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I am studying vector space from Linear Algebra Done Right by Sheldon Axler.

Going by the notion of vectors I started to visualize elements as vectors of $n$ dimension for a vector space $R^n$. I could visualize the idea of closure property w.r.t linear combination of $n$ dimensional vectors.

But I could not understand $F^S$ properly. They have taken example :

$F = R\;$ and $S = \left [ 0,1 \right ]$ and said that $R^S$ is a vector space. I could not co-relate this function space to the vector form.

Please explain a little bit about vector space in terms of function. or more formally a more generic view of vector space.

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    $\begingroup$ Where do you get stuck when checking the definition? $\endgroup$ – Jack Jul 20 '17 at 20:09
  • $\begingroup$ "Going by the notion of vectors I started to visualize elements as vectors of "n" dimension for a vector space $R^n$." This is bad if you want to understand vector spaces abstractly especially for your question. $\endgroup$ – Jack Jul 20 '17 at 20:11
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    $\begingroup$ Sum and scalar multiplication of functions is defined point wise: if $f,g \in F^S$, then $f+g \in F^S$ and $(f+g)(x) = f(x)+g(x)$ for every $x \in S$, the same for scalar multiplication, this gives the structure of vector space.Think of it as "lifting" the vector space structure of $R$ to $F^S$. $\endgroup$ – Gilberto López Jul 20 '17 at 20:13
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    $\begingroup$ An abstract vector does not need to have anything to do with ${\bf R}^n$. And in fact, in your example of ${\bf R}^{[0,1]}$, you can never visualize it as ${\bf R}^m$ for some $m$ because it is not of finite dimension at all. $\endgroup$ – Jack Jul 20 '17 at 20:14
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    $\begingroup$ Formally speaking a vector is an element of a vector space, and there are vector spaces of functions like $F^S$, matrices, "numbers" (a field is a vector space defined over itself/its subfields, $R$ is a real vector space, a $Q$ vector space), to avoid confusions think more abstractly, think vectors as elements of a vector space in general and not only as tuples. You'll see later that every vector space is a direct sum of copies of its underlying field, as many copies as its dimension. $\endgroup$ – Gilberto López Jul 20 '17 at 20:53
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There are ways in which $F^S$ relates to $F^n$, and ways in which it doesn't.

Each element of $F^S$ is determined by taking each element of $S$ and assigning it a value from $F$. This is, by definition, an $F$-valued function on $S$.

Each element of $F^n$ is an $n$-tuple of the form $(f_1, \ldots, f_n)$, with $f_i \in F$. This can be reinterpreted as an $F$-valued function on a set with $n$ elements. So there is something that makes $F^n$ resemble $F^S$.

What is very dangerous, however, is to go further and think of the function values in each element of $F^S$ as coordinates, which is not correct. Take a look at an element of $F^n$ of the form $v = (f_1, \ldots, f_n)$. Each $f_i$ is a coordinate of $v$ with respect to the basis $e_i$ ($0$ everywhere except in the $i$-th position). So in the finite case, coordinates and function values coincide. Everything works out.

However, consider an element $w \in F^S$. Its values are determined by $w(s)$ for $s \in S$, so its tempting to say that these are the coordinates of $w$ with respect to a basis of functions that take the value $1$ at particular $s$ and $0$ elsewhere. The problem with this claim is that if $S$ is an infinite set, and $w$ is a function that is everywhere nonzero, then $w$ has infinitely-many nonzero coordinates, which would mean it is expressed as an infinite sum of basis vectors. Infinite sums in linear algebra are not allowed, because they require notions of convergence conferred by calculus, notions that may or may not exist for generic vector spaces. There are ways to introduce infinite sums into linear algebra, but it's a whole different story.

So while it is true that $F^S$ is a vector space, if $S$ is an infinite set, then its basis cannot be expressed in a simple way using the values each function takes on $S$. Indeed, any such basis requires some form of the axiom of choice to describe.

All of this can be summarized in the following statement:

Let $V$ be a vector space over $F$ with a basis $\beta$. Then $V$ is isomorphic to $F^{\beta}$ if and only if $\beta$ is a finite set, a.k.a. $V$ is finite-dimensional.

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Even though I agree with most of what is written in the comments, I believe there is some similarity between $\mathbb{R}^{[0,1]}$ and $\mathbb{R}^n$. After all, $\mathbb{R}^n$ is the set of all functions from the set $\{1,\ldots,n\}$ to the real line. So if it helps, you can think of $\mathbb{R}^{[0,1]}$ as something like $\mathbb{R}^n$, only with infinitely many coordinates.

It should be mentioned that "infinitely many" is not very accurate, as there are many different levels of infinity.

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The abstract approach: you have a definition of addition and scalar multiplication that meets the definition of a vector space. Therefore all the theorems you have solved for finite dimensional vector spaces are still valid.

But this might not be very satisfying. We need some examples that you are already comfortable with that you can fall back on.

Suppose we consider function that are defined at a discrete set of points. $\{ x_1,x_2, \cdots, x_n\}$

The set of functions that map $(x_1,\cdots x_n)$ to $(f(x_1),\cdots, f(x_n))$ Makes an n-dimensional vector space.

Now let $n$ go to infinity.

Some more ways to think about it....

$P(x)$ is set of n-degree polynomials. And, hopefully you have worked with this vector space some a bit, and feel comfortable thinking about at least finite degree polynomials as vectors.

The set of functions that can be expressed as a polynomial (i.e. as a Taylor series) is a vector space.

In a similar line of thinking. The trigonometric fictions $\sin nx, \cos nx$ form the basis a set of functions over the domain $(-\pi, \pi).$ The Fourier series is an expression of these functions in this basis.

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