Chapter 1 LINEAR COMPLEMENTARITY PROBLEM, ITS ...

14 Chapter 1. Linear Complementarity Problem, Its Geometry, and Applications
where a 0
ij = aij ; (arjais)=ars, fori = r +1tom, j =1ton. As an example consider
the Gaussian pivot step in the following matrix with row 2 as the pivot row and column
3 as the pivot column. The pivot element is inside a box.
8
>:
1 ;2 10 ;4 ;1
4 6 2 ;8 ;4
;3 1 ; 1 2 3
1 ;4 2 3 0
This Gaussian pivot step transforms this matrix into
8
>:
1 ; 2 10 ; 4 ;1
4 6 2 ; 8 ;4
;1 4 0 ; 2 1
;3 ;10 0 11 4
Result 1.8 Let D be a square symmetric matrix of order n > = 2. Suppose D is PD.
Subtract suitable multiples of row 1fromeach of the other rows so that all the entries
in column 1 except the rst is transformed into zero. That is, transform
D =
2
6
4
d11 ::: d1n
d21 ::: d2n
.
.
dn1 ::: dnn
3
7
5 into D1 =
2
6
4
9
>
9
>
d11 ::: d1n
0 d22
~ ::: d2n
~
.
.
.
0 dn2
~ ::: dnn
~
by a Gaussian pivot step with row 1 as pivot row and column 1 as pivot column, clearly
~dij = dij ; d1jdi1=d11 for all i j > = 2. E1, the matrix obtained by striking o the rst
row and the rst column from D1, isalso symmetric and PD.
Also, if D is an arbitrary square symmetric matrix, it is PD i d11 > 0 and the
matrix E1 obtained as above is PD.
Proof. Since D is symmetric dij = dji for all i j. Therefore,
y T Dy =
nX
nX
i=1 j=1
= d11 y1 +
yiyjdij = d11y 2 1 +2y1
nX
j=2
d1jyj =d11
2
nX
j=2
+ X
ij> = 2
d1jyj + X
ij> = 2
yi ~ dijyj :
3
7
5
yiyjdij
Letting y1 = ;( P n
j=2 d1jyj)=d11, weverify that if D is PD, then P ij> = 2 yi ~ dijyj > 0 for
all (y2:::yn) 6= 0,which implies that E1 is PD. The fact that E1 is also symmetric
is clear since ~ dij = dij ; d1jdi1=d11 = ~ dji by the symmetry of D. If D is an arbitrary
symmetric matrix, the above equation clearly implies that D is PD i d11 > 0 and E1
is PD.