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April 30th, 2015, 12:47 AM   #1
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Basis for Complex Vector Space over the Complex and then the Real Fields

I am trying to make sense of the basis for the complex vector space C over the field of scalars F, where F = C (complex field), and then F = R (real field).

Intuitively for me I would have assumed B = {1,i} for C over F = C and B = {1} for F = R. This would yield dim(C(C)) = 2 and dim(C(R)) = 1, which, again intuitively makes sense to me. But it is the other way around. I have worked through the proofs but I still don't click it properly, I think.

So far I understand it as follows: When F = C, the vector in the vector space C, although complex a + ib, can be viewed as a single 'bag of things'. When F = R, the vector is the vector space C, still complex a + ib, is seen as being made up out of two 'things'. I am still not satisfied though.

Can anyone offer some more insights?
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April 30th, 2015, 04:00 AM   #2
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Suppose we have a basis vector $\newcommand{\i}{\mathrm i}a + b\i$ (where $a$ and $b$ are non-zero real numbers). Then with complex coefficients we can form
$$(x + y\i)(a + bi) = (ax - by) + (bx + ay)\i$$where $x$ and $y$ are real numbers.

Now, if we wish to make some vector $r + s\i$ ($r$ and $s$ both real) we have$$r = ax - by \qquad s=bx + ay$$ which means that we choose$$x = {ar-bs \over a^2 - b^2} \qquad y = {as-br \over a^2 - b^2}$$
The vector $a + b\i$ therefore spans the vector space $\mathbb C$ as required. Since this basis has only one vector, the dimension is 1.

With real coefficients we have $$a = b\i = a(1 + {b \over a}\i)$$ (where $a\ne 0$) and any real multiple of this has the imaginary part in fixed proportion ${b \over a}$ to the real part. We cannot therefore span the space $\mathbb C$ with a single vector. We can do a similar analysis for $b \ne 0$.
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Last edited by v8archie; April 30th, 2015 at 04:08 AM.
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