[proofplan] We verify the metric axioms directly from the norm axioms. Nonnegativity and definiteness follow from the corresponding norm properties. Symmetry follows by writing $x-y=-(y-x)$ and using homogeneity of the norm with the scalar $-1$. The triangle inequality follows by decomposing $x-z$ as $(x-y)+(y-z)$ and applying the norm triangle inequality. [/proofplan]

[step:Verify nonnegativity and definiteness from the norm axioms] Let $x,y\in V$. Since $\|\cdot\|$ is a norm, it takes values in $[0,\infty)$, so \begin{align*} d(x,y)=\|x-y\|\geq 0. \end{align*} Thus $d$ is nonnegative. Next, \begin{align*} d(x,y)=0 \end{align*} if and only if \begin{align*} \|x-y\|=0. \end{align*} By definiteness of the norm, this holds if and only if \begin{align*} x-y=0. \end{align*} By cancellation in the [vector space](/page/Vector%20Space) $V$, this is equivalent to $x=y$. Hence $d(x,y)=0$ if and only if $x=y$. [/step]

[step:Prove symmetry using homogeneity with the scalar $-1$]Let $x,y\in V$. Since \begin{align*} x-y=-(y-x), \end{align*} homogeneity of the norm gives \begin{align*} d(x,y)=\|x-y\|=\|-(y-x)\|=|-1|\,\|y-x\|. \end{align*} Because $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, one has $|-1|=1$. Therefore \begin{align*} d(x,y)=\|y-x\|=d(y,x). \end{align*} So $d$ is symmetric.[/step]

[guided]We want to prove that exchanging the two inputs of $d$ does not change its value. Let $x,y\in V$. By the definition of $d$, \begin{align*} d(x,y)=\|x-y\|. \end{align*} The difference $x-y$ is the negative of the reversed difference: \begin{align*} x-y=-(y-x). \end{align*} Now apply the homogeneity axiom for the norm to the scalar $-1\in\mathbb{F}$ and the vector $y-x\in V$. This gives \begin{align*} \|-(y-x)\|=|-1|\,\|y-x\|. \end{align*} Since $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$, the absolute value of $-1$ is $1$. Hence \begin{align*} \|-(y-x)\|=\|y-x\|. \end{align*} Combining these equalities, \begin{align*} d(x,y)=\|x-y\|=\|y-x\|=d(y,x). \end{align*} Thus $d$ satisfies the symmetry axiom of a metric.[/guided]

[step:Derive the triangle inequality from the norm triangle inequality] Let $x,y,z\in V$. In the vector space $V$, \begin{align*} x-z=(x-y)+(y-z). \end{align*} Using the definition of $d$ and then the triangle inequality for the norm, \begin{align*} d(x,z)=\|x-z\|=\|(x-y)+(y-z)\|\leq \|x-y\|+\|y-z\|. \end{align*} Substituting back the definition of $d$ gives \begin{align*} d(x,z)\leq d(x,y)+d(y,z). \end{align*} Therefore $d$ satisfies the triangle inequality. [/step]

[step:Conclude that the induced function is a metric] The preceding steps show that, for all $x,y,z\in V$, the function $d:V\times V\to[0,\infty)$ satisfies nonnegativity, identity of indiscernibles, symmetry, and the triangle inequality. These are exactly the metric axioms. Hence $d$ is a metric on $V$. [/step]