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Infinite dimensional counterexample

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Let V be a finite dimensional vector space and let W be a subspace of V.
    1. Then V is the direct sum of W and W' where W' denotes the orthogonal complement of W.
    2. Also, (W')' = W, i.e the orthogonal complement of the orthgonal complement of W is
    again W.

    My question is, what happens if we drop the condition that V is finite dimensional, would
    the results would be still valid? what happens with condition 1 and 2??
     
  2. jcsd
  3. Dec 11, 2008 #2

    HallsofIvy

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    Part of the problem here is that "orthogonality" depends on the specific inner product and there is no "natural" inner product on infinite dimensional vector spaces. Are you assuming a specific inner product and, if so, what?
     
  4. Dec 11, 2008 #3
    Sorry, I accidentally clicked on report.

    I read somewhere in the web that l^2(N) is such counterexample where l^2(N) is the set of all sequences of real numbers (x_1, x_2,...) such that
    sum( x_i^2 , i=1 to infinity) < infinity.

    But I don't know which subspace of l^2(N) should I consider to find where the properties fail.
     
  5. Dec 11, 2008 #4

    Dick

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    I'll give you a simpler example. Let V be the space of all continuous functions on [0,1]. Define the inner product <f,g> to be the integral of f*g over [0,1]. Let W be the subspace of all functions such that f(0)=0. The only element of W' is f=0. V obviously isn't equal to the direct sum of W and W'. They don't even span V.
     
  6. Dec 12, 2008 #5

    Dick

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    In the case of I^2(N), take W to be the subspace of all sequences with only a finite number of elements nonzero. Both of these examples have a common elements. You have a proper subspace W that is dense in V.
     
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