\) \(\ds \sqrt{ r^2 } \sqrt{ \sum_{i \mathop = 1}^n v_i^2 }\) Power of Product
\(\ds \) \(=\) \(\ds \size r \norm { \mathbf v }\)

Thus Norm Axiom $\text N 2$: Positive Homogeneity holds.

$\Box$


Norm Axiom $\text N 3$: Triangle Inequality

Follows from Triangle Inequality for Vectors in Euclidean Space.

Thus Norm Axiom $\text N 3$: Triangle Inequality holds.

$\blacksquare$


Sources

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