The title says it all. Since the scalars are often taken to be the real numbers, is it because the existence of multiplicative inverses for scalars is already assumed? It seems like there must be some deeper answer.
Why do the axioms of a vector space not include a multiplicative inverse for scalars?
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0Because the study of scalars and their properties among themselves is already studied. Read into fields and rings. – 2017-02-09
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2The scalars are a field by definition. You can relax this requirement, in which case you are dealing with a $module.$ – 2017-02-09
2 Answers
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It's because a vector space is defined over a field $F$, from which the scalars are drawn. Since $F$ is a field, all nonzero elements are guaranteed to have inverses. There is no need to repeat that in the definition of vector space.
In some familiar examples of vector spaces, $F$ is $\mathbb{R}$, as you observed, but it could be $\mathbb{C}$ or $\mathbb{Q}$, or some other field.
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It is part of the axioms, implicitly, because the scalars are assumed to be a field, not just the reals or the complex numbers.See wikipedia for the definition of a field. We can also have vector spaces defined over finite fields, e.g. We can even have "vector spaces" over unitary rings ,same axioms as for vector spaces, only just assuming the scalars are a commutative ring with $1$. These are called "modules". These do not need to have a basis ,for that we need a field structure on the scalars...