0 引言

文中符号说明：I表示适当阶数的单位矩阵；A∈C_r^m×n表示秩为r的m×n阶复矩阵；A^*和A⁺分别表示矩阵A的共轭转置和Moore-Penrose逆；‖·‖₂和‖·‖分别表示矩阵的谱范数和酉不变范数.

广义逆理论是一个应用十分广泛的数学分支，在数值线性代数、线性规划、最优化、控制论、马尔可夫链、数理统计、信号传输、微分方程等重要领域都有极其广泛的应用. 1955年PENROSE在MOORE关于广义逆的基础上提出了4个更加便于理解的方程^[1]：

设A∈C^m×n，若存在G∈C^n×m使得

(1) AGA=A;                 (2) GAG=G;

(3) (AG)^*=AG;                  (4) (GA)^*=GA,

则称G是A的Moore-Penrose逆，这4个方程称为M-P方程.

全部或部分满足M-P方程的矩阵G，称为A的广义逆.若G满足M-P方程中的第(i), …, (j)个方程，则称G为矩阵A的一个{i, …, j}-逆，记为A^{{i, …, j}}. A的所有{i, …, j}-逆的集合用A{i, …, j}表示，其中A^{1}, A^{{1, 2}}, A^{{1, 3}}和A^{{1, 4}}都是常用的广义逆，并且一般都不是唯一存在的.

广义逆的扰动研究是广义逆理论中一个非常重要的课题，WEDIN^[2]、STEWART^[3]、孙继广^[4]和WEI等^[5]等国内外专家都在此研究领域做出了重要贡献.迄今为止，广义逆扰动理论的研究成果已有很多.关于Moore-Penrose逆和Drazin逆的扰动理论可参阅文献[1-12].

鉴于{1}-逆在矩阵理论和计算中的重要作用(例如相容线性系统Ax=b的通解可以表示为x=A⁽¹⁾b+(I-A⁽¹⁾A)y)，LIU等^[12]研究了{1}-逆在保秩扰动下的连续性；WEI等^[13]和MENG等^[14]分别给出了{1}-逆在谱范数和Frobenius范数下的加法和乘法扰动界.另外，因为{1, 3}-和{1, 4}-逆在实际应用中也起着非常重要的作用，例如最小二乘问题min‖Ax-b‖₂的最小二乘解可表示为x=A^{(1, 3)}b；相容线性系统Ax=b的最小范数解可表示为x=A^{(1, 4)}b^[1]，所以MENG等^[14]研究了这两类广义逆在谱范数和Frobenius范数下的加法和乘法扰动界.

因谱范数和Frobenius范数是两类特殊的酉不变范数，因此，本文试图将文献[14]中的关于{1, 3}-和{1, 4}-逆在谱范数和Frobenius范数下的结果推广到一般的酉不变范数.对于给定的矩阵A, B∈C^m×n及A^{(1, i)}∈A{1, i}, i=3, 4，文献[14]给出了谱范数和Frobenius范数下距离A^{(1, i)}最近的矩阵B的{1, i}-逆的具体表达式，参见文献[14]theorem 3.1和theorem 3.2.

本文的主要工作为：首先证明当谱范数和Frobenius范数推广到一般的酉不变范数时，文献[14]中给出的距离A^{(1, i)}最近的矩阵B的{1, i}-逆不变；其次，利用该结果给出{1, 3}-和{1, 4}-逆在酉不变范数下的加法和乘法扰动界，所得扰动界推广和改进了文献[14]中的结果.

首先给出本文所用到的2个引理.

引理1^[1]    设矩阵A∈C_r^m×n，则

A{1, 3}={A⁺+(I-A⁺A)Z：Z∈C^n×m}.另外，设 $\mathit{\boldsymbol{A}} = \mathit{\boldsymbol{U}}\left( {\begin{array}{*{20}{c}} \mathit{\boldsymbol{ \boldsymbol{\varSigma} }}&0\\ 0&0 \end{array}} \right){\mathit{\boldsymbol{V}}^*}$ 为矩阵A的奇异值分解，其中U∈C^m×m, V∈C^n×n为酉阵，Σ=diag(σ₁, σ₂, …, σ_r)，则任意的A^{(1, 3)}可以表示为

$ {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}} = \mathit{\boldsymbol{V}}\left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}}}&0\\ \mathit{\boldsymbol{L}}&\mathit{\boldsymbol{M}} \end{array}} \right){\mathit{\boldsymbol{U}}^ * }\;且\;{\mathit{\boldsymbol{A}}^ + } = \mathit{\boldsymbol{V}}\left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}}}&0\\ \mathit{\boldsymbol{L}}&\mathit{\boldsymbol{M}} \end{array}} \right){\mathit{\boldsymbol{U}}^ * }, $

其中L, M为适当阶数的矩阵.

引理2^[14]    设矩阵X, Y∈C^m×n，如果X^*Y=0，则‖X‖≤‖X+Y‖.

1 最佳逼近的{1, 3}-和{1, 4}-逆

本节将给出B_m^{(1, i)}∈B{1, i}的具体表达式，使得对任意给定的A^{(1, i)}∈A{1, i}，满足

$ \mathop {\min }\limits_{{\mathit{\boldsymbol{B}}^{\left( {1,i} \right)}} \in \mathit{\boldsymbol{B}}\left\{ {1,i} \right\}} \left\| {{\mathit{\boldsymbol{B}}^{\left( {1,i} \right)}} - {\mathit{\boldsymbol{A}}^{\left( {1,i} \right)}}} \right\| = \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,i} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,i} \right)}}} \right\|, $

其中i=3, 4.

定理1    假设A, B∈C^m×n.对于任意给定的A^{(1, 3)}∈A{1, 3}, 存在唯一的矩阵

$ \mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} = {\mathit{\boldsymbol{B}}^ + } + \left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}} \right){\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}} \in \mathit{\boldsymbol{B}}\left\{ {1,3} \right\}, $

使得

$ \begin{array}{l} \mathop {\min }\limits_{{\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} \in \mathit{\boldsymbol{B}}\left\{ {1,3} \right\}} \left\| {{\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| = \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| = \\ \;\;\;\;\;\;\left\| {{\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\|. \end{array} $

(1)

证明    设A∈C_r^m×n和B∈C_s^m×n的奇异值分解为：

$ \mathit{\boldsymbol{A}} = \mathit{\boldsymbol{U}}\left( {\begin{array}{*{20}{c}} \mathit{\boldsymbol{ \boldsymbol{\varSigma} }}&0\\ 0&0 \end{array}} \right){\mathit{\boldsymbol{V}}^ * },\mathit{\boldsymbol{B}} = \mathit{\boldsymbol{\tilde U}}\left( {\begin{array}{*{20}{c}} {\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}&0\\ 0&0 \end{array}} \right){{\mathit{\boldsymbol{\tilde V}}}^ * }, $

其中， $\mathit{\boldsymbol{U}} = \left( {\mathop {{\mathit{\boldsymbol{U}}_1}}\limits^r \;\;\mathop {{\mathit{\boldsymbol{U}}_2}}\limits^{m - r} } \right)$ , $\mathit{\boldsymbol{\tilde U}} = \left( {\mathop {{{\mathit{\boldsymbol{\tilde U}}}_1}}\limits^s \;\;\mathop {{{\mathit{\boldsymbol{\tilde U}}}_2}}\limits^{m - s} } \right) \in {C^{m \times m}}$ ， $\mathit{\boldsymbol{V}} = \left( {\mathop {{\mathit{\boldsymbol{V}}_1}}\limits^r \;\;\mathop {{\mathit{\boldsymbol{V}}_2}}\limits^{n - r} } \right)$ , $\mathit{\boldsymbol{\tilde V}} = \left( {\mathop {{{\mathit{\boldsymbol{\tilde V}}}_1}}\limits^s \;\;\mathop {{{\mathit{\boldsymbol{\tilde V}}}_2}}\limits^{n - s} } \right) \in {C^{n \times n}}$ 都是酉阵，Σ=diag(σ₁, σ₂, …, σ_r), $\mathit{\pmb{\tilde \Sigma }} = {\rm{diag}}\left( {{{\tilde \sigma }_1},{{\tilde \sigma }_2}, \cdots ,{{\tilde \sigma }_s}} \right)$ ，且

$ {\sigma _1} \ge \cdots \ge {\sigma _r} > 0,\;\;\;\;\;{{\tilde \sigma }_1} \ge \cdots \ge {{\tilde \sigma }_s} > 0. $

由引理1知，对于给定的A^{(1, 3)}，存在适当阶数的矩阵L和M, 使得 ${\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}} = \mathit{\boldsymbol{V}}\left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}}}&0\\ \mathit{\boldsymbol{L}}&{\mathit{\boldsymbol{M}}\;\;} \end{array}} \right){\mathit{\boldsymbol{U}}^*}$ , 且任意的B^{(1, 3)}∈B{1, 3}都可以表示为

$ {\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} = \mathit{\boldsymbol{\tilde V}}\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}}&0\\ {\mathit{\boldsymbol{\tilde L}}}&{\mathit{\boldsymbol{\tilde M}}} \end{array}} \right){{\mathit{\boldsymbol{\tilde U}}}^ * }, $

其中, $\mathit{\boldsymbol{\tilde L}} \in {C^{\left( {n - s} \right) \times s}}$ 和 $\mathit{\boldsymbol{\tilde M}} \in {C^{\left( {n - s} \right) \times (m - s)}}$ 是2个任意的矩阵.利用酉不变范数的性质，可得

$ \begin{array}{l} \left\| {{\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| = \left\| {\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}}&0\\ {\mathit{\boldsymbol{\tilde L}}}&{\mathit{\boldsymbol{\tilde M}}} \end{array}} \right){{\mathit{\boldsymbol{\tilde U}}}^ * }\mathit{\boldsymbol{U}} - {{\mathit{\boldsymbol{\tilde V}}}^ * }\mathit{\boldsymbol{V}}\left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}}}&0\\ \mathit{\boldsymbol{L}}&\mathit{\boldsymbol{M}} \end{array}} \right)} \right\| = \\ \left\| {\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_1} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_1}{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_2}L}&{{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_2} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{M}}}\\ 0&0 \end{array}} \right)} \right. + \\ \left. {\left( {\begin{array}{*{20}{c}} 0&0\\ {\mathit{\boldsymbol{\tilde L\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_1} + \mathit{\boldsymbol{\tilde M\tilde U}}_2^ * {\mathit{\boldsymbol{U}}_1} - \mathit{\boldsymbol{H}}}&{\mathit{\boldsymbol{\tilde L\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_2} + \mathit{\boldsymbol{\tilde M\tilde U}}_2^ * {\mathit{\boldsymbol{U}}_2} - \mathit{\boldsymbol{\tilde V}}_2^ * {\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{M}}} \end{array}} \right)} \right\|, \end{array} $

其中, $\mathit{\boldsymbol{H}} = \mathit{\boldsymbol{\tilde V}}_2^*{\mathit{\boldsymbol{V}}_1}{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}} + \mathit{\boldsymbol{\tilde V}}_2^*{\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{L}}$ .一方面，对上述等式的右端项应用引理2得

$ \begin{array}{l} \left\| {\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_1} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_1}{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_2}L}&{{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_2} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{M}}}\\ 0&0 \end{array}} \right)} \right\| \le \\ \;\;\;\;\;\;\left\| {{\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\|. \end{array} $

另一方面，取 ${{\mathit{\boldsymbol{\tilde L}}}_m} = \mathit{\boldsymbol{HU}}_1^*{{\mathit{\boldsymbol{\tilde U}}}_1} + \mathit{\boldsymbol{\tilde V}}_2^*{\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{MU}}_2^*{{\mathit{\boldsymbol{\tilde U}}}_1}$ 和 ${{\mathit{\boldsymbol{\tilde M}}}_m} = \mathit{\boldsymbol{HU}}_1^*{{\mathit{\boldsymbol{\tilde U}}}_2} + \mathit{\boldsymbol{\tilde V}}_2^*{\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{MU}}_2^*{{\mathit{\boldsymbol{\tilde U}}}_2}$ ，容易验证

$ {{\mathit{\boldsymbol{\tilde L}}}_m}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_1} + {{\mathit{\boldsymbol{\tilde M}}}_m}\mathit{\boldsymbol{\tilde U}}_2^ * {\mathit{\boldsymbol{U}}_1} - \mathit{\boldsymbol{H}} = 0, $

且

$ {{\mathit{\boldsymbol{\tilde L}}}_m}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_2} + {{\mathit{\boldsymbol{\tilde M}}}_m}\mathit{\boldsymbol{\tilde U}}_2^ * {\mathit{\boldsymbol{U}}_2} - \mathit{\boldsymbol{\tilde V}}_2^ * {\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{M}} = 0. $

因此

$ \begin{array}{l} \mathop {\min }\limits_{{\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} \in \mathit{\boldsymbol{B}}\left\{ {1,3} \right\}} \left\| {{\mathit{\boldsymbol{B}}^{\left( {1,3} \right)}} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| = \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| = \\ \left\| {\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_1} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_1}{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_2}L}&{{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}\mathit{\boldsymbol{\tilde U}}_1^ * {\mathit{\boldsymbol{U}}_2} - \mathit{\boldsymbol{\tilde V}}_1^ * {\mathit{\boldsymbol{V}}_2}\mathit{\boldsymbol{M}}}\\ 0&0 \end{array}} \right)} \right\| = \\ \left\| {\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}}&0\\ 0&0 \end{array}} \right){{\mathit{\boldsymbol{\tilde U}}}^ * }\mathit{\boldsymbol{U}} - \left( {\begin{array}{*{20}{c}} \mathit{\boldsymbol{I}}&0\\ 0&0 \end{array}} \right){{\mathit{\boldsymbol{\tilde V}}}^ * }\mathit{\boldsymbol{V}}\left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}}}&0\\ \mathit{\boldsymbol{L}}&\mathit{\boldsymbol{M}} \end{array}} \right)} \right\| = \\ \left\| {\mathit{\boldsymbol{\tilde V}}\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}}&0\\ 0&0 \end{array}} \right){{\mathit{\boldsymbol{\tilde U}}}^ * } - \mathit{\boldsymbol{\tilde V}}\left( {\begin{array}{*{20}{c}} \mathit{\boldsymbol{I}}&0\\ 0&0 \end{array}} \right){{\mathit{\boldsymbol{\tilde V}}}^ * }\mathit{\boldsymbol{V}}\left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{ \boldsymbol{\varSigma} }}^{ - 1}}}&0\\ \mathit{\boldsymbol{L}}&\mathit{\boldsymbol{M}} \end{array}} \right)\mathit{\boldsymbol{U}}} \right\| = \\ \left\| {{\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\|, \end{array} $

其中，

$ \mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} = \mathit{\boldsymbol{\tilde V}}\left( {\begin{array}{*{20}{c}} {{{\mathit{\boldsymbol{ \boldsymbol{\tilde \varSigma} }}}^{ - 1}}}&0\\ {{{\mathit{\boldsymbol{\tilde L}}}_m}}&{{{\mathit{\boldsymbol{\tilde M}}}_m}} \end{array}} \right){{\mathit{\boldsymbol{\tilde U}}}^ * } = {\mathit{\boldsymbol{B}}^ + } + \left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}} \right){\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}. $

证毕.

注意到当且仅当G^*∈A^*{1, 3}时, G∈A{1, 4}，因此利用上述定理中的结果，容易得到下面的定理.

定理2    假设A, B∈C^m×n.对于任意给定的A^{(1, 4)}∈A{1, 4}，存在唯一的矩阵：

$ \mathit{\boldsymbol{B}}_m^{\left( {1,4} \right)} = {\mathit{\boldsymbol{B}}^ + } + {\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }} \right) \in \mathit{\boldsymbol{B}}\left\{ {1,4} \right\}, $

使得

$ \begin{array}{l} \mathop {\min }\limits_{{\mathit{\boldsymbol{B}}^{\left( {1,4} \right)}} \in \mathit{\boldsymbol{B}}\left\{ {1,4} \right\}} \left\| {{\mathit{\boldsymbol{B}}^{\left( {1,4} \right)}} - {\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}} \right\| = \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,4} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}} \right\| = \\ \;\;\;\;\;\;\left\| {{\mathit{\boldsymbol{B}}^ + } + {\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }} \right\|. \end{array} $

注记1    当定理1和定理2中的酉不变范数取为谱范数或Frobenius范数时，定理1和定理2中的结果就分别变成文献[14]中的theorem 3.1和theorem 3.2.

2 {1, 3}-和{1, 4}-逆的加法和乘法扰动界

利用上节结果，本节将给出{1, 3}-和{1, 4}-逆的加法和乘法扰动界.

定理3    设A, B=A+E∈C^m×n，rank(A)=rank(B)，且满足‖A⁺‖₂‖E‖₂ < 1.对于任意给定的A^{(1, 3)}，若B_m^{(1, 3)}由定理1给出，则下面的估计式成立：

$ \begin{array}{l} \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| \le \left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|_2^2\left\| {\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right)\mathit{\boldsymbol{E}}} \right\| + \\ \;\;\;\;{\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|_2}{\left\| {\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{E}}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| + \left( {{{\left\| \mathit{\boldsymbol{E}} \right\|}_2}\left\| \mathit{\boldsymbol{E}} \right\|} \right). \end{array} $

证明    对于给定的A^{(1, 3)}，由引理1知，存在矩阵Z∈C^n×m，使得A^{(1, 3)}=A⁺+(I-A⁺A)Z.另外，易见

$ \begin{array}{l} {\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{A}}^ + } = {\mathit{\boldsymbol{B}}^ + }\left( {\mathit{\boldsymbol{A}} - \mathit{\boldsymbol{B}}} \right){\mathit{\boldsymbol{A}}^ + } + {\mathit{\boldsymbol{B}}^ + }\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right) - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}} \right){\mathit{\boldsymbol{A}}^ + }. \end{array} $

(2)

故由式(1)、(2)及B=A+E可得

$ \begin{array}{l} \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| = \left\| {{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}\left( {{\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{A}}^ + } - \left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right)\mathit{\boldsymbol{Z}}} \right)} \right\| \le \\ \;\;\;\;\;\;\left\| { - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{A}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{E}}\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right)\mathit{\boldsymbol{Z}}} \right\| + \\ \;\;\;\;\;\;\left\| {{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right)} \right\| = \left\| {{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| + \\ \;\;\;\;\;\;\left\| {{\mathit{\boldsymbol{B}}^ + }{{\left( {{\mathit{\boldsymbol{B}}^ + }} \right)}^ * }{\mathit{\boldsymbol{E}}^ * }\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right)} \right\| \le \\ \;\;\;\;\;\;{\left\| {{\mathit{\boldsymbol{B}}^ + }} \right\|_2}{\left\| {\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{E}}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| + \\ \;\;\;\;\;\;\left\| {{\mathit{\boldsymbol{B}}^ + }} \right\|_2^2\left\| {\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right)\mathit{\boldsymbol{E}}} \right\|. \end{array} $

另外, 由定理的条件知:

$ {\left\| {{\mathit{\boldsymbol{B}}^ + }} \right\|_2} \le \frac{{{{\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|}_2}}}{{1 - {{\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|}_2}{{\left\| \mathit{\boldsymbol{E}} \right\|}_2}}}. $

从而可推得结论成立.证毕.

定理4    假设A∈C^m×n，B=D₁^*AD₂，B_m^{(1, 3)}由定理1给出，对于任意给定的A^{(1, 3)}，如果max{‖I-D₁‖₂, ‖I-D₂‖₂} < 1，则下面的估计式成立：

$ \begin{array}{*{20}{c}} {\left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,3} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| \le {{\left\| {1 - {t_1}\mathit{\boldsymbol{D}}_2^{ - 1}} \right\|}_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| + }\\ {\frac{{{{\left\| {1 - \overline {{t_1}} {\mathit{\boldsymbol{D}}_1}} \right\|}_2} + {{\left\| {1 - {t_2}\mathit{\boldsymbol{D}}_1^{ - 1}} \right\|}_2}}}{{\mathit{\Phi }\left( {{\mathit{\boldsymbol{D}}_1},{\mathit{\boldsymbol{D}}_2}} \right)}}\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|,} \end{array} $

其中，Φ(D₁, D₂)=(1-‖I-D₁‖₂)(1-‖I-D₂‖₂)，且t₁和t₂为任意复数.

证明    由B=D₁^*AD₂，可得

$ \mathit{\boldsymbol{BD}}_2^{ - 1} = \mathit{\boldsymbol{D}}_1^ * \mathit{\boldsymbol{A}},\mathit{\boldsymbol{D}}_1^{ - * }\mathit{\boldsymbol{B}} = \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{D}}_2}, $

(3)

其中D₁^-*表示D₁逆矩阵的共轭转置矩阵.利用式(3)可证得

$ {\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{BD}}_2^{ - 1}{\mathit{\boldsymbol{A}}^ + }\mathit{\boldsymbol{D}}_1^{ - * }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^ + }. $

当max{‖I-D₁‖₂, ‖I-D₂‖₂} < 1时，有

$ \begin{array}{l} \left\| {{\mathit{\boldsymbol{B}}^ + }} \right\| \le {\left\| {\mathit{\boldsymbol{D}}_2^{ - 1}} \right\|_2}{\left\| {\mathit{\boldsymbol{D}}_1^{ - 1}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\| = \\ \;\;\;\;{\left\| {{{\left( {\mathit{\boldsymbol{I}} - \left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{D}}_2}} \right)} \right)}^{ - 1}}} \right\|_2}{\left\| {{{\left( {\mathit{\boldsymbol{I}} - \left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{D}}_1}} \right)} \right)}^{ - 1}}} \right\|_2} \times \\ \;\;\;\;\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\| \le \frac{{\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|}}{{\mathit{\Phi }\left( {{\mathit{\boldsymbol{D}}_1},{\mathit{\boldsymbol{D}}_2}} \right)}}. \end{array} $

(4)

对于给定的A^{(1, 3)}，由引理1知，存在矩阵Z∈C^n×m，使得A^{(1, 3)}=A⁺+(I-A⁺A)Z.

综合式(2)和(3)，得到

$ \begin{array}{l} {\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}} = {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}\left( {{\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{A}}^ + } - \left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{A}}^ + }\mathit{\boldsymbol{A}}} \right)\mathit{\boldsymbol{Z}}} \right) = \\ \;\;\;\;\;\;\;{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}\left( {{\mathit{\boldsymbol{B}}^ + }\left( {\mathit{\boldsymbol{I}} - {t_1}\mathit{\boldsymbol{D}}_1^ * } \right)\mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}\left( {\mathit{\boldsymbol{I}} - {t_1}\mathit{\boldsymbol{D}}_2^{ - 1}} \right){\mathit{\boldsymbol{A}}^ + } + } \right.\\ \;\;\;\;\;\;\;\left. {{\mathit{\boldsymbol{B}}^ + }\left( {\mathit{\boldsymbol{I}} - {t_2}\mathit{\boldsymbol{D}}_1^{ - 1}} \right)\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^ + }} \right) - \left( {\mathit{\boldsymbol{I}} - {t_1}\mathit{\boldsymbol{D}}_2^{ - 1}} \right)\left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{A}}^ + }\mathit{\boldsymbol{A}}} \right)\mathit{\boldsymbol{Z}}} \right), \end{array} $

由上式和式(4), 可得

$ \begin{array}{l} \left\| {{\mathit{\boldsymbol{B}}^ + } - {\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| \le \\ \left( {{{\left\| {\mathit{\boldsymbol{I}} - {t_1}\mathit{\boldsymbol{D}}_1^ * } \right\|}_2} + {{\left\| {\mathit{\boldsymbol{I}} - {t_2}\mathit{\boldsymbol{D}}_1^{ - 1}} \right\|}_2}} \right)\left\| {{\mathit{\boldsymbol{B}}^ + }} \right\| + \\ {\left\| {\mathit{\boldsymbol{I}} - {t_1}\mathit{\boldsymbol{D}}_2^{ - 1}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| \le {\left\| {\mathit{\boldsymbol{I}} - {t_1}\mathit{\boldsymbol{D}}_2^{ - 1}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,3} \right)}}} \right\| + \\ \frac{{{{\left\| {\mathit{\boldsymbol{I}} - \overline {{t_1}} {\mathit{\boldsymbol{D}}_1}} \right\|}_2} + {{\left\| {\mathit{\boldsymbol{I}} - {t_2}\mathit{\boldsymbol{D}}_1^{ - 1}} \right\|}_2}}}{{\mathit{\Phi }\left( {{\mathit{\boldsymbol{D}}_1},{\mathit{\boldsymbol{D}}_2}} \right)}}\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|. \end{array} $

证毕.

再次利用G∈A{1, 4}当且仅当G^*∈A^*{1, 3}，由定理3和定理4即得{1, 4}-逆的加法和乘法扰动界：

定理5    设A, B=A+E∈C^m×n，rank(A)=rank(B)且满足‖A⁺‖₂‖E‖₂ < 1.对于任意给定的A^{(1, 4)}，若B_m^{(1, 4)}由定理2给出，则下面的估计式成立：

$ \begin{array}{l} \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,4} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}} \right\| \le \left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|_2^2\left\| {\mathit{\boldsymbol{E}}\left( {\mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{A}}^ + }\mathit{\boldsymbol{A}}} \right)} \right\| + \\ \;\;\;\;\;{\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|_2}{\left\| {\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{B}}^ + }\mathit{\boldsymbol{B}}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}} \right\| + O\left( {{{\left\| \mathit{\boldsymbol{E}} \right\|}_2}\left\| \mathit{\boldsymbol{E}} \right\|} \right). \end{array} $

定理6    假设A∈C^m×n，B=D₁^*AD₂, B_m^{(1, 4)}由定理2给出, 对于任意给定的A^{(1, 4)}，若max{‖I-D₁‖₂, ‖I-D₂‖₂} < 1，则有

$ \begin{array}{l} \left\| {\mathit{\boldsymbol{B}}_m^{\left( {1,4} \right)} - {\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}} \right\| \le {\left\| {1 - \overline {{t_1}} \mathit{\boldsymbol{D}}_1^{ - 1}} \right\|_2}\left\| {{\mathit{\boldsymbol{A}}^{\left( {1,4} \right)}}} \right\| + \\ \;\;\;\;\;\;\frac{{{{\left\| {\mathit{\boldsymbol{I}} - {t_1}{\mathit{\boldsymbol{D}}_2}} \right\|}_2} + {{\left\| {\mathit{\boldsymbol{I}} - \overline {{t_2}} \mathit{\boldsymbol{D}}_2^{ - 1}} \right\|}_2}}}{{\mathit{\Phi }\left( {{\mathit{\boldsymbol{D}}_1},{\mathit{\boldsymbol{D}}_2}} \right)}}\left\| {{\mathit{\boldsymbol{A}}^ + }} \right\|, \end{array} $

其中，Φ(D₁, D₂)=(1-‖I-D₁‖₂)(1-‖I-D₂‖₂)，t₁和t₂为任意复数.

注记2    当定理3和定理5中的酉不变范数取为谱范数或Frobenius范数时，定理3和定理5就是文献[14]中的theorem 4.1；当定理4和定理6中的酉不变范数取为谱范数或Frobenius范数且t₁=t₂=1时，定理4和定理6就变为文献[14]中的theorem 5.2.另外，注意到定理4和定理6给出的扰动界对任意复数t₁和t₂都成立，从而改进了文献[14]的结果.