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t
= l
1
−
z
1
q
1
, q
2
>
0
sdls
stls
saddle locus
= l
2
+
t
t
= l
2
−
dls
tls
ols
solution locus
mdls
t
=
0
z
2
t
= ∞
t
= −∞
mtls
t
= l
1
+
= l
1
−
t
z
1
q
2
<
0, q
1
>
0
sdls
stls
saddle locus
= l
2
+
t
= l
2
−
dls
t
tls
ols
saddle locus
mdls
t
=
0
z
2
t
= ∞
t
= −∞
mtls
t
= l
1
+
Figure 5.7
Two-dimensional case: hyperbolas and critical loci for
q
1
>
0,
q
2
>
0and
q
1
>
0,
q
2
<
0.
Remark 109 (Scheduling First Step)
To solve a GeTLS problem, a possible
technique is to begin with some iterations with
0(
i.e., the corresponding OLS
problem
)
. This method places the weight vector in the solution locus, just avoiding
entering the divergence zone.
ζ
=
If the eigenvalue decomposition of A
T
Ais
Remark 110 (An EVD Technique)
known
(
i.e., the orthogonal matrix V
)
, the vector q
≡
q
1
q
2
···
q
n
T
and the
position of the z axes can be computed. Therefore, the position of the solution
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