# Right and Left Weyl Operator Matrices in a Banach Space Setting

Authors: Aichun Liu; Junjie Huang; Alatancang Chen
Journal: Journal of Mathematics
 (2021)

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## Abstract

LetXi,Yii=1,2 be Banach spaces. The operator matrix of the form MC=AC0B acting between X1⊕X2 and Y1⊕Y2 is investigated. By using row and column operators, equivalent conditions are obtained for MC to be left Weyl, right Weyl, and Weyl for some C∈ℬX2,Y1, respectively. Based on these results, some sufficient conditions are also presented. As applications, some discussions on Hamiltonian operators are given in the context of Hilbert spaces.


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## Body

## 1. Introduction

Throughout this paper,X,Y,Z and Xi,Yii=1,2 are always reserved to denote some Banach spaces. If T is a bounded linear operator from X to Y, we write T∈ℬX,Y, and if X=Y, we write ℬX instead of ℬX,X. For T∈ℬX,Y, the range and the kernel of T are denoted by ℛT and NT, respectively. Write αT=dimNT and βT=dimX/ℛT. The sets of all left and right Fredholm operators are, respectively, defined as(1)ΦlX,Y=T∈ℬX,Y:αT<∞,ℛTistopologicalcomplemented inY,ΦrX,Y=T∈ℬX,Y:βT<∞,NTis complemented inX.They are natural extensions of Hilbert space cases since every closed subspace is complemented in the whole space. The set of Fredholm operators is defined as(2)ΦX,Y=ΦlX,Y∩ΦrX,Y=T∈ℬX,Y:αT<∞,βT<∞.An operatorT is said to be semi-Fredholm if T∈ΦlX,Y∪ΦrX,Y, for which the index is defined as indT=αT−βT. The left Weyl, right Weyl, and Weyl operators are collected in(3)WlX,Y=T∈ΦlX,Y:indT≤0,WrX,Y=T∈ΦrX,Y:indT≥0,WX,Y=T∈ΦX,Y:indT=0,respectively.An operatorT is said to be left (resp. right) invertible if there exists an operator S∈ℬY,X such that ST=IX (resp. TS=IY). If T is both left and right invertible, then T is invertible. As it is well known, the sets of all left invertible, right invertible, and invertible operators are as follows:(4)GlX,Y=T∈ℬX,Y:Tis injective,ℛTis complemented inY,GrX,Y=T∈ℬX,Y:Tis surjective,NTis complemented inX,GX,Y=T∈ℬX,Y:Tis a bijection.Again, we have the abbreviationsGlX,GrX,GX of the above classes of operators like ℬX. Now let T∈ℬX. As usual, the spectrum σT, left spectrum σlT, right spectrum σrT, Fredholm spectrum σeT, left Fredholm spectrum σleT, right Fredholm spectrum σreT, Weyl spectrum σwT, left Weyl spectrum σlwT, and right Weyl spectrum σrwT will be the collections of the numbers λ∈ℂ such that T−λI does not belong to the corresponding classes of operators, i.e., GX, GlX, GrX, ΦX, ΦlX, ΦrX, WX, WlX, and WrX, respectively.We say thatT∈ℬX is relatively regular or simply regular if there exists some S∈ℬX such that TST=T. In this case, S is called an inner generalized inverse of T. Obviously, the classes of left or right invertible, invertible, semi-Fredholm, and Fredholm operators are all regular. If M is a closed subspace in Banach space X, then M is said to be topologically complemented or simply complemented if there exists another closed subspace N of X such that M∩N=0 and X=M+N; for such complementary subspaces M and N of X, we write X=M⊕N. It is well known that T is regular if and only if ℛT and NT are complemented subspaces of X. Denote by RT and QT the complementary subspaces with NT and ℛT, respectively.For givenA∈ℬX and B∈ℬY, define(5)MC=AC0B∈ℬX⊕Y,where C∈ℬY,X is an unknown element. The spectrum and its various subdivisions of MC are considered in many papers such as [1–7] and the references therein. Although most of these papers worked in the context of Hilbert spaces, some results on the invertibility and Fredholm theory were established in Banach spaces [5–7]. In this note, we investigate the left and right Weyl spectra of upper triangular operator matrices in a Banach space setting. Our main tools are the regularity of an operator and its equivalent form, which are closely related to some appropriate space decompositions.


## 2. Preliminaries

This section is devoted to collecting some basic results. Although most of them are well-known standard results on Fredholm operators, we list them here for convenience of later proofs.Lemma 1 (see [8]).
LetT∈ℬX,Ybe a semi-Fredholm operator andS∈ℬX,Ybe a compact operator. Then, T+Sis semi-Fredholm andindT+S=indT.
The following lemma is obvious, so its proof will be omitted.Lemma 2.
Assume thatL∈ℬYandR∈ℬXhave bounded inverses defined onYandX, respectively. Then, ℛLTandℛTR (and henceℛLTR) have the same closedness asℛT, whereT∈ℬX,Y.Lemma 3 (see [9]).
ForT∈ℬX, we have(6)NT′=ℛT⊥,NT=ℛ⊥T′,ℛT⊆N⊥T′,ℛT′⊆NT⊥.
Note that the last two inclusions are strict in general, and due to the closed range theorem, equality holds here precisely whenℛT is closed.
The sets of all upper and lower semi-Fredholm operators are defined by(7)Φ+X,Y=T∈ℬX,Y:ℛTis closed andαT<∞,Φ−X,Y=T∈ℬX,Y:βT<∞,respectively. It is obvious that ΦlX,Y⊆Φ+X,Y, ΦrX,Y⊆Φ−X,Y, and ΦX,Y=Φ+X,Y∩Φ−X,Y.Lemma 4 (see [8]).
LetT∈ℬX,Ywith closed range. Then, αT′=βT,βT′=αT, and(8)T∈ΦX,Y⟺T′∈ΦY′,X′,T∈Φ+X,Y⟺T′∈Φ−Y′,X′,T∈Φ−X,Y⟺T′∈Φ+Y′,X′.Lemma 5 (see [8]).
LetT∈ℬX,YandS∈ℬY,Z.(i)
IfSTis right Fredholm, thenSis right Fredholm.(ii)
IfSTis left Fredholm, thenTis left Fredholm.(iii)
IfSTis Fredholm, thenSis right Fredholm andTis left Fredholm.Lemma 6 (see [8]).
LetT∈ℬX,YandS∈ℬY,Z. If bothSandTare upper semi-Fredholm or both are lower semi-Fredholm, thenindST=indS+indT.


## 3. Main Results and Proofs

In this section, we present the main results of this paper and their proofs. First, we establish the right Weylness ofMC.Theorem 1.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. Then, there existsC∈ℬX2,Y1such thatMCis right Weyl if and only ifBis right Fredholm, and one of the following statements is fulfilled:(i)
αB=∞, and there existsS∈ℬNB,Y1such that the row operatorASis a right Fredholm operator withindAS≥βB.(ii)
M0=A00Bis right Weyl.Proof.
LetB be right Fredholm. If assertion (ii) holds, then taking C=0 gives the conclusion. We now assume that assertion (i) is true. There are two possible cases depending on the dimension of Y1.(i)
Case 1: dimY1<∞. In this case,(9)000B:X1⊕X2⟶Y1⊕Y2,is obviously a right Weyl operator. By Lemma1, we see that MC is a right Weyl operator for any C∈ℬX2,Y1.(ii)
Case 2: dimY1=∞. In this case, ℛB is complemented in Y2 and NB is complemented in X2, since B is right Fredholm. Then, X2 and Y2 have the decompositions(10)X2=PB⊕NB,Y2=ℛB⊕QB.
Consequently,B can be written as(11)B=B1000:PB⊕NB⟶ℛB⊕QB.
Define operatorC by(12)C=0S:PB⊕NB⟶Y1.
As an operator fromX1⊕PB⊕NB to Y1⊕ℛB⊕QB, MC has the following matrix form:(13)MC=A0S0B10000.
It is not difficult to see thatℛMC is complemented in Y1⊕Y2 and NMC is complemented in X1⊕X2, and that(14)αMC=αAS,βMC=βAS+βB.
Thus,indMC=indAS−βB≥0. This shows that MC is right Weyl.
Conversely, assume that there existsC∈ℬX2,Y1 such that MC is right Weyl. Obviously, B is right Fredholm. In view of the space decompositions in (10), as an operator from X1⊕PB⊕NB to Y1⊕ℛB⊕QB, MC can be represented as(15)MC=AC1C20B10000.
Evidently,B1∈GPB,ℛB and(16)E=I−C1B1−100I000I∈GY1⊕ℛB⊕QB.
Then,(17)EMC=I−C1B1−100I000IAC1C20B10000=A0C20B10000.
IfαB<∞, then C2 is of finite rank, and hence, from (17) and Lemmas 1 and 2, it follows that M0=A00B is right Weyl, i.e., assertion (ii) is valid. We now assume αB=∞. Since MC is right Weyl, from (17), we equivalently have that ℛEMC is right Weyl. Thus, C2∈ℬNB,Y1 is exactly an operator such that the row operator AC2 is right Fredholm, and indAC2−βB=indEMC≥0, i.e., indAC2≥βB. Taking S=C2 yields assertion (i).
From Theorem1, the following perturbation result of right Weyl spectrum is obvious.Corollary 1.
LetA∈ℬXandB∈ℬYbe given operators. Then,(18)∩C∈ℬY,XσrwMC=σreB∪σrwA00B∩Ω1∪Ω2,where(19)Ω1=λ∈ℂ:αB−λI<∞,Ω2=λ∈ℂ:there is noS∈ℬNB−λI,Ysuch thatA−λISis a right Fredholm operator with indA−λIS≥βB−λI.
If having certain special properties for the given diagonal entries, one can further analyze the right Weylness ofMC on the basis of Theorem 1.Corollary 2.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. IfAis right Fredholm, then there existsC∈ℬX2,Y1such thatMCis right Weyl if and only ifM0=A00Bis right Weyl.Proof.
Assume that there existsC∈ℬX2,Y1 such that MC is right Weyl. From the assumption and the right Fredholmness of A, it follows that βA<∞ and βB<∞, and hence if αB=∞, as required in assertion (i) of Theorem 1, we still conclude that M0=A00B is right Weyl, i.e., assertion (ii) holds. By virtue of Theorem 1, the desired result is obvious.Corollary 3.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. IfAis Weyl, then there existsC∈ℬX2,Y1such thatMCis right Weyl if and only ifB∈ℬX2,Y2is right Weyl.Proof.
From the proof of Corollary2, it is sufficient to notice that M0=A00B is right Weyl if and only if B is right Weyl when A is Weyl.
Dually, we have the following left Weyl description.Theorem 2.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. Then, there existsC∈ℬX2,Y1such thatMCis left Weyl if and only ifAis left Fredholm, and one of the following statements is fulfilled:(i)
βA=∞, and there existsJ∈ℬX2,QAsuch that the column operatorJBis a left Fredholm operator withindJB≤−αA.(ii)
M0=A00Bis left Weyl.Proof.
LetA be left Fredholm. Similar to the proof of Theorem 2, it suffices to consider the case that assertion (i) is true under the condition dimX2=∞. Obviously, ℛA is complemented in Y1 and NA is complemented in X1, and hence(20)X1=PA⊕NA,Y1=ℛA⊕QA.
Then,(21)A=A1000:PA⊕NA⟶ℛA⊕QA.
Define operatorC by(22)C=0J:Y2⟶ℛA⊕QA.
At this point,MC has the following matrix form:(23)MC=A10000J00B:PA⊕NA⊕X2⟶ℛA⊕QA⊕Y2.
Evidently,ℛMC is complemented in Y1⊕Y2 and NMC is complemented in X1⊕X2. In addition, αMC=αJB+αA and βMC=βJB, which implies(24)indMC=indJB+αA≤0.
Thus,MC is left Weyl.
Conversely, assume that there existsC∈ℬX2,Y1 such that MC is left Weyl. It is clear that A is left Fredholm, and as an operator from PA⊕NA⊕X2 to ℛA⊕QA⊕Y2, MC can be written as(25)MC=A10C100C200B.
Then, with the aid of the operatorsA1∈GPA,ℛA and(26)F=I0−A1−1C10I000I∈GPA⊕NA⊕X2,we get(27)MCF=A10C100C200BI0−A1−1C10I000I=A10000C200B.
IfβA<∞, then C2 is of finite rank, which implies that M0=A00B is left Weyl, i.e., assertion (ii) is valid. We now assume that βA=∞. Since MC is left Weyl, it follows from (27) that ℛMCF is left Weyl. Thus, C2∈ℬX2,QA is a desired operator such that the column operator C2B is left Fredholm, and indC2B+αA=indMCF≤0, i.e., indC2B≤−αA. Taking J=C2 yields assertion (i).Remark 1.
Theorem2is dual to Theorem 1to some extent. However, we cannot directly resort to conjugate relationship, since Lemma 4is not valid for left and right Fredholm operators. Actually,T∈ΦlX,YimpliesT′∈ΦrY′,X′, but the converse is not true in general (see, e.g., [10] for more details).Corollary 4.
LetA∈ℬXandB∈ℬYbe given operators. Then,(28)∩C∈ℬY,XσlwMC=σleB∪σlwA00B∩Ω1′∪Ω2′,where(29)Ω1′=λ∈ℂ:βA−λI<∞,Ω2′=λ∈ℂ:there is noJ∈ℬY,QA−λIsuch thatJB−λIis a left Fredholm operator with indJB−λI≤−αA−λI.Corollary 5.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. IfBis left Fredholm, then there existsC∈ℬX2,Y1such thatMCis left Weyl if and only ifM0=A00Bis left Weyl.Corollary 6.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. IfBis Weyl, then there existsC∈ℬX2,Y1such thatMCis left Weyl if and only ifA∈ℬX2,Y2is left Weyl.
In [[6], Theorem 3.6], the author described the Weylness of MC on a Banach space. Using the row operator and the column operator in Theorems 1 and 2, we will give a different statement.Theorem 3.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. Then, there existsC∈ℬX2,Y1such thatMCis Weyl if and only ifAis left Fredholm,Bis right Fredholm, and one of the following statements is fulfilled:(i)
αB=∞=βA, and there existsS∈ℬNB,Y1andJ∈ℬX2,QAsuch thatASis right Fredholm withindAS≥βB,JBis left Fredholm withindJB≤−αA, andPQAS|NB=PQAJ|NB.(ii)
M0=A00Bis Weyl.Proof.
Assume thatA is left Fredholm and B is right Fredholm. If assertion (ii) is valid, then taking C=0 yields the desired sufficiency. We now assume that assertion (i) holds.
Obviously,αA<∞, βB<∞, and hence decompositions (10) and (20) are still satisfied. Then, the operators S and J in assertion (i) can be written as(30)S=S1Δ:NB⟶ℛA⊕QA,J=J1Δ:PB⊕NB⟶QA,where PQAS|NB=PQAJ|NB. Since AS is right Fredholm, βΔ=βAS<∞, and ℛΔ is clearly complemented in QA. Similarly, the left Fredholmness of JB implies that αΔ=αJB<∞ and NΔ is complemented in NB. Note that the assumptions indAS≥βB and indJB≤−αA in (i) imply that(31)αA+αΔ−βΔ≥βB,αΔ−βΔ−βB≤−αA,respectively. Hence,(32)indΔ=βB−αA.
Taking(33)C=0S1J1Δ:PB⊕NB⟶ℛA⊕QA,we claim that MC is Weyl.
Indeed, as an operator fromPA⊕NA⊕PB⊕NB to ℛA⊕QA⊕ℛB⊕QB,(34)MC=A100S100J1Δ00B100000,where A1∈GPA,ℛA and B1∈GPB,ℛB. Then,(35)E=I0000I−B1−1J1000I0000I∈ℬℛA⊕QA⊕ℛB⊕QB,F=I00−A1−1S10I0000I0000I∈ℬPA⊕NA⊕PB⊕NBare invertible operators such that(36)EMCF=A1000000Δ00B100000.
From Lemmas5 and 6, we infer that MC has the same semi-Fredholmness and index as the operator matrix on the right hand side of (36). Thus, ℛMC is clearly closed, and indMC=αA+αΔ−βB−βΔ, and hence indMC=0 by (32). This shows that MC is Weyl, as claimed.
Conversely, assume that there existsC∈ℬX2,Y1 such that MC is Weyl. Obviously, A is left Fredholm, and B is a right Fredholm. Then, decompositions (10) and (20) still hold true, and as an operator from PA⊕NA⊕PB⊕NB to ℛA⊕QA⊕ℛB⊕QB, MC has a new block representation:(37)MC=A10C1C200C3C400B100000,where A1∈GPA,ℛA and B1∈GPB,ℛB. Then,(38)E=I0−C1B1−100I−C3B1−1000I0000I∈GℛA⊕QA⊕ℛB⊕QB,F=I00−A1−1C20I0000I0000I∈GPA⊕NA⊕PB⊕NB,and(39)EMCF=A1000000C400B100000.
If eitherαB<∞ or βA<∞ holds, then C4 is of finite rank, which together with (39) implies that M0=A00B is Weyl, i.e., assertion (ii) is valid. We now assume that αB=∞=βA. From (39), we equivalently have that EMCF is Weyl, which implies that αA+αC4−βC4−βB=0. Taking S=0C4:NB⟶ℛA⊕QA gives a right Fredholm row operator AS with indAS=βB, while taking J=0C4:PA⊕NA⟶QA yields a left Fredholm column operator JB with indJB≤−αA. Also, PQAS|NB=PQAJ|NB. This proves assertion (i). More generally, S=C2C4∈ℬNB,Y1 and J=C3C4∈ℬX2,QA also serve as the desired operators, since A1 and B1 are invertible.
As a direct consequence of Theorem3, we have the following.Corollary 7.
LetA∈ℬXandB∈ℬYbe given operators. Then,(40)∩C∈ℬY,XσwMC=σleA∪σreB∪σwA00B∩Ω1∪Ω1′∪Ω3,whereΩ3is the collection of allλ∈ℂfor which there is not any pairS,Jof operators such thatA−λISis right Fredholm withindA−λIS≥βB−λI, JB−λIis left Fredholm withindJB−λI≤−αA−λI, andPQA−λIS|NB−λI=PQA−λIJ|NB−λI. Note thatS∈ℬNB−λI,Y,J∈ℬY,QA−λI, andΩ1,Ω1′are defined as inCorollaries 1 and 4.Corollary 8.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. IfAis right Fredholm orBis left Fredholm, then there existsC∈ℬX2,Y1such thatMCis Weyl if and only ifM0=A00Bis Weyl.proof.
Assume that there existsC∈ℬX2,Y1 such that MC is Weyl. Then, we always see that the relations αB=∞=βA do not hold, whenever A is right Fredholm or B is left Fredholm. From Theorem 3, it follows that M0=A00B is Weyl. The rest of the proof is trivial.Corollary 9.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators. IfA(resp.B) is Weyl, then there existsC∈ℬX2,Y1such thatMCis Weyl if and only ifB(resp.A) is Weyl.proof.
By symmetry of the result, we only prove the case whenA is Weyl. The sufficiency is obvious. For necessity, let MC be Weyl for some C∈ℬX2,Y1. Since A is Weyl, βA=αA<∞. Then, assertion (i) of Theorem 3 cannot be satisfied. Hence, A00B is Weyl, which is equivalent to the fact that B is Weyl.
Based on the embedded relationship of certain spaces, the sufficient conditions under whichMC is left and right Weyl in a Banach space setting are given, respectively.Definition 1.
(see6, Definition 4.2]).For two Banach spacesXandY, we say thatXcan be embedded inYand writeX≺¯Yif there exists a left invertible operatorJ:X⟶Y. Note thatX≺¯Yif and only if there exists a right invertible operatorS:Y⟶X.Theorem 4.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators such thatY1/ℛA¯⊕Y2/ℛB¯≺¯NA⊕NB. Then, there existsC∈ℬX2,Y1such thatMCis right Weyl if the following statements are fulfilled:(i)
Bis right Fredholm.(ii)
ℛAis complemented inY1.(iii)
There existsS∈ℬNB,Y1such thatASis right Fredholm.proof.
Assume that assertions (i), (ii), and (iii) hold. IfαB<∞, then AS is right Fredholm if and only if A0 is right Fredholm, and hence βA<∞ which together with the right Fredholmness of B and the relation Y1/ℛA¯⊕Y2/ℛB¯≺¯NA⊕NB implies that αA+αB≥βA+βB, i.e., indM0=indA+indB≥0. This means that M0=A00B is right Weyl, and taking C=0 gives the theorem.
Now letαB=∞. If βA<∞, again, indM0=indA+indB≥0, and C=0 is the desired operator. It remains to consider the case of βA=∞. In this case, the decompositions in (10) and the second equality of (20) are still valid, and we further have(41)AS=A1S10S2:X1⊕NB⟶ℛA⊕QA.
Obviously,A1∈ℬX1,PA is right invertible and write A1r−1 for some right inverse of A1. Since(42)A1S10S2I−A1r−1S10I=A100S2,the operator S2∈ℬNB,QA is right Fredholm. Hence,(43)NB=PS2⊕NS2,QA=ℛS2⊕QS2,with βS2<∞ and PS2≅ℛS2. Since Y1/ℛA¯⊕Y2/ℛB¯≺¯NA⊕NB, from NB/NS2≅PS2≅ℛS2, it follows that(44)QA⊕Y2/ℛB¯ℛS2≺¯NA⊕NS2.
It is clear thatαA+αS2≥βB+βS2. This together with (42) implies that(45)indAS=αA+αS2−βS2≥βB.
Therefore, operatorS is the desired one satisfying assertion (i) of Theorem 1, and hence we get the conclusion that there exists C∈ℬX2,Y1 such that MC is right Weyl.Theorem 5.
LetA∈ℬX1,Y1andB∈ℬX2,Y2be given operators such thatNA⊕NB≺¯Y1/ℛA¯⊕Y2/ℛB¯. Then, there existsC∈ℬX2,Y1such thatMCis left Weyl if the following statements are fulfilled:(i)
Ais left Fredholm.(ii)
NBis complemented inX2.(iii)
There existsJ∈ℬX2,QAsuch thatJBis left Fredholm.proof.
Assume that assertions (i), (ii), and (iii) hold. IfdimQA<∞, then JB is left Fredholm if and only if 0B is left Fredholm, and hence αB<∞ which together with the left Fredholmness of A and the relation NA⊕NB≺¯Y1/ℛA¯⊕Y2/ℛB¯ implies that αA+αB≤βA+βB, i.e., indM0=indA+indB≤0. This means that M0=A00B is left Weyl, and taking C=0 proves the theorem.
Now letdimQA=∞. If αB<∞, again, indM0=indA+indB≤0, and C=0 is the desired operator. It remains to consider the case of αB=∞. In this case, the decompositions in the first equality of (17) and (20) are still valid, and we further have(46)JB=J1J2B10:PB⊕NB⟶QA⊕Y2.
Obviously,B1∈ℬPB,Y2 is left invertible and write B1l−1 for some left inverse of B1. Since(47)I−J1B1l−10IJ1J2B10=0J2B10,the operator J2∈ℬNB,QA is left Fredholm. Hence,(48)NB=PJ2⊕NJ2,QA=RJ2⊕QJ2,with αJ2<∞ and PJ2≅ℛJ2. Since NA⊕NB≺¯Y1/ℛA¯⊕Y2/ℛB¯, from NB/NJ2≅PJ2≅ℛJ2, it follows that(49)NA⊕NJ2≺¯QA⊕Y2/ℛB¯ℛJ2.
It is clear thatαA+αJ2≤βB+βJ2. This together with (47) implies that(50)indJB=αJ2−βJ2+βB≤−αA.
Therefore, taking operatorJ above will give assertion (i) of Theorem 2, and hence this completes the proof.Remark 2.
In particular, Theorems 4 and 5hold ifAandBare regular, respectively.


## 4. Some Extensions and Applications on a Hilbert Space

In the Hilbert space setting, we have the following results as corollaries of Theorems1, 2, and 3. In what follows, ℋ,K and ℋi,Kii=1,2 are always complex separable Hilbert spaces; denote by ℋ⊕K the orthogonal sum of ℋ and K, and write nT=dimNT and dT=dimℛT⊥, as usual, for an operator T between Hilbert spaces.Corollary 10.
LetA∈ℬℋ1,K1andB∈ℬℋ2,K2be given operators. Then, there existsC∈ℬℋ2,K1such thatMCis right Weyl if and only ifB∈ℬℋ2,K2is right Fredholm, and one of the following holds:(i)
nB=∞.(ii)
M0=A00Bis right Weyl.proof.
Comparing Theorem1 with this corollary, we have to show that, under the condition nB=∞, there naturally exists an operator S∈ℬNB,K1 such that the row operator AS is right Fredholm with indAS=∞>dB. In fact, assume first that dimℋ1=∞. If dimK1=∞, then every unitary operators from NB to K1 can be chosen as the desired S. If, however, dimK1<∞, then any right invertible operator meets our choice. We now let dimℋ1<∞. At this point, decomposing NB into the orthogonal sum of two infinite dimensional closed subspaces will reduce to the case of dimℋ1=∞.
The following is a dual result of Corollary10, and it gets proved directly by Corollary 10.Corollary 11.
LetA∈ℬℋ1,K1andB∈ℬℋ2,K2be given operators. Then, there existsC∈ℬℋ2,K1such thatMCis left Weyl if and only ifA∈ℬℋ2,K2is left Fredholm, and one of the following holds:(i)
dA=∞.(ii)
M0=A00Bis left Weyl.Corollary 12.
LetA∈ℬℋ1,K1andB∈ℬℋ2,K2be given operators. Then, there existsC∈ℬℋ2,K1such thatMCis Weyl if and only ifA∈ℬℋ2,K2is left Fredholm,B∈ℬℋ2,K2is right Fredholm, and one of the following holds:(i)
nB=∞=dA.(ii)
M0=A00Bis Weyl.proof.
Assume thatA∈ℬℋ2,K2 is left Fredholm, B∈ℬℋ2,K2 is right Fredholm, and nB=∞=dA. To complete the proof, it suffices to show that there exists S∈ℬNB,Y1 and J∈ℬX2,QA such that AS is right Fredholm with indAS≥dB, JB is left Fredholm with indJB≤−nA, and PℛA⊥S|NB=PℛA⊥J|NB.
In fact, sinceA∈ℬℋ2,K2 is left Fredholm and B∈ℬℋ2,K2 is right Fredholm, ℛA,ℛB are complemented, NA,NB are complemented, and one has the following decompositions:(51)ℋ1=NA⊥⊕NA,ℋ2=NB⊥⊕NB,K1=ℛA⊕ℛA⊥,K2=ℛB⊕ℛB⊥.
SincenB=∞=dA, we have(52)NB=N1⊕N2,ℛA⊥=ℳ1⊕ℳ2,where ℳi and Nii=1,2 are infinite dimensional closed subspaces of NB and ℛA⊥, respectively. Pick an arbitrary bounded operator Δ from N2 into ℳ2 such that indΔ=dB−nA. Take(53)S=S1S2U00Δ:N1⊕N2⟶ℛA⊕ℳ1⊕ℳ2,J=J1U0J20Δ:NB⊥⊕N1⊕N2⟶ℳ1⊕ℳ2,where U is some unitary operator from N1 to ℳ1, and Si and Jii=1,2 can be chosen as certain bounded operators between the corresponding spaces. Evidently, AS is right Fredholm, JB is left Fredholm,(54)indAS=nA+indΔ=dB,indJB=indΔ−dB=−nA,and PℛA⊥S|NB=U00Δ=PℛA⊥J|NB. The proof is completed.
Recall that for an upper or lower semi-Fredholm operatorT, i.e., T∈Φ+X∪Φ−X, we can also define indT=αT−βT. Let Φ+−X be the class of all T∈Φ+X with indT≤0 and Φ+−X be the class of all T∈Φ−X with indT≥0; write Φ0X=Φ+−X∩Φ−+X, and hence Φ0X=WX. Particularly, if X is a Hilbert space, then ΦlX=Φ+X and ΦrX=Φ−X.
The next results’ proofs will be omitted. Actually, the corollaries listed above are, respectively, the reformulations of Theorems 2.1, 2.3, and 2.4 in [11], and the following are their initial forms.Corollary 13 (see11, Theorem 2.1]).
A2×2operator matrixMC=AC0B∈Φ+−H⊕Kfor someC∈ℬK,ℋif and only ifA∈Φ+Hand(55)nB<∞andnA+nB≤dA+dB,ornB=∞=dA,ifℛBis closed,dA=∞,ifℛBis not closed.Corollary 14 (see11, Theorem 2.3]).
A2×2operator matrixMC=AC0B∈Φ−+H⊕Kfor someC∈ℬK,ℋif and only ifB∈Φ−Hand(56)dA<∞and dA+dB≤nA+nB,ornB=∞=dA,ifℛAis closed,nB=∞,ifℛAis not closed.Corollary 15 (see11, Theorem 2.4]).
A2×2operator matrixMC=AC0B∈Φ0H⊕Kfor someC∈ℬK,ℋif and only ifA∈Φ+H,B∈Φ−Hand one of the cases exists:(i)
dA<∞,nB<∞, andnA+nB=dA+dB.(ii)
dA=nB=∞.
LetA∈ℬℋ. We denote by HC the operator on ℋ⊕ℋ of the form(57)HC=AC0−A∗,where C∈ℬℋ is an unknown self-adjoint operator, which is clearly a Hamiltonian operator. One can naturally find its appearance in linear Hamiltonian systems and operator Riccati equations (see, e.g., [12, 13]). As applications, we now present the analogues of Theorems 1, 2, and 3 for Hamiltonian operators.Corollary 16.
LetA∈ℬℋbe a given operator. Then, there exists a self-adjoint operatorC∈ℬℋsuch thatHCis right Weyl if and only ifAis left Fredholm.proof.
Note that, in the underlying Hilbert space,A is left Fredholm if and only if −A∗ is right Fredholm, and d−A∗=nA<∞. So, the necessity is obvious, and we now prove the sufficiency.
Ifn−A∗<∞, then dA=n−A∗<∞, and hence indA00−A∗=0, i.e., H0=A00−A∗ is Weyl and is certainly right Weyl. From condition ii of Theorem 1, the conclusion follows immediately.
Ifn−A∗=∞, since N−A∗=ℛA⊥, we see that A100I:ℋ⊕N−A∗⟶ℛA⊕ℛA⊥ is right Fredholm. Let S=0I:N−A∗⟶ℛA⊕ℛA⊥. Then, the row operator AS is right Fredholm, and indAS=nA=d−A∗. At this point, we know from the proof of Theorem 1 that C=0S:N−A∗⊥⊕N−A∗⟶ℋ is a self-adjoint operator such that HC is right Weyl.Corollary 17.
LetA∈ℬℋbe a given operator. Then, there exists a self-adjoint operatorC∈ℬℋsuch thatHCis left Weyl if and only ifAis left Fredholm.proof.
The necessity is obvious, and we now prove the sufficiency. IfdA<∞, then H0=A00−A∗ is still a Weyl operator, and from condition ii of Theorem 2, the conclusion is clear.
IfdA=∞, then 0I−A1∗0:N−A∗⊥⊕N−A∗⟶ℛA⊥⊕ℋ is left Fredholm. Let J=0I:N−A∗⊥⊕N−A∗⟶ℛA⊥. Hence, the column operator J−A∗ is left Fredholm, and indJ−A∗=−d−A∗=−nA. Thus, from the proof of Theorem 2, we see that C=0J:ℋ⟶ℛA⊕ℛA⊥ is a self-adjoint operator such that HC is left Weyl.Corollary 18.
LetA∈ℬℋbe a given operator. Then, there exists a self-adjoint operatorC∈ℬXsuch thatHCis Weyl if and only ifAis left Fredholm.proof.
In the proofs ofCorollaries 16 and 17, we actually choose the same self-adjoint operator C such that HC is Weyl. Therefore, the present corollary is a direct consequence of them.



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