Is one norm property: $p(v)=0 \iff v=0$ or positive definiteness? I'm confused as to seeing different requirements for norms.
Triangle eq. and homogeneity seem to be in all, but the last property seems to be either:
$p(v)=0 \iff v=0$ or positive definiteness
How are these related/equivalent?

It seems that positive definiteness may be able to capture the rule about norm having to be positive, while also capturing the zero element property, both at the same time?
 A: Positive definiteness of $p$ states that $p(v) = 0 \iff v = 0$ and $p(v) \geq 0$ for all $v \in V$. This is definitely required for every norm.
If you're reading the Wikipedia article, $p(v) \geq 0$ for all $v \in V$ is implicitly encoded since $p$ is a function $p: V \to [0, \infty)$, hence you only need to include $p(v) = 0 \iff v = 0$ as an additional requirement.
A: Let $V$ be an $\mathbb{R}$-vector space. A norm is a mapping $\lVert\cdot\rVert\colon V\rightarrow\mathbb{R}$, satisfying the following properties. For all $v,w\in V$, you have the triangle inequality $\lVert v+w\rVert\le\lVert v\rVert+\lVert w\rVert$. For all $\lambda\in\mathbb{R},v\in V$ you have absolute homogenity, i.e. $\lVert \lambda v\rVert=\lvert\lambda\vert\lVert v\rVert$. Then we require either of the following properties:


*

*$\lVert v\rVert=0\Rightarrow v=0$ for $v\in V$,

*$\lVert v\rVert\ge0$ for all $v\in V$ with equality iff $v=0$.


I've seen either of these two properties be referred to as positive-definiteness, so I won't name them. The second property trivially implies the former property. For the other direction, notice that if $\lVert v\rVert<0$ for some $v\in V$, you would - by absolute homogenity - have $\lVert-v\rVert=\lvert-1\rvert\lVert v\rVert=\lVert v\rVert<0$ and then - by the triangle inequality - $0=\lVert0\rVert=\lVert v+(-v)\rVert\le\lVert v\rVert+\lVert-v\rVert<0$, a contradiction, so the norm is definitely non-negative. The equivalency $\lVert v\rVert=0\Leftrightarrow v=0$ then follows from the first property as $\lVert0\rVert=\lVert\lambda0\rVert=\lvert\lambda\rvert\lVert0\rVert$ for all $\lambda\in\mathbb{R}$ and hence $\lVert0\rVert=0$.
