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Then it follows that
= H + d α i
d
u i
d u i
d
+ α i D 1 d D
d
(6.50)
ζ
ζ
ζ
r
u i
d D
d ζ
u i
d u i
d
I n + 1 + D 1 d D
d
= H + α i
u i
ζ
u i Du i
ζ
ρ
= H + α i
u i
rI n + 1 + D 1 d D
d
= H + α i ρ u i
(6.51)
ζ
Assuming that the eigenvalue decomposition of D 1 R is equal to UD α U T , it can
be deduced that
H + = UD α U T
1
1
α n + 1
U T
= U diag
, ... ,
0
, ... ,
i th position
α
α i
α i
1
Equation (6.47) yields
diag I n
ζ
D 1 d D
d ζ
1
1 ζ
2 D 1 J
=
=
,
Hence,
ρ =− rI n + 1 + diag I n
ζ
= diag 1
ζ
r I n ,
r
1
1 ζ
1
1 ζ
,
Then eq. (6.50) becomes
= α i UD α U T diag 1
ζ
r I n ,
r u i
d u i
d ζ
1
1 ζ
= α i UD α
(6.52)
To simplify eq. (6.52), the normalization of the eigenvectors must be taken into
account. Indeed, the constraint of belonging to the TLS hyperplane determines
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