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LU decomposition

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data
    0 0 1
    1 0 0
    0 1 0 decompose this matrix using LU decomposition.


    2. Relevant equations



    3. The attempt at a solution

    I took this matrix and augmented it with a zero (3 by 3) matrix. Then I performed the same row operations on both...the row operations with the purpose of making the matrix I was given an upper triangular matrix, and the zero matrix a lower triangular matrix. I ended up just switching rows twice and got the identity matrix for the upper matrix and got the matrix zero for the lower matrix...so no result. Does anyone know any other method I could use to solve this problem?

    Thank you,
    emira
     
  2. jcsd
  3. Sep 10, 2008 #2
    I believe your original LU decomposition is correct. This is a very trivial decomposition, since it is just a permuted identity matrix.
     
  4. Sep 10, 2008 #3
    But the product of the identity matrix with matrix zero is matrix zero, thus not giving me the matrix i started with. does that it mean the matrix is not decompos-able?
     
  5. Sep 10, 2008 #4
    Well, I'm not quite sure how you would have learned how to do LU decomposition, but both your upper and lower triangular matrices should have diagonals non-zero, one of which is usually normalized to be all 1's. Furthermore there is a permutation matrix (especially if you're doing anything computational).

    For example, if A is the matrix you're trying to put in LU decomposition, then you find L, U and P such that LUP = A.

    Thus let [itex] A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} [/itex]

    Then L and U are identity, and P = A (the original matrix).

    Edit: Obviously, where L is lower triangular, U is upper triangular, and P is a permuted identity matrix
     
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