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Thread: Column-Wise strict Diagonal Dominance Gaussain Elemination help please.

  1. #1
    Mar 2012

    Column-Wise strict Diagonal Dominance Gaussain Elemination help please.

    A matrix A is column-wise strictly diagonally dominant if the absolute
    value of the diagonal element in each column of A is strictly greater than
    the sum of the absolute values of the other elements in the column, i.e.,
    $\displaystyle |a_{jj}|>\sum_{i=1}^{n}|a_{ij}|$

    A: Show that after one step of Gauss elimination , the resulting matrix
    takes the form
    a_{11} & a_{12}....a_{1n}\\
    O & B
    where O is the (n-1)x 1 zero column vector and B is an (n-1)x(n-1)
    column-wise strictly diagonally dominant matrix.

    B: Hence use an inductive argument to show that the pivot element $\displaystyle a_{kk}$ at step k is always
    the element with largest absolute value in column k

    C: And then show that partial pivoting, the swapping of rows to guard against small pivot elements,
    is unnecessary.

    I have completed Part A myself however i do not know how to go about starting Part B any help would be greatly appreciated.

    Last edited by KBrosUL; Mar 27th 2012 at 07:07 AM.
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