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In my linear algebra class, many times we define vector spaces over the field $\mathbb{F}$ (i.e. $\mathbb{F}^n$) and then prove things about them. The instructor has defined $\mathbb{F}$ as "either $\mathbb{C}$ or $\mathbb{R}$". Because $\mathbb{R}$ is a subset of $\mathbb{C}$, why can we not simply define vector spaces over $\mathbb{C}$ instead of introducing new notation?

Isn't it true that a property holding over $\mathbb{C}$ immediately tell us it holds over $\mathbb{R}$ because $\mathbb{R}\subset \mathbb{C}$?

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    $\begingroup$ $\mathbb{C}$ and $\mathbb{R}$ are just two examples of fields. There are others, and the theorems you are proving hold true for any field, even though in your class you will only be considering the fields $\mathbb{C}$ and $\mathbb{R}$. $\endgroup$
    – wgrenard
    Nov 16, 2016 at 1:08
  • $\begingroup$ @GovindaDasu For that reason and one other -- vector spaces are actually defined more generally over fields (thus the $\Bbb F$), of which $\Bbb R$ and $\Bbb C$ are examples. (At least some of) The theorems you are proving likely still hold in these more general cases, as well. So once you get to a higher level class where your discussion shifts to being in terms fields as opposed to these special cases, you should hopefully be able to use most of your proofs directly from this class. $\endgroup$
    – user137731
    Nov 16, 2016 at 1:08
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    $\begingroup$ In addition to what Stefano said, there are also issues about diagonalizability over $\mathbb{C}$ vs over $\mathbb{R}$ (the characteristic polynomial may have imaginary roots!) $\endgroup$
    – oxeimon
    Nov 16, 2016 at 1:08
  • $\begingroup$ Working over algebraically closed fields does offer some nicer theory, but working with the reals probably offers a more concrete example that people are familiar with. $\endgroup$
    – Oiler
    Nov 16, 2016 at 2:12

3 Answers 3

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Ultimately the reason is similar to the reason that one wants to differentiate between fields in other areas of algebra:

  1. You may want to work with vector spaces over different fields. $\Bbb R$ and $\Bbb C$ are basically the same (apart from a crucial difference mentioned later) but you might want to work in rationals or some field extension or even a field of non-zero characteristic. These algebraic extensions are often thought of as being part vector spaces over some smaller field so it helps if you know linear algebra still works. This is also what motivates the definition of modules. It is more important in these cases that one does not leave the given field, especially as it may be bigger than $\Bbb C$.
  2. Because there are polynomials with real coefficients that do not have solutions over the reals there are lots of special cases for different types of decomposition where you basically get much stronger guarantees if you are allowed to work in the (algebraically closed) complex numbers. It is important to algebra in general though that we know when these decompositions break down in fields that aren't algebraically closed.
  3. There are different definitions for inner product and other bilinear forms when you switch between the reals and complexes. This is basically because you want to be able to compare the magnitude of things and that doesn't make so much sense over the complexes.

Furthermore, the reason is that some later things depend on the difference and so it is worth keeping track of whether or not you are allowed to go into $\Bbb C$ which causes some problems and solves some others. One may ask why a field should be associated with a vector space at all (perhaps it could only be mentioned when needed as a condition) and the answer to that is that it is a matter of fundamentals (i.e. How else can one properly define a vector space without a field) and a matter of knowing that your proofs still work if you have to carry that restriction (i.e. A specific choice of field) all the way through.

On the other hand, it is quite natural to consider $\Bbb R$-vector spaces as $\Bbb C$-vector spaces when certain properties are needed. When one writes this, one means that you should use the (natural) inclusion map to translate from one to the other and one often ends by proving that things are indeed now real again.

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  • $\begingroup$ I think you mean "reals and complexes" in 3? $\endgroup$
    – wgrenard
    Nov 16, 2016 at 1:37
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$\mathbb{R}^1$ is not a vector space over the complexes.

$\mathbb{R}^2$ can be made into a vector space over the complexes, but there are lots of different (but isomorphic) ways to do so. And as a complex vector space it would only be one dimensional.

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There are different fields: Field of rational numbers, algebraic function fields, algebraic number fields, p-adic fields...

The field is the set of scalars for the vector space. Is the division in $\mathbb{R}$ similar to that in $\mathbb{C}$? Absolute value?

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  • $\begingroup$ (1) Right but if you are dealing with just ℝ and ℂ then can you just do everything for ℂ since ℝ is a subset? OR (2) was my statement "just because a property holds over C doesn't immediately tell us it holds over R even though R is a subset of C" accurate? Only (1) or (2) can be true I believe? $\endgroup$
    – Gobi Dasu
    Nov 16, 2016 at 1:12
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    $\begingroup$ @Ruzayqat: that is not so. $\Bbb{C}$ is algebraically closed and $\Bbb{R}$ is not. This makes a significant difference to how linear algebra works out over the two fields. $\endgroup$
    – Rob Arthan
    Nov 16, 2016 at 1:22
  • $\begingroup$ @RobArthan that's what i suspected $\endgroup$
    – Gobi Dasu
    Nov 16, 2016 at 1:23
  • $\begingroup$ Deleted it! Yes you right $\endgroup$
    – Ruzayqat
    Nov 16, 2016 at 1:24

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