representations for the drazin inverses of 2 × 2 block matrices
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Applied Mathematics and Computation 217 (2010) 2833–2842
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Applied Mathematics and Computation
journal homepage: www.elsevier .com/ locate/amc
Representations for the Drazin inverses of 2 � 2 block matrices q
Li Guo a,b, Xiankun Du a,*
a School of Mathematics, Jilin University, Changchun 130012, Chinab School of Mathematics, Beihua University, Jilin 132013, China
a r t i c l e i n f o
Keywords:Block matrixDrazin inverseIndex
0096-3003/$ - see front matter � 2010 Elsevier Incdoi:10.1016/j.amc.2010.08.019
q This work was supported by ‘‘211 Project” of Jil* Corresponding author.
E-mail addresses: guomingli95@163.com (L. Guo
a b s t r a c t
Let M denote a 2 � 2 block complex matrix A BC D
� �, where A and D are square matrices,
not necessarily with the same orders. In this paper explicit representations for the Drazininverse of M are presented under the condition that BDiC = 0 for i = 0,1, . . . ,n � 1, where n isthe order of D.
� 2010 Elsevier Inc. All rights reserved.
1. Introduction
For a square matrix A 2 Cn�n, the Drazin inverse of A is a matrix AD 2 Cn�n satisfying
AAD ¼ ADA; ADAAD ¼ AD; Akþ1AD ¼ Ak
for some nonnegative integer k. It is well-known that AD always exists and is unique. If Ak+1AD = Ak for some nonnegative inte-ger k, then so does for all nonnegative integer l P k, and the smallest nonnegative integer such that the equation holds isequal to ind(A), the index of A, which is defined to be the smallest nonnegative integer such that rank (Ak+1) = rank (Ak).We adopt the convention that A0 = In, the identity matrix of order n, even if A = 0, and the index of the zero matrix is definedto be 1. We write Ap = I � AAD. For more details we refer the reader to [1,4].
The Drazin inverse is first studied by Drazin [23] in associative rings and semigroups. The Drazin inverse of complexmatrices and its applications are very important in various applied mathematical fields like singular differential equations,singular difference equations, Markov chains, iterative methods and so on [4,5,24,31,35,37].
The study on representations for the Drazin inverse of block matrices essentially originated from finding the generalexpressions for the solutions to singular systems of differential equations [3–5], and then stimulated by a problem formu-
lated by Campbell [5]: establish an explicit representation for the Drazin inverse of 2 � 2 block matrices M ¼ A BC D
� �in
terms of the blocks of the partition, where the blocks A and D are assumed to be square matrices. For a deeper discussionof applications of the Drazin inverse of a 2 � 2 block matrix, we refer the reader to [4,37]. Until now, there has been no ex-plicit formula for the Drazin inverse of general 2 � 2 block matrices. Meyer and Rose [32], and independently Hartwig andShoaf [26], first gave the formulas for block triangular matrices, and since then many less restrictive assumptions are con-sidered [2,7,9,11–14,20,22,25,26,29,32–34,36], for example,
(1) BC = 0, BD = 0 and DC = 0 (see [20]);(2) BC = 0, DC = 0 (or BD = 0) and D is nilpotent (see [25]);
. All rights reserved.
in University.
), duxk@jlu.edu.cn (X. Du).
2834 L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842
(3) BC = 0 and DC = 0 (see [14]);(4) BC = 0, BDC = 0 and BD2 = 0 (see [22]);(5) BDpC = 0, BDDD = 0 and DDpC = 0 (see [22]).
Related topics are to find representations for the Drazin inverse and the generalized Drazin inverse of the sum of twomatrices [6,27,28,34] and operator matrices on Banach spaces [8,10,15–19,21].
It is clear that the condition in (4) above implies that BDiC = 0 for any nonnegative integer i, by which our work is moti-vated. Though this condition looks like a restrictive one, it is really weaker than many ones in the literature, especially thosein (1)–(5) and [16] which considers the generalized Drazin inverses, as shown in Examples 4.1 and 4.2, respectively.
In this paper, explicit expressions for the Drazin inverse of the 2 � 2 block matrix M are provided under the condition thatBDiC = 0 for i = 0,1, . . . ,n � 1, where n is the order of D, from which many results are unified and many formulas can be de-rived, especially those in [14,20,22,25].
For notational convenience, we define a sum to be 0, whenever its lower limit is bigger than its upper limit.
2. Preliminary
Lemma 2.1. For D 2 Cn�n and matrices B, C of appropriate orders, if BDiC = 0 for i = 0,1, . . . ,n � 1, then BDkC = 0 and B(DD)kC = 0for any nonnegative integer k.
Proof. Let f(k) = kn � a1kn�1�� � ��an be the characteristic polynomial of D. By the Cayley–Hamilton theorem, f(D) = 0.
Thus
Dn ¼ a1Dn�1 þ � � � þ anI;
from which an induction on k yields BDkC = 0 for any nonnegative integer k. Since DD is expressible as a polynomial of D (see[1]), we have B(DD)kC = 0 for any nonnegative integer k. h
For simplicity of notation, we adopt the notation Ak(e) = (Ak + eI)�1 for any positive integer k, where e is a sufficiently smallpositive real number such that (Ak + eI)�1 exists.
Lemma 2.2 [30]. For A 2 Cn�n, the following statements are equivalent:(1) indðAÞ ¼ k;(2) k is the smallest nonnegative integer for which the limit lime?0ekA(e) exists.
Furthermore, for every nonnegative integer p, if p P ind (A) then
AD ¼ lime!0
Apþ1ðeÞAp:
Since all the limits in this paper are taken as e ? 0, we shall simply write ‘‘lim” instead of ‘‘lime?0”.
3. Main results
Let M denote a 2 � 2 block complex matrix A BC D
� �satisfying the following conditions:
A 2 Cm�m; D 2 Cn�n and BDiC ¼ 0; for i ¼ 0;1; . . . ;n� 1: ð3:1Þ
Then for any nonnegative integer k, a calculation yields
Mk ¼ Ak Bk
Ck Dk þ Nk
!; ð3:2Þ
where
Bkþ1 ¼ BkDþ AkB ¼ ABk þ BDk;
Ckþ1 ¼ CkAþ DkC ¼ DCk þ CAk;
Nkþ1 ¼ CkBþ NkD ¼ DNk þ CBk:
L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842 2835
Then
Bk ¼Xk�1
i¼0
AiBDk�i�1;
Ck ¼Xk�1
i¼0
DiCAk�i�1; ð3:3Þ
Nk ¼X
iþlþj¼k�2
DiCAlBDj;
where i, j and l are all nonnegative integers. It is easy to verify that for any nonnegative integer i
BkDiCk ¼ 0; BkDiNk ¼ 0; NkDiCk ¼ 0; NkDiNk ¼ 0: ð3:4Þ
Lemma 3.1. Let M ¼ A BC D
� �, which satisfies the condition (3.1) . Then for any nonnegative integer k !
MkðeÞ ¼ AkðeÞ �AkðeÞBkDkðeÞ�DkðeÞCkAkðeÞ DkðeÞ � Dk
; ð3:5Þ
where Dk = Dk(e)(Nk � CkAk(e)Bk)Dk(e).
Proof. Denote by R the right-hand side of (3.5). Observe that BkDk(e)Ck = 0, BkDk(e)Nk = 0, DkCk = 0 and DkNk = 0 by Lemma 2.1and (3.4). Then a routine calculation yields R(Mk + e I) = I, whence Mk(e) = R. h
The following lemma is of independent interest.
Lemma 3.2. Let M ¼ A BC D
� �, which satisfies the condition (3.1) . Then
indðAÞ 6 indðMÞ 6 indðAÞ þ 2indðDÞ:
Proof. By Lemma 3.1,
MðeÞ ¼AðeÞ �AðeÞBDðeÞ
�DðeÞCAðeÞ DðeÞ þ DðeÞCAðeÞBDðeÞ
� �:
For every nonnegative integer k,
ekMðeÞ ¼ ekAðeÞ �ekAðeÞBDðeÞ�ekDðeÞCAðeÞ ek DðeÞ þ DðeÞCAðeÞBDðeÞð Þ
!:
To prove the first inequality, take k = ind(M). Then by Lemma 2.2, limekM(e) exists, whence limekA(e) exists, and so ind(A) 6ind(M).
To prove the second inequality, let ind(A) = r, ind(D) = s and take k = r + 2s. Then from Lemma 2.2, the limits limekA(e) andlimekD(e) exist. Since
lim ekAðeÞBDðeÞ ¼ lim erAðeÞð ÞB e2sDðeÞ� �
;
lim ekDðeÞCAðeÞ ¼ lim e2sDðeÞ� �
C erAðeÞð Þ
and
lim ek DðeÞ þ DðeÞCAðeÞBDðeÞð Þ ¼ lim ekDðeÞ þ lim esDðeÞð ÞC erAðeÞð ÞB esDðeÞð Þ;
each exists, we see that limekM(e) exists. By Lemma 2.2, ind(M) 6 r + 2s. h
Remark 3.3. In Lemma 3.2 it is possible that ind(M) = ind(A) + 2 ind(D) or ind(M) < ind(D) as shown in Examples 4.3 and 4.4.
Lemma 3.4. Let M ¼ A BC D
� �, which satisfies the condition (3.1) . Then MD has the following form !
MD ¼ AD b
c DD þ d;
where b, c and d are matrices such that
2836 L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842
bc ¼ BDic ¼ bDiC ¼ 0; dc ¼ dDiC ¼ 0; bd ¼ BDid ¼ 0; d2 ¼ 0 ð3:6Þ
for any nonnegative integer i.
Proof. By Lemma 3.1,
Mkþ1ðeÞ ¼ Akþ1ðeÞ �Akþ1ðeÞBkþ1Dkþ1ðeÞ�Dkþ1ðeÞCkþ1Akþ1ðeÞ Dkþ1ðeÞ � Dkþ1
!;
where Dk+1 = Dk+1(e)(Nk+1 � Ck+1Ak+1(e)Bk+1)Dk+1(e). We compute
Mkþ1ðeÞMk ¼ Akþ1ðeÞAk U
W Dkþ1ðeÞDk þK
!;
where
U ¼ Akþ1ðeÞBk � Akþ1ðeÞBkþ1Dkþ1ðeÞDk;
W ¼ �Dkþ1ðeÞCkþ1Akþ1ðeÞAk þ Dkþ1ðeÞCk;
K ¼ �Dkþ1ðeÞCkþ1Akþ1ðeÞBk þ Dkþ1ðeÞNk
� Dkþ1ðeÞðNkþ1 � Ckþ1Akþ1ðeÞBkþ1ÞDkþ1ðeÞDk:
Let ind(A) = r and ind(D) = s. Take k = r + 2s. Then k P ind(M) by Lemma 3.2. Hence, from Lemma 2.2,
MD ¼ limðMkþ1 þ eIÞ�1Mk ¼ limAkþ1ðeÞAk U
W Dkþ1ðeÞDk þK
!¼ AD lim U
lim W DD þ lim K
!:
Since MD exists and is unique, the limits limU, limW and limK exist, and let their values be b, c and d, respectively. Then
MD ¼ AD b
c DD þ d
!:
By Lemma 2.1 and the representations of U, W and K, we get (3.6). h
We are now in the position to prove our main results.
Theorem 3.5. Let M ¼ A BC D
� �, which satisfies the condition (3.1) , and let ind(A) = r, ind(D) = s and k P max{ind(M), ind(D)}.
ThenD
!
MD ¼ A XY DD þ Z;
where
X ¼Xs�1
i¼0
ðADÞiþ2BDiDp þ ApXr�1
i¼0
AiBðDDÞiþ2 � ADBDD;
Y ¼Xr�1
i¼0
ðDDÞiþ2CAiAp þ DpXs�1
i¼0
DiCðADÞiþ2 � DDCAD;
Z ¼ YAX þX
iþlþj¼k�2
ðDDÞkþ1�iCAlBDjDp þ DpDiCAlBðDDÞkþ1�j� �
�Xk�1
i¼0
ðDDÞiþ2CAiðAX þ BDDÞ þ ðYAþ DDCÞAiBðDDÞiþ2� �
:
Proof. By Lemma 3.4,
MD ¼ AD b
c DD þ d
!;
where b, c and d satisfy the condition (3.6). Because MD exists and is unique, it is sufficient to prove b = X,c = Y and d = Z. Sincek P ind(M), the matrix MD is the unique one satisfying
MMD ¼ MDM;MDMMD ¼ MD;Mkþ1MD ¼ Mk:
L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842 2837
Since Bc = bC = 0, Bd = 0 and dC = 0 by (3.6), we have
MMD ¼ AAD Abþ BDD
CAD þ Dc Cbþ DDD þ Dd
!;
MDM ¼ ADA ADBþ bD
cAþ DDC cBþ DDDþ dD
!:
Thus MMD = MDM is equivalent to
Abþ BDD ¼ ADBþ bD; ð3:7ÞCAD þ Dc ¼ cAþ DDC; ð3:8ÞCbþ Dd ¼ cBþ dD: ð3:9Þ
Since bc = 0, bd = 0, dc = 0 and d2 = 0 by (3.6), we have
ðMDÞ2 ¼ ðADÞ2 ADbþ bDD
cAD þ DDc cbþ ðDDÞ2 þ DDdþ dDD
!;
Since Bc = BDDc = 0, Bd = BDDd = 0 and d2 = 0 by (3.6), we have
MðMDÞ2 ¼ AD AADbþ AbDD þ BðDDÞ2
CðADÞ2 þ DcAD þ DDDc DD þ C
!;
where C = CADb + CbDD + Dcb + DDDd + DdDD. Thus MD = M(MD)2 is equivalent to
b ¼ AADbþ AbDD þ BðDDÞ2; ð3:10Þc ¼ CðADÞ2 þ DcAD þ DDDc; ð3:11Þd ¼ CADbþ CbDD þ Dcbþ DDDdþ DdDD: ð3:12Þ
Noting that Bkc = 0, Bkd = 0, Nkc = 0 and Nkd = 0, we see that
Mkþ1MD ¼ Akþ1AD Akþ1bþ Bkþ1DD
Ckþ1AD þ Dkþ1c Dkþ1DD þ Ckþ1bþ Nkþ1DD þ Dkþ1d
!:
By Lemma 3.2, k P ind(A) and by assumption, k P ind(D). Thus Mk+1MD = Mk is equivalent to
Bk ¼ Akþ1bþ Bkþ1DD; ð3:13ÞCk ¼ Ckþ1AD þ Dkþ1c; ð3:14ÞNk ¼ Ckþ1bþ Nkþ1DD þ Dkþ1d: ð3:15Þ
Similarly, noting that bCk = 0, bNk = 0, dCk = 0 and dNk = 0, we see that MDMk+1 = Mk is equivalent to
Bk ¼ ADBkþ1 þ bDkþ1; ð3:16ÞCk ¼ cAkþ1 þ DDCkþ1; ð3:17ÞNk ¼ cBkþ1 þ DDNkþ1 þ dDkþ1: ð3:18Þ
Since Bk+1 = BkD + AkB, the Eq. (3.13) gives Ak+1b = BkDp � AkBDD, whence
AADb ¼ ðADÞkþ1BkDp � ADBDD: ð3:19Þ
Since Bk+1 = ABk + BDk, the Eq. (3.16) gives bDk+1 = ApBk � ADBDk, whence
bDDD ¼ ApBkðDDÞkþ1 � ADBDD:
From (3.7) and the last equation, we deduce that
AbDD ¼ bDDD þ ADBDD � BðDDÞ2 ¼ ApBkðDDÞkþ1 � BðDDÞ2:
Substituting (3.19) and the last equation into (3.10), we have
b ¼ ðADÞkþ1BkDp þ ApBkðDDÞkþ1 � ADBDD:
2838 L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842
It follows from (3.3) that
b ¼Xk�1
i¼0
ðADÞiþ2BDiDp þ ApXk�1
i¼0
AiBðDDÞiþ2 � ADBDD:
Since k P max{ind(M),ind(D)} P ind(A), we have
b ¼Xs�1
i¼0
ðADÞiþ2BDiDp þ ApXr�1
i¼0
AiBðDDÞiþ2 � ADBDD ¼ X:
Similarly, by (3.8), (3.11), (3.14), (3.17) and (3.3), we get
c ¼ ðDDÞkþ1CkAp þ DpCkðADÞkþ1 � DDCAD ¼Xr�1
i¼0
ðDDÞiþ2CAiAp þ DpXs�1
i¼0
DiCðADÞiþ2 � DDCAD ¼ Y :
Moreover, from (3.9), we obtain
DdDD ¼ dDDD þ ðcB� CbÞDD: ð3:20Þ
By (3.15),
DDDd ¼ ðDDÞkþ1ðNk � Ckþ1b� Nkþ1DDÞ:
Since Ck+1 = CkA + DkC and Nk+1 = CkB + NkD, it follows that
DDDd ¼ ðDDÞkþ1ðNk � CkAb� DkCb� NkDDD � CkBDDÞ ¼ ðDDÞkþ1NkDp � ðDDÞkþ1CkðAbþ BDDÞ � DDCb: ð3:21Þ
Similarly, from (3.18) and the fact that Bk+1 = ABk + BDk and Nk+1 = DNk + CBk, we obtain
dDDD ¼ ðNk � cBkþ1 � DDNkþ1ÞðDDÞkþ1 ¼ DpNkðDDÞkþ1 � ðcAþ DDCÞBkðDDÞkþ1 � cBDD: ð3:22Þ
Substituting (3.20) into (3.12), we have
d ¼ CADbþ CbDD þ Dcbþ DDDdþ dDDD þ ðcB� CbÞDD ¼ CADbþ Dcbþ cBDD þ DDDdþ dDDD:
By (3.21) and (3.22),
d ¼ CADbþ Dcb� DDCbþ ðDDÞkþ1NkDp þ DpNkðDDÞkþ1 � ðDDÞkþ1CkðAbþ BDDÞ � ðcAþ DDCÞBkðDDÞkþ1: ð3:23Þ
By (3.8),
CADbþ Dcb� DDCb ¼ cAb ¼ YAX:
Now substituting the last equation and (3.3) into (3.23), we get d = Z. h
Theorem 3.5 has the following dual version.
Theorem 3.6. Let M ¼ A BC D
� �with A 2 Cm�m and D 2 Cn�n, and let ind(A) = r, ind(D) = s and k P max{ind(M), ind(A)}. If
CAiB ¼ 0; for i ¼ 0;1; . . . ;m� 1;
then
MD ¼ AD þ Z0 X
Y DD
!;
where X and Y are as in Theorem 3.5 and
Z0 ¼ XDY þX
iþlþj¼k�2
ðADÞkþ1�iBDlCAjAp þ ApAiBDlCðADÞkþ1�j� �
�Xk�1
i¼0
ðADÞiþ2BDiðDY þ CADÞ þ ðXDþ ADBÞDiCðADÞiþ2� �
:
Proof. Let P ¼ 0 In
Im 0
� �. Then
M ¼ PD CB A
� �P�1:
L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842 2839
Since CAiB = 0 for i = 0,1, . . . ,m � 1, Theorem 3.5 gives
D C
B A
� �D
¼ DD Y
X AD þ Z0
!;
where X and Y are as in Theorem 3.5 and Z0 is as in the statement of Theorem 3.6. Hence we have
MD ¼ PD C
B A
� �D
P�1 ¼ AD þ Z0 XY DD
!;
which completes the proof. h
From Theorem 3.5 we can easily deduce some results in [14,22] and of course some results in [20,25] by simplifying theexpressions of X, Y and Z.
Corollary 3.7 [14]. Let M ¼ A BC D
� �where A and D are square matrices with ind(A) = r and ind(D) = s. If BC = 0 and DC = 0,
then
MD ¼ AD X
CðADÞ2 DD þ CADX þ CXDD
!;
where X is as in Theorem 3.5.
Proof. It is sufficient to simplify Y and Z in Theorem 3.5 to the form given here under the assumption that BC = 0 and DC = 0.Clearly Y = C(AD)2. Taking k = r + 2s + 2, we have
Z ¼ CADX þ CXk�2
i¼0
AiBðDDÞiþ3 � CXk�1
i¼0
ADAiBðDDÞiþ2 ¼ CADX þ CApXk�2
i¼0
AiBðDDÞiþ3 � CADBðDDÞ2 ¼ CADX þ CXDD: �
Corollary 3.8 [22]. Let M ¼ A BC D
� �where A and D are square matrices with ind(A) = r and ind(D) = s. If BC = 0, BDC = 0 and
BD2 = 0, then
MD ¼ AD ðADÞ3ðABþ BDÞY DD þ Z1
!;
where Y is as in Theorem 3.5 and
Z1 ¼ ðDDÞ3CBþXr�1
i¼0
ðDDÞiþ4CAiAp þ DpXs�1
i¼0
DiCðADÞiþ4 �X2
i¼0
ðDDÞiþ1CðADÞ3�i
!ðABþ BDÞ:
Proof. By Theorem 3.5 it is sufficient to prove that X = (AD)3(AB + BD) and Z = Z1. It is clear that
X ¼ ðADÞ2ðBDp þ ADBDDpÞ ¼ ðADÞ3ðABþ BDÞ:
It remains to prove Z = Z1. Taking k = r + 2s + 3, we have
Z ¼ YðADÞ2ðABþ BDÞ þXk�2
i¼0
ðDDÞiþ3CAiBþXk�3
i¼0
ðDDÞiþ4CAiBD�Xk�1
i¼0
ðDDÞiþ2CAiðADÞ2ðABþ BDÞ: ð3:24Þ
Furthermore,
YðADÞ2 ¼ DpXs�1
i¼0
DiCðADÞiþ4 � DDCðADÞ3;
Xk�2
i¼0
ðDDÞiþ3CAiBþXk�3
i¼0
ðDDÞiþ4CAiBD ¼ ðDDÞ3CBþXk�3
i¼0
ðDDÞiþ4CAiðABþ BDÞ: ð3:25Þ
2840 L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842
Substituting (3.25) into (3.24), we have
Z ¼ ðDDÞ3CBþ DpXs�1
i¼0
DiCðADÞiþ4 þXk�3
i¼0
ðDDÞiþ4CAi �Xk�1
i¼0
ðDDÞiþ2CAiðADÞ2 � DDCðADÞ3 !
ðABþ BDÞ
¼ ðDDÞ3CBþ DpXs�1
i¼0
DiCðADÞiþ4 þXr�1
i¼0
ðDDÞiþ4CAiAp � DDCðADÞ3 � ðDDÞ2CðADÞ2 � ðDDÞ3CAD
!ðABþ BDÞ ¼ Z1: �
Corollary 3.9 [22]. Let M ¼ A BC D
� �where A and D are square matrices with ind(A) = r and ind(D) = s. If BDpC = 0, BDDD = 0
and DDpC = 0, then
MD ¼AD Ps�1
i¼0ðADÞiþ2BDi
U0 DD þPs�1
i¼0Uiþ1BDi
0BBB@
1CCCA;
where for any nonnegative integer i,
Ui ¼Xr�1
j¼0
ðDDÞiþjþ2CAjAp þ DpCðADÞiþ2 �Xi
j¼0
ðDDÞjþ1CðADÞi�jþ1:
Proof. Since BDDD = 0, we have BDD = 0 and BDDDC = 0. Hence BDpC = 0 implies BC = 0, and DDpC = 0 implies DC = DDD2C.Thus BDiC = BDDDi+1C = 0 for any nonnegative integer i, that is, the condition in Theorem 3.5 is satisfied. Now it is clear thatX ¼
Ps�1i¼0 ðA
DÞiþ2BDi and Y = U0. We only need to prove that Z ¼Ps�1
i¼0 Uiþ1BDi. Take k = r + 2s + 2. Then by Theorem 3.5
Z ¼X
iþlþj¼k�2
ðDDÞkþ1�iCAlBDj þ Y �Xk�1
i¼0
ðDDÞiþ2CAi
!Xs�1
i¼0
ðADÞiþ1BDi
¼Xs�1
i¼0
Xk�2�i
j¼0
ðDDÞiþjþ3CAj
!BDi þ
Xs�1
i¼0
DpCðADÞiþ3 � DDCðADÞiþ2 �Xk�1
j¼0
ðDDÞjþ2CAjðADÞiþ1
!BDi:
Thus we only need to show that
Uiþ1 ¼Xk�2�i
j¼0
ðDDÞiþjþ3CAj þ DpCðADÞiþ3 � DDCðADÞiþ2 �Xk�1
j¼0
ðDDÞjþ2CAjðADÞiþ1:
Let us denote by W the right-hand side expression of the last equation. Then
W ¼Xr�1
j¼0
ðDDÞiþjþ3CAjAp þXk�2�i
j¼0
ðDDÞiþjþ3CAjþ1AD þ DpCðADÞiþ3 � DDCðADÞiþ2 �Xi
j¼0
ðDDÞjþ2CAjðADÞiþ1
�Xk�2�i
j¼0
ðDDÞiþjþ3CAjþ1AD ¼Xr�1
j¼0
ðDDÞiþjþ3CAjAp þ DpCðADÞiþ3 �Xiþ1
j¼0
ðDDÞjþ1CðADÞi�jþ2 ¼ Uiþ1: �
4. Examples
The following example gives a 2 � 2 block matrix M which does not satisfy the conditions given in (1)–(5) listed in Section1 but satisfies that in Theorem 3.5.
Example 4.1. Let M ¼ A BC D
� �, where A ¼ 1 �1
0 0
� �,
B ¼2 2 2�2 �2 �2
� �; C ¼
1 10 0�1 �1
0B@
1CA; D ¼
0 1 �10 1 00 �1 1
0B@
1CA:
Since DC – 0, BD2 – 0 and BDD – 0, the matrix M does not satisfy the conditions (1)–(5) given in Section 1. On the other hand,we can check that BDiC = 0 for i = 0,1,2,3. By Theorem 3.5, we obtain
L. Guo, X. Du / Applied Mathematics and Computation 217 (2010) 2833–2842 2841
MD ¼
1 �1 4 �6 40 0 0 �2 0�1 3 �12 11 �130 0 0 1 01 �3 12 �11 13
0BBBBBB@
1CCCCCCA:
The following example shows that the condition in Theorem 3.6 is not more restrictive than the ones in [16].
Example 4.2. Let M ¼ A BC D
� �with A ¼
0 0 1 00 0 0 10 0 0 00 0 0 0
0BB@
1CCA; B ¼
001�1
0BB@
1CCA; C ¼ 1 1 0 0ð Þ and D ¼ ð0 Þ. Since A2 = 0, we
have
AðI � AAdÞB ¼ AB ¼
1�100
0BBB@
1CCCA– 0;
CAðI � AAdÞ ¼ CA ¼ 0 0 1 1ð Þ – 0:
So M does not satisfy the conditions of Theorem 2.4 through Corollary 2.9 in [16]. However, CAiB = 0, i = 0,1, . . ., which showsthat M satisfies the condition of our Theorem 3.6.
Example 4.3. Let M ¼ A BC D
� �, where A ¼ ð1 Þ; B ¼ ð1 0 1 0 Þ; C ¼
0101
0BB@
1CCA; D ¼
1 0 0 00 1 0 00 0 0 00 0 0 0
0BB@
1CCA. Then BDiC = 0 for
i = 0,1,2. We compute that ind(A) = 0, ind(D) = 1 and ind(M) = 2. Thus ind(M) = ind(A) + 2 ind(D).
Example 4.4. Let M ¼ A BC D
� �, where A ¼ ð1 Þ; B ¼ ð0 0 1 Þ; C ¼
010
0@
1A; D ¼
1 0 00 0 10 0 0
0@
1A. Since BC = 0 and DC = 0, M
satisfies the condition given in Theorem 3.5. Note that ind(D) = 2 and ind(M) = 1. Thus ind(M) < ind(D).
Acknowledgements
The authors thank the referees for their helpful comments and suggestions.
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