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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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