Information Technology Reference
In-Depth Information
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)
Search WWH ::
Custom Search