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Does every independent set spans the space necessarily form a basis?

  1. Jun 2, 2012 #1
    Hi there!

    A Hilbert space E is spanned by a set S if E is generated by the element of S.

    It is well known that in the finite dimensional case that
    S spans E and S is linearly independent set iff the set S form a basis for E.

    The question is that true for the infinite dimensional case? Noting that in this case by span we mean the closure of the span.

    Thanks in advance
    LikeMath
     
  2. jcsd
  3. Jun 2, 2012 #2

    micromass

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    What do you mean by "basis" in the infinite-dimensional case??
     
  4. Jun 2, 2012 #3
    That is every member of E can be written in a uniquely linear combination of vectors from this basis. I wonder if it "basis" has another definition?
     
  5. Jun 2, 2012 #4

    micromass

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    That is commonly called a Hamel basis in infinite dimensional vector spaces.

    Something that spans the set and is linearly independent is indeed a Hamel basis.

    However, if you change the means of span to mean "the closure of the span", then this is not true anymore. For example, in [itex]\ell^2[/itex], the vectors

    [itex](1,0,0,0,...),~(0,1,0,0,...),~(0,0,1,0,...),...[/itex]

    are linearly independent and the closure of the span is [itex]\ell^2[/itex]. But it is not a Hamel basis. (however, it is a Schauder basis and an orthonormal basis).
     
  6. Jun 2, 2012 #5
    Ah ok, now I understand your point. In my question I mean Schauder basis (not the Hamel basis).

    Thank you for drawing my attention to this piont.
     
  7. Jun 2, 2012 #6

    micromass

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    Then it's still not true.
    For example, take a Hamel basis of [itex]\ell^2[/itex] (this thing exists but is very very large). Then this is linearly independent and the closure of the span is entire [itex]\ell^2[/itex]. But it's not a Schauder basis as some vectors have multiple representations as series of the basis.
     
  8. Jun 2, 2012 #7
    Ok, but the basis in your example is linearly independent in the finite sense, is not it?

    By the way are you really a high school student?

    Thx
     
  9. Jun 2, 2012 #8

    micromass

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    Indeed. So you probably propose some kind of independence in the sense

    [tex]\sum{\alpha_i e_i}=0~\Rightarrow~\alpha_i=0[/tex]

    But even then, I see a problem. It's not because an element is in the closure of the span that it can be written as a series of those elements. For example, take

    [tex]\mathcal{C}([0,1])[/tex]

    with [itex]\|~\|_\infty[/itex] (yes, I realize that this is not a Hilbert space). Then the set

    [tex]\{1,x,x^2,x^3,...\}[/tex]

    is linearly independent and its span is dense (by the Weierstrass approximation theorem). But not every continuous function can be written as [itex]\sum \alpha_nx^n[/itex]. This would imply that all continuous functions are analytic, which is not true.

    But then again, this is not a Hilbert space. What could change in a Hilbert space is that you always have an orthonormal basis (which may not be countable). So that might change things.
     
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