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Infinite-dimensional matrix multiplication

  1. Jun 4, 2007 #1
    We know that infinite-dimensional matrix multiplication in general isn't asociative. But, is there any criteria when asociativity is valid?
    thanks in advance.
     
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  3. Jun 4, 2007 #2

    Hurkyl

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    Well, isn't this the same thing as asking when you can interchange two summation symbols?
     
  4. Jun 4, 2007 #3
    Yes, I think you'r right. Can we say: we can interchange two summation symbols if both exists?
     
  5. Jun 4, 2007 #4

    Chris Hillman

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    Some buzzwords and a book

    Relevant buzzwords include Tonelli theorem and Fubini theorem. In terms of doubly indexed infinite series, these basically say that absolute convergence is the key. See Bartle, The Elements of Real Analysis, 2nd Edition, Wiley, 1976, section 36, for an undergraduate level discussion.
     
    Last edited: Jun 5, 2007
  6. Jun 5, 2007 #5

    mathwonk

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    I am puzzled. If multiplying infinite matrices corresponds to composing linear maps, as in the finite dimensional case, then it seems it would be associative, since composition is so.

    I guess these matrices do not correspond to linear maps in the algebraic sense, as in that case there would be a finiteness condition on the number of non zero entries in the columns.
     
    Last edited: Jun 5, 2007
  7. Jun 5, 2007 #6

    NateTG

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    Consider, for example:

    [tex]a_{ij}=\frac{1}{2^{i+j}}[/tex]
    [tex]b_{ij}=1[/tex]
    [tex]c_{ij}=\frac{1}{i+j}[/tex]

    Some products will be convergent, and some will be divergent.
     
  8. Jun 5, 2007 #7

    mathwonk

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    one needs to be a little more precise. in linear algebra, convergence is not an issue, only in analysis.

    so one needs to say what subject one is working in, and what one means by "associative".

    the matrices you gave do not represent maps in terms of a basis in the linear algebra sense.
     
  9. Jun 5, 2007 #8

    NateTG

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    I meant that the elements in of the matrix are divergent sums.
     
  10. Jun 5, 2007 #9

    Chris Hillman

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    Yes, indeed, thanks. I have made the correction.
     
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