Schur Complement of con-s-k-ep Matrices

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1 Advances in Linear Algebra & Matrix heory - Published Online March ( Schur Complement of con-s-k-ep Matrices Bagyalakshmi Karuna Nithi Muthugobal Ramanujan Research Centre Department of Mathematics Government Arts College (Autonomous) Kumbakonam India bkn.math@gmail.com Received February 8 ; revised March 8 ; accepted March 5 ABSRAC Necessary sufficient conditions for a schur complement of a con-s-k-ep matrix to be con-s-k-ep are determined. Further it is shown that in a con-s-k-ep r matrix every secondary sub matrix of rank r is con-s-k-ep r. We have also discussed the way of expressing a matrix of rank r as a product of con-s-k-ep r matrices. Necessary sufficient conditions for products of con-s-k-ep r partitioned matrices to be con-s-k-ep r are given. Keywords: con-s-k-ep Matrices; Partitioned Matrices; Schur Complements. Introduction Let Cn n be the space of n n complex matrices of order n. Let C n be the space of all complex n-tuples. For A Cn n let A A A * A S S A A R(A) N(A) ρ(a) denote the conjugate transpose conjugate transpose secondary transpose conjugate secondary transpose Moore-Penrose inverse range space null space rank of A respectively. A solution X of the equation AXA = A is called generalized inverses of A is denoted by A. If A Cn n then the unique solution of the equations AXA * * = A XAX = X [ AX] AX [ XA] XA [] is called the moore penrose inverse of A is denoted by A. A matrix A is called con-s-k-ep r if A r N(A) = N (A VK) or R(A) = R (KVA ). hroughout this paper let k be the fixed product of disjoint transposition in S n = { n} K be the associated permute- tion matrix. Let us define the function kx xk xk xkn. A matrix A = (a ij ) C nxn is s-k symmetric if aij an k j nk i for i j = n. A matrix AC nxn is said to be con-s-k-ep if it satisfies s the condition Ax A k( x) or equivalently N(A) = N (A VK). In addition to that A is con-s-k-ep KVA is con-ep or AVK is con-ep A is con-s-k-ep A is con-s-k-ep. Moreover A is said to be con-s-k-ep r if A is con-s-k-ep of rank r. For further properties of con-s-k-ep matrices one may refer []. In this paper we derive the necessary sufficient conditions for a schur complement of a con-s-k-ep matrix to be con-s-k-ep. Further it is shown that in a cons-k-ep r matrix every secondary submatrix of rank r is con-s-k-ep r. We have also discussed the way of expressing a matrix of rank r as a product of con-s-k-ep r matrices. Necessary sufficient conditions for products of con-s-k-ep r partitioned matrices to be con-s-k-ep r are given. In this sequel we need the following theorems. heorem. [] Let AB C then nxn N A N B R B R A B BA A for all A A{} ) N A N B R B R A B AA B for all A A {} heorem. [3] A B Let M C D then ) M M A CA M A A A B M A CA A B M A N A NC N A NB N M A N C N M A N B. Also N A M NC M D A BM A M A DC M D N A N B N M A N C N M A N B N D N B NC N D N C N M D N B N M D When M A A PA M B PA. A B then M C CA B A PC B PC Copyright SciRes. ALAM

2 B. K. N. MUHUGOBAL where P AA BB A A A C C. heorem.3 [4] Let A B Cn n U Cn n be any nonsingular matrix then RA ( ) RB ( ) RUAU RUBU ) ) N( A) N( B) NUAU NUBU. Schur Complements of con-s-k-ep Matrices In this section we consider a r r matrix M Partitioned in the form A B M (.) C D where A B C D are all square matrices. If a partitioned matrix M of the form. is con-s-k-ep then in general the schur complement of C in M that is (M/C) is not con-s-k-ep. Here necessary sufficient conditions for (M/C) to be con-s-k-ep are obtained for the class M C M C analogous to that of results in [5]. Now we consider the matrix M A M B M C M B S (.) the matrix formed by the Schur complements of M over A B C D respectively. his is also a partitioned matrix. If a partitioned matrix S of the form. is con-sk-ep then in general Schur complement of (M/C) in S that is [S/(M/C)] is not con-s-k-ep. Here the necessary sufficient conditions for [S/(M/C)] to be con-s-k-ep are obtained for the class S M C S M C analogous to that of results in [5] As an application a decomposition of a partitioned matrix into a sum of con-s-k-ep r matrices is obtained. Further it is shown that in a con-s-k-ep r matrix every secondary sub matrix of rank r is con-s-k-ep r. hroughout this section let k = k k with. K K (.3) K where K K are the permutation matrices relative to k k let V be the permutation matrix with units in its secondary diagonal of order r r partitioned in such a way that v V v (.4) heorem.5 Let S be a matrix of the form. with NM C NM A N S/ M C N( M D ) then the following are equivalent: ) S is a con-s-k-ep r matrix with k = k k V= ν. ν ) (M/C) is a con-s-k-ep S/ M C is con-s-k -EP. N M C N M D N N S M C M A. 3) Both the matrices M C M C M D M A S M C S M C are con-s-k-ep r. Proof: Since S is con-s-k-ep r with k=k k KVS is Con-EP K o K where K K are permutation o K o ν matrices associated with k k V. ν o I Consider M AM C Q P O I M DS M C I O I O S M C L. M C O Clearly P Q are non singular. Now K O O ν I M AM C I O O S M C KVPQL O K ν O O I M DS M C I M C O O K I M A M C M D S M C M A M C O S M C ν Kν O M D S M C I M C O Kν M C Kν M D S M C S M C KνM AM C M C Kν S M C M AM C M D S M C S M C Copyright SciRes. ALAM

3 B. K. N. MUHUGOBAL 3 Since NM C have N M A - M A= M AM C M C by heorem. we - that is Since NS M C N M D K ν M A =K ν M A M C M C. we have by heorem. M D M DS M C S M C. hat is Kν M D Kν M DS M C S M C. Also Kν S M C M AM C M D S M C S M C K νm B. Since S M C M B M A M C M D therefore K ν M C KVPQL K ν M A K νm D K νm B O Kν M A M B K ν O M C M D B K O O ν M A M O K ν O M C M D KVS hus KVS is factorized as KVS = KVPQL. Hence ρ( KVS) ρ( L) N( KVS) N( L). But S is con-s-k-ep. herefore KVS is con-ep (By heorem. []). N( KVS) N( KVS) N( L) N( S VK) herefore by using heorem. again we get SVK SVKLL holds for every L. We choose SVK L as L M B M D O M C S M C O M A M B O ν K O M C M D ν O O K M A M C O O M C M A M D M B As the equation (at the bottom of this page). since ρ Kν M C ρ Kν M C ρ M C ρ M C N M C N M C Hence (M/C) is con-s-k-ep. From M D M D M C M C is follows that NMC ( ) NMD N M C N M D (using (M/C) is con-s-k-ep r ). herefore NM C NM D. After substituting M B S M C M A M C M D using M A M A S M C S M C in M B M B S M C S M C M C M A SVK SVKLL M D M B M C M A O M C O S M C M D M B S M C O M C O M C M C M C A S M C S M C M D M C M C B S M C S M C M C M C M C M C K νm C Kν M C M C M C NM C N K νm C NM C Copyright SciRes. ALAM

4 4 B. K. N. MUHUGOBAL We get M B M B S M C S M C S S M C M A M C M B M C M AM C M B S M C S M C S MC M A M C M B M C M C M C S S S M AM C M B M C M C S M C S M C S M C S M C N S M C N S M C By heorem. since ρ K ν S M C ρ S M C ρ S M C we get N K ν S M C N S M C M C N S M C N S M C S is con-s-k -EP r. Further M A M A S M C S M C N S M C N M A N K ν S M C N M A N S M C N M A N S M C N M A hus ) holds ) ). Since NM A NM C NM D N M C N S M C N M D N S M C N M A holds according to the assumption by applying heorem. KVS is given by the formula KVS Kν M C Kν M CKν M D Kν M CKν M D Kν S MC Kν S M C Kν M A Kν M C Kν S MC Kν M AKν M C Kν S MC S S S M C M AKν M C S M C Kν M C M C M D M C M A Kν M C M C M D M C KVSKVS Kν M CKν M C Kν M CM C Kν M CM C M DS M C M DS M C M A Kν M C Kν M D S M C Kν M DS M C M AKν M C KνM AM C KνM AKν M C KνM AM C M DS M C M DS M C M DKν M C KνM BS M C KνM BS M C M AKν M C (.6) Copyright SciRes. ALAM

5 B. K. N. MUHUGOBAL 5 According to heorem. the assumptions N(M/C) N(M/A) N M C N M D S M C is invariant for every choice of M C Hence S Kν M B Kν M C herefore Kν S KνM C Kν M C Kν M D M C M D KνM A Kν M C Kν M D KνM BKν S M C Kν M B Kν M A Kν M C Kν S K ν M B M C M D Kν M B M C M AM C M D M B S M C Further using K ν M A S S Kν MC Kν M C Kv M A K ν M D K ν M C K ν M C K v M D. hat is K ν M A Kν S M C S M C Kν M A S KKν S M C M C M A M A S M CS M C M A Kν M D Kν M CM C Kν M D Kν M CM C M D M D M CM C M D KVS KVS reduces to the form As the Equation (a) below. Again using K ν M D S S Kν M D Kν M C Kν M C KνM A KνM A Kν M C Kν M C that is M D M D S M C S M C M A M AM C M C KVSKVS reduces to the form As the Equation (b) below. Since M C is con-s-k -EP Kν M C is con-ep. herefore we have KvM C KvM C KvM C Similarly since S M C K ν M C have Kν S M C Kν S M C KνM C Kν S M C hus KVS KVS KVS KVS KVSS VK S VKKVS KVSS VK S S KVSS S SKV S is con-s-k-ep (by heorem. []). hus ) holds ) 3) KνM C KνM A Kν S M C is con-ep if only if Kν M C K ν S M C are con-ep. herefore K ν is con-s-k -EP r. We M C M A S M C K ν KVS KVS KVS KVS Kν M CKν M C Kν M CKν M C Kν S MC Kν S MC Kν S MC Kν S MC (a) (b) Copyright SciRes. ALAM

6 6 B. K. N. MUHUGOBAL is con-ep if only if Kν con-ep. M C M A S M C M C is con-s-k -EP S M C Further NM C NM A N S M C NM D Also K νm D K ν S M C M C K ν M C are is con-s-k-ep if only if is con-s-k -EP. Kν M C is con-ep if only if K ν S M C herefore only if M C M D S M C con-ep. is con-s-k-ep if M C is con-s-k -EP S M C con-s-k -EP further NM C is N M D N S M C N M D. his proves the equivalence of ) 3). he proof is complete. heorem.7 Let S be a matrix of the form (.) with N M C N M D N S M C N M A then the following are equivalent. ) S is con-s-k-ep with k = k k where K ν K V K ν ) M C is con-s-k -EP. Further S M C is con-s-k -EP. Further NM C NM A NS M C NM D 3) Both the matrices M C M D S M C M C M A S M C are con-s-k-ep. Proof his follows from heorem.5 from the fact that S is con-s-k-ep S is con-s-k-ep. In particular when M D M A we got the following. Corollary.8 M A M B Let S = M C M A NM A N M C with N S M C N M A. hen the following are equivalent. ) S is a con-s-k-ep matrix. ) (M/C) is con-s-k -EP S M C is con-sk -EP. M C 3) he matrix M A S M C is con-s-k- EP. Remark.9 he conditions taken on S in heorem.6 heo- rem.7 are essential. his is illustrated in the following example. A B Let M C D A B C D M M A M B M C M D S M A M B M C M D S K V KV Now KVS KVS is symmetric of rank 3 KVS is con-ep S is con-s-k-ep. M C M BM DM C M A S M A M B Copyright SciRes. ALAM

7 B. K. N. MUHUGOBAL 7 M D M C S M C 3 Hence K ν S M C that is S is con-ep M C is con-s-k -EP. Also M C Kν M C is M C is con-s-k - con-ep. K ν M C is con-ep EP. Moreover NM C NM A N M D N M C. But NS M D N M D N S M C NM A. Further M C KV M A S M C is not con-ep. herefore M C M D S M C is not con-s-k-ep. hus the heorem.5 the heorem.7 as well as the corollary.8 fail. Remarks. We conclude from heorem.5 heorem.7 that for a con-s-k-ep matrix of the form. k = k k k where K ν ν the following k ν are equivalent. NM C NM A. N S M C NM D NM D N M C N S M C N M A. However this fails if we omit the condition that S is con-s-k-ep. For example A B Let M where C D A B C M ABCD M A M B M C M D M A M B M C M D S S K V KVS is not con-ep. herefore S is not con-s-k-ep. Here Kν M C is con-ep. M C is con-s-k-ep. K νm D K ν M D K ν M D M D M D A νm C νm A νm C νm D He nce S M C M C. Now. D is i nd ependent of the choice of Copyright SciRes. ALAM

8 8 B. K. N. MUHUGOBAL S M C M B M A M C M D M B M A M D M C S M C Kν S M C is not con-ep. S Also M C is not con-s-k -EP. N S M C N M D N S M C N M D. Let S be of the form. with ρs ρm C r. hen S is con-s-k-ep r K V are of the form.3.4 if only if (M/C) is con-s-k -EP r. But M A M C M C M D. Proof Let S be of the form. let k= k k with k ν K ν then k ν K ν M C KVS K ν M A K νm D K νm B Since ρs ρm C r. ρ KVS ρ K ν M C r by [ 6] NM C NM A NM C KVS Kν MC K ν S M C S M C N M D hus. holds while. fails. Remark.3. It is clear by Remark. that for a con-s-k-ep mar- By heorem. these relation equivalent to trix S formula.6 gives KVS if only if either KνM A K ν M AM C. or. holds. Corollary.4 Kν M D Kν M CM C M D Let S be a matrix of the form. with K V are of the forms.3 an d.4 respectively for which KVS is KνM B KνM AM C M D given by the formula then S is con-s-k-ep if only if Let us consider the matrices both (M/C) S M C con-s-k-ep. I M AM C Proof P his follows em.5 using I from heor Remark.3. Now we proceed to prove the most important I M C M D heorem. Q L I M C heorem.5 KVPLQ K ν I M C I K ν I M A M C I M C M D Kν M A M C M C I M C M D Kν M C I K ν M AM CM C M AM CM C M C M C Kν M C Kν M CM C M D KνM AM CM C KνM AM C M D Kν M C Kν M D KνM A KνM B K ν M A M B K ν M C M D KVS Kν M C M C M C M D Copyright SciRes. ALAM

9 B. K. N. MUHUGOBAL 9 hus KVS can be factorized as KVS = KVPLQ. Since KVP = (KVQ). We have KVP VK=Q. herefore KVS KVPLKVP VK KVPLKV KVP KVPKVLKVP [since LVK = KVL]. Since (M/C) is con-s-k -EP r. We have k v(m/c) is con-ep r. herefore N( L) NL VK (heorem. of []) NKVL NKVL By heorem.3 N KVPKVLKVP N KVPKVL KVP NKVS N ( KVS) NS N S VK S is con-s-k-ep (heorem. of []). Since ρs r S is con-s-k-ep r. Conversely let us assume that S is con-s-k-ep r. Since S is con-s-k-ep r KVS is con-ep r. Since KVS = KVPLQ one choice of KVS Q P VK KVS is con-ep M C NKVS N ( KVS) By heorem. KVS KVS KVS KVS. hat is M C K νm D K νm B K ν K ν M A K νm D K νm B K ν M C K ν M A Q M C K νm D Kν M C P VK K ν M A K ν M B As the equation (at the bottom of this page). or conversely Kν M C Kν M C M C M C KνM C Kν M C M C M D Kν M C Kν M C M C M C From it follows that N M C N K ν M C N M C N M C M C is con-s-k-ep. ρ M C Since r. From M C is con-s-k-ep r. KνM A Kν M C M C M D it follows that. Now M C K ν M A M D M D M C M CM C M D M C M D M D M C K ν M C M C Kν M C (By theorem. [ ] ) M C M C M C K ν K ν M A M C M C M D M AM C K ν M C M D M AM C M C M D Mark.6 When (M/A) is non singlular KV(M/A) is automati- cally con-ep r (M/A) is con-s-k-ep r heorem.5 reduces to the following. Corollary.7 Let S be of the form. with C non singular ρ[ S] ρm C. hen S is con-s-k-ep with K = k k ν ν M AM C ν. M C M D Kν M A M C M C M C M C M D Kν M C Kν M C Kν Kν MD Kν MB Kν MD MC MC Kν M C M C M D Copyright SciRes. ALAM

10 B. K. N. MUHUGOBAL Remark.8 When k(i) = i we have K = K = I then the heorem.5 reduces to the result for con-s-ep matrices. When KV = I then heorem.5 reduces to heorem 3 of [5]. Remark.9 heorem.5 fails if we relax the condition on the rank of S. For example let us consider the matrix S K given in Remark. ρ[ KVS] ρ[ S]. But ρ KV M C ρ M C M A ρ ( KVS) ρ K ν M A ρ ( S) ρ. KVS is not con-ep herefore S is not Con-s-k-EP. However KV M C herefore (M/C) is con-s-k -EP M C M D. M AM C M C is con-ep. hus the theorem fails. Corollary. Les S be a r x r matrix of rank r. hus S is con-s-k-ep r with K = K K where K ν V = every secondary sub K ν matrix of S of rank r is con-s-k-ep r. Proof Suppose S is con-s-k-ep r matrix then KVS is an con-ep r matrix by heorem. []. Let Kν M C be any Principal submatrix of KVS such that ρ[ KVS] ρ K ν M C r then there exists a permutation matrix P such that Kν M C Kν M D KVS PKVS P KνM A KνM B with ρkvs ρ K νm C r. By [4] KVS is con-ep r. Now we conclude from heorem.5 that Kν M C is con-ep r. hat is (M/C) is con-s-k -EPr Since [M/C] is arbitrary it follows that every secondary submatrix of rank r is con-s-k-ep r. he converse is obvious. he conditions under which a partitioned matrix is decomposed into complementary sum S of con-sk-ep matrices are given. S S called comple- mentary summs of S if S = S + S ρ S ρ S ρ S. heorem. Let S be of the form. with ρ S ρ M C ρ S M C ( where S M C M B M A M C M D K is of the form.3 V is of the form.4. If (M/C) is con-s-k -EP S M C is con-s-k -EP matrices such that M A M C M C M D M D S M C S M C M C then S can be decomposed into complementary summs of con-s-k-ep matrices. Proof Let us consider the matrices S S M C M CM C M D M AM C M C M AM C M D M A I M C M C ) aking in to account that I M C M C S M C N M AM C M A M D M C N M C vk N M A M C M C vk. C M DM C M CM C M D M AM C M DM AM C M CM C M CM C M D M AM C M DM AM C M CM C M D M AM C M DM AM C M D S M C M A M C M D M A M Copyright SciRes. ALAM

11 We obtain by [6] that ρ S ρ M C Since (M/C) is con-s-k -EP. C M C M AM C M AM C M C M D M A M C M M C M C M C M C M C M D We have by heorem.5 that is S is con-s-k -EP. Since ρs ρm C ρ S M C heorem of [6] gives N S M C NI M CM C M D I M C M C M D S M C B. K. N. MUHUGOBAL N S M C N M C I M C M C I M C M C M D S M C I M C M C herefore S S M C. hus by [7] we get ρs ρ S M C ρ S ρ S ρ S Further using.. M CM CK ν K νm C M C We obtain I M CM C S M C M A S M C v A I M CM C S M C v AI M CM C Kν S M C v AKν M CM C K ν ( S M C M A Kν K ν M C M C S M C M AKν I M C M C A I M C M C S M C M REFERENCES [] S. Krishnamoorthy K. Gunasekaran B. K. N. Muthugobal con-s-k-ep Matries Journal of Mathematical Sciences Engineering Applications Vol. 5 No. pp [] C. R. Rao S. K. Mitra Generalized Inverse of Matrices Its Applications Wiley Sons New York 97. [3] R. Penrose On Best Approximate Solutions of Linear Matrix Equations Mathematical Proceedings of the Cambridge Philosophical Society Vol. 5 No. 959 pp [4]. S. Baskett I. J. Katz heorems on Products of EP r Matrices Linear Algebra Its Applications Vol. No. 969 pp hus [5] A. R. Meenakshi On Schur Complements in an EP Matrix Periodica Mathematica Hungarica Periodica Mathematica Hungarica Vol. 6 No pp [6] D. H. Carlson E. Haynesworth. H. Markham A Generalization of the Schur Complement by Means of the Moore-Penrose Inverse SIAM Journal on Applied Mathematics Vol. 6 No. 974 pp [7] A. B. Isral. N. E. Greviue Generalized Inverses heory Applications Wiley Sons New York 974. [8] S. Krishnamoorthy K. Gunasekaran B. K. N. Muthugobal On Sums of con-s-k-ep Matrix hai Journal of Mathematics in Press. Copyright SciRes. ALAM