THE EXPLICIT EXPRESSION OF THE DRAZIN INVERSE OF SUMS OF TWO MATRICES AND ITS APPLICATION

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1 italian journal of pure an applie mathematics n (45 6) 45 THE EXPLICIT EXPRESSION OF THE DRAZIN INVERSE OF SUMS OF TWO MATRICES AND ITS APPLICATION Xiaoji Liu Liang Xu College of Science Guangxi University for Nationalities Nanning PR China Yaoming Yu College of Eucation Shanghai Normal University Shanghai 0034 PR China Abstract In this paper, we give explicit expressions of (P ± Q) of two matrices P an Q, in terms of P, Q, P an Q, uner the conition that P Q P, an apply the result to fining an explicit representation for the Drazin inverse of a block matrix A B uner some conitions C D Keywors: Drazin inverse, group inverse, representation, block matrix, iempotent matrix AMS Classification: 5A09 Introuction Recently, the representations of the Drazin inverse of matrices have been wiely investigate (see, for example,, 0,, 5, 6, 7 an the literature mentione below) In, Meyer an Rose presente a representation for the Drazin A B inverse of in terms of sub-blocks A, B, D an their Drazin inverses 0 D An in later, Campbell an Meyer suggeste to fin an explicit representation for the Drazin inverse of a block matrix in relation to its sub-blocks Since then, a lot of special cases of this problem have been stuie (see, for example, 3, Corresponing authors s: xiaojiliu7@6com (X Liu, Tel ), Liangxu@6com (L Xu), yaomingyu@monasheu (Y Yu)

2 46 xiaoji liu, liang xu, yaoming yu 4, 5, 6, 7, 8, 3 an references therein) But no one can solve the general problem up to the present These investigations motivate us to eal with a representation for the Drazin inverse of a block matrix by exploiting an explicit expression of the Drazin inverse of sums of two matrices The paper is organize as follows In this section, we will introuce some notions an lemmas In Section, we will present these explicit expressions of ifferences an sums of two matrices P an Q uner the conitions P Q P In Section 3, we will euce an explicit representation A B for the Drazin inverse of the block matrix uner the conitions C D AB 0 an D CB in terms of its sub-blocks an A an D Throughout this paper the symbol C m n stans for the set of m n complex matrices an I C n n stans for the unit matrix Let A C n n, the Drazin inverse, enote by A, of matrix A is efine as the unique matrix satisfying A k A A k, A AA A, AA A A where k In(A) is the inex of A In particular, if In(A), then A is calle the group inverse, enote by A g, of A (see, 9, 4) Apparently, if A is nonsingular, then In(A) 0, otherwise In(A), especially In(0) If A is nilpotent, then A 0 If X is nonsingular an B XAX, then B XA X For convenience, we write A π I AA an A B () M C D where A C n n, B C n m, C C m n an D C m m Also, we nee two functions: the ceiling function x, the smallest integer greater than or equal to x, an the floor function x, the largest integer less than or equal to x Now, we list an euce some lemmas Lemma, Theorem 3 Let M be given by () with C 0, t In(A) an l In(D) Then A S M 0 D where t S A π A n BD n l A n BD n D π A BD Lemma 8, Theorem 3 Let M be given by () with D 0, t In(A) an r In(BC) If AB 0, then XA (BC) B M CX 0

3 the explicit expression of the razin inverse of sums 47 where r X (BC) π (BC) n A n t (BC) n A n A π Lemma 3 Let P, Q C n n with s In(P ) an h In(Q) If P Q 0, then (i) for any positive integer n, () (ii)0, Theorem n (P Q) n P n Q n Q i P n i s (P Q) i h Q n P n P π Q π Q n P n Proof (i) When n, obviously, (P Q) P Q QP hols Assume that () hols for k, namely, (P Q) k P k Q k k Q i P k i, k Then for k, k (P Q) k (P k Q k Q i P k i )(P Q) P k Q k i Hence, by inuction, () hols for any positive integer n Remark : If P r 0 in Lemma 3(ii), then (P Q) r The following lemma generalizes, Theorem 784(iii) Lemma 4 Let P C n m, Q C m n, then (P Q) P (QP ) Q i k Q i P k i i Q n P n Lemma 5 Let P, Q C n n If P Q P, then for any positive integer n, (3) (Q P ) n Q n (Q P ) Proof When n, obviously, (3) hols Assume that (3) hols for k, namely, (Q P ) k Q k (Q P ) Then for k, (Q P ) k Q k (Q P )(Q P ) Q k (Q P ) Hence, by inuction, (3) hols for any positive integer n The Drazin inverse of ifferences an sums of two matrices In this section, we will investigate how to express I Q(W P ) k for any positive integer k, an (P ± Q) uner some conitions We begin with the following theorem

4 48 xiaoji liu, liang xu, yaoming yu Theorem Let P, Q C n n with In(P ) s If P Q P an Q is iempotent, then (i) () (ii) (I Q) (I Q P k ) s k P kn P π QP k, if k < s; (I Q)P π QP k, if k s () I Q (QP ) k I Q QP k Proof (i) Since Q Q, h In(I Q) From P Q P, we have P k (I Q) 0 for any positive integer k, an (P k ) π P π Thus, by Lemma 3(ii), we get (3) s 0 I Q P k (I Q) P kn P π QP k where s 0 InP k Note that n s k implies kn s an that when kn s, P kn P π 0 So (3) becomes I Q P k (I Q) s k P kn P π QP k Consequently, () hols (ii) Replacing P with QP in the proof of (i) yiels l k (4) I Q (QP ) k (I Q) (QP ) kn (QP ) kn (QP ) Q(QP ) k where l In(QP ) k Since P Q P, by Lemma 4, (QP ) Q(P Q) P QP P QP In general, by inuction, we can easily show that for any positive integer k, (5) (QP ) k QP k By the above equation an Lemma 4, P (P Q) P (QP ) Q P QP Q P Q Since (I Q)(QP ) i 0 for any positive integer i, by (4) an (5), we reach () Now, we present the expression of (P ± Q)

5 the explicit expression of the razin inverse of sums 49 Theorem Let P, Q C n n with In(P ) s an In(Q) r If P Q P an h In(P Q), then h r an (i) (6) (P Q) Q P Q Q (P Q) (ii) (7) (P Q)(P Q) Q (Q P ) (iii) (8) (P Q) Q s (h) Q π Q h P h h n Q n P n P π Q π (n) Q n P n In particular, when h r, (9) (P Q) Q s r n Q n P n P π Q π (n) Q n P n Proof Since In(P ) s, there exists a nonsingular matrix W such that P 0 P P W W 0 (0) an P 0 P W W where P is nonsingular an P is nilpotent with P s 0 Partitioning W QW conformably with W P W, we have Q W Q Q 4 Q 3 Q W Since P Q P, we can euce P Q, Q 4 0 an P Q P Thus P P Q W W 0 () an Q Q 3 P Q W W H (Q ) where H is some matrix obtaine by using Lemma Since P Q k P k 0, k max{s, }, we have (Q s P ) Q s (P Q s )P 0, P (Q ) P Q s (Q ) s 0, (Q s P ) 0 By (3) an Remark, therefore, we have (P Q ) s ( ) s (Q s P Q s ) ( ) s (Q ) s (Q ) s P

6 50 xiaoji liu, liang xu, yaoming yu an then, by Lemma 5, (P Q ) (P Q ) s (P Q ) s ( ) s (Q ) s (Q ) s P (Q s P Q s ) (Q ) P (Q ), (P Q ) (Q ) P (Q ) P (Q ) (Q ) (Q ) P (Q ) (Q ) (Q ) 3 P By Lemma, we get that (P Q) W (P Q ) Q 3 (P Q ) () W W (Q ) P (Q ) Q 3 (Q ) P (Q ) W (Q P Q )P π (Q Q3 P )P π QP P Q P Q (Q P Q )P P (Q Q3 P )(QP P P P QP P ) Q P Q (Q P Q )P P Q (QP P P P ) Q P Q Q (P Q) Since In(Q) r, Q Q r Q r By (3) an (), (P Q) (P Q) r Q (P Q)r3 ( ) r Q Qr (P Q) Hence In(P Q) In(Q P ) r (ii) By (i) an Lemma 5, ( ) r Q r (P Q) (P Q) r (P Q)(P Q) (P Q) (P Q) Q (P Q) Q (Q P ) (iii) By (i) an (ii), (Q P ) Q (Q P )(Q P ) Q 3 (Q P ) Generally, for any positive integer n, by inuction, (3) (Q P ) n Q n (Q P ) From P Q P, it follows easily that P (P Q) 0 an P Q k P k for

7 the explicit expression of the razin inverse of sums 5 k Thus, by Lemma an (3), (7) an (3), we have (P Q) P (Q P ) s h (Q P ) n (P ) n P π (Q P ) π (Q P ) n (P ) n s n Q n (Q P )P n P π (Q π Q P ) P h s n (Q n P n Q n P n )P π (Qπ Q P )P Q π h s Q P π n n Q π h (n) Q n (Q P )P n s n Q n P n P π n n (n) Q n P n n n (n) Q n (Q P )P n n Q n P n P π (Qπ Q P )P h Q π Q Q s P π n Q n P n P π Q s h h Q π Q h P h Q π n h n Q n P n P π Q π n (n) Q n P n When h r, (9) follows from (8) (n) Q n P n Q π P Q π P (n) Q n P n (h) Q π Q h P h Remark If h 0 in Theorem, then P Q is nonsingular an therefore P 0 since P Q P Similarly, if s 0, then P is nonsingular an so P Q Thus Theorem is trivial for the two special cases If P is iempotent in Theorem, then P k P π 0, k 0 So we have the following result Corollary Let P, Q C n n with r In(Q) an h In(P Q) If P Q P an P is iempotent, then h r an (P Q) Q Q h P Q π (n) Q n P (h) Q π Q h P In particular, when h r, (P Q) Q Q r P Q π (n) Q n P

8 5 xiaoji liu, liang xu, yaoming yu Since If P an Q are both iempotent in Theorem, then (P Q) Q 4 P 3 4 QP ( (P Q) (P Q) (P Q) P Q ) QP P Q, we have the result below Corollary Let P, Q C n n be iempotent If P Q P, then (P Q) QP Q an (P Q) g Q 4 P 3 4 QP Theorem 3 Let P, Q C n n If P Q P an QP Q, then (P Q) 4 (P Q ) an (P Q) 0 Proof Premultiplying QP Q by Q yiels Q P QQ an then Q P P QQ P So, by Lemma 4, we get QP QP P Q (P Q) Q P (QP ) Q Q 3 Q 4 Q QQ Q P, an then Q QP Q Thus for n, (4) (5) Q P n (I P P ) (Q Q P P )P n 0, (I QQ )Q n P (I QQ )Q n Q P 0 Since P Q P, by Theorem (iii), we have (8) an put (4) an (5) in (8) As a result, (P Q) Q 4 (I Q Q)P Q 4 (P Q ) 4 (P Q ) Since (P Q) P P Q QP Q 0, (P Q) 0 3 The Drazin inverse of block matrices In this section, we turn our attention to the representation for the Drazin inverse of a -by- block matrix M given by (), in terms of its sub-blocks an A an D, by Theorem ( A Theorem 3 Let M be given by () with t In(A), s In B ( ) C 0 A 0 q In ( 0 B 0 D ), h In C D ), an r In(BC) If D CB

9 the explicit expression of the razin inverse of sums 53 an AB 0, then h s an (3) M L K(, r) S(t) 0 BD q n n BK(n, r)a 0 K(n, r) 0 3 ( h 4 )G(0, t) 6 ( h h 4 )G(, t) (n) G(n, n) (h) G(h, h ) G(h, t) In particular, when h s, (3) M where m nk L K(, r) S(t) 0 BD n n q n BK(n, r)a 0 K(n, r) 0 3 ( s 4 )G(0, t) 6 ( s s 4 )G(, t) (n) G(n, n) 0 whenever m < k an n (33) (34) (35) (36) L (I BD C)A BK(, r )A BD S(t)A BD S(t), S(n) n G(n, m) m K(n, m) D π D k CA k A π, BD S(m)A 0 D S(m)A 0 BD3 S(m )A 0 D S(m )A 0 D nk CA nk, n is even;, n is o, Proof Assume that Y0 Y W Y 0 Y 0 Y Y 0 : W W where Y 0 Y 0 So, for k, W k W k Y k, W k 0 (Y Y ) k Y (Y Y ) k Y 0 (Y Y ) k 0 0 (Y Y ) k,

10 54 xiaoji liu, liang xu, yaoming yu an then (37) where H an (37), (38) 0 0 W k W n k W k W n k (Y Y ) k Y0 n k 0, (Y Y ) k Y Y n k HW j W k 0, for j >, partitione conformably to W For n, by Lemma 3(i) n HW n H(W n W n WW i n i ) HW n n HW n H Rewrite M as k M i A B C 0 (Y Y ) k Y Y n k 0 B 0 D n k : P Q, HW k W n k From the conitions that D CB an AB 0, it follows P Q Q, namely Q T P T (Q T )( where the) symbol F T stans for the transpose of a matrix F A 0 Note that In whenever square matrix A A 3 A or A is singular by, Theorem So q If A is nonsingular, then B 0 since AB 0 Thus D 0 an then D is singular So h Therefore, applying Theorem to M T, we have h In(P Q) In(P ) s an (39) M q P Q π (h) Q h P h P π h n Q n P n (n) Q n P n P π Obviously, for n, 0 Q n BDn 0 D n, Q n 0 BDn 0 D n I, Q π BD 0 D π By Lemma 4, ( ) (30) C BC (3) (3) C ( ) ( ) BC BC C(BC) B D CB D D, ( ) CB C DC, D C(BC) π D C(I (BC) BC) D C D DC 0

11 the explicit expression of the razin inverse of sums 55 So, by Lemma an (3), XA (BC) B P CX 0 (BC), P π π XA 0 CXA D π where X ( ) π r ( ) m BC BC A m m0 t m0 ( ) m BC A m A π an AX A A 0 Now, taking W P, we have W an W accoringly, Y 0 A, Y B, an Y C Thus, Taking H Q n (Y Y ) k in (38) for n yiels Q n P n P π Q n W n P π (33) Q n W n P π Q n W n P π ( ) k CB D k n 0 BDn 0 D n k n BDn k4 k n 0 B C 0 D k CA n k 0, an, P π CA n k (BC) π XA 0 D n k3 CA n k (BC) π XA 0 BDn k CA n k A π 0 D n k CA n k A π 0 since A(BC) π XA AA π an k n/ In orer to continue conveniently the above computation, we write V (n) n D n k CA n k A π, n, an consier it in terms of the parity of n as follows If n w, w, then since we have S((w ) ) V (w ) w w w D k CA k A π D w k CA w k A π D S((w ) )A w i0 D k CA k A π, D i4 CA i A π

12 56 xiaoji liu, liang xu, yaoming yu If n w, w, then V (w) w D w k CA w k A π D S(w)A w D i3 CA i A π i0 Hence, V (n) an then for n, G(n, n) BD V (n) 0 (34) V (n) 0 { D S(n)A, n is even; D S(n )A, BDn k n is o CA n k A π 0 D n k CA n k A π 0 We also nee to eal with Q n W n P π Likewise, we consier it in terms of the parity of n as follows When n k, Q k W k P π 0 BDk 0 D k BD CXA 0 D CXA 0 ( BC)k 0 0 D k, (BC) π XA 0 CXA D π an when n k, by (3), 0 Q k W k P π BDk 0 ( 0 D k BC)k B D k C 0 BD3 CXA 0 D CXA 0 By (30) an (3), (BC) π XA 0 CXA D π Thus D CXA D C D CXA D t D S(t)A, D t t ( ) k BC A k A π D k CA k A π D D t t D k CA k A π D S(t )A D k CA k A π D k CA k A π (35) Q n W n P π G(n, t)

13 the explicit expression of the razin inverse of sums 57 Hence, putting (34) an (35) in (33) yiels (36) Q n P n P π G(n, n) G(n, t) where n Especially, (37) Q h P h P π G(h, h ) G(h, t) By (3), 4 Q P π 8 Q P P π 0 BD (BC) π XA D CXA D π 0 BD3 A B 8 0 D (BC) π XA 0 C 0 CXA BD CXA 0 BD3 CXA 0 4 D CXA 0 8 D CXA 0 (38) G(0, t) 3 G(, t) D π h (39) Hence, we can obtain, by (36) an (38), (n) Q n P n P π 4 Q P π h 8 Q P P π n (n) G(n, n) G(n, t) h h (n) G(n, n) (n) G(n, t) n In particular, when h < 3, the first sum h n 0 in (39) Note that by (35), G(, t) G(3, t) G(k, t) an G(0, t) G(, t) G(k, t) Thus (30) h (n) G(n, t) h (k) G(k, t) h (k3) G(k, t) 3 ( 4 h )G(0, t) 6 ( 4 h )G(, t) XA 0 0 (BC) B Next, taking W P, we have W an W CX 0 an, accoringly, Y 0 XA, Y (BC) B, an Y CX Thus, an, by AB 0 an Lemma 4, (XA) i X(AX) i A XA i A, for i, Y Y CX(BC) B C ( ) ( ) BC (BC) B CB D

14 58 xiaoji liu, liang xu, yaoming yu Note that Q π Q n I BD 0 I DD 0 BDn 0 D n 0 BDn D π 0 D n D π Therefore, taking H Q n in (38) for n yiels (3) Q π Q n P n Q π Q n W n n 0 BDn D π 0 D n D π k Q π Q n W n D k CX(XA) n k 0 BDn D π CX AA n 0 D n D π CX AA n 0 Consier Q π Q n W n as follows When n k, 0 Q π Q n W k BDn D π 0 (Y Y ) k Y 0 D n D π D ky 0 0 An when n k, Hence (3) Q π Q n W k By (30) an (3), 0 BDn D π 0 D n D π Q π Q n W n 0 (Y Y ) k 0 0 D k 0 r X (BC) π A (BC) π t BDk CA k BC( BC)m A k A π an then k (BC) π A r BDπ k (BC) π A r BDπ k D k CA k D k CA k t B D k4 CA k A π BD S(t) (33) r CX D π CA D π D k r D π D k CA k K(, r) S(t), D k CA k S(t) D D S(t)

15 the explicit expression of the razin inverse of sums 59 (34) D π CX K(, r), (35) XA (BC) π A BDπ r k Dk CA k BD S(t)A ( I BD C) A BDπ r Dk CA k3 BD S(t)A ( I BD C) A BK(, r )A BD S(t)A, (36) D π CX A K(, r)xa K(, r)a By Lemma 4, (33) an (34), I Q π P BD XA (BC) B 0 D π CX 0 XA BD CX (BC) B (37) D π CX 0 XA BD S(t) BD, K(, r) 0 an, by (3), (3) an (36) (38) Q π Q n P n BDn K(, r)a A n 0 D n K(, r)a A n 0 BK(n, r)a 0 K(n, r) 0 Consequently, by (37), (38),(33) an (35), (39) where q P Q π n Q n P n XA BD XA BD S(t) CX 0 q n BK(n, r)a 0 n K(n, r) 0 L BD K(, r) S(t) 0 BD K(, r) 0 q n n BK(n, r)a 0 K(n, r) 0 L (I BD C)A BK(, r )A BD S(t)A BD S(t) Especially, when q, the sum q n 0 in (39)

16 60 xiaoji liu, liang xu, yaoming yu Putting (37), (39), (30), (3) an (39) in (39) yiels (3) If h s, then (39) becomes M q P Q π s n Q n P n (n) Q n P n P π So, putting (39), (30), (3) an (39) in the above equation yiels (3) Aing some restrictions to D, we have the following results Corollary 33 Let M be given by () If D CB 0 an AB 0, then (330) M A BCA 3 BDCA4 0 CA DCA3 0 Proof Assume that inices of matrices mentione are the same as those in Theorem 3 Since D 0, then D 0 Putting D 0 in (33) (36), we have S(n) 0, K(, m) CA, G(n, m) 0, K(, m) DCA 3, L A BCA 3, K(n, m) 0 (n 3) Substituting the above equations in (3) yiels (330) ( A Corollary 34 Let M be given by () with t In(A), s In B ) ( C 0 A 0 q In ( 0 B 0 D AB 0, then h s an an h In M H BD (I D)CA S(t) 0 (33) C D ), ) If D D CB an (h) G(h, h ) G(h, t) 3 ( h 4 )G(0, t) 6 ( h h 4 )G(, t) (n) G(n, n) In particular, when h s, (33) M H BD (I D)CA S(t) 0 3 ( s 4 )G(0, t) n 6 ( s s 4 )G(, t) (n) G(n, n) n

17 the explicit expression of the razin inverse of sums 6 where H (I BDC)A qb(i D)CA 3 BS(t)A BS(t), S(n) n DCA k A π, BS(m)A 0, n is even; S(m)A 0 G(n, m) BS(m )A 0, n is o S(m )A 0 Proof Since D D, then In(D) an D D an then K(n, m) 0, n, an K(, m) (I D)CA Clearly, (333) S(n) (334) G(k, m) BS(m)A 0 S(m)A 0 n DCA k A π, ; G(k, m) BS(m )A 0 S(m )A 0, (335) L (I BDC)A B(I D)CA 3 BS(t)A BS(t) By, Theorem, q Thus when q, q BK(n, r)a n 0 B(I D)CA3 0 K(n, r) 0 n However when q, q n 0 Therefore, Define H L (q )B(I D)CA 3 (336) (I BDC)A qb(i D)CA 3 BS(t)A BS(t) Putting (333) (336) in (3) an (3), respectively, yiels (33) an (33) Acknowlegements The authors woul like to thank the referee for very etaile comments an suggestions on our previous manuscript This work was supporte by the Guangxi Natural Science Founation (03GXNSFAA09008), the Key Project of Eucation Department of Guangxi (00ZD03), Project supporte by the National Science Founation of China (36009), an Science Research Project 03 of the China-ASEAN Stuy Center (GuangxiScience Experiment Center) of Guangxi University for Nationalities

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