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LDL factorization

  1. Dec 1, 2005 #1
    Find the [itex] LDL^T [/itex] factorization of this matrix

    [tex] \left(\begin{array}{ccc}{2&-1&0\\-1&2&-1\\0&-1&2\end{array}\right) [/tex]

    now i can find the L matrix by gaussian elimination
    that yields
    [tex]L = \left(\begin{array}{ccc}{1&0&0\\\frac{-2}{3}&1&0\\0&\frac{-1}{2}&1\end{array}\right) [/tex]
    [tex] D = \left(\begin{array}{ccc}{\frac{1}{4}&0&0\\0&\frac{1}{3}&0\\0&0&\frac{1}{2}\end{array}\right) [/tex]

    i am pretty sure about the ansswer since i checked my working many times.
    However this is not the answer at the back of the book! In fact i am not even close!
    What am i doing wrong?? Can anyone please help me iwth this?
    Thank you for your help!
  2. jcsd
  3. Dec 1, 2005 #2


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    Science Advisor

    It's hard to tell what you did wrong when you did wrong when you don't tell us what you did! I did a quick "column reduction" to get L and didn't get any like you got.
  4. Dec 1, 2005 #3
    well i got those answers by Gaussian Elimination
    this is what i did

    [tex] \left(\begin{array}{ccc}{2&-1&0\\-1&2&-1\\0&-1&2\end{array}\right) [/tex]

    R3 + 2R2
    [tex] \left(\begin{array}{ccc}{2&-1&0\\-2&3&0\\0&-1&2\end{array}\right) [/tex]

    [tex] \left(\begin{array}{ccc}{4&0&0\\-2&3&0\\0&-1&2\end{array}\right) [/tex]

    and my textbook says that that the D matrix is formed by dividing the square terms of the lower matrix formed and multiply that by the elementary matrix yielding
    [tex] D = \left(\begin{array}{ccc}{\frac{1}{4}&0&0\\0&\frac{ 1}{3}&0\\0&0&\frac{1}{2}\end{array}\right) [/tex]
  5. Dec 2, 2005 #4
    can anyone tell me what i have done wrong? my answer is not even close to the tedxxt book's answer. However all my steps with the row reductions are correct, as you can see.

    I was told that i was not supposed to use row reduction to get the lower matrix? SO what do i do then?
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