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Linear algebra+ linear operators

  1. May 22, 2008 #1
    1. The problem statement, all variables and given/known data

    In [tex]R^{3}[/tex] ||x||= [tex]a_{1}[/tex]*|[tex]x_{1}[/tex]|+ [tex]a_{2}[/tex]*|[tex]x_{2}[/tex]|+ [tex]a_{3}[/tex]*|[tex]x_{3}[/tex]|. where [tex]a_{i}[/tex]>0

    What is ||A||(indused norm = sup||Ax|| as ||x||=1). (Suppose we know the coeffisients of the matrix/operator A)??


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 22, 2008 #2

    Dick

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    If A is given by a matrix, it's the square root of the largest eigenvalue of AA* (*=hermitian conjugate or transpose in the real case).
     
  4. May 22, 2008 #3
    Thanks, but do you know how to proove it?
     
  5. May 22, 2008 #4
    By the way, do you want to say that it does nod depend on a_i??
     
  6. May 22, 2008 #5
    (i just need the real case)
     
  7. May 22, 2008 #6

    Dick

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    |Av|^2=v^(T)*A^T*A*v. A^T*A is a symmetric matrix, so it has a complete orthogonal set of eigenvectors. Can you show the eigenvalues are nonnegative? Now expand a general v in terms of those eigenvectors and go from there...
     
  8. May 22, 2008 #7

    Dick

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    Hey, that's not a matrix operator at all, is it? You just want to maximize the inner product of (a1,a2,a3) and x. That's much easier. Think Cauchy-Schwarz.
     
  9. May 22, 2008 #8
    Unfortunately I do not understand why

    |Av|^2=v^(T)*A^T*A*v is true.. Why does it not depend on a_i??
    2. I don't think I know how to proove that the eigenvalues are nonnegative or where to go from there..
     
  10. May 22, 2008 #9
    I think that it is a matrix operator..
     
  11. May 22, 2008 #10

    Dick

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    What you wrote is the dot product of (a1,a2,a3) with (x1,x2,x3). That's a matrix operator only in the sense it's a 1x3 matrix. You want to maximize it. Will you look up the Cauchy-Schwarz inequality. Please?
     
  12. May 23, 2008 #11
    The thing is that the dot product does not have the modules..
    Even if I do maximize it.. I do not see where to go from there..
     
  13. May 23, 2008 #12
    Cauchy Shwartz: (x,y)<=||x||*||y||
     
  14. May 23, 2008 #13

    Dick

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    Yes, and add "with equality holding only when y is a multiple of x". That's the part you want.
     
  15. May 24, 2008 #14

    HallsofIvy

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    Dick, I think you have misinterpreted the question. What he wrote was the definition of norm (not an inner product between (a1, a2, a3) and (x1, x2, x3)- a1, a2, and a3 are fixed for all x) THEN he asked
    "What is ||A||(indused norm = sup||Ax|| as ||x||=1). (Suppose we know the coeffisients of the matrix/operator A)??"

    In other words, what norm does that defined norm on the vector space induce on the linear operators.

    The definition of "induced norm" of A is ||A||= lub ||Ax|| for all x with norm 1.
     
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