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where
E
=
r
u
,[
A
;
b
]
T
[
A
;
b
]
(4.25)
and
k
=
∂
E
/∂
u
n
+
1
u
n
+
1
=−
1
. In the critical points
x
c
;−
1
T
is
E
TLS
(
u
)
of the
RQ,
1
corresponding to the singular values
σ
c
of [
A
;
b
] (i.e., the eigenvalues
c
of [
A
;
b
]
T
[
A
;
b
]) the Hessian matrix is
λ
c
=
σ
A
T
A
−
σ
2
1
+
x
c
x
c
2
A
T
b
c
I
n
H
TLS
=
b
T
A
b
T
b
−
σ
c
A
T
A
−
σ
c
I
n
x
c
A
T
A
−
σ
2
c
I
n
2
2
1
+
x
c
x
c
n
1
syst
.(
1
.
18
)
=
b
T
A
b
T
Ax
c
(4.26)
n
1
So
H
TLS
is singular because the
(
n
+
1
)
th column is a linear combination (by
x
c
)ofthefirst
n
columns. The
n
×
n
submatrix
1
+
x
c
x
c
A
T
A
−
σ
c
I
n
2
h
TLS
=
2
(4.27)
is positive definite in the critical point corresponding to the minimum
eigenvalue. Indeed, respecting the notation of Section 1.3, given the EVD of
A
T
A
=
V
V
T
,where
=
diag
σ
1
,
σ
2
,
...
,
σ
n
, eq. (4.27) yields
1
+
x
c
x
c
V
−
σ
c
I
n
V
T
2
h
TLS
=
2
(4.28)
Recalling the interlacing property (Theorem 6),
σ
1
≥
σ
1
≥···≥
σ
n
≥
σ
n
≥
σ
n
+
1
(4.29)
σ
n
> σ
if the TLS existence condition
1
is valid, then
in the TLS hyperplane,
all critical points are a maximum (negative semidefinite Hessian matrix), saddles
(indefinite Hessian matrix), and a minimum (positive definite Hessian matrix)
.
The submatrix
h
TLS
is the Hessian matrix of the TLS cost function,
n
+
T
E
TLS
(
x
)
=
(
Ax
−
b
)
(
Ax
−
b
)
1
+
x
T
x
=
E
TLS
(
u
)
(4.30)
n
. Indeed,
as a function of
x
∈
1
+
x
T
x
A
T
(
Ax
−
b
)
−
xE
TLS
2
∇
x
E
TLS
(
x
)
=
(4.31)
1
At the critical point the RQ assumes the value of the corresponding eigenvalue (see Proposition
43).
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