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Vectors are an abstract notion. If you have two sets and two operations that satisfy the definition of a vector space, then you have a vector space; and we refe

Vectors are an abstract notion. If you have two sets and two operations that satisfy the definition of a vector space, then you have a vector space; and we refer to elements of the vector set as "vectors" within that vector space. The observation here is that set of real value functions, combined with the set of real numbers, and the natural notion of function addition and multiplication by a real number satisfies the definition of a vector space. As a result all the results of linear algebra can be applied to real valued functions. It is true that any vector space is isomorphic to a vector space whose vectors are functions. Linear algebra does make a lot of usage of that result, but it is different from what the article is discussing.

ttoinou 1y ago
I agree but we're using functions for different things here. Yes some specific families of functions can be treated as vector spaces. In this article it seems like the author is pretending to take all real->real functions and treating them as if they are a vector space, whatever the content of the functions, quote : we’ve built a vector space of functions and later he admits it is impossible Ideally, we could express an arbitrary function f as a linear combination of these basis functions. However, there are uncountably many of them—and we can’t simply write down a sum over the reals. Still, considering their linear combination is illustrative: They are uncountable because they are aleph1
- 998244353 1y ago
  The set of all real->real functions is still a vector space. This vector space also has a basis (even if it is not as useful): there is a (uncountably infinite) subset of real->real functions such that every function can be expressed as a linear combination of a finite number of these basis functions, in exactly one way. There isn't a clean way to write down this basis, though, as you need to use Zorn's lemma or equivalent to construct it.
  - ttoinou 1y ago
    I'd love to read more about that, he's not talking about that at all in this article though
    - lmm 1y ago
      He doesn't need to talk about it (though you might like to look up the notorious Hilbert's Basis Theorem); it happens to be the case that any vector space has a basis, but even if you don't know that, a vector space is still a vector space and its elements are still vectors.
      - ttoinou 1y ago
        The choice of basis is very important for the applications, which is what the article is about
    - tel 1y ago
      If you're familiar with Zorn's Lemma, the construction is just to order bases by inclusion and to consider chains created by noting that there must be an independent dimension and adding it inductively. You can upper bound each of these chains by unioning the members of the chain (which preserves linear independence). By Zorn's Lemma that means there is a maximal linearly independent system and if an element existed outside of that system's span it would contradict that maximality.
  - seanhunter 1y ago
    If you restrict yourself to Lebesque-integrable functions, can’t you take the complex Fourier transform of the function and call the terms of the Fourier series a basis, with the coefficients being the components of the vector of the function? This is a bit above my current mathematical paygrade, so forgive me if I’m not expressing the idea accurately but I’m learning a lot both from the article and the ensuing discussion - hopefully you understand what I’m getting at. I think what I may be asking is “Does the complex Fourier transform make a Hilbert space?” but I might be wrong both about that and about that being the right question.
- gizmo686 1y ago
  It is not required for vector spaces to have a basis. As it turns out, the claim that every vector space has a basis is equivalent to the axiom of choice, which seems well beyond the scope of the article. However, the particular vector space in question (functions from R to R) does have a basis, which the author describes. That basis is not as useful as a basis typically is for finite dimensional (or even countably unfitine dimensional) vector spaces, but it still exists.
  - ttoinou 1y ago
    But it's the article talking about vectors as a sequence of reals and having a basis, then extending that to infinite sequences of reals. The author is playing on multiple definitions of vector to produce a "woaw that's cool" effect, and that's bad maths
    - Sharlin 1y ago
      There is only one definition of "vector space" (up to isomorphism anyway), and that's what the author uses. You'll note that he doesn't talk about bases at all, the assumption of a basis is entirely in your mind. The entire point of the article is that the ℝ→ℝ function space is a vector space. A vector space is not required to have a basis, but assuming the axiom of choice, every vector space does have (at least) one, including that of ℝ→ℝ functions.
      - ttoinou 1y ago
        The choice of basis is very important for the applications. It doesnt matter that function could be vector spaces in theory without constructing a base, the article is about "hey look at functions they're as easy to operate in as this thing you already know with 3D vectors"
    - MarkusQ 1y ago
      You are misreading the article. At a guess, you have come into it "knowing what a vector is" from physics or computer science. Specifically, a vector is not "a sequence of reals", it is any object that obeys a certain set of properties. He is not "extending" your definition, he's trying to teach you the actual definition, which is more general than the examples you are used to. (Imagine/remember what it feels like when children first learn that the integers aren't the only number, there are also fractions, then irrationals, then complex numbers...this is a very similar situation). With that in mind, you may want to reread the text and pay attention to the definitions he is using, and not assume that your definitions are the whole story.

ttoinou 1y ago

I agree but we're using functions for different things here. Yes some specific families of functions can be treated as vector spaces. In this article it seems like the author is pretending to take all real->real functions and treating them as if they are a vector space, whatever the content of the functions, quote : we’ve built a vector space of functions and later he admits it is impossible Ideally, we could express an arbitrary function f as a linear combination of these basis functions. However, there are uncountably many of them—and we can’t simply write down a sum over the reals. Still, considering their linear combination is illustrative: They are uncountable because they are aleph1

998244353 1y ago
The set of all real->real functions is still a vector space. This vector space also has a basis (even if it is not as useful): there is a (uncountably infinite) subset of real->real functions such that every function can be expressed as a linear combination of a finite number of these basis functions, in exactly one way. There isn't a clean way to write down this basis, though, as you need to use Zorn's lemma or equivalent to construct it.
- ttoinou 1y ago
  I'd love to read more about that, he's not talking about that at all in this article though
  - lmm 1y ago
    He doesn't need to talk about it (though you might like to look up the notorious Hilbert's Basis Theorem); it happens to be the case that any vector space has a basis, but even if you don't know that, a vector space is still a vector space and its elements are still vectors.
    - ttoinou 1y ago
      The choice of basis is very important for the applications, which is what the article is about
  - tel 1y ago
    If you're familiar with Zorn's Lemma, the construction is just to order bases by inclusion and to consider chains created by noting that there must be an independent dimension and adding it inductively. You can upper bound each of these chains by unioning the members of the chain (which preserves linear independence). By Zorn's Lemma that means there is a maximal linearly independent system and if an element existed outside of that system's span it would contradict that maximality.
- seanhunter 1y ago
  If you restrict yourself to Lebesque-integrable functions, can’t you take the complex Fourier transform of the function and call the terms of the Fourier series a basis, with the coefficients being the components of the vector of the function? This is a bit above my current mathematical paygrade, so forgive me if I’m not expressing the idea accurately but I’m learning a lot both from the article and the ensuing discussion - hopefully you understand what I’m getting at. I think what I may be asking is “Does the complex Fourier transform make a Hilbert space?” but I might be wrong both about that and about that being the right question.
gizmo686 1y ago
It is not required for vector spaces to have a basis. As it turns out, the claim that every vector space has a basis is equivalent to the axiom of choice, which seems well beyond the scope of the article. However, the particular vector space in question (functions from R to R) does have a basis, which the author describes. That basis is not as useful as a basis typically is for finite dimensional (or even countably unfitine dimensional) vector spaces, but it still exists.
- ttoinou 1y ago
  But it's the article talking about vectors as a sequence of reals and having a basis, then extending that to infinite sequences of reals. The author is playing on multiple definitions of vector to produce a "woaw that's cool" effect, and that's bad maths
  - Sharlin 1y ago
    There is only one definition of "vector space" (up to isomorphism anyway), and that's what the author uses. You'll note that he doesn't talk about bases at all, the assumption of a basis is entirely in your mind. The entire point of the article is that the ℝ→ℝ function space is a vector space. A vector space is not required to have a basis, but assuming the axiom of choice, every vector space does have (at least) one, including that of ℝ→ℝ functions.
    - ttoinou 1y ago
      The choice of basis is very important for the applications. It doesnt matter that function could be vector spaces in theory without constructing a base, the article is about "hey look at functions they're as easy to operate in as this thing you already know with 3D vectors"
  - MarkusQ 1y ago
    You are misreading the article. At a guess, you have come into it "knowing what a vector is" from physics or computer science. Specifically, a vector is not "a sequence of reals", it is any object that obeys a certain set of properties. He is not "extending" your definition, he's trying to teach you the actual definition, which is more general than the examples you are used to. (Imagine/remember what it feels like when children first learn that the integers aren't the only number, there are also fractions, then irrationals, then complex numbers...this is a very similar situation). With that in mind, you may want to reread the text and pay attention to the definitions he is using, and not assume that your definitions are the whole story.