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I am getting a bit confused with the terminology here. I understand that a field means some set of scalars like real numbers but why do we need a field for a vector space? Are not the numerical values used to define a vector are inherent properties of the vector space? Why do we term it like "vector space over real field"? If not numbers, what are the other fields possible for a vector space because you obviously numbers to define the value of magnitude and direction of the vectors?

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    $\begingroup$ Your title is somewhat confused: one does not define a vector space in terms of a vector space over some field (which would be kind of circular); instead we define vector spaces while referring to some field that was fixed beforehand. $\endgroup$ – Marc van Leeuwen Sep 6 '15 at 15:52
  • $\begingroup$ math.stackexchange.com/questions/2775822/… $\endgroup$ – alfC May 11 '18 at 17:28
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A vector space is defined as a quadruple $(V,K,+,\cdot)$ where $V$ is a set whose elements are called ''vectors'' , $K$ is a field, $+\colon V\times V \rightarrow V$ and $\cdot\colon K\times V \rightarrow V$ are operations that satisfy a suitable set of axioms (see here).

For the same set $V$ we can define different vector spaces changing the field $F$ and the difference can be dramatic.

The simpler case is for $V=\mathbb{R}$. If we take $K=\mathbb{R}$ with the usual addition and multiplication, we have a vector space of dimension $1$, but if we take $K=\mathbb{Q}$ we have a vector space that has a uncountable basis so its dimension is infinite (Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?).

Another, less dramatic, example, is if $V=\mathbb{C}^n$: if we choose $K= \mathbb{C}$ than we have a vector space (with the usual operations) over $\mathbb{C}$ of dimension $n$, if we choose $K= \mathbb{R}$ we can obtain a space of dimension $2n$.

Again, if $K=\mathbb{Q}$ the change is more relevant and we have a space of infinite (not countable) dimension.

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  • $\begingroup$ @AdityaKumarGupta You seem to think that vectors can always be represented by tuples of scalars. The example here ($\mathbb R$ as a vector space over $\mathbb Q$) shows that this is not true. $\endgroup$ – user1551 Sep 6 '15 at 16:51
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First you cannot define magnitude and direction for all vector spaces: these notions correspond to so-called *normed and/or inner product vector spaces*.

Next one has very naturally have to consider vector spaces over $\mathbf C$, even in problems concerning real spaces. I'll mention Fourier series, or second order linear differential equations.

In arithmetic, one has to consider vector spaces over $\mathbf Q$ or algebraic number fields or $p$-adic numbers.

I once heard a well-known mathematician say cryptography, a domain closely related to arithmetic, is but the theory of finite-dimensional vector spaces over finite fields.

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Are not the numerical values used to define a vector are inherent properties of the vector space?

Yes, they are. If you use other numerical values (like complex numbers, or $\mathbb F_2$), you get another vector space (for example $\mathbb F_2^3$ instead of $\mathbb R^3$). However you can also get a different vector spaces using the same field (for example, $\mathbb R^2$ and $\mathbb R^3$). Now it turns out that vector spaces over the same field are in some sense compatible with each other, in a way that vector spaces over different fields generally are not. For example, it's no problem to define a linear function $f:\mathbb R^2\to\mathbb R^3$. However there are no linear functions $f:\mathbb F_2^2\to\mathbb R^2$. Therefore it makes sense to always explicitly say on which field the vector space is based.

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    $\begingroup$ Between vector space over unrelated fields, one cannot even define the notion of linear map, since the definition requires the same scalars to be used in both spaces (more precisely, the linear map must commute with any scalar multiplication applied in both spaces). $\endgroup$ – Marc van Leeuwen Sep 6 '15 at 15:47
  • $\begingroup$ @MarcvanLeeuwen: You're right; I'll edit my answer. $\endgroup$ – celtschk Sep 6 '15 at 16:11
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Not all vector fields have "magnitutes". In order to have a magnitude of a vector, you need to have a norm defined on the vector space, and typically, only vector spaces over $\mathbb R$ or $\mathbb C$ have norms defined on them.

We have to say "vector space over real field" because we need to be careful about what we are doing at all times. For example, $[1,0,0]$ is an element of both $\mathbb R^3$ which is a vector space over $\mathbb R$, and an element of $\mathbb C^3$ which is a vector space over $\mathbb C$.

Furthermore, the fact that $\mathbb C$ is itself a vector space over $\mathbb R$ means that any vector space $V$ over $\mathbb C$ can also be seen as a vector space over $\mathbb R$, but of a different dimension.

You can also have vector spaces over other fields, like $\mathbb Z_p$ where $p$ is prime, and not define any norm there.

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