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Hence,
(
∞
,
e
n
+
1
)
is an eigenpair with an infinite simple eigenvalue. The same
type of analysis as before estimates the remaining eigenpairs. From eq. (6.2), it
follows that
A
T
AA
T
b
b
T
Ab
T
b
u
1
u
n
+
1
I
n
u
1
u
n
+
1
0
=
2
γ
0
T
0
A
T
Au
1
A
T
bu
n
+
1
+
=
2
γ
u
1
⇒
b
T
Au
1
+
b
T
bu
n
+
1
=
0
A
T
Au
1
−
A
T
b
b
T
b
−
1
b
T
Au
1
=
2
γ
u
1
u
n
+
1
=−
b
T
b
−
1
b
T
Au
1
⇒
A
T
I
m
−
b
b
T
b
−
1
b
T
Au
1
⇒
=
2
γ
u
1
(6.13)
Equation (6.13) can be solved for
u
1
=
0(i.e.,
u
parallel to
e
n
+
1
), as seen previ-
ously. To estimate the other eigenpairs, it can be derived from eq. (6.13) :
A
T
P
b
Au
1
=
2
γ
u
1
I
m
−
b
b
T
b
−
1
b
T
is the symmetric projection matrix (
P
⊥
2
where
P
b
=
P
b
) that projects the column space of
A
into the orthogonal complement of
b
.
Hence,
=
b
A
T
P
⊥
b
Au
1
=
2
γ
u
1
⇒
A
T
P
⊥
b
P
b
Au
1
=
2
γ
u
1
⇒
P
b
A
T
P
b
A
u
1
=
2
γ
u
1
(6.14)
It can be concluded that if the singular values and right singular vectors of
matrix
P
b
A
are given by
π
i
and
v
i
, respectively, for
i
=
1,
...
,
n
(the singular
values are sorted in decreasing order), the required eigenpairs of the DLS case
are given by the
n
couples
π
i
where
ρ
i
=−
b
T
b
−
1
b
T
A
v
i
. It can
be deduced that the first
n
components of the DLS eigenvectors are parallel to
the
v
i
singular vectors. In particular, the right singular vector associated with
the smallest singular value of
P
b
A
corresponds to the DLS solution (see [51]).
Indeed, it corresponds to the last eigenpair, scaled by
ρ
min
=−
1.
2
,
v
i
ρ
i
6.1.2 General Case
In the case 0
<ζ <
1,
R
can be diagonalized by a
D
-orthogonal transformation
[178]. Indeed,
Z
T
Z
D
=
(6.15)
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