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LU decomposition

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

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    2. Relevant equations

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    3. The attempt at a solution
    I think the key to the solution is the matrix A being non-singular but I can't see how.
     
  2. jcsd
  3. Oct 9, 2008 #2
    Suppose we have two different LU decompositions, A = LU and A=L'U'. Because A is non-singular L, U, L' and U' are all non-singular and invertible. This implies that [tex]U = L^{-1}L'U'[/tex]. Now you should be able to show that [tex]I = L^{-1}L'[/tex] in order to preserve upper-triangularity.
     
  4. Oct 9, 2008 #3
    Aha, very smart!

    Well...from my instructors note the L is defined in a product of M-matrices(see picture) but if that's the case then L and L' have different M's (for we presume L and L' to be different).

    If I invert a upper triangular matrix it remains upper triangular so I don't see what you mean by "preserving upper-triangularity".
     
  5. Oct 9, 2008 #4
    The M's are not necessary as far as I can see only that L and L' are lower-triangular and U and U' are upper-triangular, and I mean preserve upper-triangularity of U' to U.
     
  6. Oct 9, 2008 #5
    So to preserve upper triangularity of U w.r.t. U' L-1 L' should be I?
     
  7. Oct 9, 2008 #6
    It needs to be diagonal. It remains to prove that that diagonal matrix must be the identity.
     
    Last edited: Oct 9, 2008
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