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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

3 Answers 3

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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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    @Hardy: $\mathbb{R}$ is uncountable, but any finite-dimensional vector space over $\mathbb{Q}$ is countable (exercise).2012-01-28
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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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    Thanks mate. i appreciate your help it has been very educational.2012-01-30