LEAST-SQUARES ADJUSTMENTS - GPS Satellite Surveying - page 152

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with a cofactor matrix for the residuals

P − 1 AN − 1 A T

Q v = − (4.337)

Co mpute the trace

Tr I AN − 1 A T P Tr ( Q v P ) = −

Tr N − 1 A T PA

(4.338)

= n −

= n − u

A more general expression is obtained by noting that the matrix AN − 1 A T P is idem-
potent. The trace of an idempotent matrix equals the rank of that matrix. Thus,
Tr AN − 1 A T P =

[15

R A T PA =

R( A ) = r ≤ u (4.339)

Lin
—
1.3
——
Nor
PgE

Thus, from Equations (4.338) and (4.339)

Tr ( Q v P ) = Tr ( PQ v ) = n − R( A ) (4.340)

By denoting the diagonal element of the matrix Q v P by r i , we can write

n

r i = n − R( A ) (4.341)

i = 1

[15

The sum of the diagonal elements of Q v P equals the degree of freedom. The element
r i is called the redundancy number for the observation i . It is the contribution of the
i th observation to the degree of freedom. If the weight matrix P is diagonal, which is
usually the case when original observations are adjusted, then

=

r i q i p i (4.342)

where q i is the diagonal element of the cofactor matrix Q v , and p i denotes the weight
of the i th observation. Equation (4.337) implies the inequality

1
p i

0 ≤ q i ≤ (4.343)

Multiplying by p i gives the bounds for the redundancy numbers,

0 ≤ r i ≤ 1 (4.344)

Considering the general relation

Q = Q

b −

Q v (4.345)

a