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Math Help - LU-decomposition

  1. #1
    MHF Contributor Swlabr's Avatar
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    LU-decomposition

    The LU-decomposition of a matrix A can be used to solve the system of equations Ax=v for the vector x. But...why would you do this? Is it more computationally effective? Or is the reason more subtle than that?...
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Suppose for example Ax=v with A invertible and you know L and U. The system Ly=v is triangular and let y_0 be its solution. The system Ux=y_0 is also triangular and let x_0 be its solution. Then, Ax_0=L(Ux_0)=Ly_0=v . That is, we have solved Ax=v by means of two triangular systems.


    Fernando Revilla
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    Suppose for example Ax=v with A invertible and you know L and U. The system Ly=v is triangular and let y_0 be its solution. The system Ux=y_0 is also triangular and let x_0 be its solution. Then, Ax_0=L(Ux_0)=Ly_0=v . That is, we have solved Ax=v by means of two triangular systems.


    Fernando Revilla
    Yes, I know this. My question, however, is why you would bother to do it this way? Why not just triangularize your matrix and go from there?
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  4. #4
    A Plied Mathematician
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    The benefit is more when you have multiple systems of the form Ax = b_{k}, for multiple b_{k}'s. Then, once you've done the LU factorization once, solving each system doesn't require any more row reductions, just back substitutions, which are fast. If you were going to solve only one system, then for exact methods, you'd do Gaussian elimination with back substitution, and for iterative methods (such as with very large, sparse matrices), you'd do something fancy like Gauss-Seidel (but more advanced).
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