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A T Au 1 + A T bu n + 1 = 0
b T Au 1 + b T bu n + 1 = 2 γ u n + 1
u 1 =− A T A 1 A T bu n + 1
b T A A T A 1 A T bu n + 1 + b T bu n + 1 = 2 γ u n + 1
b T I A A T A 1 A T bu n + 1 = 2 γ u n + 1
(6.7)
Two solutions are possible. The first requires that u n + 1 =
0 and is another proof
of the existence of the
(
, u )
eigenpairs. The second requires that u n + 1 =
0
(i.e., u not parallel to u )and
2 b T I A A T A 1 A T b =
2 P A b
1
1
1
2
2
2 b T P A b =
γ =
(6.8)
I m A A T A 1 A T is the symmetric projection matrix ( P 2
A
where P A
=
P A ) that projects into the orthogonal complement of the column space of A .
From eq. (6.4) it follows that for
=
ζ =
0,
x OLS = A T A 1 A T b = A + b
(6.9)
which is the well-known OLS solution. Recalling that γ
corresponds to the value
of the error function (5.6), we have
1
2
T
1
2
2
γ OLS = E OLS ( x OLS ) =
( Ax OLS b )
( Ax OLS b ) =
Ax OLS b
2 A A T A 1 A T b b
2 P A b
2
2 =
2
2
1
1
=
(6.10)
which confirms the result (6.8).
Summarizing: There are n infinite eigenvalues whose corresponding eigenvec-
tors do not intersect the TLS hyperplane and one eigenvalue, γ OLS , corresponding
to the eigenvector which is is the minimum of the GeTLS EXIN cost function
and intersects the TLS hyperplane in the OLS solution (6.9).
6.1.1.2 Case
ζ =
1 (DLS) If ζ = 1, it follows that
D = 2 I n
0
(6.11)
0 T
0
1
γ
1
γ
u = e n + 1 De n + 1 =
0
Re n + 1 =
0
=
0
γ =∞
(6.12)
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