Applied Mathematics and Computation 220 (2013) 439–445

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate/amc

Identities concerning the reverse order law forthe Moore–Penrose inverse

0096-3003/$ - see front matter � 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.amc.2013.05.076

⇑ Corresponding author.E-mail addresses: ndincic@hotmail.com (N.C. Dincic), dragan@pmf.ni.ac.rs (D.S. Djordjevic).

1 The authors are supported by the Ministry of Science, Republic of Serbia, Grant No. 174007.

Nebojša C. Dincic ⇑, Dragan S. Djordjevic 1

Faculty of Sciences and Mathematics, University of Niš, P.O. Box 224, 18000 Niš, Serbia

a r t i c l e i n f o

Keywords:Moore–Penrose inverseWeighted Moore–Penrose inverseReverse order law

a b s t r a c t

We prove some identities related to the reverse order law for the Moore–Penrose inverse ofoperators on Hilbert spaces, extending some results from (Y. Tian, S. Cheng, Linear Multi-linear Algebra 52 (2004)) and (R. E. Cline, SIAM Rev., Vol. 6, No. 1 (1964)) to infinite dimen-sional settings.

� 2013 Elsevier Inc. All rights reserved.

1. Introduction

If S is a semigroup with the unit 1, and if a; b 2 S are invertible, then the equality ðabÞ�1 ¼ b�1a�1 is called the reverse orderlaw for the ordinary inverse. It is well-known that the reverse order law does not hold for various classes of generalized in-verses. Notice that the classical result for the reverse order law for the Moore–Penrose inverse is proved in [8] for complexmatrices, and in [1,2,9] for bounded linear operators on Hilbert spaces (meaning ðABÞy ¼ ByAy if and only if: AyA commuteswith BB�, and A�A commutes with BBy). Latter, the corresponding result for elements in a ring with involution is proved in[10]. A significant number of papers treat the sufficient or equivalent conditions such that the reverse order law holds insome sense (see [4–7,11]).

The reverse order law has applications in the numerical computation of the Moore–Penrose inverse of a product of oper-ators. In this paper we specialize our investigation to certain identities related to the Moore–Penrose inverse of closed-rangebounded linear operators on Hilbert spaces.

Let X and Y be Hilbert spaces, and let LðX;YÞ denote the set of all linear bounded operators from X to Y. We abbreviateLðXÞ ¼ LðX;XÞ. For A 2 LðX;YÞ we denote by NðAÞ and RðAÞ, respectively, the null-space and the range of A. An operatorB 2 LðY;XÞ is an inner inverse of A, if ABA ¼ A holds. In this case A is inner invertible, or relatively regular. It is well-knownthat A is inner invertible if and only if RðAÞ is closed in Y. The Moore–Penrose inverse of A 2 LðX;YÞ is the operatorAy 2 LðY ;XÞ which satisfies the Penrose equations

AAyA ¼ A; AyAAy ¼ Ay; ðAAyÞ� ¼ AAy; ðAyAÞ� ¼ AyA:

The Moore–Penrose inverse of A exists if and only if RðAÞ is closed in Y, and in this case Ay is unique.Let M 2 LðYÞ and N 2 LðXÞ be positive and invertible operators, and let A 2 LðX;YÞ have a closed range. Then there exists

the unique operator B 2 LðY;XÞ such that the following equations are satisfied:

ABA ¼ A; BAB ¼ B; ðMABÞ� ¼ MAB; ðNBAÞ� ¼ NBA:

Such B is denoted by AyM;N and it is known as the weighted Moore–Penrose inverse of A with respect to the weights M and N.

440 N.C. Dincic, D.S. Djordjevic / Applied Mathematics and Computation 220 (2013) 439–445

There is one very useful result linking the ordinary and the weighted Moore–Penrose inverse [10, Theorem 5]:

AyM;N ¼ N�12 M

12AN�

12

� �yM

12:

The paper is organized as follows. In the rest of Introduction we formulate some auxiliary results. In Section 2 we prove someidentities related to various mixed-type reverse order rules for the Moore–Penrose inverse of products of Hilbert space oper-ators with closed ranges. We also consider one classical identity proved by Cline in [3]. Present paper is an extension of re-sults from [11,3] to infinite dimensional settings. Recall that the results in [11] are obtained mostly using the finitedimensional methods. In this paper we use operator matrices on arbitrary Hilbert spaces.

Lemma 1.1. Let A 2 LðX;YÞ have a closed range. Then A has a matrix decomposition with respect to the orthogonaldecompositions of spaces X ¼ RðA�Þ � N ðAÞ and Y ¼ RðAÞ � N ðA�Þ:

A ¼A1 0

0 0

" #:RðA�Þ

N ðAÞ

" #!

RðAÞ

N ðA�Þ

" #;

where A1 is invertible. Moreover,

Ay ¼A�1

1 0

0 0

" #:RðAÞ

N ðA�Þ

" #!

RðA�Þ

N ðAÞ

" #:

The proof is straightforward.

Lemma 1.2. Let A 2 LðX;YÞ be a closed range operator and let X ¼ X1 � X2 and Y ¼ Y1 � Y2 be orthogonal decompositions withclosed subspaces. Then A has the following matrix representations:

(a)

A ¼A1 A2

0 0

� �:

X1

X2

� �!

RðAÞN ðA�Þ

� �;

where D ¼ A1A�1 þ A2A�2 maps RðAÞ into itself and D > 0 (D is positive and invertible). Also,

Ay ¼A�1D�1 0

A�2D�1 0

" #:RðAÞN ðA�Þ

� �!

X1

X2

� �:

(b)

A ¼A1 0

A2 0

" #:RðA�Þ

N ðAÞ

" #!

Y1

Y2

" #;

where D ¼ A�1A1 þ A�2A2 maps RðA�Þ into itself and D > 0 (D is positive and invertible). Also,

Ay ¼D�1A�1 D�1A�2

0 0

" #:

Y1

Y2

" #!

RðA�Þ

N ðAÞ

" #:

Here A1;A2 may be different operators in (a) and (b).The previous lemma is proved in [6]. The following result is well-known, and it can be found in [9].

Lemma 1.3. Let A 2 LðY ; ZÞ and B 2 LðX; YÞ have closed ranges. Then AB has a closed range if and only if AyABBy has a closedrange.

The following lemma appears to be useful later, when we deal with the weighted Moore–Penrose inverses.

Lemma 1.4 [6]. Let X and Y be Hilbert spaces, let C 2 LðX;YÞ have a closed range, and let D 2 LðYÞ be Hermitian and invertible.Then RðDCÞ ¼ RðCÞ if and only if DCCy ¼ CCyD.

Lemma 1.5. Let W;X;Y ; Z be Hilbert spaces, let P 2 LðX;YÞ;Q 2 LðY; ZÞ and R 2 LðW;XÞ be operators such that P;Q ;QP; PR haveclosed ranges. If Q and R are invertible, then:

(a) ðPðQPÞyÞy ¼ QPPy;(b) ððPRÞyPÞy ¼ PyPR.

N.C. Dincic, D.S. Djordjevic / Applied Mathematics and Computation 220 (2013) 439–445 441

Proof. For (a) we verify that the operators A ¼ PðQPÞy and B ¼ QPPy satisfy the Penrose equations. As a sample we check thatABA ¼ A:

ABA ¼ PðQPÞyQPPyPðQPÞy ¼ Q�1QPðQPÞyQPðQPÞy ¼¼ Q�1QPðQPÞy ¼ PðQPÞy ¼ A:

The remaining equations are checked analogously. Equation (b) is verified in a similar manner. Notice that from (a) and (b)we conclude that PðQPÞy and ðPRÞyP have closed ranges.

2. Main results

In this section we prove the results concerning the mixed-type reverse-order law for the Moore–Penrose inverse of aproduct of two and multiple Hilbert space operators with close ranges.

Theorem 2.1. Let X;Y ; Z;W be Hilbert spaces, and let A 2 LðZ;WÞ;B 2 LðY ; ZÞ and C 2 LðX;YÞ be the operators, such thatA;B;C;AB;ABC have closed ranges. Then the following hold:

(a) ðABÞy ¼ ðAyABÞyðABByÞy;(b) ðABÞy ¼ ½ðAyÞ�B�

yðByAyÞ�½AðByÞ��

y;

(c) ðABCÞy ¼ ðAyABCÞyBðABCCyÞy;(d) ðABCÞy ¼ ½ðABÞyABC�yBy½ABCðBCÞy�y;(e) ðABCÞy ¼ ½ðABByÞyABC�

yB½ABCðByBCÞy�

y;

(f) ðABCÞy ¼ ½ðAyÞ�BC�yðAyÞ�BðCyÞ�½ABðCyÞ��

y;

(g) ðABCÞy ¼ f½AðByÞ��yABCg

yB�BB�fABC½ðByÞ�C�

ygy;

(h) ðABCÞy ¼ f½ðABÞy��Cgy½ðABÞy��By½ðBCÞy��fA½ðBCÞy��g

y.

Proof. According to lemmas from the previous section, it is easy to conclude that operators A;B and C have the followingmatrix representations with the respect to the appropriate decompositions of spaces:

A ¼A1 A2

0 0

� �:RðBÞN ðB�Þ

� �!

RðAÞN ðA�Þ

� �;

where D ¼ A1A�1 þ A2A�2 is invertible and positive in LðRðAÞÞ. Then

Ay ¼A�1D�1 0

A�2D�1 0

" #:RðAÞN ðA�Þ

� �!

RðBÞN ðB�Þ

� �:

Moreover,

B ¼B1 00 0

� �:RðB�ÞN ðBÞ

� �!

RðBÞN ðB�Þ

� �;

where B1 is invertible. Then

By ¼ B�11 00 0

" #:RðBÞN ðB�Þ

� �!RðB�ÞN ðBÞ

� �:

Finally,

C ¼C1 0C2 0

� �:RðC�ÞN ðCÞ

� �!RðB�ÞN ðBÞ

� �;

where E ¼ C�1C1 þ C�2C2 is invertible and positive in LðRðC�ÞÞ. Then

Cy ¼ E�1C�1 E�1C�20 0

" #:RðB�ÞN ðBÞ

� �!RðC�ÞN ðCÞ

� �:

(a): Notice that RðAyABÞ ¼ AyAðRðBÞÞ ¼ AyAðRðBByÞÞ ¼ RðAyABByÞ is closed according to Lemma 1.3. Also, RðB�A�Þ is closed.

Again, from Lemma 1.3 and RððABByÞ�Þ ¼ RððB�ÞyB�A�Þ ¼ RððB�ÞyB�A�ðA�ÞyÞ ¼ RððAyABByÞ�Þ, it follows that RðABByÞ isclosed. Now, using matrix forms of A and B, we have:

442 N.C. Dincic, D.S. Djordjevic / Applied Mathematics and Computation 220 (2013) 439–445

ABBy ¼A1 00 0

� �; ðABByÞy ¼ Ay1 0

0 0

" #; AyAB ¼

A�1D�1A1B1 0

A�2D�1A1B1 0

" #;

ðAyABÞy ¼ ððAyABÞ�ðAyABÞÞyðAyABÞ� ¼ ðB�1A�1D�1A1B1Þ

yB�1A�1D�1A1 ðB�1A�1D�1A1B1Þ

yB�1A�1D�1A2

0 0

" #:

Therefore, ðABÞy ¼ ðAyABÞyðABByÞy is equivalent to

ðA1B1Þy ¼ ðB�1A�1D�1A1B1ÞyB�1A�1D�1A1Ay1;

which is further equivalent to

ðA1B1Þy ¼ ðD�1=2A1B1ÞyD�1=2A1Ay1:

The last equality follows by checking Penrose equations; as a sample we check the second one:

ðA1B1ÞyA1B1ðA1B1Þy ¼ ðD�1=2A1B1ÞyD�1=2A1Ay1A1B1ðD�1=2A1B1Þ

yD�1=2A1Ay1 ¼ ðD

�1=2A1B1ÞyD�1=2A1Ay1 ¼ ðA1B1Þy:

(b): Notice that RðððAyÞ�BÞ�Þ ¼ RðB�AyÞ ¼ RðB�A�Þ ¼ RððABÞ�Þ is closed, so RððAyÞ�BÞ is closed. Also, RðAðByÞ�Þ ¼

RðAððB�BÞyB�Þ�Þ ¼ RðABðB�BÞyÞ ¼ ABðRððB�BÞyÞÞ ¼ ABðRðB�ÞÞ ¼ AðRðBÞÞ ¼ RðABÞ is closed. Again, using matrix forms

of A and B, we have that ðABÞy ¼ ½ðAyÞ�B�yðByAyÞ�½AðByÞ��

yis equivalent to the following:

ðA1B1Þy ¼ ðD�1A1B1ÞyD�1A1ðB�1

1 Þ�ðA1ðB�1

1 Þ�Þy:

The last equality can easily be proved by checking the Penrose equations.(c): Notice that RððAyABCÞ�Þ ¼ ðBCÞ�ðRðAyAÞÞ ¼ ðBCÞ�ðRðA�ÞÞ ¼ RððABCÞ�Þ is closed. Also, RðABCCyÞ ¼ ABðRðCCyÞ ¼

ABðRðCÞÞ ¼ RðABCÞ is closed. Now we show that ðABCÞy ¼ ðAyABCÞyBðABCCyÞy. First we compute factors appearing onthe right side. Denote:

T ¼ AyABC ¼A�1D�1A1B1C1 0

A�2D�1A1B1C1 0

!:

Now,

Ty ¼ ðT�TÞyT� ¼XA1 XA2

0 0

� �;

where

X ¼ ðC�1B�1A�1D�1A1B1C1ÞyC�1B�1A�1D�1:

Let S be the following:

S ¼ ABCCy ¼ A1B1C1E�1C�1 A1B1C1E�1C�20 0

!:

It is easy to find:

Sy ¼ S�ðSS�Þy ¼C1E�1C�1B�1A�1ðA1B1C1E�1C�1B�1A�1Þ

y0

C2E�1C�1B�1A�1ðA1B1C1E�1C�1B�1A�1Þy

0

0@ 1A:

Therefore, the statement (c) is equivalent to:

ðA1B1C1Þy ¼ ðC�1B�1A�1D�1A1B1C1ÞyC�1B�1A�1D�1A1 � B1C1E�1C�1B�1A�1ðA1B1C1E�1C�1B�1A�1Þ

y;

i.e.,

ðA1B1C1Þy ¼ ðD�1=2A1B1C1ÞyD�1=2A1B1C1E�1=2ðA1B1C1E�1=2Þy:

This formula can be proved on analogous way as in (a).(f): Notice that RðððAyÞ�BCÞ

�Þ ¼ ðBCÞ�ðRðAyÞÞ ¼ ðBCÞ�ðRðA�ÞÞ ¼ RððABCÞ�Þ is closed, so RððAyÞ�BCÞ is closed. Also,

RðABðCyÞ�Þ ¼ ABðRððCyÞ�ÞÞ ¼ ABðRðCÞÞ ¼ RðABCÞ is closed. An easy computation shows that

ðABCÞy ¼ ½ðAyÞ�BC�yðAyÞ�BðCyÞ�½ABðCyÞ��

y

N.C. Dincic, D.S. Djordjevic / Applied Mathematics and Computation 220 (2013) 439–445 443

is equivalent to:

ðA1B1C1Þy ¼ ðD�1A1B1C1ÞyD�1A1B1C1E�1ðA1B1C1E�1Þy:

This equality follows a standard argument.So far we have proved four identities. Now we use (c), to show that (d), (e) and (g) are satisfied. Also, we use (f) to prove that(h) holds.

(d): An easy computation shows that (d) is equivalent to the following:

ðA1B1C1Þy ¼ ½ðA1B1ÞyA1B1C1�yB�1

1 ½A1B1C1ðB1C1Þy�y:

If we put: A0 ¼ A1B1; B0 ¼ B�11 ; C0 ¼ B1C1, then (d) becomes already proven identity (c) for operators A0;B0;C0. For the com-

pleteness, notice that the following operator ranges are closed:

RðA0Þ ¼ RðABÞ; RðB0Þ ¼ RðB�Þ; RðC 0Þ ¼ RðBCÞ;RðA0B0Þ ¼ RðAÞ; RðB0C 0Þ ¼ RðCÞ; RðA0B0C 0Þ ¼ RðABCÞ:

(e): An easy computation shows that (e) is equivalent to the following:

ðA1B1C1Þy ¼ ½Ay1A1B1C1�yB1½A1B1C1Cy1�

y:

The last identity is proved in (c).(g): An easy computation shows that

ðABCÞy ¼ f½AðByÞ��yABCg

yB�BB�fABC½ðByÞ�C�

ygy

is equivalent to the following:

ðA1B1C1Þy ¼ f½A1ðB�1Þ�1�yA1B1C1g

yB�1B1B�1fA1B1C1½ðB�1Þ

�1C1�ygy:

We put: A00 :¼ A1ðB�1Þ�1; B00 :¼ B�1B1B�1; C00 :¼ ðB�1Þ

�1C1. Now we have that the following operator ranges are closed:

RðA00Þ ¼ A1ðRððB�1Þ�1ÞÞ ¼ RðABÞ; RðB00Þ ¼ RðB�Þ;

RðC 00Þ ¼ RðBCÞ; RðA00B00Þ ¼ RðA1B1B�1Þ ¼ RðABÞ;RððB00C 00Þ�Þ ¼ RððB�BCÞ�Þ ¼ C�ðRðB�BÞÞ ¼ RððBCÞ�Þ;RðA00B00C 00Þ ¼ RðABCÞ:

So, conditions of the identity (c) are satisfied. Hence, (g) follows from (c).(h): An easy computation shows that (h) is equivalent to the following:

ðA1B1C1Þy ¼ f½ðA1B1Þy��C1g

y½ðA1B1Þy�

�B�1

1 ½ðB1C1Þy��fA1½ðB1C1Þy�

�gy:

If we put: A000 :¼ A1B1; B000 :¼ B�11 ; C000 :¼ B1C1, then (h) becomes already proven identity (f). For the completeness, notice that

the following operator ranges are closed:

RðA000Þ ¼ RðABÞ; RðB000Þ ¼ RðB�Þ; RðC 000Þ ¼ RðBCÞ;RðA000B000Þ ¼ RðAÞ; RðB000C000Þ ¼ RðCÞ; RðA000B000C 000Þ ¼ RðABCÞ: �

Remark 2.1. The existence of the Moore–Penrose inverses of various operators in the previous theorem, follows from theclosedness of operator ranges RðAÞ;RðBÞ;RðCÞ;RðABÞ;RðBCÞ;RðABCÞ. This fact is important, and it is explained in details.The same is true for the rest of theorems.

The next two corollaries are immediate consequences of Theorem 2.1(a).

Corollary 2.1. Let X;Y; Z be Hilbert spaces, and let A 2 LðY ; ZÞ;B 2 LðX;YÞ be operators, such that A;B;AB have closed ranges. IfAyAB ¼ B and ABBy ¼ A, then ðABÞy ¼ ByAy.

Corollary 2.2. Let P and Q be two orthogonal projectors, i.e. P2 ¼ P ¼ P� and Q2 ¼ Q ¼ Q �. Then ðPQÞyis an idempotent.Moreover, all other corollaries from [11] are also true with some slight changes in their formulations.If U;V are operators acting on the same space, then ½U;V � ¼ UV � VU is the usual notation for their commutator.

Theorem 2.2. Let X;Y ; Z be Hilbert spaces, and let A 2 LðY; ZÞ;B 2 LðX;YÞ be operators, such that A;B;AB have closed ranges. LetM 2 LðZÞ and N 2 LðXÞ be positive and invertible operators. Then the weighted Moore–Penrose inverse of AB with respect to M andN satisfies the following two identities:

444 N.C. Dincic, D.S. Djordjevic / Applied Mathematics and Computation 220 (2013) 439–445

(a) ðABÞyM;N ¼ ðAyABÞyI;NðABByÞyM;I;

(b) ðABÞyM;N ¼ ½ðAyM;IÞ

�B�y

M�1 ;NðByI;NAyM;IÞ

�½AðByI;NÞ��y

M;N�1 .

Proof. By using well-known relation AyM;N ¼ N�12 M

12AN�

12

� �yM

12, it is easy to obtain that (a) is equivalent to:

M12ABN�

12

� �y¼ AyABN�

12

� �yM

12ABBy

� �y: ð1Þ

Let us denote: eA ¼ M12A; eB ¼ BN�

12. We prove the following:

M�12 ~A

� �y¼ ~AyM

12:

The last statement holds if and only if M�12eAeAyM1

2 is Hermitian, which is equivalent to ½M; eAeAy� ¼ 0. Using Lemma 1.4, the lastexpression is equivalent to RðMeAÞ ¼ RðeAÞ, which is valid, because of the invertibility of the Hermitian operator M. Analo-gously we prove that:

ðeBN12Þy¼ N�

12eBy:

Now, (1) becomes:

ðeAeBÞy ¼ ðeAyeAeBÞyðeAeBeByÞy;

which is already proven identity in Theorem 2.1(a).

Analogously, we prove the statement (b). h

Theorem 2.3. Let X;Y ; Z;W be Hilbert spaces, and let A 2 LðZ;WÞ; B 2 LðY ; ZÞ;C 2 LðX;YÞ be operators, such thatA;B; C;AB;BC;ABC have closed ranges. Let M 2 LðWÞ and N 2 LðXÞ be positive and invertible operators. Then the weightedMoore–Penrose inverse of ABC with respect to M and N satisfies the following identities:

(a) ðABCÞyM;N ¼ ðAyABCÞyI;NBðABCCyÞyM;I;

(b) ðABCÞyM;N ¼ ððABÞyABCÞyI;NByðABCðBCÞyÞyM;I;

(c) ðABCÞyM;N ¼ ððABByÞyABCÞy

I;NBðABCðByBCÞyÞyM;I;

(d) ðABCÞyM;N ¼ ½ðAyM;IÞ

�BC�

y

M�1 ;NðAyM;IÞ

�BðCyI;NÞ

�½ABðCyI;NÞ��y

M;N�1 ;

(e) ðABCÞyM;N ¼ f½AðByÞ��

yABCg

y

I;NB�BB�fABC½ðByÞ�C�ygy

M;I;

(f) ðABCÞyM;N ¼ f½ðABÞyM;I��Cgy

M�1 ;N½ðABÞyM;I�

�By½ðBCÞyI;N �

� � fA½ðBCÞyI;N��gy

M;N�1 .

Proof. The proof in all cases is similar to the proof of Theorem 2.2. First, we transform all weighted Moore–Penrose inverses

to the ordinary ones, then we put: eA ¼ M12A; eB ¼ B; eC ¼ CN�

12, and apply Lemma 1.4. After that, all cases reduce to already-

proven identities from Theorem 2.1. h

Some more general identities can also be derived from previous theorems.

Theorem 2.4. Let X;Y; Z be Hilbert spaces, and let A 2 LðY ; ZÞ;B 2 LðX;YÞ be operators, such that A;B;AB have closed ranges. LetM 2 LðZÞ;N 2 LðXÞ and P 2 LðYÞ be positive and invertible operators. Then the weighted Moore–Penrose inverse of AB withrespect to M and N satisfies the following identity:

ðABÞyM;N ¼ ðAyI;PABÞy

P;NðABByP;IÞ

yM;P:

Proof. The proof is similar to the proof of Theorem 2.2. First, we transform all weighted Moore–Penrose inverses to theordinary ones, which gives:

M12ABN�

12

� �y¼ AP�

12

� �yABN�

12

� �yM

12AB P

12B

� �y� �y:

If we put: eA ¼ M12AP�

12; eB ¼ P

12BN�

12, and then apply Lemma 1.4, this statement reduces to the already-proven identity from

Theorem 2.1(a). h

The following theorem can be proven similarly.

N.C. Dincic, D.S. Djordjevic / Applied Mathematics and Computation 220 (2013) 439–445 445

Theorem 2.5. Let X;Y; Z;W be Hilbert spaces, and let A 2 LðZ;WÞ;B 2 LðY ; ZÞ;C 2 LðX;YÞ be operators, such thatA;B;C;AB;BC;ABC have closed ranges. Let M 2 LðWÞ;N 2 LðXÞ; P 2 LðYÞ;Q 2 LðZÞ be positive and invertible operators. Thenthe weighted Moore–Penrose inverse of ABC with respect to M and N satisfies the following identities:

(a) ðABCÞyM;N ¼ ðAyI;PABCÞy

P;NBðABCCyQ ;IÞ

yM;Q

;

(b) ðABCÞyM;N ¼ ððABÞyI;Q ABCÞyQ ;N

ByP;Q ðABCðBCÞyP;IÞyM;P

;

(c) ðABCÞyM;N ¼ ððABByP;IÞyM;P

ABCÞy

P;NBðABCðByI;Q BCÞy

Q ;NÞy

M;Q.

Now, we return to one classical matrix identity from [3]. Our next theorem shows that the result from [3] holds forbounded linear Hilbert space operators.

Theorem 2.6. Let X;Y ; Z be Hilbert spaces, let A 2 LðY; ZÞ;B 2 LðX;YÞ be operators such that A;B;AB have closed ranges. Then:

ðABÞy ¼ ðAyABÞyðABðAyABÞyÞy:

Proof. Using a method described in Theorem 2.1 (and decompositions and matrix forms for A and B) we conclude thatðABÞy ¼ ðAyABÞyðABðAyABÞyÞ

yis equivalent to the following (D is positive and invertible as in Lemma 1.2):

ðA1B1Þy ¼ D�12A1B1

� �yðA1B1ðD�

12A1B1Þ

yÞy;

which is, by Lemma 1.5, further equivalent to:

ðA1B1Þy ¼ D�12A1B1

� �yD�

12A1B1ðA1B1Þy:

We check directly all four Penrose equations, so we have the proof. h

3. Conclusion remarks

We proved several identities related to the reverse order law for the Moore–Penrose inverse of bounded linear operatorson Hilbert spaces. The assumption of closed-range operator is essential for the existence of the Moore–Penrose inverse. Weused operator matrices which follow naturally from orthogonal decompositions of Hilbert spaces. This method seems to beeffective in investigating the infinite dimensional case. On the other hand, corresponding results for complex matrices wereproved in [3,11] using finite dimensional methods, mostly properties of the rank of a complex matrix. It will be challengingto examine whether similar/analogous statements hold for some fi; j; kg-inverses (fi; j; kg � f1;2;3;4g) instead of the Moore–Penrose inverses.

Acknowledgement

We are grateful to the referee for valuable comments and suggestions which improve readability of the paper.

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