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Matrices and norms

  1. Dec 17, 2007 #1
    I'm trying to understand one step in the following proof to the following problem:

    Last edited: Dec 17, 2007
  2. jcsd
  3. Dec 17, 2007 #2
    I'm not sure if this will help you explicitly, but there's a little identity that might contribute to your solution

    We know that

    [tex] \displaystyle\left( \sum_i a_i \right) ^2 \geq 0 [/tex]

    by non-negativity of a square, with equality holding iff the summand is identically 0. Furthermore, we can expand this to

    [tex] \displaystyle\left( \sum_i a_i \right) ^2 = \sum_i a_i^2 + 2 \sum_{i<j} a_i a_j [/tex]


    [tex]\sum_i a_i^2 + 2 \sum_{i<j} a_i a_j \geq 0 [/tex]

    Also, this can probably be generalized but off the top of my head I'm not too sure how, but

    [tex] (x-y)^2 \geq 0 [/tex]

    [tex]\Rightarrow x^2+y^2-2xy \geq 0 [/tex]

    [tex]\Rightarrow 2xy \leq x^2+y^2 [/tex]
  4. Dec 17, 2007 #3
    Thanks, but I already knew all that. So far I have

    [si [sj(A_ij)y_j]^2]^(1/2) <= [si [sjN y_j]^2]^(1/2) = nN [[sj(y_j)]^2]^(1/2)

    but leaves me wondering what to do with [sj(y_j)]^2
  5. Dec 17, 2007 #4
    Then the only thing left that I can think of using is the identity above that I gave, namely

    [tex] \displaystyle\left( \sum_i a_i \right) ^2 = \sum_i a_i^2 + 2 \sum_{i<j} a_i a_j[/tex]

    Then as long as you can make a non-negativity argument about

    [tex] \sum_{i<j} a_i a_j [/tex]

    You'll be good to go
  6. Dec 17, 2007 #5
    Still can't get it. I'm just trying to understand one step in the following proof to the following problem:

  7. Dec 17, 2007 #6
    In all honesty, I'm wondering if there just isn't a typo. It seems like the exponents should all be nested one parentheses earlier, though I might be missing something
  8. Dec 17, 2007 #7
    It's an online solution by a professor (whom I don't know personally) to Spivak's "Calculus on Manifolds". If it is a typo, what is the proper way to finish it off? I've checked that there is no mistake before the inequality sign.

    I'm thinking that his n should be mn
    I think I've corrected the professor's solution, and I think his M is supposed to be N[mn]^(1/2)
    Last edited: Dec 17, 2007
  9. Dec 17, 2007 #8


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    I think this is correct. He probably made the same mistake I did and treated both summations as if they run up to n, and not that one runs up to n and the other to m.
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