Different equivalent approaches to the geodetic reference system

Abstract

A new approach to the determination of least-squares solutions for the rank deficient model is presented, which, in place of the classical minimal constraints, utilizes the expression of the excess parameters as a function of a subset of unknown parameters describing the model without ambiguities. The role of the lack of reference system definition in the resulting rank deficiency is explained, first for the original non-linear model case, and is specialized next to the linearized model. The whole set of solutions is parameterized in terms of two matrices defining the linearized relation from the model describing parameters to the remaining ones. Particular values are obtained for the case of solution with minimal trace of its covariance matrix, as well for the solution for minimum norm. Finally, the connection with the existing classical approach is established, while the approach is further elaborated in terms of the full-rank decompositions of the model design matrix.

Appendices

Appendix A: Derivation of the minimum trace of covariance solution

To minimize

$$ \phi = tr\left[ {Q_{0} (I + K^{T} K)} \right], $$

(45)

where

$$ Q_{0} = (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (I + K^{T} \varGamma^{\rm T} )^{ - 1} , $$

(46)

we need to set

$$ H_{ij} \equiv \frac{\partial \phi }{{\partial K_{ij} }} = 0,\quad \forall i,j. $$

(47)

The formal rule for differentials gives

$$ d\phi = \sum\limits_{i} {\sum\limits_{j} {\frac{\partial \phi }{{\partial K_{ij} }}dK_{ij} } } = \sum\limits_{i} {\sum\limits_{j} {H_{ij} dK_{ij} } } = \sum\limits_{j} {\sum\limits_{i} {(H^{T} )_{ji} dK_{ij} } } = \sum\limits_{j} {(H^{T} dK)_{jj} } = {\text{tr}}(H^{T} dK). $$

(48)

Therefore, the minimization problem can be solved by determining the matrix $H = H(K)$, such that $d\phi = {\text{tr}}(H^{T} dK)$, and then solve the equation $H(K) = 0$ for $K.$

From $0 = dI = d(MM^{ - 1} ) = dMM^{ - 1} + MdM^{ - 1}$ follows that $dM^{ - 1} = - M^{ - 1} dMM^{ - 1}$, and hence,

$$ d\{ (I + \varGamma K)^{ - 1} \} = - (I + \varGamma K)^{ - 1} \varGamma dK(I + \varGamma K)^{ - 1} , $$

(49)

$$ d\{ (I + K^{T} \varGamma^{\rm T} )^{ - 1} \} = - (I + K^{T} \varGamma^{\rm T} )^{ - 1} dK^{T} \varGamma^{\rm T} (I + K^{T} \varGamma^{\rm T} )^{ - 1} . $$

(50)

Consequently

$$ \begin{aligned} dQ_{0} = d\{ (I + \varGamma K)^{ - 1} \} N_{0}^{ - 1} (I + K^{T} \varGamma^{\rm T} )^{ - 1} + (I + \varGamma K)^{ - 1} N_{0}^{ - 1} d\{ (I + K^{T} \varGamma^{\rm T} )^{ - 1} \} \hfill \\ = - (I + \varGamma K)^{ - 1} \varGamma dK\,Q_{0} - Q_{0} \,dK^{T} \varGamma^{\rm T} (I + K^{T} \varGamma^{\rm T} )^{ - 1} . \hfill \\ \end{aligned} $$

(51)

The differential of the target function $\phi$ becomes

$$ \begin{aligned} d\phi = {\text{tr}}\left[ {dQ_{0} (I + K^{T} K) + Q_{0} \,dK^{T} K + Q_{0} K^{T} dK} \right] \hfill \\ = {\text{tr}}\left[ { - (I + \varGamma K)^{ - 1} \varGamma dK\,Q_{0} (I + K^{T} K) - Q_{0} \,dK^{T} \varGamma^{\rm T} (I + K^{T} \varGamma^{\rm T} )^{ - 1} (I + K^{T} K) + Q_{0} dK^{T} K + Q_{0} K^{T} dK} \right]. \hfill \\ \end{aligned} $$

(52)

Using the well-known properties ${\text{tr}}(M^{T} ) = {\text{tr}}(M)$ and ${\text{tr}}({\text{AB}}) = {\text{tr}}({\text{BA}}),$ we easily arrive at

$$ d\phi = \,tr\left\{ {2Q_{0} \left[ {K^{T} - (I + K^{T} K)(I + \varGamma K)^{ - 1} \varGamma } \right]dK} \right\}. $$

(53)

Comparison with (48) shows that

$$ H = 2\left\{ {K - \varGamma^{T} (I + K^{T} \varGamma^{T} )^{ - 1} (I + K^{T} K)} \right\}Q_{0} , $$

(54)

and since $\det Q_{0} \ne 0$, the solution system $H(K) = 0$ reduces to

$$ K = \varGamma^{T} (I + K^{T} \varGamma^{T} )^{ - 1} (I + K^{T} K), $$

(55)

which is obviously satisfied by

$$ K = \varGamma^{T} . $$

(56)

To assess the uniqueness of this solution, we transpose and rearrange (55) into

$$ \varGamma = (I + \varGamma {\rm K})(I + K^{T} K)^{ - 1} K^{T} = (I + K^{T} K)^{ - 1} K^{T} + \varGamma {\rm K}(I + K^{T} K)^{ - 1} K^{T} . $$

(57)

Solving for $\varGamma$ yields

$$ \varGamma = (I + K^{T} K)^{ - 1} K^{T} \left[ {{\rm I} - {\rm K}(I + K^{T} K)^{ - 1} K^{T} } \right]^{ - 1} , $$

(58)

and upon using the easy to prove matrix identity $(I + KK^{T} )^{ - 1} = {\rm I} - {\rm K}(I + K^{T} K)^{ - 1} K^{T}$

$$ \varGamma = (I + K^{T} K)^{ - 1} K^{T} (I + KK^{T} ) = (I + K^{T} K)^{ - 1} (K^{T} + K^{T} KK^{T} ) = (I + K^{T} K)^{ - 1} (I + K^{T} K)K^{T} = K^{T} . $$

(59)

Appendix B: Derivation of the minimum norm solution

The target function to be minimized is as follows:

$$ \phi = \left\| x \right\|^{2} = \left[ {\begin{array}{*{20}c} {x_{0} } \\ {x_{c} } \\ \end{array} } \right]^{T} \left[ {\begin{array}{*{20}c} {x_{0} } \\ {x_{c} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {x_{0} } \\ {Kx_{0} + k} \\ \end{array} } \right]^{T} \left[ {\begin{array}{*{20}c} {x_{0} } \\ {Kx_{0} + k} \\ \end{array} } \right] = x_{0}^{T} x_{0} + (Kx_{0} + k)^{T} (Kx_{0} + k) = \mathop {\hbox{min} }\limits_{K,k} , $$

(60)

where also $x_{0}$ depends on $K$ and $k$ through $x_{0} = (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k)$. Taking into account that

$$ \begin{aligned} \frac{{\partial (I + \varGamma K)^{ - 1} }}{{\partial K_{ij} }} &= - (I + \varGamma K)^{ - 1} \frac{\partial (I + \varGamma K)}{{\partial K_{ij} }}(I + \varGamma K)^{ - 1} \\&= - (I + \varGamma K)^{ - 1} \varGamma \frac{\partial K}{{\partial K_{ij} }}(I + \varGamma K)^{ - 1} \hfill \\ &= - (I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} , \hfill \\ \end{aligned} $$

(61)

where $e_{j}$, $\varepsilon_{i}$ are vectors with elements $(e_{j} )_{k} = \delta_{jk}$ and $(\varepsilon_{i} )_{k} = \delta_{ik}$, we compute first the partial derivatives

$$ \frac{{\partial x_{0} }}{{\partial K_{ij} }} = \frac{{\partial (I + \varGamma K)^{ - 1} }}{{\partial K_{ij} }}N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k) = - (I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k), $$

(62)

$$ \frac{{\partial (Kx_{0} + k)}}{{\partial K_{ij} }} = \frac{\partial K}{{\partial K_{ij} }}x_{0} + K\frac{{\partial x_{0} }}{{\partial K_{ij} }} = e_{j} \varepsilon_{i}^{T} x_{0} - K(I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k). $$

(63)

$$ \frac{{\partial x_{0} }}{\partial k} = - (I + \varGamma K)^{ - 1} N_{0}^{ - 1} N_{0} \varGamma = - (I + \varGamma K)^{ - 1} \varGamma , $$

(64)

$$ \frac{{\partial (Kx_{0} + k)}}{\partial k} = K\frac{{\partial x_{0} }}{\partial k} + I = - K(I + \varGamma K)^{ - 1} \varGamma + I. $$

(65)

To obtain the minimum norm solution, we must set the derivatives of the target function $\phi$ with respect to the vector $k$ and all elements of the matrix $K$ equal to zero. For the elements of $K$, we obtain

$$ \begin{aligned} \frac{1}{2}\frac{\partial \phi }{{\partial K_{ij} }} & = x_{0}^{T} \frac{{\partial x_{0} }}{{\partial K_{ij} }} + (Kx_{0} + k)^{T} \frac{{\partial (Kx_{0} + k)}}{{\partial K_{ij} }} \\ = - x_{0}^{T} (I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k) \\ + (Kx_{0} + k)^{T} \left\{ {e_{j} \varepsilon_{i}^{T} x_{0} - K(I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k)} \right\} \\ = - x_{0}^{T} (I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k) \\ + (Kx_{0} + k)^{T} e_{j} \varepsilon_{i}^{T} x_{0} - (Kx_{0} + k)^{T} K(I + \varGamma K)^{ - 1} \varGamma e_{j} \varepsilon_{i}^{T} (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k) \\ = - e_{j}^{T} \varGamma^{T} (I + \varGamma K)^{ - T} x_{0} (u_{0} - N_{0} \varGamma k)^{T} N_{0}^{ - 1} (I + \varGamma K)^{ - T} \varepsilon_{i} \\ + e_{j}^{T} (Kx_{0} + k)x_{0}^{T} \varepsilon_{i} - e_{j}^{T} \varGamma^{T} (I + \varGamma K)^{ - T} K^{T} (Kx_{0} + k)(u_{0} - N_{0} \varGamma k)^{T} N_{0}^{ - 1} (I + \varGamma K)^{ - T} \varepsilon_{i} = 0, \\ \end{aligned} $$

(66)

which for all $i$ and $j$ leads to

$$ \begin{aligned} - \varGamma^{T} (I + \varGamma K)^{ - T} x_{0} (u_{0} - N_{0} \varGamma k)^{T} N_{0}^{ - 1} (I + \varGamma K)^{ - T} \hfill \\ + (Kx_{0} + k)x_{0}^{T} - \varGamma^{T} (I + \varGamma K)^{ - T} K^{T} (Kx_{0} + k)(u_{0} - N_{0} \varGamma k)^{T} N_{0}^{ - 1} (I + \varGamma K)^{ - T} = 0. \hfill \\ \end{aligned} $$

(67)

From $x_{0} = (I + \varGamma K)^{ - 1} N_{0}^{ - 1} (u_{0} - N_{0} \varGamma k)$ follows that $(u_{0} - N_{0} \varGamma k) = N_{0} (I + \varGamma K)x_{0}$, which replaced above gives

$$ - \varGamma^{T} (I + \varGamma K)^{ - T} x_{0} x_{0}^{T} + (Kx_{0} + k)x_{0}^{T} - \varGamma^{T} (I + \varGamma K)^{ - T} K^{T} (Kx_{0} + k)x_{0}^{T} = 0. $$

(68)

For the elements of $k$, we obtain

$$ \frac{1}{2}\frac{\partial \phi }{\partial k} = x_{0}^{T} \frac{{\partial x_{0} }}{\partial k} + (Kx_{0} + k)^{T} \frac{{\partial (Kx_{0} + k)}}{\partial k} = - x_{0}^{T} (I + \varGamma K)^{ - 1} \varGamma + (Kx_{0} + k)^{T} (I - K(I + \varGamma K)^{ - 1} \varGamma ) = 0. $$

(69)

Thus, the minimum norm solution is obtained for the values of $K$ and $k$, which satisfy the system

$$ - \varGamma^{T} (I + \varGamma K)^{ - T} x_{0} x_{0}^{T} + (Kx_{0} + k)x_{0}^{T} - \varGamma^{T} (I + \varGamma K)^{ - T} K^{T} (Kx_{0} + k)x_{0}^{T} = 0, $$

(70)

$$ - \varGamma^{T} (I + \varGamma K)^{ - T} x_{0} + [I - \varGamma^{T} (I + \varGamma K)^{ - T} K^{T} ](Kx_{0} + k) = 0. $$

(71)

It is easy to see that this system is satisfied by $K = \varGamma^{T}$ and $k = 0.$ Indeed, for these values, the right-hand side of (B11) becomes

$$ - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} x_{0} x_{0}^{T} + \varGamma^{T} x_{0} x_{0}^{T} - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} \varGamma \varGamma^{T} x_{0} x_{0}^{T} = - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} (I + \varGamma \varGamma^{T} )x_{0} x_{0}^{T} + \varGamma^{T} x_{0} x_{0}^{T} = 0, $$

(71)

while the right-hand side of (B12) becomes

$$ \begin{aligned} - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} x_{0} + \left[ {I - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} \varGamma } \right]\varGamma^{T} x_{0} & = - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} x_{0} + \varGamma^{T} x_{0} - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} \varGamma \varGamma^{T} x_{0} \\ = \varGamma^{T} x_{0} - \varGamma^{T} (I + \varGamma \varGamma^{T} )^{ - 1} (I + \varGamma \varGamma^{T} )x_{0} = 0, \\ \end{aligned} $$

(72)

and this completes the proof.

Appendix C: Proof of equivalence of new and classical solutions

Since the minimum trace and the minimum norm solutions are simply special cases ($C \to E$,$k \to 0$) of the general least-squares solution, we need to prove the equivalence of new and classical formulas only for the general case.

To compare the new and classical old solution, which we will now symbolize with $\hat{x}_{\text{NEW}}$ and $\hat{x}_{\text{OLD}}$, respectively, for the sake of distinction, we will rewrite (4) in the form

$$ \hat{x}_{\text{NEW}} = \left[ {\begin{array}{*{20}c} {\hat{x}_{0} } \\ {\hat{x}_{c} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {(I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k)} \\ {K(I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k) + k} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} I \\ K \\ \end{array} } \right](I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k) + \left[ {\begin{array}{*{20}c} 0 \\ k \\ \end{array} } \right]. $$

(73)

Instead of comparing (73) with the classical solution

$$ \hat{x}_{\text{OLD}} = (N + CC^{T} )^{ - 1} u + E(C^{T} E)^{ - 1} d, $$

(74)

we will multiply them both from the left with the same non-singular matrix $(N + CC^{T} ),$ and we will compare instead

$$ (N + CC^{T} )\hat{x}_{NEW} = (N + CC^{T} )\left[ {\begin{array}{*{20}c} I \\ K \\ \end{array} } \right](I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k) + (N + CC^{T} )\left[ {\begin{array}{*{20}c} 0 \\ k \\ \end{array} } \right], $$

(75)

with

$$ (N + CC^{T} )\hat{x}_{OLD} = u + (N + CC^{T} )E(C^{T} E)^{ - 1} d. $$

(76)

To this purpose, we note that

$$ N = \left[ {\begin{array}{*{20}c} {A_{0}^{T} PA_{0} } & {A_{0}^{T} PA_{c} } \\ {A_{c}^{T} PA_{0} } & {A_{c}^{T} PA_{c} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {A_{0}^{T} PA_{0} } & {A_{0}^{T} PA_{0} \varGamma } \\ {\varGamma^{T} A_{0}^{T} PA_{0} } & {\varGamma^{T} A_{0}^{T} PA_{0} \varGamma } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {N_{0} } & {N_{0} \varGamma } \\ {\varGamma^{T} N_{0} } & {\varGamma^{T} N_{0} \varGamma } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]N_{0} \left[ {\begin{array}{*{20}c} I & \varGamma \\ \end{array} } \right], $$

(77)

and

$$ u = \left[ {\begin{array}{*{20}c} {A_{0}^{T} Py^{ob} } \\ {A_{c}^{T} Py^{ob} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {A_{0}^{T} Py^{ob} } \\ {\varGamma^{T} A_{0}^{T} Py^{ob} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {u_{0} } \\ {\varGamma^{T} u_{0} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {u_{0} } \\ { - E_{c}^{ - T} E_{0}^{T} u_{0} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]u_{0} , $$

(78)

while recalling that $K = - C_{c}^{ - T} C_{0}^{T} ,$ $k = C_{c}^{ - T} d,$ $\varGamma = - E_{0} E_{c}^{ - 1}$

$$ C^{T} \left[ {\begin{array}{*{20}c} I \\ K \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {C_{0}^{T} } & {C_{c}^{T} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} I \\ { - C_{c}^{ - T} C_{0}^{T} } \\ \end{array} } \right] = C_{0}^{T} - C_{c}^{T} C_{c}^{ - T} C_{0}^{T} = 0, $$

(79)

$$ C^{T} \left[ {\begin{array}{*{20}c} 0 \\ k \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {C_{0}^{T} } & {C_{c}^{T} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 0 \\ {C_{c}^{ - T} d} \\ \end{array} } \right] = C_{c}^{T} C_{c}^{ - T} d = d, $$

(80)

$$ \left[ {\begin{array}{*{20}c} I & \varGamma \\ \end{array} } \right]E = \left[ {\begin{array}{*{20}c} I & { - E_{0} E_{c}^{ - 1} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {E_{0} } \\ {E_{c} } \\ \end{array} } \right] = E_{0} - E_{0} E_{c}^{ - 1} E_{c} = 0. $$

(81)

Considering the above relations, direct calculation gives

$$ \begin{aligned} (N + CC^{T} )\hat{x}_{\text{NEW}} &= N\left[ {\begin{array}{*{20}c} I \\ K \\ \end{array} } \right](I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k) \\ &\,\,+ CC^{T} \left[ {\begin{array}{*{20}c} I \\ K \\ \end{array} } \right](I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k) + N\left[ {\begin{array}{*{20}c} 0 \\ k \\ \end{array} } \right] + CC^{T} \left[ {\begin{array}{*{20}c} 0 \\ k \\ \end{array} } \right] \\ &= \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]N_{0} \left[ {\begin{array}{*{20}c} I \varGamma \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} I \\ K \\ \end{array} } \right](I + \varGamma K)^{ - 1} (N_{0}^{ - 1} u_{0} - \varGamma k) + \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]N_{0} \left[ {\begin{array}{*{20}c} I \varGamma \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 0 \\ k \\ \end{array} } \right] + Cd \\ &= \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]N_{0} (N_{0}^{ - 1} u_{0} - \varGamma k) + \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]N_{0} \varGamma k + Cd \\ &= \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]u_{0} + Cd. \\ \end{aligned} $$

(82)

Noting that (77) and (81) imply the well-known relation $NE = 0,$ we obtain

$$ (N + CC^{T} )\hat{x}_{OLD} = u + (N + CC^{T} )E(C^{T} E)^{ - 1} d = u + NE(C^{T} E)^{ - 1} d + CC^{T} E(C^{T} E)^{ - 1} d = u + Cd, $$

(83)

Which, in view of (78), becomes

$$ (N + CC^{T} )\hat{x}_{\text{OLD}} = \left[ {\begin{array}{*{20}c} I \\ {\varGamma^{T} } \\ \end{array} } \right]u_{0} + Cd. $$

(84)

Comparing (82) with (84), it follows that $(N + CC^{T} )\hat{x}_{\text{NEW}} = (N + CC^{T} )\hat{x}_{\text{OLD}}$, and hence, $\hat{x}_{\text{NEW}} = \hat{x}_{\text{OLD}} ,$ which proves the equivalence of the two solutions.