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Math Help - Congruent Diagonalization

  1. #1
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    Congruent Diagonalization



    So, in the answer I don't understand how they obtained

    L= \begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 1 & -1 & 1 \end{pmatrix}

    I don't think this "L" here is the lower triangular matrix used in the LU factorization of A. Because I followed the LU decomposition algorithm and ended up with

    L= \begin{pmatrix}1 & 0 & 0 \\ 3 & 8 & 0 \\ 1 & 4 & 5 \end{pmatrix}

    So where did they get that matrix from? Any explanation is very much appreciated.

    PS: I'm sorry this was meant to go to the Linear Algebra section!
    Last edited by demode; May 13th 2010 at 08:52 PM.
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  2. #2
    MHF Contributor

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    On the contrary, that is precisely the "L" in the "LU" form.

    Here's how I would get L: 'zero' out the area above the diagonal using row operations and keeping track of the row operations.

    We can change \begin{pmatrix}1 & 3 & 1 \\ 3 & 8 & 4 \\ 1 & 4 & 5\end{pmatrix} to
     \begin{pmatrix}1 & 3 & 1 \\ 0 & -1 & 1 \\ 0 & 1 & 4\end{pmatrix}
    by "subtract 3 times the first row from the second" and "subtract 1 times the first row from the third"

    We can change that matrix to
    \begin{pmatrix}1 & 3 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 5\end{pmatrix}
    in "upper triangular" form by "add the second row to the third row".

    Now, do the opposite operations, in the opposite order, to the identity matrix.

    That is, first "subtract the second row from the third row" to get
    \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1\end{pmatrix}

    and then "add 1 times the first row to the third" and "add 3 times the first row to the second" to get
    \begin{pmatrix}1 & 0 & 0 \\3 & 1 & 0 \\1 & -1 & 1\end{pmatrix}

    Now you can check that
    \begin{pmatrix}1 & 0 & 0 \\3 & 1 & 0 \\1 & -1 & 1\end{pmatrix}\begin{pmatrix}1 & 3 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 5\end{pmatrix}= \begin{pmatrix}1 & 3 & 1 \\ 3 & 8 & 4 \\ 1 & 4 & 5\end{pmatrix}
    or LU= A.
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