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eq. (4.54) is rewritten in the simpler form
z
T
−
γ
I
n
z
=
g
(γ )
(4.56)
The study of this hyperconic family requires analysis of the behavior of the
known term
g
(γ )
vs.
γ
.For
γ
=
0,
)
=
b
T
A
A
T
A
−
1
A
T
b
−
b
T
b
=−
b
T
A
A
T
A
−
1
A
T
g
(
0
−
I
n
T
A
A
T
A
−
1
A
T
−
I
n
b
≤
0
(4.57)
where the equal sign is valid only for a compatible system of linear equations.
For an incompatible system, the negativeness of
g
(
0
)
implies that no hyperconics
of the family in eq. (4.56) exist for
0.
Definition 89
By introducing the vector q
≡
q
1
q
2
···
q
n
T
γ
=
=
V
T
A
T
b
,
eq.
(
4.55
)
becomes
n
q
i
+
γ
−
b
T
b
=
q
T
(
−
γ
I
n
)
−
1
q
+
γ
−
b
T
b
g
(γ )
=
(4.58)
λ
−
γ
i
i
=
1
Definition 90 (Convergence Keys)
The components q
i
of the vector q will be
called the convergence keys because they play a fundamental role in the analysis
of the TLS and GTLS domain of convergence.
Recalling the definition of
g
in eqs. (1.19) and (1.2), it follows that
q
=
V
T
A
T
b
=
V
T
V
T
U
T
b
=
T
U
T
b
=
g
(4.59)
2
n
and therefore eq. (4.58) coincides with eq. (1.20) for
1
. It can be con-
cluded that eq. (4.58) is a version of the TLS secular equation [74,98] and that
this zero of
g
γ
=
σ
+
gives the level of the TLS solution [eq. (1.21)].
Disregard the nonrealistic case
q
n
(γ )
=
0. A direct inspection of eq. (4.58) estab-
lishes that the term
g
strictly increases from the negative value at
γ
=
0to
∞
for
γ
→
λ
n
.
Then one and only one value
γ
min
of
γ
exists in the interval
(
0,
λ
n
)
such that
g
(γ
min
)
=
0. Consequently, the equilevel hyperconics in eq. (4.56) are
hyperellipsoids [(
n
−
1)-dimensional ellipsoids]
∀
γ
∈
(γ
min
,
λ
n
)
. These hyperel-
lipsoids are of the same type and dimensionality of the intersection between the
cone of the saddle nearer the minimum and the hyperplane
ε
N
=
ε
n
+
1
=
const
=
0intheMCA
(
n
−
1
)
-dimensional space (see Section 2.5.1.2). For
γ
=
γ
min
,the
hyperconic collapses into a unique point
z
=
y
=
0. This point corresponds to
the unique minimum of
E
TLS
(
x
)
with position given by
(γ )
)
=
A
T
A
min
I
n
−
1
A
T
b
=
A
T
A
n
+
1
I
−
1
A
T
b
2
x
=
x
c
(γ
−
γ
−
σ
(4.60)
min
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