# Vector Spaces are Free Objects

Warning: I know little linear algebra and my assertions below may all be incorrect. I am interested in lists --i.e., free monoids-- and my interest has led me to [finite-dimensional] vector spaces.

Let π be any field, then we can freely equip any [finite] set S with a π-vector space $$F(S) = (S β π)$$, a set of functions where vector addition is performed pointwise, and likewise for the other vector operations. Moreover any function $$S β |π±|$$ from $$S$$ to the underlying set of a vector space π± can be extended to a linear operator $$F(S) β π±$$ that behaves the same on elements of $$S$$ construed as vectors of $$F(S)$$.

That is, $$F$$ is a βfree functorβ: It constructs the least vector space that contains a (copy of a) given set.

It is known that every vector space π₯ admits a basis Ξ² and so is isomorphic to $$F(Ξ²)$$ --which is essentially the dual space of π₯. That is, every vector space is free; i.e., is in the image of functor $$F$$.

Of-course we have to choose a basis to realise a vector space as a free object; there is no canonical basis.

( Aside: With this in-hand, we can easily prove equi-dimensional spaces are necessarily isomorphic: That they're equi-dimensional means their basis are in bijection, but free objects are unique up to unique isomorphism and so their F-images must be unique up to unique isomorphism as well. That is, $$π₯ β π^{dim π±} = π^{dim π±} β π¦$$. Neat stuff! )

Questions:

1. Intuitively why is it that vector spaces admit basis.

• I'm not interested in a proof.
• For example, why is it that monoids or rings are not always generated by some set, but vector spaces are. The definition of a vector space does not immediately give rise to a basis.
2. All [finite dimensional] vector spaces admit basis and so the objects of π½π¬πͺ, the category of finite dimensional vector spaces and linear operators, are all images of a free functor. (Assuming we have a way to choose basis.)

• Is there a name for categories whose objects are all isomorphic to an image of a free functor?
• What other examples of such categories are there?
• Are there any constructions that produce such categories.
3. Lists over a set $$S$$ are isomorphic to $$β_{n β β} SβΏ$$ and provide the free monoid over a given set $$S$$. Why is it that lists, whence monoids, are far more prevalent than their vector space counterparts within computing science.

• It is probably worth noting that the claim that every vector space can be given a basis requires the Axiom of Choice (or at least goes beyond ZF). In a similar vein, the claim fails in constructive foundations and I believe it fails much more severely. Sep 21, 2018 at 20:21
• It's at best difficult to explain something "intuitively" because any intuitive explanation will only work if you have the necessary intuitions. Different people have different intuitions. In a sense, you're asking for someone else to do the work of understanding for you. For finite-dimensional vector spaces, the process is fairly easy... pick any nonzero vector, then keep picking a vector not in the span of previously picked vectors. After a finite number of steps you will stop and at that point you have a basis. Sep 21, 2018 at 20:24
• Intuitively, you can say that this process works for infinite vector spaces, you just need an infinite or even uncountable number of steps. But that's just a fuzzy way of describing Zorn's lemma. Sep 21, 2018 at 20:26
• @DietrichEpp Why must that process necessarily terminate? --in the finite dimensional setting. Sep 22, 2018 at 11:12
• By definition, a finite-dimensional vector space is one with a finite basis. If it didnβt terminate it would be an infinite dimensional vector space. Sep 23, 2018 at 0:23