Weighted Drazin inverse of a modified matrix

Transcription

1 Functional Analysis, Approximation Computation 6 (2) (2014), Published by Faculty of Sciences Mathematics, University of Niš, Serbia Available at: Weighted Drazin inverse of a modified matrix Tanja Totić a a Hemijsko - tehnološka škola, Ćirila i Metodija bb, Kruševac, Serbia Abstract. We present conditions under which the weighted Drazin inverse of a modified matrix A CWD d,w WB can be expressed in terms of the weighted Drazin inverse of A the generalized Schur complement D BWA d,w WC. The results extend the earlier works about the Drazin inverse. 1. Introduction The Drazin inverse the weighted Drazin inverse are very useful because of their various applications which can be found in [1,2,4,9]. Let C n n denote the set of n n complex matrices. For A C m m, the smallest nonnegative integer k such that rank(a k+1 ) = rank(a k ) is called the index of A, is denoted by k = ind(a). Let A C m m with ind(a) = k, X C m m be a matrix such that A k+1 X = A k, XAX = X, AX = XA, (1) then X is called the Drazin inverse of A, is denoted X = A d. In particular, when ind(a) = 1, the matrix X which is satisfying (1) is called the group inverse of A, is denoted by X = A #. Let A C m n, W C n m with ind(aw) = k, X C m n be a matrix such that (AW) k+1 XW = (AW) k, XWAWX = X, AWX = XWA, (2) then X is called the W-weighted Drazin inverse of A, is denoted by X = A d,w. In particular when A is an square matrix W = I, where I is the identity matrix with proper size, (2) coincides with (1), A d,w = A d. Wei [11] studied the expressions of the Drazin inverse of a modified square matrix A CB. Chen Xu [3] discussed some representations for the weighted Drazin inverse of a modified rectangular matrix A CB under some conditions. These results can be applied to update finite Markov chains. In [5] Cvetković-Ilić, Ljubisavljević present expressions for the Drazin inverse of generalized Schur complement A CD d B in terms of Drazin inverse of A the generalized Schur complement D BA d C Mathematics Subject Classification. 15A09 Keywords. Weighted Drazin inverse; Modified matrix. Received: 12 February 2014; Accepted: 8 June 2014 Communicated by Dragan S. Djordjević address: tanja.rakocevic.totic@gmail.com (Tanja Totić)

2 Tanja Totić / FAAC 6 (2) (2014), Dopazo, Martínez-Serrano [6], Mosić [8] Shakoor, Yang, Ali [10] give representations for the Drazin inverses of a modified matrix A CD d B under new conditions to generalize some results in the literature. Recently Zhang Du give representations for the Drazin inverse of the generalized Schur complement A CD d B in terms of the Drazin inverses of A the generalized Schur complement D BA d C under less weaker conditions, which generalized results of [5,6,8,10,11]. These results extends the formula of Sherman-Morrison-Woodbury type (A CD 1 B) 1 = A 1 + A 1 C(D BA 1 C) 1 BA 1 where the matrices A, D the Schur complement D BA 1 C are invertible. Throughout this papper, let A, B, C, D C m n W C n m, (AW) π = I AWA d,w W. The generalized Schur complements will be denoted by S = A CWD d,w WB Z = D BWA d,w WC. In this paper we present conditions under which the weighted Drazin inverse of a modified matrix A CWD d,w WB can be expressed in terms of the weighted Drazin inverse of A the generalized Schur complement D BWA d,w WC. The results extend the earlier works about the Drazin inverse. 2. Weighted Drazin inverse of a modified matrix Some conclusions in [12] are obtained directly from the results. Let A, B, C, D C m n W C n m. Throughout this section we use the following notations: Lemma 2.1. If (AW) π CWD d,w WB = 0, then where S A = SWAWA d,w k = ind(aw). S = A CWD d,w WB, Z = D BWA d,w WC, (3) K = A d,w WC, H = BWA d,w (4) (AW) π = I AWA d,w W, (DW) π = I DWD d,w W. (5) (SW) d = (S A W) d + k 1 ((S A W) d ) i+2 SW(AW) i (AW) π (6) Proof. Since (AW) π CWD d,w WB = 0, we have (AW) π (A S) = 0 or alternatively (AW) π A = (AW) π S. Now we can obtain that (AW) π S A = (AW) π SWAWA d,w = (AW) π AWAWA d,w = (I AWA d,w W)AWAWA d,w = AWAWA d,w AWA d,w WAWAWA d,w = AWAWA d,w AWAWA d,w WAWA d,w = AWAWA d,w AWAWA d,w = 0. Since ((AW) π ) 2 = (AW) π (AW) π AW = AW(AW) π, then (SW(AW) π ) i = SW(AW) π SW(AW) π...sw(aw) π = SW(AW) π AW(AW) π...aw(aw) π = SWAW(AW) π...aw(aw) π =... = SW(AW) i 1 (AW) π,

3 Tanja Totić / FAAC 6 (2) (2014), for any positive integer i. Since (AW) k (AW) π = (AW) k (I AWA d,w W) = (AW) k (AW) k+1 A d,w W = (AW) k (AW) k = 0, we have SW(AW) π is nilpotent, k ind(sw(aw) π ) k + 1 so (SW(AW) π ) d = 0 (SW(AW) π ) π = I. Let ind(sw(aw) π ) = s. By [7, Theorem 2.1] for P = SW(AW) π Q = S A W we have PQ = SW(AW) π S A W = 0, P + Q = SW(AW) π + S A W = SW(I AWA d,w W) + SWAWA d,w W = SW, (SW) d = (SW(AW) π + S A W) d = s 1 ((S A W) d ) i+1 (SW(AW) π ) i (SW(AW) π ) π = (S A W) d + s ((S A W) d ) i+1 SW(AW) i 1 (AW) π i=1 = (S A W) d + s 1 ((S A W) d ) i+2 SW(AW) i (AW) π. Since s 1 k s (AW) i (AW) π = 0 for any i k, we get (SW) d = (S A W) d + k 1 ((S A W) d ) i+2 SW(AW) i (AW) π. Let (AW) e = AWA d,w W M = A d,w + KWZ d,w WH. It is not difficult to prove MW = (AW) e MW = MW(AW) e S A W = S A W(AW) e, where S A = SWAWA d,w. If (AW) π CWD d,w WB = 0 from Lemma 2.1 is satisfied then we have or (AW) π S A = 0, (I AWA d,w W)SWAWA d,w = 0, SWAWA d,w = AWA d,w WSWAWA d,w, S A W = (AW) e S A W. Now we give the following result. Lemma 2.2. Let S A = SWAWA d,w M = A d,w +KWZ d,w WH, then the following statements are equivalent: KW(DW) π Z d,w WHW = KWD d,w W(ZW) π HW; (7) AWA d,w WS A WMW = AWA d,w W; (8) MWAWA d,w WS A W = AWA d,w W; (9) KW(ZW) π D d,w WHW = KWZ d,w W(DW) π HW. (10)

4 Furthermore, (AWA d,w WS A W) # = MW. Proof. Firstly, we have Tanja Totić / FAAC 6 (2) (2014), AWA d,w WS A W = AWA d,w WSWAWA d,w W = AWA d,w W(A CWD d,w WB)WAWA d,w W = AWA d,w WAWAWA d,w W AWA d,w WCWD d,w WBWAWA d,w W = AWAWA d,w WAWA d,w W AWKWD d,w WBWA d,w WAW = AWAWA d,w W AWKWD d,w WBWA d,w WAW = AWA d,w WAW AWKWD d,w WBWA d,w WAW AWA d,w WS A WMW = (AWA d,w WAW AWKWD d,w WBWA d,w WAW)(A d,w W + KWZ d,w WHW) = AWA d,w WAWA d,w W + AWA d,w WAWKWZ d,w WHW AWKWD d,w WBWA d,w WAWA d,w W AWKWD d,w WBWA d,w WAWKWZ d,w WHW = AWA d,w W + AWKWZ d,w WHW AWKWD d,w WBWA d,w W AWKWD d,w WBWKWZ d,w WHW = AWA d,w W + AWKWZ d,w WHW AWKWD d,w WHW AWKWD d,w W(D Z)WZ d,w WHW = AWA d,w W + AWKW[Z d,w W D d,w W D d,w W(D Z)WZ d,w W]HW = AWA d,w W + AWKW[(DW) π Z d,w W D d,w W(ZW) π ]HW. From this it follows (7) is equivalent to (8). Similarly, (9) is equivalent to (10). Let us prove that (8) implies (9). Let (AW) e = AWA d,w W. Now (AW) e S A WMW = (AW) e i.e., (AW) e S A W(AW) e MW = (AW) e, by [12, Lemma 2.3] we have (AW) e MW(AW) e (AW) e S A W(AW) e = (AW) e or MW(AW) e S A W = (AW) e. Similarly (9) implies (8). Thus, the statements (8) (9) are equivalent. If any of the four conditions is satisfied, then Hence, ((AW) e S A W) # = MW. MW(AW) e S A W = (AW) e S A WMW, MW(AW) e S A WMW = MW(AW) e = MW ((AW) e S A W) 2 MW = (AW) e S A W(AW) e S A WMW = (AW) e S A W(AW) e = (AW) e S A W. Theorem 2.1. If (AW) π CWD d,w WB = 0 KW(DW) π Z d,w WHW = KWD d,w W(ZW) π HW, then (SW) d = (A d,w + KWZ d,w WH)W + k 1 ((A d,w + KWZ d,w WH)W) i+2 SW(AW) i (AW) π (11)

5 Tanja Totić / FAAC 6 (2) (2014), S d,w = ((SW) d ) 2 S; or alternatively (SW) d = (A d,w + A d,w WCWZ d,w WBWA d,w )W k 1 ((A d,w + A d,w WCWZ d,w WBWA d,w )W) i+1 A d,w WCWZ d,w WBW(AW) i (AW) π + k 1 where k = ind(aw). ((A d,w + A d,w WCWZ d,w WBWA d,w )W) i+1 A d,w WCW(Z d,w W(DW) π (ZW) π D d,w W)BW(AW) i, (12) Proof. Since (AW) π CWD d,w WB = 0, then S A W = (AW) e S A W. Using Lemma 2.1 Lemma 2.2 we have Substituting M we get (11). Since S A;d,W = M = A d,w + KWZ d,w WH, (SW) d = MW + k 1 (MW) i+2 SW(AW) i (AW) π. (A d,w + KWZ d,w WH)WSW(AW) π = (A d,w W + KWZ d,w WHW)(AW CWD d,w WBW)(AW) π = ((AW) e KWD d,w WBW + KWZ d,w WBW(AW) e KWZ d,w W(D Z)WD d,w WBW)(AW) π = KWD d,w WBW(AW) π KWZ d,w W(D Z)WD d,w WBW(AW) π = KW(Z d,w W(DW) π (ZW) π D d,w W)BW(AW) π KWZ d,w WBW(AW) π = KW(Z d,w W(DW) π (ZW) π D d,w W)BW KWZ d,w WBW(AW) π, we have (12). By Theorem 2.1, when A, B, C D are square W = I, we can get directly some results in [12]. Corollary 2.1. Let A, B, C, D C m m W = I in (3),(4),(5). Suppose A π CD d B = 0 KD π Z d H = KD d Z π H then S d = A d + KZ d H + k 1 (A d + KZ d H) i+2 SA i A π. Corollary 2.2. If (AW) π CWD d,w WB = 0, CW(DW) π Z d,w WB = 0 CWD d,w W(ZW) π B = 0, then (SW) d = (A d,w + A d,w WCWZ d,w WBWA d,w )W k 1 ((A d,w + A d,w WCWZ d,w WBWA d,w )W) i+1 A d,w WCWZ d,w WBW(AW) i (AW) π + k 1 ((A d,w + A d,w WCWZ d,w WBWA d,w )W) i+1 A d,w WCW(Z d,w W(DW) π (ZW) π D d,w W)BW(AW) i,

6 Tanja Totić / FAAC 6 (2) (2014), where k = ind(aw). Corollary 2.3. If (AW) π CWD d,w WB = 0 (DW) π = (ZW) π, then (SW) d = (A d,w + A d,w WCWZ d,w WBWA d,w )W ((A d,w + A d,w WCWZ d,w WBWA d,w )W) i+1 A d,w WCWZ d,w WBW(AW) i (AW) π, k 1 where k = ind(aw). The following theorem can be proved similary to Theorem 2.1. Theorem 2.2. If CWD d,w WBW(AW) π = 0 KW(ZW) π D d,w WHW = KWZ d,w W(DW) π HW then (SW) d = (A d,w + KWZ d,w WH)W + k 1 (AW) i (AW) π SW((A d,w + KWZ d,w WH)W) i+2. References [1] A. Ben-Israel, T.N.E. Greville, Generalized Inverses: Theory Applications, 2ed., Springer, New York, [2] S.L. Campbell, C.D. Meyer, Generalized Inverse of Linear Transformations, Dover, New York, [3] J. Chen, Z. Xu, Representations for the weighted Drazin inverse of a modified matrix, Appl. Math. Comput. 203 (2008) [4] R.E. Cline, T.N.E. Greville, A.Drazin inverse for rectangular matrices. Linear Algebra Appl.29, (1980) [5] D.S. Cvetković-Ilić, J. Ljubisavljević, A note on the Drazin inverse of a modified matrix. Acta Math. Sci. Ser. B 32 (2012) [6] E. Dopazo, M.F. Martinez-Serrano, On deriving the Drazin inverse of a modified matrix, Linear Algebra Appl. 438 (2013) [7] R.E. Hartwig, G. Wang, Y.Wei, Some additive results on Drazin inverse, Linear Algebra Appl 322 (2001) [8] D.Mosić, Some results on the Drazin inverse of a modified matrix, Calcolo 50 (2013) [9] V.Rakočević, Y.Wei, A weighted Drazin inverse applications. Linear Algebra Appl. 350 (2002) [10] A.Shakoor, H.Yang, I.Ali, Some representations for the Drazin inverse of a modified matrix, Calcolo, Published online: 8 October 2013 [DOI /s ]. [11] Y. Wei, The Drazin inverse of a modified matrix, Appl. Math. Comput. 125 (2002) [12] D.Zhang, X.Du, Representations for the Drazin inverse of the generalized Schur complement,