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Questions about a basis and others for the vector space R^X (functions from X->R)

  1. Feb 3, 2012 #1
    The finite case is fine, as a vector space it is easy to show that R^X is isomorphic to R^n.

    What about when X is infinite? I believe it is true in general that dim(R^X) = #(X), which I hope holds in the infinite case too. I know that the set given by B={b_x; x in X} defined as b_x(y) = delta_xy gives a linearly independent set in R^X. But as far as I am aware, the dimension of a vector space is the cardinality of a basis. Since this is not a basis (I found by googling), it does not give the dimension of the space correct?

    If not, how can one construct a basis of R^X? Though I don't care too much about what the basis is, just the size of it.
     
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  3. Feb 3, 2012 #2

    morphism

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    The dimension of R^X isn't going to be equal to #X. For example, if X is countably infinite, then it's not too hard to show that dim R^X is uncountable. I suspect that, for general infinite X, dim R^X = 2^#X, but haven't tried to prove this. There'll probably some set theoretic mumbo jumbo involved in computing dim R^X (e.g. [STRIKE]I used the axiom of choice to show that dim R^X is uncountable if X is countably infinite - though I don't know if that was necessary[/STRIKE] edit: it wasn't necessary*).

    *: I suppose the uncountability of dim R^X follows more easily from the Baire category theorem, which IIRC doesn't need choice (?). The argument I have in mind uses the well-known fact (which can be proved using Baire) that an infinite-dimensional Banach space is uncountably-infinite-dimensional. To see how this applies, simply note that R^X will contain, e.g., the Banach space [itex]\ell^2[/itex] of square-summable real sequences as a subspace, so dim R^X >= dim [itex]\ell^2[/itex] is uncountable.
     
    Last edited: Feb 3, 2012
  4. Feb 3, 2012 #3
    I see, that's a shame. Thanks for the reply though.
     
  5. Feb 3, 2012 #4

    micromass

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    It can be proven that

    [itex]dim(\mathbb{R}^X)=|\mathbb{R}^X|[/itex].
     
  6. Feb 3, 2012 #5
    Only when X is infinite though, correct?
     
  7. Feb 3, 2012 #6

    micromass

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    Correct. A proof can be found (essentially) in "Lectures in Abstract Algebra II" by Jacobson Chapter IX.
     
  8. Feb 3, 2012 #7

    morphism

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    Okay, at least that agrees with my intuition for X countably infinite (because then |R^X|=2^|X|). How difficult is the proof, micromass? I don't have access to the book.
     
  9. Feb 3, 2012 #8

    micromass

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    I haven't looked at the proof, but it seems quite technical. It requires two lemma's to prove. One lemma is a result from Kaplansky and Erdos. So you were right about the set theoretic mumbo jumbo :tongue2:
     
  10. Feb 3, 2012 #9

    morphism

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    Ah..go figure!
     
  11. Feb 3, 2012 #10
    Hmm. So say i had any finite dimensional space A contained in R^X (where X can actually be infinite), say dimension n. Then I know that it is isomorphic to R^n, which is then again isomorphic to functions from some set Y to R, where |Y|=n. So if I'm thinking about this right, then A ~ R^Y. (I think Y could be chosen arbitrarily amongst the elements of X, as they are just place holders of some sort, right?)

    If the above is okay, then I'm pretty sure A does not have to equal R^Y, but can I conclude that A = R^Z where |Z|=|Y|=n? They are both subspaces of R^X, which, since they are isomorphc leads me to believe they should be related in this way, but I'm not sure of the mechanics.
     
  12. Feb 3, 2012 #11
    Okay, regarding the above, I get an interesting (wrong?) result. So A is an n-dim subspace of R^X, and A~R^Y for |Y|=n.

    If θ is an isomorphism between R^Y and A, then if R^Y has basis {f1,f2,...fn} we can map this to a basis {a1,...an} of A to get that θ(f1)=a1 is still a function from Y to F.

    So: if this is correct, then any n-dim subspace A actually is the set of functions from Y to R for any subset Y of size n. So there is no dependence on the subset chosen? Something must be wrong with my reasoning.
     
  13. Feb 4, 2012 #12

    morphism

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    I think you're confusing isomorphism and equality. A subspace of R^X consists of functions defined on X, so it won't be equal to a set of functions defined on Y, even if it really looks like one.
     
  14. Feb 4, 2012 #13

    Deveno

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    for example, one can construct the set of (real) polynomial functions of degree n-1 or less defined on the interval [0,1].

    this, in turn is isomorphic to the space of all real polynomials of degree n-1 or less defined on all of R, which in turn is isomorphic to Rn (and a point in Rn could be considered as a function f:{1/n,2/n,...,n/n} → R with xj = f(j/n)).

    but a point in Rn is not the same thing as a real polynomial with deg < n defined on [0,1].

    or: it is well-know that the functions:

    1,ex,e2x,e3x,....,e(n-1)x are linearly independent on any real interval (erm, of non-zero length), so the subspace of RI spanned by them is isomorphic to the subspace of RI spanned by {1,x,x2,....x(n-1)}.

    but clearly, exponential functions are not polynomials.

    the invariant here is the dimension, which is why dimensional arguments are so prominent in linear algebra. and to explicitly demonstrate an isomorphism, we "choose a basis". this is pretty straight-forward when our domain set (the X in FX) is finite. when it's infinite (and so might be uncountable), it becomes much more complicated (perhaps you can see why with sets with a "rich" structure, the axiom of choice is required to demonstrate a basis. if X is uncountable, there's "too many subsets", that is, it's not clear how to decide on a basis (generating subset)).

    if you ever hit upon a constructive algorithm for finding a basis for RX where X is uncountable, i'm sure you'll get invited to all the most fashionable mathematical parties.
     
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