LEt $V = \{ (x,y) : x,y \in \mathbb{C} \} $. The addition and scalar multiplication are the usual ones. We know $V$ is a vector space over $\mathbb{C}$ with dimension $2$. My question: Why does $V$ is a vector space space over $R$ and $\mathbb{Q}$ also ?? I am having difficulties understanding this. Why the dimension of $V$ is $4$ if $V$ is a vector space over the reals? and why is it inifinite if it is over $\mathbb{Q}$ ??
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$\mathbb{C}^2\cong\mathbb{R}^4$,
For verifying the dimension of $\mathbb{R}$ over $\mathbb{Q}$ is infinite, see Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?
One way to prove is that any finite subset of $\mathbb{R}$ cannot span the whole space just using rational coefficients.
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Any $(x,y)\in V$ can be written as a linear combination with real coefficients in the following way: $$\eqalign{(x,y) &=(a+bi,c+di)\cr &=a(1,0)+b(i,0)+c(0,1)+d(0,i)\ .\cr}$$ Therefore the four vectors on the RHS span $V$. Also, they are linearly independent (I'll leave you to check this for yourself); so they form a basis for $V$ as a vector space over $\Bbb R$; so $V$ has dimension $4$.
David
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Gotcha! how about the case over the rationals? – ILoveMath Oct 21 '14 at 00:08
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The rational case is harder to prove rigorously. You could use countability arguments, if you have seen that sort of thing. – David Oct 21 '14 at 00:10