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I am reading through an introductory linear algebra textbook. My understanding is that a vector space, defined as a set, is defined over a given field.

If $S = \{ (a_1, a_2): a_1, a_2 \in R \}$, then is $R$ necessarily the field over which the vector space $S$ is defined? If not, why? Please give a counterexample.

Thank you.

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Your $S$ is just a set, unless you define the vector space operations and the scalar multiplication. There is an obvious definition for this set which will make it a two dimensional $S$ vector space, namely by defining addition as $$((a_1, a_2), (b_1, b_2)) \mapsto (a_1+a_2, b_1+b_2)$$ and scalar multiplication by $$(\lambda,(a_1, a_2))\mapsto (\lambda a_1, \lambda a_2)$$ with $\lambda, a_i , b_i\in R$.

But if, e.g., $R= \mathbb{R}$ you can make the set $S$ a one dimensional complex vector space if you define the operations appropriately. Addition would be the same, but multiplication would be defined as $$(z, (a_1, a_2) )\mapsto (z_1a_1-z_2a_2, z_1a_2 + z_2a_1)$$ for $z\in \mathbb{C}$ and $a_i\in \mathbb{R}$. (This is just the complex product of $z_1+iz_2$ with $a_1+ia_2$ written down component wise). With this definition of scalar multiplication the set $S$ is then a one dimensional complex vector space.

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    You can also go the other way and let $S = \Bbb C^2$, while it is really a four-dimensional real space. – 2017-01-05
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    @Arthur yes, there are several variations possible. – 2017-01-05
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    I'm a novice, so your response goes over my head. Can you please rephrase it for a beginner in linear algebra? – 2017-01-05
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    @ThePointer do you know the real and complex number systems and are you aware of the fact that these are fields? – 2017-01-05
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    Yes, that is fine. – 2017-01-05
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    @I'll edit my answer – 2017-01-05
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    So although the elements of our vector space may be in a certain domain (say, real numbers), we can still define the field to be different (say, the complex numbers)? In which case, our operations with scalars would also be different? – 2017-01-05
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    And in addition to my comment above, wouldn't this invalidate the set as a vector space, since the operations would not satisfy the requirements for a vector space? – 2017-01-05
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    @ThePointer yes, sure. – 2017-01-05
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    Ok, I think I understand. Thank you for your assistance. – 2017-01-05