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 $||a – b|| = ||b – a||$.
Best Answer
$$\|a-b\|=\|(-1)(b-a)\|=|-1|\cdot\|b-a\|=\|b-a\|$$