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In Linear Algebra, assume you have:

  • A field Q
  • A vector space V which is ontop of Q.
  • A vector v which belongs to V.

Could v be composed of values not contained in Q? For example, could this be a valid value for v: v = ( sqrt(2), 1 )?


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I assume that you mean the field of rational numbers $\mathbb Q$ - not any field - ecause you say that $\sqrt{2}$ is not an element of it? – born Mar 6 '13 at 9:33
up vote 5 down vote accepted

A vector space $V$ over a field $F$ is an abstract set, whose elements can be anything at all, such that addition is defined on that set as well as scalar multiplication, satisfying the vector space axioms. A given vector does not consist of anything in particular and it can be anything. As an extreme example, choose, for every real number $r\in \mathbb R$, an elephant and call that elephant $e_r$. Now define addition of elephants by $e_r+e_s=e_{r+s}$ and a scalar multiplication by $s\cdot e_r=e_{sr}$. Then the set of elephants $\{e_r\mid r\in \mathbb R\}$ is a vector space over $\mathbb R$. A vector in that vector space is an elephant.

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-1 for possibly making the OP feel bad. +1 because you made me laugh. – bubba Mar 6 '13 at 10:03
I really hope nobody felt bad because of this example. I often use elephants when I teach abstract algebra to drive the point that vectors (group elements or whatever) really can be anything at all. Of course, no elephants were harmed in the process, so animal lovers should not be offended either ;) – Ittay Weiss Mar 6 '13 at 10:06
A "more mathy" example for the OP is $\mathbb{C}$ over $\mathbb{R}$. The imaginey unit $i$ is in the vector space but not in the underlying real field. But I do like elephants...high five. – Fixed Point Mar 6 '13 at 11:13

The things in $V$ can be anything you like, provided they satisfy the vector space axioms. They don't have to be "tuples" or "arrays", so they are not necessarily "composed" of anything. They could be functions, for example.

It might be helpful if you think of a vector space as two things: (1) an abelian group $G$ with an "addition" operation, and (2) a field $F$. $G$ and $F$ are completely unrelated, except for the fact that they are connected by a "scalar multiplication" operation that has certain properties.

As a specific example, let's take $G$ to be $\mathbb R^2$, with addition defined in the obvious way, and let $F$ be $\mathbb Q$. Then, if we define scalar multiplication by $q*(x,y) = (q*x, q*y)$ for $q \in F$ and $(x,y) \in G$, we get a vector space V (I think). The pair $(\sqrt 2, 1)$ is a perfectly legal member of $G$, and hence it's a perfectly legal "vector" in the vector space V.

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>they could be functions ... or elephants :-) – bubba Mar 6 '13 at 9:37

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