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Linear Operators in Hilbert Space - A Dense Question

  1. Mar 5, 2010 #1
    Let H be a Hilbert space and let S be the set of linear operators on H. Is there a proper subset of S that is dense in S?
     
  2. jcsd
  3. Mar 6, 2010 #2

    mathman

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    S is a vector space, so it must have a basis. Let T be a subset of S defined by linear combinations of basis elements with rational coefficients. T is a proper subset of S and is dense in S.
     
  4. Mar 6, 2010 #3
    The question assumes that H (and thus S) is not a zero space, of course.

    A more trivial solution would be to consider S \ {0}.
     
  5. Jul 12, 2010 #4
    Not if his meaning of "basis" refers to the usual definition of the basis of a vector space. (That is, B is a basis of S iff every element of S is uniquely expressed as a finite linear combination of elements of B.) Of course, as stated, the problem is rather trivial.

    Way to bring up an old thread ;)
     
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