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1.8 MIXED OLS-TLS PROBLEM
If
n
1
columns of the
m
×
n
data matrix
A
are
known exactly
, the problem is called
mixed OLS-TLS
[98]. It is natural to require that the TLS solution not perturb
the exact columns. After some column permutations in
A
such that
A
=
[
A
1
;
A
2
],
where
A
1
∈
n
2
, perform
n
1
Householder transformations
Q
on the matrix [
A
;
B
] (QR factorization) so that
m
×
n
1
m
×
is made of the exact
n
1
columns and
A
2
∈
R
11
R
12
R
1
b
0
n
1
R
22
R
2
b
m
−
n
1
[
A
1
;
A
2
;
B
]
=
Q
(1.46)
n
1
n
−
n
1
d
where
R
11
is a
n
1
×
n
1
upper triangular matrix. Then compute the TLS solution
X
2
X
2
of
R
22
X
≈
R
2
b
.
yields the last
n
−
n
1
components of each solution
X
=
X
1
;
X
2
T
,
X
1
of the solution matrix
vector
x
i
.Tofindthefirst
n
1
rows
R
12
X
2
. Thus, the
entire
method amounts to a
preprocessing step, a TLS problem, an OLS problem, and a
postprocessing
step
(inverse row permutations) [72].
(OLS)
R
11
X
1
solve
=
R
1
b
−
Theorem 29 (Closed-Form Mixed OLS-TLS Solution)
Let rank A
1
=
n
1
;
denote by
σ
(
respectively,
σ
)
the smallest
[
respectively,
(
n
2
+
1
)
th
]
singular
value of R
22
(
respectively,
[
R
22
;
R
2
b
]);
assume that
the smallest singular
σ
> σ
, then the mixed OLS-TLS
values
σ
=
σ
n
2
+
1
=···=
σ
n
2
+
d
coincide. If
solution is
A
T
A
−
σ
2
00
0
−
1
X
=
A
T
B
(1.47)
I
n
2
Proof.
It is a special case of [97, Th. 4].
1.9 ALGEBRAIC COMPARISONS BETWEEN TLS AND OLS
Comparing (1.29) with the LS solution
X
=
A
T
A
−
1
A
T
B
(1.48)
shows that
σ
n
+
1
completely determines the difference between the solutions.
Assuming that
A
is of full rank,
σ
n
+
1
=
0 means that both solutions coincide
(
AX
≈
B
compatible or underdetermined). As
σ
n
+
1
deviates from zero, the set
AX
≈
B
becomes more and more incompatible and the differences between the
TLS and OLS solutions become deeper and deeper.
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