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

$\displaystyle

\begin{smallmatrix}

a_{11} & a_{12}....a_{1n}\\

O & B

\end{smallmatrix}

$

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.

Thanks,

Kevin