Vector Space DimensionsDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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