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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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