Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following theorem is given in Metric Spaces by O'Searcoid

Theorem: Suppose $V$ is a normed linear space. Then the function $d$ defined on $V \times V$ by $(a,b) \to ||a-b||$ is a metric on $V$

Three conditions of a metric are fairly straight-forward.

By the definitions of a norm, I know that $||x|| \ge 0$ and only holds with equality if $x=0$. Thus $||a-b||$ is non-negative and zero if and only if $a=b$.

The triangle inequality of a normed linear space requires: $||x+y|| \le ||x|| + ||y||$. Let $x = a - b$ and $y = b - c$. Then $||a - c|| \le || a - b || + || b - c||$ satisfying the triangle inequality for a metric space.

What I am having trouble figuring out is symmetry. The definition of a linear space does not impose any condition of a symmetry. I know from the definition of a linear space that given two members of $V$, $u$ and $v$ they must be commutative, however, I do not see how that could extend here.

Thus what I would like to request help with is demonstrating $||a - b|| = ||b - a||$.

share|improve this question
    
Well, which clause in the definition of a norm have you not used? –  Chris Eagle Jun 29 '13 at 0:25
    
@PeterTamaroff I knew it had to be something really simple. Thanks! –  GovEcon Jun 29 '13 at 0:26

2 Answers 2

up vote 2 down vote accepted

$$\|a-b\|=\|(-1)(b-a)\|=|-1|\cdot\|b-a\|=\|b-a\|$$

share|improve this answer

Use $\lVert c\cdot x\rVert =|c|\lVert x\rVert$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.