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Find a diagonal matrix D such that the tridiagonal matrix T

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    For a symmetric matrix B=B' where ' is the transpose.

    3. The attempt at a solution

    Since we know that for a symmetric matrix,
    B = B'

    I attempted to substitude that in and tried to solve for D.

    DTD-1 = (D-1)'T'D
    DT = D-1T'DD
    D = D-1T'DDT-1

    At this point I am stumped since there are D's on both sides, what am I suppose to do?

    Thanks a lot for your help.

    Code (Text):
  2. jcsd
  3. Dec 10, 2008 #2


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    Staff Emeritus
    Science Advisor

    Seems to me you could just "write it out". Let x[subn[/sub] be the nth diagonal entry in D. Then 1/xn is the corresponding entry in D-1. Now, actually write out the entries for the product DTD-1. What must be true in order for that to be symmetric? If you are not sure how to do that, try it with 2 by 2 and 3 by 3 matrices first.

    Notice, by the way, that saying "bici+1> 0" is exactly the same as saying "bi/ci+1> 0" or even "ci+1/bi> 0" because the only thing that is relevant is the sign: all three just say that neither of bi nor ci+1 is 0 and they have the same sign.
  4. Dec 10, 2008 #3
    Hi HallsofIvy,

    Thanks so much for your answer, I actually did it last night. Maybe you can have a look at my result and confirm:

    In the end I got that the diagonal entries of D are:

    di+1 = di[tex]\sqrt{b_i/c_{i+1}}[/tex]

    The square root is why bici+1 > 0

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