Decomposition of a ring induced by minus partial order

Electronic Journal of Linear Algebra Volume 23 Volume 23 (2012) Article 72 2012 Decomposition of a ring induced by minus partial order Dragan S Rakic rakicdragan@gmailcom Follow this and additional works at: http://repositoryuwyoedu/ela Recommended Citation Rakic Dragan S (2012) "Decomposition of a ring induced by minus partial order" Electronic Journal of Linear Algebra Volume 23 DOI: https://doiorg/1013001/1081-38101573 This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information please contact scholcom@uwyoedu

DECOMPOSITION OF A RING INDUCED BY MINUS PARTIAL ORDER DRAGAN S RAKIĆ Abstract The minus partial order linear algebraic methods have proven to be useful in the study of complex matrices This paper extends study of minus partial orders to general rings It is shown that the condition a < b where < is the minus partial order defines two triples of orthogonal idempotents and thus a decomposition of a ring into a direct sum of abelian groups Hence several well-known results concerning minus partial order on real and complex matrices are generalized Analogous decompositions for rectangular matrices over a ring and for Banach space operators are obtained The equivalent conditions for the invariance of ab (1) a under the choices of b (1) are also obtained An original inspiration for this work came from the study of minus partial order on complex matrices and from linear algebra methods Key words Minus partial order Generalized inverse Idempotent Von Neumann regular ring Operator matrices Peirce decomposition AMS subject classifications 15A09 06A06 16U99 1 Introduction Throughout the paper R denotes a ring with identity 1 The minus partial order on a ring R introduced by Hartwig [7] is defined by a < b if there exists an x R such that (11) ax = bx xa = xb and axa = a Our aim is to find an equivalent condition for a < b which will enable us to obtain some properties of minus partial order For this purpose we look how this problem is solved for real and complex matrices It is known that for real or complex matrices A and B the condition A < B is equivalent to the existence of non-singular matrices S and T such that (12) A = S I A 0 0 T and B = S I A 0 0 0 I B A 0 T Received by the editors on July 18 2012 Accepted for publication on November 22 2012 Handling Editor: Bryan L Shader Faculty of Sciences and Mathematics University of Niš Višegradska 33 PO Box 224 18000 Niš Serbia (rakicdragan@gmailcom) Supported by the Ministry of Education and Science Government of the Republic of Serbia grant no 174007 1040

Decomposition of a Ring Induced by Minus Partial Order 1041 where I A and I B A are identity matrices The above simultaneous diagonalization proves very useful in obtaining many of the properties of minus partial order For more details concerning minus partial order on matrices see the monograph [13] and the references given there The proofs in [13] are mostly based on linear algebra and finite dimensional methods Of course we cannot use these techniques when a b are elements of arbitrary ring We are interested in finding the matrix forms of ab R which are analogous to those given in (12) when a < b We will show in our main Theorem 35 that the condition a < b where a b are regular elements naturally defines two triples of orthogonal idempotents (e i ) and (f i ) i = 123 with sums equal to 1 This yields the decomposition of the ring R into a direct sum of abelian groups (13) R = such that a e 1 Rf 1 and b a e 2 Rf 2 3 e i Rf j ij=1 Using this result we will generalize a considerable number of well-known results for real and complex matrices For example when a < b we can easily characterize all elements x appearing in (11) and the class of all idempotents p and q such that a = pb = bq Thesamenotionofdecompositionenableustofindequivalentconditions for the invariance of ab (1) a under the choices of b (1) b{1} Also it is proven that the conditions a < b b{1} a{1} and b{12} a{1} are equivalent which is known for matrices (see [12 17]) Another advantage of using the decomposition (13) lies in the fact that the same idea can be applied to rectangular matrices over a ring and even to the class of bounded operators on Banach spaces This concepts will be discussed in the last section 2 Preliminaries An element a R is called von Neumann regular (regular for short) if there exists x R satisfying axa = a The element x is called a generalized inverse of a If axa = a and xax = x then x is called a reflexive generalized inverse of a If x 1 and x 2 are generalized inverses of a then x 1 ax 2 is a reflexive generalized inverse of a We denote by a{1} the set of all generalized inverses of a and by a{12} the set of all reflexive generalized inverses of a The set of all regular elements of R is denoted by R (1) Some properties of generalized inverses in a ring was studied in [2 4 15] A ring R is von Neumann regular (regular for short) if every element of R is regular For details concerning regular rings we refer the reader to [5] Note that a

1042 DS Rakić ring of all n n complex matrices is regular Definition 21 Let ab R Then a is below b under the minus partial order denoted by a < b if a R (1) and ax = bx xa = xb for some x a{1} The idempotents ef R are orthogonal if ef = fe = 0 The idempotents e 1 e 2 e n R are called orthogonal if they are mutually orthogonal An equality 1 = e 1 +e 2 + +e n where e 1 e 2 e n R are orthogonal idempotents is called a decomposition of the identity of the ring R Remark 22 Let 1 = e 1 + +e m and 1 = f 1 + +f n be two decompositions of the identity of a ring R For any x R we have x = 1 x 1 = (e 1 + +e m )x(f 1 + +f n ) = m i=1 j=1 n e i xf j It is not difficult to verify that above sum defines a decomposition of R into a direct sum of abelian groups e i Rf j := {e i xf j : x R}: (21) R = m i=1 j=1 n e i Rf j It is convenient to write x as a matrix x 11 x 12 x 1n x 21 x 22 x 2n x = x m1 x m2 x mn e f where x ij = e i xf j e i Rf j Similarly R = n i=1 j=1 m f i Re j and any y R can be written in a matrix form y = [y ij ] where y ij = f i ye j f i Re j i = 1n j = 1m By the orthogonality of idempotents involved one can use usual matrix rules in order to add and multiply x and y When m = n and e i = f i i = 1n the decomposition (21) is known as the two-sided Peirce decomposition of the ring R [8]

Decomposition of a Ring Induced by Minus Partial Order 1043 3 The minus partial order Before moving to the minus partial order we consider another order which is associated to the minus partial order If ab R then we say that a is below b under the space pre-order denoted by a < s b if ar br and Ra Rb This definition is analogous to the definition of space pre-order on complex matrices (see [11]) in which case A < s B if R(A) R(B) and R(B ) R(A ) where R(A) denotes the column space of matrix A and A is the conjugate transpose of A It is easily seen that < s is pre-order and that a < b implies a < s b The following result is well-known in the matrix case [1] Theorem 31 Let R be a regular ring and ab R Then the following conditions are equivalent: (i) a < s b; (ii) a = bb (1) a = ab (1) b for all b (1) b{1}; (iii) a = bb (1) ab (1) b for all b (1) b{1}; and (iv) ab (1) a is invariant under the all choices of b (1) b{1} Proof If a = 0 or b = 0 then the theorem holds Suppose that a 0 and b 0 (i) = (ii): Since a < s b we have ar br so there exists an x R such that a = bx Hence a = bb (1) bx = bb (1) a for all b (1) b{1} Similarly a = ab (1) b for all b (1) b{1} (ii) = (iii) is trivial (iii) = (iv): Fix h b{1} For every b (1) b{1} we have which does not depend on b (1) Then (iv) = (i): Fix h b{1} and set ab (1) a = (bhahb)b (1) (bhahb) = bhahbhahb e 1 = bh e 2 = 1 bh f 1 = hb f 2 = 1 hb 1 = e 1 +e 2 and 1 = f 1 +f 2 are two decompositions of the identity of the ring R If b (1) b{1} then f 1 b (1) e 1 = hbb (1) bh = hbh (= f 1 he 1 ) If f 1 b (1) e 1 = hbh then b(f 1 b (1) e 1 )b = b(hbh)b Thus

1044 DS Rakić bb (1) b = b Therefore b (1) b{1} if and only if [ hbh (31) b (1) x12 = x 21 x 22 ] where x ij f i Re j are arbitrary Now ab (1) a = [af 1 af 2 ] 1 f [ hbh x12 x 21 x 22 ] = ahbha+ax 21 a+ax 12 a+ax 22 a [ e1 a e 2 a does not depend on x 21 x 12 x 22 Setting x 12 = x 22 = 0 it follows that ax 21 a = 0 for all x 21 f 2 Re 1 ie af 2 xe 1 a = 0 for all x R Multiplying this equation by e 1 from the left and by f 2 from the right we obtain (e 1 af 2 )x(e 1 af 2 ) = 0 Since R is regular we can choose x = (e 1 af 2 ) (1) (e 1 af 2 ){1} Hence e 1 af 2 = 0 Similarly e 2 af 1 = 0 and e 2 af 2 = 0 so we conclude that a = e 1 af 1 = bhahb This implies a < s b Note that the regularity of R is used only in the part (iv) (i) and the set b{1} is characterized by (31) Lemma 32 Let b R (1) (R is not necessarily regular) Under the notation of the proof of Theorem 31 the set of all reflexive generalized inverses of b is given by [ ] hbh (32) b (12) x12 = x 21 x 21 bx 12 where x 12 f 1 Re 2 and x 21 f 2 Re 1 are arbitrary Proof We already know that b (12) must be of the form (31) We have x 22 = x 21 bx 12 because [ hbh x12 x 21 x 22 ] f f ] e 1 [ ] [ ] hbh = b (12) = b (12) bb (12) x12 b 0 = x 21 x 22 0 0 e f [ ] [ ] [ ] f1 0 hbh x12 hbh x12 = = x 21 b 0 x 21 x 22 x 21 x 21 bx 12 If b (12) is given by (32) then it is easy to check that b (12) b{12} b (12) It is stated in [12] that if R is a regular ring then a < b is equivalent to b{1} a{1} with the additional hypothesis a brb Using the direct sum partial

Decomposition of a Ring Induced by Minus Partial Order 1045 order as an intermediate step it is proved in[3] that the equivalence still holds without the additional hypothesis We show that a stronger result is in fact true Theorem 33 Let R be a regular ring and ab R Then the following conditions are equivalent: (i) a < b; (ii) b{1} a{1}; and (iii) b{1 2} a{1} Proof (i) = (ii): This is well-known However we prove it here for completeness As a < b there exists a (1) a{1} such that aa (1) = ba (1) and a (1) a = a (1) b For any b (1) b{1} we have ab (1) a = aa (1) ab (1) aa (1) a = aa (1) bb (1) ba (1) a = aa (1) ba (1) a = a (ii) = (iii) is trivial (iii) = (i): Lemma 32 shows that a = ab (12) a = ahbha+ax 21 a+ax 12 a+ax 21 bx 12 a for all x 12 f 1 Re 2 and x 21 f 2 Re 1 As in the proof of Theorem 31 we get (33) e 1 af 2 = e 2 af 1 = 0 and a = ahbha = af 1 he 1 a By (33) we obtain a = (e 1 +e 2 )(af 1 he 1 a)(f 1 +f 2 ) = e 1 (af 1 he 1 a)f 1 = e 1 af 1 = e 1 a = af 1 Let x = f 1 hahe 1 Then x = he 1 af 1 h = he 1 ah = haf 1 h = hah and he 1 = f 1 h = hbh b{12} a{1} Thus axa = a and ax = ahe 1 ah = ah = e 1 ah = bhah = bx xa = haf 1 ha = ha = haf 1 = hahb = xb

1046 DS Rakić By definition a < b holds It is known that minus partial order is a partial order on R when R is regular Reflexivity and transitivity follows from Theorem 33 If a < b and b < a then a{1} b{1} and there exists a (1) a{1} such that aa (1) = ba (1) and a (1) a = a (1) b So a = aa (1) a = ba (1) b = b and < is a partial order Note that 0 < b for all b in R and that a < 0 a = 0 Lemma 34 Let ab R (1) The following conditions are equivalent: (i) a < b; (ii) a = ab (1) a = ab (1) b = bb (1) a for all b (1) b{1}; (iii) b a = (b a)b (1) (b a) = (b a)b (1) b = bb (1) (b a) for all b (1) b{1}; and (iv) b a < b Proof (i) = (ii): The proof is similar to that of Theorem 33 (i) (ii) (ii) = (i): Let x = b (1) ab (1) Then x a{1} and Likewise xa = xb and hence a < b ax = ab (1) ab (1) = ab (1) = bb (1) ab (1) = bx (ii) (iii) is a matter of direct computation (iii) (iv) follows from the equivalence of (i) and (ii) Note that the word all can be replaced by some in the conditions (ii) and (iii) of Lemma 34 From Lemma 34 we see that the condition a < b is symmetric in a and b a Next if A and B are complex matrices then [7] (34) A < B rank(b) = rank(a)+rank(b A) and the latter condition is symmetric in A and B A The element x a{1} that appears in Definition 21 does not imply the symmetry since (b a)x = 0 bx But from Lemma 34 we see that elements of b{1} somehow point out that symmetry For this reason we start from arbitrary but fixed h b{1} in order to obtain the decomposition of R induced by the condition a < b Remark 22 is crucial for our main result which follows Theorem 35 Let ab R (1) Then the following conditions are equivalent: (i) a < b; (ii) There exist decompositions of the identity of the ring R (35) 1 = e 1 +e 2 +e 3 1 = f 1 +f 2 +f 3

Decomposition of a Ring Induced by Minus Partial Order 1047 with respect to which a and b have the following matrix forms: a 0 0 a 0 0 (36) a = b = 0 b a 0 e f Proof The cases a = 0 or b = 0 are trivial (i) = (ii): Fix h b{1} and set e f (37) e 1 = ah e 2 = (b a)h e 3 = 1 bh f 1 = ha f 2 = h(b a) f 3 = 1 hb Since a < b Lemma 34 shows that and It follows that (ah)(ah) = (ah)(bh) = (bh)(ah) = ah (bh)(bh) = bh (ha)(ha) = (ha)(hb) = (hb)(ha) = ha (hb)(hb) = hb 1 = e 1 +e 2 +e 3 1 = f 1 +f 2 +f 3 are two decompositions of the identity of the ring R and From e 1 af 1 = ahaha = a e 2 (b a)f 2 = (b a)h(b a)h(b a) = b a we conclude that a and b have the matrix forms given by (36) (ii) = (i): Fix a (1) a{1} and set x = f 1 a (1) e 1 It is easily seen that x a{1} ax = bx and xa = xb When it is the case as in Theorem 35 we say that the decompositions (35) where idempotents are defined by (37) are standard decompositions Theorem 36 Let ab R (1) such that a 0 b 0 and a < b Let e i f i i = 13 be defined by (37) Then there exist unique elements x a f 1 Re 1 and x b a f 2 Re 2 such that (38) ax a = e 1 x a a = f 1 (b a)x b a = e 2 x b a (b a) = f 2

1048 DS Rakić Furthermore with respect to the standard decompositions a{1} is given by (39) a (1) = [x ij ] where x 11 = x a and x ij f i Re j (ij) (11) are arbitrary and the set b{1} is given by (310) b (1) = where x ij f i Re j are arbitrary Proof Let x a 0 x 13 0 x b a x 23 x 31 x 32 x 33 (311) x a := f 1 he 1 = hah x b a := f 2 he 2 = h(b a)h One can check that these elements satisfy (38) The proof of the uniqueness is left to the reader Let b (1) = [f i b (1) e j ] b{1} be arbitrary From a 0 0 0 b a 0 e f = b = bb (1) b = ab (1) a ab (1) (b a) 0 (b a)b (1) a (b a)b (1) (b a) 0 it follows that ab (1) a = a Multiplying this equation by x a from the both sides yields f 1 b (1) e 1 = x a Also multiplying f 1 b (1) e 1 = x a by a from the both sides yields ab (1) a = a Similarly ab (1) (b a) = 0 f 1 b (1) e 2 = 0 (b a)b (1) a = 0 f 2 b (1) e 1 = 0 and (b a)b (1) (b a) = b a f 2 b (1) e 2 = x b a Therefore the set b{1} is given by (310) The characterization of a{1} can be proved analogously By the end of the section we will follow the notation of Theorems 35 and 36 Remark 37 Let ab R (1) such that a < b Suppose that h b{1} is regular Fix r 1 h{1} and set r = b+e 3 r 1 f 3 Then r h{1} Let It is not difficult to show that e 1 = e 1 e 2 = e 2 e 3 = (r b)h e 4 = 1 rh f 1 = f 1 f 2 = f 2 f 3 = h(r b) f 4 = 1 hr e f 1 = e 1 + +e 4 1 = f 1 + +f 4

Decomposition of a Ring Induced by Minus Partial Order 1049 are two decompositions of the identity of the ring R Also f 3 (h hbh)e 3 = h hbh It follows that a b and h have the following matrix forms with respect to these decompositions: a = h = a 0 0 0 e f b = x a 0 x b a 0 0 0 0 h hbh 0 0 f e a 0 b a 0 0 0 0 e f Moreover there exists unique x e 3 Rf 3 (namely x = e 3 rf 3 = rhr b) such that (h hbh)x = f 3 and x(h hbh) = e 3 Remark 38 We survey some known characterizations of minus partial order Let ab R (1) such that (312) a < b If p 1 = e 1 p 2 = 1 e 1 q 1 = f 1 q 2 = 1 f 1 then 1 = p 1 +p 2 and 1 = q 1 +q 2 are two decompositions of the identity of the ring R and by Theorem 35 it follows [ ] a 0 (313) b = 0 b a This is equivalent to a = p 1 bq 1 and b a = p 2 bq 2 Multiplying the latter equation by p 1 from the left gives p 1 (b a) = 0 so a = p 1 b Similarly a = bq 1 On the other hand suppose that there exist idempotents p 1 and q 1 such that (314) a = p 1 b = bq 1 Let x = q 1 a (1) p 1 where a (1) a{1} We have a < b because axa = aq 1 a (1) p 1 a = aa (1) a = a ax = bq 1 x = bx and xa = xp 1 b = xb Obviously (313) implies p q (315) (316) (317) br = ar (b a)r Rb = Ra R(b a) ar (b a)r = {0} = Ra R(b a) Note that condition (315) is used as a definition of direct sum partial order The equivalence of (312) and (315) was proved in [3] Lemma 3 and equivalence of(315) (317) was proved in [9] Theorem 1 Thus (312) (317) are equivalent Some other characterizations of minus partial order on regular semigroup can be found in [14]

1050 DS Rakić Let a{1} b = {x a{1} : ax = bxxa = xb} and a{12} b = {x a{12}: ax = bxxa = xb} In the next theorem we obtain explicit representations of a{1} b and a{12} b For the case when a b are complex matrices see [10] and [11] Theorem 39 Let ab R (1) such that a < b Then (i) a{1} b = {b (1) b (1) (b a)b (1) : b (1) b{1}}; (ii) a{12} b = {b (1) ab (1) : b (1) b{1}} = {b (12) ab (12) : b (12) b{12}} Proof (i): Let us denote the set on the right-hand side of (i) by S Since a < b we have that a b and b (1) have the representations given by (36) and (310) respectively It follows that x S if and only if (318) x = b (1) b (1) (b a)b (1) = x a 0 x 13 x 31 0 x 33 for certain elements x 13 f 1 Re 3 x 31 f 3 Re 1 and x 33 f 3 Re 3 (x 33 = x 33 x 32 (b a)x 23 ) Atrivialverificationshowsthataxa = a ax = bxandxa = xb ie x a{1} b Assume now that x a{1} b Then x a{1} and hence x = [x ij ] where x 11 = x a From ax = bx and xa = xb we obtain x 12 = x 21 = x 22 = x 23 = x 32 = 0 One can check that x = b (1) b (1) (b a)b (1) S where b (1) = x a 0 x 13 0 x b a 0 x 31 0 x 33 (ii): The proof of (ii) is similar We obtain that x a{12} b is given by: x = where x 13 and x 31 are arbitrary x a 0 x 13 x 31 0 x 31 ax 13 The following theorem is generalization of Theorem 356 in [13] where a and b are complex matrices Theorem 310 Let ab R (1) such that a < b Then

Decomposition of a Ring Induced by Minus Partial Order 1051 (i) For any a (1) a{1} b there exists b (1) b{1} such that (319) b (1) a = a (1) a ab (1) = aa (1) ; (ii) For any b (1) b{1} there exists a (1) a{1} b such that (319) holds Proof (i): Proof of Theorem 39 shows that a (1) a{1} b has matrix representation given in (318) Then (319) holds for b (1) = x a 0 x 13 0 x b a x 23 x 31 x 32 x 33 where x 23 f 2 Re 3 x 32 f 3 Re 2 and x 33 f 3 Re 3 are arbitrary (ii): Any b (1) b{1} is of the form (310) The element a (1) = x a 0 x 13 x 31 0 x 33 where x 33 f 3 Re 3 is arbitrary has desired properties As we pointed out in Remark 38 for ab R (1) the condition a < b is equivalent to a = pb = bq where pq R are some idempotents We characterize the class of all such idempotents The following results are analogous to Theorems 3513 3518 in [13] where it is considered the case when a and b are complex matrices All of them can be proved using matrix forms (36) and identities (38) Theorem 311 Let ab R (1) such that a < b Then the class of all idempotents p R satisfying a = pb is given by (320) p = e 1 0 x 13 (e 3 p 33 ) 0 0 x 23 p 33 0 0 p 33 where p 33 e 3 Re 3 is some idempotent and x 13 e 1 Re 3 x 23 e 2 Re 3 are arbitrary Proof If p is of the given form then p is an idempotent and a = pb Let p be an idempotent such that a = pb Suppose that p = [p ij ] e e ij = 13 with respect to standard decomposition 1 = e 1 +e 2 +e 3 From a = pb using (36) and (38) we obtain p 11 = e 1 and p 12 = p 21 = p 22 = p 31 = p 32 = 0 Condition p = p 2 implies p 23 = p 23 p 33 p 13 = p 13 + p 13 p 33 and p 33 = p 2 33 Hence p 13 = x 13 (e 3 p 33 ) and p 23 = x 23 p 33 where x 13 e 1 Re 3 x 23 e 2 Re 3 are arbitrary e e

1052 DS Rakić In the same manner we obtain the following theorem Theorem 312 Let ab R (1) such that a < b Then the class of all idempotents q R such that a = bq is given by f 1 0 0 q = (f 3 q 33 )x 31 q 33 x 32 q 33 where q 33 f 3 Rf 3 is some idempotent and x 31 f 3 Rf 1 x 32 f 3 Rf 2 are arbitrary Remark 313 Let f f e 1 = e 1 e 2 = e 2 e 3 = p 33 e 4 = e 3 p 33 f 1 = f 1 f 2 = f 2 f 3 = q 33 f 4 = f 3 q 33 where p 33 and q 33 are idempotents appearing in Theorems 311 and 312 Then 1 = e 1 + +e 4 1 = f 1 + +f 4 are decompositions of the identity of the ring R It is clear that and a = p = a = q = a 0 0 e f e 1 0 0 x 14 0 0 x 23 0 0 0 e 3 0 0 a 0 0 f 1 0 0 x 32 f 3 0 x 41 b = e e e f b = f f a 0 0 0 b a 0 a 0 b a 0 0 0 for some x 14 e 1 Re 4 x 23 e 2 Re 3 x 32 f 3 Rf 2 and x 41 f 4 Rf 1 Corollary 314 Let ab R (1) such that a < b Then the class of all idempotents p such that a = pb and pr = ar is given by {aa (1) : a (1) a{1} b } The e f e f

Decomposition of a Ring Induced by Minus Partial Order 1053 class of all idempotents q such that a = bq and Rq = Ra is given by {a (1) a : a (1) a{1} b } Proof Since a = e 1 a and e 1 = ax a we have ar = e 1 R By (320) we see that p is idempotent such that a = pb and pr = ar = e 1 R if and only if p 33 = 0 if and only if p = e 1 0 x 13 where x 13 e 1 Re 3 is arbitrary From (318) we see that aa (1) where a (1) a{1} b has the above form The second characterization can be obtained in the same manner Corollary 315 Let ab R (1) such that a < b Every idempotent p satisfying a = pb can be written as p = p 1 + p 2 where p 1 is an idempotent such that a = p 1 b p 1 R = ar and p 2 is an idempotent such that p 1 p 2 = p 2 p 1 = p 2 a = p 2 b = 0 Every idempotent q satisfying a = bq can be written as q = q 1 + q 2 where q 1 is an idempotent such that a = bq 1 Rq 1 = Ra and q 2 is an idempotent such that q 1 q 2 = q 2 q 1 = aq 2 = bq 2 = 0 Proof According to Theorems 311 312 and Corollary 314 we can take e 1 0 x 13 (e 3 p 33 ) p 1 = p 2 = 0 0 x 23 p 33 and 0 0 p 33 q 1 = f 1 0 0 (f 3 q 33 )x 31 0 0 e e f f q 2 = e e 0 q 33 x 32 q 33 Suppose now that A is an algebra with identity 1 over a field K Obviously the algebra A is a ring (A+ ) and the related concepts and results are preserved in the passage from R to A Let A < B where AB C n n are complex matrices with ranks a and b respectively and let c 1 c 2 C c 2 0 c 1 +c 2 0 In [18] it is proved that c 1 A+c 2 B is invertible if and only if B is invertible and (c 1 A+c 2 B) 1 = (c 1 +c 2 ) 1 B 1 +(c 1 2 (c 1 +c 2 ) 1 )[(0 I n a )B(0 I n a )] holds where ( ) is the Moore-Penrose inverse of ( ) The next theorem shows that the same result is valid when ab A (1) We avoid the use of Moore-Penrose inverse in our formula Theorem 316 Let ab A (1) such that a < b and let c 1 c 2 K c 2 0 e e f f

1054 DS Rakić c 1 +c 2 0 Then c 1 a+c 2 b is invertible if and only if b is invertible Furthermore (321) (c 1 a+c 2 b) 1 = c 1 2 b 1 +((c 1 +c 2 ) 1 c 1 2 )b 1 ab 1 = c 1 2 b 1 +((c 1 +c 2 ) 1 c 1 2 )a(1) where a (1) a{1} b Proof Since a < b by Theorem 35 we have the following representations with respect to standard decompositions: b = a 0 0 0 b a 0 e f c 1 a+c 2 b = (c 1 +c 2 )a 0 0 0 c 2 (b a) 0 Since c 2 0 c 1 + c 2 0 it follows that b is invertible if and only if c 1 a + c 2 b is invertible if and only if e 3 = f 3 = 0 In this case e f 1 = e 1 +e 2 1 = f 1 +f 2 are two decompositions of the identity of A and with respect to these decompositions we have [ ] [ b 1 xa 0 = (c 1 a+c 2 b) 1 (c1 +c 2 ) 1 ] x a 0 = 0 x b a 0 c 1 2 x b a so the formula (321) can be easily checked As b is invertible and a < b it follows from Theorem 39 that a{1} b = {b 1 ab 1 } 4 Applications In this section we indicate how the concepts and results from previous sections can be extended to matrices with entries in a ring R and to Banach space operators We want to point out the universality of the idea of matrix representation of elements of the various structures Minus partial order for matrices with entries in a ring As before R denotes a ring with identity 1 The set of all m n matrices with entries from R will be denoted by M m n (R) For any A M m n (R) we will denote by CS(A) := {Aξ : ξ M n 1 (R)} and RS(A) := {ξa : ξ M 1 m (R)} the column space of A and row space of A respectively If A M m n (R) we say that A is regular matrix if there is a matrix X M n m (R) such that AXA = A in which case we call X the generalized inverse ofa IfAXA = AandXAX = X then X isareflexive generalized inverse of A Of course A{1} (A{12}) stands for the set of all generalized (reflexive generalized) inverses of A Let M m n(r) (1) denotes the set of all regular matrices in M m n (R)

Decomposition of a Ring Induced by Minus Partial Order 1055 According to von Neumann if R is a regular ring then M n n (R) is also a regular ring Moreover every matrix A M m n (R) over a regular ring R is regular; see Theorem 35 in [2] For AB M m n (R) minus partial order is defined analogously as in the ring case and space pre-order is defined by A < s B if (41) CS(A) CS(B) and RS(A) RS(B) The idea from Remark 22 can be applied to M m n (R) Let I m = E 1 + +E r and I n = F 1 + + F s be decompositions of the identity of the rings M m m (R) and M n n (R) respectively where I m M m m (R) and I n M n n (R) are identity matrices For any X M m n (R) we have X = I m XI n = (E 1 + +E r )X(F 1 + +F s ) = r i=1 j=1 s E i XF j To avoid repetition we only note that conclusions analogous to those from Remark 22 are true in the present case As the reader might already have guessed except Theorem 316 all results from the previous section remain valid in the present setting Some comments are still necessary Depending on the context the ring R is replaced by M m n (R) M m m (R) M n m (R) or M n n (R) Also the set R (1) is replaced by M m n(r) (1) or M n m(r) (1) but in the statements where the regularity of R is assumed this assumption remains the same (due to the von Neumann result stated above) Conditions xrρyr and RxρRy where ρ { =} are replaced by CS(X)ρCS(Y) and RS(X)ρRS(Y) respectively Since R has identity for XY M m n (R) the condition CS(X) CS(Y) is equivalent to X = YZ for some Z M n n (R) Also RS(X) RS(Y) is equivalent to X = ZY for some Z M m m (R) Except Theorem 316 the proofs of all statements in the present setting proceed along the same lines as the proofs of corresponding statements in the previous setting Lemma 32 in present setting was originally proved in [6] In the same paper it is shown that if R is a regular prime ring and ABC M m n (R) then CS(C) CS(B) and RS(A) RS(B) if and only if AB (12) C is invariant under the choices of B (12) B{12} In Theorem 31 in the present setting we only require that R is regular but we consider only invariance of AB (1) A Minus partial order for operators on Banach spaces We now consider how the concept can be extended to Banach space operators Let B(XY) denote the set of all bounded linear operators from Banach space X

1056 DS Rakić to Banach space Y Also B(X) = B(XX) We use N(A) and R(A) to denote null space and range of A B(XY) respectively An operator A B(XY) is regular if there exists an operator A (1) B(YX) such that AA (1) A = A The operator A (1) is called generalized inverse of A It is well known that A B(XY) is regular if and only if R(A) is closed and complemented in Y and N(A) is complemented in X If AA (1) A = A and A (1) AA (1) = A (1) then A (1) is reflexive generalized inverse of A Of course A{1} and A{12} stands for the sets of all generalized and reflexive generalized inverses of A B (1) (XY) denotes the set of all regular operators from B(XY) For AB B(XY) A < B if A is regular and there exists A (1) A{1} such that AA (1) = BA (1) and A (1) A = A (1) B; and A < s B if R(A) R(B) and N(B) N(A) Let B B (1) (XY) Note that R(B) = R(BB (1) ) because R(B) = R(BB (1) B) R(BB (1) ) R(B) Also R(A) R(B) A = BB (1) A because R(A) R(B) = R(BB (1) ) = N(I BB (1) ) so (I BB (1) )A = 0 and the opposite direction is trivial Similarly N(B) = N(B (1) B) and N(B) N(A) A = AB (1) B Because of this the conditions arρbr and RaρRb where ρ { =} are analogous to the conditions R(A)ρR(B) and N(B)ρN(A) respectively A bounded idempotent E B(X) is called a projection If we consider algebra B(X) as a ring with identity I X B(X) then the notion of decomposition of the identity of B(X) makes sense Lemma 41 If E 1 E 2 B(X) are two projections such that E 1 E 2 = E 2 E 1 = 0 then E 1 +E 2 is projection and R(E 1 +E 2 ) = R(E 1 ) R(E 2 ) Proof It is clear that E 1 +E 2 is a projection and that R(E 1 +E 2 ) R(E 1 )+ R(E 2 ) If x R(E 1 ) then 0 = E 2 E 1 x = E 2 x so x = (E 1 + E 2 )x R(E 1 + E 2 ) Similarly R(E 2 ) R(E 1 + E 2 ) If x R(E 1 ) R(E 2 ) then x = E 1 x = E 2 x and hence x = E 2 1 x = E 1E 2 x = 0 Corollary 42 Let (42) I X = E 1 + +E n be a decomposition of the identity of B(X) and let S {12n} Then i S E i is a projection and ( ) (43) R E i = R(E i ) i S i S

Decomposition of a Ring Induced by Minus Partial Order 1057 In particular n (44) X = R(E i ) i=1 Proof Without loss of generality we can assume that S = {12k} 1 k n Using Lemma 41 (43) follows easily by induction on k Equation (44) follows from (43) by S = {12n} Note that from (43) it follows that i S R(E i) is closed and complemented in X Also it follows that N(E j ) = n i=1i j R(E i) Let I X = F 1 + +F n and I Y = E 1 + +E m be decompositions of the identity of B(X) and B(Y) respectively Suppose that A B(XY) Then (45) A = m i=1 j=1 n E i AF j and so for x = x 1 + +x n R(F 1 ) R(F n ) = X we have (46) Ax = m i=1 j=1 n A ij x j where operator A ij : R(F j ) R(E i ) is defined by A ij x j := E i AF j x j = E i AF j x It is clear that A ij B(R(F j )R(E i )) Now suppose that A ij B(R(F j )R(E i )) and let A : X Y be defined by (46) It follows that Ax mm( x 1 + + x n ) = mm( F 1 x + + F n x ) mm( F 1 + + F n ) x where M = max{ A ij : i = 1mj = 1n} Therefore A B(XY) if and only if A ij B(R(F j )R(E i )) In this case (46) ie (45) can be rewritten in the matrix form A = A 11 A m1 A 1n A mn : R(F 1 ) R(F n ) R(E 1 ) R(E m ) We can now restate Theorems 35 and 36 in the new setting Suppose that AB B (1) (XY) A < B and let E i and F i be defined by (37) Since R(E 1 ) = R(AH) =

1058 DS Rakić R(A) R(E 2 ) = R((B A)H) = R(B A) and R(F 3 ) = R(I X HB) = N(HB) = N(B) we have (47) A = B = C A 0 0 : C A 0 0 0 C B A 0 R(F 1 ) R(F 2 ) N(B) : R(F 1 ) R(F 2 ) N(B) R(A) R(B A) R(E 3 ) R(A) R(B A) R(E 3 ) forsomeoperatorsc A B(R(F 1 )R(A)) andc B A B(R(F 2 )R(B A)) Theorem 36 actually says that C A and C B A are invertible The matrix forms in the above are originally obtained in [16] but in the different way It is now clear that all results coming after Theorem 36 are still valid in the present setting The crucial assumption in Theorems 31 and 33 is that R is a regular ring We cannot require the same condition for Banach space operators But it is shown in [16] that these theorems still hold in the present setting The only requirement is that A and B are regular operators Suppose that AB B (1) (XY) and A < B Therefore R(B) = R(BH) = R(E 1 +E 2 ) = R(E 1 ) R(E 2 ) = R(A) R(B A) Conversely suppose that (48) R(B) = R(A) R(B A) Then R(A) R(B) so A = BB (1) A Hence A AB (1) A = BB (1) A AB (1) A = (B A)B (1) A AlsoA(I B (1) B) = (A B)(I B (1) B) SinceR(A) R(B A) = {0} we have A = AB (1) A = AB (1) B By Lemma 34 we conclude that A < B Suppose now that X and Y are finite dimensional vector spaces ie suppose that A and B are m n complex matrices Note that (48) is equivalent to rank(b) = rank(a) + rank(b A) so we obtain (34) Let us look at (47) Decompositions C n = R(F 1 ) R(F 2 ) N(B) and C m = R(A) R(B A) R(E 3 ) induce changes of bases in C n and C m respectively Thus there exist non-singular matrices S 1 and T such that A = S 1 C A 0 0 T and B = S 1 C A 0 0 0 C B A 0 T Setting S = S 1 C A 0 0 0 C B A 0 0 0 I

Decomposition of a Ring Induced by Minus Partial Order 1059 we obtain representations (12) from the beginning of the paper Acknowledgment This is part of the author s PhD Thesis which is written under the supervision of Professor Dragan S Djordjević at the University of Niš The author wishes to express his thanks to Professor Djordjević for suggesting the problem and for many stimulating conversations The author would also like to thank the anonymous reviewer for constructive comments that improved the presentation of this paper REFERENCES [1] JK Baksalary F Pukelsheim and GP Styan Some properties of matrix partial orderings Linear Algebra Appl 119:57 85 1989 [2] KPS Bhaskara Rao The Theory of Generalized Inverses Over Commutative Rings Taylor and Francis London 2002 [3] B Blackwood SK Jain KM Prasad and AK Srivastava Shorted operators relative to a partial order in a regular ring Comm Algebra 37(11):4141 4152 2009 [4] DS Djordjević and Y Wei Outer generalized inverses in rings Comm Algebra 33:3051 3060 2005 [5] KR Goodearl Von Neumann Regular Rings Monographs and Studies in Mathematics Vol 4 Pitman Publishing Limited London 1979 [6] RE Hartwig 1-2 inverses and the invariance of BA + C Linear Algebra Appl 11:271 275 1975 [7] RE Hartwig How to partially order regular elements Math Japonica 25(1):1 13 1980 [8] N Jacobson Structure of Rings American Mathematical Society Providence 1968 [9] SK Jain and KM Prasad Right-left symmetry of ar br = (a+b)r in regular rings Pure Appl Algebra 133:141 142 1998 [10] SK Mitra Infimum of a pair of matrices Linear Algebra Appl 105:163 182 1988 [11] SK Mitra Matrix partial orders through generalized inverses: unified theory Linear Algebra Appl 148:237 263 1991 [12] SK Mitra The minus partial order and the shorted matrix Linear Algebra Appl 83:1 27 1986 [13] SK Mitra P Bhimasankaram and SB Malik Matrix Partial Orders Shorted Operators and Applications Series in Algebra Vol 10 World Scientific Singapore 2010 [14] H Mitsch A natural partial order for semigroups Proc Amer Math Soc 97(3):384 388 1986 [15] B Načevska and DS Djordjević Inner generalized inverses with prescribed idempotents Comm Algebra 39:634 646 2011 [16] DS Rakić and DS Djordjević Space pre-order and minus partial order for operators on Banach spaces Aequationes Math doi:101007/s00010-012-0133-2 2012 [17] P Sambamutry Characterization of a matrix by its subclass of G-inverses Sankhyā Ser A 49(3):412 414 1987 [18] M Tošić and DS Cvetković-Ilić Invertibility of a linear combination of two matrices and partial orderings Appl Math Comput 218:4651 4657 2012