5_xie.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 5, No. 4, 2012, 480-491 ISSN 1307-5543 – www.ejpam.com The Drazin Inverses of Combinations of Two Idempotents Tao Xie∗, Kezheng Zuo Math Department, Hubei Normal University, Hubei, Huangshi, 435002, China Abstract. By using the methods of splitting operator’s matrix into blocks and space decompositions, the existence and calculation formulas of Drazin inverse of the combinations aP + bQ+ cPQ+ dQP of two idempotent operators P and Q on a Hilbert space are obtained under the conditions PQP = 0, PQP = P and PQP = PQ respectively. These generalized the related results of Deng’s work, which characterized the Drazin inverse of the sum and difference of two idempotents. 2010 Mathematics Subject Classifications: 15A09, 47A05 Key Words and Phrases: Idempotent operator; Drazin inverse; combination 1. Preliminaries Let H be a Hilbert space, the set of all bounded linear operators on H is denoted by B(H ). For an operator T ∈ B(H ), N (T ) and R(T ) denote the null space and the range of T , respectively. An operator P ∈ B(H ) is said to be idempotent if P2 = P. If P satisfies P2 = P = P∗ then P is called orthogonal projector, where P∗ is the conjugate operator of P ∈ B(H ). Let T ∈ B(H ), if there exists an operator T D ∈ B(H ) and nonnegative integer k such that T T D = T DT, T DT T D = T D, T k+1T D = T k, then T D is called a Drazin inverse of T . The least integer k such that the above identities are hold is called the index of T , which is denoted by ind(T ) = k. Specifically, if k = 0, then T is invertible and T D = T−1. For Drazin invertible operator T ∈ B(H ), the Drazin inverse T D of T is unique [13]. The set of all idempotents in B(H ) is invariant under similarity, that is, if P is an idempo- tent operator and S ∈ B(H ) is an invertible operator, then S−1PS is also an idempotent oper- ator. Moreover the Drazin invertibility is also invariant under similarity, that is, if T is Drazin invertible and S is invertible, then S−1TS is Drazin invertible and (S−1TS)D = S−1T DT . Two facts are well known on a Hilbert space, one is that the orthogonal operator P is Drazin in- vertible and PD = P, another is that for any idempotent operator P, there exists an invertible ∗Corresponding author. Email addresses: xietao_1294�163. om (T. Xie), xiangzuo28�yahoo. n (K. Zuo) http://www.ejpam.com 480 c© 2012 EJPAM All rights reserved. T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 481 operator S such that S−1PS is an orthogonal projector [13]. In the following discussion, given two idempotent operators P and Q onH , without loss of generality, we may assume that P is orthogonal. The concept of a Drazin inverse was shown to be very useful in various applied mathe- matical settings which can be found in references [2, 7, 9, 10]. The problem of finding the Drazin inverse (P ± Q)D of the sum and difference of two idempotents P and Q was first considered by Drazin in 1958 in his celebrated paper [3]. Herein, it was proved that (P +Q)D = PD +QD provided PQ = QP = 0. The general question of how to express (P+Q)D as a function of P,Q, PD,QD, without side condition, is very difficult and remains open [8]. In 2009, Deng extended Drazin’s result to the three different cases (i)PQP = 0; (ii)PQP = P; (iii)PQP = PQ, see [4]. These cases are useful in several applications, such as in the splitting of operators and iteration theory. Zhang and Wu discussed the Drazin inverse of the linear combinations of two idempotents in a Banach algebras and represent the Drazin inverse as a function of P, Q, PQ, QP, PQP, QPQ [14]. In 2010, Zuo considered a special combination aP+ bQ− cPQ of two idempotent matrices over complex numbers, and obtained that r(aP + bQ− cPQ) = ( r(P −Q), whenc = a+ b r(P +Q), whenc 6= a+ b, where r(A) represents the rank of the matrix A [15]. Later, Xie and Zuo found that the Fredholmness of aP + bQ− cPQ is independent of choices of scalars a, b, c ∈ C with ab 6= 0 [12]. After that, Liu, Wu and Yu discussed the group invertibility of combinations of two idempotents and represent the group inverse as a function of P, Q, PQ, QP, PQP, QPQ [11]. Under the above works, we consider the Drazin invertibility of combinations aP + bQ + cPQ + dQP of two idempotent operators P and Q on H . Under the conditions PQP = 0, PQP = P and PQP = PQ, the representations for the Drazin inverse of aP + bQ+ cPQ+ dQP as a functions of P, Q, PQ, QP, PQP, QPQ are obtained by using the technique of splitting matrices into blocks and space decompositions. The following two Lemmas which were proved for a bounded linear operator [5] and for arbitrary elements in a Banach algebra [1]. Lemma 1. Let A∈ B(X ), B ∈ B(Y ) and C ∈ B(Y, X ). If A and B are Drazin invertible, then M = � A C 0 B � , N = � B 0 C A � T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 482 are Drazin invertible and M D = � AD X 0 BD � , N D = � BD 0 X AD � , where X = (AD)2[ ∑∞ i=0(A D)iCBi](I − BBD) + (I −AAD)[ ∑∞ i=0 AiC(BD)i](BD)2 −ADCBD. Lemma 2. Let A∈ B(X ), B ∈ B(Y ) and C ∈ B(Y, X ). If A is invertible and Bk = 0, then M = � A 0 C B � are Drazin invertible and M D = � A−1 0 X 0 � , where X = ∑k−1 i=0 Bk−1−iCAi−k−1. Lemma 3 (see [6]). Let A, B ∈ B(H ). Then the following conditions are equivalent. (i) R(B) ⊆R(A); (ii) There exists D ∈ B(H ) such that B = AD. 2. Main results Theorem 1. Let P and Q be two idempotents in B(H ), and a, b, c, d ∈ C, ab 6= 0. If PQP = 0, then aP + bQ+ cPQ+ dQP is Drazin invertible and (aP + bQ+ cPQ+ dQP)D = 1 a P + 1 b Q− ( 1 a + 1 b + c ab )PQ− ( 1 a + 1 b + d ab )QP + ( 1 a + 2 b + c ab + d ab + cd ab2 )QPQ. Proof. Let P and Q be two idempotent operators in B(H ). With out loss of generality, we assume that P is an orthogonal projector. By Lemma 3, the condition PQP = 0 implies that R(QP)⊆N (P) and R(QP)⊆R(Q). Observing that Q(R(QP)⊕R(P))⊆R(QP), the space H can be decomposed as H =R(QP)⊕R(P)⊕ (R(QP)⊥⊖R(P)). Then P and Q can be represented as P =    0 0 0 0 I 0 0 0 0    , Q =    I Q12 Q13 0 0 Q23 0 0 Q33    , T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 483 where R(QP) denotes the closure ofR(QP). On the other hand, Q2 = Q gives that Q2 33 = Q33 and R(QP)⊥⊖R(P) =R(Q33)⊕R(Q33) ⊥. It follows that P and Q can be written as P =      0 0 0 0 0 I 0 0 0 0 0 0 0 0 0 0      , Q =      I Q12 Q′13 Q′′13 0 0 Q′23 Q′′23 0 0 I Q′′33 0 0 0 0      , under the space decompositionH =R(QP)⊕R(P)⊕R(Q33)R(Q33) ⊥. The idempotency of Q implies that Q′23Q′′33 = Q′′23, Q12Q′23 +Q′13 = 0, Q12Q′′23 +Q′13Q′′33 = 0. Direct calculations show that aP + bQ+ cPQ+ dQP =      bI (b+ d)Q12 bQ′13 bQ′′13 0 aI (b+ c)Q′23 (b+ c)Q′′23 0 0 bI bQ′′33 0 0 0 0      It is clear that the condition a, b 6= 0 implies the invertibility of    bI (b+ d)Q12 bQ′13 0 aI (b+ c)Q′23 0 0 bI    on R(QP)⊕R(P)⊕R(Q33) and its inverse is     1 b I − b+d ab Q12 −[ (b+c)(b+d)+ab ab2 ]Q′13 0 1 a I − b+c ab Q′23 0 0 1 b I     . Moreover,     1 b I − b+d ab Q12 −[ (b+c)(b+d)+ab ab2 ]Q′13 0 1 a I − b+c ab Q′23 0 0 1 b I     2   bQ′′13 (b+ c)Q′′23 bQ′′33    =     1 b Q′′13 − [ (b+c)(b+d) ab2 + 2 b ]Q′13Q′′33 −( 1 a + c ab )Q′′23 1 b Q′′33     . Applying B = 0 to the formula of representing Drazin inverse of upper triangle block matrix in Lemma 1, we have (aP + bQ+ cPQ+ dQP)D = T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 484       1 b I − b+d ab Q12 − (b+c)(b+d)+ab ab2 Q′13 1 b Q′′13 − [ (b+c)(b+d) ab2 + 2 b ]Q′13Q′′33 0 1 a I − b+c ab Q′23 − b+c ab Q′′23 0 0 1 b I 1 b Q′′33 0 0 0 0       . Moreover, through direct calculations, we have PQ =      0 0 0 0 0 0 Q′23 Q′′23 0 0 0 0 0 0 0 0      , QP =      0 Q12 0 0 0 0 0 0 0 0 0 0 0 0 0 0      , and QPQ =      0 0 Q12Q′23 Q12Q′′23 0 0 0 0 0 0 0 0 0 0 0 0      . Therefore, (aP + bQ+ cPQ+ dQP)D = 1 a P + 1 b Q− ( 1 a + 1 b + c ab )PQ− ( 1 a + 1 b + d ab )QP +( 1 a + 2 b + c ab + d ab + cd ab2 )QPQ. Now we can derive some special cases from Theorem 1. These results are also the cases of Theorem 2.1 in [4]. Corollary 1. Let P and Q be two idempotents in B(H ). Assume that PQP = 0, then the following statements hold. (i) (P +Q)D = P +Q− 2(PQ+QP) + 3QPQ. (ii) (P −Q)D = P −Q−QPQ. If either of the stronger condition PQ = 0 or QP = 0 is satisfied, then by Theorem 1, we obtain the following results. Corollary 2. Let P and Q be two idempotents in B(H ) and a, b ∈ C, ab 6= 0. Then the following statements hold. (i) If QP = 0, then (aP + bQ)D = 1 a P + 1 b Q− ( 1 a + 1 b )PQ. (ii) If PQ = 0, then (aP + bQ)D = 1 a P + 1 b Q− ( 1 a + 1 b )QP. Next we discuss the Drazin inverse of aP + bQ+ cPQ + dQP under the assumption that PQP = P. T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 485 Theorem 2. Let P and Q be two idempotents in B(H ), then for any a, b, c, d ∈ C, ab 6= 0, the combinations aP + bQ+ cPQ + dQP are Drazin invertible under the condition PQP = P. The Drazin inverses of aP + bQ+ cPQ+ dQP can be represented as following: (i) If a+ b+ c + d 6= 0, then (aP + bQ+ cPQ+ dQP)D = (a+ c)(a+ d) (a+ b+ c + d)3 P + 1 b Q+ (b+ c)(a+ c) (a+ b+ c + d)3 PQ + (a+ d)(b+ d) (a+ b+ c + d)3 QP + [ (b+ c)(b+ d) (a+ b+ c + d)3 − 1 b ]QPQ. (ii) If a+ b+ c + d = 0, then (aP + bQ+ cPQ+ dQP)D = 1 b (Q−QPQ). Proof. If PQP = P, then P and Q can be written as P = � I 0 0 0 � , Q = � I Q1 Q2 Q3 � under the space decomposition of H = R(P)⊕R(P)⊥. The idempotency of Q yields that Q1Q2 = 0, Q1Q3 = 0, Q3Q2 = 0 and Q2Q1 + Q2 3 = Q3. It follows that R(Q2) ⊆ N (Q1), R(Q2)⊆N (Q3), R(Q3)⊆N (Q1). With respect to the space decomposition H =R(Q1)⊕R(Q1) ⊥ ⊕R(Q2)⊕R(Q2) ⊥, P and Q can be represented as P =      I 0 0 0 0 I 0 0 0 0 0 0 0 0 0 0      , Q =      I 0 0 Q′11 0 I 0 0 Q21 Q22 0 Q31 0 0 0 Q32      , where Q11Q32 = 0,Q2 32 = Q32 and Q21Q11+Q31Q32 = Q31. So, under the space decomposition of H =R(Q1)⊕R(Q1) ⊥ ⊕R(Q2)⊕R(Q32)⊕R(Q32) ⊥, the operators P and Q can then be further written as P =        I 0 0 0 0 0 I 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0        , Q =        I 0 0 0 Q′′11 0 I 0 0 0 Q21 Q22 0 Q′31 Q′′31 0 0 0 I Q′′32 0 0 0 0 0        , where Q21Q′′11 +Q′31Q′′32 = Q′′31. (i) If a+ b+ c + d 6= 0, then aP+bQ+cPQ+dQP =        (a+ b+ c + d)I 0 0 0 (b+ c)Q′′11 0 (a+ b+ c + d)I 0 0 0 (b+ d)Q21 (b+ d)Q22 0 bQ′31 bQ′′31 0 0 0 bI bQ′′32 0 0 0 0 0        . T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 486 Since b 6= 0, let a′ = a b , c′ = c b , d ′ = d b , then we consider the following combination a′P+Q+c′PQ+d ′QP =        (a′ + 1+ c′ + d ′)I 0 0 0 (1+ c′)Q′′11 0 (a′ + 1+ c′ + d ′)I 0 0 0 (1+ d ′)Q21 (1+ d ′)Q22 0 Q′31 Q′′31 0 0 0 I Q′′32 0 0 0 0 0        . Let S =        I 0 0 0 1+c′ (a′+1+c′+d′) Q′′11 0 I 0 0 0 0 0 0 I Q′′32 0 0 I 0 Q′′32 0 0 0 0 I        , then S−1 =        I 0 0 0 − 1+c′ (a′+1+c′+d′) Q′′11 0 I 0 0 0 0 0 0 I −Q′′32 0 0 I 0 −Q′′32 0 0 0 0 I        . Direct calculation shows that S(a′P +Q+ c′PQ+ d ′QP)S−1 =        (a′+ 1+ c′ + d ′)I 0 0 0 0 0 (a′+ 1+ c′ + d ′)I 0 0 0 0 0 I 0 0 (1+ d ′)Q21 (1+ d ′)Q22 Q′31 0 a′−c′d′ (a′+1+c′+d′) Q21Q′′11 0 0 0 0 0        . It follows that (a′P +Q+ c′PQ)D = S−1(S(a′P +Q+ c′PQ)S−1)DS =         1 (a′+1+c′+d′) I 0 0 0 1+c′ (a′+1+c′+d′)2 Q′′11 0 1 (a′+1+c′+d′) I 0 0 0 1+d′ (a′+1+c′+d′)2 Q21 1+d′ (a′+1+c′+d′)2 Q22 0 Q′31 (1+c′)(1+d′) (a′+1+c′+d′)3 Q21Q′′11 +Q′31Q′′32 0 0 0 I Q′′32 0 0 0 0 0         = (a′ + c′)(a′+ d ′) (a′ + 1+ c′ + d ′)3 P +Q+ (1+ c′)(a′+ c′) (a′ + 1+ c′ + d ′)3 PQ+ (1+ d ′)(a′+ d ′) (a′+ 1+ c′ + d ′)3 QP +[ (1+ c′)(1+ d ′) (a′ + 1+ c′ + d ′)3 − 1]QPQ. T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 487 Moreover, since (cT )D = 1 c T D holds for any c 6= 0 and any Drazin invertible operator T ∈ B(H ). Hence (aP + bQ+ cPQ+ dQP)D = [b(a′P +Q+ c′PQ+ d ′QP)]D = 1 b (a′P +Q+ c′PQ+ d ′QP)D = (a+ c)(a+ d) (a+ b+ c + d)3 P + 1 b Q+ (b+ c)(a+ c) (a+ b+ c + d)3 PQ + (a+ d)(b+ d) (a+ b+ c + d)3 QP +[ (b+ c)(b+ d) (a+ b+ c + d)3 − 1 b ]QPQ. (ii) If a+ b+ c + d = 0, then (aP + bQ+ cPQ+ dQP)D =        0 0 0 0 (b+ c)Q′′11 0 0 0 0 0 (b+ d)Q21 (b+ d)Q22 0 bQ′31 bQ′′31 0 0 0 bI bQ′′32 0 0 0 0 0        D = 1 b        0 0 0 0 0 0 0 0 0 0 0 0 0 Q′31 Q′31Q′′32 0 0 0 I Q′′32 0 0 0 0 0        = 1 b (Q−QPQ). Now we can derive some special cases from Theorem 2. These results are the special cases of Theorem 2.3 in [4]. Corollary 3. Let P and Q be two idempotents in B(H ). Assume that PQP = P, then the following statements hold. (i) (P +Q)D = 1 8 P +Q+ 1 8 (PQ+QP)− 7 8 QPQ. (ii) (P −Q)D = QPQ−Q. If the stronger condition QP = P is satisfied, then by Theorem 2, we can also derive the formulaes of Drazin inverses of linear combinations of P and Q. Corollary 4. Let P and Q be two idempotents in B(H ), and a, b ∈ C, ab 6= 0. If QP = P, then (aP + bQ)D = ( a (a+b)2 P + 1 b Q+ [ b (a+b)2 − 1 b ]PQ, when a+ b 6= 0 1 b (Q− PQ), when a+ b = 0. T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 488 Next we discuss the Drazin inverse of aP + bQ+ cPQ + dQP under the assumption that PQP = PQ. Theorem 3. Let P and Q be two idempotents in B(H ), then for any a, b, c, d ∈ C, ab 6= 0, the combinations aP + bQ+ cPQ+ dQP are Drazin invertible under the condition PQP = PQ. The Drazin inverses of aP + bQ+ cPQ+ dQP can be represented as following: (i) If a+ b+ c + d 6= 0, then (aP + bQ+ cPQ+ dQP)D = 1 a P + 1 b Q+ [ 1 (a+ b+ c + d) − b+ d (a+ b+ c + d)2 − 1 a ]PQ − ( a+ b+ d ab )QP +[ b+ d (a+ b+ c + d)2 + b+ d ab ]QPQ. (ii) If a+ b+ c + d = 0, then (aP + bQ+ cPQ+ dQP)D = 1 a P + 1 b Q− 1 a PQ − ( 1 a + 1 b + d ab )QP + b+ d ab QPQ. Proof. If PQP = PQ, then P and Q can be written as P = � I 0 0 0 � , Q = � Q1 0 Q2 Q3 � under the space decomposition of H = R(P)⊕R(P)⊥. The idempotency of Q yields that Q2 1 = Q1, Q2 3 = Q3, Q3Q2 = 0 and Q2Q1 +Q2 3 = Q2. With respect to the space decomposition H =R(Q1) ⊥ ⊕R(Q1)⊕R(Q ∗ 3)⊕R(Q ∗ 3), P and Q can be further represented as P =      I 0 0 0 0 I 0 0 0 0 0 0 0 0 0 0      , Q =      0 0 0 0 Q11 I 0 0 Q21 0 I 0 Q23 Q24 Q31 0      , where Q24Q11 +Q31Q21 = Q23. (i) If a+ b+ c + d 6= 0, then aP + bQ+ cPQ+ dQP =      aI 0 0 0 (b+ c + d)Q11 (a+ b+ c + d)I 0 0 (b+ d)Q21 0 bI 0 (b+ d)Q23 (b+ d)Q24 bQ31 0      . Since ab 6= 0 and a+ b+ c + d 6= 0 then the submatrix    aI 0 0 (b+ c + d)Q11 (a+ b+ c + d)I 0 (b+ d)Q21 0 bI    T. Xie, K. Zuo / Eur. J. Pure Appl. Math, 5 (2012), 480-491 489 of aP + bQ+ cPQ+ dQP is invertible and it’s inverse is     1 a I 0 0 − b+c+d a(a+b+c+d) Q11 1 a+b+c+d I 0 − b+d ab Q21 0 1 b I     By using the results of Lemma 2 we have (aP + bQ+ cPQ+ dQP)D =       1 a I 0 0 0 − b+c+d a(a+b+c+d) Q11 1 a+b+c+d I 0 0 − b+d ab Q21 0 1 b I 0 X b+d (a+b+c+d)2 Q24 1 b Q31 0       , where X = − b+d ab Q23 + [ b+d ab + b+d a2 − (b+d)(b+c+d) (a+b+c+d) ( 1 a2 + 1)]Q24Q11. The coefficients of P, Q, PQ, QP, QPQ in the expression of (aP + bQ+ cPQ + dQP)D can be obtained by solving some linear equations. Then we have (aP + bQ+ cPQ+ dQP)D = 1 a P + 1 b Q+ [ 1 (a+ b+ c + d) − b+ d (a+ b+ c + d)2 − 1 a ]PQ −( a+ b+ d ab )QP + [ b+ d (a+ b+ c + d)2 + b+ d ab ]QPQ. (ii) If a+ b+ c + d = 0, then aP + bQ+ cPQ+ dQP =      aI 0 0 0 −aQ11 0 0 0 (b+ d)Q21 0 bI 0 (b+ d)Q23 (b+ d)Q24 bQ31 0      . Let S =      I 0 0 0 0 0 I 0 0 I 0 0 0 0 0 I      , then S(aP + bQ+ cPQ+ dQP)S−1 =      aI 0 0 0 (b+ d)Q21 bI 0 0 −aQ11 0 0 0 (b+ d)Q23 bQ31 (b+ d)Q24 0      . (aP + bQ+ cPQ+ dQP)D =S−1[S(aP + bQ+ cPQ)S−1]DS REFERENCES 490 =       1 a I 0 0 0 − 1 a Q11 0 0 0 − b+d ab Q21 0 1 b I 0 − b+d ab Q31Q21 0 1 b Q31 0       = 1 a P + 1 b Q− 1 a PQ− ( 1 a + 1 b + d ab )QP + b+ d ab QPQ. Now we can derive some special cases from Theorem 3. These results are also special cases of Theorem 2.6 in [4]. Corollary 5. Let P and Q be two idempotents in B(H ). Assume that PQP = PQ, then the following statements hold. (i) (P +Q)D = P +Q− 3 4 PQ− 2QP + 5 4 QPQ. (ii) (P −Q)D = P −Q− PQ+QPQ. We can also derive the formulaes of Drazin inverses of linear combinations of P and Q under the condition PQP = PQ. Corollary 6. Let P and Q be two idempotents in B(H ). Assume that PQP = PQ, then the following statements hold. (aP + bQ)D =    1 a P + 1 b Q+ [ 1 a+b − b (a+b)2 − 1 a ]PQ − a+b ab QP + [ b (a+b)2 + 1 a ]QPQ, whe n a+ b 6= 0 1 a (P −Q− PQ+QPQ), when a+ b = 0. ACKNOWLEDGEMENTS The paper is supported by the Key Research Project and Youth Re- search Project of Educational Department of Hubei Province(D20122202) and (B20122203) of China. References [1] G.N. Castro , J.J. Koliha. New additive results for the g-Drazin inverse. Preceedings of the Royal Society of Edinburgh, 134(1):1085-1097, 2004. [2] S.L. Campbell, C.D. Meyer. Generalized inverse of linear transformations. London: Pit- man Press, 1979. [3] M.P. Drazin. Pseudoinverse in associative rings and semigroups. American Mathematical Monthly, 65:506-514, 1958. [4] Chunyuan Deng. The Drazin inverses of sum and difference of idempotents. Linear Al- gebra and its Applications, 430: 1282-1291, 2009. REFERENCES 491 [5] D.S. Djordjrvic, P.S. Stanimirovic. On the generalized Drazin inverse and generalized resolvent. Czechoslovak Mathematical Journal, 126: 671-634, 2001. [6] R.G. Douglas. On majorization factorization and range inclusion of operators in Hilbert space. Proceedings of the American Mathematical Society, 17: 413-416, 1966. [7] R.E. Hartwig, J. Levine. Applications of the Drazin inverse to the Hill cryptographic system. Crytologia, 5:67-77,1981. [8] R.E. Hartwig, G.R. Wang, Y. Wei. Some additive results on Drazin inverse. Linear Algebra and its Applications, 322:207-217,2001. [9] C.D. Meyer. The condition number of a finite Markov chains and perturbation bounds for the limiting probabilities. SIMA Journal on Algebraic Discrete Methods, 1:273-283,1980. [10] B. Simeon, C. Fuhrer, P. Rentrop. The Drazin inverse in multibody system dynamics. Numerische Mathematik, 64:521-536, 1993. [11] Xiaoji Liu, Lingling Wu, Yaoming Yu. The group inverse of the combinations of two idempotent matrices. Linear and Multilinear Algebra, 59(1):101-115, 2011. [12] T. Xie, K. Zuo. Fredholmness of combinations of two idempotents. European Journal of Pure and Applied Mathematics, 3(4):678-685, 2010. [13] G. Wang, Y. Wei, S. Qiao. Generalized inverse: theory and computations. Graduate Series in Mathematics, Beijing: Science Press, 2004. [14] Shifang Zhang, Junde Wu. The Drazin inverse of the linear combinations of two idempo- tents in the Banach algebras. Linear Algebra and its Applications, 436:3132-3138, 2012.. [15] Kezheng Zuo. Nonsingularity of the difference and the sum of two idempotents matrices, Linear Algebra and its Applications, 433:476-482, 2010.