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Equalities of ideals associated with twoprojections in rings with involutionJulio Benítez a & Dragana Cvetković-Ilić ba Instituto de Matemática Multidisciplinar , Universitat Politècnicade València , Camino de Vera s/n, Valencia 46022 , Spainb Faculty of Sciences and Mathematics, Department ofMathematics , University of Nis , Nis 18000 , SerbiaPublished online: 11 Dec 2012.

To cite this article: Julio Benítez & Dragana Cvetković-Ilić (2013) Equalities of ideals associatedwith two projections in rings with involution, Linear and Multilinear Algebra, 61:10, 1419-1435,DOI: 10.1080/03081087.2012.743026

To link to this article: http://dx.doi.org/10.1080/03081087.2012.743026

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Equalities of ideals associated with two projections in rings

with involution

Julio Benıteza and Dragana Cvetkovic-Ilicb*

aInstituto de Matematica Multidisciplinar, Universitat Politecnica de Valencia, Caminode Vera s/n, Valencia 46022, Spain; bFaculty of Sciences and Mathematics,

Department of Mathematics, University of Nis, Nis 18000, Serbia

Communicated by R.B. Bapat

(Received 5 August 2012; final version received 16 October 2012)

In this article we study various right ideals associated with two projections(self-adjoint idempotents) in a ring with involution. Results of O.M.Baksalary, G. Trenkler, R. Piziak, P.L. Odell and R. Hahn aboutorthogonal projectors (complex matrices which are Hermitian and idem-potent) are considered in the setting of rings with involution. New proofsbased on algebraic arguments, rather than finite-dimensional and ranktheory, are given.

Keywords: rings with involution; projections; Moore–Penrose inverse

AMS Subject classifications: 16W10; 16D25

1. Introduction

Throughout this article, the symbolR will denote a unital ring (1 will be its unit) withan involution. Let us recall that an involution in a ring R is a map a � a* in R suchthat (aþ b)*¼ a*þ b*, (ab)*¼ b*a* and (a*)*¼ a for any a, b2R. The word‘projection’ will be reserved for an element q of R which is self-adjoint andidempotent, that is q*¼ q¼ q2. With each element a2R we associate an image idealaR¼ {ax: x2R}, and a kernel ideal a�¼ {x2R: ax¼ 0}.

This article studies some ideals and functions depending on two projections of aring with an involution. The results given here generalize to several considered in[1,15]. It is worth noting that in the proofs of those results, matrix theory is used(specifically, rank theory and singular value decomposition). We believe that simplerand algebraic proofs (our proofs only use algebraic reasoning) gives a greater insightof the problems considered here. We will notably consider the case of a *-reducingring (where there is an implication a*a¼ 0) a¼ 0 for all a2R). One of the mostimportant tools in this case is the Moore–Penrose inverse theory, which will bequickly revised together with useful results in Section 2. Section 3 then studiesinvertibility of pþ q and p� q in rings with involution, together with ideal properties.Section 4 finally focuses on *-reducing rings. In this case, existence of particularMoore–Penrose inverses leads to certain ideal equalities.

*Corresponding author. Email: dragana@pmf.ni.ac.rs

� 2012 Taylor & Francis

http://dx.doi.org/10.1080/03081087.2012.743026

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2. Moore–Penrose inverse and auxiliary lemmas

The link between generalized inverses and range or kernel ideals is not new, see for

instance [4,12] in the case of semigroups, [13] in rings or [5,6] in C*-algebras.

The study of sums and differences of idempotents has notably been studied in

connection with the Drazin inverse (see e.g. [10] and the references therein). In this

article, we use the involutive structure of the ring and properties of the

Moore–Penrose inverse to study particular idempotents, projections (self-adjoint

idempotents).It can be proved that for any a2R, there is at most one ay 2R such that

aaya ¼ a, ayaay ¼ ay, ðaayÞ� ¼ aay, ðayaÞ� ¼ aya:

[3,5–7,16]. If there exists such ay we say that a is Moore–Penrose invertible and ay is

the Moore–Penrose inverse of a. The subset of R composed of all Moore–Penrose

invertible elements is denoted by Ry. We write R�1 as the set of all invertible

elements in R.We say that a2R is relatively regular if there exists b2R such that aba¼ a.

In this case b is called an inner generalized inverse of a. A known result

(see Theorem 1.4.11 of [3]) is the following: let R be a ring with involution obeying

the Gelfand–Naimark property. Then a2R is Moore–Penrose invertible if and

only if a is relatively regular. Let us recall that a ring R with involution has the

Gelfand–Naimark property if 1þx*x2R�1 for all x2R. It is known that any

C*-algebra has the Gelfand–Naimark property, see also [5,9].An element a2R is left *-cancellable if a*ax¼ a*ay implies ax¼ ay.

Analogously, a2R is right *-cancellable if xaa*¼ yaa* implies xa¼ ya. Finally,

a2R is *-cancellable if it is both left and right *-cancellable. A ring R is called

*-reducing if every element of R is *-cancellable. Let us remark that any C*-algebra

is a *-reducing ring.We use the following notation: if X, Y�R, then

X ? Y () 8 ðx, yÞ 2X� Y, x�y ¼ 0:

Observe that if R is *-reducing and if X 6¼ ø 6¼Y, then X ? Y implies X\Y¼ {0}.Let x2R and let p2R be an idempotent (p¼ p2). Then we can write

x ¼ pxpþ pxð1� pÞ þ ð1� pÞxpþ ð1� pÞxð1� pÞ

and use the notations

x11 ¼ pxp, x12 ¼ pxð1� pÞ, x21 ¼ ð1� pÞxp, x22 ¼ ð1� pÞxð1� pÞ:

Every projection p2R induces a matrix representation which preserves the

involution in R, namely x2R can be represented by means of the following matrix:

x ¼pxp pxð1� pÞ

ð1� pÞxp ð1� pÞxð1� pÞ

� �p

¼x11 x12x21 x22

� �p

: ð2:1Þ

From now on, for an arbitrary projection p, we shall denote p ¼ 1� p.

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Two elementary facts derived from the definition of the Moore–Penrose inverseare the following:

(i) Let a2R. Then a2Ry , a*2Ry, and under this situation one has(a*)y¼ (ay)*.

(ii) Let a2Ry. Then ay 2Ry and (ay)y¼ a.

The following simple result also will be useful.

LEMMA 2.1 Let R be a ring with involution and a2R. Then

(i) If a2Ry, then a*a, aa*2Ry and

ða�aÞy ¼ ayða�Þy, ðaa�Þy ¼ ða�Þyay, ay ¼ ða�aÞya� ¼ a�ðaa�Þy,

a� ¼ ayaa� ¼ a�aay:

(ii) If R is *-reducing, then a*a2Ry) a2Ry and aa*2Ry) a2Ry.

Proof The proof of (i) is a consequence of direct computations. We will prove onlythe first implication of (ii) since to prove the other one, it is sufficient to make thesame argument for a* instead of a. Assume that a*a2Ry, and let x¼ (a*a)y a*.Observe that the Moore–Penrose inverse of a self-adjoint Moore–Penrose invertibleelement is again self-adjoint, and thus, (a*a)y is self-adjoint. Now (ax)*¼[a(a*a)ya*]*¼ a(a*a)ya*¼ ax; xa¼ (a*a)ya*a is self-adjoint; xax¼ (a*a)ya*a(a*a)y

a*¼ (a*a)ya*¼ x. Finally, a*axa¼ a*a(a*a)ya*a¼ a*a, and sinceR is *-reducing, weget axa¼ a. g

A simple consequence of Lemma 2.1 is the following: let x2Ry be self-adjoint.Then xxy¼ xyx and x is the commuting (or group) inverse of x (see e.g. [5]). In fact,xxy¼ x(x*x)yx*¼ x*(xx*)yx¼ xyx. For a better insight on the formulae of Lemma2.1, commutation and cancellation properties, see [12]. For the class of elements x ina C*-algebra such that xxy¼ xyx, the reader is refferred to [2,8]. More generally,elements admitting both a group inverse and a Moore–Penrose inverse are discussedin [14] in the case of a ring.

Let p and q be two projections in a ring R with involution. Then

p ¼p 00 0

� �p

and q ¼a bb� d

� �p

, ð2:2Þ

where

a ¼ pqp, b ¼ pqð1� pÞ, d ¼ ð1� pÞqð1� pÞ: ð2:3Þ

LEMMA 2.2 Let p, q2R be projections given by (2.2). Then

(i) a¼ a2þ bb*,(ii) b¼ abþ bd,(iii) d¼ d2þ b*b.

Proof All the equalities follow from the condition q¼ q2. g

The following result is a generalization of Lemma 3 (v)-(x) from [1]:

LEMMA 2.3 LetR be a *-reducing ring. If p, q2R are projections given by (2.2), thenthe following hold:

(i) If a is Moore–Penrose invertible, then aayb¼ b.

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(ii) If 1� a is Moore–Penrose invertible, then (1� a)(1� a)yb¼ (p� a)(p� a)yb¼ b.

(iii) If d is Moore–Penrose invertible, then bddy¼ b.(iv) If 1� d is Moore–Penrose invertible, then b(1� d)(1� d)y¼ b(1� p� d)

(1� p� d)y¼ b.(v) If a and 1� d are Moore–Penrose invertible, then ayb¼ b(1� p� d)y¼

b(1� d)y.(vi) If 1� a and d are Moore–Penrose invertible, then bdy¼ (1� a)yb¼ (p� a)yb.

Proof (i): Since a¼ pqp¼ pq(pq)* is Moore–Penrose invertible, by Lemma 2.1,we have that pq is Moore–Penrose invertible and

aayb ¼ pqð pqÞ�ð pqð pqÞ�Þypqð1� pÞ

¼ pqð pqÞ�ðð pqÞ�Þyð pqÞypqð1� pÞ

¼ pqð pqÞypqð pqÞypqð1� pÞ

¼ b:

To prove (ii), it is sufficient to use former item (i) for projections p and 1� q. If weuse item (i) for projections 1� p and q, we get ddyb*¼ b*, and (iii) follows by taking *in both sides. Item (iv) follows by using item (i) for projections 1� p and 1� q.

(v): Observe that since q is self-adjoint, then the representation of q given in (2.2)implies that a is self-adjoint, hence aay¼ aya. By condition (ii) of Lemma 2.2 and by(i), it follows that ay b¼ ay(abþ bd)¼ aaybþ aybd¼ bþ ay bd. Hence,

b ¼ aybð1� d Þ ¼ aybð1� p� d Þ:

Multiplying the last equality from the left side by (1� d)y and using (iv), we get thatb(1� d)y¼ ayb. Similarly, ayb¼ b(1� p� d)y. The proof of (vi) follows by using item(v) for projections p and 1� q. g

LEMMA 2.4 LetR be a *-reducing ring. If p, q2R are projections and q is partitionedas in (2.2), then

(i) If 1� a and d are Moore–Penrose invertible, then a� bdyb*¼ 1� (1� a)(1� a)y¼ p� (p� a)y(p� a).

(ii) If a and 1� d are Moore–Penrose invertible, then d� b*ayb¼ 1� (1� d)(1� d)y¼ 1� p� (1� p� d)(1� p� d)y.

(iii) If 1� a and d are Moore–Penrose invertible, then dþ b*(1� a)yb¼dþ b*(p� a)yb¼ ddy.

(iv) If a and 1� d are Moore–Penrose invertible, then aþ b(1� d)yb*¼aþ b(1� p� d)yb*¼ aay.

Proof (i): As we pointed out in the proof of item (v), Lemma 2.3, we have that a isself-adjoint, hence 1� a is again self-adjoint, and thus, (1� a)y(1� a)¼ (1� a)(1� a)y. Now we have

1� a ¼ ð1� aÞyð1� aÞð1� aÞ ¼ ð1� aÞyð1� aÞ � ð1� aÞyð1� aÞa:

By Lemma 2.3 (vi), Lemma 2.2 (i) and the previous computation, we get thatbdyb*¼ (1� a)ybb*¼ (1� a)y(1� a)a¼ (1� a)y(1� a)� (1� a). Hence, a� bdyb*¼1� (1� a)–(1� a)y. The proofs of (ii)–(iv) are similar. g

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3. Projections in ring

THEOREM 3.1 Let R be a ring and p, q2R be projections. Then

ð p� qÞR ¼ pqR�? pqR:

Proof It is evident that pqR ? pqR. Now, we will prove that ð p� qÞR ¼ pqRþ

pqR. Take any z2 (p� q)R. We have that z¼ (p� q)x for some x2R. If we take

y¼ (1� 2q)x, we get ð p� qÞx ¼ pqyþ pqy, so ð p� qÞR � pqRþ pqR. For arbitrary

a2 pqRþ pqR, we have that a ¼ pqy1 þ pqy1, for some y1, y22R and

a ¼ pqy1 þ pqy2 ¼ ð p� qÞð y1 � qy1 � qy2Þ,

i.e., a2 (p� q)R. Hence pqRþ pqR � ð p� qÞR. The proof is complete. g

In [11, Th. 4.2], it was characterized when the difference of two projections is

invertible. Former Theorem 3.1 permits another characterization.

COROLLARY 3.1 LetR be a ring and p, q2R be projections. Then p� q is invertible if

and only if pqR�? pqR ¼ R.

THEOREM 3.2 Let R be a ring with involution and p, q2R be projections. The

following statements are equivalent:

(i) pþ q2R�1.(ii) There exists h2R such that 1¼ phþ q(1� h) and (1� h)p¼ hq.(iii) There exists h2R such that h¼ ph, q(1� h)¼ 1� h and (1� h)p¼ hq.

The element h in conditions (ii) and (iii) is unique and it satisfies h¼ p(pþ q)�1.

Proof

(i)) (ii): Define h¼ p(pþ q)�1. Obviously, one has (1� h)p¼ hq. Since (pþ q)p¼ pþ

qþ 2qp� q(pþ q), we have (pþ q)h¼ 1þ 2qh� q, which implies 1¼ phþ q(1� h).

(ii)) (i): From (1� h)p¼ hq we get p¼ (pþ q)h*. Now,

ð pþ qÞð1� h� h� þ 2h�hÞ ¼ phþ qð1� hÞ ¼ 1:

Since pþ q and 1� h� h*þ 2h*h are self-adjoint, by taking * in (pþ q)

(1� h� h*þ 2h*h)¼ 1 we get (1� h� h*þ 2h*h)(pþ q)¼ 1.

(ii)) (iii): Since we have proved (ii)) (i), we get pþ q2R�1. The second condition

of (ii) leads to h¼ p(pþ q)�1. Thus ph¼ h, and substituting this into 1¼ phþ q(1� h)

leads to 1¼ hþ q(1� h), i.e., q(1� h)¼ 1� h.

(iii)) (ii): It is evident. g

THEOREM 3.3 Let R be a unital ring and p, q2R be projections. The following

statements are equivalent:

(i) There exists h2R such that h¼ ph and q(1� h)¼ 1� h.(ii) R¼ pRþ qR.

Proof

(i) ) (ii) follows from 1¼ hþ (1� h)¼ phþ q(1� h)2 pRþ qR.

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(ii) ) (i): Let x, y2R be such that 1¼ pxþ qy and let us denote h¼ px. Now,

ph¼ h and q(1� h)¼ q(1� px)¼ qqy¼ qy¼ 1� px¼ 1� h. g

As a corollary of Theorem 3.2, Theorem 3.3 and [11, Theorem 4.4], we have the

following:

COROLLARY 3.2 Let R be a ring with involution and p, q2R be projections. The

following statements are equivalent:

(1) pþ q2R�1 and h¼ p(pþ q)�1 is idempotent.(2) R¼ pR � qR.(3) p� q2R�1.(4) There exists idempotent k such that kR¼ pR and (1� k)R¼ qR.

Proof

(1)) (2) h¼ p(pþ q)�1) (1� h)¼ q(pþ q)�1 and hR¼ pR, (1� h)R¼ qR. h2¼ h

then implies R¼ hR� (1� h)R¼ pR � qR.

(2)) (3)) (4) by [11, Theorem 4.4].(4)) (1) By Lemma 4.1, we have that k satisfies pk¼ k, kp¼ p, q(1� k)¼ k and

(1� k)q¼ q. It follows that (1� k)p¼ 0¼ kq and k satisfies condition (iii) of

Theorem 3.2.

4. Projections in *-reducing rings

We shall need the following simple lemma:

LEMMA 4.1 Let x, y2R.

(1) If x, y and xy are self-adjoint, then yx¼ xy.(2) Self-adjoint x is an invertible if and only if there exists y2R such that xy¼ 1 if

and only if xR¼R.(3) If e and f are idempotents, then eR¼ fR if and only if ef¼ f and fe¼ e.(4) If p, q are projectors and pR¼ qR, then p¼ q.

Proof The proofs of (1) and (2) are trivial.(3) Let e, f2R be two idempotents such that eR¼ fR. Since e2 eR¼ fR, then

exists t2R such that e¼ ft, so fe¼ e. By reversing the roles of e and f we have ef¼ f.(4) It follows by (3). g

The following result, which will be of major importance in the sequel, gives

sufficient conditions for the Moore–Penrose invertibility of several elements in a

*-reducing ring of the form f(p, q), where p and q are two projections and f is a

polynomial in two non-commuting variables.

THEOREM 4.1 Let R be a *-reducing ring. If p, q2R are projections, then

(i) If pqp and pqp are Moore–Penrose invertible, then pþ q is Moore–Penrose

invertible and

ð pþ qÞð pþ qÞy ¼ pþ pqðpqpÞy:

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(ii) If pqp is Moore–Penrose invertible, then pq is Moore–Penrose invertible and

ðpqÞy ¼ qðpqpÞy:

(iii) If pqp and pqp are Moore–Penrose invertible, then p� q is Moore–Penrose

invertible and

ð p� qÞð p� qÞy ¼ ð p� pqpÞð p� pqpÞy þ pqðpqpÞy:

(iv) If pqp is Moore–Penrose invertible, then pq� qp is Moore–Penrose

invertible and

ð pq� qpÞð pq� qpÞy ¼ pqð pqpÞy þ ð pqpÞyqp:

Proof

(i) Let us suppose that the projections p and q are represented as in (2.2). By

hypothesis one has that p� a, d2Ry. Since 1� a¼ (p� a)þ (1� p) and

p� a, p2Ry (observe that since p is a projection, obviously p2Ry and

py ¼ p) we get 1� a2Ry. Let

x ¼1

2pþ ð p� aÞð p� aÞy� �

� bdy � dyb� þ 2dy � ddy: ð4:1Þ

We shall prove that x¼ (pþ q)y by verifying the four conditions of the

Moore–Penrose invertibility. We shall decompose x as in (2.1). Obviously we

have

px¼x11þx12 and qx¼ ax11þbx21þax12þbx22þb�x11þdx21þb�x12þdx22,

where

x11 ¼1

2pþ ð p� aÞð p� aÞy� �

, x12 ¼ �bdy, x21 ¼ �d

yb�, x22 ¼ 2dy � ddy:

Let us remark that p� a is self-adjoint. Then

p� a ¼ ð p� aÞð p� aÞð p� aÞy ¼ ð p� aÞð p� aÞy � að p� aÞð p� aÞy,

and thus, by utilizing Lemma 2.4 (i) we get

ð pþ aÞ pþ ð p� aÞð p� aÞy� �

¼ pþ ð p� aÞð p� aÞy þ aþ að p� aÞð p� aÞy

¼ 2 ð p� aÞð p� aÞy þ a� �

¼ 2 pþ bdyb�� �

:

Thus,

x11 þ ax11 þ bx21 ¼1

2ð pþ aÞ pþ ð p� aÞð p� aÞy

� �� bdyb� ¼ p:

Observe that Lemma 2.3 (ii) in conjunction with Lemma 2.3 (vi) can be

written bdy� b¼ abdy. Hence by Lemma 2.3 (iii), we get

x12 þ ax12 þ bx22 ¼ �bdy � abdy þ bð2dy � ddyÞ ¼ 0:

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Lemma 2.3 (ii) and the self-adjointness of a imply b*(p� a)(p� a)y¼ b*.Furthermore, from the definition of b given in (2.3) we trivially get b*p¼ b.Now, Lemma 2.3 (iii) yields

b�x11 þ dx21 ¼1

2b� pþ ð p� aÞð p� aÞy� �

� ddyb� ¼ 0:

Since q is self-adjoint, the representation of q given in (2.2) yields that d isself-adjoint, hence ddy¼ dyd. In view of 2.2 (iii), we have

b�x12 þ dx22 ¼ ðd2 � d Þdy þ 2ddy � d ¼ ddy:

The above computations show that

ð pþ qÞx ¼ pþ ddy: ð4:2Þ

Thus, (pþ q)x is self-adjoint. Since x, pþ q, and (pþ q)x are self-adjoint,Lemma 4.1 (1) permits to get that x(pþ q)¼ (pþ q)x. By Lemma 2.3 (iii) and(4.2) we easily have (pþ q)x(pþ q)¼ pþ q and x(pþ q)x¼ x.Now, since d¼ (1� p)q(1� p), it is evident that (i) holds.

(ii) Since ðpqÞðpqÞ� ¼ pqp2Ry, by Lemma 2.1 we have pq2Ry and

ðpqÞy ¼ ðpqÞ� pqðpqÞ�ð Þy¼ qpðpqpÞy ¼ qðpqpÞy:

By computation we get that pqðpqÞy ¼ pqpðpqpÞy. Now ðpqÞypq is self-adjoint,hence

ðpqÞypq ¼ ðpqÞypq� ��

¼ qp ðpqÞy� ��

¼ qp ðpqÞ�½ y¼ qpðqpÞy,

and thus, by Lemma 2.1,

pq ¼ pqðpqÞypq ¼ pqpðqpÞy ¼ pqpðqpÞ� qpðqpÞ�½ y¼ pqðqpÞ� qpðqpÞ�½

y¼ pqðqpÞy:

(iii) Let us denote z ¼�ð p�aÞð p�aÞy �bdy

�dyb� �ddy

�p. By a direct computation and Lemma

2.2 (iii), Lemma 2.3 (ii), (iii), (vi) and Lemma 2.4 (i) we get(p� q)z¼ (p� a)(p� a)yþ ddy is self-adjoint. Since z is self-adjoint we getthat z(p� q)¼ (p� q)z. By Lemma 2.3 (ii), (iii) we get (p� q)z(p� q)¼ p� qand z(p� q)z¼ z. Thus, p� q2Ry and z¼ (p� q)y.

(iv) Observe that pq� qp¼ b� b*. Since b ¼ pqp2Ry and by ¼ b�ðbb�Þy 2 pRp,by a direct verification of the four Moore–Penrose equations andusing Lemma 2.1 we get (pq� qp)y¼ by� (b*)y and (pq� qp)(pq� qp)y¼ bbyþ by b. g

COROLLARY 4.1 Let R be a *-reducing ring. If p, q2R are projections such that pqpis Moore–Penrose invertible, then pq is also Moore–Penrose invertible and(pq)y¼ q(pqp)y.

Proof It follows from Lemma 4.1 (ii) by changing p by p. g

The following result is a generalization of the result given in [15, Th. 3, Th. 4] forthe matrix case:

LEMMA 4.2 Let R be a *-reducing ring and let p, q2R be projections such that pqpand pqp are Moore–Penrose invertible. Then

(i) x ¼ pþ pðpqÞy is a projection and xR¼ pRþ qR.(ii) y ¼ p� pð pqÞy is a projection and yR¼ pR\ qR.

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Proof

(i) By Theorem 4.1 (i) and (ii) we have x ¼ pþ pðpqÞy ¼ pþ pqðpqpÞy ¼

ð pþ qÞð pþ qÞy, which implies that x is a projection. Since x¼ p(pþ q)yþ

q(pþ q)y, it is evident that xR� pRþ qR. Observe that pqðpqÞy ¼

pqðpqÞy� ��

¼ ðpqÞy� ��

qp, which yields xp ¼ ð pþ pqðpqÞyÞ p ¼ p. From

pþ q¼ (pþ q)(pþ q)y(pþ q)¼ x(pþ q)¼ xpþ xq and xp¼ x we deduce

xq¼ q. Finally, pRþ qR¼ xpRþ xqR� xR.(ii) First of all, we must prove that the definition of y is meaningful, in other

words, we must prove that pq2Ry; but this follows easily from p� pqp2Ry,

ð pqÞð pqÞ� ¼ pqp ¼ p� pqp and Lemma 2.1. Since y ¼ p� pqð pqÞy, we have

that y is self-adjoint. Using that ð pqÞyp ¼ ð pqÞy, we trivially get y2¼ y.

Let us remark that we can write y ¼ p� pqð pqÞyp. Evidently,

y2 pR, so yR� pR. Since yq ¼ ð p� pqð pqÞypÞq ¼ 0, we have yq¼ y.

By taking * in the last equality we have qy¼ y, hence yR� qR.

Thus yR� pR\ qR.

To prove pR\ qR� yR, take arbitrary z2 pR\ qR. Observe that

pqð pqÞy ¼ pqð pqÞy� ��

¼ ð pqÞy� ��

qp ¼ ð pqÞ�½ yqp ¼ ðqpÞyqp:

Since z¼ pz¼ qz, it follows that yz ¼ ð p� ðqpÞyqpÞz ¼ z. Hence, pR\ qR� yR.

g

Next four results continue the study of the sum pþ q and the ideals (pþ q)R,

(pþ q)� under the additional assumption that pqp and pqp are Moore–Penrose

invertible.

THEOREM 4.2 Let R be a *-reducing ring and p, q2R be two projections such that

pqp and pqp are Moore–Penrose invertible. Then

(i) ð pþ qÞR ¼ pqpR�? pR.(ii) ð pþ qÞR ¼ pqR�? pR.(iii) (pþ q)R¼ pRþ qR.(iv) (pþ q)R¼ (p� q)R�? (pR\ qR).(v) ð pþ qÞ� ¼ ðpqÞ� \ p�.(vi) (pþ q)�¼ p� \ q�.

Proof

(i) It is evident that pqpR?pR. By Lemma 2.1 (i), we have that

pþ q ¼ pqþ pð1þ qÞ ¼ pqðpqÞypqþ pð1þ qÞ ¼ pqpðpqpÞypqþ pð1þ qÞ,

which implies that ð pþ qÞR � pqpRþ pR. To prove the opposite inclusion,

take arbitrary z2 pqpRþ pR. We have z ¼ pqpxþ py for some x, y2R.

By Theorem 4.1 (i), we get (pþ q)(pþ q)yz¼ z, so z2 (pþ q)R. Hence

ð pþ qÞR ¼ pqpRþ pR.

(ii) Since for any Moore–Penrose invertible a2R, one has aR¼ aayR,

by Theorem 4.1 (ii), we have pqR ¼ pqpR. Now, the assertion follows

by item (i).

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(iii) The inclusion � is evident. To prove the opposite, let us use the notationsgiven in (2.3) and let us demonstrate pqþ ddyq¼ q. In fact, Lemma 2.3 (iii)

yields ddyb*¼ b* and now

pqþ ddyq ¼ pqþ ddyð1� pÞq ¼ pqþ ddyðb� þ d Þ ¼ pqþ bþ d ¼ q:

If x2 pRþ qR, then exist u, v2R such that x¼ puþ qv. By Theorem 4.1 (i),one gets

ð pþ qÞð pþ qÞyx ¼ ð pþ ddyÞð puþ qvÞ ¼ puþ pqvþ ddyqv ¼ puþ qv ¼ x:

Thus, x¼ (pþ q)(pþ q)yx2 (pþ q)R.(iv) By Theorem 4.1 (i) and (iii) and (2.3), it follows that (pþ q)(pþ q)y¼ pþ ddy

and (p� q)(p� q)y¼ (p� a)(p� a)yþ ddy. By Lemma 4.2 (ii), it follows that

y ¼ p� pð pqÞy is a projection and yR¼ pR\ qR. If in Theorem 4.1 (ii), wereplace p with p and q with q, we get that ð pqÞy ¼ qð pqpÞy, soy¼ p� (p� a)(p� a)y. Now, since p� (p� a)(p� a)y and (p� a)(p� a)yþ

ddy are commuting projections whose product is equal to zero, we have

ð pþ qÞR ¼ ð pþ ddyÞR

¼ ðð p� aÞð p� aÞy þ ddyÞ þ ð p� ð p� aÞð p� aÞyÞ� �

R

¼ ð p� aÞð p� aÞy þ ddy� �

R�?�p� ð p� aÞð p� aÞy

R

¼ ð p� qÞR �? ð pR\ qRÞ:

(v �): Let x2 (pþ q)�. By Theorem 4.1 (i) and by employing the notations givenin 2.2 we have 0¼ (pþ q)y(pþ q)x¼ (pþ ddy)x¼ pxþ ddyx; which by

premultiplying by p and p we get 0¼ px and 0¼ ddyx, respectively.Notice that 0¼ ddyx , dx¼ 0. Finally,pqx ¼ ðb� þ d Þx ¼ b�xþ dx ¼ pqpxþ dx ¼ 0.

(v ): Let x2 ðpqÞ� \ p�. we have 0 ¼ pqx ¼ ðb� þ d Þx ¼ ðpqpþ d Þx ¼ dx. FromTheorem 4.1 (i) we get (pþ q)(pþ q)yx¼ (pþ ddy)x¼ pxþ ddyx¼ 0, whichleads to (pþ q)x¼ 0.

(vi): The inclusion p� \ q�� (pþ q)� is obvious. Let x2 (pþ q)�. As in the proof of

(i �) we get px¼ 0. Now, (pþ q)x¼ 0 leads to qx¼ 0. g

Observe that Theorem 4.2 (ii) and (vi) generalize to [15, Cor. 2]. Remark thatTheorem 6 and Theorem 8 from [1] are equivalent (replacing P with I�Q and Q withP in Theorem 6, we get Theorem 8). The following corollary follows immediatelyfrom Theorem 4.2.

COROLLARY 4.2 Let R be a *-reducing ring and p, q2R be two projections such thatpqp and pqp are Moore–Penrose invertible. Then the following statements areequivalent:

(i) pþ q is invertible,(ii) R ¼ pqR�? pR,(iii) R ¼ pqpR�? pR,(iv) R¼ pRþ qR.

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THEOREM 4.3 Let R be a *-reducing ring and p, q2R be two projections such that

pqp and pqp are Moore–Penrose invertible. Then

(i) ð1� p qÞR ¼ ð pþ qÞR,(ii) ð1� p qÞ� ¼ ð pþ qÞ�.

Proof

(i) By Theorem 4.1 (i), using the notations given by (2.3), we get that

(pþ q)(pþ q)y¼ pþ ddy. On the other side, since 1� p q ¼ pþ b� þ d, by

Lemma 2.3 it is easy to check that ð1� p qÞy ¼ dy � b�ð p� aÞy þ p, so

ð1� p qÞð1� p qÞy ¼ pþ ddy. Hence, ð1� p qÞð1� p qÞy ¼ ð pþ qÞð pþ qÞy

which is equivalent to ð1� p qÞR ¼ ð pþ qÞR.(ii) By Theorem 4.2 (v), we have ð pþ qÞ� ¼ ðpqÞ� \ p�. Let us prove

ð1� p qÞ� ¼ ðpqÞ� \ p�: If x2 ð1� p qÞ�, then p qx ¼ x, which yields

pqx¼ pxþ qx, hence px¼ 0 and (1� p)qx¼ 0, in other words,

x2 ðpqÞ� \ p�. The opposite inclusion is evident. g

THEOREM 4.4 Let R be a *-reducing ring and p, q2R be two projections such that

pqp and pqp are Moore–Penrose invertible. Then

(i) (pq� qp)R� (pþ q)R.(ii) (pþ q)�� (pq� qp)�.

Proof

(i) By Theorem 4.1 (i), we have that ð pþ qÞð pþ qÞy ¼ pþ pqðpqpÞy. Now, by

Lemma 2.3 (iii) using the notations from (2.3), we have that

ð pþ qÞð pþ qÞyð pq� qpÞ ¼ ð pþ pqðpqpÞyÞð pq� qpÞ

¼ pq� pqp� pqðpqpÞypqp

¼ pqð1� pÞ � pqpðpqpÞypqp

¼ b� ddyb�

¼ b� b�

¼ pq� qp,

which implies that (pq� qp)R� (pþ q)R.

(ii) It follows from Theorem 4.2 (vi). g

Note that the inclusions in the previous theorem can be strict, as shown by the

case q¼ 1� p. Following theorems study further the image ideal (pq� qp)R together

with the ideals (p� q)R and (1� p� q)R.

THEOREM 4.5 Let R be a *-reducing ring and p, q2R be two projections such that

p q p, pqp, pqp, pqp and pqp are Moore–Penrose invertible. Then

(i) (p� q)(p� q)yþ (1� p� q)(1� p� q)y� (pq� qp)(pq� qp)y¼ 1,(ii) (p� q)Rþ (1� p� q)R¼R,(iii) (p� q)(p� q)y(1� p� q)(1� p� q)y¼ (pq� qp)(pq� qp)y.

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Proof

(i) Let x¼ (p� q)(p� q)yþ (1� p� q)(1� p� q)y� (pq� qp)(pq� qp)y. By the

proof of Theorem 4.1 we have (p� q)(p� q)y¼ (p� a)(p� a)yþ ddy.

By doing the same computation having replaced p by p we get

(1� p� q)(1� p� q)y¼ (1� p� d)(1� p� d)yþ aay. By Theorem 4.1 (iv)

we get (pq� qp)(pq� qp)y¼ bbyþ byb. Thus,

x ¼ ð p� aÞð p� aÞy þ ddy þ ð1� p� d Þð1� p� d Þy

þ aay � bby � byb:

Denote y¼ aayþ (p� a)(p� a)y� bby. Since a is self-adjoint, we get that the

elements of the set {a, ay, p� a, (p� a)y} are mutually commuting by

Lemma 4.1. Now it is easy to prove by the definition of the Moore–Penrose

inverse that [a(p� a)]y¼ ay(p� a)y. By Lemma 2.2 (i), we have

bby ¼ bb�ðbb�Þy ¼ ða� a2Þða� a2Þy ¼ að p� aÞ½að p� aÞy

¼ aayð p� aÞð p� aÞy, ð4:3Þ

so (p� a)(p� a)y� bby¼ (1� aay)(p� a)(p� a)y. Using that (1� aay)(p� a)

(p� a)y is a self-adjoint idempotent we get

y ¼ aay þ ð1� aayÞð p� aÞð p� aÞy

¼ aay þ ð1� aayÞ�ð1� aayÞð p� aÞð p� aÞy

�¼ aay þ ð1� aayÞ

�ð1� aayÞð p� aÞð p� aÞy

�y:

By Lemma 4.2 (i), it follows that y is a projection and yR¼ aayRþ

(p� a)(p� a)yR¼ aRþ (p� a)R¼ pR. So, y¼ p. Similarly, we get that

ddy þ ð1� p� d Þð1� p� d Þy � byb ¼ 1� p: ð4:4Þ

Hence x¼ 1.

(ii) It follows from Theorem 3.1.(iii) By the proof of the item (i) of this theorem we have that

ð p� qÞð p� qÞyð1� p� qÞð1� p� qÞy

¼

�ð p� aÞð p� aÞy þ ddy

�ð1� p� d Þð1� p� d Þy þ aay

¼ ð p� aÞð p� aÞyaay þ ddyð1� p� d Þð1� p� d Þy:

As in (i), we have that bby¼ aay(p� a)(p� a)y and similarly byb¼ ddy

(1� p� d)(1� p� d)y. Again, the proof of (i) distills (pq� qp)

(pq� qp)y¼ bbyþ byb. The proof is complete. g

THEOREM 4.6 Let R be a *-reducing ring and p, q2R be two projections such that

pqp, pqp, pqp and p q p are Moore–Penrose invertible. The following conditions are

equivalent:

(i) (pq� qp)R¼ (p� q)R,(ii) (1� p� q)R¼R,

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(iii) p¼ pq(pq)y, q¼ qp(qp)y.

Proof

(i), (ii): By Theorem 4.1 (iii) and (iv) using the notations given in (2.3), we have that

(i) is equivalent to

ð p� aÞð p� aÞy ¼ bby and ddy ¼ byb: ð4:5Þ

Similarly, (1� p� q)R¼R is equivalent to (1� p� q)(1� p� q)y¼ 1. If in

Theorem 4.1 (iii) we replace p with p, we have that (1� p� q)(1� p� q)y¼

(1� p� d)(1� p� d)yþ aay. Hence, (ii) is equivalent to (1� p� d)(1� p� d)yþ

aay¼ 1, i.e.

ð1� p� d Þð1� p� d Þy ¼ 1� p and aay ¼ p: ð4:6Þ

By (4.3), we have that bby¼ aay(p� a)(p� a)y. Now, multiplying the equality given

in Lemma 2.4 (i) by aay from the left side and using Lemma 2.3 (i), we get that

p� ð p� aÞð p� aÞy ¼ aay � aayð p� aÞð p� aÞy:

Now, it is evident that (p� a)(p� a)y¼ bby if and only if aay¼ p. Analogously, using

(4.4), we get that ddy¼ byb if and only if (1� p� d)(1� p� d)y¼ 1� p. Hence, (i)

and (ii) are equivalent.

(ii), (iii): By Corollary 4.1 and Lemma 2.3 (i), we have that (iii) is equivalent to

p¼ pq(pq)y¼ pq(pqp)y¼ aay and q¼ qp(qp)y¼ (pq)ypq¼ qp(pqp)ypq¼ aþ bþ b*þ

b*ayb. Now, by Lemma 2.3 (ii), we get that (iii) is equivalent to p¼ aay and

(1� p� d)(1� p� d)y¼ 1� p, which together with (4.6) implies that (ii) and (iii) are

equivalent. g

THEOREM 4.7 Let R be a *-reducing ring and p, q2R be two projections such that

pqp, pqp, pqp, p q p and pqp are Moore–Penrose invertible. The following conditions are

equivalent:

(i) (pq� qp)R¼R,(ii) (p� q)R¼R, (1� p� q)R¼R,(iii) ð pRþ qRÞ \ ðpRþ qRÞ ¼ R, ð pRþ qRÞ \ ðpRþ qRÞ ¼ R,(iv) pRþ qR ¼ R, pRþ qR ¼ R, pRþ qR ¼ R, pRþ qR ¼ R.

Proof

(i)) (ii): Assume (pq� qp)R¼R. By Theorem 2.1 (iv) we have bbyþ byb¼ 1, which

implies bby¼ p and byb ¼ p. Obviously p� a2 pR, which shows (p� a)(p� a)yR¼

(p� a)R� pR. By Lemma 2.2 (ii) we get p¼ bby¼ (p� a)(p� a)y bby, which proves

pR� (p� a)(p� a)yR. Since pR¼ (p� a)(p� a)yR, then p¼ (p� a)(p� a)y. Also,

from byb ¼ p and Lemma 2.2 (iii), we get pddy ¼ p, hence ddy ¼ p. From

Theorem 2.1 (iii) we get ð p� qÞð p� qÞy ¼ ð p� aÞð p� aÞy þ ddy ¼ pþ p ¼ 1,

which yields (p� q)R¼R. Furthermore, since (pq� qp)R¼ (p� q)R, by Theorem

4.6 we get (1� p� q)R¼R.

(ii)) (i) is evident in view of Theorem 4.6.

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(ii) , (iv): By Theorem 4.1 (iii), we have (p� q)(p� q)y¼ (p� a)(p� a)yþ ddy and

(1� p� q)(1� p� q)y¼ (1� p� d)(1� p� d)yþ aay. Hence (ii) is equivalent to the

following

ð p� aÞð p� aÞy ¼ p, ddy ¼ 1� p, aay ¼ p

and ð1� p� d Þð1� p� d Þy ¼ 1� p:

From Lemma 4.2, we can conclude that pRþ qR¼R if and only if pðpqÞy ¼ p.

By Theorem 4.1 (ii), we have that pðpqÞy ¼ ddy. Hence, pRþ qR¼R if and only if

ddy¼ 1� p. Similarly, changing p, q with 1� p, 1� q, respectively, we get the

following:

ð1� pÞR þ qR ¼ R, aay ¼ p,

pRþ ð1� qÞR ¼ R , ð1� p� d Þð1� p� d Þy ¼ 1� p,

ð1� pÞR þ ð1� qÞR ¼ R , ð p� aÞð p� aÞy ¼ p:

(iii) , (iv) is evident. g

THEOREM 4.8 Let R be a *-reducing ring and p, q2R be two projections such that

pqp is Moore–Penrose invertible. Then

(i) ð pq� qpÞR ¼ pqpR�? pqpR,(ii) ð pq� qpÞ� ¼ ð pqpÞ� \ ðpqpÞ�.

Proof (i) By Theorem 4.1 (iv), (pq� qp)R¼ (pq� qp)(pq� qp)yR¼ (bbyþ byb)R.

Since, bby 2 pR p and byb2 pRp, we get that (bby)(byb)¼ (byb)(bby)¼ 0 which implies

that (pq� qp)R¼ bbyRþ bybR¼ bRþ b*R and furthermore that this sum is

orthogonal.

(ii): By Theorem 4.1 (iv) and by Lemma 2.1 we have

ð pq� qpÞx ¼ 0() ð pq� qpÞyð pq� qpÞx ¼ 0 () ðbby þ bybÞx ¼ 0

() bbyx ¼ 0 and bybx ¼ 0 () byx ¼ 0 and bx ¼ 0

() b�x ¼ 0 and bx ¼ 0: g

Let us remark that when a and b are Hermitian commuting elements of a ring with

involution, the reverse order law for the Moore–Penrose inverse holds, i.e.

(ab)y¼ byay (see, e.g. [3, Th. 6.3.2]).

THEOREM 4.9 Let R be a *-reducing ring and p, q2R be two projections such that

pqp, p� pqp and pqp are Moore–Penrose invertible. Then

(i) pqRþ (p� q)R¼ (pþ q)R,(ii) pqR\ (p� q)R¼ {0} if and only if pq is a projection.

Proof

(i) By Lemma 4.2, we have that

x ¼ pqð pqÞy þ ð1� pqð pqÞyÞ�ð1� pqð pqÞyÞð p� qÞð p� qÞy

y

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is a projection and that xR¼ pqRþ (p� q)R. By Corollary 4.1, we have

that (pq)y¼ q(pqp)y, which using the notations introduced in (2.3) implies that

pq(pq)y¼ pqp(pqp)y¼ aay. Also, by Theorem 4.1 (iii), we have that

(p� q)(p� q)y¼ (p� a)(p� a)yþ ddy. Now,

x ¼ aay þ ð1� aayÞ�ð1� aayÞðð p� aÞð p� aÞy þ ddyÞ

y

¼ aay þ ð1� aayÞ�ð1� aayÞð p� aÞð p� aÞy þ ddy

y

¼ aay þ ð1� aayÞ�ð p� aayÞð p� aÞy þ ddy

y:

Since (p� aay)(p� a)y 2 pRp and ddy 2 pRp, we have that ((p� aay)

(p� a)yþ ddy)y¼ ((p� aay)(p� a)y)yþ (ddy)y¼ ((p� aay) (p� a)y)yþ zddy.

Since aay¼ paay¼ aay p, it follows that p� aay is a projection, which implies

that (p� aay)y¼ p� aay. Now, using the reverse order law for the Moore–

Penrose inverse and the fact that a, ay commute, we get that

((p� aay)(p� a)y)y¼ (p� a)(p� aay)¼ p� a� aayþ aaay¼ p� aay. Hence,

x ¼ aay þ ð1� aayÞð p� aay þ ddyÞ ¼ pþ ddy:

Now, by Theorem 4.1 (i), it follows that x¼ (pþ q)(pþ q)y, i.e.

xR¼ (pþ q)R.

(ii) From Lemma 4.2, we have that

y ¼ pqð pqÞy � pqð pqÞy�pqð pqÞy

�1� ð p� qÞð p� qÞy

y

is a projection and that yR¼ pqR\ (p� q)R. As in the proof of item (i), we

have that

y ¼ aay � aay aay 1� ð p� aÞð p� aÞy � ddy� �� �y

¼ aay � aay�aay � aayð p� aÞð p� aÞy

y:

By Lemma 2.4 (i) and Lemma 2.3 (i), we have that aay� aay(p� a)

(p� a)y¼ p� (p� a)(p� a)y which is a projection, so we get that

y ¼ aay � aay�p� ð p� aÞð p� aÞy

¼ aayð p� aÞð p� aÞy:

By (4.3), we have that y¼ bby. Now, pqR\ (p� q)R¼ 0 if and only if y¼ 0 if

and only if b¼ 0 which is equivalent to pq¼ qp. Finally, since p, q are

projections, pq¼ qp is equivalent to the fact that pq is a projection. g

THEOREM 4.10 Let R be a *-reducing ring and p, q2R be two projections such that

pqp and pqp are Moore–Penrose invertible. Then the following conditions are

equivalent:

(i) pR\ qR¼ {0},(ii) (p� q)R¼ pRþ qR,(iii) pqR�? pqR ¼ pRþ qR.

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Proof By Lemma 4.2 (ii), it follows that pR\ qR¼ {0} is equivalent to p ¼ pð pqÞy.Now, by using the representations of p, q given in (2.2) we have

pð pqÞy ¼ pð pqÞ� pqð pqÞ�½ y¼ pqpð pqpÞy ¼ ð p� aÞð p� aÞy:

Hence, (i) is equivalent to p¼ (p� a)(p� a)y.For the element x defined in item (i) of Lemma 4.2, we have that

x ¼ pþ pqðpqÞy ¼ pþ pqðpqÞ� pqðpqÞ�½ y¼ pþ pqpðpqpÞy ¼ pþ ddy: ð4:7Þ

By (4.7) and Theorem 4.1 (iii) one has

ð p� qÞð p� qÞy ¼ ð p� aÞð p� aÞy þ ddy ¼ ð p� aÞð p� aÞy þ x� p:

Thus, (i) is equivalent to (p� q)(p� q)y¼ x.

(i)) (ii): If (p� q)(p� q)y¼ x, then by Lemma 4.2 (i), one has (p� q)R¼ (p� q)(p� q)yR¼ xR¼ pRþ qR.

(ii)) (i): From the hypothesis and Lemma 4.2 (i), we get (p� q)(p� q)yR¼ xR.Since both (p� q)(p� q)y and x are projections we get (p� q)(p� q)y¼ x.

(ii) , (iii): This part follows by Theorem 3.1. g

Acknowledegments

The authors thank the anonymous reviewer for hisnher useful suggestions, which helped toimprove the original version of this article. The second author is supported by GrantNo. 174007 of the Ministry of Science, Technology and Development, Republic of Serbia.

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