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### Vector Space Dimensions

```
Date: 12/09/97 at 01:24:18
From: Dan Dumitrescu
Subject: Vector spaces

Hi,

Why does C^2 considered as a vector space over the complex numbers
have dimension 2, but as a vector space over the real numbers have
dimension 4?

I've been trying to figure this out for a good while now, but to no
avail. Thank you for your help.

Dan
```

```
Date: 12/09/97 at 06:56:44
From: Doctor Pete
Subject: Re: Vector spaces

Hi,

The short answer to your question is that C is a field of dimension 2
over the reals. More to the point, every element in C corresponds to
an ordered pair (x,y) where x, y are real, where addition and
multiplication on these pairs are defined as follows:

(a,b) + (c,d) = (a+c,b+d)

Then (a,b) = a+bi, and in particular, (1,0) = 1, (0,1) = i. Clearly
this implies dim(C) = 2 over R, and hence dim(C^2) = 4 over R. But
dim(C^2) over C is 2, because a basis B of C^2 can be {(1,0),(0,1)}
over C, since over C we may multiply (1,0) by any complex number z to
obtain (z,0), and similarly, (0,1) by w to obtain (0,w), and hence
every element in C^2 is obtainable as a (complex) linear combination
of elements of B. However, this fails when we consider C^2 over R,
because we can no longer multiply elements from B by complex numbers
in our linear combinations.  Rather, a basis
D = {(1,0),(i,0),(0,1),(0,i)} fits the bill, or if we prefer,

D = {((1,0),(0,0)),((0,1),(0,0)),((0,0),(1,0)),((0,0),(0,1))}.

So with this example, we see that the question of what the dimension
of a vector space is depends on what space we consider this dimension
to be over, which is the same thing as asking what types of numbers we
allow in our "linear combinations." Usually, we calculate dimension
over R, but not always, as in this case.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra
High School Linear Algebra

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