Global Positioning System Reference
In-Depth Information
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The matrix on the left side of (4.170) is a nonsingular matrix if the conditions (4.169)
are linearly independent; i.e., the
(u
u
matrix
B
has full row rank, and the rows
are linear-independent of the rows of the design matrix
A
. A general expression for
the inverse is obtained from
A
T
PA B
T
BO
−
r)
×
Q
B
S
T
SR
IO
OI
=
(4.171)
This matrix equation gives the following four equations of submatrices:
A
T
PAQ
B
+
B
T
S
=
I
(4.172)
A
T
PAS
T
B
T
R
+
=
O
(4.173)
[12
BQ
B
=
O
(4.174)
Lin
—
2
——
Lon
PgE
BS
T
=
I
(4.175)
Th
e solution of these equations requires the introduction of the
(u
u
matrix
E
,
wh
ose rows span the null space of the design matrix
A
or the null space of the normal
m
atrix. According to (A.53), there is a matrix
E
such that
A
T
PA
E
T
−
r)
×
=
O
(4.176)
[12
or
AE
T
EA
T
=
O
or
=
O
(4.177)
Be
cause the rows of
B
are linearly independent of the rows of
A
, the
(u
r)
m
atrix
BE
T
has full rank and thus can be inverted. Multiplying (4.172) by
E
from the
lef
t and using (4.177), we get
−
r)
×
(u
−
=
EB
T
−
1
E
S
(4.178)
Th
is expression also satisfies (4.175). Substituting
S
into (4.173) gives
A
T
PAE
T
BE
T
−
1
B
T
R
+
=
O
(4.179)
Because of (4.176), this expression becomes
B
T
R
=
O
(4.180)
Be
cause
B
has full rank, it follows that the matrix
R
=
O
. Thus,
A
T
PA B
T
BO
−
1
E
T
BE
T
−
1
Q
B
=
(4.181)
EB
T
−
1
E
O
