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Defining scalar product from norm

  1. Dec 11, 2008 #1

    mma

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    Euclidean norm is defined usually as|v|2= g(v,v), where g is a nondegenerate, positive definite, symmetric bilinear form. But how can make it backwards? What properties must norm have that g(v,w) = (|v+w|2 - |v|2 - |w|2)/2 be a positive definite, symmetric bilinear form?
     
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  3. Dec 11, 2008 #2

    lurflurf

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    The parrallelogram identity
    (1/2)(|x+y|^2+|x-y|^2)=|x|^2+|y|^2
    Watch out
    (1/2)(|v+w|^2 - |v|^2 - |w|^2)
    is not the form of an inner product over other fields or with different conjugations
    for example the complex inner product is
    <v|w>=(1/2)(|v+w|^2 - |v|^2 - |w|^2)+(i/2)(|v+iw|^2 - |v|^2 - |w|^2)
    where i*i=-1
     
  4. Dec 11, 2008 #3

    mma

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    Cool! Thanks! Now I see that this is a theorem of von Neumann and Jordan. Is the proof very long? I haven't found it anywhere, only remarks that it is complicated enough. Do you know if it is available somewhere online?
     
  5. Dec 12, 2008 #4

    lurflurf

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    If you have access tohttp://www.jstor.org (perhaps from a school of library) you can access these

    A Characterization of Inner Product Spaces
    Neil Falkner
    The American Mathematical Monthly, Vol. 100, No. 3 (Mar., 1993), pp. 246-249 (article consists of 4 pages)
    Published by: Mathematical Association of America

    On Inner Products in Linear, Metric Spaces
    P. Jordan and J. V. Neumann
    The Annals of Mathematics, Second Series, Vol. 36, No. 3 (Jul., 1935), pp. 719-723 (article consists of 5 pages)
    Published by: Annals of Mathematics

    or you could track down hardcopy at a univercity library.

    The proof is a few pages, you may be able to do it yourself. Here is a start
    4<x|y>+4<z|y>=2|x+y|^2+2|z+y|-2|x|^2-4|y|^2-2|z|^2
    =|x+2y+z|^2+|x+y|^2-4|y|^2
    =4<x+z|y>

    Homogeneity is harder, it is obvious that
    <r x|y>=r <x|y>
    when r is rational
    when r is real more is needed (like Cauchy-Swartz)
     
  6. Dec 12, 2008 #5

    mma

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    Thank you!
     
  7. Dec 13, 2008 #6

    mma

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    Sorry, but this seems false (subststitute for example y=z=0, x<>0).

    I made a derivation, perhaps isn't too elegant, but I hope that good:

    [tex]4<x|y>+4<z|y>[/tex] applying the defiition of scalar product:

    [tex]=2|x+y|^2+2|z+y|^2 -2|x|^2-4|y|^2-2|z|^2[/tex] applyig the parallelogram law to (x+y) and (z+y):

    [tex]= |x+2y+z|^2 + |x-z|^2 -2|x|^2-4|y|^2-2|z|^2[/tex] applyig the parallelogram law to (x+y+z) and (y):

    [tex]= 2|x+y+z|^2 - |x+z|^2 + 2|y|^2 + |x-z|^2 -2|x|^2-4|y|^2-2|z|^2[/tex] applyig the parallelogram law to x and z:

    [tex]= 2|x+y+z|^2 - |x+z|^2 + 2|y|^2 + 2|x|^2 + 2|y|^2 - |x+z|^2 -2|x|^2-4|y|^2-2|z|^2[/tex]

    [tex]=2|x+y+z|^2-2|x+z|^2-2|y|^2[/tex] applying the defiition of scalar product:

    [tex]=4<x+z|y>[/tex]
     
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