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Math Help - Column-Wise strict Diagonal Dominance Gaussain Elemination help please.

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    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.,
    |a_{jj}|>\sum_{i=1}^{n}|a_{ij}|

    A: Show that after one step of Gauss elimination , the resulting matrix
    takes the form
    <br />
 \begin{smallmatrix} <br />
  a_{11} & a_{12}....a_{1n}\\<br />
  O & B<br />
\end{smallmatrix} <br />
    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 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.

    Thanks,
    Kevin
    Last edited by KBrosUL; March 27th 2012 at 07:07 AM.
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