# Developing Reverse Order Law for the Moore–Penrose Inverse with the Product of Three Linear Operators

Authors: Yang Qi; Liu Xiaoji; Yu Yaoming
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we study the reverse order law for the Moore–Penrose inverse of the product of three bounded linear operators in Hilbert spaces. We first present some equivalent conditions for the existence of the reverse order lawABC†=C†B†A†. Moreover, several equivalent statements of ℛAA∗ABC=ℛABC and ℛC∗CABC∗=ℛABC∗ are also deducted by the theory of operators.


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## Body

## 1. Introduction

LetℒX,Y denote the set of all bounded linear operators from Hilbert space X to Hilbert space Y. Especially, it is briefly written as ℒX when Y=X. For A∈ℒX,Y, the symbols NA, ℛA, and A† denote null space, range, and Moore–Penrose inverse of A, respectively.The Moore–Penrose inverse ofA∈ℒX,Y is the operator A†∈ℒY,X that satisfies the Penrose equations, i.e.,(1)AA†A=A;A†AA†=A†;AA†∗=AA†;A†A∗=A†A.If the Moore–Penrose inverseA† of A exists, then it is unique. A† exists if and only if ℛA is closed in Y (see [1, 18] for details).The reverse order law for the generalized inverse of a product of matrices or bounded linear operators yields a class of interesting fundamental problems in the theory of the generalized inverse of matrices or operators. It has been widely studied since the middle 1960s. Greville [7] first discussed the reverse order law for the Moore–Penrose inverse of the product of two matrices AB†=B†A† and gave a necessary and sufficient condition for the law. Hartwig et al. [8] studied the reverse order law for the Moore–Penrose inverse of the product of three matrices and derived the equivalent conditions for the existence of the reverse order law. Ji and Wei [10] studied the weighted Moore–Penrose inverse and extended the fundamental theorem of linear algebra for matrix spaces to tensor spaces. Recently, Panigrahy and Mishra presented some properties for the weighted Moore–Penrose inverse for an arbitrary order tensor via the Einstein product [12]; they also proposed the expression for the Moore–Penrose inverse of the product of two tensors via the Einstein product [13, 14]. Until now, some reverse order laws for the known generalized inverses have been widely studied and extended into a complex number field, a Hilbert space, or an associative ring (see [2, 3, 5, 6, 9, 11, 15–17, 19–23] and references therein).Recently, based on the results in [8], Dinčić and Djordjević [4] extended some finite dimensional results to infinite dimensional settings and obtained new results related to the mixed-type reverse order law for the Moore–Penrose inverse of various products of three operators in Hilbert space. This motivates us to have a further study concerning the reverse order law of three operators in Hilbert space and to extend some results in [4, 8].In the remainder of this section, we will give some denotations. In Section2, we will present several lemmas. In Section 3, we will investigate some equivalent conditions for the existence of ABC†=C†B†A†. Moreover, several equivalent statements of ℛAA∗ABC=ℛABC and ℛC∗CABC∗=ℛABC∗ are also deducted by the theory of operators.Now, we describe several denotations. IfP,Q∈ℒX, then the symbol P,Q stands for the commutator of P and Q, i.e., P,Q=defPQ−QP. Also, for A∈ℒX,Y, the symbols A−∗=defA∗−1 and A∗†=defA∗† are obtained. Throughout this paper, we always assume that A∈ℒX2,X1,B∈ℒX3,X2,andC∈ℒX4,X3, where Xi,i=1,…,4, denote arbitrary Hilbert spaces. And then, ABC will be well-defined in what follows.


## 2. Lemmas

First of all, we list two lemmas, which come from [4, 6], and the first is a rewritten version of Lemmas 1.1 and 1.2 in [6].Lemma 2.1 (see [6]).
LetK∈ℒX,Y have a closed range. Let X1 and X2 be closed and mutually orthogonal subspaces of X such that X=X1⊕X2 (here, ⊕ represents direct sum). Let Y1 and Y2 be closed and mutually orthogonal subspaces of Y such that Y=Y1⊕Y2. Then, the operator K has the following matrix representations with respect to the orthogonal sums of subspaces X=X1⊕X2=ℛK∗⊕NK and Y=ℛK⊕NK∗=Y1⊕Y2:(a)
Representations with respect to the orthogonal sums of subspacesX=X1⊕X2=ℛK∗⊕NK are(2)K=K1K200:X1X2⟶ℛKNK∗,whereDK=K1K1∗+K2K2∗ maps ℛK into itself and DK>0 (meaning DK>0 is invertible). Also,(3)K†=K1∗DK−10K2∗DK−10,KK†=I000.(b)
Representations with respect to the orthogonal sums of subspacesY=ℛK⊕NK∗=Y1⊕Y2 are(4)K=K10K20:ℛK∗NK⟶Y1Y2,whereDK=K1∗K1+K2∗K2 maps ℛK∗ into itself and DK>0 (meaning DK>0 is invertible). Also,(5)K†=DK−1K1∗DK−1K2∗00,K†K=I000.Here,Ki,i=1,2, denote different operators in any of these two cases.In particular,(6)K=K1000:ℛK∗NK⟶ℛKNK∗,where K1 is invertible. Also,(7)K†=K1−1000:ℛKNK∗⟶ℛK∗NK.Lemma 2.2 (see [4]).
LetXandY be Hilbert spaces. Let B∈ℒX,Y have a closed range, and let H∈ℒY be Hermitian and invertible. Then, ℛHB=ℛB if and only if H,BB†=0.Next, the following lemma can guarantee the existence of the Moore–Penrose inverses of various operators appearing in what follows.Lemma 2.3.
LetA,C, and ABC have closed ranges. Then A∗A†, CC∗†, A†ABC†, A∗ABC†, ABCC††, ABCC∗†, A†ABCC∗†, A∗ABCC††, A∗ABCC∗†, and A†ABCC†† exist.Proof.
SinceℛC=ℛCC†=ℛCC∗, ℛABCC†=ℛABCC∗=ℛABC. Furthermore, we have CC∗†, ABCC††, and ABCC∗† exist since C and ABC have closed ranges. Then,(8)ℛCC∗B∗A∗A†∗=ℛCC∗B∗A∗⇔ℛA†ABCC∗is closed⇔ℛABCC∗=ℛABCis closed.
Hence,A†ABCC∗† exists.
Similarly, we can obtainA∗A†, A†ABC†, A∗ABC†, A†ABCC∗†, A∗ABCC††, A∗ABCC∗†, and A†ABCC†† exist.
Finally, the following two lemmas are used for proving some results in the next section. Here, the symbolA,B=AB−BA, which denotes the commutator of A and B.Lemma 2.4.
LetAandC be invertible and B have a closed range and AB†, BC†, and ABC† exist. Then,(i)
AB†=B†A−1 if and only if A∗A,BB†=0 if and only if A∗A−1,BB†=0 if and only if AA∗,ABAB†=0(ii)
BC†=C−1B† if and only if CC∗,B†B=0 if and only if CC∗−1,B†B=0 if and only if C∗C,BCBC†=0(iii)
ABC†=C−1B†A−1 if and only if A∗A,BB†=0 and CC∗,B†B=0 if and only if A∗A−1,BB†=0 and CC∗−1,B†B=0Proof.
Obviously,ℛBC=ℛB due to the invertibility of C. Then, BC has a closed range since ℛB is closed. Hence, BC† exists. So does AB∗. And then, AB† exists. Similarly, ABC† exists. Hence,(i)
ClearlyB†A−1ABB†A−1=B†A−1, ABB†A−1AB=AB, and B†A−1AB=B†A−1AB∗.
Note thatABB†A−1∗=ABB†A−1 holds if and only if BB†A∗A=A∗ABB†, namely, A∗A,BB†=0, which is equivalent to A∗A−1,BB†=0. Hence, the proposition is true by the definition of the Moore–Penrose inverse.
The proofs of (ii) and (iii) are similar.Lemma 2.5.
LetAandC be Hermitian and invertible and B have a closed range. Then, for arbitrary rational numbers n and p, we have the following properties:(i)
A,An+pBAn+pB†=0 if and only if A,AnBAnB†=0(ii)
C,BCn+p†BCn+p=0 if and only if C,BCn†BCn=0(iii)
A,BB†=0 implies A2,BB†=0(iv)
C,B†B=0 implies C2,B†B=0Proof.
For arbitrary rational numbersk, AkB† and BCk† exist by Lemma 2.4, since Ak and Ck are obviously invertible.(i)
By Lemma2.2, A,An+pBAn+pB†=0 if and only if ℛAn+pB=ℛAAn+pB if and only if A−pℛAn+pB=A−pℛAAn+pB, i.e., ℛAnB=ℛAAnB if and only if A,AnB†AnB=0.(iii)
By Lemma2.2, A,BB†=0 leads to ℛAB=ℛB. Thus, ℛA2B=AℛB=ℛB and therefore A2,BB†=0.
The proofs of (ii) and (iv) are similar.Remark 1.
If matrixA∈Rn is Hermitian and invertible, let λ1,…,λn be eigenvalues for A; then, the matrix A is similar to a diagonal matrix. And, it can be written as follows:(9)A=QΛQT=Qλ10⋯00λ2⋯0⋮⋮⋱⋮00⋯λnQT,where Q is a unitary matrix. Therefore, we have(10)Ap=QΛpQT=Qλ1p0⋯00λ2p⋯0⋮⋮⋱⋮00⋯λnpQT,where p denotes arbitrary rational numbers.


## 3. Reverse Order Laws

In this section, we will prove the result concerning the reverse order law for the Moore–Penrose inverse of various products of bounded linear operators in Hilbert spaces.Firstly, we shall show some equivalent conditions for the existence of the reverse order lawABC†=C†B†A†, which extends Theorem 1 in [8].Theorem 3.1.
LetA,B,C, and ABC have closed ranges. Then, the following statements are equivalent:(i)
ABC†=C†B†A†(ii)
A†ABCC††=CC†B†A†A, A∗ABCC†B†A†A, and CC†B†A†ABCC∗ are Hermitian(iii)
A∗ABCC††=CC†B†A†A∗†, A∗ABCC†B†A†A, and CC†B†A†ABCC∗ are Hermitian(iv)
A†ABCC∗†=C∗†C†B†A†A, A∗ABCC†B†A†A, and CC†B†A†ABCC∗ are Hermitian(v)
A∗ABCC∗†=C∗†C†B†A†A∗†, A∗ABCC†B†A†A, and CC†B†A†ABCC∗ are HermitianProof.
By Lemma2.3, the Moore–Penrose inverses ABC†, A†ABCC††, A∗ABCC††, and A†ABCC∗† exist.
SetK=C in Lemma 2.1(b). By Lemma 2.1(b),(11)C=C10C20:ℛK∗NK⟶Y1Y2,where DC=C1∗C1+C2∗C2 maps ℛK∗ into itself and DC>0,(12)CC†=C1DC−1C1∗C1DC−1C2∗C2DC−1C1∗C2DC−1C2∗,C†C=I000.
SetK=B in Lemma 2.1(b). Then,(13)B=B10B20,ABC=A1B1C1000,ABC†=A1B1C1†000,A†ABCC∗=B1C1C1∗B1C1C2∗00.
SinceA†ABCC∗ has a closed range, by Lemma 2.1, we can obtain that(14)D=B1C1C1∗B1C1C1∗∗+B1C1C2∗B1C1C2∗∗=B1C1DCC1∗B1∗=B1C1DC1/2DC1/2C1∗B1∗=B1C1DC1/2B1C1DC1/2∗.
Thus,D is invertible. Furthermore, we have(15)B1C1∗D−1=DC−1/2∗B1C1DC1/2∗D−1=DC−1/2B1C1DC1/2†,and then,(16)A†ABCC∗†=B1C1C1∗∗D−10B1C1C2∗∗D−10=C1DC−1/2B1C1DC1/2†0C2DC−1/2B1C1DC1/2†0.
Similarly, we can obtain(17)A†ABCC††=C1DC−1/2B1C1DC−1/2†0C2DC−1/2B1C1DC−1/2†0,(18)A∗ABCC††=C1DC−1/2A1∗A1B1C1DC−1/2†0C2DC−1/2A1∗A1B1C1DC−1/2†0,(19)A∗ABCC∗†=C1DC−1/2A1∗A1B1C1DC1/2†0C2DC−1/2A1∗A1B1C1DC1/2†0.
In addition, sinceA∗ABCC†B†A†A is Hermitian, we have(20)A∗ABCC†B†A†A=A1∗A1B1000C1DC−1C1∗DB−1B1∗0C2DC−1C1∗DB−1B1∗0,=A1∗A1B1C1DC−1C1∗DB−1B1∗000,(21)A1∗A1B1C1DC−1C1∗DB−1B1∗=B1C1DC−1C1∗DB−1B1∗∗A∗A.
And,CC†B†A†ABCC∗ is Hermitian, so(22)C1DC−1C1∗DB−1B1∗B1C1C1∗C1DC−1C1∗DB−1B1∗B1C1C2∗C2DC−1C1∗DB−1B1∗B1C1C1∗C2DC−1C1∗DB−1B1∗B1C1C2∗=C1C1∗B1∗B1DB−1C1DC−1C1∗C1C1∗B1∗B1DB−1C1DC−1C2∗C2C1∗B1∗B1DB−1C1DC−1C1∗C2C1∗B1∗B1DB−1C1DC−1C2∗.
Multiplying byC∗ and C, respectively, on the left and right sides of equation (22) yields(23)DC−1C1∗DB−1B1∗B1C1=C1∗DB−1B1∗B1C1∗DC−1.(i)⇔(ii): similar to the proof of Theorem 1 in [8].(ii)⇒(iii): by (12), we have(24)CC†B†A†A=C1DC−1C1∗C1DC−1C2∗C2DC−1C1∗C2DC−1C2∗DB−1B1∗DB−1B2∗00I000,=C1DC−1C1∗DB−1B1∗0C2DC−1C1∗DB−1B1∗0.
Then,A†ABCC††=CC†B†A†A implies C∗A†ABCC††=C∗CC†B†A†A, and therefore, by (17) and (24),(25)B1C1DC−1/2†=DC−1/2C1∗DB−1B1∗.
By (21) and (25),(26)A1∗A1B1C1DC−1/2B1C1DC−1/2†=A1∗A1B1C1DC−1C1∗DB−1B1∗=B1C1DC−1C1∗DB−1B1∗∗A1∗A1=B1C1DC−1/2B1C1DC−1/2†∗A1∗A1,and then,(27)A1∗A12B1C1DC−1/2B1C1DC−1/2†=B1C1DC−1/2B1C1DC−1/2†A1∗A12.
As a result, by Lemma2.4(i), we have(28)A1∗A1B1C1DC−1/2†=B1C1DC−1/2†A1∗A1−1.
Hence, by (18), (25), and (28), we can obtain(29)A∗ABCC††=C1DC−1/2B1C1DC−1/2†A1∗A1−10C2DC−1/2B1C1DC−1/2†A1∗A1−10=C1DC−1C1∗DB−1B1∗A1∗A1−10C2DC−1C1∗DB−1B1∗A1∗A1−10=C1DC−1C1∗C1DC−1C2∗C2DC−1C1∗C2DC−1C2∗DB−1B1∗DB−1B2∗00A1∗A1−1000=CC†B†A†A∗†.(iii)⇒ (v): since(30)CC†B†A†A∗†=C1DC−1C1∗DB−1B1∗A1∗A1−10C2DC−1C1∗DB−1B1∗A1∗A1−10,A∗ABCC††=CC†B†A†A∗† implies(31)A1∗A1B1C1DC−1/2†=DC−1/2C1∗DB−1B1∗A1∗A1−1.By (18), (24), and (31), we have(32)DC−1A1∗A1B1C1DC−1/2†A1∗A1B1C1DC−1/2=DC−1/2C1∗DB−1B1∗B1C1DC−1/2=DC−1/2C1∗DB−1B1∗B1C1∗DC−3/2=A1∗A1B1C1DC−1/2†A1∗A1B1C1DC−1/2∗DC−1andDC−2,A1∗A1B1C1DC−1/2†A1∗A1B1C1DC−1/2=0. Consequently, by Lemma 2.4(ii), we have(33)A1∗A1B1C1DC1/2†=DC−1A1∗A1B1C1DC−1/2†.Combined with (19), (31), and (33), we can further obtain(34)A∗ABCC∗†=C1DC−1/2DC−3/2C1∗DB−1B1∗A1∗A1−10C2DC−1/2DC−3/2C1∗DB−1B1∗A1∗A1−10=C1DC−10C2DC−10DC−1C1∗DC−1C2∗00DB−1B1∗DB−1B2∗00A1∗A1−1000=C∗†C†B†A†A∗†.(v)⇒ (iv): since(35)C∗†C†B†A†A∗†=C1DC−2C1∗DB−1B1∗A1∗A1−10C2DC−2C1∗DB−1B1∗A1∗A1−10,using (19) and A∗ABCC∗†=C∗†C†B†A†A∗†, we can obtain(36)A1∗A1B1C1DC1/2†=DC−3/2C1∗DB−1B1∗A1∗A1−1.Combining (21) and (36), we have(37)A1∗A1−1A1∗A1B1C1DC1/2A1∗A1B1C1DC1/2†=B1C1DC−1C1∗DB−1B1∗A1∗A1−1=A1∗A1−1B1C1DC−1C1∗DB−1B1∗∗=A1∗A1B1C1DC1/2A1∗A1B1C1DC1/2†∗A1∗A1−1,and then,A1∗A1−2,A1∗A1B1C1DC1/2A1∗A1B1C1DC1/2†=0. Hence, by Lemma 2.4(i), we have(38)A1∗A1B1C1DC1/2†=B1C1DC1/2†A1∗A1−1,and therefore, by (16), (38), and (36), we have(39)A†ABCC∗†=C1D2−1/2A1∗A1B1C1DC1/2†A1∗A10C2D2−1/2A1∗A1B1C1DC1/2†A1∗A10=C1DC−2C1∗DB−1B1∗0C2DC−2C1∗DB−1B1∗0=C1DC−10C2DC−10DC−1C1∗DC−1C2∗00DB−1B1∗DB−1B2∗00I000=C∗†C†B†A†A.(iv)⇒ (ii): since(40)C∗†C†B†A†A=C1DC−2C1∗DB−1B1∗0C2DC−2C1∗DB−1B1∗0,by (16), A†ABCC∗†=C∗†C†B†A†A implies(41)B1C1DC1/2†=DC−3/2C1∗DB−1B1∗.Combining (23) and (41), we have(42)DCB1C1DC1/2†B1C1DC1/2=DC1/2C1∗DB−1B1∗B1C1DC1/2=DC1/2C1∗DB−1B1∗B1C1∗DC−1/2=B1C1DC1/2†B1C1DC1/2∗DC.and then, by Lemma2.4(ii), we can obtain that(43)B1C1DC−1/2†=DCB1C1DC1/2†.Accordingly, by (17), (41), and (43), we have(44)A†ABCC††=C1DC1/2B1C1DC1/2†0C2DC1/2B1C1DC1/2†0=C1DC−1C1∗DB−1B1∗0C2DC−1C1∗DB−1B1∗0=C1DC−1C1∗C1DC−1C2∗C2DC−1C1∗C2DC−1C2∗DB−1B1∗DB−1B2∗00I000=CC†B†A†A.
In [4], Dinčić and Djordjević showed some equivalent statements of ℛAA∗ABC=ℛABC and ℛC∗CABC∗=ℛABC∗. Here, we will present other equivalent statements.Theorem 3.2.
LetA,C, and ABC have closed ranges. Then, the following statements are equivalent:(i)
A†ABCC∗†=C∗†ABC†A(ii)
A∗ABCC††=CABC†A∗†(iii)
C†A†ABCC∗†A∗=C∗C†ABC†AA∗(iv)
C∗A∗ABCC††A†=C∗CABC†AA∗†(v)
ABC†=C∗A†ABCC∗†A†(vi)
ABC†=C†A∗ABCC††A∗(vii)
ℛAA∗ABC=ℛABC and ℛC∗CABC∗=ℛABC∗Proof.
By Lemma2.3, we can obtain that the Moore–Penrose inverse ABC†, A†ABCC∗†, and A∗ABCC†† exist.i⇒iii: multiplying C† and A∗ on the left and the right side of i, respectively, we then have(45)C†A†ABCC∗†A∗=C∗C†ABC†AA∗,whereC∗C†=C†C∗†.iii⇒v: multiplying C∗C and AA∗† on the left and the right side of iii, respectively, we thus have(46)C∗CC†A†ABCC∗†A∗AA∗†=C∗CC∗C†ABC†AA∗AA∗†,namely,(47)C∗A†ABCC∗†A†=C†CABC†AA†.Thus, by (12), v obviously holds.v⇒i: multiplying C∗† and A on the left and the right side of v, respectively, then we have(48)C∗†ABC†A=CC†A†ABCC∗†A†A.By (12) and (16), CC†A†ABCC∗†=A†ABCC∗† and then i easily holds.v⇔vii: by (16), we have(49)C∗A†ABCC∗†A†=DC1/2B1C1DC1/2†A1−1000.Therefore,v is equivalent to(50)B1C1DC1/2†=A1−1A1B1C1DC1/2†=DC−1/2A1B1C1†A1.By Lemma2.4(iii), we can obtain that (50) holds if and only if the following equations are satisfied:(51)A1B1C1A1B1C1†A1A1∗=A1A1∗A1B1C1A1B1C1†,DCA1B1C1†A1B1C1=A1B1C1†A1B1C1DC.Namely,(52)AA∗,ABCABC†=0,C∗C,ABC†ABC=0.On the other hand, by Lemma2.2, (v) is equivalent to vii.vi⇔vii: the equation A1∗A1B1C1DC−1/2†=DC1/2A1B1C1†A1−∗ can be deduced and then, the equivalence between vi and vii is similarly proved.
In addition, we can deduce a key equation, similar to (16), and therefore obtain the equivalence of ii, iv, and vi.
By Theorems 2.1 and 2.2 in [4], we can prove the following theorem, similar to the proof of Theorem 3.2.Theorem 3.3.
LetA,C, and ABC have closed ranges. Then, the following statements are equivalent:(i)
A∗ABCC∗†=C∗†ABC†A∗†(ii)
A†ABCC††=CABC†A(iii)
C†A∗ABCC∗†A†=C∗C†ABC†AA∗†(iv)
C∗A†ABCC††A∗=C∗CABC†AA∗(v)
ABC†=C†A†ABCC††A†(vi)
ABC†=C∗A∗ABCC∗†A∗(vii)
ℛAA∗ABC=ℛABC and ℛC∗CABC∗=ℛABC∗
Next, we will further deduce some statements, each of which is equivalent toℛAA∗ABC=ℛABC and ℛC∗CABC∗=ℛABC∗.Theorem 3.4.
LetA,C, and ABC have closed ranges. Then, the following statements are equivalent:(i)
A†ABC†A†=C†ABCC††(ii)
A†ABC†A∗=C†A†∗BCC††(iii)
A†ABC†A†=C∗ABCC∗†(iv)
A∗ABC†A∗=C†ABCC††(v)
A†ABC∗††A†=C†ABCC††(vi)
A∗ABC∗††A∗=C†ABCC∗††(vii)
A∗A†BC†A†=C∗A†∗BCC∗†(viii)
ℛAA∗ABC=ℛABCandℛC∗CABC∗=ℛABC∗Proof.
By Lemma2.3, we can obtain that A∗A†, CC∗†, A†ABC†, A∗ABC†, ABCC††, ABCC∗†, A∗ABC∗††, A∗A†BC†, and ABCC∗†† exist. By (12), we have(53)ABCC†=A1B1C1DC−1C1∗A1B1C1DC−1C2∗00.
SinceABCC† has a closed range, by Lemma 2.1, we have(54)D=A1B1C1DC−1C1∗A1B1C1DC−1C1∗∗+A1B1C1DC−1C2∗A1B1C1DC−1C2∗∗=A1B1C1DC−1C1∗B1∗A1∗=A1B1C1DC−1/2A1B1C1DC−1/2∗,which is invertible, and(55)A1B1C1DC−1∗D−1=DC−1/2∗A1B1C1DC−1/2∗D−1=DC−1/2A1B1C1DC−1/2†,and then,(56)ABCC††=C1DC−1/2A1B1C1DC−1/2†0C2DC−1/2A1B1C1DC−1/2†0.
Similarly, we can get that(57)A∗†BCC††=C1DC1/2A1−∗B1C1DC−1/2†0C2DC1/2A1−∗B1C1DC−1/2†0,(58)ABCC∗†=C1DC−1/2A1B1C1DC1/2†0C2DC−1/2A1B1C1DC1/2†0,(59)ABCC∗††=C1DC−1/2A1B1C1DC−3/2†0C2DC−1/2A1B1C1DC−3/2†0,(60)A∗†BCC∗†=C1DC−1/2A1−∗B1C1DC1/2†0C1DC−1/2A1−∗B1C1DC1/2†0,(61)A†ABC†=B1C1†000,(62)A∗ABC†=A1∗A1B1C1†000,(63)A†ABC∗††=B1C1DC−1†000,(64)A∗ABC∗††=A1∗A1B1C1DC−1†000,(65)A∗A†BC†=A1∗A1−1B1C1†000.(i)⇒(viii): using (56) and (61), we have(66)A†ABC†A†=B1C1†A1−1000,C†ABCC††=DC−1/2A1B1C1DC−1/2†000.Now, (i) holds; i.e.,A†ABC†A†=C†ABCC†† holds if and only if(67)A1B1C1DC−1/2†=DC1/2B1C1†A1−1.By Lemma2.4, (67) is equivalent to(68)A1∗A1,B1C1B1C1†=0,DC,B1C1†B1C1=0.SinceA1 is invertible, by Lemma 2.4, we have(69)B1C1†=A1B1C1†A1⇔A1∗A1,B1C1B1C1†=0⇔A1A1∗,AB1C1AB1C1†=0.In addition,DC,B1C1†B1C1=0 is equivalent to DC,A1B1C1†A1B1C1=0 whenever B1C1†A1−1=A1B1C1†. From this, we can get that (68) is equivalent to(70)A1A1∗,A1B1C1A1B1C1†=0,DC,A1B1C1†A1B1C1=0,namely,(71)AA∗,ABCABC†=0,C∗C,ABC†ABC=0.By Lemma2.2, (71) holds if and only if(72)ℛAA∗ABC=ℛABC,ℛC∗CABC∗=ℛABC∗.Namely, (i) is equivalent toviii.(viii)⇒(vii): statement (viii) implies that(73)AA∗,ABCABC†=0,C∗C,ABC†ABC=0,which is equivalent to(74)A1A1∗,A1B1C1A1B1C1†=0,DC,A1B1C1†A1B1C1=0.Thus,(75)A1A1∗−1,A1B1C1A1B1C1†=0,DC,A1B1C1†A1B1C1=0,and then, the left equation in (75) implies(76)A1A1∗−1,A1−∗B1C1A1−∗B1C1†=0.By Lemma2.5, we have(77)A1A1∗2,A1−∗B1C1A1−∗B1C1†=0.Hence, by Lemma2.4, we have A1B1C1†=A1−∗B1C1†A1A1∗−1. From this, the right equation in (75) becomes DC,A1−∗B1C1†A1−∗B1C1=0 and then(78)DC−1,A1−∗B1C1DC1/2†A1−∗B1C1DC1/2=0.Therefore, we haveA1−∗B1C1†=DC1/2A1−∗B1C1DC1/2†. Consequently, (76) becomes(79)A1A1∗2,A1−∗B1C1DC1/2A1−∗B1C1DC1/2†=0.Thus, combining Lemma2.4, (78), and (79), we have(80)A1∗A1−1B1C1†=DC1/2A1−∗B1C1DC1/2†A1.(vii)⇒(vi): by (60) and (65), (vii) holds if and only if(81)A1∗A1−1B1C1†=DC1/2A1−∗B1C1DC1/2†A1.Then, (81) is equivalent to(82)A1∗A1−1,A1−∗B1C1DC1/2A1−∗B1C1DC1/2†=0,DC−1,A1−∗B1C1DC1/2†A1−∗B1C1DC1/2=0.By Lemmas2.4 and 2.5, the first equation in (82) is equivalent to(83)A1∗A1−1,A1−∗B1C1DC1/2A1−∗B1C1DC1/2†=0⇔A1A1∗,A1−∗B1C1DC1/2A1−∗B1C1DC1/2†=0⇔A1A1∗,A1B1C1DC1/2A1B1C1DC1/2†=0.Thus,(84)A1A1∗2,A1B1C1DC1/2A1B1C1DC1/2†=0,and therefore,A1−∗B1C1DC1/2†=A1B1C1DC1/2†A1A1∗.The second equation in (82) can be further rewritten as follows:(85)DC−1,A1B1C1DC1/2†A1B1C1DC1/2=0⇔DC2,A1B1C1DC−3/2†A1B1C1DC−3/2=0⇔A1B1C1DC1/2†=DC−2A1B1C1DC−3/2†.Consequently, (82) becomes(86)A1A1∗,A1B1C1DC−3/2A1B1C1DC−3/2†=0,DC,A1B1C1DC−3/2†A1B1C1DC−3/2=0,and then,A1∗A1B1C1DC−1†=DC−1/2A1B1C1DC−3/2†A1−∗, namely, (vi) holds.(vi)⇔(v): by (59) and (64), we can obtain that (vi) is equivalent to(87)A1∗A1B1C1DC−1†=DC−1/2A1B1C1DC−3/2†A1−∗,or, equivalently,(88)A1A1∗,A1B1C1DC−3/2A1B1C1DC−3/2†=0,DC,A1B1C1DC−3/2†A1B1C1DC−3/2=0.By Lemma2.4 and 2.5, the right equation in (88) is equivalent to(89)DC,A1B1C1DC−1†A1B1C1DC−1=0⇔A1B1C1DC−3/2†=DC1/2A1B1C1DC−1†.By Lemma2.5, the left equation in (88) is equivalent to(90)A1A1∗,A1B1C1DC−1A1B1C1DC−1†=0⇔A1∗A1,B1C1DC−1B1C1DC−1†=0⇔A1B1C1DC−1†=B1C1DC−1†A1−1.As a result, (88) is equivalent to(91)A1A1∗,B1C1DC−1B1C1DC−1†=0andDC,B1C1DC−1†B1C1DC−1=0,if and only ifA1B1C1DC−1/2†=DC−1/2B1C1DC−1†A−1, namely, (v) holds.(v)⇒(iv): by (56) and (63), (v) holds if and only if(92)A1B1C1DC−1/2†=DC−1/2B1C1DC−1†A−1.Using Lemma2.4, we have(93)A1∗A1,B1C1DC−1B1C1DC−1†=0,DC,B1C1DC−1†B1C1DC−1=0.By Lemma2.5, the right equation in (93) is equivalent to DC,B1C1†B1C1=0 and implies DC2,B1C1†B1C1=0. Thus, by Lemma 2.4, we have B1C1DC−1†=DCB1C1†. Accordingly, (93) becomes(94)A1∗A1,B1C1B1C1†=0,DC−1,B1C1†B1C1=0.Similarly, combining Lemmas2.5 and 2.4, from the left equation in (94), it follows that B1C1†=A1∗A1B1C1†A1∗A1.Consequently, (94) becomes(95)A1∗A1−1,A1∗A1B1C1A1∗A1B1C1†=0,DC−1,A1∗A1B1C1†A1∗A1B1C1=0,if and only ifA1B1C1DC−1/2†=DC1/2A1∗A1B1C1†A1∗, namely, (iv) holds.(iv)⇒(iii): by (56) and (62), (iv) holds if and only if(96)A1B1C1DC−1/2†=DC1/2A1∗A1B1C1†A1∗,A1∗A1,A1∗A1B1C1A1∗A1B1C1†=0,DC,A1∗A1B1C1†A1∗A1B1C1=0.Notice that the left equation in (96) holds if and only if A1∗A1,B1C1B1C1†=0 with Lemma 2.5. Hence, A1∗A12,B1C1B1C1†=0 and then, by Lemma 2.4, we have A1∗A1B1C1†=B1C1†A1∗A1−1. Thus, (96) becomes(97)A1∗A1,B1C1B1C1†=0,DC,B1C1†B1C1=0.Hence,A1B1C1DC1/2†=DC−1/2B1C1†A1−1 and therefore, (iii) holds.(iii)⇒(ii): by (58) and (61), (iii) holds if and only if(98)A1B1C1DC1/2†=DC−1/2B1C1†A1−1.Similarly,A1∗A1,B1C1B1C1†=0 and DC,B1C1†B1C1=0 if and only if A1∗A1−1,B1C1B1C1†=0 and DC−1,B1C1†B1C1=0. Therefore,(99)A1∗−1B1C1DC−1/2†=DC1/2B1C1†A1∗if and only if (ii) holds.(ii)⇒(i): by (57) and (61), (ii) is equivalent to(100)A1∗−1B1C1DC−1/2†=DC1/2B1C1†A1∗.Therefore, by Lemma2.4, we have(101)A1∗A1,B1C1B1C1†=0,DC,B1C1†B1C1=0,if and only ifA1B1C1DC−1/2†=DC1/2B1C1†A1−1, i.e., (i) holds.
By Theorems3.1–3.3, we can obtain the following theorem.Theorem 3.5.
LetA,C, and ABC have closed ranges. The following statements are equivalent:(i)
ABC†=C†B†A†(ii)
A∗†BC†=C†B†A∗(iii)
ABC∗††=C∗B†A†(iv)
ABC†=C†A†ABCC††A† and A†ABCC††=CC†B†A†A(v)
ABC†=C†A∗ABCC††A∗ and A∗ABCC††=CC†B†A†A∗†(vi)
ABC†=C∗A†ABCC∗†A† and A†ABCC∗†=C∗†C†B†A†A(vii)
ABC†=C∗A∗ABCC∗†A∗ and A∗ABCC∗†=C∗†C†B†A†A∗†(viii)
ABC†=A†ABC†A† and A†ABC†=C†B†A†A(ix)
ABC†=C†ABCC††andABCC††=CC†B†A†(x)
ABC†=A∗ABC†A∗ and A∗ABC†=C†B†A†A∗†(xi)
ABC†=C∗ABCC∗† and ABCC∗†=C∗†C†B†A†(xii)
ABCC††=CC†B†A† and A†ABC†=C†B†A†A(xiii)
ABCC∗†=C∗†C†B†A† and A∗ABC†=C†B†A†A∗†Proof.
By Lemma2.3, those Moore–Penrose inverses exist.(i)⇔(ii): if (i) holds, then C†B†A∗A∗†BC=C†B†A†ABC=ABC†ABC and then(102)C†B†A∗A∗†BCC†B†A∗=ABC†ABCABC†AA∗=C†B†A†AA∗=C†B†A∗.The other Penrose equations are similarly verified. Hence, (ii) holds.Conversely, the proof is same.(i)⇔(iii): this is similar to the proof above.(i)⇒(iv): by Theorem 3.1, we have A†ABCC††=CC†B†A†A and(103)C†A†ABCC††A†=C†CC†B†A†AA†=C†B†A†=ABC†.(iv)⇒(v): by Theorems 3.2 and 3.3, we have ABC†=C†A∗ABCC††A∗ and(104)A∗ABCC††=CABC†A∗†=CABC†AA†A∗†=A†ABCC††A†A∗†=CC†B†A†AA†A∗†=CC†B†A†A∗†.(v)⇒(vi) and (vi)⇒(vii): This is similar to the proof of “(iv)⇒ (v).”(vii)⇒(i): The proof is obvious.(i)⇒(viii): if (i) holds, then we can obtain that A†ABCC†B†A†A=A†ABCABC†A=I000 is Hermitian by (13). Thus, it is easily verified that A†ABC†=C†B†A†A holds by the definition of the Moore–Penrose inverse. Therefore,(105)A†ABC†A†=C†B†A†AA†=ABC†.Conversely, the following is apparent:(i)⇔(ix), (i)⇔(x), and (i)⇔(xi): these are analogous to the proof above.(i)⇔(xii): since (i), (viii), and (ix) are equivalent, (i) implies (xii).
Conversely,ABCC†B†A†=ABCC†ABCC†† and then,(106)ABCC†B†A†ABC=ABCC†ABCC††ABCC†C=ABCC†C=ABC.
Thus, the first and third Penrose equations are satisfied. Similarly, the other Penrose equations are satisfied:(i)⇔(xiii): the proof is the same as the above.



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