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Vector Spaces of Infinite Dimension

  1. Jun 19, 2011 #1
    I was hoping you guys could help me in understanding some vector spaces of infinite dimension. My professor briefloy touched n them (class on linear algebra), but moved on rather quickly since they are not our primary focus.

    He gave me the example of the closed unit interval where f(x) is continuous over I. From my studies in analysis, I understand how there exists an infinite amount of points within the interval, and am having trouble on connecting this fact to the dimension of the space.

    Can anyone offer me a more complete explanation on this? I'd also love to hear some other infinite dimension spaces and their prospective applications. I don't like only grasping things of R^n ;).

    Thanks so much!
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  3. Jun 19, 2011 #2


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    A vector space is said to be n-dimensional if it contains a linearly independent set with n members, but none with n+1 members, and is said to be infinite-dimensional if it for each positive integer n contains a linearly independent set with n members.

    He wasn't saying that [0,1] is infinite dimensional. He was saying that the set of functions from [0,1] into ℝ, with the standard vector space structure*, is infinite-dimensional. This is easy to see. For each non-negative integer k, define [itex]u_k(x)=x^k[/itex] for all x in [0,1]. Then for each positive integer n, [itex]\{u_0,\dots,u_{n-1}\}[/itex] is a linearly indepdendent set with n members.

    *) Strictly speaking, that set of functions is just a set, not a vector space. Let's call the set V. If we for each a in ℝ and each f,g in V, define f+g and af by [tex]\begin{align} &(f+g)(x)=f(x)+g(x)\\ & (af)(x)=a(f(x))\end{align}[/tex] for all x in [0,1], then the triple (V,addition, scalar multiplication) is a vector space, with underlying set V. There are many vector spaces with the same underlying set. (Just define addition and scalar multiplication differently). I mentioned the "standard vector space structure" to indicate that I was talking about this particular vector space.

    It is however completely standard to be sloppy and refer to V as a vector space even though it isn't. It won't cause confusion in this case, because there's hardly ever any reason to define addition and scalar multiplication differently.
    Last edited: Jun 19, 2011
  4. Jun 19, 2011 #3


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    R^n is naturally the space of all real valued functions on the set {1,....,n}. That is why R^n has dimension n. The set of all real valued functions on an infinite set has infinite dimension.

    If S is any set, even the subspace of all real valued function on S that only take non zero values at a finite number of points of S, already has dimension equal to the cardinality of S.
  5. Jun 19, 2011 #4
    You are already very familiar with one infinite dimensional vector space.

    Consider the set of all polynomial expressions of one variable with real coefficients. For example [itex]x^2 + 1[/itex], or [itex]5x^7 - 3x^6 + sqrt(2)x^5 -47x^4[/itex]. The collection of all such polynomials is denoted [itex]\mathbb{R}[x][/itex].

    Given two such polynomials we can add them by adding the corresponding coefficients. There are additive inverses. We can also multiply by scalars, for example [itex]3 * (x^2 - 6) = 3x^2 -18[/itex].

    I'll leave it to you to verify that [itex]\mathbb{R}[x][/itex] with this addition and scalar multiplication is a vector space. It would be instructive for you to work through the list of vector space axioms and convince yourself that [itex]\mathbb{R}[x][/itex] is indeed a vector space.

    What is a basis for this space? I claim that the set of vectors

    B = {[itex]1, x, x^2, x^3[/itex], ...}

    is a basis. You should convince yourself that:

    a) Any polynomial in [itex]\mathbb{R}[x][/itex] can be written as a finite linear combination of elements in B. In other words, B spans [itex]\mathbb{R}[x][/itex]; and

    b) No proper subset of B spans [itex]\mathbb{R}[x][/itex].

    So [itex]\mathbb{R}[x][/itex] is an infinite-dimensional vector space that is very familiar to us from high school math.

    The example of the set of all continuous functions on [0,1] is also a vector space. Again, you should walk down the list of axioms and convince yourself that this is so. It's infinite dimensional, but it's impossible for us to visualize a basis.
    Last edited: Jun 19, 2011
  6. Jun 20, 2011 #5
    Thanks for the explanation! I wrote it in my own words to make sure I'm grasping the root of problem properly. Mind confirming me?

    All functions produce an image space. This space has a dimension, based on the series of functions f_n(x).
    Where S = { f_n(x)=x_n | n ∈ ℕ } for all x in [0, 1]. S is of infinite dimension because it is closed under scalar addition:
    Let a_k,b_k ∈ [0, 1] for f(a_k) = a_k^n and f(b_k) = b_k^n so f(a_k^n) + f((b_k^n) = a_k^n + b_k^n =( a_k + b_k)^n= f(a+b) Q.E.D.
    Also closed under scalar multiplication. Let c ∈ ℝ. f_n(cx) = cx_n = c(f+n(x)) Q.E.D
    Therefore, S is a vector space and of infinite dimension.
  7. Jun 20, 2011 #6


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    Not sure what you mean, but I would just say for any set X, the set of all functions from X into ℝ can be turned into a vector space by the standard definitions of addition and scalar multiplication.

    [itex]f_n[/itex] is a function. [itex]f_n(x)[/itex] is a member of its codomain, i.e. it's a real number. The set you have in mind is probably [itex]S=\{f_n|n\in\mathbb N\}[/itex], where each [itex]f_n[/itex] is defined by [itex]f_n(x)=x^n[/itex] for all x.

    The term is "addition", not "scalar addition". The "scalars" associated with our vector space are real numbers, and addition is an operation that takes two functions to a function. It doesn't involve any scalars (i.e. real numbers). S is not closed under addition. Note e.g. that [itex]f_1+f_2[/itex] isn't equal to [itex]f_n[/itex] for any n. S is however an infinite set, but that only means that it has infinitely many members.

    If you want to show that S is closed under addition, you need show that if f,g are in S, then f+g is in S.

    S is not. Note e.g. that [itex]2f_3[/itex] isn't equal to [itex]f_n[/itex] for any n.

    Don't you mean [itex]f_n(cx)=(cx)^n=c^nx^n[/itex]? Anyway, if you want to show that S is closed under scalar multiplication, you need to show that if c is in ℝ and f is in S, then cf is in S. (Note that this operation does involve a scalar. Hence the name "scalar" multiplication).

    It's not a vector space, so it doesn't have a dimension. It is however a basis of a vector space.
  8. Jun 20, 2011 #7
    Disconfirmed. You are not understanding what the space of continuous functions is.

    Given any two functions f and g, we define a new function f+g by saying

    (f+g)(x) = f(x) + g(x) for any point x.

    For example if f(x) = e^x and g(x) = sin(x), then (f+g)(3) = e^3 + sin(3).

    Likewise you can define pointwise multiplication by scalars.

    What you need to show is that the set of all such continuous functions, under pointwise addition and scalar multiplication, is a vector space. To do that, you must go down each of the vector space axioms and show that it's true about the set of continuous functions.

    The example given that S = {1, x, x^2, x^3, ...} is linearly independent, shows that the dimension of this space is infinite. However note that that set of functions is not a basis, so we don't know exactly what the dimension is from this argument. For example, e^x is not in the span of S, so S is not a basis. S is a basis for the vector space of polynomial functions as I showed earlier. The space of all continuous functions is more complicated than that.

    I believe you've allowed the argument involving S to cloud your thinking about the problem. It would be better to start from scratch by understanding the definition of the set of continuous functions with pointwise operations, and then verifying each of the vector space axioms.

    That makes no sense. "Image space" is not a term in common use and has no meaning here.

    No, it's not. x^2 + x^3 is not a function in {1, x, x^2, ...} so this set is not closed under addition.

    You need to go back and revisit this till you understand it. Not because infinite-dimensional spaces are important in your first linear algebra class; but because you need to get a better grasp on what a vector space is even in the finite-dimensional case.
  9. Jun 21, 2011 #8
    Thank you all for your input. So it happens that I typed my argument into Microsoft Word and when I copy and pasted it, the revisions I made were incorrect. For example, the set S should have read x^n not x_n. My wording was a bit poor at times, by scalar addition I was implying a faster method of (af+g)(x)=af(x)+g(x) (proving both simultaneously). I believe my professors have just instructed me in such a way there is a gap in our communication.

    Thanks for all the help. It is much appreciated!
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