Do all vectors belong to a vector space?

If we were given the vector $(1,1,1)$, say, we know immediately that it belongs to the vector space $\mathbb{R}^3$ (and infinitely many others). But, if we take this vector of dolphins:

or some other bizarre contrived vector: Does it belong to a vector space?

An element of a vector space.

This suggests the answer to the above question is yes; if they weren't in a vector space, they wouldn't be vectors. But I'm not convinced by this sentence alone.

• While one vector space can be contained in another, an author needs to take care in clarifying for the reader which vector space is being referenced. For example, $(1,1,1)$ might be in $\mathbb{R}^3$ for an article's purpose, but another author might mean this to be a vector in $\mathbb{Q}^3$ or $\mathbb{C}^3$, or possibly in a vector space over a finite field. Without more context the reader cannot be sure. – hardmath Aug 29 '13 at 13:49
• A tuple is only a vector if its elements all come from a field $\mathbb{F}$, in which case it is an element of the vector space $\mathbb{F}^n$. Without some field structure on e.g. the set of aquatic mammals, your dolphin example is not a vector. – Anthony Carapetis Aug 29 '13 at 13:53

A vector space always exists over a certain field. In the case of your dolphins, for it to be part of a vector space, two things would have to happen:

1. You would have to define the $+$ and $\times$ operators for the field of dolphins, define a "zero dolphin", and prove that dolphins abide by all of the field axioms.

2. You would have show that the set of all dolphin vectors is closed on vector addition, and scalar multiplication.

Otherwise, it is not a vector, but an ordered set.

• Thanks for that; I think that's clear. As it stands, my vector of dolphins is an ordered set, but if I were to identify a field $\mathbb{F}$ of dolphins (which wouldn't be hard; take a large enough field and relabel the elements as dolphins), then the ordered set would belong to the vector space $\mathbb{F}^5$ (in this case), and thus be a vector. (It'd probably get more set-theoretic in the infinite case, but I think the principle is the same.) – Rebecca J. Stones Aug 29 '13 at 14:09
• @Rebecca Yes, spot on. And you're welcome :) – Ataraxia Aug 29 '13 at 16:13

If a vector belongs to a vector space, its components must belong to a field $K$. This is not the case for dolphins. A similar question is, even using numers instead of dolphins: is the set $\lbrace 2 \rbrace \subset \mathbb{R}$ on the real line a real vector space ? But then it should contain $\lbrace 0\rbrace$, which it does not. On the other hand, can we define $2$ as the zero element ?

A vector, by definition, is an element of vector space, which is a set of vectors satisfying eight axioms. So no matter what the element is, even dolphin, it is a vector if and only if all elements in consideration form a vector space.

For example, which is the simplest, the "dolphin" is a vector if the vector space on $\mathcal{Z}_2$ is $\{0,''dolphin''\}$ and $''dolphin''+''dolphin''=0$