近几十年来,算子广义逆的研究一直备受关注.广义逆的理论和方法不仅是许多数学分支的基本工具,而且在数理统计、最优化控制、数值分析、常微分方程等领域有着广泛的应用^[1].为方便描述,下面我们首先介绍一些基本概念和记号.

设$ B(H) $是复Hilbert空间$ H $上有界线性算子全体组成的集合. $ {\Bbb C}^{n\times n} $表示复数域$ {\Bbb C} $上$ n\times n $矩阵全体组成的集合.设$ A\in B(H) $, $ A^{*}, {\cal R}(A) $, $ {\cal N}(A) $分别表示$ A $的伴随算子,值域和零空间.设$ A\in B(H) $,若$ A+A^{*}\geq 0 $,则称$ A $是实正的.若$ A = A^{*} $,则称$ A $是自伴的.若对于任意的$ x\in H $都有$ \langle Ax, x\rangle\geq0 $,则称$ A $为正算子. $ B(H)^{+} $表示$ H $上正算子全体组成的集合.若$ A = A^{*} = A^{2} $,则称$ A $是投影,用$ P_{M} $表示值域是闭子空间$ M $的投影.

设$ A\in B(H) $.如果存在$ X\in B(H) $满足下列4个算子方程

(1.1) $ \begin{eqnarray} &&AXA = A, \end{eqnarray} $

(1.2) $ \begin{eqnarray} &&XAX = X, \end{eqnarray} $

(1.3) $ \begin{eqnarray} &&(AX)^{*} = AX, \end{eqnarray} $

(1.4) $ \begin{eqnarray} &&(XA)^{*} = XA, \end{eqnarray} $

则称$ X $为$ A $的Moore-Penrose逆,记作$ A^{\dagger} $.等式(1.1)–(1.4)称为Penrose方程.注意到$ AA^{\dagger} = P_{\overline{{{\cal R}(A)}}}, A^{\dagger}A = P_{\overline{{{\cal R}(A^{*})}}} $.对于集合$ K\subseteq\{1, 2, 3, 4\} $,如果$ X\in B(H) $满足Penrose方程$ (j), \forall j\in K $,则称$ X $是$ A $的$ K $ -逆.用$ A_{\geq}^{(i, j, \cdots, k)} $表示$ A $的一个非负$ \{i, j, \cdots, k\} $ -逆,所有的非负$ \{i, j, \cdots, k\} $ -逆的集合记作$ A_{\geq}{\{i, j, \cdots, k\}} $.注意到$ A_{\geq}\{1, 3\} = \{X\in B(H): AX = AA^{\dagger}, X\geq0\} $, $ A_{\geq}\{1, 4\} = \{X\in B(H): XA = A^{\dagger}A, X\geq0\} $.

在文献[2]中, Liu和Yang给出了矩阵$ A\in{\Bbb C}^{n\times n} $的自伴$ \{1, 3\} $ -逆, $ \{1, 4\} $ -逆, $ \{1, 2, 3\} $ -逆, $ \{1, 2, 4\} $ -逆, $ \{1, 3, 4\} $ -逆存在的充要条件.在文献[3-4]中,作者得到了矩阵的实正和非负广义逆存在性的等价刻画.对于$ A\in {\mathbb C}^{n\times n} $, Liu和Fang给出了$ A $的实正$ \{1, 2, 3\} $ -逆, $ \{1, 2, 4\} $ -逆, $ \{1, 3, 4\} $ -逆存在的充要条件^[3]. Nikolov和Cvetković-Ilić给出了$ A $的实正$ \{1, 3\} $ -逆, $ \{1, 4\} $ -逆, $ \{1, 3, 4\} $ -逆存在的充要条件,同时,他们也得到了$ A_{\geq}^{(1, 3)}, A_{\geq}^{(1, 4)}, A_{\geq}^{(1, 3, 4)} $存在的充要条件^[4].本文将其推广到无限维复Hilbert空间$ H $上,主要利用算子分块技巧研究闭值域算子$ A\in B(H) $的非负广义逆,给出$ A_{\geq}^{(1, 3)}, A_{\geq}^{(1, 4)}, A_{\geq}^{(1, 3, 4)} $存在的充要条件以及它们的一般形式.同时,本文也得到$ A $的非负$ \{1, 3\} $ -逆存在与非负$ \{1, 2, 3\} $ -逆存在是等价的,非负$ \{1, 4\} $ -逆存在与非负$ \{1, 2, 4\} $ -逆存在是等价的.

这一部分主要讨论闭值域算子的非负广义逆.为了得到主要结论,首先给出下面的几个引理.

引理2.1^[5]  设$ A, B\in B(H) $,则下列条件是等价的:

$ (1) $$ {\cal R}(B)\subseteq {\cal R}(A) $;

$ (2) $存在常数$ \lambda > 0 $使得$ BB^{*}\leq \lambda AA^{*} $;

$ (3) $存在算子$ C\in B(H) $使得$ AC = B $.

引理2.2^[6]  设$ H = H_{1}\oplus H_{2} $是Hilbert空间, $ A_{11}\in B(H_{1}), A_{12}\in B(H_{2}, H_{1}), A_{21}\in B(H_{1}, H_{2}), A_{22}\in B(H_{2}) $,则算子矩阵$ A = \left(\begin{array}{cc} A_{11} & A_{12}\\ A_{21} &A_{22} \\ \end{array}\right) $为$ H $上的正算子当且仅当下列条件同时成立:

$ (1) $$ A_{ii}\geq0, i = 1, 2 $;

$ (2) $$ A_{21} = A_{12}^{*} $;

$ (3) $存在压缩算子$ D\in B(H_{2}, H_{1}) $使得$ A_{12} = A_{11}^{\frac{1}{2}}DA_{22}^{\frac{1}{2}} $,其中$ A_{ii}^{\frac{1}{2}} $是$ A_{ii}(i = 1, 2) $的正的平方根,压缩算子$ D $是指$ \parallel D\parallel\leq1 $.

引理2.3^[7]  设$ A, B, C, D\in B(H) $且$ {\cal R}(A) $是闭的, $ M = \left(\begin{array}{cc} A & B\\ C & D \\ \end{array}\right)\in B(H\oplus H) $,则$ M $是$ H\oplus H $上的正算子当且仅当下列条件同时成立:

$ (1) $$ A = A^{*}, D = D^{*}, C = B^{*} $;

$ (2) $$ A\geq0 $;

$ (3) $$ AA^{\dagger}B = B $;

$ (4) $$ D-CA^{\dagger}B\geq0 $.

引理2.4^[8]  设$ A\in B(H) $且$ {\cal R}(A) $是闭的,则按空间分解$ H = {\cal R}(A^{*})\oplus {\cal N}(A) $,有

$ A = \left( \begin{array}{cc} A_{1} & 0\\ A_{2} & 0 \\ \end{array}\right), A^{\dagger} = \left( \begin{array}{cc} D^{-1}A_{1}^{*} & D^{-1}A_{2}^{*} \\ 0 & 0 \\ \end{array}\right), $

其中$ D = A_{1}^{*}A_{1}+A_{2}^{*}A_{2}\in B({\cal R}(A^{*})) $是可逆正算子.

文献[4]讨论了有限维Hilbert空间$ H $上有界线性算子的非负广义逆,对于$ A\in B(H) $, Nikolov和Cvetković-Ilić给出了$ A_{\geq}^{(1, 3)}, A_{\geq}^{(1, 4)}, A_{\geq}^{(1, 3, 4)} $存在的充要条件,下面我们将其推广到无限维Hilbert空间$ H $上.

定理2.1  设$ A\in B(H) $且$ {\cal R}(A) $是闭的,则下列条件等价:

$ (1) $$ A_{\geq}^{(1, 3)} $存在;

$(2) $$ A_{\geq}^{(1, 2, 3)} $存在;

$ (3) $$ (A^{\dagger})^{2}A\geq0 $且$ {\cal R}((A^{\dagger})^{2}A) = {\cal R}(A^{*}) $;

$ (4) $$ A^{2}A^{\dagger}\geq0 $且$ {\cal R}(A^{2}A^{\dagger}) = {\cal R}(A) $;

$(5) $$ (A^{\dagger})^{2}A\geq0 $且存在$ r > 0 $使得$ (A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*}\geq0 $;

$(6) $$ A^{2}A^{\dagger}\geq0 $且存在$ r > 0 $使得$ A^{2}A^{\dagger}-rAA^{\dagger}(I-A^{\dagger}A)AA^{\dagger}\geq0 $.

且当以上条件成立时,有

(2.1) $ \begin{eqnarray} A_{\geq}\{1, 3\} & = &\{A^{\dagger}+(A^{\dagger})^{*}-(A^{\dagger})^{2}A+(I-A^{\dagger}A)Y(I-A^{\dagger}A): Y\in B(H)^{+}, \\ & &(I-A^{\dagger}A)Y(I-A^{\dagger}A)\geq((A^{\dagger})^{*}-(A^{\dagger})^{2}A)((A^{\dagger})^{2}A)^{\dagger}(A^{\dagger}-(A^{\dagger})^{2}A) \}. \end{eqnarray} $

证   $ (1)\Rightarrow(3) $设$ X\in A_{\geq}\{1, 3\} $.由引理2.4可知,按空间分解$ H = {\cal R}(A^{*})\oplus {\cal N}(A) $,我们有

$ A = \left( \begin{array}{cc} A_{1} & 0\\ A_{2} & 0 \\ \end{array}\right), A^{\dagger} = \left( \begin{array}{cc} D^{-1}A_{1}^{*} & D^{-1}A_{2}^{*} \\ 0 & 0 \\ \end{array}\right), $

其中$ D = A_{1}^{*}A_{1}+A_{2}^{*}A_{2}\in B({\cal R}(A^{*})) $是可逆正算子.令$ X = \left(\begin{array}{cc} X_{11} & X_{12} \\ X_{12}^{*} & X_{22} \\ \end{array}\right)\in B({\cal R}(A^{*})\oplus {\cal N}(A)) $,则由$ AX = AA^{\dagger} $得

(2.2) $ \begin{eqnarray} &&A_{1}X_{11} = A_{1}D^{-1}A_{1}^{*}, \end{eqnarray} $

(2.3) $ \begin{eqnarray} &&A_{1}X_{12} = A_{1}D^{-1}A_{2}^{*}, \end{eqnarray} $

(2.4) $ \begin{eqnarray} && A_{2}X_{11} = A_{2}D^{-1}A_{1}^{*}, \end{eqnarray} $

(2.5) $ \begin{eqnarray} &&A_{2}X_{12} = A_{2}D^{-1}A_{2}^{*}, \end{eqnarray} $

从而由(2.2)和(2.4)式得到$ X_{11} = D^{-1}A_{1}^{*} $,由(2.3)和(2.5)式得到$ X_{12} = D^{-1}A_{2}^{*} $.于是

$ X = \left( \begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & X_{22} \\ \end{array}\right). $

由$ X\geq0 $和引理2.2知$ D^{-1}A_{1}^{*}\geq0 $且$ {\cal R}(D^{-1}A_{2}^{*})\subseteq {\cal R}((D^{-1}A_{1}^{*})^{\frac{1}{2}}) $.又由引理2.1知$ {\cal R}(D^{-1}A_{1}^{*})\subseteq {\cal R}((D^{-1}A_{1}^{*})^{\frac{1}{2}}) $.于是

$ {\cal R}(A^{\dagger}) = {\cal R}(D^{-1}A_{1}^{*})+{\cal R}(D^{-1}A_{2}^{*})\subseteq{\cal R}((D^{-1}A_{1}^{*})^{\frac{1}{2}})\subseteq {\cal R}(A^{*}) = {\cal R}(A^{\dagger}), $

从而$ {\cal R}((D^{-1}A_{1}^{*})^{\frac{1}{2}}) = {\cal R}(A^{*}) $.因为$ {\cal R}(A) $是闭的当且仅当$ {\cal R}(A^{*}) $是闭的,所以$ {\cal R}((D^{-1}A_{1}^{*})^{\frac{1}{2}}) $是闭的.注意到对于任意的$ B\in B(H)^{+} $, $ {\cal R}(B^{\frac{1}{2}}) $是闭的当且仅当$ {\cal R}(B) = {\cal R}(B^{\frac{1}{2}}) $.因此

$ {\cal R}(D^{-1}A_{1}^{*}) = {\cal R}((D^{-1}A_{1}^{*})^{\frac{1}{2}}) = {\cal R}(A^{*}). $

容易计算

$ (A^{\dagger})^{2}A = \left( \begin{array}{cc} D^{-1}A_{1}^{*} &0 \\ 0 & 0 \\ \end{array}\right), $

从而$ {\cal R}((A^{\dagger})^{2}A) = {\cal R}(D^{-1}A_{1}^{*}) = {\cal R}(A^{*}) $.

$ (3)\Rightarrow(1) $设$ (A^{\dagger})^{2}A\geq0 $,且$ {\cal R}((A^{\dagger})^{2}A) = {\cal R}(A^{*}) $,则

$ ((A^{\dagger})^{2}A)((A^{\dagger})^{2}A)^{\dagger} = ((A^{\dagger})^{2}A)^{\dagger}((A^{\dagger})^{2}A) = A^{\dagger}A. $

于是

$ \begin{eqnarray*} A(A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger} & = &AA^{\dagger}A(A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}\\ & = &A((A^{\dagger})^{2}A)((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}\\ & = &AA^{\dagger}AA^{\dagger}\\ & = &AA^{\dagger}. \end{eqnarray*} $

因此$ (A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger} $是算子方程$ AX = AA^{\dagger} $的正解,从而$ (A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger} A^{\dagger}\in A_{\geq}\{1, 3\} $.

$ (2)\Rightarrow(1) $显然.

$ (1)\Rightarrow(2) $设$ A_{\geq}^{(1, 3)} $存在,则由$ (3)\Rightarrow (1) $的证明过程可知$ (A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}\in A_{\geq}\{1, 3\} $,而且

$ \begin{eqnarray*} (A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}A(A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger} & = &(A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}(A^{\dagger})^{2}A((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}\\ & = &(A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}, \end{eqnarray*} $

从而$ (A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}\in A_{\geq}\{2\} $.因此$ (A^{\dagger})^{*}((A^{\dagger})^{2}A)^{\dagger}A^{\dagger}\in A_{\geq}\{1, 2, 3\} $.

$ (1)\Rightarrow(5) $设$ X\in A_{\geq}\{1, 3\} $,则按空间分解$ H = {\cal R}(A^{*})\oplus {\cal N}(A) $, $ A, A^{\dagger} $如引理2.4所示,从而由$ (1)\Rightarrow(3) $的证明过程可知$ X = \left(\begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & X_{22} \\ \end{array}\right)\geq0 $,其中$ X_{22}\in B({\cal N}(A)) $是正算子.取$ M = \parallel X_{22}\parallel $,则

$ \left( \begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & M \\ \end{array}\right)\geq\left( \begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & X_{22} \\ \end{array}\right)\geq0, $

从而由引理2.3可知$ D^{-1}A_{1}^{*}-rD^{-1}A_{2}^{*}A_{2}D^{-1}\geq0 $,其中$ r = \frac{1}{M} > 0 $.于是

$ (A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*} = \left( \begin{array}{cc} D^{-1}A_{1}^{*}-rD^{-1}A_{2}^{*}A_{2}D^{-1} &0 \\ 0 & 0 \\ \end{array}\right)\geq0. $

$ (5)\Rightarrow(1) $设$ (A^{\dagger})^{2}A\geq0 $且存在$ r > 0 $使得$ (A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*}\geq0 $.按空间分解$ H = {\cal R}(A^{*})\oplus {\cal N}(A) $, $ A, A^{\dagger} $如引理2.4所示.令

$ X_{0} = \left( \begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} &\frac{1}{r} \\ \end{array}\right). $

由$ (A^{\dagger})^{2}A\geq0 $知$ D^{-1}A_{1}^{*}\geq0 $,由$ (A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*}\geq0 $知$ D^{-1}A_{1}^{*}-rD^{-1}A_{2}^{*}A_{2}D^{-1}\geq0 $,从而由引理2.3可知$ X_{0}\geq0 $.另外容易验证$ AX_{0} = AA^{\dagger} $.因此$ X_{0}\in A_{\geq}\{1, 3\} $.

$ (3)\Rightarrow(4) $设$ (A^{\dagger})^{2}A\geq0 $,且$ {\cal R}((A^{\dagger})^{2}A) = {\cal R}(A^{*}) $,则$ A^{2}A^{\dagger} = A(A^{\dagger})^{2}AA^{*}\geq0 $.注意到

$ \begin{eqnarray*} A^{2}A^{\dagger}(A^{\dagger})^{*}((A^{\dagger})^{^{2}}A)^{\dagger}A^{\dagger} & = &A(A^{\dagger})^{*}((A^{\dagger})^{^{2}}A)^{\dagger}A^{\dagger}\\ & = &AA^{\dagger}A(A^{\dagger})^{*}((A^{\dagger})^{^{2}}A)^{\dagger}A^{\dagger}\\ & = &A(A^{\dagger})^{2}A((A^{\dagger})^{^{2}}A)^{\dagger}A^{\dagger}\\ & = &AA^{\dagger}AA^{\dagger}\\ & = &AA^{\dagger}, \end{eqnarray*} $

从而$ {\cal R}(A) = {\cal R}(AA^{\dagger})\subseteq{\cal R}(A^{2}A^{\dagger}) $.由引理2.1知$ {\cal R}(A^{2}A^{\dagger})\subseteq{\cal R}(A) $.于是$ {\cal R}(A^{2}A^{\dagger}) = {\cal R}(A) $.

$ (4)\Rightarrow(3) $设$ A^{2}A^{\dagger}\geq0 $且$ {\cal R}(A^{2}A^{\dagger}) = {\cal R}(A) $,则$ (A^{\dagger})^{2}A = A^{\dagger}A^{2}A^{\dagger}(A^{\dagger})^{*}\geq0 $.注意到

$ \begin{eqnarray*} (A^{\dagger})^{2}AA^{*}(A^{2}A^{\dagger})^{\dagger}A & = &A^{\dagger}A^{*}(A^{2}A^{\dagger})^{\dagger}A\\ & = &A^{\dagger}AA^{\dagger}A^{*}(A^{2}A^{\dagger})^{\dagger}A\\ & = &A^{\dagger}(A^{2}A^{\dagger})(A^{2}A^{\dagger})^{\dagger}A\\ & = &A^{\dagger}AA^{\dagger}A\\ & = &A^{\dagger}A, \end{eqnarray*} $

从而$ {\cal R}(A^{*}) = {\cal R}(A^{\dagger}A)\subseteq {\cal R}((A^{\dagger})^{2}A) $.由引理2.1知$ {\cal R}((A^{\dagger})^{2}A)\subseteq{\cal R}(A^{\dagger}) = {\cal R}(A^{*}) $.于是$ {\cal R}((A^{\dagger})^{2}A) = {\cal R}(A^{*}) $.

$ (5)\Leftrightarrow(6) $我们已经证明$ A^{2}A^{\dagger}\geq0 $当且仅当$ (A^{\dagger})^{2}A\geq0 $.因此,对于任意的$ r > 0 $,有

$ A^{2}A^{\dagger}-rAA^{\dagger}(I-A^{\dagger}A)AA^{\dagger} = A((A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*})A^{*} $

和

$ (A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*} = A^{\dagger}(A^{2}A^{\dagger}-rAA^{\dagger}(I-A^{\dagger}A)AA^{\dagger})(A^{\dagger})^{*}, $

从而$ (A^{\dagger})^{2}A-rA^{\dagger}(I-A^{\dagger}A)(A^{\dagger})^{*}\geq0 $和$ A^{2}A^{\dagger}-rAA^{\dagger}(I-A^{\dagger}A)AA^{\dagger}\geq0 $是等价的.

下面证明(2.1)式成立.设$ X\in A_{\geq}\{1, 3\} $.由$ (1)\Rightarrow(3) $的证明过程知

$ \begin{eqnarray*} X& = &\left( \begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & X_{22} \\ \end{array}\right)\\ & = &\left( \begin{array}{cc} D^{-1}A_{1}^{*} &0 \\ 0 & 0 \\ \end{array}\right)+ \left( \begin{array}{cc} 0 &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & 0 \\ \end{array}\right) +\left( \begin{array}{cc} 0 &0 \\ 0 & X_{22} \\ \end{array}\right)\\ & = &(A^{\dagger})^{2}A+A^{\dagger}(I-A^{\dagger}A)+(I-A^{\dagger}A)(A^{\dagger})^{*}+(I-A^{\dagger}A)Y(I-A^{\dagger}A)\\ & = &A^{\dagger}+(A^{\dagger})^{*}-(A^{\dagger})^{2}A+(I-A^{\dagger}A)Y(I-A^{\dagger}A), \end{eqnarray*} $

其中$ Y\in B(H)^{+} $.另外,设$ Y = \left(\begin{array}{cc} X_{11} &X_{12} \\ X_{12}^{*} & X_{22} \\ \end{array}\right) $.由$ X\geq0 $知

$ \begin{eqnarray*} & &(I-A^{\dagger}A)Y(I-A^{\dagger}A)-((A^{\dagger})^{*}-(A^{\dagger})^{2}A)((A^{\dagger})^{2}A)^{\dagger}(A^{\dagger}-(A^{\dagger})^{2}A)\\ & = &\left( \begin{array}{cc} 0 &0 \\ 0 & X_{22}-A_{2}D^{-1}(A_{1}D^{-1})^{-1}D^{-1}A_{2}^{*} \\ \end{array}\right)\geq0. \end{eqnarray*} $

因此

$ \begin{eqnarray*} A_{\geq}\{1, 3\} &\subseteq&\{A^{\dagger}+(A^{\dagger})^{*}-(A^{\dagger})^{2}A+(I-A^{\dagger}A)Y(I-A^{\dagger}A): Y\in B(H)^{+}, \\ & &(I-A^{\dagger}A)Y(I-A^{\dagger}A)\geq((A^{\dagger})^{*}-(A^{\dagger})^{2}A)((A^{\dagger})^{2}A)^{\dagger}(A^{\dagger}-(A^{\dagger})^{2}A) \}. \end{eqnarray*} $

反之,用类似的方法我们可以证明

$ \begin{eqnarray*} A_{\geq}\{1, 3\} &\supseteq&\{A^{\dagger}+(A^{\dagger})^{*}-(A^{\dagger})^{2}A+(I-A^{\dagger}A)Y(I-A^{\dagger}A): Y\in B(H)^{+}, \\ & &(I-A^{\dagger}A)Y(I-A^{\dagger}A)\geq((A^{\dagger})^{*}-(A^{\dagger})^{2}A)((A^{\dagger})^{2}A)^{\dagger}(A^{\dagger}-(A^{\dagger})^{2}A) \}. \end{eqnarray*} $

证毕.

因为$ X $是$ A $的$ \{1, 3\} $ -逆当且仅当$ X^{*} $是$ A^{*} $的$ \{1, 4\} $ -逆,所以由定理2.1容易得到下面的定理.

定理2.2  设$ A\in B(H) $且$ {\cal R}(A) $是闭的,则下列条件等价:

$(1) $$ A_{\geq}^{(1, 4)} $存在;

$(2) $$ A_{\geq}^{(1, 2, 4)} $存在;

$(3) $$ A(A^{\dagger})^{2}\geq0 $且$ {\cal R}(A(A^{\dagger})^{2}) = {\cal R}(A) $;

$(4) $$ A^{\dagger}A^{2}\geq0 $且$ {\cal R}(A^{\dagger}A^{2}) = {\cal R}(A^{*}) $;

$(5) $$ A(A^{\dagger})^{2}\geq0 $且存在$ r > 0 $使得$ A(A^{\dagger})^{2}-r(A^{\dagger})^{*}(I-AA^{\dagger})A^{\dagger}\geq0 $;

$(6) $$ A^{\dagger}A^{2}\geq0 $且存在$ r > 0 $使得$ A^{\dagger}A^{2}-rA^{\dagger}A(I-AA^{\dagger})A^{\dagger}A\geq0 $.

且当以上条件成立时,有

$ \begin{eqnarray*} A_{\geq}\{1, 4\} & = &\{A^{\dagger}+(A^{\dagger})^{*}-A(A^{\dagger})^{2}+(I-AA^{\dagger})Y(I-AA^{\dagger}): Y\in B(H)^{+}, \\ & &(I-AA^{\dagger})Y(I-AA^{\dagger})\geq(A^{\dagger}-A(A^{\dagger})^{2})(A(A^{\dagger})^{2})^{\dagger}((A^{\dagger})^{*}-A(A^{\dagger})^{2})\}. \end{eqnarray*} $

接下来,我们给出$ A_{\geq}^{(1, 3, 4)} $存在的充要条件.

定理2.3  设$ A\in B(H) $且$ {\cal R}(A) $是闭的,则下列条件等价:

$(1) $$ A_{\geq}^{(1, 3, 4)} $存在;

$(2) $$ A^{\dagger} = (A^{\dagger})^{2}A\geq0 $.

且当以上条件成立时,有

(2.6) $ A_{\geq}\{1, 3, 4\} = \{A^{\dagger}+(I-AA^{\dagger})Y(I-AA^{\dagger}): Y\in B(H)^{+}\}. $

证   $ (1)\Rightarrow(2) $设$ X\in A_{\geq}\{1, 3, 4\} $.设$ A, A^{\dagger} $如引理2.4所示,则由$ AX = AA^{\dagger} $知

$ X = \left( \begin{array}{cc} D^{-1}A_{1}^{*} &D^{-1}A_{2}^{*} \\ A_{2}D^{-1} & X_{22} \\ \end{array}\right), $

其中$ D^{-1}A_{1}^{*}\in B({\cal R}(A^{*})) $是可逆正算子, $ X_{22}\in B({\cal N}(A)) $是正算子且满足

$ X_{22}\geq A_{2}D^{-1}(A_{1}D^{-1})^{-1}D^{-1}A_{2}^{*}. $

由$ XA = A^{\dagger}A $知

(2.7) $ A_{2}D^{-1}A_{1}+X_{22}A_{2} = 0. $

(2.7)式中取$ X_{22} = A_{2}D^{-1}(A_{1}D^{-1})^{-1}D^{-1}A_{2}^{*} $,有

(2.8) $ A_{2}D^{-1}A_{1}+A_{2}D^{-1}(A_{1}D^{-1})^{-1}D^{-1}A_{2}^{*}A_{2} = 0. $

由$ I = A_{1}^{*}A_{1}D^{-1}+A_{2}^{*}A_{2}D^{-1} = A_{1}^{*}D^{-1}A_{1}^{*}+A_{2}^{*}A_{2}D^{-1} $知$ A_{1}D^{-1}A_{1}+D^{-1}A_{2}^{*}A_{2} = I, $从而

(2.9) $ A_{1} = (A_{1}D^{-1})^{-1}-(A_{1}D^{-1})^{-1}D^{-1}A_{2}^{*}A_{2}. $

由(2.8)和(2.9)式可得$ A_{2}D^{-1}(A_{1}D^{-1})^{-1} = 0 $,从而$ A_{2} = 0 $.于是

$ A^{\dagger} = (A^{\dagger})^{2}A = \left( \begin{array}{cc} D^{-1}A_{1}^{*} &0 \\ 0 & 0\\ \end{array}\right)\geq0. $

$ (2)\Rightarrow(1) $设$ A, A^{\dagger} $如引理2.4所示,则由$ A^{\dagger} = (A^{\dagger})^{2}A $知$ D^{-1}A_{2}^{*} = 0 $,从而$ A_{2} = 0 $.于是$ A = \left(\begin{array}{cc} A_{1} & 0 \\ 0 & 0\\ \end{array}\right), $$ A^{\dagger} = \left(\begin{array}{cc} D^{-1}A_{1}^{*} & 0 \\ 0 & 0\\ \end{array}\right). $令$ X = \left(\begin{array}{cc} D^{-1}A_{1}^{*} & 0 \\ 0 & X_{22}\\ \end{array}\right), $其中$ X_{22}\in B({\cal N}(A)) $是任意正算子.容易计算$ AX = AA^{\dagger}, XA = A^{\dagger}A $且$ X\geq0 $,从而$ X\in A_{\geq}\{1, 3, 4\} $.

由条件(2)和(2.1)式容易验证(2.6)式成立.证毕.