半正定Hermitian矩阵伪Schur补的商公式

楼嫏嬛

引用本文:
Citation:

半正定Hermitian矩阵伪Schur补的商公式

    作者简介: 楼嫏嬛(1978−),女,浙江人,硕士,讲师,主要从事矩阵理论方面的研究. E-mail:loulanghuan@163.com;
  • 中图分类号: O151.21

The quotient formulas on generalized Schur complements of positive semidefinite Hermitian matrices

  • CLC number: O151.21

  • 摘要: 利用半正定Hermitian矩阵及其伪Schur补的性质研究半正定Hermitian矩阵伪Schur补的商性质. 通过分块矩阵的计算与比对,将矩阵Schur补的商公式推广到半正定Hermitian矩阵的伪Schur补上. 并以此为基础,得出半正定Hermitian矩阵伪Schur补的{1}广义逆也满足商公式以及其Moore-Penrose广义逆满足商公式的条件.
  • [1] Schur I. Potenzreihn in inner des heitskreises[J]. Journal fur die Reine und Angewandte Mathematik, 1917, 147: 205-232.
    [2] Ben-Israel A, Greville T N E. Generalized inverse: theory and application[M]. 2nd ed. New York: Springer-verlag New York Inc, 2003.
    [3] Wang G R, Wei Y M, Qiao S Z. Generalized inverse: theory and computations[M]. Beijing: Science Press, 2004.
    [4] Zhang F Z. The Schur complement and its applications[M]. New York: Springer-verlag New York Inc, 2005.
    [5] Crabtree D, Haynsworth E V. An identity for the Schur complement of a matrix[J]. Proceedings of the American Mathematical Society, 1969, 22: 364-366. DOI:  10.1090/S0002-9939-1969-0255573-1.
    [6] Wang B Y, Zhang F Z. Schur complements and matrix inequalities of Hadamard products[J]. Linear and Multilinear Algebra, 1997, 43(1~3): 315-326.
    [7] Liu J Z. Some löwner partial orders of Schur complements and Kronecker products of matrices[J]. Linear Algebra and its Applications, 1999, 291(1): 143-149.
    [8] Baksalary J K, Baksalary O M, Szulc T. Properties of Schur complements in partitioned idempotent matrices[J]. Linear Algebra and its Applications, 2004, 379: 303-318. DOI:  10.1016/S0024-3795(03)00546-9.
    [9] Zhou J H, Wang G R. Block idempotent matrices and generalized Schur complement[J]. Journal of Applied Mathematics, 2007, 188: 246-256.
    [10] 刘喜富. 广义Schur补为零时分块矩阵的Drazin逆[J]. 华东交通大学学报, 2014, 31(1): 98-101. DOI:  10.3969/j.issn.1005-0523.2014.01.018. Liu X F. The Drazin inverses of block matrices with zero generalized Schur complement[J]. Journal of East China Jiaotong University, 2014, 31(1): 98-101.
    [11] 郭美华, 刘丁酉. 分块2次幂零矩阵的广义Schur补[J]. 武汉大学学报: 理学版, 2015, 61(6): 563-567. Guo M H, Liu D Y. Generalized Schur complement on 2-nilpotent partitioned matrix[J]. Journal of Wuhan University: Natural Sciences Edition, 2015, 61(6): 563-567.
    [12] 朱永林. 分块幂等矩阵中广义Schur补的幂等性[J]. 数学的实践与认识, 2014, 44(6): 295-297. Zhu Y L. Idempotency of the generalized Schur complement in partitioned idempotent matrices[J]. Mathematics in Practice and Theory, 2014, 44(6): 295-297.
    [13] Albert A. Conditions for positive and nonnegative definiteness in terms of pseudo inverse[J]. SIAM Journal of Applied Mathematics, 1969, 17(1): 434-440.
    [14] Wang B Y, Zhang X P, Zhang F Z. Some inequalities on generalized Schur complements[J]. Linear Algebra and its Applications, 1999, 302-303(1): 163-172.
    [15] 楼嫏嬛. 半正定矩阵Kronecker积的伪Schur补的几点注记[J]. 云南大学学报: 自然科学版, 2017, 39(5): 719-726. DOI:  10.7540/j.ynu.20160644. Lou L H. Some results on generalized Schur complements for the Kronecker product of positive semidefinite matrices[J]. Journal of Yunnan University: Natural Sciences Edition, 2017, 39(5): 719-726.
    [16] Ouellette D V. Schur complements and statistics[J]. Linear Algebra and its Applications, 1981, 36: 187-295. DOI:  10.1016/0024-3795(81)90232-9.
    [17] Jürgen G. The Moore-Penrose inverse of a partitioned nonnegative definite matrix[J]. Linear Algebra and its Applications, 2000, 321: 113-121. DOI:  10.1016/S0024-3795(99)00073-7.
  • [1] 楼嫏嬛 . 半正定矩阵Kronecker积的伪Schur补的几点注记. 云南大学学报(自然科学版), 2017, 39(5): 719-726. doi: 10.7540/j.ynu.20160644
    [2] 袁晖坪米玲 . 广义行(列)酉对称矩阵的满秩分解及其Moore-Penrose逆. 云南大学学报(自然科学版), 2009, 31(5): 439-443 .
    [3] 王国栋杨洋陆正福 . 二次剩余码的自同构群. 云南大学学报(自然科学版), 2004, 26(2): 93-97.
    [4] 王国栋叶锐陆正福 . 一般有限域GF(pm)上线性码的自同构群. 云南大学学报(自然科学版), 2004, 26(1): 11-14,19.
    [5] 王国栋罗裕梅 . GF(2m)上线性码的自同构群的进一步研究. 云南大学学报(自然科学版), 2002, 24(2): 85-87.
    [6] 李耀堂刘庆兵 . 非Hermite正定矩阵与正稳定矩阵. 云南大学学报(自然科学版), 2002, 24(3): 164-166.
    [7] 李艳艳 . Nekrasov矩阵逆的无穷范数改进的估计式. 云南大学学报(自然科学版), 2018, 40(4): 632-637. doi: 10.7540/j.ynu.20170774
    [8] 朱艳李耀堂 . Nekrasov矩阵的逆矩阵的无穷范数新的上界估计式. 云南大学学报(自然科学版), 2017, 39(1): 13-17. doi: 10.7540/j.ynu.20160382
    [9] 袁晖坪 . 关于行(列)反对称矩阵的Schur分解. 云南大学学报(自然科学版), 2008, 30(6): 549-552.
    [10] 孙丽英许兴业 . H-矩阵的刻化及一类实矩阵逆的上下界估计. 云南大学学报(自然科学版), 2005, 27(4): 285-288.
    [11] 许洁赵微孙玉祥 . 广义对角占优矩阵的实用新判定. 云南大学学报(自然科学版), 2014, 36(5): 637-641. doi: 10.7540/j.ynu.20130580
    [12] 赵仁庆熊昌明李耀堂 . 块H-矩阵的判定及其逆的无穷大范数的上界. 云南大学学报(自然科学版), 2011, 33(2): 125-130, .
    [13] 周平李耀堂 . 弱链对角占优B-矩阵线性补问题解的误差界估计式的改进. 云南大学学报(自然科学版), 2020, 42(): 1-8. doi: 10.7540/j.ynu.20190656
    [14] 李映辉张荣华 . LR-逆半群的半直积. 云南大学学报(自然科学版), 2006, 28(3): 188-191.
    [15] 张文专石磊 . 数据矩阵条件指数的影响评价. 云南大学学报(自然科学版), 2004, 26(4): 292-296,300.
    [16] 李甦尹泽华王鹏 . 基于压缩感知的测量矩阵研究. 云南大学学报(自然科学版), 2015, 37(S1): 22-. doi: 10.7540/j.ynu.20150115a
    [17] 张海岛贺西平王维鸽 . 夹角型超声变幅杆的矩阵解析法. 云南大学学报(自然科学版), 2016, 38(2): 225-231. doi: 10.7540/j.ynu.20150576
    [18] 袁晖坪 . 行(列)反对称矩阵的奇异值分解. 云南大学学报(自然科学版), 2007, 29(5): 449-452,458.
    [19] 周平李耀堂 . M-矩阵及非负矩阵Hadamard积和Fan积的特征值界的估计. 云南大学学报(自然科学版), 2012, 34(1): 9-14,25.
    [20] 伍俊良陈香萍邹黎敏 . 2个四元数正规矩阵的同时对角化问题. 云南大学学报(自然科学版), 2009, 31(3): 222-226 .
  • 加载中
计量
  • 文章访问数:  564
  • HTML全文浏览量:  315
  • PDF下载量:  22
  • 被引次数: 0
出版历程
  • 收稿日期:  2019-11-05
  • 录用日期:  2020-01-19
  • 网络出版日期:  2020-04-11
  • 刊出日期:  2020-05-01

半正定Hermitian矩阵伪Schur补的商公式

    作者简介:楼嫏嬛(1978−),女,浙江人,硕士,讲师,主要从事矩阵理论方面的研究. E-mail:loulanghuan@163.com
  • 西安邮电大学 理学院,陕西 西安 710121

摘要: 利用半正定Hermitian矩阵及其伪Schur补的性质研究半正定Hermitian矩阵伪Schur补的商性质. 通过分块矩阵的计算与比对,将矩阵Schur补的商公式推广到半正定Hermitian矩阵的伪Schur补上. 并以此为基础,得出半正定Hermitian矩阵伪Schur补的{1}广义逆也满足商公式以及其Moore-Penrose广义逆满足商公式的条件.

English Abstract

  • 在有关矩阵理论和计算的研究中,Schur补理论占有重要地位. Schur补概念最早是由Schur在1917年提出的[1],它的提出大大推动了数学领域的发展,尤其是作为处理大规模矩阵计算的有效工具,被广泛应用于矩阵理论、统计学、应用数学等领域[2-3],同时Schur补在导出矩阵、行列式、迹、范数、特征值、奇异值不等式和控制不等式的过程中也扮演着非常重要的角色[4]. 国内外许多学者都致力于矩阵Schur补的研究工作,Haynsworth利用Schur补,得到了著名的Haynsworth不等式和Schur补的商公式[5];王伯英等得出了一些有关正定矩阵Hadamard积的Schur补的矩阵不等式[6];刘建洲研究了非奇异矩阵Kronecker积的Schur补问题等[7].

    将广义逆理论与Schur补概念结合可以衍生出伪Schur补这一新的概念,这种伪Schur补经常出现在系数矩阵为分块矩阵的线性方程组最小二乘解问题中,并在矩阵理论、统计分析、数值计算、区域分解方法、线性控制等问题的研究中也有大量应用,是现代许多研究领域不可缺少的工具. 关于矩阵伪Schur补的研究是近年来矩阵研究的重要问题之一,同样取得了一些重要结果[8-10].

    由于分块矩阵在许多矩阵问题的研究中都起到了非常重要的作用,所以分块矩阵的伪Schur补一直是一个重要的研究课题. 文献[11]研究了分块2次幂零矩阵的伪Schur补仍为2次幂零矩阵的问题;文献[12]讨论了分块幂等矩阵的伪Schur补的幂等性. 本文以分块矩阵为主要工具,讨论半正定Hermitian矩阵的伪Schur补与其主子矩阵的伪Schur补之间的关系,将Schur补的商公式推广到半正定Hermitian矩阵的伪Schur补上,并以此为基础,得出半正定Hermitian矩阵伪Schur补的{1}广义逆也满足商公式以及其Moore-Penrose广义逆满足商公式的条件. 同时也得到了有关半正定Hermitian矩阵伪Schur补的一些偏序结果.

    文中用 ${{\bf{C}}^{m \times n}}$ 表示所有 $m \times n$ 复矩阵构成的集合,${{{H}}_{{n}}}$ 表示所有 $n \times n$ 阶Hermitian矩阵之集,${{H}}_{{n}}^ \geqslant $ 表示所有半正定矩阵之集. 若 ${{A}} \in {{H}}_{{n}}^ \geqslant $,记 $A \geqslant O$;若 ${{A}} - {{B}} \in {{H}}_{{n}}^ \geqslant $,记 $A \geqslant B$. 对任意的 ${{A}} \in {{\bf{C}} ^{m \times n}}$${A^ * }$ 表示矩阵 $A$ 的共轭转置矩阵,$r(A)$ 表示矩阵 $A$ 的秩. ${I_n}$ 表示 $n \times n$ 阶单位矩阵.${e_i}$ 表示第 $i$ 个分量等于1的 $n$ 维单位向量,${\tilde e_i}$ 表示第 $i$ 个分量等于1的 $r$ 维单位向量.

    • ${{A}} \in {{\bf{C}} ^{m \times n}}$,对任意的 ${{X}} \in {{\bf{C}} ^{n \times m}}$,如果满足 $(1)\;\;AXA = A$$(2)\;\;XAX = X$$(3){\rm{ (}}AX{)^ * } = AX$$(4){\rm{ (}}XA{)^ * } = XA$ 中的一个或几个方程,都称为 $A$ 的广义逆矩阵. 若 $X$ 只满足第一个方程,则称 $X$$A$ 的{1}广义逆,记为 ${A^ - }$;若 $X$ 满足全部四个方程,则称 $X$$A$ 的Moore-Penrose广义逆,记为 ${A^ + }$.

      设分块矩阵 $A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$,则称 ${A_{22}} - {A_{21}}A_{11}^ + {A_{12}}$ 为矩阵 ${A_{11}}$$A$ 中的伪Schur补,记作 $A/{A_{11}}$. 显然,若 ${A_{11}}$ 为方阵且非奇异,则 $A_{11}^ + = A_{11}^{ - 1}$,此时伪Schur补即为一般意义下的Schur补.

      事实上,可以定义矩阵 $A$ 的任意子矩阵在 $A$ 中的伪Schur补. 比如:$A/{A_{22}} = {A_{11}} - {A_{12}}A_{22}^ + {A_{21}}$. 若 ${A_{11}}$$A$ 分别为 $r$ 阶,$n$ 阶方阵,取置换矩阵 ${{P = }}({{{e}}_{{{r}} + 1}},{{{e}}_{{{r}} + 2}}, \cdots ,{{{e}}_{{{r}} + ({{n}} - {{r}})}},{{{e}}_1}, \cdots ,{{{e}}_{{r}}}) \in {{\bf{C}} ^{n \times n}}$,由置换矩阵的性质得 ${P^ * }AP = \left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right)$,于是 $({P^ * }AP)/{A_{22}} = {A_{11}} - {A_{12}}A_{22}^ + {A_{21}} = A/{A_{22}}$. 这表明矩阵 $A$ 的任意 $r$ 阶主子矩阵的伪Schur补均可以通过置换矩阵 $P$ 化为 ${P^ * }AP$$r$ 阶顺序主子矩阵的伪Schur补.

      有关Schur补的许多结论可以推广到伪Schur补上. 但是由于在Moore-Penrose广义逆下,${A^ + }A \leqslant I$,所以伪Schur补又不仅仅是Schur补的简单推广.

      文献[13]指出若分块矩阵 ${{A = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}} \end{array}} \right) \in {{{H}}_{{n}}}$,其中 ${A_{11}}$${A_{22}}$ 均为方阵,则

      $\qquad A \geqslant O \Leftrightarrow {A_{11}} \geqslant O,\;\;{A_{12}} = {A_{11}}A_{11}^ + {A_{12}},\;\;A/{A_{11}} \geqslant O.$

      类似地有

      $\qquad A \geqslant O \Leftrightarrow {A_{22}} \geqslant O,\;\;{A_{12}} = {A_{12}}A_{22}^ + {A_{22}},\;\;A/{A_{22}} \geqslant O.$

      利用(1)式不难证明如下结论:

      ${{A = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}} \\ {{{{A}}_{21}}}&{{{{A}}_{22}}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant $${A_{11}}$$A$$r$ 阶主子矩阵,$U = \left( {\begin{array}{*{20}{c}} {{I_r}}&O \\ Y&{{I_{n - r}}} \end{array}} \right)$,其中 $Y$ 为任意的 $(n - r) \times r$ 矩阵,则

      $\qquad (UA{U^ * })/{A_{11}} = A/{A_{11}}.$

      类似地,设 ${{A = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}} \\ {{{{A}}_{21}}}&{{{{A}}_{22}}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant $${A_{22}}$$A$$r$ 阶主子矩阵,$U = \left( {\begin{array}{*{20}{c}} {{I_{n - r}}}&Y \\ O&{{I_r}} \end{array}} \right)$,其中 $Y$ 为任意的 $(n - r) \times r$ 矩阵,则

      $\qquad (UA{U^ * })/{A_{22}} = A/{A_{22}}.$

      ${{A = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant $,则它的 $n$ 个特征值均非负,记其特征值按如下顺序排列:${\lambda _1}(A) \geqslant {\lambda _2}(A) \geqslant \cdots \geqslant$$ {\lambda _n}(A)$. 文献[14]指出对 $A$$r$ 阶主子矩阵 ${A_{11}}$,有

      $\qquad {\lambda _i}(A) \geqslant {\lambda _i}({A_{22}}) \geqslant {\lambda _i}(A/{A_{11}}) \geqslant {\lambda _{i + r}}(A),\;\;\;\;\;\;i = 1,2, \cdots ,n - r. $

    • 引理1 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}} \\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}} \\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant $$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$M = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{B_1}} \\ {{C_1}}&D \end{array}} \right)$$H$$n - r + {r_1}$ 阶主子矩阵,则 $A/{A_{11}}$$M/{A_{11}}$ 都是 $H/{A_{11}}$ 的主子矩阵.

      证明 记 $C = ({C_1}\;\;\;{C_2})$$B = \left( \begin{array}{l} {B_1} \\ {B_2} \\ \end{array} \right)$,因为 $H$ 是半正定Hermitian矩阵,所以 $B = {C^ * }$. 那么

      $\qquad H/{A_{11}} = \left( {\begin{array}{*{20}{c}} {{A_{22}}}&{C_2^ * } \\ {{C_2}}&D \end{array}} \right) - \left( \begin{array}{l} {A_{21}} \\ {C_1} \\ \end{array} \right)A_{11}^ + ({A_{12}}\;\;C_1^ * )= \left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&{C_2^ * - {A_{21}}A_{11}^ + C_1^ * } \\ {{C_2} - {C_1}A_{11}^ + {A_{12}}}&{M/{A_{11}}} \end{array}} \right). $

      (6)式表明 $A/{A_{11}}$$M/{A_{11}}$ 都是 $H/{A_{11}}$ 的主子矩阵. 证毕.

      若记 $N = \left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{B_2}} \\ {{C_2}}&D \end{array}} \right)$,同理可证 $A/{A_{22}}$$N/{A_{22}}$ 都是 $H/{A_{22}}$ 的主子矩阵;$M/D$$N/D$ 都是 $H/D$ 的主子矩阵.

      引理2[15] 设 ${{A}} \in {{\bf{C}} ^{n \times n}}$${{U}},{{V}} \in {{\bf{C}} ^{n \times n}}$ 是酉矩阵,则 ${(UAV)^ + } = {V^ * }{A^ + }{U^ * }$.

      定理1 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,则

      $\qquad (H/{A_{11}})/(A/{A_{11}}) = (H/{A_{22}})/(A/{A_{22}}) = H/A.$

      证明 取 ${U_1} = \left( {\begin{array}{*{20}{c}} {{I_r}}&O \\ { - C{A^ + }}&{{I_{n - r}}} \end{array}} \right)$$C = ({C_1}\;\;\;{C_2})$,则依据(1)式有

      $\qquad {U_1}HU_1^ * = \left( {\begin{array}{*{20}{c}} {{I_r}}&O \\ { - C{A^ + }}&{{I_{n - r}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} A&{{C^ * }} \\ C&D \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{I_r}}&{{{( - C{A^ + })}^ * }} \\ O&{{I_{n - r}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} A&O \\ O&{H/A} \end{array}} \right).$

      类似地,取 ${U_2} = \left( {\begin{array}{*{20}{c}} {{I_{{r_1}}}}&O&O \\ { - {A_{21}}A_{11}^ + }&{{I_{r - {r_1}}}}&O \\ O&O&{{I_{n - r}}} \end{array}} \right)$${U_3} = \left( {\begin{array}{*{20}{c}} {{I_{{r_1}}}}&O \\ { - \left( \begin{array}{l} {A_{21}} \\ {C_1} \\ \end{array} \right)A_{11}^ + }&{{I_{n - {r_1}}}} \end{array}} \right)$,且注意到 $\left( \begin{array}{l} {A_{21}} \\ {C_1} \\ \end{array} \right) =$$ \left( \begin{array}{l} {A_{21}} \\ {C_1} \\ \end{array} \right)A_{11}^ + {A_{11}}$,有

      $\qquad {U_2}{U_1}HU_1^ * U_2^ * = {U_2}\left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}&O \\ {{A_{21}}}&{{A_{22}}}&O \\ O&O&{H/A} \end{array}} \right)U_2^ * = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&O&O \\ O&{A/{A_{11}}}&O \\ O&O&{H/A} \end{array}} \right),$

      $\qquad {U_3}HU_3^ * = {U_3}\left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}&{C_1^ * } \\ {{A_{21}}}&{{A_{22}}}&{C_2^ * } \\ {{C_1}}&{{C_2}}&D \end{array}} \right)U_3^ * = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&O \\ O&{H/{A_{11}}} \end{array}} \right).$

      不难发现 ${U_3}$ 可逆,$U_3^{ - 1} = \left( {\begin{array}{*{20}{c}} {{I_{{r_1}}}}&O \\ {\left( \begin{array}{l} {A_{21}} \\ {C_1} \\ \end{array} \right)A_{11}^ + }&{{I_{n - {r_1}}}} \end{array}} \right)$,于是 $H = U_3^{ - 1}\left( {\begin{array}{*{20}{c}} {{A_{11}}}&O \\ O&{H/{A_{11}}} \end{array}} \right){(U_3^{ - 1})^ * }$.

      $U = {U_2}{U_1}U_3^{ - 1}$,经计算得 $U = \left( {\begin{array}{*{20}{c}} {{I_{{r_1}}}}&O&O \\ O&{{I_{r - {r_1}}}}&O \\ Y&{ - {{(C{A^ + })}_2}}&{{I_{n - r}}} \end{array}} \right)$,这里 $Y = - {(C{A^ + })_1} - {(C{A^ + })_2}{A_{21}}A_{11}^ + + {C_1}A_{11}^ + $,而 ${(C{A^ + })_1}$${(C{A^ + })_2}$ 分别表示矩阵 $C{A^ + }$ 的前 ${r_1}$ 列和后 $r - {r_1}$ 列. 将 $U$ 代入(9)式有

      $\qquad U\left( {\begin{array}{*{20}{c}} {{A_{11}}}&O \\ O&{H/{A_{11}}} \end{array}} \right){U^ * } = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&O&O \\ O&{A/{A_{11}}}&O \\ O&O&{H/A} \end{array}} \right).$

      $S = \left( \begin{array}{l} O \\ Y \\ \end{array} \right)$$T = \left( {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&O \\ { - {{(C{A^ + })}_2}}&{{I_{n - r}}} \end{array}} \right)$,于是

      $\qquad U\left( {\begin{array}{*{20}{c}} {{A_{11}}}&O \\ O&{H/{A_{11}}} \end{array}} \right){U^ * } = \left( {\begin{array}{*{20}{c}} {{I_{{r_1}}}}&O \\ S&T \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{A_{11}}}&O \\ O&{H/{A_{11}}} \end{array}} \right){\left( {\begin{array}{*{20}{c}} {{I_{{r_1}}}}&O \\ S&T \end{array}} \right)^ * } = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{11}}{S^ * }} \\ {S{A_{11}}}&{S{A_{11}}{S^ * } + T(H/{A_{11}}){T^ * }} \end{array}} \right).$

      结合(11)式与(12)式可得

      $\qquad S{A_{11}} = O,$

      $\qquad T(H/{A_{11}}){T^ * } = \left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right).$

      由于 $\left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right)/(A/{A_{11}}) = H/A$,利用(3)式得

      $\qquad (H/{A_{11}})/(A/{A_{11}}) = H/A.$

      考虑取置换矩阵 ${{P = }}({{{e}}_{{{{r}}_1} + 1}},{{{e}}_{{{{r}}_1} + 2}}, \cdots ,{{{e}}_{{{{r}}_1} + ({{r}} - {{{r}}_1})}},{{{e}}_1},{{{e}}_2}, \cdots ,{{{e}}_{{{{r}}_1}}},{{{e}}_{{{r}} + 1}}, \cdots ,{{{e}}_{{{r}} + ({{n}} - {{r}})}}) \in {{\bf{C}} ^{n \times n}}$,则 ${P^ * }HP =$$ \left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}}&{{B_2}} \\ {{A_{12}}}&{{A_{11}}}&{{B_1}} \\ {{C_2}}&{{C_1}}&D \end{array}} \right)$. 由上述推导过程可得

      $\qquad [({P^ * }HP)/{A_{22}}]/\Bigg[\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right)/{A_{22}}\Bigg] = ({P^ * }HP)/\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right).$

      其中 $({P^ * }HP)/{A_{22}} = H/{A_{22}}$$\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right)/{A_{22}} = A/{A_{22}}$. 而另一方面,

      $\qquad ({P^ * }HP)/\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right) = D - ({C_2}\;\;\;\;{C_1}){\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right)^ + }{({C_2}\;\;\;\;{C_1})^ * },$

      取置换矩阵 ${{\tilde P = }}({{{\tilde e}}_{{{{r}}_1} + 1}},{{{\tilde e}}_{{{{r}}_1} + 2}}, \cdots ,{{{\tilde e}}_{{{{r}}_1} + ({{r}} - {{{r}}_1})}},{{{\tilde e}}_1},{{{\tilde e}}_2}, \cdots ,{{{\tilde e}}_{{{{r}}_1}}}) \in { {\bf{C}}^{r \times r}}$,则 $({C_2}\;\;\;\;{C_1}){\tilde P^ * } = ({C_1}\;\;\;\;{C_2})$. 由于置换矩阵 $\tilde P$ 是酉矩阵,所以根据引理2有

      $\qquad \begin{array}{*{20}{l}} {({P^ * }HP)/\left( {\begin{array}{*{20}{c}} {{A_{22}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{11}}} \end{array}} \right) = D - ({C_2}\;\;\;\;{C_1}){[{\tilde P^ * }\left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)\tilde P]^ + }{({C_2}\;\;\;\;{C_1})^ * }}=\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \quad \quad D - ({C_2}\;\;\;\;{C_1}){\tilde P^ * }{\left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)^ + }\tilde P{({C_2}\;\;\;\;{C_1})^ * }}=\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \quad \quad D - ({C_1}\;\;\;\;{C_2}){\left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)^ + }{({C_1}\;\;\;\;{C_2})^ * } = H/A.} \end{array}$

      于是(16)式表明

      $\qquad (H/{A_{22}})/(A/{A_{22}}) = H/A.$

      综合(15)式与(17)式得(7)式成立. 证毕.

      同理可证

      $\qquad (H/{A_{11}})/(M/{A_{11}}) = (H/D)/(M/D) = H/M;$

      $\qquad (H/{A_{22}})/(N/{A_{22}}) = (H/D)/(N/D) = H/N.$

      定理1表明半正定Hermitian矩阵 $H$ 的任意 $r$ 阶($r \geqslant 2$)主子矩阵在 $H$ 中的伪Schur补均满足商公式. 从而将一般意义下的Schur补商公式推广至半正定Hermitian矩阵的伪Schur补上,从证明过程可以看出矩阵 $H$ 的半正定性必不可少.

      推论1 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$M = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{B_1}}\\ {{C_1}}&D \end{array}} \right)$$H$$n - r + {r_1}$ 阶主子矩阵,则

      $\qquad {\lambda _i}(H) \geqslant {\lambda _i}(H/{A_{11}}) \geqslant {\lambda _i}(M/{A_{11}}) \geqslant {\lambda _i}(H/A) \geqslant {\lambda _{i + r - {r_1}}}(H/{A_{11}}) \geqslant {\lambda _{i + r}}(H),\;i = 1,2, \cdots ,n - r; $

      $\qquad {\lambda _i}(H) \geqslant {\lambda _i}(H/{A_{22}}) \geqslant {\lambda _i}(N/{A_{22}}) \geqslant {\lambda _i}(H/A) \geqslant {\lambda _{i + {r_1}}}(H/{A_{22}}) \geqslant {\lambda _{i + r}}(H),\;i = 1,2, \cdots ,n - r. $

      证明 由(5)~(7)式可得

      $\qquad {\lambda _i}(H) \geqslant {\lambda _i}(H/{A_{11}}) \geqslant {\lambda _i}(M/{A_{11}}) \geqslant {\lambda _i}(H/A) \geqslant {\lambda _{i + r - {r_1}}}(H/{A_{11}}),\;\;\;\;i = 1,2, \cdots ,n - r. $

      ${\lambda _{i + r - {r_1}}}(H/{A_{11}})$ 继续使用(5)式有 ${\lambda _{i + r - {r_1}}}(H/{A_{11}}) \geqslant {\lambda _{i + r}}(H)$$i = 1,2, \cdots ,n - r$. 故(20)式成立. 同理可证(21)式. 证毕.

      推论1表明半正定Hermitian矩阵 $H$ 的任意 $r$ 阶($r \geqslant 2$)主子矩阵 $X$$H$ 中的伪Schur补的特征值交错位于 $X$ 的任意主子矩阵在 $H$ 中的伪Schur补的特征值之间,将Smith关于半正定Hermitian矩阵Schur补的特征值交错结论推广到了半正定Hermitian矩阵的伪Schur补上.

      推论2 设 ${{{H}}_1}{{ = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_{11}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_{12}}}\\ {{{{C}}_{11}}}&{{{{C}}_{12}}}&{{{{D}}_1}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$${{{H}}_2}{{ = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_{21}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_{22}}}\\ {{{{C}}_{21}}}&{{{{C}}_{22}}}&{{{{D}}_2}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$${H_i}$($i = 1,2$)的 $r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,则

      $\qquad [({H_1} + {H_2})/(2{A_{11}})]/[2 \cdot (A/{A_{11}})] = ({H_1} + {H_2})/(2A).$

      证明 注意到 $(2A)/(2{A_{11}}) = 2 \cdot (A/{A_{11}})$,根据定理1得

      $\qquad [({H_1} + {H_2})/(2{A_{11}})]/[(2A)/(2{A_{11}})] = [({H_1} + {H_2})/(2{A_{11}})]/[2 \cdot (A/{A_{11}})] = ({H_1} + {H_2})/(2A).$

      证毕.

      推论2可以推广至 $n$ 个半正定Hermitian矩阵的和的情形:

      推论3 设 ${{{H}}_{{i}}}{{ = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_{{{i}}1}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_{{{i}}2}}}\\ {{{{C}}_{{{i}}1}}}&{{{{C}}_{{{i}}2}}}&{{{{D}}_{{i}}}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$${H_i}$($i = 1,2, \cdots ,n$)的 $r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,则

      $\qquad [({H_1} + {H_2}{\rm{ + }} \cdots {\rm{ + }}{H_n})/(n{A_{11}})]/[n \cdot (A/{A_{11}})] = ({H_1} + {H_2}{\rm{ + }} \cdots {\rm{ + }}{H_n})/(nA).$

      推论4 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$n$ 为任意正整数,则

      $\qquad [(nH)/(n{A_{11}})]/[n \cdot (A/{A_{11}})] = n(H/A).$

      定理2 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$M = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{B_1}}\\ {{C_1}}&D \end{array}} \right)$$H$$n - r + {r_1}$ 阶主子矩阵,记 $Q = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{C_1}}&{{C_2}} \end{array}} \right)$,则

      $\qquad H/A = M/{A_{11}} - (Q/{A_{11}}) \cdot {(A/{A_{11}})^ + } \cdot {(Q/{A_{11}})^ * };$

      $\qquad H/M = A/{A_{11}} - {(Q/{A_{11}})^ * } \cdot {(M/{A_{11}})^ + } \cdot (Q/{A_{11}}).$

      证明 由(6)式得 $(H/{A_{11}})/(A/{A_{11}}) = M/{A_{11}} - ({C_2} - {C_1}A_{11}^ + {A_{12}}){(A/{A_{11}})^ + }{({C_2} - {C_1}A_{11}^ + {A_{12}})^ * }$,经计算知 $Q/{A_{11}} = {C_2} - {C_1}A_{11}^ + {A_{12}}$,根据定理1即得 $H/A = M/{A_{11}} - (Q/{A_{11}}) \cdot {(A/{A_{11}})^ + } \cdot {(Q/{A_{11}})^ * }$.

      同理可得(26)式. 证毕.

      推论5 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$M = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{B_1}}\\ {{C_1}}&D \end{array}} \right)$$H$$n - r + {r_1}$ 阶主子矩阵,则

      $\qquad M/{A_{11}} \geqslant H/A;$

      $\qquad A/{A_{11}} \geqslant H/M.$

      证明 因为半正定Hermitian矩阵的主子矩阵、伪Schur补及Moore-Penrose广义逆仍然是半正定Hermitian矩阵,所以 $(Q/{A_{11}}) \cdot {(A/{A_{11}})^ + } \cdot {(Q/{A_{11}})^ * } = M/{A_{11}} - H/A \geqslant O$,即 $M/{A_{11}} \geqslant H/A$.

      同理可得(28)式. 证毕.

      事实上,由(18)~(19)式还可以得出 $N/{A_{22}} \geqslant H/A$$A/{A_{22}} \geqslant H/N$$N/D \geqslant H/M$$M/D \geqslant H/N$. 这表明半正定Hermitian矩阵 $H$ 的任意 $r$ 阶($r \geqslant 2$)主子矩阵 $X$$H$ 中的伪Schur补与 $X$ 的任意主子矩阵在 $H$ 的某个特定主子矩阵中的伪Schur补存在偏序.

      定理3 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,则 ${(H/A)^ - }$${(H/M)^ - }$${(H/{A_{11}})^ - }$ 的主子矩阵,且对 $H/{A_{11}}$ 的一部分{1}广义逆有

      $\qquad {(H/{A_{11}})^ - }/{(H/A)^ - } = {(A/{A_{11}})^ - };$

      $\qquad {(H/{A_{11}})^ - }/{(H/M)^ - } = {(M/{A_{11}})^ - }.$

      证明 记 $Q = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{C_1}}&{{C_2}} \end{array}} \right)$,注意到 $H/{A_{11}} \geqslant O$,故由(1)式和(25)式有

      $\qquad \left(\!\! {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&O \\ { - (Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{{I_{n - r}}} \end{array}} \!\! \right)\left( \!\!{\begin{array}{*{20}{c}} {A/{A_{11}}}&{{{(Q/{A_{11}})}^ * }} \\ {Q/{A_{11}}}&{M/{A_{11}}} \end{array}} \!\! \right){\left(\!\! {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&O \\ { - (Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{{I_{n - r}}} \end{array}} \right)^ * } = \left(\!\! {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right),$

      $\qquad \left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&{{{(Q/{A_{11}})}^ * }} \\ {Q/{A_{11}}}&{M/{A_{11}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&O \\ {(Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{{I_{n - r}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right){\left( {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&O \\ {(Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{{I_{n - r}}} \end{array}} \right)^ * },$

      $\left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&{{{(Q/{A_{11}})}^ * }} \\ {Q/{A_{11}}}&{M/{A_{11}}} \end{array}} \right)$ 的全体{1}广义逆可以表示为

      $\qquad \left( {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&{ - {{(A/{A_{11}})}^ + }{{(Q/{A_{11}})}^ * }} \\ O&{{I_{n - r}}} \end{array}} \right){\left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right)^ - }\left( {\begin{array}{*{20}{c}} {{I_{r - {r_1}}}}&O \\ { - (Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{{I_{n - r}}} \end{array}} \right).$

      注意到 $\left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right)$ 的全体{1}广义逆具有 $\left( {\begin{array}{*{20}{c}} {{{(A/{A_{11}})}^ - }}&X \\ Y&{{{(H/A)}^ - }} \end{array}} \right)$ 的形式,这里 $X,Y$ 分别满足 $(A/{A_{11}})X(H/A) = O$$(H/A)Y(A/{A_{11}}) = O$. 计算可知 ${(H/A)^ - }$${(H/{A_{11}})^ - }$ 的主子矩阵.

      又因为 $\left( {\begin{array}{*{20}{c}} {{{(A/{A_{11}})}^ - }}&O \\ O&{{{(H/A)}^ - }} \end{array}} \right)$$\left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&O \\ O&{H/A} \end{array}} \right)$ 的一部分{1}广义逆,所以 $H/{A_{11}}$ 的一部分{1}广义逆可以表示为

      $\qquad \left( {\begin{array}{*{20}{c}} {{{(A/{A_{11}})}^ - } + {{(A/{A_{11}})}^ + }{{(Q/{A_{11}})}^ * }{{(H/A)}^ - }(Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{ - {{(A/{A_{11}})}^ + }{{(Q/{A_{11}})}^ * }{{(H/A)}^ - }} \\ { - {{(H/A)}^ - }(Q/{A_{11}}){{(A/{A_{11}})}^ + }}&{{{(H/A)}^ - }} \end{array}} \right),$

      且对于这部分 ${(H/{A_{11}})^ - }$${(H/{A_{11}})^ - }/{(H/A)^ - } = {(A/{A_{11}})^ - }$.

      同理可得(30)式. 证毕.

      定理3指出对于半正定Hermitian矩阵 $H$ 的伪Schur补 $H/{A_{11}}$ 的一部分{1}广义逆来说,商公式依旧成立. 我们知道,在许多重要应用场合,往往不必得到一个矩阵的全部{1}广义逆,所以定理 3的结论是有意义的.

      引理3[16] 设 ${{A = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}} \end{array}} \right) \in {{\mathop{\bf{C}}\nolimits} ^{m \times n}}$,若 $r(A) = r({A_{11}}) + r({A_{22}})$,则

      $\qquad \begin{aligned}{A^ + } = &\left( {\begin{array}{*{20}{c}} {A_{11}^ + + A_{11}^ + {A_{12}}{{(A/{A_{11}})}^ + }{A_{21}}A_{11}^ + }&{ - A_{11}^ + {A_{12}}{{(A/{A_{11}})}^ + }} \\ { - {{(A/{A_{11}})}^ + }{A_{21}}A_{11}^ + }&{{{(A/{A_{11}})}^ + }} \end{array}} \right)= \\ &\left( {\begin{array}{*{20}{c}} {{{(A/{A_{22}})}^ + }}&{ - {{(A/{A_{22}})}^ + }{A_{12}}A_{22}^ + } \\ { - A_{22}^ + {A_{21}}{{(A/{A_{22}})}^ + }}&{A_{22}^ + + A_{22}^ + {A_{21}}{{(A/{A_{22}})}^ + }{A_{12}}A_{22}^ + } \end{array}} \right).\end{aligned}$

      定理4 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$M = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{B_1}}\\ {{C_1}}&D \end{array}} \right)$$H$$n - r + {r_1}$ 阶主子矩阵,若 $r(M/{A_{11}}) = r(H/A)$,则

      $\qquad {(H/{A_{11}})^ + }/{(H/A)^ + } = {(A/{A_{11}})^ + };$

      $\qquad {(H/{A_{11}})^ + }/{(H/M)^ + } = {(M/{A_{11}})^ + }.$

      证明 记 $Q = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{C_1}}&{{C_2}} \end{array}} \right)$,则由(6)式得 $H/{A_{11}} = \left( {\begin{array}{*{20}{c}} {A/{A_{11}}}&{{{(Q/{A_{11}})}^ * }} \\ {Q/{A_{11}}}&{M/{A_{11}}} \end{array}} \right)$$H/{A_{11}} \geqslant O$. 对于半正定Hermitian矩阵 $H$,利用(8)式考虑 $H$ 的秩得 $r(H) = r(A) + r(H/A)$,则

      $\qquad r(M/{A_{11}}) = r(H/A)\Leftrightarrow r(M/{A_{11}}) + r(A/{A_{11}}) = r(H/{A_{11}}). $

      故由定理1与引理3得

      $\qquad {(H/{A_{11}})^{\rm{ + }}} = \left( {\begin{array}{*{20}{c}} {{{\left( {A/{A_{11}}} \right)}^{\rm{ + }}}{\rm{ + }}{{\left( {A/{A_{11}}} \right)}^{\rm{ + }}}{{(Q/{A_{11}})}^ * }{{(H/A)}^ + }(Q/{A_{11}}){{\left( {A/{A_{11}}} \right)}^{\rm{ + }}}}&{ - {{\left( {A/{A_{11}}} \right)}^{\rm{ + }}}{{(Q/{A_{11}})}^ * }{{(H/A)}^ + }}\\ { - {{(H/A)}^ + }(Q/{A_{11}}){{\left( {A/{A_{11}}} \right)}^{\rm{ + }}}}&{{{(H/A)}^ + }} \end{array}} \right).$

      对半正定Hermitian矩阵 ${(H/{A_{11}})^ + }$,计算得(34)式成立.

      同理可得(35)式. 证毕.

      定理4指出在一定条件下,半正定Hermitian矩阵 $H$ 的任意 $r$ 阶($r \geqslant 2$)主子矩阵在 $H$ 中的伪Schur补的Moore-Penrose广义逆也满足商公式.

      定理5 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,$C = ({C_1}\;\;\;\;{C_2})$,则

      $\qquad C{A^ + }{C^ * } = C\left( {\begin{array}{*{20}{c}} {{A_{11}}^ + + {A_{11}}^ + {A_{12}}{{(A/{A_{11}})}^ + }{A_{21}}{A_{11}}^ + }&{ - {A_{11}}^ + {A_{12}}{{(A/{A_{11}})}^ + }} \\ { - {{(A/{A_{11}})}^ + }{A_{21}}{A_{11}}^ + }&{{{(A/{A_{11}})}^ + }} \end{array}} \right){C^ * };$

      $\qquad C{A^ + }{C^ * } = C\left( {\begin{array}{*{20}{c}} {{{(A/{A_{22}})}^ + }}&{ - {{(A/{A_{22}})}^ + }{A_{12}}{A_{22}}^ + } \\ { - {A_{22}}^ + {A_{21}}{{(A/{A_{22}})}^ + }}&{{A_{22}}^ + + {A_{22}}^ + {A_{21}}{{(A/{A_{22}})}^ + }{A_{12}}{A_{22}}^ + } \end{array}} \right){C^ * }.$

      证明 由定理1有

      $\qquad (H/{A_{11}})/(A/{A_{11}}) = H/A = D - C{A^ + }{C^ * }.$

      另一方面,

      $\qquad \begin{split} (H/{A_{11}})/(A/{A_{11}}) =& M/{A_{11}} - ({C_2} - {C_1}A_{11}^ + {A_{12}}){(A/{A_{11}})^ + }{({C_2} - {C_1}A_{11}^ + {A_{12}})^ * }=\\ &D - {C_1}A_{11}^ + C_1^ * - {C_2}{(A/{A_{11}})^ + }C_2^ * + {C_2}{(A/{A_{11}})^ + }{A_{21}}A_{11}^ + C_1^ * + {C_1}A_{11}^ + {A_{12}}{(A/{A_{11}})^ + }C_2^ * -\\ &{C_1}A_{11}^ + {A_{12}}{(A/{A_{11}})^ + }{A_{21}}A_{11}^ + C_1^ *=\\ & D - ({C_1}\;\;{C_2})\left( {\begin{array}{*{20}{c}} {{A_{11}}^ + + {A_{11}}^ + {A_{12}}{{(A/{A_{11}})}^ + }{A_{21}}{A_{11}}^ + }&{ - {A_{11}}^ + {A_{12}}{{(A/{A_{11}})}^ + }} \\ { - {{(A/{A_{11}})}^ + }{A_{21}}{A_{11}}^ + }&{{{(A/{A_{11}})}^ + }} \end{array}} \right)\left( \begin{array}{l} C_1^ * \\ C_2^ * \\ \end{array} \right). \end{split} $

      由(40)~(41)式即得(38)式.

      同理可得(39)式. 证毕.

      推论6 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,若 $C = ({C_1}\;\;\;{C_2})$ 为可逆矩阵或列满秩矩阵,则 $A$ 可逆且

      $\qquad {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{A_{11}}^{ - 1} + {A_{11}}^{ - 1}{A_{12}}{{(A/{A_{11}})}^{ - 1}}{A_{21}}{A_{11}}^{ - 1}}&{ - {A_{11}}^{ - 1}{A_{12}}{{(A/{A_{11}})}^{ - 1}}} \\ { - {{(A/{A_{11}})}^{ - 1}}{A_{21}}{A_{11}}^{ - 1}}&{{{(A/{A_{11}})}^{ - 1}}} \end{array}} \right);$

      $\qquad {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{{(A/{A_{22}})}^{ - 1}}}&{ - {{(A/{A_{22}})}^{ - 1}}{A_{12}}{A_{22}}^{ - 1}} \\ { - {A_{22}}^{ - 1}{A_{21}}{{(A/{A_{22}})}^{ - 1}}}&{{A_{22}}^{ - 1} + {A_{22}}^{ - 1}{A_{21}}{{(A/{A_{22}})}^{ - 1}}{A_{12}}{A_{22}}^{ - 1}} \end{array}} \right).$

      证明 若 $C = ({C_1}\;\;\;{C_2})$ 为可逆矩阵或列满秩矩阵,则 $r(C) = r$. 但是由(1)式,$C = C{A^ + }A$,于是 $r(C) = r(C{A^ + }A) \leqslant r(A) \leqslant r$. 这样 $r(A) = r$,半正定Hermitian矩阵 $H$ 的主子矩阵 $A$ 可逆,从而 ${A_{11}}$ 可逆,由 ${A_{11}}$ 可逆得 $A/{A_{11}}$ 也可逆. 于是(42)式成立.

      同理可得(43)式. 证毕.

      引理4[14] 设 ${{A = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$${A_{11}}$$A$$r$ 阶主子矩阵,${{C = }}\left( {\begin{array}{*{20}{c}} {{{{C}}_{11}}}&{{{{C}}_{12}}}\\ {{{{C}}_{21}}}&{{{{C}}_{22}}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$${C_{11}}$$C$$r$ 阶主子矩阵. 若记 $X$$CA{C^ * }$$r$ 阶顺序主子矩阵,则

      $\qquad (CA{C^ * })/X \leqslant (C/{C_{11}}) \cdot {A_{22}} \cdot ({C^ * }/{C_{11}}).$

      引理5[17] 设 ${{A = }}\left( {\begin{array}{*{20}{c}}{{{{A}}_{11}}}&{{{{A}}_{12}}}\\{{{{A}}_{21}}}&{{{{A}}_{22}}}\end{array}} \right) \in {{H}}_{{n}}^ \geqslant$,则

      $\qquad{A^ + } = \left( {\begin{array}{*{20}{c}} {A_{11}^ + + A_{11}^ + {A_{12}}{S^ \sim }{A_{21}}A_{11}^ + }&{ - A_{11}^ + {A_{12}}{S^ \sim }} \\ { - {S^ \sim }{A_{21}}A_{11}^ + }&{{S^ \sim }} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - A_{11}^ + (FZ + {Z^ * }{F^ * })A_{11}^ + }&{A_{11}^ + {Z^ * }} \\ {ZA_{11}^ + }&O \end{array}} \right).$

      这里

      $\qquad {S^ \sim } = (Z\;\;I)\left( {\begin{array}{*{20}{c}} {A_{11}^ + + A_{11}^ + {A_{12}}{S^ + }{A_{21}}A_{11}^ + }&{ - A_{11}^ + {A_{12}}{S^ + }} \\ { - {S^ + }{A_{21}}A_{11}^ + }&{{S^ + }} \end{array}} \right){(Z\;\;I)^ * },$

      $\qquad Z = (I - {S^ + }S){A_{21}}A_{11}^ + {[I + A_{11}^ + {A_{12}}(I - {S^ + }S){A_{21}}A_{11}^ + ]^{ - 1}},$

      $\qquad S = {A_{22}} - {A_{21}}A_{11}^ + {A_{12}} = A/{A_{11}}.$

      进一步地,${S^ \sim }$$S$ 的一个{1}广义逆.

      定理6 设 ${{H = }}\left( {\begin{array}{*{20}{c}} {{{{A}}_{11}}}&{{{{A}}_{12}}}&{{{{B}}_1}}\\ {{{{A}}_{21}}}&{{{{A}}_{22}}}&{{{{B}}_2}}\\ {{{{C}}_1}}&{{{{C}}_2}}&{{D}} \end{array}} \right) \in {{H}}_{{n}}^ \geqslant$$A = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{22}}} \end{array}} \right)$$H$$r$ 阶主子矩阵,${A_{11}}$$A$${r_1}$ 阶($1 \leqslant {r_1} < r$)主子矩阵,则

      $\qquad {(A/{A_{11}})^ + } \leqslant {(A/{A_{11}})^ - },$

      这里 ${(A/{A_{11}})^ - }$${A^ + }$ 的主子矩阵.

      证明 特别地取 $C = ({C_1}\;\;\;{C_2}) = \left( {\begin{array}{*{20}{c}} {{O_{{r_1} \times {r_1}}}}&{{O_{{r_1} \times (r - {r_1})}}} \\ {{O_{(r - {r_1}) \times {r_1}}}}&{{I_{(r - {r_1}) \times (r - {r_1})}}} \end{array}} \right)$,则由定理5有

      $\qquad \begin{aligned} C{A^ + }{C^ * } =& \left( {\begin{array}{*{20}{c}} O&O \\ O&I \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{A_{11}}^ + + {A_{11}}^ + {A_{12}}{{(A/{A_{11}})}^ + }{A_{21}}{A_{11}}^ + }&{ - {A_{11}}^ + {A_{12}}{{(A/{A_{11}})}^ + }} \\ { - {{(A/{A_{11}})}^ + }{A_{21}}{A_{11}}^ + }&{{{(A/{A_{11}})}^ + }} \end{array}} \right)\left( {\begin{array}{*{20}{c}} O&O \\ O&I \end{array}} \right) = \\ &\left( {\begin{array}{*{20}{c}} O&O \\ O&{{{(A/{A_{11}})}^ + }} \end{array}} \right).\end{aligned}$

      注意到 $C \in {{H}}_{{n}}^ \geqslant $,由引理4和引理5得

      $\qquad {(A/{A_{11}})^ + } = (C{A^ + }{C^ * })/O \leqslant (C/O) \cdot {(A/{A_{11}})^ - } \cdot ({C^ * }/O) = {(A/{A_{11}})^ - },$

      这里 ${(A/{A_{11}})^ - }$ 由(46),(47)式确定. 证毕.

      例1 设 ${{H = }}\left( {\begin{array}{*{20}{c}} 1&1&{ - 1}&0&1&0 \\ 1&1&{ - 1}&0&1&0 \\ { - 1}&{ - 1}&1&0&{ - 1}&0 \\ 0&0&0&3&0&0 \\ 1&1&{ - 1}&0&2&0 \\ 0&0&0&0&0&0 \end{array}} \right) \in {{H}}_6^ \geqslant $${{A}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&1&{ - 1}&0 \\ 1&1&{ - 1}&0 \\ { - 1}&{ - 1}&1&0 \\ 0&0&0&3 \end{array}} \right)$$H$$4$ 阶主子矩阵,其中

      $\qquad \begin{aligned} &{{{A}}_{11}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&1 \\ 1&1 \end{array}} \right),\;{{{A}}_{22}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&3 \end{array}} \right),\;{{D}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&0 \end{array}} \right),\;{{C}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&1&{ - 1}&0 \\ 0&0&0&0 \end{array}} \right),\;\\ &{{M}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&1&1&0 \\ 1&1&1&0 \\ 1&1&2&0 \\ 0&0&0&0 \end{array}} \right),\;{{Q}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&1&{ - 1}&0 \\ 1&1&{ - 1}&0 \\ 1&1&{ - 1}&0 \\ 0&0&0&0 \end{array}} \right). \end{aligned}$

      计算可得

      $\qquad {{A/}}{{{A}}_{11}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 0&0 \\ 0&3 \end{array}} \right),\; {{M/}}{{{A}}_{11}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right),\; {{H/A}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right),\; {{H/M}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 0&0 \\ 0&3 \end{array}} \right),\; {{H/}}{{{A}}_{11}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 0&0&0&0 \\ 0&3&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{array}} \right). $

      可见 ${{A/}}{{{A}}_{11}}$${{M/}}{{{A}}_{11}}$${{H/}}{{{A}}_{11}}$ 的主子矩阵,且

      $\qquad {{(H/}}{{{A}}_{11}}{{)/(A/}}{{{A}}_{11}}{{) = }}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right){{ = H/A}},\;{{(H/}}{{{A}}_{11}}{{)/(M/}}{{{A}}_{11}}{{) = }}\left( {\begin{array}{*{20}{c}} 0&0 \\ 0&3 \end{array}} \right){{ = H/M}}. $

      经计算 ${{H/}}{{{A}}_{11}}$ 的全体{1}广义逆为 $\left( {\begin{array}{*{20}{c}} {{g_{33}}}&{\dfrac{1}{3}{g_{31}}}&{{g_{32}}}&{{g_{34}}} \\ {{g_{13}}}&{\dfrac{1}{3}}&0&{{g_{14}}} \\ {{g_{23}}}&0&1&{{g_{24}}} \\ {{g_{43}}}&{\dfrac{1}{3}{g_{41}}}&{{g_{42}}}&{{g_{44}}} \end{array}} \right)$,这里 ${g_{ij}}{\rm{ }}(i,j = 1,2,3,4)$ 为任意实数.$A/{A_{11}}$ 的全体{1}广义逆为 $\left( {\begin{array}{*{20}{c}} {{k_{11}}}&{\dfrac{1}{3}{k_{12}}} \\ {{k_{21}}}&{\dfrac{1}{3}} \end{array}} \right)$,这里 ${k_{ij}}{\rm{ }}(i,j = 1,2)$ 为任意实数.$H/A$ 的全体{1}广义逆为 $\left( {\begin{array}{*{20}{c}} 1&{{c_{12}}} \\ {{c_{21}}}&{{c_{22}}} \end{array}} \right)$,这里 ${c_{ij}}{\rm{ }}(i,j = 1,2)$ 为任意实数.${(H/A)^ - }$${(H/{A_{11}})^ - }$ 的主子矩阵. 注意到 $Q/{A_{11}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right)$,特别地,取 $H/{A_{11}}$ 的部分{1}广义逆为 ${(H/{A_{11}})^ - } = \left( {\begin{array}{*{20}{c}} {{g_{33}}}&{\dfrac{1}{3}{g_{31}}}&0&0 \\ {{g_{13}}}&{\dfrac{1}{3}}&0&0 \\ 0&0&1&{{g_{24}}} \\ 0&0&{{g_{42}}}&{{g_{44}}} \end{array}} \right)$ 时,有 ${(H/{A_{11}})^ - }/{(H/A)^ - } = {(A/{A_{11}})^ - }$.

      $r(M/{A_{11}}{\rm{) = }}r(H/A)$${(H/{A_{11}}{\rm{)}}^ + }{\rm{ = }}\left( {\begin{array}{*{20}{c}} 0&0&0&0 \\ 0&{\dfrac{1}{3}}&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{array}} \right)$${(H/A)^ + }{\rm{ = }}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right)$${(A/{A_{11}})^ + }{\rm{ = }}\left( {\begin{array}{*{20}{c}} 0&0 \\ 0&{\dfrac{1}{3}} \end{array}} \right)$${(H/A)^ + }$${(H/{A_{11}})^ + }$ 的主子矩阵,且 ${(H/{A_{11}})^ + }/{(H/A)^ + } = \left( {\begin{array}{*{20}{c}} 0&0 \\ 0&{\dfrac{1}{3}} \end{array}} \right) = {(A/{A_{11}})^ + }$.

      ${A^ + } = \left( {\begin{array}{*{20}{c}} {\dfrac{1}{9}}&{\dfrac{1}{9}}&{ - \dfrac{1}{9}}&0 \\ {\dfrac{1}{9}}&{\dfrac{1}{9}}&{ - \dfrac{1}{9}}&0 \\ { - \dfrac{1}{9}}&{ - \dfrac{1}{9}}&{\dfrac{1}{9}}&0 \\ 0&0&0&{\dfrac{1}{3}} \end{array}} \right)$$A/{A_{11}}$ 的一个{1}广义逆为 $\left( {\begin{array}{*{20}{c}} {\dfrac{1}{9}}&0 \\ 0&{\dfrac{1}{3}} \end{array}} \right)$,是 ${A^ + }$ 的主子矩阵,且有

      $\qquad {(A/{A_{11}})^ - } = \left( {\begin{array}{*{20}{c}} {\dfrac{1}{9}}&0 \\ 0&{\dfrac{1}{3}} \end{array}} \right) \geqslant \left( {\begin{array}{*{20}{c}} 0&0 \\ 0&{\dfrac{1}{3}} \end{array}} \right) = {(A/{A_{11}})^ + }. $

参考文献 (17)

目录

    /

    返回文章
    返回