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New representation of Drazin inverse for a block complex matrix

Huanyin Chen  and  Marjan Sheibani∗ Department of Mathematics
Hangzhou Normal University
Hang -zhou, China ¡[email protected]¿ Farzanegan Campus, Semnan University, Semnan, Iran ¡[email protected]¿
Abstract.

We present a new formula for the Drazin inverse of the sum of two complex matrices under weaker conditions with perturbations. By using this additive results, we establish new representations for the Drazin inverse of 2×22\times 2 block complex matrix (ABCD)\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right) under certain new conditions involving perturbations. These extend many known results, e.g., Yang and Liu (J. Comput. Applied Math., 235(2011), 1412–1417), Dopazo and Martinez-Serrano (Linear Algebra Appl., 432(2010), 1896–1904).

Key words and phrases:
Drazin inverse; additive formula; spectral idempotent; block matrix.
2010 Mathematics Subject Classification:
15A09, 15A10.
∗Corresponding author

1. Introduction

Let ℂn×n{\mathbb{C}}^{n\times n} be the set of all n×nn\times n matrices over the complex field ℂ{\mathbb{C}}, and let A∈ℂn×nA\in{\mathbb{C}}^{n\times n}. The Drazin inverse of AA is the unique matrix AD∈ℂn×nA^{D}\in{\mathbb{C}}^{n\times n} satisfying the following relations:

A​AD=AD​A,AD=AD​A​AD​and​Am=Am+1​ADAA^{D}=A^{D}A,A^{D}=A^{D}AA^{D}~{}\mbox{and}~{}A^{m}=A^{m+1}A^{D}

for some m∈ℕm\in{\mathbb{N}}. As is well known, AA has Drazin inverse if and only if r​a​n​k​(Am)=r​a​n​k​(Am+1)rank(A^{m})=rank(A^{m+1}) for some m∈ℕm\in{\mathbb{N}}. The Drazin invertibility of a complex matrix is attractive. It has widespread applications in singular differential equations, Markov chains, and iterative methods (see [1]). Many authors have studied such problems from many different views, e.g.,  [1, 7, 9, 10, Z1, 13].

Let P,Q∈ℂn×nP,Q\in{\mathbb{C}}^{n\times n}. It is attractive to give the explicit formula of (P+Q)D(P+Q)^{D}. The formula was firstly obtained by Drazin in the case of P​Q=Q​P=0PQ=QP=0. In 2001, Hartwig et al. gave the formula when P​Q=0PQ=0 in  [H]. In 2011, the additive formula was given under the new condition P​Q2=0,P​Q​P=0PQ^{2}=0,PQP=0 (see  [9, Theorem 2.1]). This is an elementary result and it deduce many wider conditions under which the Drazin inverse is expressed, e.g.,  [7, 10]. We denote by AπA^{\pi} the eigenprojection of AA corresponding to the eigenvalue 0 that is given by Aπ=I−A​ADA^{\pi}=I-AA^{D}. The aim of this paper is to generalize elementary result. In Section 2, we extend  [9, Theorem 2.1] and present a representation of (P+Q)D(P+Q)^{D} in the case of P​Q2=0PQ^{2}=0 and P​Q​P​(P​Q)π=0PQP(PQ)^{\pi}=0.

Let M=(ABCD)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right), where A∈ℂm×m,B∈ℂm×n,C∈ℂn×m,A\in{\mathbb{C}}^{m\times m},B\in{\mathbb{C}}^{m\times n},C\in{\mathbb{C}}^{n\times m}, D∈ℂn×nD\in{\mathbb{C}}^{n\times n}. It is of interesting to find the Drazin inverse of the block complex matrix MM. This problem is quite complicated and was expensively studied by many authors, see for example [3, 4, 9, 10]. In Section 3, we apply these computational formulas to give the Drazin inverse of a block complex matrix MM. If B​C​B=0,D​C​B=0,B​C​A​(B​C)π=0BCB=0,DCB=0,BCA(BC)^{\pi}=0 and D​C​A​(B​C)π=0DCA(BC)^{\pi}=0, we establish the representation of MDM^{D}, which also extend the result of Yang and Liu (see [9, Theorem 3.1]).

2. additive results

In this section, we will give a new additive formula for the Drazin inverse under a weaker condition, for which the following has been developed. We begin with

Lemma 2.1.

Let P,Q∈ℂn×nP,Q\in{\mathbb{C}}^{n\times n}. If P​Q=0PQ=0, then

(P+Q)D=∑i=0∞(QD)i+1​Pi​Pπ+∑i=0∞Qi​Qπ​(PD)i+1.(P+Q)^{D}=\sum\limits_{i=0}^{\infty}(Q^{D})^{i+1}P^{i}P^{\pi}+\sum\limits_{i=0}^{\infty}Q^{i}Q^{\pi}(P^{D})^{i+1}.
Proof.

See [10, Lemma 1.2].∎

Lemma 2.2.

Let M=(EIF0),E,F∈ℂn×nM=\left(\begin{array}[]{cc}E&I\\ F&0\end{array}\right),E,F\in{\mathbb{C}}^{n\times n}. If F​E​Fπ=0FEF^{\pi}=0, then

MD=(ΓΔΛΞ)(∗)\begin{array}[]{lll}M^{D}&=&\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(*)\end{array}

where

Γ=∑i=0∞(ED)2​i+1​Fi​Fπ+(Eπ−∑j=1∞(ED)2​j​Fj)​Fπ​E​FD+∑i=1∞[(Eπ−∑j=1∞(ED)2​jFj)FiFπE+EπE2​iFπEFFD+∑k=0i−2EπE2​k+2Fi−k−1FπE]FDCi+∑i=1∞[EπE2​i−1FπEFFD−∑j=0∞(ED)2​j+1Fi+jFπE+∑k=0i−2EπE2​k+1Fi−k−1FπE]Ai,Δ=−FD−ED​Fπ​E​FD−∑i=1∞(ED)2​i+1​Fi​Fπ​E​FD+∑i=0∞(ED)2​i​Fi​Fπ−[Eπ−∑j=1∞(ED)2​jFj]FπEFDEFD+∑i=1∞[(Eπ−∑j=1∞(ED)2​jFj)FiFπE+EπE2​iFπEFFD+∑k=0i−2EπE2​k+2Fi−k−1FπE]FDDi+∑i=1∞[EπE2​i−1FπEFFD−∑j=0∞(ED)2​j+1Fi+jFπE+∑k=0i−2EπE2​k+1Fi−k−1FπE]Bi,Λ=F​FD+∑i=1∞Fi​Fπ​E​Ai,Ξ=−F​FD​E​FD+∑i=1∞Fi​Fπ​E​B;\begin{array}[]{lll}\Gamma&=&\sum\limits_{i=0}^{\infty}(E^{D})^{2i+1}F^{i}F^{\pi}+(E^{\pi}-\sum\limits_{j=1}^{\infty}(E^{D})^{2j}F^{j})F^{\pi}EF^{D}\\ &+&\sum\limits_{i=1}^{\infty}[(E^{\pi}-\sum\limits_{j=1}^{\infty}(E^{D})^{2j}F^{j})F^{i}F^{\pi}E+E^{\pi}E^{2i}F^{\pi}EFF^{D}\\ &+&\sum\limits_{k=0}^{i-2}E^{\pi}E^{2k+2}F^{i-k-1}F^{\pi}E]F^{D}C_{i}+\sum\limits_{i=1}^{\infty}[E^{\pi}E^{2i-1}F^{\pi}EFF^{D}\\ &-&\sum\limits_{j=0}^{\infty}(E^{D})^{2j+1}F^{i+j}F^{\pi}E+\sum\limits_{k=0}^{i-2}E^{\pi}E^{2k+1}F^{i-k-1}F^{\pi}E]A_{i},\\ \Delta&=&-F^{D}-E^{D}F^{\pi}EF^{D}-\sum\limits_{i=1}^{\infty}(E^{D})^{2i+1}F^{i}F^{\pi}EF^{D}+\sum\limits_{i=0}^{\infty}(E^{D})^{2i}F^{i}F^{\pi}\\ &-&[E^{\pi}-\sum\limits_{j=1}^{\infty}(E^{D})^{2j}F^{j}]F^{\pi}EF^{D}EF^{D}+\sum\limits_{i=1}^{\infty}[(E^{\pi}-\sum\limits_{j=1}^{\infty}(E^{D})^{2j}F^{j})F^{i}F^{\pi}E\\ &+&E^{\pi}E^{2i}F^{\pi}EFF^{D}+\sum\limits_{k=0}^{i-2}E^{\pi}E^{2k+2}F^{i-k-1}F^{\pi}E]F^{D}D_{i}\\ &+&\sum\limits_{i=1}^{\infty}[E^{\pi}E^{2i-1}F^{\pi}EFF^{D}-\sum\limits_{j=0}^{\infty}(E^{D})^{2j+1}F^{i+j}F^{\pi}E\\ &+&\sum\limits_{k=0}^{i-2}E^{\pi}E^{2k+1}F^{i-k-1}F^{\pi}E]B_{i},\\ \Lambda&=&FF^{D}+\sum\limits_{i=1}^{\infty}F^{i}F^{\pi}EA_{i},\\ \Xi&=&-FF^{D}EF^{D}+\sum\limits_{i=1}^{\infty}F^{i}F^{\pi}EB_{;}\\ \end{array}
A1=−FD​E,B1=(FD)2+FD​E2,C1=FD+F​FD​E2,D1=−FD​E−F​FD​E​FD−F​FD​E3;\begin{array}[]{lll}A_{1}&=&-F^{D}E,\\ B_{1}&=&(F^{D})^{2}+F^{D}E^{2},\\ C_{1}&=&F^{D}+FF^{D}E^{2},\\ D_{1}&=&-F^{D}E-FF^{D}EF^{D}-FF^{D}E^{3};\end{array}
Ai+1=FD​Ai−FD​E​Ci,Bi+1=FD​Bi−FD​E​Di,Ci+1=−F​FD​E​Ai+(FD+F​FD​E2)​Ci,Di+1=−F​FD​E​Bi+(FD+F​FD​E2)​Di.\begin{array}[]{lll}A_{i+1}&=&F^{D}A_{i}-F^{D}EC_{i},\\ B_{i+1}&=&F^{D}B_{i}-F^{D}ED_{i},\\ C_{i+1}&=&-FF^{D}EA_{i}+(F^{D}+FF^{D}E^{2})C_{i},\\ D_{i+1}&=&-FF^{D}EB_{i}+(F^{D}+FF^{D}E^{2})D_{i}.\end{array}
Proof.

Construct Ai,Bi,Ci,Di​(i∈ℕ)A_{i},B_{i},C_{i},D_{i}(i\in{\mathbb{N}}) as in the preceding, we easily check that

(0FDF​FD−F​FD​E​FD)2​i+1=(AiBiCiDi),\begin{array}[]{rll}\left(\begin{array}[]{cc}0&F^{D}\\ FF^{D}&-FF^{D}EF^{D}\end{array}\right)^{2i+1}&=&\left(\begin{array}[]{cc}A_{i}&B_{i}\\ C_{i}&D_{i}\end{array}\right),\end{array}
(0FDF​FD−F​FD​E​FD)2​i+2=(0FDF​FD−F​FD​E​FD)​(AiBiCiDi)=(FD​CiFD​DiF​FD​Ai−F​FD​E​FD​CiF​FD​Bi−F​FD​E​FD​Di).\begin{array}[]{ll}&\left(\begin{array}[]{cc}0&F^{D}\\ FF^{D}&-FF^{D}EF^{D}\end{array}\right)^{2i+2}\\ =&\left(\begin{array}[]{cc}0&F^{D}\\ FF^{D}&-FF^{D}EF^{D}\end{array}\right)\left(\begin{array}[]{cc}A_{i}&B_{i}\\ C_{i}&D_{i}\end{array}\right)\\ =&\left(\begin{array}[]{cc}F^{D}C_{i}&F^{D}D_{i}\\ FF^{D}A_{i}-FF^{D}EF^{D}C_{i}&FF^{D}B_{i}-FF^{D}EF^{D}D_{i}\end{array}\right).\end{array}

As the g-Drazin and Drazin inverse for a complex matrix coincide with each other, we obtain the result by [11, Theorem 2.3].∎

We now apply the preceding lemmas to obtain the following useful result.

Theorem 2.3.

Let P,Q∈ℂn×nP,Q\in{\mathbb{C}}^{n\times n}. If P​Q2=0PQ^{2}=0 and P​Q​P​(P​Q)π=0PQP(PQ)^{\pi}=0, then

(P+Q)D=(I,Q)​[(PP​QIQ)D]2​(PI),(P+Q)^{D}=(I,Q)\big{[}\left(\begin{array}[]{cc}P&PQ\\ I&Q\end{array}\right)^{D}\big{]}^{2}\left(\begin{array}[]{c}P\\ I\end{array}\right),

where

(PP​QIQ)D=∑i=0∞(LD)i+1​Ki​Kπ+∑i=0∞Li​Lπ​(KD)i+1,\left(\begin{array}[]{cc}P&PQ\\ I&Q\end{array}\right)^{D}=\sum\limits_{i=0}^{\infty}(L^{D})^{i+1}K^{i}K^{\pi}+\sum\limits_{i=0}^{\infty}L^{i}L^{\pi}(K^{D})^{i+1},
L=(000Q),LD=(000QD),K=(PP​QI0),KD=(PII0)​(HD)2​(I00P​Q),\begin{array}[]{c}L=\left(\begin{array}[]{cc}0&0\\ 0&Q\end{array}\right),L^{D}=\left(\begin{array}[]{cc}0&0\\ 0&Q^{D}\end{array}\right),\\ K=\left(\begin{array}[]{cc}P&PQ\\ I&0\end{array}\right),K^{D}=\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)(H^{D})^{2}\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right),\end{array}
HD=(ΓΔΛΞ),E=P,F=P​Q,H^{D}=\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right),E=P,F=PQ,

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi constructed as in (∗)(*).

Proof.

Clearly, we have P+Q=(I,Q)​(PI)P+Q=(I,Q)\left(\begin{array}[]{c}P\\ I\end{array}\right). By using the Cline’s formula (see [8, Lemma 1.5]), we have

(P+Q)D=(I,Q)​[(PP​QIQ)D]2​(PI).(P+Q)^{D}=(I,Q)\big{[}\left(\begin{array}[]{cc}P&PQ\\ I&Q\end{array}\right)^{D}\big{]}^{2}\left(\begin{array}[]{c}P\\ I\end{array}\right).

Moreover, we have

(PP​QIQ)=K+L,\left(\begin{array}[]{cc}P&PQ\\ I&Q\end{array}\right)=K+L,

where

K=(PP​QI0),L=(000Q).K=\left(\begin{array}[]{cc}P&PQ\\ I&0\end{array}\right),L=\left(\begin{array}[]{cc}0&0\\ 0&Q\end{array}\right).

Let

H=(PIP​Q0),E=P,F=P​Q.H=\left(\begin{array}[]{cc}P&I\\ PQ&0\end{array}\right),E=P,F=PQ.

Then F​E​Fπ=P​Q​P​(P​Q)π=0FEF^{\pi}=PQP(PQ)^{\pi}=0. In view of Lemma 2.2,

HD=(ΓΔΛΞ),H^{D}=\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right),

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi are constructed as in (∗)(*).

Obviously, we check that

K=(PII0)​(I00P​Q),H=(I00P​Q)​(PII0).\begin{array}[]{lll}K&=&\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right),\\ H&=&\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right)\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right).\end{array}

By using Cline’s formula, we have

KD=(PII0)​(HD)2​(I00P​Q).K^{D}=\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)(H^{D})^{2}\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right).

Since K​L=0KL=0, it follows by Lemma 2.1 that

(PP​QIQ)D=∑i=0∞(LD)i+1​Ki​Kπ+∑i=0∞Li​Lπ​(KD)i+1,\left(\begin{array}[]{cc}P&PQ\\ I&Q\end{array}\right)^{D}=\sum\limits_{i=0}^{\infty}(L^{D})^{i+1}K^{i}K^{\pi}+\sum\limits_{i=0}^{\infty}L^{i}L^{\pi}(K^{D})^{i+1},

where

(KD)i=(PII0)​(HD)i+1​(I00P​Q),Kπ=I2−(PII0)​HD​(I00P​Q),\begin{array}[]{rll}(K^{D})^{i}&=&\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)(H^{D})^{i+1}\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right),\\ K^{\pi}&=&I_{2}-\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)H^{D}\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right),\\ \end{array}
Ki​Kπ=(PII0)​Hi−1​Hπ​(I00P​Q)​(i≥1).K^{i}K^{\pi}=\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)H^{i-1}H^{\pi}\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right)(i\geq 1).

This completes the proof.∎

For future use, we now record the following.

Corollary 2.4.

Let P,Q∈ℂn×nP,Q\in{\mathbb{C}}^{n\times n}. If Q2=0Q^{2}=0 and P​Q​P​(P​Q)π=0PQP(PQ)^{\pi}=0, then

(P+Q)D=(I,Q)​[KD+L​(KD)2]2​(PI),E=P,F=P​Q,L=(000Q),KD=(PII0)​(ΓΔΛΞ)2​(I00P​Q),\begin{array}[]{c}(P+Q)^{D}=(I,Q)\big{[}K^{D}+L(K^{D})^{2}\big{]}^{2}\left(\begin{array}[]{c}P\\ I\end{array}\right),E=P,F=PQ,\\ L=\left(\begin{array}[]{cc}0&0\\ 0&Q\end{array}\right),K^{D}=\left(\begin{array}[]{cc}P&I\\ I&0\end{array}\right)\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right)^{2}\left(\begin{array}[]{cc}I&0\\ 0&PQ\end{array}\right),\end{array}

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi are constructed as in (∗)(*).

Proof.

Since K​L=0KL=0 and L2=0L^{2}=0, it follows by Theorem ??? that

(PP​QIQ)D=(K+L)D=KD+L​(KD)2.\begin{array}[]{lll}\left(\begin{array}[]{cc}P&PQ\\ I&Q\end{array}\right)^{D}&=&(K+L)^{D}\\ &=&K^{D}+L(K^{D})^{2}.\end{array}

Therefore we obtain the result by Theorem 2.3.∎

As a direct consequence of Theorem 2.3, we derive

Corollary 2.5.

(see [9, Theorem 2.1]) Let P,Q∈ℂn×nP,Q\in{\mathbb{C}}^{n\times n}. If P​Q2=0PQ^{2}=0 and P​Q​P=0PQP=0, then

(P+Q)D=∑i=0∞(QD)i+1​Pi​Pπ+∑i=0∞Qi​Qπ​(PD)i+1+∑i=0∞Qi​Qπ​(PD)i+2​Q+∑i=0∞(QD)i+3​Pi+1​Pπ−QD​PD​Q−(QD)2​P​PD​Q.\begin{array}[]{lll}(P+Q)^{D}&=&\sum\limits_{i=0}^{\infty}(Q^{D})^{i+1}P^{i}P^{\pi}+\sum\limits_{i=0}^{\infty}Q^{i}Q^{\pi}(P^{D})^{i+1}\\ &+&\sum\limits_{i=0}^{\infty}Q^{i}Q^{\pi}(P^{D})^{i+2}Q+\sum\limits_{i=0}^{\infty}(Q^{D})^{i+3}P^{i+1}P^{\pi}\\ &-&Q^{D}P^{D}Q-(Q^{D})^{2}PP^{D}Q.\end{array}

3. block complex matrices

The purpose of this section is to use the preceding additive results to give some representations for the Drazin inverse of block matrix MM. We come now to extend  [9, Theorem 3.1] as follows.

Theorem 3.1.

Let M=(ABCD)∈ℂ(m+n)×(m+n)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in{\mathbb{C}}^{(m+n)\times(m+n)}. If B​C​B=0,D​C​B=0,B​C​A​(B​C)π=0BCB=0,DCB=0,BCA(BC)^{\pi}=0 and D​C​A​(B​C)π=0DCA(BC)^{\pi}=0, then

MD=(I0000IC0)​[KD+L​(KD)2]2​(AB0DI00I),E=(AB0D),F=(B​C0D​C0),L=(00000000000000C0),KD=(ABI00D0II0000I00)​(ΓΔΛΞ)2​(I0000I0000B​C000D​C0),\begin{array}[]{c}M^{D}=\left(\begin{array}[]{cccc}I&0&0&0\\ 0&I&C&0\end{array}\right)\big{[}K^{D}+L(K^{D})^{2}\big{]}^{2}\left(\begin{array}[]{cc}A&B\\ 0&D\\ I&0\\ 0&I\end{array}\right),\\ E=\left(\begin{array}[]{cc}A&B\\ 0&D\end{array}\right),F=\left(\begin{array}[]{cc}BC&0\\ DC&0\end{array}\right),\\ L=\left(\begin{array}[]{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&C&0\end{array}\right),\\ K^{D}=\left(\begin{array}[]{cccc}A&B&I&0\\ 0&D&0&I\\ I&0&0&0\\ 0&I&0&0\end{array}\right)\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right)^{2}\left(\begin{array}[]{cccc}I&0&0&0\\ 0&I&0&0\\ 0&0&BC&0\\ 0&0&DC&0\end{array}\right),\end{array}

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi are constructed as in (∗)(*).

Proof.

Clearly, we have M=P+QM=P+Q, where

P=(AB0D),Q=(00C0).P=\left(\begin{array}[]{cc}A&B\\ 0&D\end{array}\right),Q=\left(\begin{array}[]{cc}0&0\\ C&0\end{array}\right).

Then

Q2=0,P​Q​P​(P​Q)π=0.Q^{2}=0,PQP(PQ)^{\pi}=0.

Therefore we complete the proof by Corollary 2.4.∎

As a consequence, we can extend [4, Theorem 2.2] as follows:

Corollary 3.2.

Let M=(ABCD)∈ℂ(m+n)×(m+n)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in{\mathbb{C}}^{(m+n)\times(m+n)}. If B​C​B=0,B​D​C=0,B​D2=0BCB=0,BDC=0,BD^{2}=0 and B​C​A​(B​C)π=0BCA(BC)^{\pi}=0, then

MD=Qπ​∑i=0∞Qi​(PD)i+1+∑i=0∞(QD)i+1​Pi​Pπ+Qπ​∑i=0∞Qi​(PD)i+2​Q+∑i=0∞(QD)i+3​Pi+1​Pπ​Q−QD​PD​Q−(QD)2​P​PD​Q,\begin{array}[]{rll}M^{D}&=&Q^{\pi}\sum\limits_{i=0}^{\infty}Q^{i}(P^{D})^{i+1}+\sum\limits_{i=0}^{\infty}(Q^{D})^{i+1}P^{i}P^{\pi}+Q^{\pi}\sum\limits_{i=0}^{\infty}Q^{i}(P^{D})^{i+2}Q\\ &+&\sum\limits_{i=0}^{\infty}(Q^{D})^{i+3}P^{i+1}P^{\pi}Q-Q^{D}P^{D}Q-(Q^{D})^{2}PP^{D}Q,\\ \end{array}
P=(ABC0),Q=(000D),PD=(I0000IC0)​[KD+L​(KD)2]2​(AB00I00I),E=(AB00),F=(B​C000),L=(00000000000000C0),KD=(ABI0000II0000I00)​(ΓΔΛΞ)2​(I0000I0000B​C00000),\begin{array}[]{c}P=\left(\begin{array}[]{cc}A&B\\ C&0\end{array}\right),Q=\left(\begin{array}[]{cc}0&0\\ 0&D\end{array}\right),\\ P^{D}=\left(\begin{array}[]{cccc}I&0&0&0\\ 0&I&C&0\end{array}\right)\big{[}K^{D}+L(K^{D})^{2}\big{]}^{2}\left(\begin{array}[]{cc}A&B\\ 0&0\\ I&0\\ 0&I\end{array}\right),\\ E=\left(\begin{array}[]{cc}A&B\\ 0&0\end{array}\right),F=\left(\begin{array}[]{cc}BC&0\\ 0&0\end{array}\right),\\ L=\left(\begin{array}[]{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&C&0\end{array}\right),\\ K^{D}=\left(\begin{array}[]{cccc}A&B&I&0\\ 0&0&0&I\\ I&0&0&0\\ 0&I&0&0\end{array}\right)\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right)^{2}\left(\begin{array}[]{cccc}I&0&0&0\\ 0&I&0&0\\ 0&0&BC&0\\ 0&0&0&0\end{array}\right),\end{array}

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi are constructed as in (∗)(*).

Proof.

Let

P=(ABC0),Q=(000D).P=\left(\begin{array}[]{cc}A&B\\ C&0\end{array}\right),Q=\left(\begin{array}[]{cc}0&0\\ 0&D\end{array}\right).

Since B​D​C=0BDC=0 and B​D2=0BD^{2}=0, we verify that P​Q2=0PQ^{2}=0 and P​Q​P=0PQP=0. In view of Corollary 2.5,

MD=Qπ​∑i=0∞Qi​(PD)i+1+∑i=0∞(QD)i+1​Pi​Pπ+Qπ​∑i=0∞Qi​(PD)i+2​Q+∑i=0∞(QD)i+3​Pi+1​Pπ​Q−QD​PD​Q−(QD)2​P​PD​Q,\begin{array}[]{rll}M^{D}&=&Q^{\pi}\sum\limits_{i=0}^{\infty}Q^{i}(P^{D})^{i+1}+\sum\limits_{i=0}^{\infty}(Q^{D})^{i+1}P^{i}P^{\pi}\\ &+&Q^{\pi}\sum\limits_{i=0}^{\infty}Q^{i}(P^{D})^{i+2}Q+\sum\limits_{i=0}^{\infty}(Q^{D})^{i+3}P^{i+1}P^{\pi}Q\\ &-&Q^{D}P^{D}Q-(Q^{D})^{2}PP^{D}Q,\\ \end{array}

This complete the proof by applying Theorem 3.1 to PP.∎

The following example illustrates that Corollary 3.4 is a nontrivial generalization of [4, Theorem 2.2].

Example 3.3.

Let M=(ABCD)∈M4​(ℂ)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in M_{4}({\mathbb{C}}), where

A=(0100),B=(0200),C=(010−1),D=(−1100).A=\left(\begin{array}[]{cc}0&1\\ 0&0\end{array}\right),B=\left(\begin{array}[]{cc}0&2\\ 0&0\end{array}\right),C=\left(\begin{array}[]{cc}0&1\\ 0&-1\end{array}\right),D=\left(\begin{array}[]{cc}-1&1\\ 0&0\end{array}\right).

Then B​C​B=0,B​D​C=0,B​D2=0BCB=0,BDC=0,BD^{2}=0 and B​C​A​(B​C)π=0BCA(BC)^{\pi}=0 while B​C≠0BC\neq 0. Clearly, we compute that

AD=0,DD=(−1100).A^{D}=0,D^{D}=\left(\begin{array}[]{cc}-1&1\\ 0&0\end{array}\right).

Construct E,F,PE,F,P and QQ as in Theorem 3.1. Obviously, we have Q2=0Q^{2}=0. Hence, by Corollary 3.2, we obtain the exact value MDM^{D}:

MD=PD+Q​(PD)2=(0000000002−110000).\begin{array}[]{lll}M^{D}&=&P^{D}+Q(P^{D})^{2}\\ &=&\left(\begin{array}[]{cccc}0&0&0&0\\ 0&0&0&0\\ 0&2&-1&1\\ 0&0&0&0\\ \end{array}\right).\end{array}

In a similar way as it was done in Theorem 3.1, using the another splitting, we have

Theorem 3.4.

Let M=(ABCD)∈ℂ(m+n)×(m+n)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in{\mathbb{C}}^{(m+n)\times(m+n)}. If A​B​C=0,C​B​C=0,A​B​D​(C​B)π=0ABC=0,CBC=0,ABD(CB)^{\pi}=0 and C​B​D​(C​B)π=0CBD(CB)^{\pi}=0, then

MD=(I00B0I00)​[KD+L​(KD)2]2​(A0CDI00I),E=(A0CD),F=(0A​B0C​B),L=(00000000000B0000),KD=(A0I0CD0II0000I00)​(ΓΔΛΞ)2​(I0000I00000A​B000C​B),\begin{array}[]{c}M^{D}=\left(\begin{array}[]{cccc}I&0&0&B\\ 0&I&0&0\end{array}\right)\big{[}K^{D}+L(K^{D})^{2}\big{]}^{2}\left(\begin{array}[]{cc}A&0\\ C&D\\ I&0\\ 0&I\end{array}\right),\\ E=\left(\begin{array}[]{cc}A&0\\ C&D\end{array}\right),F=\left(\begin{array}[]{cc}0&AB\\ 0&CB\end{array}\right),\\ L=\left(\begin{array}[]{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&0&B\\ 0&0&0&0\end{array}\right),\\ K^{D}=\left(\begin{array}[]{cccc}A&0&I&0\\ C&D&0&I\\ I&0&0&0\\ 0&I&0&0\end{array}\right)\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right)^{2}\left(\begin{array}[]{cccc}I&0&0&0\\ 0&I&0&0\\ 0&0&0&AB\\ 0&0&0&CB\end{array}\right),\end{array}

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi are constructed as in (∗)(*).

Proof.

Write M=P+QM=P+Q, where

P=(A0CD),Q=(0B00).P=\left(\begin{array}[]{cc}A&0\\ C&D\end{array}\right),Q=\left(\begin{array}[]{cc}0&B\\ 0&0\end{array}\right).

Then one checks that

Q2=0,P​Q​P​(P​Q)π=0.Q^{2}=0,PQP(PQ)^{\pi}=0.

Therefore we complete the proof by using Corollary 2.4.∎

As a consequence of the preceding theorem, we now derive

Corollary 3.5.

Let M=(ABCD)∈ℂ(m+n)×(m+n)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in{\mathbb{C}}^{(m+n)\times(m+n)}. If C​A​B=0,C​B​C=0,A2​B=0CAB=0,CBC=0,A^{2}B=0 and C​B​D​(C​B)π=0CBD(CB)^{\pi}=0, then

MD=Qπ​∑i=0∞Qi​(PD)i+1+∑i=0∞(QD)i+1​Pi​Pπ+P​∑i=0∞(QD)i+2​Pi​Pπ+P​∑i=0∞Qπ​Qi+1​(PD)i+3−P​QD​PD−P​Q​QD​(PD)2,\begin{array}[]{rll}M^{D}&=&Q^{\pi}\sum\limits_{i=0}^{\infty}Q^{i}(P^{D})^{i+1}+\sum\limits_{i=0}^{\infty}(Q^{D})^{i+1}P^{i}P^{\pi}+P\sum\limits_{i=0}^{\infty}(Q^{D})^{i+2}P^{i}P^{\pi}\\ &+&P\sum\limits_{i=0}^{\infty}Q^{\pi}Q^{i+1}(P^{D})^{i+3}-PQ^{D}P^{D}-PQQ^{D}(P^{D})^{2},\\ \end{array}
P=(A000),Q=(0BCD),\begin{array}[]{c}P=\left(\begin{array}[]{cc}A&0\\ 0&0\end{array}\right),Q=\left(\begin{array}[]{cc}0&B\\ C&D\end{array}\right),\end{array}
QD=(I00B0I00)​[KD+L​(KD)2]2​(00CDI00I),E=(00CD),F=(000C​B),L=(00000000000B0000),KD=(00I0CD0II0000I00)​(ΓΔΛΞ)2​(I0000I000000000C​B),\begin{array}[]{c}Q^{D}=\left(\begin{array}[]{cccc}I&0&0&B\\ 0&I&0&0\end{array}\right)\big{[}K^{D}+L(K^{D})^{2}\big{]}^{2}\left(\begin{array}[]{cc}0&0\\ C&D\\ I&0\\ 0&I\end{array}\right),\\ E=\left(\begin{array}[]{cc}0&0\\ C&D\end{array}\right),F=\left(\begin{array}[]{cc}0&0\\ 0&CB\end{array}\right),\\ L=\left(\begin{array}[]{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&0&B\\ 0&0&0&0\end{array}\right),\\ K^{D}=\left(\begin{array}[]{cccc}0&0&I&0\\ C&D&0&I\\ I&0&0&0\\ 0&I&0&0\end{array}\right)\left(\begin{array}[]{cc}\Gamma&\Delta\\ \Lambda&\Xi\\ \end{array}\right)^{2}\left(\begin{array}[]{cccc}I&0&0&0\\ 0&I&0&0\\ 0&0&0&0\\ 0&0&0&CB\end{array}\right),\end{array}

where Γ,Δ,Λ\Gamma,\Delta,\Lambda and Ξ\Xi are constructed as in (∗)(*).

Proof.

Write M=P+QM=P+Q, where

P=(A000),Q=(0BCD).P=\left(\begin{array}[]{cc}A&0\\ 0&0\end{array}\right),Q=\left(\begin{array}[]{cc}0&B\\ C&D\end{array}\right).

Then we have

P2​Q=0,Q​P​Q=0.P^{2}Q=0,QPQ=0.

In view of [9, Theorem 2.2], we have

MD=Qπ​∑i=0∞Qi​(PD)i+1+∑i=0∞(QD)i+1​Pi​Pπ+P​∑i=0∞(QD)i+2​Pi​Pπ+P​∑i=0∞Qπ​Qi+1​(PD)i+3−P​QD​PD−P​Q​QD​(PD)2.\begin{array}[]{lll}M^{D}&=&Q^{\pi}\sum\limits_{i=0}^{\infty}Q^{i}(P^{D})^{i+1}+\sum\limits_{i=0}^{\infty}(Q^{D})^{i+1}P^{i}P^{\pi}+P\sum\limits_{i=0}^{\infty}(Q^{D})^{i+2}P^{i}P^{\pi}\\ &+&P\sum\limits_{i=0}^{\infty}Q^{\pi}Q^{i+1}(P^{D})^{i+3}-PQ^{D}P^{D}-PQQ^{D}(P^{D})^{2}.\end{array}

Applying Theorem 2.3 to QQ, we complete the proof.∎


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