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  大地测量与地球动力学  2021, Vol. 41 Issue (2): 157-161  DOI: 10.14075/j.jgg.2021.02.009

引用本文  

谢波, 刘连旺. 应用分块矩阵求逆解算约束秩亏间接平差模型[J]. 大地测量与地球动力学, 2021, 41(2): 157-161.
XIE Bo, LIU Lianwang. Calculation of Rank-Defective Indirect Adjustment Model with Constraint by Inverting Block Matrix[J]. Journal of Geodesy and Geodynamics, 2021, 41(2): 157-161.

项目来源

安徽省高校自然科学重点研究项目(KJ2019A1114);安徽省高校质量工程项目(2018jxtd037)。

Foundation support

Natural Science Foundation of Colleges and Universities in Anhui Province, No.KJ2019A1114; Quality Engineering Project of Colleges and Universities in Anhui Province, No.2018jxtd037.

第一作者简介

谢波,副教授,主要从事测量数据处理和精密工程测量研究,E-mail: shie73@163.com

About the first author

XIE Bo, associate professor, majors in measurement data processing and precision engineering survey, E-mail:shie73@163.com.

文章历史

收稿日期:2020-05-03
应用分块矩阵求逆解算约束秩亏间接平差模型
谢波1     刘连旺2     
1. 合肥职业技术学院建筑工程学院,合肥市姥山南路1号,238000;
2. 桂林理工大学博文管理学院,桂林市雁山街317号,541006
摘要:在通常的四分块矩阵求广义逆矩阵和凯利逆矩阵公式基础上,分析左上角子矩阵为秩亏、右下角子矩阵为零的特殊四分块矩阵的凯利逆矩阵存在条件,应用广义逆矩阵法和矩阵变换法推导该类特殊四分块矩阵的凯利逆矩阵显性表达公式,并用于解算约束秩亏间接平差模型的参数估计。实验数据表明,当满足存在性条件时,应用分块矩阵求逆公式解算约束秩亏间接平差模型的结果与间接平差模型的解算结果一致,表明推导的显性表达公式具有可行性。
关键词测量平差分块矩阵广义逆矩阵约束秩亏间接平差模型

在经典测量平差模型中,将附有限制条件的间接平差模型作为概括平差函数模型[1]具有列立误差方程式规律性强、参数平差值即为结果、便于编程计算等优点[2],但该模型中满秩的误差方程包含已知数据,约束方程也由已知数据建立,会混淆已知数据和观测数据2类不同性质的数据。若将平差系统中所有点作为未知参数,利用观测数据建立误差方程,用已知数据建立约束方程或基准方程,两者联合组成约束秩亏间接平差模型,不仅可有效区分观测数据和已知数据,并且在理解参考系效应和内部噪声、评定观测值的内部符合精度、分析已知数据对平差结果的影响[3-4]等方面具有明显优势。因此,约束秩亏间接平差模型被广泛应用于变形监测、近景摄影测量、GPS测量、地球参考框架建立[5-8]等大地测量数据处理中。

国内学者对解算约束秩亏间接平差模型的方法进行过大量研究[9-12],本文在这些研究基础上进一步分析约束秩亏间接平差模型的法方程系数四分块矩阵结构,研究其凯利逆矩阵的存在条件和显性表达式,并验证公式的正确性,为解算约束秩亏间接平差模型提供简单准确的方法。

1 特殊分块矩阵求逆 1.1 引理[13]

引理1   设Nu×u非负定矩阵,Cs×u矩阵,N'bb=N+CTCN'cc=CN'bb-CT,则:

$ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^ - } = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^ - - \mathit{\boldsymbol{N'}}_{bb}^ - {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^ - \mathit{\boldsymbol{CN'}}_{bb}^ - }&{\mathit{\boldsymbol{N'}}_{bb}^ - {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^ - }\\ {\mathit{\boldsymbol{N'}}_{cc}^ - \mathit{\boldsymbol{CN'}}_{bb}^ - }&{\mathit{\boldsymbol{N'}}_{cc}^ - {{\mathit{\boldsymbol{N'}}}_{cc}} - \mathit{\boldsymbol{N'}}_{cc}^ - } \end{array}} \right] \end{array} $ (1)

引理2   设实矩阵$ \boldsymbol{A}=\left[\begin{array}{ll} \boldsymbol{A}_{11} & \boldsymbol{A}_{12} \\ \boldsymbol{A}_{21} & \boldsymbol{A}_{22} \end{array}\right]$,若A可逆且$\left|\boldsymbol{A}_{11}\right| \neq 0, \boldsymbol{B}=\boldsymbol{A}_{22}-\boldsymbol{A}_{21} \boldsymbol{A}_{11}^{-1} \boldsymbol{A}_{12} $,则:

$ {\mathit{\boldsymbol{A}}^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{A}}_{11}^{ - 1} + \mathit{\boldsymbol{A}}_{11}^{ - 1}{\mathit{\boldsymbol{A}}_{12}}{\mathit{\boldsymbol{B}}^{ - 1}}{\mathit{\boldsymbol{A}}_{21}}\mathit{\boldsymbol{A}}_{11}^{ - 1}}&{ - \mathit{\boldsymbol{A}}_{11}^{ - 1}{\mathit{\boldsymbol{A}}_{12}}{\mathit{\boldsymbol{B}}^{ - 1}}}\\ { - {\mathit{\boldsymbol{B}}^{ - 1}}{\mathit{\boldsymbol{A}}_{21}}\mathit{\boldsymbol{A}}_{11}^{ - 1}}&{{\mathit{\boldsymbol{B}}^{ - 1}}} \end{array}} \right] $ (2)
1.2 存在性条件

设存在四分块矩阵$\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right] $Nu×u非负定矩阵,R(N)=r < uCs×u矩阵。如果四分块矩阵的凯利逆存在,则必须满足[14]

1) R(C)=s,表示C为行满秩矩阵。如果C行秩亏,R(C⋮0)必行秩亏,$ R\left(\begin{array}{l} \boldsymbol{C} \\ 0 \end{array}\right)$必列秩亏,四分块矩阵$ \left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]$必定行和列秩亏,则该四分块矩阵为奇异矩阵。

2) $ R\left(\begin{array}{l} \boldsymbol{N} \\ \boldsymbol{C} \end{array}\right)=u$,表示矩阵$ \left[ \begin{array}{l} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array} \right]$为列满秩矩阵。根据Nu×u非负定矩阵,Cs×u矩阵,则$ \left[ \begin{array}{l} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array} \right]$为(u+su矩阵。如果$ \left[ \begin{array}{l} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array} \right]$列秩亏,四分块矩阵$\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right] $必定列秩亏,且为奇异矩阵,则必有$ \left[ \begin{array}{l} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array} \right]$列满秩,即满足$R\left(\begin{array}{l} \boldsymbol{N} \\ \boldsymbol{C} \end{array}\right)=u $。该结论等价于:

$ R\left( {\begin{array}{*{20}{l}} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array}} \right) \le R(\mathit{\boldsymbol{N}}) + R(\mathit{\boldsymbol{C}}) $ (3)

$ u \le r + s{\rm{ 或 }}u - r \le s $ (4)

式(4)表明,C的行数大于等于N的秩亏数。

1.3 定理及证明

设存在四分块矩阵$\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right] $Nu×u非负定矩阵,R(N)=r < uCs×u矩阵。当满足上述存在性条件1)和2)时,凯利逆矩阵可表示为:

$ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^{ - 1}} = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^{ - 1} - \mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{\mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}}\\ {\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{{\mathit{\boldsymbol{I}}_s} - \mathit{\boldsymbol{N'}}_{cc}^{ - 1}} \end{array}} \right] \end{array} $ (5)

本文将通过广义逆矩阵法和矩阵变换法2种方法验证该表达式。

1.3.1 广义逆矩阵法

Nu×u非负定矩阵,Cs×u矩阵,N'bb=N+CTC,可得:

$ {\mathit{\boldsymbol{N'}}_{bb}} = \mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}} = \left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{I}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array}} \right] $ (6)

式中,[I CT]为u×(u+s)行满秩矩阵,$ \left[ \begin{array}{l} \mathit{\boldsymbol{N}}\\ \mathit{\boldsymbol{C}} \end{array} \right]$为(u+su列满秩矩阵。

根据矩阵分解性质[15],存在u×u满秩矩阵A,使得行满秩矩阵:

$ \left[ \begin{array}{*{35}{l}} \mathit{\boldsymbol{I}} & {{\mathit{\boldsymbol{C}}}^{\text{T}}} \\ \end{array} \right]=\underset{u\times u}{\mathop{\mathit{\boldsymbol{A}}}}\,\left[ \begin{array}{*{35}{l}} \underset{u\times u}{\mathop{\mathit{\boldsymbol{I}}}}\, & \underset{u\times s}{\mathop{0}}\, \\ \end{array} \right] $ (7)

式中,$ {\mathop {\mathit{\boldsymbol{I}}}\limits_{u \times u}} $u单位矩阵,$ \mathop 0\limits_{u \times s} $u×s零元素矩阵。

同理,存在u×u满秩矩阵B,使得列满秩矩阵:

$ \left[ \begin{matrix} \mathit{\boldsymbol{N}} \\ \mathit{\boldsymbol{C}} \\ \end{matrix} \right]=\left[ \begin{matrix} \underset{u\times u}{\mathop{\mathit{\boldsymbol{I}}}}\, \\ \underset{s\times u}{\mathop{0}}\, \\ \end{matrix} \right]\underset{u\times u}{\mathop{\mathit{\boldsymbol{B}}}}\, $ (8)

所以,

$ \begin{align} & {{{\mathit{\boldsymbol{{N}'}}}}_{bb}}=\mathit{\boldsymbol{N}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}\mathit{\boldsymbol{C}}=\left[ \begin{array}{*{35}{l}} \mathit{\boldsymbol{I}} & {{\mathit{\boldsymbol{C}}}^{\text{T}}} \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} \mathit{\boldsymbol{N}} \\ \mathit{\boldsymbol{C}} \\ \end{array} \right]= \\ & \underset{u\times u}{\mathop{\mathit{\boldsymbol{A}}}}\,\left[ \begin{array}{*{35}{l}} \underset{u\times u}{\mathop{\mathit{\boldsymbol{I}}}}\, & \underset{u\times s}{\mathop{0}}\, \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} \underset{u\times u}{\mathop{\mathit{\boldsymbol{I}}}}\, \\ \underset{s\times u}{\mathop{0}}\, \\ \end{array} \right]\underset{u\times u}{\mathop{\mathit{\boldsymbol{B}}}}\,=\underset{u\times u}{\mathop{\mathit{\boldsymbol{A}}}}\,\underset{u\times u}{\mathop{\mathit{\boldsymbol{B}}}}\, \\ \end{align} $ (9)

由于AB均为u×u满秩矩阵,N'bb必满秩可逆,且

$ R\left( {{{\mathit{\boldsymbol{N'}}}_w}} \right) = R\left( {\mathit{\boldsymbol{N'}}_{bb}^{ - 1}} \right) = u $ (10)

上述推算过程和结果表明,s×u行满秩矩阵和u×s列满秩矩阵相乘后可得s×s满秩矩阵。

根据矩阵秩基本性质[16]可知,任意矩阵左乘列满秩矩阵和右乘行满秩矩阵,该矩阵的秩不变。因此,s×u矩阵C右乘u×u行满秩矩阵N'bb-1得到的s×u矩阵CN'bb-1和矩阵C的秩均为s,所以CN'bb-1s×u行满秩矩阵。

s×u行满秩矩阵CN'bb-1u×s列满秩矩阵CT相乘得到s×s满秩矩阵CN'bb-1CT,因此,N'cc=CN'bb-1CT必满秩可逆。

根据广义逆的定义,凯利逆为广义逆的特殊逆,因此凯利逆N'bb-1N'cc-1必满足引理1,将之代入式(1)可得:

$ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^{ - 1}} = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^ - - \mathit{\boldsymbol{N'}}_{bb}^ - {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^ - \mathit{\boldsymbol{CN'}}_{bb}^ - }&{\mathit{\boldsymbol{N'}}_{bb}^ - {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^ - }\\ {\mathit{\boldsymbol{N'}}_{cc}^ - \mathit{\boldsymbol{CN'}}_{bb}^ - }&{\mathit{\boldsymbol{N'}}_{cc}^ - {{\mathit{\boldsymbol{N'}}}_{cc}} - \mathit{\boldsymbol{N'}}_{cc}^ - } \end{array}} \right] = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^{ - 1} - \mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{\mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}}\\ {\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{{\mathit{\boldsymbol{I}}_s} - \mathit{\boldsymbol{N'}}_{cc}^{ - 1}} \end{array}} \right] \end{array} $ (11)

得证。

1.3.2 矩阵变换法

$ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{Q}}_{11}}}&{{\mathit{\boldsymbol{Q}}_{12}}}\\ {{\mathit{\boldsymbol{Q}}_{21}}}&{{\mathit{\boldsymbol{Q}}_{22}}} \end{array}} \right]\\ {\rm{或}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{{\bf{Q}}_{11}}}&{{\mathit{\boldsymbol{Q}}_{12}}}\\ {{\mathit{\boldsymbol{Q}}_{21}}}&{{\mathit{\boldsymbol{Q}}_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{I}}_u}}&{\mathop 0\limits_{u \times s} }\\ {\mathop 0\limits_{s \times u} }&{{\mathit{\boldsymbol{I}}_s}} \end{array}} \right] \end{array} $ (12)

则:

$ \mathit{\boldsymbol{N}}{{\mathit{\boldsymbol{Q}}}_{11}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}{{\mathit{\boldsymbol{Q}}}_{21}}={{\mathit{\boldsymbol{I}}}_{u}} $ (13)
$ \mathit{\boldsymbol{N}}{{\mathit{\boldsymbol{Q}}}_{12}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}{{\mathit{\boldsymbol{Q}}}_{22}}=\underset{u\times s}{\mathop{0}}\, $ (14)
$ \mathit{\boldsymbol{C}}{{\mathit{\boldsymbol{Q}}}_{11}}=\underset{s\times u}{\mathop{0}}\, $ (15)
$ \mathit{\boldsymbol{C}}{{\mathit{\boldsymbol{Q}}}_{12}}={{\mathit{\boldsymbol{I}}}_{s}} $ (16)

在式(15)左乘CT与式(13)相加可得:

$ \left( \mathit{\boldsymbol{N}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}\mathit{\boldsymbol{C}} \right){{\mathit{\boldsymbol{Q}}}_{11}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}{{\mathit{\boldsymbol{Q}}}_{21}}={{\mathit{\boldsymbol{I}}}_{u}} $ (17)

同理,将式(16)左乘CT再与式(14)相加可得:

$ \left( \mathit{\boldsymbol{N}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}\mathit{\boldsymbol{C}} \right){{\mathit{\boldsymbol{Q}}}_{12}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}\left( {{\mathit{\boldsymbol{Q}}}_{22}}-{{\mathit{\boldsymbol{I}}}_{s}} \right)=\underset{u\times s}{\mathop{0}}\, $ (18)

联立式(15)~式(18),用矩阵形式可表示为:

$ \left[ \begin{matrix} \mathit{\boldsymbol{N}}+{{\mathit{\boldsymbol{C}}}^{\text{T}}}\mathit{\boldsymbol{C}} & {{\mathit{\boldsymbol{C}}}^{\text{T}}} \\ \mathit{\boldsymbol{C}} & 0 \\ \end{matrix} \right]\left[ \begin{array}{*{35}{c}} {{\mathit{\boldsymbol{Q}}}_{11}} & {{\mathit{\boldsymbol{Q}}}_{12}} \\ {{\mathit{\boldsymbol{Q}}}_{21}} & {{\mathit{\boldsymbol{Q}}}_{22}}-{{\mathit{\boldsymbol{I}}}_{s}} \\ \end{array} \right]=\left[ \begin{matrix} {{\mathit{\boldsymbol{I}}}_{u}} & \underset{u\times s}{\mathop{0}}\, \\ \underset{s\times u}{\mathop{0}}\, & {{\mathit{\boldsymbol{I}}}_{s}} \\ \end{matrix} \right] $ (19)

式(19)四分块矩阵$\left[\begin{array}{cc} \mathit{\boldsymbol{N}}+\mathit{\boldsymbol{C}}^{\mathrm{T}} \mathit{\boldsymbol{C}} & \mathit{\boldsymbol{C}}^{\mathrm{T}} \\ \mathit{\boldsymbol{C}} & 0 \end{array}\right] $C为行满秩矩阵,N+CTC为满秩矩阵,因此其凯利逆矩阵存在。根据引理2及式(2)可得:

$ \begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{Q}}_{11}}}&{{\mathit{\boldsymbol{Q}}_{12}}}\\ {{\mathit{\boldsymbol{Q}}_{21}}}&{{\mathit{\boldsymbol{Q}}_{22}} - {\mathit{\boldsymbol{I}}_s}} \end{array}} \right] = {\left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^{ - 1}} = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^{ - 1} - \mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{\mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}}\\ {\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{{\mathit{\boldsymbol{I}}_s} - \mathit{\boldsymbol{N'}}_{cc}^{ - 1}} \end{array}} \right] \end{array} $ (20)

可得:

$ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{l}} {{\mathit{\boldsymbol{Q}}_{11}}}&{{\mathit{\boldsymbol{Q}}_{12}}}\\ {{\mathit{\boldsymbol{Q}}_{21}}}&{{\mathit{\boldsymbol{Q}}_{22}}} \end{array}} \right] = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^{ - 1} - \mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{\mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}}\\ {\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{{\mathit{\boldsymbol{I}}_s} - \mathit{\boldsymbol{N'}}_{cc}^{ - 1}} \end{array}} \right] \end{array} $ (21)

得证。

2 约束秩亏间接平差模型参数的应用 2.1 约束秩亏间接平差模型的参数估计

约束秩亏间接平差模型可表示为:

$ \underset{n\times 1}{\mathop{\mathit{\boldsymbol{V}}}}\,=\underset{n\times u}{\mathop{\mathit{\boldsymbol{B}}}}\,\underset{u\times 1}{\mathop{{\mathit{\boldsymbol{\hat{X}}}}}}\,-\underset{n\times 1}{\mathop{\mathit{\boldsymbol{L}}}}\, $ (22)
$ \underset{s\times u}{\mathop{\mathit{\boldsymbol{C}}}}\,\underset{u\times 1}{\mathop{{\mathit{\boldsymbol{\hat{X}}}}}}\,=\underset{s\times 1}{\mathop{{{\mathit{\boldsymbol{W}}}_{X}}}}\, $ (23)

式(22)为秩亏间接平差模型的误差方程,Vn维观测误差向量,E(V)=0,D(V)=σ2P-1B为误差方程的n×u系数矩阵,R(B)=r < uu为未知参数个数,${\mathit{\boldsymbol{\hat X}}} $u维参数向量,L为误差方程的n维常数向量。式(23)为约束方程,C为约束方程的s×u系数矩阵,R(C)=su-rWX为约束方程的s维常数向量。

上述参数估计为约束条件下的极值求解问题,按照拉格朗日乘数法构造函数:

$ {\varphi = {\mathit{\boldsymbol{V}}^{\rm{T}}}\mathit{\boldsymbol{PV}} + 2{\mathit{\boldsymbol{K}}^{\rm{T}}}\left( {\mathit{\boldsymbol{C\hat X}} - {\mathit{\boldsymbol{W}}_X}} \right)} $ (24)

式中,K为对应于约束方程的s×1常数向量。解算该函数的极小值,只需令$\frac{\partial \varphi}{\partial \hat{\boldsymbol{X}}}=0, \frac{\partial \varphi}{\partial \boldsymbol{K}}=0 $,得:

$ {{\mathit{\boldsymbol{B}}^{\rm{T}}}\mathit{\boldsymbol{PB\hat X}} - {\mathit{\boldsymbol{B}}^{\rm{T}}}\mathit{\boldsymbol{PL}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{K}} = 0} $ (25)
$ {\mathit{\boldsymbol{C\hat X}} - {\mathit{\boldsymbol{W}}_X} = 0} $ (26)

$\boldsymbol{N}=\boldsymbol{B}^{\mathrm{T}} \boldsymbol{P} \boldsymbol{B}, \boldsymbol{W}=\boldsymbol{B}^{\mathrm{T}} \boldsymbol{P} \boldsymbol{L} $,用矩阵形式可表示为:

$ {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {\mathit{\boldsymbol{\hat X}}}\\ \mathit{\boldsymbol{K}} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{W}}\\ {{\mathit{\boldsymbol{W}}_X}} \end{array}} \right] = 0} $ (27)

则:

$ {\left[ {\begin{array}{*{20}{l}} {\mathit{\boldsymbol{\hat X}}}\\ \mathit{\boldsymbol{K}} \end{array}} \right] = {{\left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]}^{ - 1}}\left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{W}}\\ {{\mathit{\boldsymbol{W}}_X}} \end{array}} \right]} $ (28)

式中,Nu×u阶秩亏矩阵,R(N)=R(B)=t < u,四分块矩阵无法应用通常的用分块矩阵求凯利逆的方法求解逆矩阵,但在满足存在条件1)和2)的情况下,采用本文推导的该类特殊四分块矩阵求解逆矩阵可得:

$ \begin{array}{l} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {\begin{array}{*{20}{l}} {\mathit{\boldsymbol{\hat X}}}\\ \mathit{\boldsymbol{K}} \end{array}} \right] = {\left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{N}}&{{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ \mathit{\boldsymbol{C}}&0 \end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{W}}\\ {{\mathit{\boldsymbol{W}}_X}} \end{array}} \right] = \\ \left[ {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{N'}}_{bb}^{ - 1} - \mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{\mathit{\boldsymbol{N'}}_{bb}^{ - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N'}}_{cc}^{ - 1}}\\ {\mathit{\boldsymbol{N'}}_{cc}^{ - 1}\mathit{\boldsymbol{CN'}}_{bb}^{ - 1}}&{{\mathit{\boldsymbol{I}}_s} - \mathit{\boldsymbol{N'}}_{cc}^{ - 1}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} \mathit{\boldsymbol{W}}\\ {{\mathit{\boldsymbol{W}}_X}} \end{array}} \right] \end{array} $ (29)

展开得:

$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{X}} = \left( {{\mathit{\boldsymbol{N}}^{\prime - 1}} - \mathit{\boldsymbol{N}}_{bb}^{\prime - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{N}}_{cc}^{\prime - 1}\mathit{\boldsymbol{C}}{\mathit{\boldsymbol{N}}^{\prime - 1}}} \right)\mathit{\boldsymbol{W}} + }\\ {\mathit{\boldsymbol{N}}_{bb}^{\prime - 1}{\mathit{\boldsymbol{C}}^{\rm{T}}}{\mathit{\boldsymbol{N}}^{\prime - 1}}{\mathit{\boldsymbol{W}}_X} = \{ {{(\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}})}^{ - 1}} - }\\ {{{\left( {\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}}} \right)}^{ - 1}}{\mathit{\boldsymbol{C}}^{\rm{T}}}{{\left[ {\mathit{\boldsymbol{C}}\left( {\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}}} \right){\mathit{\boldsymbol{C}}^{\rm{T}}}} \right]}^{ - 1}}}\\ {\left. {\mathit{\boldsymbol{C}}{{\left( {\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}}} \right)}^{ - 1}}} \right\}{\mathit{\boldsymbol{B}}^{\rm{T}}}\mathit{\boldsymbol{PL}} + {{\left( {\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}}} \right)}^{ - 1}}{\mathit{\boldsymbol{C}}^{\rm{T}}}}\\ {{{\left[ {\mathit{\boldsymbol{C}}\left( {\mathit{\boldsymbol{N}} + {\mathit{\boldsymbol{C}}^{\rm{T}}}\mathit{\boldsymbol{C}}} \right){\mathit{\boldsymbol{C}}^{\rm{T}}}} \right]}^{ - 1}}{\mathit{\boldsymbol{W}}_X}} \end{array} $ (30)

式(30)即为约束秩亏间接平差模型参数估计的一般显性表达式。

2.2 数值算例

图 1所示的水准网中,已知水准点AB的高程分别为HA=237.483 m、HB=233.868 m,为求P1P2点的高程,进行水准测量,表 1为测量高差和水准路线。分别采用约束秩亏间接平差模型和间接平差模型解算待求点P1P2的高程,并对结果进行比较。

图 1 水准网 Fig. 1 Leveling network

表 1 观测数据 Tab. 1 Observation data
2.2.1 约束秩亏间接平差模型解算

ABP1P2点的高程为未知参数$ \hat{x}_{1}$$\hat{x}_{2} $$ \hat{x}_{3}$$ \hat{x}_{4}$,将所有观测值的改正数表示为未知参数的误差方程:

$ \begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{v_1}}\\ {{v_2}}\\ {{v_3}}\\ {{v_4}}\\ {{v_5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&0&1&0\\ 0&0&{ - 1}&1\\ { - 1}&0&0&1\\ 0&{ - 1}&1&0\\ 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\hat x}_1}}\\ {{{\hat x}_2}}\\ {{{\hat x}_3}}\\ {{{\hat x}_4}} \end{array}} \right] - \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {\begin{array}{*{20}{r}} {3.782}\\ { - 9.640}\\ { - 5.835}\\ {7.384}\\ { - 2.270} \end{array}} \right] \Leftrightarrow \mathit{\boldsymbol{V}} = \mathit{\boldsymbol{B\hat X}} - \mathit{\boldsymbol{L}} \end{array} $

式中,未知参数的系数矩阵B为列秩亏,符号表示等价的矩阵形式。

AB为已知点,可以组成限制条件方程:

$ \begin{array}{*{20}{l}} {\left[ {\begin{array}{*{20}{l}} 1&0&0&0\\ 0&1&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{{\hat x}_1}}\\ {{{\hat x}_2}}\\ {{{\hat x}_3}}\\ {{{\hat x}_4}} \end{array}} \right] = }\\ {\left[ {\begin{array}{*{20}{l}} {237.483}\\ {233.868} \end{array}} \right] \Leftrightarrow \mathit{\boldsymbol{C\hat X}} = {\mathit{\boldsymbol{W}}_X}} \end{array} $

按观测距离定义观测值的权重,并令1 km的观测高差为单位权,则观测值的权阵为:

$ \mathit{\boldsymbol{P}} = {\mathop{\rm diag}\nolimits} \left[ {\begin{array}{*{20}{l}} {1/2.0}&{1/1.0}&{1/2.0}&{1/2.0}&{1/2.0} \end{array}} \right] $

通过分析可知,上述误差方程式系数矩阵B秩亏数为1,限制条件方程系数C为行满秩,条件方程个数为2,大于秩亏数,满足存在性条件。

将已知参数BPLCWX代入本文推导的约束秩亏间接平差模型参数估计公式(30),计算可得:

$ \begin{array}{*{20}{l}} {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{\left[ {\begin{array}{*{20}{l}} {{{\hat x}_1}}&{{{\hat x}_2}}&{{{\hat x}_3}}&{{{\hat x}_4}} \end{array}} \right]}^{\rm{T}}} = }\\ {{{\left[ {\begin{array}{*{20}{l}} {237.483}&{233.868}&{241.260}&{231.622} \end{array}} \right]}^{\rm{T}}}} \end{array} $

式中,$ \hat{x}_{1}$$\hat{x}_{2} $为已知点AB的高程,$ \hat{x}_{3}$$\hat{x}_{4} $为待求点的高程。

2.2.2 间接平差模型解算

考虑AB为已知点,设P1P2点的高程为未知参数$ \hat{x}_{3}$$\hat{x}_{4} $,将所有观测值的改正数表示为未知参数的误差方程:

$ \begin{array}{c} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {\begin{array}{*{20}{c}} {{v_1}}\\ {{v_2}}\\ {{v_3}}\\ {{v_4}}\\ {{v_5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ { - 1}&1\\ 0&1\\ 1&0\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\hat x}_3}}\\ {{{\hat x}_4}} \end{array}} \right] - \\ \left[ {\begin{array}{*{20}{c}} {241.265}\\ { - 9.640}\\ {231.648}\\ {241.252}\\ {231.598} \end{array}} \right] \Leftrightarrow \mathit{\boldsymbol{V}} = \mathit{\boldsymbol{B\hat X}} - \mathit{\boldsymbol{L}} \end{array} $

同理,按观测距离定义观测值的权重。

根据最小二乘原则,将已知参数BPL代入,解算得:

$ \left[ {\begin{array}{*{20}{l}} {{{\hat x}_3}}\\ {{{\hat x}_4}} \end{array}} \right] = {\left( {{\mathit{\boldsymbol{B}}^{\rm{T}}}\mathit{\boldsymbol{PB}}} \right)^{ - 1}}{\mathit{\boldsymbol{B}}^{\rm{T}}}\mathit{\boldsymbol{PL}} = \left[ {\begin{array}{*{20}{l}} {241.260}\\ {231.622} \end{array}} \right] $

由上述计算过程和结果可知,约束秩亏间接平差模型的分块矩阵求逆公式和间接平差模型对未知点P1P2的参数估计$ \hat{x}_{3}$$\hat{x}_{4} $的结果完全一致,证明本文推导的分块矩阵求逆矩阵公式具有可行性。

3 结语

1) 本文对左上角子矩阵秩亏、右下角子矩阵为零的特殊四分块矩阵进行分析,在满足左下角子矩阵为行满秩及行数大于左上角子矩阵秩亏数的条件下,证明其存在凯利逆矩阵,该存在条件可为自由网附加条件法的平差提供理论依据。

2) 在通常的四分块矩阵求广义逆矩阵和凯利逆矩阵公式的基础上,利用广义逆矩阵法和矩阵变换法推导该类特殊四分块矩阵凯利逆矩阵的显性表达式。该公式可用于解算约束秩亏间接平差模型的参数估计,为约束秩亏间接平差模型的解算提供简单准确的方法,具有理论和应用价值。

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Calculation of Rank-Defective Indirect Adjustment Model with Constraint by Inverting Block Matrix
XIE Bo1     LIU Lianwang2     
1. School of Civil Engineering and Architecture, Hefei Technology College, 1 South-Laoshan Road, Hefei 238000, China;
2. Bowen College of Management, Guilin University of Technology, 317 Yanshan Street, Guilin 541006, China
Abstract: Based on the general formula of generalized inverse matrix and Kelly inverse four block matrix, we analyze the existence conditions of a special kind of Kelly inverse four block matrix with a rank defective sub-block matrix in the upper left corner and a zero sub-block matrix in the lower right corner. The explicit expression formula of a Kelly inverse matrix of this kind is derived using the generalized inverting matrix method and the matrix transformation method, and is applied to the parameter estimation of rank-defective indirect adjustment model with constraints. The results show that when existence conditions are satisfied, the solution of rank-defective indirect adjustment model with constraints by inverting block matrix is consistent with that of the indirect adjustment model. This proves that the derived explicit expression formula is feasible.
Key words: survey adjustment; block matrix; generalized inverse matrix; rank-defective indirect adjustment model with constraint