IMFM
Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia
Preprint series Vol. 51 (2013), 1191
ISSN 2232-2094
MINUS PARTIAL ORDER IN RICKART RINGS
Dragan S. Djordjevic Dragan S. Rakic Janko Marovt
Ljubljana, October 16, 2013
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Minus partial order in Rickart rings1
Dragan S. Djordjevica'1, Dragan S. Rakic a'1'*, Janko Marovtb'2
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a University of Niš, Faculty of Sciences and Mathematics, Visegradska 33, P.O. Box 95,
18000 Niš, Serbia
b University of Maribor, Faculty of Economics and Business, Razlagova 14, SI-2000
Maribor, Slovenia,
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Abstract
The minus partial order is already known for complex matrices and bounded linear operators on Hilbert spaces. We extend this notion to Rickart rings, and thus we generalize some well-known results.
Keywords: Rickart ring, minus partial order 2000 MSC: 16B99, 47A05
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1. Introduction and motivation
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Let A be a ring with the unit 1. If M C A, then the right annihilator of M is denoted by M° = {x E A : (Vm E M) mx = 0}, and the left annihilator of M is denoted by °M = {x E A : (Vm E M) xm = 0}. M° is the right ideal of A, and °M is a left ideal of A. Particularly, if a E A and M = {a}, then we shortly use a° = {a}° and °a = °{a}.
The set of idempotents of A is denoted by A^ = {p E A : p2 = p}.
*This work is a part of a bilateral project between Serbia and Slovenia: "Preservers, operator and matrix equations with applications".
* Corresponding author.
Email addresses: dragan@pmf.ni.ac.rs (Dragan S. Djordjevic ), rakic.dragan@gmail.com (Dragan S. Rakic ), janko.marovt@uni-mb.si (Janko Marovt )
xThe authors are supported by the Ministry of Education, Science and Technological Development of Serbia, under grant no. 174007.
2The author is supported by the Ministry of Science of Slovenia, under grant no. BI-RS/12-13-001.
Preprint submitted to Journal of Mathematical Analysis and ApplicationsOctober 15, 2013
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A ring A is a Rickart ring, if for every a G A there exist some p, q G A' such that a° = pA and °a = Aq. Note that if A is a Rickart ring, then A has a unity element. The proof is similar to that used for Rickart *-rings [1].
Let H be a Hilbert space, L(H) the set of all bounded linear operators on H, and let R(A) and N(A) denote the range and the null-space of A G L(H). If M C H, we will denote by M the norm-closure of M in H. For a finite dimensional Hilbert space H Hartwig [3] defined a partial order in L(H) in the following way:
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A X B if and only if rank(B — A) = rank(B) — rank(A).
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He also observed that there exists another equivalent definition of this order, namely
A X B if and only if A-A = A-B and AA- = BA-,
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where A- is a generalized inner inverse of A, i.e. AA-A = A. The partial order X is thus usually called the minus partial order.
In [6] Semrl extended the minus partial order in L(H) for an arbitrary Hilbert space H. The notion of a rank of an operator (equivalently, a rank of a finite complex matrix) can not be applied for bounded linear operators on general Hilbert spaces. Moreover, A G L(H) has a generalized inner inverse if and only if its image is closed. Since Semrl could not use the notion of rank of an operator and since he did not want to restrict his attention only to closed range operators, he found a new approach how to extend the minus partial order. He introduced another equivalent definition of the minus partial order: for A, B G L(H), where H is a finite dimensional Hilbert space, we have A X B if and only if there exist idempotent operators P, Q G L(H) such that R(P) = R(A), N(Q) = N(A), PA = PB and AQ = BQ. Recall that the range of an idempotent operator P G L(H), where H can be a general Hilbert space, is closed. Using the same equations, only adding the closure on R(A) Semrl extended the concept of the minus partial order in L(H) for an arbitrary Hilbert space H:
Definition 1.1. Let H be a Hilbert space and let A, B G L(H). Then A X B if and only if there exists idempotents P, Q G (L(H))' such that the following hold:
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(1)	R(P) = R(A);
(2)	N(A) = N(Q);
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(3)	PA = PB;
(4)	AQ = BQ.
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Semrl proved that ^ is indeed a partial order in B(H). Moreover, it is proved in [6] that Semrl's definition is a proper extension of Hartwig's definition of the minus partial order of matrices. Also, in [5], the minus partial order is generalized on Banach space operators which have generalized inverses.
We prove the following result, which allows us to consider the algebraic version of the minus partial order. First, we need the following auxiliary statement.
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Lemma 1.2. Let H,K,L,N be Hilbert spaces, Ai E L(H,L) and A2 E L(K,L). Then the following statements are equivalent:
(1)	For every B E L(L,N) the following equivalence holds: (BAi = 0 and BA2 = 0) if and only if B = 0;
(2)	R(Ai)+ R(A2) = L.
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Proof. (1) (2): Suppose that R(Ai) + R(A2) = L. Then there exists a non-trivial closed subspace Li of L, such that R(Ai) + R(A2) © Li = L. Let Bi E L(Li,N) be any non-zero bounded linear operator, and define B E L(L, N) as follows:
Bx = 10, x gR(Ai)+ R(A2), 1 Bix, x G L\.
Obviously, B = 0, BA1 = 0 and BA2 = 0.
(2) (1): Obvious.	□
Definition 1.3. Let H be a Hilbert space and let A,B G L(H). Then we write A -< B if and only if there exist idempotent operators P,Q G L(H) such that the following hold:
(1)	◦ A = L(H) ■ (I - P);
(2)	A° = (/ - Q) ■ L(H);
(3)	PA = PB;
(4)	AQ = BQ.
Theorem 1.4. The minus partial order given by Definition 1.1 is on L(H) equivalent to the order given by Definition 1.3.
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Proof. First, let us prove that Definition 1.1 implies Definition 1.3. Let ^ be the order defined with Definition 1.1 and suppose A ^ B, A,B e L(H). Let P be a projection from H onto R(A) and let I — Q be a projection from H onto N(A). This is the same choice of projections as in [6], so statements (3) and (4) of this theorem hold.
Let D e L(H). Then D(1 — P)A = 0, since R(A) C R(P) = N(I — P). Thus, L(H) ■ (I — P) C °A. On the other hand, suppose that D e °A. Then DA = 0 and R(P) C N(D) since N(D) is closed. Thus, N(P*) D R(D*), or equivalently R(1 — P*) D R(D*). By the Douglas theorem [2], there exists some C e L(H) such that D* = (I — P*)C, implying D = C*(/ — P). Hence, °A = L(H) ■ (I — P).
If D e L(H), since R(1 — Q) = N(A), we get A(1 — Q)D = 0. Thus, (I — Q) ■ L(H) C A°. On the other hand, let D e A°. Then R(D) C N(A) = R(1—Q). Again, using the Douglas theorem [2], we conclude that there exists some C e L(H) such that D = (I — Q)C. It follows that A° = (I — Q) -L(H).
Let us now prove that Definition 1.3 implies Definition 1.1. Suppose A ^ B, A, B e L(H), where ^ is the order defined with Definition 1.3.
From (1) we get (I — P)A = 0, so R(A) C N(I — P) = R(P), and consequently R(A) C R(P). Since H = R(P) ©N(P), every operator from L(H) has a 2 x 2 matrix form with respect to this decomposition. Particularly, from R(A) C R(P) we obtain the following:
A
Ai A2' 00
R(P)'
N(P)

R(P)
N(P)
Now we use the fact °A = L(H) ■ (I — P). Notice that B e L(H) ■ (I — P) if and only if B = B(1 — P). If
B
Bi B2
B3 B4
R(P)'
N(P)

R(P)
N(P)
then B = B(1 — P) is equivalent to B1 = 0 and B3 = 0. On the other hand, BA = 0 if and only if
Bi Ai B1A2" B3 Ai B3A2
0.
So we have the equivalence:
(B1A1 = 0, B1A2 = 0, B3A1 = 0, B3A2 = 0)
(Bi = 0,B3 = 0).
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From Lemma 1.2 we know that R(A 1) + R^) = R(P). Since Ai and A2 act on different subspaces, we actually have R(A) = R(P).
Now, from the condition (2) of this theorem, using the result we have just proved, the following hold:
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A° = (I - Q) • L(H) ^ °(A*) = L(H) • (I - Q*) ^ R(Q*) = R(A*)
^ R(1 - Q) = N(Q) = N(A).
Hence, A X B.	□
2. Results in rings
Previous Theorem 1.4 suggests the following definition of the minus partial order. Since some preliminary results can be proved in a general setting, we shall in this section use that A is a ring with the unit 1.
Definition 2.1. Let A be a ring with the unit 1, and let a, b E A. Then we write a ^ b if and only if there exists idempotents p, q E A^ such that the following hold:
(1)	°a = A(1 - p);
(2)	a° = (1 - q)A;
(3)	pa = pb;
(4)	aq = bq.
We call ^ the minus partial order on A. In the next section we will prove CO	that when A is a Rickart ring, ^ is indeed a partial order.
Notice that from (1) we obtain (1 — p)a = 0 so a = pa. Similarly, a = aq. We need some auxiliary results.
Lemma 2.2. Let p, q E AV Then
(i)	A(1 — p) = °p;
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(ii)	(1 — q)A = q°.
Proof. We have A(1 — p) C °p, since (1 — p)p = 0. Suppose that for u E A, up = 0. Then u = u(1 — p) E A(1 — p). The proof of (ii) is similar.	□
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It follows that we can replace the conditions (1) and (2) of Definition 2.1 by the conditions °a = °p and a° = q° respectively.
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Lemma 2.3. Let p G A^ and a G Then (i) (°p)° = pA;
Proof. (i): By Lemma 2.2,
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(°p)° = (A(1 - p))° = {u G A : (Vx g A) x(1-p)u = 0} = {u G A : (l-p)u = 0} = pA
(ii): The "only if" part follows from Lemma 2.2. Now, suppose that (°a)° = (°p)° i.e. (°a)° = pA. Let u G °a. As p G pA = (°a)° we have up = 0 so u = u(1 — p) G A(1 — p). On the other hand, suppose that u G A(1 — p) i.e. up = 0. As a G (°a)° = pA we have a = pa so ua = upa = 0.	□
In the same manner we obtain the following lemma.
Lemma 2.4. Let q G A' and a G A. Then
(i)	°(q°) = Aq;
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(ii)	a° = (1 — q)A ^ °(a°) = °(q°).
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It follows that we can replace the conditions (1) and (2) of Definition 2.1 by the conditions (°a)° = pA and °(a°) = Aq respectively.
Our definition of order X is a proper extension of well known partial order on the set of idempotents.
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Theorem 2.5. Let a,b G A'. Then a X b if and only if ab = ba = a.
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Proof. Suppose that a, b G A' and ab = ba = a. By Lemma 2.2 we have °a = A(1 — a), a° = (1 — a)A and by assumption aa = ab, aa = ba so a X b. Now suppose that a X b. There exist p, q G A' as in Definition 2.1 so ab = (pa)b = (pb)b = pb = a and ba = b(aq) = b(bq) = bq = a.	□
Recall that von Neumann regular ring is a ring A such that for every a G A there exists an x G A such that axa = a. The following theorem shows that, when A is a von Neumann regular ring, X order coincides with well known minus partial order which is defined by a < b if there exists an x G A such that ax = bx and xa = xb where axa = a. Thus, the minus partial order in von Neumann regular ring is defined in the same way as in L(H) where H is a finite dimensional Hilbert space.
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Theorem 2.6. Suppose that A is a von Neumann regular ring with the unit 1 and let a, b G A. Then a ^ b if and only if a < b.
Proof. Suppose that a ^ b and let p, q G A* be as in Definition 2.1. Since A is von Neumann regular ring, there exists an x G A such that axa = a. Set y = qxp. We have aya = a(qxp)a = axa = a, ay = aqxp = bqxp = by, ya = qxpa = qxpb = yb so a < b.
Now suppose that a < b. There exists an x G A such that axa = a, ax = bx, xa = xb. Set p = ax and q = xa. Then p G A* and 1 — p G °a. On the other hand if u G °a then up = u(ax) = 0, so u = u(1 — p), °a = A(1 — p). Moreover, pa = axa = axb = pb. Similarly, q G A*, a° = (1 — q)A, aq = bq, so a ^ b.	□
Since we can not use decompositions of spaces induced by projections, we have to use idempotents appropriately.
Remark 1. We say that equality 1 = e1 + e2 + ■ ■ ■ + en, where ei,e2,..., en G A*, is a decomposition of the identity of the ring A if ei and ej are orthogonal for i = j, i.e. e^ej = 0 for i = j. Let 1 = e1 + ■ ■ ■ + en and 1 = f1 + ■ ■ ■ + fn be two decompositions of the identity of a ring A. For any x G A we have
x
1 ■ x ■ 1 = (ei +-----+ en)x(fi +-----+ fn)=^2 e^x/j
i,j=1
and above sum defines a decomposition of A into a direct sum of groups:
n
A = 0 eiA/j.	(2.1)
i,j=1
It is convenient to write x as a matrix
xn ■
x
xn1
x1n
xn
ex/
where xj = e^xfj G e^A/j
If x = (xij)ex/ and y = (yj)ex/, then it is obvious that x + y = (xj + yj)ex/. Moreover, if 1 = g1 + ■ ■ ■ + gn is a decomposition of the identity
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of A and z = (zj)/xg, then, by the orthogonality of idempotents involved,
xz = ( xikzkj J . Thus, if we have decompositions of the identity of
A, then the usual algebraic operations in A can be interpreted as simple operations between appropriate n x n matrices over A.
When ei = fi, i = 1, n, the decomposition (2.1) is known as the two-sided Peirce decomposition of the ring A, [4]. When n = 2, e1 = p and f1 = q then we write
x11 x12 X21 X22 pxq
x
We prove the following result.
Theorem 2.7. Let A be a ring with the unit, and let a,b G A. Then a ^ b if and only if there exists idempotents p,q G A' such that the following three conditions hold:
(1) a
a1 0
0 0
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(2)	If z G Ap and za1
(3)	If z G qA and a1z
and b =
a1 0
0
b1
0, then z 0, then z
px q 0; 0.
Proof. Suppose that a ^ b and let p, q G A' be corresponding idempotents. As we have seen a = pa = aq = paq, so
a1 0 00
Let
From p(b — a) = 0 we get
b4
bs
b2 b1
px q
px q
p	0		b4 — a1	b2	
0	0	px p	bs	b1	px q
p(b4 — a1) pb2
0
0
0,
px q
implying that p(b4 — a1) = 0 and pb2 = 0. Since pa1 = a1, pb4 = b4, and pb2 = b2, we get a1 = b4 and b2 = 0. Analogously, from (b — a)q = 0 we get bs = 0. Thus, the statement (1) of this theorem is proved.
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Since a
we get za = 0, so z G °a = A(1 — p) = °p, i.e.
In order to prove the statement (2), suppose that z G Ap and zai = 0. ai 0 00
z = zp = 0. The statement (3) can be proved proved in the same manner.
Now, we suppose that there exists idempotents p, q G A' such that statements (1) - (3) of this theorem hold. We immediately obtain
pxq
p(a—b)
p	0		00
0	0	px p	0 —bi
0 and (a—b)q
pxq
0	0		q	0
0	—bi	px q	0	0
qxq
Now, we prove that °a = A(1 — p). If y G A(1 — p), then y It is easy to see that ya = 0. Thus, we established A(1 — p) C °a.
To prove the opposite inclusion, suppose that z G °a. Then z = and
0 = za =
0 y2
0 y4
qxp
zi z2
z3 z4
qxp
ziai 0 z3a1 0
qx q
We conclude ziai — z3ai = 0. Since z1, z3 G Ap, (2) shows that z1 = z3 = 0. "0 z2_
Thus, z
0 z4
G A(1 — p). Hence, we proved °a C A(1 — p).
qx p
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In the same manner we can prove that a° = (1 — q)A.
3. Minus partial order in Rickart rings
The idempotents in Definition 2.1 need not be unique. Write
LP(a) := {p gA' : °a = A(1 — p)}, RP(a) := {q G A' : a° = (1 — q)A}.
When A is Rickart ring then LP(a) and RP(a) are nonempty. Lemma 2.2 gives characterizations LP(a) = {p G A' : °a = °p} and RP(a) = {q G A' : a° = q°}.
Lemma 3.1. Let a G A, p G LP(a) and q G RP(a). (Such idempotents exist if A is a Rickart ring.) Then
(i) LP(a)
p p i 00
: pi G pA(1 — p)
px p
0
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(ii) RP(a)
q 0
qi 0
: qi G (1 - q)Aq
qxq
Proof. From (1 — p)a = 0 = a(1 — q) we conclude that a
ai 0 00
. If
pxq
P'
P Pi 00
then p'2 = p' and
pxp
(1 — p')a It is easily seen that u =
0	-Pi		ai	0
0	1 - p	px p	0	0
(3.1)
pxq
u1 u2
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G °a if and only if u1 = u3 = 0. From
px p
0	«2		P	Pi
0	u4	px p	0	0
= 0 we conclude °a C °p'. Now, (3.1) and Lemma 2.2
px p
give °a = A(1 — p'), that is p' G LP3(a).
Suppose now that p'
P2 Pi P3 P4
G LP(a). Then °p' = °a = °p, so
px p
P - P2 -Pi
- P 3 1 - P - P4
	P	0		P - P2	0
px p	0	0	px p	- P3	0
0 = (1 — p')p
L —P3 1 — p — P4J ,
and hence p2 = p and p3 = 0. From °a C °p' it follows 0
px p
0	«2		P	Pi		0	«2P4
0	u4	px p	0	P4	px p	0	u4p4
px p
so u4p4 = 0, for every u4 G (1 — p)A(1 — p). Setting u4 = 1 — p we get p4 = 0. Thus, the statement (i) of the theorem is proved. In the same manner we can prove the statement (ii).	□
Corollary 3.2. Let a,b G A. Suppose that a ^ b and let p, q G A be corresponding idempotents. Then
{p' G LP(a) : a = p'b} =
{q' G RP(a) : a = bq'} = where bi is as in Theorem 2.7.
P Pi 00
q 0'
qi 0
: pi G pA(1 - p),pibi = 0^ (3.2)
px p
: qi G (1 - q)Aq, biqi = 0
qx q
0
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Proof. We will prove only the equality (3.2); the proof of the other one is analogous. Since a ^ b, Theorem 2.7 gives
a1 0 0 0
b
pxq
ai 0 0 bi
pxq
If p' belongs to the set on the right hand side of (3.2) then, by Lemma 3.1, p' e LP(a). Also,
p'b =
p	pi		ai	0		ai	pibi		ai	0
0	0	px p	0	bi	px q	0	0	px q	0	0
pxq
To prove the opposite inclusion, suppose that p' e LP(a) and a = p'b.
Lemma 3.1 leads to p =
ai 0 00
p pi 00
px q
Now, a = p'b gives
px p
a = p'b
ai pibi 00
px q
so pi bi = 0.
□
However, to prove that ^ is actually a partial order, we need the assumption that A is Rickart ring.
We now prove the main result of this section.
Theorem 3.3. Let A be a Rickart ring. Then ^ is a partial order in A.
Proof. Since A is a Rickart ring, for any a e A there exist idempotents p,q e A", such that °a = A(1 — p) and a° = (1 — q)A. Now the reflexivity of ^ follows.
To prove the antisymmetry, suppose that a ^ b and b ^ a. Then
ai 0 00
b
px q
ai 0 0 bi
(3.3)
px q
and there exist r,s e A" such that b = ra = as. Let r have
a0
= b = ra =
ri r 2
r3 r4
We
px p
ai 0 0 bi
px q
ri ai 0 r3ai 0
px q
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so bi = 0 and hence a = b.
We have to show transitivity. Let a ^ b and b ^ c. Then there exist idempotents p,q,r, s E A^ such that a and b have the matrix forms as in (3.3), °ai = °p, ai = q° and °b = A(1 — r) = °r, b° = (1 — s)A = s°, b = rc = cs. Suppose that
r=	r1	r2		and		c=		c1	c2
	r3	r4	px p					c3	c4
									
p — r1		—r2			a1	0"			r
—r3	1—	p—	r4	px p	0	b1	px q		
pxq
Since
0 = (1—r)b
°a1 = °p shows that 0 = (p — r1)p = p — r1 and 0 = r3p = r3. Also, r2b1 = 0. From b = rc we conclude that
(p — r1)a1 —r2b1
—r3a1 (1 — p — r4)b1
pxq
a1	0		p	r2		c1	c2	
0	b1	px q	0	r4	px p	c3	c4	px q
c1 + r2c3 c2 + r2c4 r4c3	r4c4
pxq
so
Let p'
Since,
p r2 00
px p
a1 = c1 + r2c3 and 0 = c2 + r2c4. From Corollary 3.2 it follows that
°a = A(1 — p') and a = p'b.
(3.4)
(3.5)
p'c
c1 + r2c3 c2 + r2c4 00
px q
(3.4) shows that
p c = a
Similar consideration shows that if s
s1 s2 s3 s4
than s1
(3.6)
q, s2 = 0 and
that for q' =
q0
s3 0
qx q
we have
qx q
a° = (1 — q')A, a = bq' = cq'. By definition, from (3.5)-(3.7) we obtain that a ^ c.
(3.7)
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Moreover, from the proof of Theorem 3.3 it follows that if a X b and b X c then there exist common idempotents p' and q' showing that a X b and a X c.
Theorem 3.4. Let A be a Rickart ring and a, b G A. Then a X b if and only if there exist decompositions of the identity of the ring A
1 = ei + e2 + es, 1 = / + /2 + /3 such that the following five conditions hold:
(1) a
ai 0 0 0 0 0 0 0 0
and b =
ex/
ai 0 0 0 bi 0 0 0 0
ex /
(2)	If z G Ae1 and za1 = 0, then z = 0;
(3)	If z G f1 A and a1z = 0, then z = 0;
(4)	If z G Ae2 and zb1 = 0, then z = 0;
(5)	If z G f2A and b1z = 0, then z = 0.
Proof. "If" part follows from Theorem 2.7. Now, suppose that a X b. By Theorem 2.7 we have
ai 0 00
and b
pxq
ai 0 0 bi
pxq
such that z = 0 whenever z G Ap and za1 = 0, and z = 0 whenever z G qA and a1z = 0. A is a Rickart ring so there exist r, s G A' such that °&1 = A(1 - r) = °r and = (1 - s)A = s°. Notice that pr = 0. Indeed, since b1 G (1 — p)A(1 — q), we have pb1 = 0 and therefore p G °b1 = °r. Let r' := r — rp(1 — r) = r — rp = r(1 — p). We have r' G A' since r/2 = (r — rp)(r—rp) = r—rp = r'. Our next claim is that °b1 = °r'. If ub1 = 0 then ur = 0 so ur' = 0. On the other hand, (1 — r')b1 = (1 — r + rp)rb1 = 0, due to b1 = rb1. Thus
°&1 = °r' = A(1 — r').	(3.8)
Next, pr = 0 implies pr' = 0. Moreover, r'p = r(1 — p)p = 0. Set e1 = p, e2 = r', and e3 = 1 — p — r'. Then 1 = e1 + e2 + e3 is decomposition of the
a
identity of the ring A and from (3.8) it follows that zbi = 0 implies z = 0 when z G Ae2.
Now, set f1 = q, /2 = (1 — q)s and /3 = 1 — f1 — /2. With similar consideration we can show that 1 = f1 + /2 + /3 is the decomposition of the identity of the ring A and that condition (5) of this theorem holds. Of course, statements (2) and (3) are satisfied by Theorem 2.7 since e1 = p and f1 = q. As °b1 = A(1 — r) and b1 = (1 — s)A we have e2bf2 = r(1 — p)b(1 — q)s = rb1s = b1. The statement (1) is proved.	□
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Note that the statements (1)-(5) of the previous theorem are equivalent
to
ei e LP(a), e2 e LP(b — a), / e RP(a), /2 e RP(b — a).
Corollary 3.5. Suppose that A is a Rickart ring and a, b e A. Then a ^ b if and only if b — a ^ b.
We conclude this section with one more characterization of minus partial order.
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Theorem 3.6. Let A be a Rickart ring and a, b e A. Then a < b if and only if there exists idempotents e1 e LP(a), e2 e LP(b — a), /1 e RP(a), /2 e RP(b — a) such that e1 e2 = 0 and /2/1 = 0.
Proof. The "only if" part follows from Theorem 3.4. In order to prove "if" part suppose that e1 e LP(a), e2 e LP(b — a) and e1e2 = 0. Then e1a = a and e1b = e1a + e1(b — a) = a+e1e2(b — a) = a. Similarly, from /2/1 = 0 where /1 e RP(a), /2 e RP(b — a) it follows that a/1 = a and b/1 = a/1 + (b — a)/1 = a + (b — a)/2/1 = a. By definition, a ^ b.	□
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Under the notation of Theorem 3.4 it is easy to check that e1 + e2 G LP(b) and /1 + /2 G RP(b). Also, one can show
(e1 + e2)A = e1A © e2A and A(/ + /2) = A/1 © A/2. We have not proof for the opposite implication:
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If there exist e1 e LP(a), e2 e LP(b — a), e3 e LP(b), /1 e RP(a), /2 e RP(b — a), /3 e RP(b) and if
e3A = eiA© e2A and A/3 = A/i ©A/2,
then a ^ b. This can be suggested as an open problem.
References
[1]	S. K. Berberian, Baer *-rings, Springer-Verlag, New York, 1972.
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[2]	R. Douglas, On majorization, factorization, and range inclusion of operators on Hilbert space, Proc. Amer. Math. Soc. 17 (1966), 413-415.
[3]	R. E. Hartwig, How to partialy order regular elements, Math. Japon. 25 (1980), 1-13.
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[4]	N. Jacobson, Structure of rings, Amer. Math. Soc., 1968
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[5]	D. S. Rakic, D. S. Djordjevic, Space pre-order and minus partial order for operators on Banach spaces, Aequationes Math. Publisged online first: 28. June 2012. doi: 10.1007/s00010-012-0133-2
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[6]	P. Semrl, Automorphisms of B(H) with respect to minus partial order, J. Math. Anal. Appl. 369 (2010), 205-213.
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Corresponding author Dragan S. Rakic
University of Nis, Faculty of Sciences and Mathematics, Visegradska 33, P.O. Box 95, 18000 Nis, Serbia
E-mail: rakic.dragan@gmail.com Telephone number: +381646541008 Fax: +38118533014
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