在代数系统中, 一个具有形式$ \widehat{a}=a+\varepsilon a_0 $的元素称为对偶元素, 其中$ a $和$ a_0 $分别称为$ \widehat{a} $的实部和对偶部. $ \varepsilon $称为对偶元, 满足法则$ \varepsilon \neq 0 $, $ 0\varepsilon=\varepsilon0=0 $, $ 1\varepsilon=\varepsilon1=\varepsilon $和$ {\varepsilon}^2=0 $. 对偶数的概念最早由Clifford在1873年提出, 而对偶数的应用最早来源于Study^[1]在1901年用对偶数来表示空间几何学中的对偶角. 在过去几十年里, 对偶数被广泛应用于多个领域, 如运动学分析^[2]、机器人^[3]、刚体运动分析^[4]等.

作为对偶数概念的推广, 元素为对偶数的矩阵称为对偶矩阵. 令$ A $和$ A_0 $是两个$ m\times n $的实矩阵, 则$ \widehat{A}=A+\varepsilon A_0 $称为$ m\times n $的对偶矩阵, 且$ A $和$ A_0 $分别称为$ \widehat{A} $的实部和对偶部. 由于在许多运动学与机械系统领域的问题中需要求解线性对偶方程组, 与复矩阵和实矩阵类似的是, 对偶广义逆是研究对偶线性方程组的解与最小二乘解的重要工具, 所以对偶矩阵及其对偶广义逆广泛应用于这些领域. 然而, 对偶矩阵的对偶广义逆的许多重要性质不同于复矩阵和实矩阵的广义逆, 例如, 复矩阵的Moore-Penrose逆总是存在的, 但对偶矩阵的对偶Moore-Penrose逆却有可能不存在. 对偶广义逆的存在性对于运动学与机械系统领域中产生的对偶线性方程组的数值解问题有重要影响, 例如, 一个线性对偶方程组通常需要分裂成实部和对偶部, 且实部和对偶部所对应的两个实系数线性方程组需要独立计算, 但若该线性对偶方程组的系数对偶矩阵的对偶广义逆存在, 则计算过程可在一步完成, 从而可以提高计算的效率. 基于此原因, 讨论对偶矩阵的对偶广义逆的存在性及表示不仅具有重要理论意义, 同时在运动学和机械系统等领域也有应用价值.

对偶矩阵的对偶广义逆是近年来的一个研究热点, 关于各种对偶广义逆的存在性与表示有大量研究, 如文献[5-8]研究了对偶Moore-Penrose逆的存在性与表示, 文献[9]研究了对偶群逆的存在性与表示, 文献[10]介绍了对偶指标的概念, 并研究了对偶核逆的存在性与表示. 本文将研究长方对偶矩阵的加权对偶群逆的存在性与表示, 给出长方对偶矩阵的加权对偶群逆存在的充分必要条件, 并在加权对偶群逆存在时给出其显式表示, 同时推广对偶群逆的相关结论.

需要指出的是, 复矩阵的加权群逆已有相关研究. 2007年, 岑建苗^[11]给出了长方矩阵加权群逆的定义, 研究了长方矩阵加权群逆的存在性, 给出了长方矩阵加权群逆存在的一些充分必要条件以及计算公式. 之后长方矩阵的加权群逆得到关注和研究, 陈永林^[12]给出了加权群逆一些新的存在条件与表示, 并研究了Cline和Greville定义的加权群逆的存在性与表示. 武玲玲等^[13]研究了长方矩阵加权群逆的扰动. 胡春梅等^[14]给出了计算长方矩阵加权群逆的三种迭代法. 胡振英等^[15]讨论了加权群逆的有限算法. 此外, 长方矩阵加权群逆的相关结论也被推广到Banach空间有界线性算子上(见文献[16-18]).

记$ {\mathbb{C}}^{m\times n} $和$ {\mathbb{R}}^{m\times n} $分别为所有$ m\times n $复矩阵和实矩阵组成的集合, $ {\mathbb{D}}^{m\times n} $为所有$ m\times n $对偶矩阵组成的集合. 对矩阵$ A\in {\mathbb{C}}^{m\times n} $, $ r(A) $表示$ A $的秩. $ n $阶方阵$ A $的指标，记为Ind$ (A) $, 是指满足$ r(A^k )=r(A^{k+1}) $的最小非负整数$ k $. $ I_n $表示$ n $阶单位矩阵, 在不引起混淆的情况下简记为$ I $.

下面给出加权群逆的定义.

定义2.1^[11]  令$ A\in {\mathbb{C}}^{m\times n} $, $ W\in {\mathbb{C}}^{n\times m} $. 如果存在$ X\in {\mathbb{C}}^{m\times n} $使得

$ \begin{eqnarray*} AWXWA=A, \; \; XWAWX=X, \; \; AWX=XWA, \end{eqnarray*} $

则称$ X $为$ A $的加$ W $权群逆, 记为$ A_W^{\#} $. 若$ A\in {\mathbb{C}}^{n\times n} $且$ W=I_n $, 则$ X $即为$ A $的群逆$ A^{\#} $.

类似的, 可以给出长方对偶矩阵的加权对偶群逆的定义.

定义2.2   令$ \widehat{A}\in {\mathbb{D}}^{m\times n} $, $ \widehat{W}\in {\mathbb{D}}^{n\times m} $. 如果存在$ \widehat{X}\in {\mathbb{D}}^{m\times n} $使得

$ \begin{eqnarray*} \widehat{A}\widehat{W}\widehat{X}\widehat{W}\widehat{A}=\widehat{A}, \; \; \widehat{X}\widehat{W}\widehat{A}\widehat{W}\widehat{X}=\widehat{X}, \; \; \widehat{A}\widehat{W}\widehat{X}=\widehat{X}\widehat{W}\widehat{A}, \end{eqnarray*} $

则称$ \widehat{X} $为$ \widehat{A} $的加$ \widehat{W} $权对偶群逆, 记为$ \widehat{A}_{\widehat{W}}^{\#} $. 若$ \widehat{A}\in {\mathbb{D}}^{n\times n} $且$ \widehat{W}=I_n $, 则$ \widehat{X} $即为$ \widehat{A} $的对偶群逆$ {\widehat{A}}^{\#} $.

引理2.1^[12]  设$ A\in {\mathbb{C}}^{m\times n} $, $ W\in {\mathbb{C}}^{n\times m} $. 则$ A $的加$ W $权群逆$ A_W^{\#} $存在当且仅当$ AW $和$ WA $的群逆均存在并满足$ r(AW)=r(WA)=r(A) $. 若$ A_W^{\#} $存在, 则它是唯一的, 并且可表示为$ A_W^{\#}=(AW)^{\#}A(WA)^{\#}=A[(WA)^{\#}]^2=[(AW)^{\#}]^2A. $

引理2.2  设矩阵$ A\in {\mathbb{C}}^{m\times n} $, $ W\in {\mathbb{C}}^{n\times m} $, $ r(A)=r $. 若$ A_W^{\#} $存在，则存在可逆矩阵$ P\in {\mathbb{C}}^{m\times m} $和$ Q\in {\mathbb{C}}^{n\times n} $使得

$ \begin{eqnarray} A=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}, \; \; \; W=Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}, \end{eqnarray} $

(2.1)

其中$ A_1 $和$ W_1 $均为$ r $阶可逆矩阵.

此时, $ A_W^{\#} $具有表示

$ \begin{eqnarray} A_W^{\#}=P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}. \end{eqnarray} $

(2.2)

证  若$ A_W^{\#} $存在, 则由引理2.1, $ AW $和$ WA $的群逆均存在, 且由于$ AW $和$ WA $有相同的秩和非零特征值, 从而由$ AW $和$ WA $的若尔当标准形可知存在可逆矩阵$ P\in {\mathbb{C}}^{m\times m} $和$ Q\in {\mathbb{C}}^{n\times n} $使得$ AW=P\left[\begin{array}{cc}C&0\\0&0\end{array}\right]P^{-1}, \; \; \; WA=Q\left[\begin{array}{cc}D&0\\0&0\end{array}\right]Q^{-1}, $其中$ C $和$ D $均为$ r $阶可逆矩阵.

不妨将$ A $和$ W $分别表示为$ A=P\left[\begin{array}{cc}A_1&A_2\\A_3&A_4\end{array}\right]Q^{-1}, \; \; \; W=Q\left[\begin{array}{cc}W_1&W_2\\W_3&W_4\end{array}\right]P^{-1}, $其中$ A_1 $和$ W_1 $均为$ r $阶方阵.

由$ (AW)A=A(WA) $可得

$ \begin{eqnarray*} P\left[\begin{array}{cc}CA_1&CA_2\\0&0\end{array}\right]Q^{-1}=P\left[\begin{array}{cc}A_1D&0\\A_3D&0\end{array}\right]Q^{-1}, \end{eqnarray*} $

于是$ CA_2=0 $, $ A_3D=0 $. 因为$ C $和$ D $均为可逆矩阵, 所以$ A_2=0 $, $ A_3=0 $.

类似的, 由$ (WA)W=W(AW) $可得$ W_2=0 $和$ W_3=0 $. 于是由

$ \begin{eqnarray*} AW=P\left[\begin{array}{cc}A_1W_1&0\\0&A_4W_4\end{array}\right]P^{-1}=P\left[\begin{array}{cc}C&0\\0&0\end{array}\right]P^{-1} \end{eqnarray*} $

可得$ A_1 W_1=C $. 由于$ C $为可逆矩阵, 所以$ A_1 $和$ W_1 $均为$ r $阶可逆矩阵.

此外, 由于$ r=r(A)=r(A_1)+r(A_4) $且$ r(A_1)=r $, 所以$ A_4=0 $. 这就证明了$ A $和$ W $具有(2.1)中的表示.

若$ A_W^{\#} $存在, 则由引理2.1,

$ \begin{eqnarray*} A_W^{\#}&=&(AW)^{\#}A(WA)^{\#}\\&=&P\left[\begin{array}{cc}C^{-1}&0\\0&0\end{array}\right]P^{-1}P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1} Q\left[\begin{array}{cc}D^{-1}&0\\0&0\end{array}\right]Q^{-1}\\&=&P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}. \end{eqnarray*} $

引理2.3^[19]  令$ A\in {\mathbb{C}}^{m\times n} $, $ B\in {\mathbb{C}}^{m\times m} $, $ C\in {\mathbb{C}}^{n\times n} $, 其中$ B $和$ C $的指标均为1, 则

$ \begin{eqnarray*} r\left[\begin{array}{cc}A&B\\C&0\end{array}\right]=r(B)+r(C)+r[(I-BB^{\#})A(I-CC^{\#})]. \end{eqnarray*} $

本节研究长方对偶矩阵的加权对偶群逆的存在性与表示, 给出加权对偶群逆存在的充分必要条件, 并在加权对偶群逆存在时给出其表示.

首先给出一个对偶矩阵是某个长方对偶矩阵的加权对偶群逆的充分必要条件.

引理3.1   令$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{m\times n} $, $ \widehat{W}=W+\varepsilon W_0\in {\mathbb{D}}^{n\times m} $, 其中$ A, \; A_0\in {\mathbb{R}}^{m\times n} $, $ W, \; W_0\in {\mathbb{R}}^{n\times m} $. 则对偶矩阵$ \widehat{G}=G+\varepsilon G_0\in {\mathbb{D}}^{m\times n} $是$ \widehat{A} $的加$ \widehat{W} $权对偶群逆当且仅当$ G=A_W^{\#} $且

$ \begin{eqnarray} AWGWA_0+AWGW_0A+AWG_0WA+AW_0GWA+A_0WGWA=A_0, \end{eqnarray} $

(3.1)

$ \begin{eqnarray} GWAWG_0+GWAW_0G+GWA_0WG+GW_0 AWG+G_0WAWG=G_0, \end{eqnarray} $

(3.2)

$ \begin{eqnarray} AWG_0+AW_0G+A_0WG=GWA_0+GW_0A+G_0WA. \end{eqnarray} $

(3.3)

证  由$ \widehat{A}=A+\varepsilon A_0 $, $ \widehat{W}=W+\varepsilon W_0 $, $ \widehat{G}=G+\varepsilon G_0 $可得

$ \begin{eqnarray*} \widehat{A}\widehat{W}\widehat{G}\widehat{W}\widehat{A}=AWGWA+\varepsilon(AWGWA_0+AWGW_0A+AWG_0WA+AW_0 GWA+A_0WGWA), \\ \widehat{G}\widehat{W}\widehat{A}\widehat{W}\widehat{G}=GWAWG+\varepsilon(GWAWG_0+GWAW_0G+GWA_0WG+GW_0 AWG+G_0WAWG), \\ \widehat{A}\widehat{W}\widehat{G}=AWG+\varepsilon(AWG_0+AW_0 G+A_0WG), \; \; \widehat{G}\widehat{W}\widehat{A}=GWA+\varepsilon(GWA_0+GW_0 A+G_0WA). \end{eqnarray*} $

若$ \widehat{G}=G+\varepsilon G_0 $是$ \widehat{A}=A+\varepsilon A_0 $的加$ \widehat{W} $权对偶群逆, 则由定义2.2可得$ \widehat{A}\widehat{W}\widehat{G}\widehat{W}\widehat{A}=\widehat{A} $, $ \widehat{G}\widehat{W}\widehat{A}\widehat{W}\widehat{G}=\widehat{G} $, $ \widehat{A}\widehat{W}\widehat{G}=\widehat{G}\widehat{W}\widehat{A} $. 比较这三个等式的实部和对偶部可得

$ \begin{eqnarray*} AWGWA=A, \; GWAWG=G, \; AWG=GWA\quad \text{和}\qquad\qquad\qquad\qquad \\ AWGWA_0+AWGW_0 A+AWG_0 WA+AW_0 GWA+A_0 WGWA=A_0, \\ GWAWG_0+GWAW_0 G+GWA_0 WG+GW_0 AWG+G_0 WAWG=G_0, \\ AWG_0+AW_0 G+A_0 WG= GWA_0+GW_0 A+G_0 WA. \end{eqnarray*} $

于是由定义2.1可知$ A $的加$ W $权群逆存在且$ G=A_W^{\#} $. 此外, 等式(3.1)–(3.3)也成立.

反之, 若$ A $的加$ W $权群逆存在且$ G=A_W^{\#} $, 等式(3.1)–(3.3)也成立, 则由定义2.2直接可得$ \widehat{G}=G+\varepsilon G_0\in {\mathbb{D}}^{m\times n} $是$ \widehat{A}=A+\varepsilon A_0 $的加$ \widehat{W} $权对偶群逆, 证毕.

文献[11]说明若$ A $的加$ W $权群逆存在, 则一定唯一. 下面证明该结论也适用于加权对偶群逆.

定理3.1   设$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{m\times n} $, $ \widehat{W}=W+\varepsilon W_0\in {\mathbb{D}}^{n\times m} $, 其中$ A, \; A_0\in {\mathbb{R}}^{m\times n} $, $ W, \; W_0\in {\mathbb{R}}^{n\times m} $. 若$ \widehat{A} $的加$ \widehat{W} $权对偶群逆存在, 则是唯一的.

证  由引理3.1, 若$ \widehat{A} $的加$ \widehat{W} $权对偶群逆存在, 则$ \widehat{A}_{\widehat{W}}^{\#} $具有形式$ \widehat{A}_{\widehat{W}}^{\#}=A_W^{\#}+\varepsilon R $. 假设$ A_W^{\#}+\varepsilon R_1 $和$ A_W^{\#}+\varepsilon R_2 $是$ \widehat{A} $的两个加$ \widehat{W} $权对偶群逆, 下面证明$ R_1=R_2 $.

由(3.1)式可得

$ \begin{eqnarray*} AWA_W^{\#} WA_0+AWA_W^{\#} W_0 A+AWR_1 WA+AW_0 A_W^{\#} WA+A_0 WA_W^{\#} WA=A_0, \\ AWA_W^{\#}WA_0+AWA_W^{\#} W_0 A+AWR_2 WA+AW_0 A_W^{\#} WA+A_0 WA_W^{\#}WA=A_0. \end{eqnarray*} $

两式相减可得

$ \begin{eqnarray} AW(R_1-R_2)WA=0. \end{eqnarray} $

(3.4)

同样的，由(3.2)式可得

$ \begin{eqnarray*} A_W^{\#} WAWR_1+A_W^{\#} WAW_0 A_W^{\#}+A_W^{\#} WA_0 WA_W^{\#}+A_W^{\#} W_0 AWA_W^{\#}+R_1 WAWA_W^{\#}=R_1, \\ A_W^{\#} WAWR_2+A_W^{\#} WAW_0 A_W^{\#}+A_W^{\#} WA_0 WA_W^{\#}+A_W^{\#} W_0 AWA_W^{\#}+R_2 WAWA_W^{\#}=R_2. \end{eqnarray*} $

两式相减可得

$ \begin{eqnarray} A_W^{\#} WAW(R_1-R_2 )+(R_1-R_2)WAWA_W^{\#}=R_1-R_2. \end{eqnarray} $

(3.5)

最后, 由(3.3)式可得

$ \begin{eqnarray*} AWR_1+AW_0 A_W^{\#}+A_0 WA_W^{\#}=A_W^{\#} WA_0+A_W^{\#} W_0 A+R_1 WA, \\ AWR_2+AW_0 A_W^{\#}+A_0 WA_W^{\#}=A_W^{\#} WA_0+A_W^{\#} W_0 A+R_2 WA. \end{eqnarray*} $

两式相减可得

$ \begin{eqnarray} AW(R_1-R_2 )=(R_1-R_2 )WA. \end{eqnarray} $

(3.6)

在(3.4)式右边乘以$ WA_W^{\#} $并利用(3.6)式可得

$ \begin{eqnarray*} AW(R_1-R_2 )WAWA_W^{\#}=(R_1-R_2)WAWAWA_W^{\#}=(R_1-R_2 )WAWA_W^{\#}WA=(R_1-R_2)WA=0. \end{eqnarray*} $

再将$ AW(R_1-R_2)=(R_1-R_2)WA=0 $代入(3.5)式即可得$ R_1-R_2=0 $, 证毕.

下面给出矩阵$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{m\times n} $的加$ \widehat{W} $权对偶群逆存在的一些充分必要条件, 并在$ \widehat{A}_{\widehat{W}}^{\#} $存在时给出其显示表示.

定理3.2   设$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{m\times n} $, $ \widehat{W}=W+\varepsilon W_0\in {\mathbb{D}}^{n\times m} $, 其中$ A, \; A_0\in {\mathbb{R}}^{m\times n} $, $ W, \; W_0\in {\mathbb{R}}^{n\times m} $. 则下面命题等价:

(ⅰ) $ \widehat{A}_{\widehat{W}}^{\#} $存在;

(ⅱ) $ A_W^{\#} $存在, 且$ \widehat{A}=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1} $, $ \widehat{W}=Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}+\varepsilon Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1} $, 其中$ P, \; Q, \; A_1, \; W_1 $均为可逆矩阵;

(ⅲ) $ A_W^{\#} $存在且$ (I-A_W^{\#}WAW)A_0(I-WAWA_W^{\#})=0 $;

(ⅳ) $ A_W^{\#} $存在且$ r\left[\begin{array}{cc}A_0&AW\\WA&0\end{array}\right]=2r(A) $.

此外, 若$ \widehat{A}_{\widehat{W}}^{\#} $存在, 则

$ \begin{eqnarray} \widehat{A}_{\widehat{W}}^{\#}=A_W^{\#}+\varepsilon R, \end{eqnarray} $

(3.7)

其中

$ \begin{eqnarray} R&=&[(AW)^{\#}]^2A_0(I-WAWA_W^{\#})+(I-A_W^{\#}WAW)A_0[(WA)^{\#}]^2 \\ &&-A_W^{\#}WAW_0A_W^{\#}-A_W^{\#}W_0AWA_W^{\#}-(AW)^{\#}A_0(WA)^{\#}. \end{eqnarray} $

(3.8)

证  (ⅰ)$ \Rightarrow $(ⅱ): 若$ \widehat{A}_{\widehat{W}}^{\#} $存在, 则由引理3.1, $ \widehat{A}_{\widehat{W}}^{\#} $的实部为$ A_W^{\#} $, 即$ A_W^{\#} $存在. 不妨设$ \widehat{A}_{\widehat{W}}^{\#}=A_W^{\#}+\varepsilon R $.

因为$ A_W^{\#} $存在, 从而由引理2.2可知$ A $和$ W $具有(2.1)中的表示, $ A_W^{\#} $具有(2.2)中的表示. 令$ A_0=P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&{\tilde{A}}_4\end{array}\right]Q^{-1}, \; W_0=Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1}, \; R=P\left[\begin{array}{cc}R_1&R_2\\R_3&R_4\end{array}\right]Q^{-1}. $则由(3.1)式有$ AWA_W^{\#}WA_0+AWA_W^{\#}W_0A+AWRWA+AW_0A_W^{\#}WA+A_0WA_W^{\#}WA=A_0 $, 即

$ \begin{eqnarray*} &&P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\0&0\end{array}\right]Q^{-1} +P\left[\begin{array}{cc}W_1^{-1}{\tilde{W}}_1A_1&0\\0&0\end{array}\right]Q^{-1} +P\left[\begin{array}{cc}A_1W_1R_1W_1A_1&0\\0&0\end{array}\right]Q^{-1}\\ &&+P\left[\begin{array}{cc}A_1{\tilde{W}}_1W_1^{-1}&0\\0&0\end{array}\right]Q^{-1} +P\left[\begin{array}{cc}{\tilde{A}}_1&0\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}\\ &=&P\left[\begin{array}{cc}2{\tilde{A}}_1+W_1^{-1}{\tilde{W}}_1A_1+A_1W_1R_1W_1A_1+A_1{\tilde{W}}_1W_1^{-1}&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}\\ &=&P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&{\tilde{A}}_4\end{array}\right]Q^{-1}. \end{eqnarray*} $

由上式可得$ {\tilde{A}}_4=0 $. 所以,

$ \begin{eqnarray*} \widehat{A}=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}, \widehat{W}=Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}+\varepsilon Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1}, \end{eqnarray*} $

其中$ P, Q, A_1, W_1 $均为可逆矩阵.

(ⅱ) $ \Rightarrow $(ⅲ):

若$ \widehat{A}=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}, \widehat{W}=Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}+\varepsilon Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1} $, 其中$ P, \; Q, \; A_1, \; W_1 $均为可逆矩阵, 则

$ \begin{eqnarray*} (I-A_W^{\#}WAW)A_0(I-WAWA_W^{\#})=P\left[\begin{array}{cc}0&0\\0&I\end{array}\right]P^{-1} P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1} Q\left[\begin{array}{cc}0&0\\0&I\end{array}\right]Q^{-1}=0. \end{eqnarray*} $

(ⅲ) $ \Rightarrow $(ⅰ): 若$ A_W^{\#} $存在, 由引理2.2, $ A $和$ W $分别具有(2.1)中的表示, $ A_W^{\#} $具有(2.2)中的表示. 令$ \widehat{A}=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&{\tilde{A}}_4\end{array}\right]Q^{-1}, \widehat{W}=Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}+\varepsilon Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1} $, 其中$ P, Q, A_1, W_1 $均为可逆矩阵. 则由$ (I-A_W^{\#}WAW)A_0(I-WAWA_W^{\#})=0 $不难得到$ {\tilde{A}}_4=0 $.

令

$ \begin{eqnarray*} \widehat{G}&=&P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\\&+&\varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}. \end{eqnarray*} $

则直接计算可得

$ \begin{eqnarray*} \widehat{A}\widehat{W}\widehat{G}&=&\left(P\left[\begin{array}{cc}A_1W_1&0\\0&0\end{array}\right]P^{-1}+\varepsilon P\left[\begin{array}{cc} A_1{\tilde{W}}_1+{\tilde{A}}_1W_1&A_1{\tilde{W}}_2+{\tilde{A}}_2W_4\\{\tilde{A}}_3W_1&0\end{array}\right]P^{-1}\right)\\ &\times& \left(P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\right.\\ &+&\left.\varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}\right)\\ &=&P\left[\begin{array}{cc}W_1^{-1}&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}-W_1^{-1}{\tilde{W}}_1W_1^{-1}&(A_1W_1)^{-1} {\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-1}&0\end{array}\right]Q^{-1}, \end{eqnarray*} $

$ \begin{eqnarray*} \widehat{G}\widehat{W}\widehat{A}&=&\left(P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\right.\\ &+&\left.\varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}\right)\\ &\times& \left(Q\left[\begin{array}{cc}W_1A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon Q\left[\begin{array}{cc} W_1{\tilde{A}}_1+{\tilde{W}}_1A_1&W_1{\tilde{A}}_2\\{\tilde{W}}_3A_1+W_4{\tilde{A}}_3&0\end{array}\right]Q^{-1}\right)\\ &=&P\left[\begin{array}{cc}W_1^{-1}&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}-W_1^{-1}{\tilde{W}}_1W_1^{-1}&(A_1W_1)^{-1} {\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-1}&0\end{array}\right]Q^{-1}. \end{eqnarray*} $

由此可见, $ \widehat{A}\widehat{W}\widehat{G}=\widehat{G}\widehat{W}\widehat{A} $.

此外,

$ \begin{eqnarray*} \widehat{A}\widehat{W}\widehat{G}\widehat{W}\widehat{A}&=&\left(P\left[\begin{array}{cc}W_1^{-1}&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}-W_1^{-1}{\tilde{W}}_1W_1^{-1}&(A_1W_1)^{-1} {\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-1}&0\end{array}\right]Q^{-1}\right)\\ &\times& \left(Q\left[\begin{array}{cc}W_1A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon Q\left[\begin{array}{cc} W_1{\tilde{A}}_1+{\tilde{W}}_1A_1&W_1{\tilde{A}}_2\\{\tilde{W}}_3A_1+W_4{\tilde{A}}_3&0\end{array}\right]Q^{-1}\right)\\ &=&P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}\\ &=&\widehat{A}, \end{eqnarray*} $

$ \begin{eqnarray*} &&\widehat{G}\widehat{W}\widehat{A}\widehat{W}\widehat{G}\\&=&\left(P\left[\begin{array}{cc}W_1^{-1}&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}-W_1^{-1}{\tilde{W}}_1W_1^{-1}&(A_1W_1)^{-1} {\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-1}&0\end{array}\right]Q^{-1}\right)\\ &&\times \left(Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}+\varepsilon Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1}\right) \times \left(P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\right.\\ &&+ \left.\varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}\right) \\ &=&\left(P\left[\begin{array}{cc}I&0\\0&0\end{array}\right]P^{-1}+\varepsilon P\left[\begin{array}{cc}0&W_1^{-1}{\tilde{W}}_2+(A_1W_1)^{-1}{\tilde{A}}_2 W_4\\{\tilde{A}}_3A_1^{-1}&0\end{array}\right]P^{-1}\right) \times \left(P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\right.\\ &&+ \left.\varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}\right) \end{eqnarray*} $

$ \begin{eqnarray*} &=& P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\\ &&+ \varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}\\ &=&\widehat{G}. \end{eqnarray*} $

由定义2.2, $ \widehat{A} $的加$ \widehat{W} $权对偶群逆存在且$ \widehat{A}_{\widehat{W}}^{\#}=\widehat{G} $.

又因为

$ \begin{eqnarray*} A_W^{\#}WAW_0A^{\#}=P\left[\begin{array}{cc}W_1^{-1}{\tilde{W}}_1(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}, A_W^{\#}W_0AWA_W^{\#}=P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}{\tilde{W}}_1W_1^{-1}&0\\0&0\end{array}\right]Q^{-1}, \end{eqnarray*} $

$ \begin{eqnarray*} (AW)^{\#}A_0(WA)^{\#}&=&P\left[\begin{array}{cc}(A_1W_1)^{-1}{\tilde{A}}_1(W_1A_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}, \end{eqnarray*} $

$ \begin{eqnarray*} [(AW)^{\#}]^2A_0(I-WAWA_W^{\#})&=&P\left[\begin{array}{cc}(A_1 W_1)^{-2}&0\\0&0\end{array}\right]P^{-1} P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1} Q\left[\begin{array}{cc}0&0\\0&I\end{array}\right]Q^{-1}\\ &=&P\left[\begin{array}{cc}0&(A_1W_1)^{-2}{\tilde{A}}_2\\0&0\end{array}\right]Q^{-1}, \\ (I-A_W^{\#}WAW)A_0[(WA)^{\#}]^2&=&P\left[\begin{array}{cc}0&0\\0&I\end{array}\right]P^{-1}P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1} Q\left[\begin{array}{cc}(W_1A_1)^{-2}&0\\0&0\end{array}\right]Q^{-1}\\ &=&P\left[\begin{array}{cc}0&0\\{\tilde{A}}_3(W_1A_1)^{-2}&0\end{array}\right]Q^{-1}. \end{eqnarray*} $

所以, $ \widehat{A}_{\widehat{W}}^{\#}=\widehat{G}=A_W^{\#}+ \varepsilon\{[(AW)^{\#}]^2A_0(I-WAWA_W^{\#})+(I-A_W^{\#}WAW)A_0[(WA)^{\#}]^2 -A_W^{\#}WAW_0A_W^{\#}-A_W^{\#}W_0AWA_W^{\#}-(AW)^{\#}A_0(WA)^{\#}\} $, 这就证明了(3.7)和(3.8)是成立的.

(ⅲ) $ \Leftrightarrow $(iv): 若$ A_W^{\#} $存在, 则由引理2.1, $ AW $和$ WA $的群逆均存在且$ r(AW)=r(WA)=r(A) $. 再由引理2.3可得

$ \begin{eqnarray*} r\left[\begin{array}{cc}A_0&AW\\WA&0\end{array}\right]&=&r(AW)+r(WA)+r\{[I-(AW)(AW)^{\#}]A_0[I-(WA)(WA)^{\#}]\}\\ &=&2r(A)+r\{[I-(AW)(AW)^{\#}]A_0[I-(WA)(WA)^{\#}]\}\\ &=&2r(A) \end{eqnarray*} $

当且仅当$ [I-(AW)(AW)^{\#}]A_0[I-(WA)(WA)^{\#}]=0 $. 又因为$ (AW)(AW)^{\#}=A_W^{\#}WAW $, $ (WA)(WA)^{\#}=WAWA_W^{\#} $, 所以(ⅲ)等价于(ⅳ). 证毕.

当$ \widehat{A}\in {\mathbb{D}}^{n\times n} $且$ \widehat{W}=I $, 即$ W=I $, $ W_0=0 $时, 由定理3.2可得到对偶群逆的存在性刻画及表示.

推论3.1^[9]  设$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{n\times n} $, 其中$ A, \; A_0\in {\mathbb{R}}^{n\times n} $. 则下面命题等价:

(ⅰ) $ \widehat{A} $的对偶群逆存在;

(ⅱ) Ind$ (A)=1 $且$ \widehat{A}=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]P^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1& {\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]P^{-1} $, 其中$ P $和$ A_1 $均为可逆矩阵;

(ⅲ) $ (I-AA^{\#})A_0(I-AA^{\#})=0 $;

(ⅳ) Ind$ (A)=1 $且$ r\left[\begin{array}{cc}A_0&A\\A&0\end{array}\right]=2r(A) $.

此外, 若$ {\widehat{A}}^{\#} $存在, 则

$ \begin{eqnarray*} {\widehat{A}}^{\#}=A^{\#}+\varepsilon[(A^{\#})^2A_0(I-AA^{\#})+(I-AA^{\#})A_0(A^{\#})^2-A^{\#}A_0A^{\#}]. \end{eqnarray*} $

例3.1   令$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{3\times 4} $, $ \widehat{W}=W+\varepsilon W_0\in {\mathbb{D}}^{4\times 3} $, 其中

$ \begin{eqnarray*} A=\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&0&0\end{array}\right], \; A_0=\left[\begin{array}{cccc}1&-1&2&1\\2&1&0&0\\1&3&0&0\end{array}\right], \; W=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}\right], \; W_0=\left[\begin{array}{ccc}2&6&-1\\0&2&1\\1&2&-1\\0&0&3\end{array}\right]. \end{eqnarray*} $

则

$ \begin{eqnarray*} AW=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right], WA=\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array}\right]. \end{eqnarray*} $

不难看出, $ AW $和$ WA $的指标均为1, 从而它们的群逆均存在. 又因为$ r(AW)=r(WA)=r(A)=2 $, 于是由引理2.1可知$ A_W^{\#} $存在, 且

$ \begin{eqnarray*} A_W^{\#}=A[(WA)^{\#}]^2=\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&0&0\end{array}\right]. \end{eqnarray*} $

此外, 由于

$ \begin{eqnarray*} r\left[\begin{array}{cc}A_0&AW\\WA&0\end{array}\right]=r\left[\begin{array}{ccccccc}1&-1&2&1&1&0&0\\2&1&0&0&0&1&0\\1&3&0&0&0&0&0\\1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&0\end{array}\right]=4=2r(A), \end{eqnarray*} $

从而由定理3.2可知$ {\widehat{A}}_{\widehat{W}}^{\#} $存在, 且

$ \begin{eqnarray*} \widehat{A}_{\widehat{W}}^{\#}&=&A_W^{\#}+\varepsilon \{[(AW)^{\#}]^2A_0(I-WAWA_W^{\#})+(I-A_W^{\#}WAW)A_0[(WA)^{\#}]^2\nonumber \\ &-&A_W^{\#}WAW_0A_W^{\#}-A_W^{\#}W_0AWA_W^{\#}-(AW)^{\#}A_0(WA)^{\#}\}\\ &=&\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&0&0\end{array}\right]+\varepsilon \left[\begin{array}{cccc}-5&-11&2&1\\-2&-5&0&0\\1&3&0&0\end{array}\right]. \end{eqnarray*} $

最后给出$ \widehat{A}_{\widehat{W}}^{\#} $基于$ (\widehat{A}\widehat{W})^{\#} $和$ (\widehat{W}\widehat{A})^{\#} $的表示.

定理3.3   令$ \widehat{A}=A+\varepsilon A_0\in {\mathbb{D}}^{m\times n} $, $ \widehat{W}=W+\varepsilon W_0\in {\mathbb{D}}^{n\times m} $, 其中$ A, \; A_0\in {\mathbb{R}}^{m\times n} $, $ W, \; W_0\in {\mathbb{R}}^{n\times m} $. 若$ \widehat{A}_{\widehat{W}}^{\#} $存在, 则$ (\widehat{A}\widehat{W})^{\#} $和$ (\widehat{W}\widehat{A})^{\#} $均存在, 且

$ \begin{eqnarray*} \widehat{A}_{\widehat{W}}^{\#}=(\widehat{A}\widehat{W})^{\#}\widehat{A}(\widehat{W}\widehat{A})^{\#}=[(\widehat{A}\widehat{W})^{\#}]^2\widehat{A} =\widehat{A}[(\widehat{W}\widehat{A})^{\#}]^2. \end{eqnarray*} $

证  若$ \widehat{A}_{\widehat{W}}^{\#} $存在, 则由定理3.2, $ A_W^{\#} $存在, 于是由引理2.1可知$ (AW)^{\#} $和$ (WA)^{\#} $均存在. 此外, $ \widehat{A} $和$ \widehat{W} $分别具有如下表示:

$ \begin{eqnarray*} \widehat{A}=P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}, \widehat{W}=Q\left[\begin{array}{cc}W_1&0\\0&W_4\end{array}\right]P^{-1}+\varepsilon Q\left[\begin{array}{cc}{\tilde{W}}_1&{\tilde{W}}_2\\{\tilde{W}}_3&{\tilde{W}}_4\end{array}\right]P^{-1}. \end{eqnarray*} $

于是

$ \begin{eqnarray*} \widehat{A}\widehat{W}&=&P\left[\begin{array}{cc}A_1W_1&0\\0&0\end{array}\right]P^{-1}+\varepsilon P\left[\begin{array}{cc} A_1{\tilde{W}}_1+{\tilde{A}}_1W_1&A_1{\tilde{W}}_2+{\tilde{A}}_2W_4\\{\tilde{A}}_3W_1&0\end{array}\right]P^{-1}, \\ \widehat{W}\widehat{A}&=&Q\left[\begin{array}{cc}W_1A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon Q\left[\begin{array}{cc} W_1{\tilde{A}}_1+{\tilde{W}}_1A_1&W_1{\tilde{A}}_2\\W_4{\tilde{A}}_3+{\tilde{W}}_3A_1&0\end{array}\right]Q^{-1}. \end{eqnarray*} $

因为$ AW $和$ WA $分别是$ \widehat{A}\widehat{W} $和$ \widehat{W}\widehat{A} $的实部, 且二者的指标均为1, 于是由推论3.1以及上面$ \widehat{A}\widehat{W} $和$ \widehat{W}\widehat{A} $的表示可知$ (\widehat{A}\widehat{W})^{\#} $和$ (\widehat{W}\widehat{A})^{\#} $均存在, 且

$ \begin{eqnarray*} (\widehat{A}\widehat{W})^{\#}=&&P\left[\begin{array}{cc}(A_1W_1)^{-1}&0\\0&0\end{array}\right]P^{-1}\\&+&\varepsilon P\left[\begin{array}{cc} -(A_1W_1)^{-1}(A_1{\tilde{W}}_1+{\tilde{A}}_1W_1)(A_1W_1)^{-1}&(A_1W_1)^{-2}(A _1{\tilde{W}}_2+{\tilde{A}}_2W_4)\\{\tilde{A}}_3W_1(A_1W_1)^{-2}&0\end{array}\right]P^{-1}, \end{eqnarray*} $

$ \begin{eqnarray*} (\widehat{W}\widehat{A})^{\#}=Q\left[\begin{array}{cc}(W_1A_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}+\varepsilon Q\left[\begin{array}{cc} -(W_1A_1)^{-1}(W_1{\tilde{A}}_1+{\tilde{W}}_1A_1)(W_1A_1)^{-1}&(W_1A_1)^{-2}W_1{\tilde{A}}_2\\(W_4{\tilde{A}}_3+{\tilde{W}}_3A_1)(W_1A_1)^{-2}&0\end{array}\right]Q^{-1}. \end{eqnarray*} $

直接计算可得

$ \begin{eqnarray*} (\widehat{A}\widehat{W})^{\#}\widehat{A}(\widehat{W}\widehat{A})^{\#}&=&\left(P\left[\begin{array}{cc}(A_1W_1)^{-1}&0\\0&0\end{array}\right]P^{-1}\right.\\ &&+\varepsilon P\left.\left[\begin{array}{cc} -(A_1W_1)^{-1}(A_1{\tilde{W}}_1+{\tilde{A}}_1W_1)(A_1W_1)^{-1}&(A_1W_1)^{-2}(A_1{\tilde{W}}_2+{\tilde{A}}_2W_4)\\{\tilde{A}}_3W_1(A_1W_1)^{-2}&0\end{array}\right]P^{-1}\right)\\ &&\times \left(P\left[\begin{array}{cc}A_1&0\\0&0\end{array}\right]Q^{-1}+\varepsilon P\left[\begin{array}{cc}{\tilde{A}}_1&{\tilde{A}}_2\\{\tilde{A}}_3&0\end{array}\right]Q^{-1}\right) \times \left(Q\left[\begin{array}{cc}(W_1A_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\right.\\ &&+\varepsilon \left.Q\left[\begin{array}{cc} -(W_1A_1)^{-1}(W_1{\tilde{A}}_1+{\tilde{W}}_1A_1)(W_1A_1)^{-1}&(W_1A_1)^{-2}W_1{\tilde{A}}_2\\(W_4{\tilde{A}}_3+{\tilde{W}}_3A_1)(W_1A_1)^{-2}&0\end{array}\right]Q^{-1}\right)\\ &=&P\left[\begin{array}{cc}(W_1A_1W_1)^{-1}&0\\0&0\end{array}\right]Q^{-1}\\ &&+\varepsilon P\left[\begin{array}{cc}-W_1^{-1}({\tilde{W}}_1 W_1^{-1}A_1^{-1}+A_1^{-1}{\tilde{A}}_1A_1^{-1}+A_1^{-1}W_1^{-1}{\tilde{W}}_1)W_1^{-1}&(A_1W_1)^{-2}{\tilde{A}}_2\\{\tilde{A}}_3(W_1A_1)^{-2}&0 \end{array}\right]Q^{-1}\\ &=&\widehat{A}_{\widehat{W}}^{\#}. \end{eqnarray*} $

类似的, 可证明$ \widehat{A}_{\widehat{W}}^{\#}=[(\widehat{A}\widehat{W})^{\#}]^2\widehat{A} =\widehat{A}[(\widehat{W}\widehat{A})^{\#}]^2. $