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The title pretty much says it all; is a vector space meaningless without a choice of basis which has been predefined? I've been reading through my algebra book and it says something like, take a vector $v$ and let $x$ be its coordinate with respect to a particular basis. My question is, how can you even take a vector before you assign a basis to its vector space? I know similar questions may have been asked here, but they all involve something about the Axiom of Completeness, which I have not learned about yet. Thanks so much for your help!

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    $\begingroup$ For example, the set $V$ of all $m \times n$ matrices $A$ such that $\text{trace}(A) = 0$ is a vector space. We have not yet specified a basis for $V$. $\endgroup$ – littleO Aug 14 at 0:22
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    $\begingroup$ Linear algebra is largely a collection of conceptual rules that relate abstractly defined subspaces of vector spaces. Like the image of the adjoint of a linear map is the annihilator of its kernel. The examples often rely on bases, and the means of proof are very concrete, but the content is abstract. $\endgroup$ – Charlie Frohman Aug 14 at 0:26
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    $\begingroup$ Indeed, in an infinite-dimensional vector space, in general you can't write down an explicit basis at all, as the existence of a basis depends on the axiom of choice. $\endgroup$ – Bungo Aug 14 at 0:35
  • $\begingroup$ @Bungo Whoa, I didn't know that! I guess even in math everything is somewhat relative? $\endgroup$ – DavidNiu Aug 14 at 0:37
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    $\begingroup$ "even in math everything is somewhat relative" I don't know what that means. But I like the enthusiasm. $\endgroup$ – littleO Aug 14 at 0:44
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Any set of objects that satisfies the relevant axioms is a vector space. Not all vectors are tuples of scalars, although that’s certainly what you work with most in introductory courses. For instance, the set of real-valued functions defined on the interval $[0,1]$ forms a vector space over the reals. I’d be hard-pressed to even define a basis for this space at all.

Now, for a finite-dimensional space over some field $\mathbb K$, choosing a basis amounts to defining an isomorphism between the space and $\mathbb K^n$, which is why you can focus on tuples of scalars when studying finite-dimensional vector spaces.

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  • $\begingroup$ Hmm, that makes sense, this might be a dumb question but I suppose the set which the vector space is defined over has be predefined? $\endgroup$ – DavidNiu Aug 14 at 0:28
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    $\begingroup$ @DavidNiu Not a dumb question at all. A vector space does involve two sets: the vectors proper, and the scalars (along with two operations with which the latter set makes a field). $\endgroup$ – amd Aug 14 at 0:31
  • $\begingroup$ okay, and I take that these spaces and subspaces of vectors can be kind of recursively defined, and if you keep going you eventually arrive at a set of axioms from which you just have to assume things exist? Does that even make sense? $\endgroup$ – DavidNiu Aug 14 at 0:35
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    $\begingroup$ @DavidNiu Not quite. Unlike the axioms of a geometry, with which you build a structure from the ground up, the vector space axioms codify the properties that the two sets and associated operations must have in order to be called a vector space. They are the common “conceptual rules” that Charlie Frohman talks about in his comment. $\endgroup$ – amd Aug 14 at 0:40
  • $\begingroup$ Okay, that makes sense, I guess I have a lot more math to learn . . . anyway, thanks so much for answering my question! $\endgroup$ – DavidNiu Aug 14 at 0:42

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