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Using the following notation:
d u 1
d ζ
d U
d ζ
d u n + 1
d ζ
=
···
0
τ
···
τ
1,2
i , n
+
1
τ
0
···
τ
2,1
2, n
+
1
T =
.
.
. . .
0
τ n + 1,1
τ n + 1,2
···
0
the following main result is obtained:
d U
d
= η UT
(6.55)
ζ
6.3.2.1 Analysis of Matrix T From definition (6.54) it is deduced that
τ f , g =− α g
α f τ g , f
(6.56)
6.4 RANK ONE ANALYSIS AROUND THE TLS SOLUTION
b ]givenby USW T
Consider the SVD of matrix [ A
;
and the corresponding
diag σ
1 .
W T ,where
2
2
n
decomposition of matrix R given by W
=
1 ,
...
,
σ
+
It follows that
D 1 Ru
D 1 W
W T u
= γ
u
= γ
u
(6.57)
Defining n = W T u , we obtain
W T D 1 W n = γ n
(6.58)
that is, n = γ n ,where
= W T D 1 W = W T D 1 RW
(6.59)
Decomposing the orthonormal matrix W as W 1
w
,where
T
n
w
1 is its last
T
n
+
+
1
row, after some simple mathematics we obtain
I n + 1
xy T
2
1 2 ζ
1
=
(6.60)
ζ
ζ
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