Fredholm^[1]于1903年首次提出了广义逆的概念,并称之为"伪逆".随着广义逆理论和应用研究的不断深入,又产生了各种不同类型的广义逆,例如Moore-Penrose逆^[2-3]、Drazin逆^[4]、群逆、K -逆($ K\in\{1, 2, 3, 4\})\mbox{、} $ Bott-Duffin逆^[5]、core逆^[6]、Mary逆^[7]以及DMP逆^[8]等.由于广义逆理论在算子理论、微分方程、控制论、马尔科夫链、人工智能以及模式识别等方面都有着重要的应用,因此,多年来,广义逆一直是人们研究的热点,学者们分别从复空间、Hilbert空间、Banach代数、半群以及环上等来研究广义逆,而且成果不断涌现.

对于可逆的算子或矩阵$ P, Q $而言, $ (PQ)^{-1} = Q^{-1}P^{-1} $,这便是普通逆的逆序律.但对于广义逆,其逆序律却不一定成立.于是,学者们纷纷对各种广义逆的逆序律进行了探讨.文献[9-12]分别在复空间, Hilbert空间和环上讨论了Moore-Penrose逆的逆序律成立的条件;文献[13-16]分别在复空间, Hilbert空间, Banach代数和环上研究了Drazin逆的逆序律成立的充分必要条件;文献[17-18]在半群、环和Hilbert空间上分别给出了群逆的逆序律成立的等价条件;文献[19-21]分别在复空间和Hilbert空间上讨论了K -逆的逆序律.

本文采用空间分解的方法在某些交换条件下给出了Banach空间中两个有界线性算子的Drazin逆的逆序律成立的等价条件,这推广了文献[16]中的结果.

为便于叙述,文中通篇采用以下记号:设$ {\cal X, Y} $是复Banach空间,记$ B({\cal X, Y}) $是从$ {\cal X} $到$ {\cal Y} $的所有有界线性算子的集合; $ B({\cal X}) $是从$ {\cal X} $到$ {\cal X} $的所有有界线性算子的集合. $ {\Bbb N} $表示数集$ \{1, 2, \cdots\} $.

定义1.1  对于算子$ T\in{\cal B}({\cal X}) $,若存在$ T^{D}\in{\cal B}({\cal X}) $使得方程组

$ \begin{eqnarray} \label{dengshi01} T^{D}T = TT^{D}, \ \ T^{D}TT^{D} = T^{D}, \ \ T^{k+1}T^{D} = T^{k} \ \mbox{对某个非负整数$k$} \end{eqnarray} $

成立,则称$ T $是Drazin可逆的,其中$ T^{D} $称为$ T $的Drazin逆,并称最小的非负整数$ k $为$ T $的指标,记为$ {{\rm ind}}(T) $.当$ T $是Drazin可逆时,记$ T^{\pi} = I-TT^D $,则$ T $在空间分解$ {\cal X} = {\cal N}(T^{\pi})\oplus {\cal R}(T^{\pi}) $下具有矩阵形式$ T = \left(\begin{array}{cc} T_1 & 0 \\ 0 & N_1 \end{array}\right), $其中$ T_1 $是可逆的, $ N_1^{{{\rm ind}}(T)} = 0 $,且$ T^D = \left(\begin{array} {cc}T_1^{-1} & 0 \\ 0 & 0 \end{array}\right) $.

引理1.2^{[22, Theorem 5.5]}  设$ P, Q\in{\cal B}({\cal X}) $是Drazin可逆的,且$ PQ = QP $,那么$ P, Q, $$ P^D, $$ Q^D $都是可交换的,且$ (PQ)^D = Q^DP^D = P^DQ^D. $

引理1.3  令$ A\in{\cal B}({{\cal X}}) $, $ D\in{\cal B}({{\cal Y}}) $, $ C\in{\cal B}({\cal X, Y}) $,且$ M_1 = \left(\begin{array}{cc} A\ & 0\\ C\ & D \end{array}\right), $$ M_2 = \left(\begin{array}{cc} D\ & C\\ 0\ & A \end{array}\right). $

$ \rm(1) $^{[23, Lemma 2.4]}若$ A, D $和$ M_1 $中有两个算子是Drazin可逆的,则第三个算子也是Drazin可逆的.

$ \rm(2) $^{[24, Theorem 5.1]}若$ A, D $是Drazin可逆的, $ {{\rm ind}}(A) = s $且$ {{\rm ind}}(D) = t $,则

$ M_1^D = \left(\begin{array}{cc} A^D & 0\\ X& D^D \end{array}\right), \ \ \ \ M_2^D = \left(\begin{array}{cc} D^D & X\\ 0& A^D \end{array}\right), $

其中$ X = \sum\limits_{i = 0}^{s-1} (D^D)^{i+2}CA^iA^\pi+D^\pi\sum\limits_{i = 0}^{t-1} D^iC(A^D)^{i+2}-D^DCA^D. $

引理1.4  ^{[25, Theorem 4.2]}令$ A\in{\cal B}({\cal X}) $, $ D\in{\cal B}({\cal Y}) $, $ B\in{\mathcal B}({\cal Y}, {\cal X}) $, $ C\in{\cal B}({\cal X, Y}) $,且$ N = \left(\begin{array}{cc} A\ & B\\ C\ & D \end{array}\right). $若$ A, D $是Drazin可逆的,且$ BC = 0, BD = 0, $则$ N $是Drazin可逆的,且

$ N^D = \left(\begin{array}{cc} A^D {\quad} & ( A^D)^2B\\ X{\quad} & D^D + D^DXB + XA^DB \end{array}\right), $

其中$ X $同引理1.3.

下面给出讨论逆序律时所需的必要结论.

定理2.1  设$ N, Q\in{\cal B}({\cal X}) $,且$ N^2QN = NQN^2 $.当$ k\in{{\Bbb N}}, \ n = 2, 3, \cdots $时,下列结论成立.

$ \rm(1) $若$ NQN^2Q = NQNQN $,则

$ (N^kQ)^n = N^{k n-1}QNQ^{n-1}, {\qquad} (N^kQN)^n = N^{n(k+1)-1}QNQ^{n-1}. $

$ \rm(2) $若$ N^2Q^2 = NQNQ $,则

$ (N^kQ)^n = N^{nk}Q^{n}, {\qquad} (N^kQN)^n = N^{n(k+1)-1}Q^{n}N. $

证  只证明$ \rm(1) $, $ \rm(2) $的证明类似.

$ \rm(1) $由$ N^2QN = NQN^2 $和$ NQN^2Q = NQNQN $,有$ NQN^kQN = N^kQNQN = N^{k+1}QNQ. $那么

$ \begin{eqnarray*} (N^kQ)^n& = &(N^kQ)^{n-2}N^kQN^kQ = ( N^kQ)^{n-2}N^{2k-1}QNQ\\ & = &(N^kQ)^{n-3}N^kQN^{2k-1}QNQ = (N^kQ)^{n-3}N^{3k-1}QNQ^2\\ & = &\cdots = N^{nk-1}QNQ^{n-1} \end{eqnarray*} $

且

$ \begin{eqnarray*} (N^kQN)^n& = &(N^kQN)^{n-2}N^kQN^{k+1}QN = (N^kQN)^{n-2}N^{2k+1}QNQ\\ & = &(N^kQN)^{n-3}N^kQN^{2k+2}QNQ = (N^kQN)^{n-3}N^{3k+2}QNQ^2\\ & = &\cdots = N^{n(k+1)-1}QNQ^{n-1}. \end{eqnarray*} $

证毕.

下面给出本文的主要结果.

定理2.2  设$ P, Q\in{\cal B}({\cal X}), $$ P $为Drazin可逆,且$ P^2QP = PQP^2 $.记

$ {{\cal M}} = \{P^k Q, P^kQP, PP^DQ, P^DQP, QPP^D, PP^DPQ\}, \ k\in{\Bbb N}. $

若$ P, Q $满足下列条件之一

$ \rm(i) $$ PQPQ = QPQP $; $ \rm(ii) $$ QP^2Q = QPQP $; $ \rm(iii) $$ P^2Q^2 = PQPQ $,

则当$ {\cal M} $中任一元素是Drazin可逆时, $ {\cal M} $中其它元素都是Drazin可逆的.

证  令$ {{\rm ind}}(P) = r $,则在空间分解$ {\cal H} = {\cal N}(P^{\pi})\oplus {\cal R}(P^{\pi}) $下, $ P $具有如下形式

(2.1) $ \begin{eqnarray} P = \left(\begin{array}{cc} P_1{\quad} & 0 \\ 0{\quad} & N_1 \end{array}\right), \end{eqnarray} $

其中$ P_1 $是可逆的, $ N_1^r = 0 $,且$ P^D = \left(\begin{array}{cc} P_1^{-1}\ & 0 \\ 0\ & 0 \end{array}\right). $令$ Q $在上述空间分解下的矩阵形式为

(2.2) $ \begin{equation} Q = \left(\begin{array}{c c} Q_1{\quad} & Q_2 \\ Q_3{\quad} & Q_4 \end{array}\right). \end{equation} $

由$ P^2QP = PQP^2 $,得到

(2.3) $ \begin{equation} P_1^{2} Q_1P_1 = P_1Q_1P_1^{2}, \ \ \ \ N_1^{2} Q_4N_1 = N_1Q_4N_1^{2} \end{equation} $

和

(2.4) $ \begin{equation} P_1^{2} Q_2N_1 = P_1Q_2N_1^{2}, \ \ \ \ N_1^{2} Q_3P_1 = N_1Q_3P_1^{2}. \end{equation} $

注意到$ P_1 $是可逆的,根据(2.3)和(2.4)式,有

(2.5) $ \begin{equation} P_1Q_1 = Q_1P_1 \end{equation} $

和

$ Q_2N_1 = P_1^{-1}Q_2N_1^2, \ \ \ \ N_1Q_3 = N_1^2Q_3P_1^{-1}, $

进而, $ Q_2N_1 = P_1^{-r}Q_2N_1^{r+1}, \ N_1Q_3 = N_1^{r+1}Q_3P_1^{-r}. $显然

(2.6) $ \begin{eqnarray} Q_2N_1 = 0, \ \ \ N_1Q_3 = 0. \end{eqnarray} $

因此

(2.7) $ \begin{eqnarray} P^{k}Q = \left(\begin{array}{cc} P_1^{k}Q_1\ & P_1^{k}Q_2 \\ 0\ & N_1^{k}Q_4 \end{array}\right), \ \ \ \ \ P^{k}QP = \left(\begin{array}{cc} P_1^{k+1}Q_1\ & 0 \\ 0\ & N_1^{k}Q_4N_1 \end{array}\right), k\in{\Bbb N}. \end{eqnarray} $

若$ P, Q $满足条件$ \rm(i) $,即$ PQPQ = QPQP $,则有

(2.8) $ \begin{eqnarray} \left(\begin{array}{cc} P_1 Q_1P_1Q_1\ & P_1 Q_1P_1Q_2 \\ 0\ & N_1Q_4N_1Q_4 \end{array}\right) = \left(\begin{array}{c c} Q_1P_1Q_1P_1\ & 0 \\ Q_3P_1Q_1P_1\ & Q_4N_1Q_4N_1 \end{array}\right), \end{eqnarray} $

于是

(2.9) $ \begin{equation} N_1Q_4N_1Q_4 = Q_4N_1Q_4N_1. \end{equation} $

由(2.5)式以及引理1.2可知$ P_1^{k}Q_1, P_1^{k+1}Q_1 $是Drazin可逆的充分必要条件是$ Q_1 $是Drazin可逆的.而(2.3)和(2.9)式知$ N_1Q_4N_1Q_4N_1 = N_1Q_4N_1^2Q_4 $,进而由定理2.1$ \rm(1) $表明$ N_1^{k}Q_4 $和$ N_1^{k}Q_4N_1 $都是幂零的.因此,根据引理1.3$ \rm(1) $, $ P^{k}Q, P^{k}QP $是Drazin可逆的充分必要条件是$ Q_1 $是Drazin可逆的.

对于$ P^DPQ $, $ P^DQP $, $ QP^DP $以及$ PP^DPQ $,经过计算,得到$ P^DPQ = \left(\begin{array}{cc} Q_1\ & Q_2 \\ 0\ & 0 \end{array}\right) $, $ QP^DP = \left(\begin{array}{cc} Q_1\ & 0 \\ Q_3\ & 0 \end{array}\right) $, $ P^DQP = \left(\begin{array}{cc} P_1^{-1}Q_1P_1\ & 0 \\ 0\ & 0 \end{array}\right) $, $ PP^DPQ = \left(\begin{array}{cc} P_1Q_1\ & P_1Q_2 \\ 0\ & 0 \end{array}\right) $.显然, $ P^DPQ $, $ P^DQP $, $ QP^DP $和$ PP^DPQ $是Drazin可逆的充分必要条件是$ Q_1 $是Drazin可逆的.

若$ P, Q $满足条件$ \rm(ii) $,即$ QP^2Q = QPQP $,则有

(2.10) $ \begin{eqnarray} \left(\begin{array}{cc} Q_1P_1^2Q_1\ & Q_1P_1^2Q_2 \\ Q_3P_1^2Q_1\ & Q_3P_1^2Q_2 + Q_4N_1^2Q_4 \end{array}\right) = \left(\begin{array}{cc} Q_1P_1Q_1P_1\ & 0 \\ Q_3P_1Q_1P_1\ & Q_4N_1Q_4N_1 \end{array}\right), \end{eqnarray} $

即有

$ Q_3P_1^2Q_2 + Q_4N_1^2Q_4 = Q_4N_1Q_4N_1, $

上式两边同时左乘$ N_1 $,再由(2.6)式,可得

(2.11) $ \begin{eqnarray} N_1Q_4N_1^2Q_4 = N_1Q_4N_1Q_4N_1. \end{eqnarray} $

这样, (2.3)和(2.11)式以及定理2.1$ \rm(1) $表明$ N_1^{k}Q_4 $和$ N_1^{k}Q_4N_1 $都是幂零的.余下的证明类似于条件$ \rm(i) $.

若$ P, Q $满足条件$ \rm(iii) $,即$ P^2Q^2 = PQPQ $,则有

(2.12) $ \begin{eqnarray} \left(\begin{array}{cc} P_1^2 Q_1^2 + P_1^2 Q_2Q_3\ &P_1^2 Q_1Q_2+P_1^2 Q_2Q_4 \\ N_1^2Q_4Q_3&N_1^2Q_4^2 \end{array}\right) = \left(\begin{array}{cc} P_1 Q_1P_1Q_1\ & P_1 Q_1P_1Q_2 \\ 0 &N_1Q_4N_1Q_4 \end{array}\right), \end{eqnarray} $

那么

(2.13) $ \begin{equation} N_1^2Q_4^2 = N_1Q_4N_1Q_4. \end{equation} $

这样, (2.3)和(2.13)式以及定理2.1$ \rm(2) $表明$ N_1^{k}Q_4 $和$ N_1^{k}Q_4N_1 $都是幂零的.余下的证明类似于条件$ \rm(i) $.

综上,定理得证.

类似于定理2.2,可得到如下结论.

定理2.3  设$ P, Q\in{\cal B}({\cal X}), $$ Q $为Drazin可逆,且$ QPQ^2 = Q^2PQ $.记

$ {{\cal N}} = \{PQ^k, QPQ^k, QQ^DP, Q^DPQ, PQQ^D, QQ^DQP\}, \ k\in{\Bbb N}. $

若$ P, Q $满足下列条件之一: $ \rm(i) $$ PQ^2P = QPQP $; $ \rm(ii) $$ P^2Q^2 = PQPQ $,则当$ {\cal N} $中任一元素是Drazin可逆时, $ {\cal N} $中其它元素都是Drazin可逆的.

定理2.4  设$ P, Q\in{\cal B}({\cal X}), $$ P $为Drazin可逆,且$ P^2QP = PQP^2 $.若$ P, Q $还满足$ PQPQ = QPQP $或$ QP^2Q = QPQP $,则当$ {\cal M} $中任一元素是Drazin可逆时, $ PQ $是Drazin可逆的,且$ (PQ)^D = (PQ)^DPP^D = PP^D(PQ)^D $.进一步,当$ Q $是Drazin可逆时,有

$ \rm(i) $若$ PP^DPQ = PQPP^D $,则下列逆序律等价

$ \rm(a) $$ (PQ)^D = Q^DP^D $;

$ \rm(b) $$ (PQ)^DP = Q^DP^DP $;

$ \rm(c) $$ (P^DPQ)^D = Q^DP^DP $;

$ \rm(d) $$ P(PQ)^D = Q^DP^DP $;

$ \rm(e) $$ P^DQ^D = Q^DP^D $.

$ \rm(ii) $若$ P^DPQ = QPP^D $,则逆序律成立:$ (PQ)^D = Q^DP^D = P^DQ^D $.

证  根据定理2.2的证明知$ PQ $是Drazin可逆的, $ PQ = \left(\begin{array}{cc} P_1Q_1\ & P_1Q_2 \\ 0\ & N_1Q_4 \end{array}\right), $且$ N_1Q_4 $是幂零的,从而,由引理1.3得

$ (PQ)^D = \left(\begin{array}{cc} (P_1Q_1)^D\ & X_1 \\ 0\ &0 \end{array}\right), $

其中$ X_1 = \sum\limits_{i = 0}^{{{\rm ind}}(N_1Q_4)-1} ((P_1Q_1)^D)^{i+2}P_1Q_2(N_1Q_4)^i. $从(2.8)和(2.10)式,均能得到

$ P_1 Q_1P_1Q_2 = 0, $

由$ P_1 $的可逆性以及(2.5)式,便得到

(2.14) $ \begin{equation} Q_1Q_2 = 0, \end{equation} $

同时,注意到$ (P_1Q_1)^D = ((P_1Q_1)^D)^2P_1Q_1 $,这样得到$ X_1 = 0 $.从而

(2.15) $ \begin{eqnarray} (PQ)^D = \left(\begin{array}{cc} (P_1Q_1)^D \ & 0 \\ 0\ &0 \end{array}\right) = \left(\begin{array}{cc} Q_1^D P_1 ^D\ & 0 \\ 0\ &0 \end{array}\right) = (PQ)^DPP^D = PP^D(PQ)^D. \end{eqnarray} $

$ \rm(i) $若$ PP^DPQ = PQPP^D $,则$ Q_2 = 0 $,此时$ Q = \left(\begin{array}{cc} Q_1\ & 0 \\ Q_3\ & Q_4 \end{array}\right). $由已知条件$ Q $是Drazin可逆的,且根据定理2.2的证明知$ Q_1 $是Drazin可逆的,于是,由引理1.3知$ Q_4 $是Drazin可逆的.进而,由引理1.3知$ Q^D = \left(\begin{array}{cc} Q_1^D\ & 0 \\ X_2\ & Q_4^D \end{array}\right). $其中

(2.16) $ \begin{eqnarray} X_2 = \sum\limits_{i = 0}^{{{\rm ind}}(Q_1)-1} (Q_4^D)^{i+2}Q_3Q_1^iQ_1^\pi+Q_4^\pi\sum\limits_{i = 0}^{{{\rm ind}}(Q_4)-1} Q_4^iQ_3(Q_1^D)^{i+2}-Q_4^DQ_3Q_1^D. \end{eqnarray} $

由此, $ Q^D P^D = \left(\begin{array}{cc} Q_1^DP_1^{-1}\ & 0 \\ X_2P_1^{-1}\ & 0 \end{array}\right). $再根据(2.15)式,显然, $ (PQ)^D = Q^DP^D $的充分必要条件是$ X_2 = 0 $.经过简单计算,有

$ (PQ)^D P = P (PQ)^D = (P^DPQ)^D = \left(\begin{array}{cc} Q_1^D\ & 0 \\ 0\ & 0 \end{array}\right), $

$ Q^DP^DP = \left(\begin{array}{cc} Q_1^D\ & 0 \\ X_2\ & 0 \end{array}\right), {\quad} P^DQ^D = \left(\begin{array}{cc} P_1^{-1}Q_1^D\ & 0 \\ 0\ & 0 \end{array}\right). $

综上,结论$ \rm(i) $得证.

$ \rm(ii) $若$ P^DPQ = QPP^D $,则$ Q_2 = Q_3 = 0 $,此时, $ Q = \left(\begin{array}{cc} Q_1\ & 0 \\ 0\ & Q_4 \end{array}\right). $那么, $ Q^D = \left(\begin{array}{cc} Q_1^D\ & 0 \\ 0\ & Q_4^D \end{array}\right). $显然, $ (PQ)^D = P^DQ^D = Q^DP^D $.

类似于定理2.4,可得到如下结论.

定理2.5  设$ P, Q\in{\cal B}({\cal X}), $$ Q $为Drazin可逆.若$ QPQ^2 = Q^2PQ $且$ PQ^2P = QPQP $,则当$ {\cal N} $中任一元素是Drazin可逆时, $ PQ $是Drazin可逆的,且$ (PQ)^D = (PQ)^DQQ^D = QQ^D(PQ)^D $.进一步,当$ P $是Drazin可逆时,有

$ \rm(i) $若$ Q^DQPQ = PQQ^DQ $,则下列逆序律等价

$ \rm(a) $$ (PQ)^D = Q^DP^D $;

$ \rm(b) $$ (PQ)^DQ = QQ^DP^D $;

$ \rm(c) $$ (PQQ^D)^D = QQ^DP^D $;

$ \rm(d) $$ Q(PQ)^D = QQ^DP^D $;

$ \rm(e) $$ P^DQ^D = Q^DP^D $.

$ \rm(ii) $若$ PQQ^D = QQ^DP $,则逆序律成立: $ (PQ)^D = Q^DP^D = P^DQ^D $.

定理2.6  设$ P, Q\in{\cal B}({\cal X}), $$ P $为Drazin可逆, $ P^2QP = PQP^2 $,且$ P^2Q^2 = PQPQ $.若$ {\cal M} $中任一元素是Drazin可逆的,则$ PQ $是Drazin可逆的,且$ (PQ)^D = PP^D(PQ)^D $.进一步,当$ P^{\pi}Q $是Drazin可逆时,下列逆序律等价

$ \rm(i) $$ (PQ)^D = Q^DP^D $;

$ \rm(ii) $$ (P^DPQ)^D = Q^DP^DP $;

$ \rm(iii) $$ P(PQ)^D = Q^DP^DP $;

$ \rm(iv) $$ P^DQ^D = Q^DP^D $.

证  根据定理2.2的证明知$ PQ $是Drazin可逆的, $ PQ = \left(\begin{array}{cc} P_1Q_1\ & P_1Q_2 \\ 0\ & N_1Q_4 \end{array} \right), $且$ N_1Q_4 $是幂零的,从而,由引理1.3得

(2.17) $ \begin{eqnarray} (PQ)^D = \left(\begin{array}{cc} (P_1Q_1)^D\ & X_3 \\ 0\ &0 \end{array} \right), \end{eqnarray} $

其中

$ X_3 = \sum\limits_{i = 0}^{{{\rm ind}}(N_1Q_4)-1} ((P_1Q_1)^D)^{i+2}P_1Q_2(N_1Q_4)^i = P_1^{-1}(Q_1^D)^2Q_2. $

另外, $ (PQ)^D = PP^D(PQ)^D $是显然的.

根据$ Q $的分解(2.2)以及$ P^{\pi} Q $是Drazin可逆的,知$ Q_1 $$ Q_4 $均是Drazin可逆的.再由(2.12)式以及(2.5)式得到

(2.18) $ \begin{equation} Q_2Q_3 = 0, \ \ \ \ Q_2Q_4 = 0, \end{equation} $

因此,根据引理1.4, $ Q $是Drazin可逆的,且

$ Q^D = \left(\begin{array}{cc} Q_1^D\ & (Q_1^D)^2Q_2 \\ X_2\ & Q_4^D+Q_4^DX_2Q_2+X_2Q_1^DQ_2 \end{array} \right), $

其中$ X_2 $同(2.17)式.由此, $ Q^D P^D = \left(\begin{array}{cc} Q_1^DP_1^{-1}\ & 0 \\ X_2P_1^{-1}\ & 0 \end{array} \right). $再根据(2.17)式,显然, $ (PQ)^D = Q^DP^D $的充分必要条件是$ X_2 = 0 $且$ Q_1^D Q_2 = 0 $.经过简单计算,有

$ P (PQ)^D = (P^DPQ)^D = \left(\begin{array}{cc} Q_1^D\ & (Q_1^D )^2Q_2 \\ 0\ & 0 \end{array} \right), $

$ Q^DP^DP = \left(\begin{array}{cc} Q_1^D\ & 0 \\ X_2 & 0 \end{array} \right), \ \ P^DQ^D = \left(\begin{array}{cc} P_1^{-1}Q_1^D\ & P_1^{-1}(Q_1^D)^2Q_2 \\ 0\ & 0 \end{array} \right). $

综上,结论得证.

类似于定理2.6,可得到如下结论.

定理2.7  设$ P, Q\in{\cal B}({\cal X}), $$ Q $为Drazin可逆. $ QPQ^2 = Q^2PQ $,且$ P^2Q^2 = PQPQ $.若$ {\cal N} $中任一元素是Drazin可逆的,则$ PQ $是Drazin可逆的,且$ (PQ)^D = (PQ)^DQQ^D $.进一步,当$ PQ^{\pi} $是Drazin可逆时,下列逆序律等价

$ \rm(i) $$ (PQ)^D = Q^DP^D $;

$ \rm(ii) $$ (PQQ^D)^D = QQ^DP^D $;

$ \rm(iii) $$ (PQ)^DQ = QQ^DP^D $;

$ \rm(iv) $$ P^DQ^D = Q^DP^D $.

由定理2.6,可得到下列推论.

推论2.8^[16]  设$ P, Q\in{\cal B}({\cal X}), $$ P, Q $为Drazin可逆(或$ P, P^\pi Q $为Drazin可逆),且$ P^2Q = PQP $.若$ {\cal M} $中任一元素是Drazin可逆的,则$ PQ $是Drazin可逆的, $ (PQ)^D = PP^D(PQ)^D $,且下列逆序律等价

$ \rm(i) $$ (PQ)^D = Q^DP^D $;

$ \rm(ii) $$ (P^DPQ)^D = Q^DP^DP $;

$ \rm(iii) $$ P(PQ)^D = Q^DP^DP $;

$ \rm(iv) $$ P^DQ^D = Q^DP^D $;

$ \rm(v) $$ (PQ)^DP = Q^DP^DP $.

证  根据定理2.6的证明,当$ P^2Q = PQP $时, $ Q_2 = 0 $.此时,只需注意到$ PQ, P^\pi Q, Q $三者中任意两个是Drazin可逆的,则第三个也是Drazin可逆的.

推论2.9  设$ P, Q\in{\cal B}({\cal X}), $$ P, Q $为Drazin可逆(或$ P, P^\pi Q $为Drazin可逆), $ PQP = QP^2 = P^2Q $.若$ {\cal M} $中任一元素是Drazin可逆的,则$ PQ $是Drazin可逆的,且逆序律成立: $ (PQ)^D = P^DQ^D = Q^DP^D $.

证  根据定理2.6的证明,由$ PQP = QP^2 $得$ Q_3 = 0 $;再由$ QP^2 = P^2Q $得$ Q_2 = 0 $.此时, $ (P^DPQ)^D = Q^DP^DP $显然成立.根据定理2.6得证.

由定理2.7,可得到下列推论.

推论2.10^[16]  设$ P, Q\in{\cal B}({\cal X}), $$ P, Q $为Drazin可逆（或$ PQ^{\pi}, Q $为Drazin可逆),且$ PQ^2 = QPQ $.若$ {\cal N} $中任一元素是Drazin可逆的,则$ PQ $是Drazin可逆的, $ (PQ)^D = (PQ)^DQQ^D $,且下列逆序律等价

$ \rm(i) $$ (PQ)^D = Q^DP^D $;

$ \rm(ii) $$ (PQQ^D)^D = QQ^DP^D $;

$ \rm(iii) $$ (PQ)^DQ = QQ^DP^D $;

$ \rm(iv) $$ P^DQ^D = Q^DP^D $;

$ \rm(v) $$ Q(PQ)^D = QQ^DP^D $.