1. ## 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
$
\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 $a_{kk}$ 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