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Matrix question

  1. Nov 29, 2006 #1
    I have a 3x3 symmetrical matrix A:


    1, 8^.5, 0
    8^.5, 1, 8^.5
    0, 8^.5, 1

    that I need to find A^5 for. Is there a method aside from brute force multiplication to do so?
     
  2. jcsd
  3. Nov 29, 2006 #2

    radou

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    A fact which should be useful is that the product of symmetric matrices is symmetric itself.
     
  4. Nov 30, 2006 #3
    Ok, how does that help?
     
  5. Nov 30, 2006 #4

    radou

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    Homework Helper

    It reduces the number of multiplications you have to do. But it's still a semi-brute multiplication method. Hope someone passes by with a better suggestion perhaps. :tongue2:
     
  6. Nov 30, 2006 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since that is a symmetric matrix, it can be "diagonalized". That is there exist an invertible matrix C and a diagonal matrix D such that D= CAC-1. Then A= C-1DC. Then A5= C-1D5C. D is the matrix having the eigenvalues of A (which are easy to find) on its main diagonal, 0s off the diagonal. C is a matrix having the corresponding eigenvectors of A as columns.
     
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