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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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