Chapter 1 LINEAR COMPLEMENTARITY PROBLEM, ITS ...

16 Chapter 1. Linear Complementarity Problem, Its Geometry, and Applications
is PD. So the determinant of F is nonzero. In a similar manner we conclude that the
determinant ofany PD-matrix is nonzero. For 0 < 0 for all 0 < = < = 1 because F is PD. So F ( )
is a PD matrix for all 0 < = < = 1. So from the above argument f( ) 6= 0 for any
satisfying 0 < = < = 1. Clearly, f(0) = 1, and f(1) = determinant of F . If f(1) < 0
by Theorem 1.1 there exists a satisfying 0 < < 1 and f( ) = 0, a contradiction.
Hence f(1) 6< 0. Hence the determinant of F cannot be negative. Also it is nonzero.
Hence the determinant of F is strictly positive.
Theorem 1.3 If F is a PD matrix, whether it is symmetric or not, all principal
subdeterminants of F are strictly positive.
Proof. This follows from Result 1.2 and Theorem 1.2.
Theorem 1.4 If F is a PSD matrix, whether it is symmetric or not, its determinant
is nonnegative.
Proof. For 0 < = < = 1, de ne F ( ), f( ) as in the proof of Theorem 1.2. Since I is
PD, and F is PSD F ( ) is a PD matrix for 0 < = < 1. f(0) = 1, and f(1) is the
determinant of F . If f(1) < 0, there exists a satisfying 0 < :
d11 ::: d1n d1n+1
.
dn1 ::: dnn dnn+1
dn+11 ::: dn+1n dn+1n+1
.
.
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>
D =
8
>:
d11 ::: d1n
.
.
dn1 ::: dnn
be symmetric matrices. H is of order n +1 and D is a principal submatrix of H.
So dij = dji for all i, j = 1 to n +1. Let x 2 R n , d = (d1n+1:::dnn+1) T , and
Q(x) =x T Dx+2d T x + dn+1n+1. Suppose D is a PD matrix. Let x = ;D ;1 d. Then
x is the point which minimizes Q(x) over x 2 R n , and
Q(x )= (determinant ofH) / (determinant of D). (1:12)
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