320 Nonlinear Programming II: Unconstrained Optimization Techniques
and hence
∇Q(X 1 ) −∇Q(X 2 )=A(X 1 −X 2 ) (6.27)
IfSis any vector parallel to the hyperplanes, it must be orthogonal to the gradients
∇Q(X 1 ) nda ∇Q(X 2 ) Thus.
ST∇Q(X 1 )=STAX 1 +STB= 0 (6.28)
ST∇Q(X 2 )=STAX 2 +STB= 0 (6.29)
By subtracting Eq. (6.29) from Eq. (6.28), we obtain
STA(X 1 −X 2 )= 0 (6.30)
HenceSand(X 1 −X 2 ) rea A-conjugate.
The meaning of this theorem is illustrated in a two-dimensional space in Fig. 6.7.
IfX 1 andX 2 are the minima ofQobtained by searching along the directionSfrom two
Figure 6.7 Conjugate directions.