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)
     (a,b) * (c,d) = (ac-bd,ad+bc).

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
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