Chapter 1 LINEAR COMPLEMENTARITY PROBLEM, ITS ...

20 Chapter 1. Linear Complementarity Problem, Its Geometry, and Applications
Proof. Let D = F +F T . D is symmetric and by Result 1.9, D is PSD. For all x 2 R n ,
x T Dx =2x T Fx. So x T Dx =0too. We wish to prove that Dx =0. Let x 2 R n . For
all real numbers (x + x) T D(x + x) > = 0, that is
2 x T Dx +2 x T Dx > = 0 (1:16)
since x T Dx = 0. If x T Dx = 0, by taking = 1 and then ;1 in (1.16), we conclude
that x T Dx = 0. If x T Dx 6= 0, since D is PSD, x T Dx > 0. In this case, from (1.16)
we conclude that 2x T Dx > = ; x T Dx for > 0, and 2x T Dx < = ; x T Dx for < 0.
Taking to be a real number of very small absolute value, from these we conclude
that x T Dx must be equal to zero in this case. Thus whether x T Dx =0,orx T Dx > 0,
we have x T Dx = 0. Since this holds for all x 2 R n , we must have x T D = 0, that is
Dx =0.
Algorithm for Testing Positive De niteness
Let F = (fij) be a given square matrix of order n. Find D = F + F T . F is PD i
D is. To test whether F is PD, we can compute the n principal subdeterminants of
D determined by the subsets f1g f1 2g:::f1 2:::ng. F is PD i each of these n
determinants are positive, by Theorem 1.8. However, this is not an e cient method
unless n is very small, since the computation of these separate determinants is time
consuming.
We now describe a method for testing positive de niteness of F which requires at
most n Gaussian pivot steps on D along its main diagonal hence the computational
e ort required by this method is O(n 3 ). This method is based on Result 1.8.
(i) If any of the principal diagonal elements in D are nonpositive, D is not PD.
Terminate.
(ii) Subtract suitable multiples of row 1 from all the other rows, so that all the entries
in column 1 and rows 2 to n of D are transformed into zero. That is, transform
D into D1 as in Result 1.8. If any diagonal element in the transformed matrix,
D1, is nonpositive, D is not PD. Terminate.
(iii) In general, after r steps we will have a matrix Dr of the form:
2
6
4
d11 d12 ::: d1n
0 d22
~ ::: d1n
~
.
0 ..
.
drr ::: drn
.
.
.
0
.
.
dr+1r+1
^
.
.
::: dr+1n
^
.
.
0 0 0 dnr+1
^ ::: dnn
^
Subtract suitable multiples of row r +1in Dr from rows i for i >r+1,so that
all the entries in column r +1and rows i for i > r +1 are transformed into 0.
3
7
5
: