2
$\begingroup$
  1. Is there a vector space $V$ over field $\mathbb F$ where the components of the vectors in $V$ are not in $\mathbb F$? I think yes, because you can think of $\mathbb C$ as an $\mathbb R$-vector space.

  2. Is there an example where the field $\mathbb F$ is not a subset of the "value set" of the components of $V$? I think yes, as you can think of $\mathbb R$ as a $\mathbb C$-vector space. Does an example exist using usual addition and multiplication?

Hence, can you say that there isn't a relationship which is easy to describe (wihtout the axioms defining vector space) between $\mathbb F$ and $V$?

$\endgroup$
3
  • $\begingroup$ The positive real numbers are a vector space over the real numbers where vector addition is real multiplication, and scalar multiplication is real exponentiation. The reals are not a subset of the positive reals. $\endgroup$ May 31, 2022 at 15:55
  • $\begingroup$ That said, every finite dimensional (and assuming the Axiom of Choice, every) vector space over $F$ is isomorphic to a suitable space of (almost null) tuples of elements of $F$. $\endgroup$ May 31, 2022 at 15:57
  • 1
    $\begingroup$ Given a vector space $V$ over a field $\Bbb F$ and a vector $v\in V$, what do you mean by the components of $v$? And what are the components of $V$? $\endgroup$ May 31, 2022 at 15:57

1 Answer 1

4
$\begingroup$

Vectors in a vector space do not have "components" until you introduce a basis (a coordinate system).

The set of points in the ordinary Euclidean plane with a fixed point specified as the origin and a fixed unit length is a vector space over the real numbers: addition is the parallelogram law and scalar multiplication is stretching lengths. Once you choose two points that form a nondegenerate triangle with the origin to specify two coordinate axes you can represent any point as a pair of real numbers.

When you think of the complex numbers as a two dimensional real vector space, each element is represented as a pair of real numbers when you use $1$ and $i$ to determine the real and imaginary axes.

When you think of the complex numbers as a one dimensional vector space over itself, each complex number is represented by itself.

The set of real valued functions with domain the unit interval is a vector space over the real field: it's straightforward to define addition and scalar multiplication. Representing those functions with "coordinates" is a subtle problem you may encounter in more advanced courses.

$\endgroup$
3
  • $\begingroup$ "The set of points in the ordinary Euclidean plane with a fixed point specified as the origin and a fixed unit length ...". You don't even need to specify length to get a vector space! $\endgroup$
    – Ruy
    May 31, 2022 at 16:20
  • $\begingroup$ @Ruy True. It's a little easier to visualize scalar multiplication if you do specify a unit length. $\endgroup$ May 31, 2022 at 16:24
  • $\begingroup$ Agreed. My point is that if you want a real vector space whose vectors are not at all related to real numbers you might not want to mention length. $\endgroup$
    – Ruy
    May 31, 2022 at 16:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .