# Complex vector spaces

I was wondering if someone could please shed some light on how the following two vector spaces are different. The examples are from Paul Halmos's "Finite Dimensional vector spaces" book and the author asks the reader to verify that they are different, but i am not so sure if i understand the two examples well enough to spot all the differences.

(1) Let $\mathbb C^1$ be the set of all complex numbers; if we interpret "$x + y$" and "$ax$" as ordinary complex numerical addition and multiplication, $\mathbb C^1$ becomes a complex vector space.

(2) If, in the set $\mathbb C$ of all complex numbers, addition is defined as usual and multiplication of a complex number by a real number is defined as usual, then $\mathbb C$ becomes a real vector space.

Is the difference between these two vector spaces examples only that in the first the scalar field is also the set $\mathbb C$ and in the second one the scalar field is $\mathbb R$ and hence only the multiplication and distributive properties would behave differently? What about dimensions of the two examples?

Any Help would be highly appreciated.

Cheers Hardy

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They are vector spaces over different fields. The first is a one-dimensional vector space over $\mathbb{C}$ ($\{ 1 \}$ is a basis) and the second is a two-dimensional vector space over $\mathbb{R}$ ($\{ 1, i \}$ is a basis).

This might have you wondering what exactly the difference is between the two perspectives. The point is that when you only consider real vector spaces, "multiplication by $i$" is not a well-defined operation. In other words, a complex vector space is precisely a real vector space together with a "multiplication by $i$" operation or, more formally, with a (real-)linear operator $J : V \to V$ such that $J^2 = -I$ (where $I$ is the identity).

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 @Zev: no, I think that was already a problem. – Qiaochu Yuan Jan 28 '12 at 6:02 Hm, I thought I remembered a single backslash working correctly before. Perhaps it was just wishful thinking though. – Zev Chonoles♦ Jan 28 '12 at 6:03 so to clarify as an example 2 + i belongs to the first vector space where as (2, 3i) is an example of the second ? Thank you i really appreciate your help. Although i am still a little confused as the multiplication of scalars in R would required for the second coordinate in second example. for instance in the axiom a(bx) = (ab)x if x was (2, 3i) in the second example do n't we get (2ab, 3abi) ? – Hardy Jan 28 '12 at 6:08 @Hardy: what elements belong to the two spaces are irrelevant; one can identify them as sets if one desires. The point is the structures that exist on each, which is extra data that one specifies. – Qiaochu Yuan Jan 28 '12 at 6:20 How did u determine the basis for these vector spaces ? I know the definition of basis, and the 2 properties, linear independence and that all vectors can be expressed as sum of the basis vectors, though i am struggling at finding basis ? Any help would be highly appreciated. – Hardy Jan 28 '12 at 6:25

As you mention they are vector spaces over different fields. For example, $\mathbb C^1$ is $1$-dimensional as a vector space over $\mathbb C$ while $\mathbb C^1$ is a two-dimensional vector space over $\mathbb R$.

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In the comments on Qiaochu's answer, OP asks how one finds a basis for a given (finite dimensional) vector space. Here's one approach.

You pick a nonzero vector. You look at all the multiples of that vector. If that gives you the entire vector space, you're done: that one vector form a basis.

If the multiples of that vector don't give you the entire vector space, pick some vector you don't get as a multiple of the first one, and look at all the linear combinations of the two vectors you have selected - the "span" of the two vectors. If that span is the entire vector space, you're done: that pair of vectors forms a basis.

If the span of the two vectors isn't the whole vector space, pick a vector that isn't in the span, and look at the span of the three vectors you have selected. I could write out a few more iterations of this procedure, but I hope by now the idea is clear; just keep picking vectors not yet in the span until the span is everything, at which point you have a basis.

For practice, find a basis for the complex numbers as a vector space over the real numbers. Then throw that basis away, and find an entirely different one. Repeat until you're confident that you have no more to learn from this exercise.

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Cool Thanks for that bud, I suppose for the exercise in your comment. I think there maybe infinite basis of dimension 2 as one could pick (1, i), (2, i).....(3, 10i) any possible combination of (R, Ri) as basis. I think i get the point. I am also starting to think a vector space formed by vectors in R or C with the field Q has finite dimensions as originally i thought the denominators of the rationals can only come from scalars, but of course Q is a subset of R and C,hence we would have basis {1} and {1, i} for R and C cases respectively. please let me know if i have it wrong. Cheers – Hardy Jan 28 '12 at 22:52
Yes, there are infinitely many different 2-element bases for $\bf C$ as a vector space over $\bf R$. They don't all look like the ones you mention, e.g., $\\{2+3i,4+5i\\}$ is another such. But both $\bf R$ and $\bf C$ are infinite-dimensional as vector spaces over the rationals. Using the rationals as scalars, the only multiples of 1 you get are the rationals, so you don't get $\sqrt2$, $\pi$, etc. – Gerry Myerson Jan 28 '12 at 23:08
$\bf C$ is a 2-dimensional space over $\bf R$ and it has a 2-element basis as a vector space over $\bf R$. $\bf C$ is an infinite-dimensional vector space over $\bf Q$, and it doesn't have a 2-element basis, or any finite basis, over $\bf Q$. A vector space has two things, a set $V$ of vectors and a field $F$, and you can't say what the dimension of $V$ is unless you know which field $F$ is intended. The same set (e.g., $\bf C$) can be a vector space over many different fields, and the dimension can depend on which field you have in mind. – Gerry Myerson Jan 29 '12 at 2:00
You will continue to be confused, as long as you express yourself so carelessly. Don't say "there maybe infinite basis for it" when what you really mean is "there may be infinitely many bases for it." Also, an infinite-dimensional vector space is guaranteed to have infinitely many different bases. – Gerry Myerson Jan 29 '12 at 8:51
"there maybe infinite basis for it" simply isn't English, so it's hard to tell what it means, but the juxtaposition of "infinite" and "basis" suggests you are talking about a single basis with infinitely many elements. "there may be infinitely many bases for it" says nothing about how big any one basis might be, just that there are infinitely many of them. $R-Q$ has no place in a basis, as the set of all irrationals is not linearly independent over the rationals - indeed, the subset $\lbrace\pi,-\pi\rbrace$ is not linearly independent over the rationals. – Gerry Myerson Jan 30 '12 at 0:00