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Homework Help: Vector subspace

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Let F be the field of all real numbers and let V be the set of all sequences (a1,a2,....a_n,...), a_i in F, where equality, addition, and scalar multiplication are defined component-wise.

    (a) Prove that V is a vector space over F
    (b) Let W={(a1, a2,....,a_n,...) in V | lim a_n = 0 as n-->inf}. Prove that W is a subspace of V.
    (c) Let U={(a1,...,a_n,...) in V | summation of (a_i)^2 is finite, i evaluated from 1 to inf}. Prove that U is a subspace of V and is contained in W.

    3. The attempt at a solution
    I know that in order for W to be a subspace of V, W must form a vector space over F under the operations of V. I've already proved (a). Do I need to know the limit of a_n to prove (b) or is that just for (c)? It seems like proving (b) is pretty similar to (a), right? Any tips on proving (c)?
  2. jcsd
  3. Jan 31, 2009 #2
    In part b, the lim a_n = 0 as n --> inf just means that only those a_n that are contained in V that meet the criteria lim a_n = 0 as n --> inf are contained in the set.

    So to prove part b, you just need to show that addition and scalar multiplication are closed in the subspace.

    Sorry, I'm not too sure about part C (I don't want to guess and tell you wrong either).
  4. Jan 31, 2009 #3
    Suppose {a_i} is a sequence in U such that the a_i's do not tend to 0 as i increases without bound. Consider the infinite series Sum[(a_i)^2] = Sum[b_i] where each b_i is positive and does not tend to 0. Suppose the sum converges to L, which means for each r > 0, there exists a number N so that | L - Sum[b_i] | < r where i ranges from 0 to n for all n > N. If the terms b_i are never negative and never tend to 0, can this condition be satisfied? (Note that the condition that the b_i's tend to 0 is that for all d > 0, there is some N so that |b_i| < d for all i > N.)
  5. Feb 1, 2009 #4
    I don't really understand, but I would say that the condition cannot be satisfied because if the terms b_i never tend to 0 then the summation must diverge.
  6. Feb 1, 2009 #5
    If you can show that rigorously, then you have shown that each element of U is necessarily an element of W, and is thus contained in W. The only thing left is to show that sums of elements in U remain in U, and so do scalar multiples, which is the easy part.
  7. Feb 1, 2009 #6
    How would I show that the sum of any multiples of elements of U is still in U? Wouldn't I have to show that the sum of every sequence^2 is finite?
  8. Feb 2, 2009 #7
    As an example, suppose {a_i} is an element of U. Then the series (a_1)^2 + (a_2)^2 + ... converges. Consider the element s*{a_i} defined to be {s*a_i}, so we now consider the series (s*a_1)^2 + (s*a_2)^2 + ... = s^2*(a_1)^2 + s^2*(a_2)^2 + ... = s^2*((a_1)^2 + (a_2)^2 + ...) = s^2*A where A is the number that the original series converges to, showing constructively that the component-wise defined multiple also converges. Now you just have to consider what happens to a component-wise sum.
    As for the latter question, you would already have shown that by showing that U is contained in W (by contradiction). (Need more of a hint?)
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