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Symmetric positive definite matrix

  1. Jun 30, 2013 #1
    1. The problem statement, all variables and given/known data

    In a symmetric positive definite matrix, why does max{|aij|}=max{|aii|}

    2. Relevant equations

    |aij|≤(aii+ajj)/2

    3. The attempt at a solution

    max{|aij|}≤max{(aii+ajj)/2

    max{|aij|}≤max{aii/2}+max{ajj/2}

    max{|aij|}≤[itex]\frac{1}{2}[/itex]max{aii}+[itex]\frac{1}{2}[/itex]max{ajj}

    then I am stuck! :(
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 30, 2013 #2

    jbunniii

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    Try applying ##a_{ii} \leq |a_{ii}| \leq \max |a_{ii}|## to the original inequality.
     
  4. Jun 30, 2013 #3
    But how does that help with the equality I need?
     
  5. Jun 30, 2013 #4

    Ray Vickson

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    Assume that the word "symmetric" is always present in the following. Remember that a positive-definite matrix is, by definition, one that renders the quadratic form
    [tex] Q(x) = x^T A x = \sum_i a_{ii} x_i^2 + 2 \sum_{i < j} a_{ij} x_i x_j [/tex]
    positive for all vectors ##x## that are not the zero vector. You are being asked to show that for a positive-definite matrix, the largest element occurs on the diagonal (in terms of magnitudes, that is). So, assume the contrary: that the largest element is not on the diagonal. By re-labelling if necessary we can assume that ##|a_{12}|## is the largest element. Now look at vectors of the form ##\tilde{x} = (x_1, x_2, 0, 0, \ldots,0)^T##, which give
    [tex] Q(\tilde{x}) = a_{11} x_1^2 + 2 a_{12} x_1 x_2 + a_{22}x_2^2.[/tex] Now try to show that if ##|a_{12}| > a_{11}## and ##|a_{12}| > a_{22}## then we can find ##(x_1,x_2) \neq (0,0)## that makes ##Q(\tilde{x}) < 0##.
     
  6. Jun 30, 2013 #5
    By proving that it is not greater in that instance still doesn't prove they are equal, does it?
     
  7. Jun 30, 2013 #6

    Ray Vickson

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    I have no idea what you are talking about. I am just trying to derive a contradiction by assuming an off-diagonal element is maximal.
     
  8. Jun 30, 2013 #7
    I am trying to prove that max{|aij|} = max{|aii|}

    Maybe I am going about it incorrectly and that is the problem. Where would you begin to try to prove these are equal?
     
  9. Jun 30, 2013 #8

    jbunniii

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    I can't do the problem for you. What do you get when you plug in the fact that ##a_{ii} \leq \max |a_{ii}|## into ##|a_{ij}| \leq (a_{ii} + a_{jj}) / 2##?
     
  10. Jun 30, 2013 #9
    |aij|≤(aii +ajj)/2≤(max{aii} +ajj)/2 Is this what you mean?

    So I shouldn't do the steps I did to get to? max{|aij|}≤[itex]\frac{1}{2}[/itex]max{aii}+[itex]\frac{1}{2}[/itex]max{ajj}
     
  11. Jul 1, 2013 #10

    jbunniii

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    In fact, it should be ##\max |a_{ij}| \leq (\max |a_{ii}| + \max |a_{jj}|) / 2##.

    What can you say about ##\max |a_{ii}|## and ##\max |a_{jj}|##?
     
  12. Jul 1, 2013 #11
    Aren't they equal and I get that max{|aij|}≤max{aii}
     
  13. Jul 1, 2013 #12

    jbunniii

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    You dropped the absolute values on the right hand side. It should be ##\max |a_{ij}| \leq \max |a_{ii}|##. Can you see why this gives you the result you need?
     
  14. Jul 1, 2013 #13
    That has been my question, how can I show/prove that they are equal?
     
  15. Jul 1, 2013 #14

    jbunniii

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    If you can show the opposite inequality, that will suffice, right? In other words, is it true that
    $$\max |a_{ii}| \leq \max |a_{ij}|$$
     
  16. Jul 1, 2013 #15

    Ray Vickson

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    No need for absolute values on the right. If A is positive-definite we MUST have ##a_{ii} > 0## for all ##i##. If we weaken the assumption to positive-semidefinite, we still have ##a_{ii} \geq 0## for all ##i##.
     
  17. Jul 1, 2013 #16

    jbunniii

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    True, good point. For this problem, it might be clearer to use the absolute values anyway, even if they are redundant for the diagonal elements.
     
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