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Hi i'm having trouble with this homework question:

We view $\mathbb{C}^2=\{(w,z) : w,z\in \mathbb{C} \}$ as a vector space over $\mathbb{C},\mathbb{R},\mathbb{Q}$ (complex numbers, real numbers and rational numbers) let $$x_1=(i,0), x_2=(\sqrt2, \sqrt5), x_3=(0,1), x_4=(i \sqrt3,\sqrt3), x_5=(1,3)\in \mathbb{C}^2.$$ Determine the $\dim_F(Span(x_1,x_2,x_3,x_4,x_5))$ for $F=\mathbb{C},\mathbb{R},\mathbb{Q}$.

So what I've done so far is Gaussian elimination on the matrix of whom columns are $x_1,x_2,x_3,x_4,x_5$. From this I think that the number of basis vectors would be 2 so the dimension is 2. Does this mean it would be 2 for all of them as I have always taken the number of basis vectors to equal the dimension. I think that this is wrong though.

Many thanks for your help.

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That's correct over $\mathbb C$, as the scalars are complex and you must have used complex scalars to row-reduce.

But first, how were you told to express $(i, 0)$ as a vector over $\mathbb R$? One needs two real coordinates to express a complex number $z = a+bi$, where $z\in\mathbb C$ and $a, b\in\mathbb R$; so over $\mathbb R$ the way I would write it, your vector $x_1$ becomes $(0, 1, 0, 0)$, $x_2 = (\sqrt 2, 0, \sqrt 2, 0)$, and so on. $\mathbb C^2$ is 4-dimensional over the reals, so do your Gaussian elimination again, but this time, only using real scalars.

The next thing you have to consider is that real numbers may be independent of each other over $\mathbb Q$, so even more vectors may be linearly independent if the only scalars allowed are rational. I would write the vectors in the same way as over $\mathbb R$. But for example, is there any rational scalar $c$ for which $\sqrt 2 = c * 3$? Use this to determine if $x_5$ is in the span of $x_2$, and so on. (I.e. do Gaussian elimination again, but now with only rational scalars.)

As this is a homework problem I'll leave it there, but I hope this gives you a path forward.

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  • $\begingroup$ Yes this does help a lot thank you $\endgroup$ – Thomas Mar 22 '17 at 18:51

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