Right and Left Weyl Operator Matrices in a Banach Space Setting

Let Xi,Yi(i � 1, 2) be Banach spaces. *e operator matrix of the form MC � A C 0 B 􏼢 􏼣 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, rightWeyl, andWeyl for someC ∈B(X2,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.


Introduction
roughout this paper, X, Y, Z and X i , Y i (i � 1, 2) are always reserved to denote some Banach spaces. If T is a bounded linear operator from X to Y, we write T ∈ B(X, Y), and if X � Y, we write B(X) instead of B(X, X). For T ∈ B(X, Y), the range and the kernel of T are denoted by R(T) and N(T), respectively. Write α(T) � dim N(T) and β(T) � dim(X/R(T)). e sets of all left and right Fredholm operators are, respectively, defined as ey are natural extensions of Hilbert space cases since every closed subspace is complemented in the whole space. e set of Fredholm operators is defined as An operator T is said to be semi-Fredholm if T ∈ Φ l (X, Y) ∪ Φ r (X, Y), for which the index is defined as ind(T) � α(T) − β(T). e left Weyl, right Weyl, and Weyl operators are collected in respectively.
An operator T is said to be left (resp. right) invertible if there exists an operator S ∈ B(Y, X) such that ST � I X (resp. TS � I Y ). 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: G(X, Y) � T ∈ B(X, Y): T is a bijection .

(4)
Again, we have the abbreviations G l (X), G r (X), G(X) of the above classes of operators like B(X). Now let T ∈ B(X). As usual, the spectrum σ(T), left spectrum σ l (T), right spectrum σ r (T), Fredholm spectrum σ e (T), left Fredholm spectrum σ le (T), right Fredholm spectrum σ re (T), Weyl spectrum σ w (T), left Weyl spectrum σ lw (T), and right Weyl spectrum σ rw (T) will be the collections of the numbers λ ∈ C such that T − λI does not belong to the corresponding classes of operators, i.e., G(X), G l (X), G r (X), Φ(X), Φ l (X), Φ r (X), W(X), W l (X), and W r (X), respectively.
We say that T ∈ B(X) is relatively regular or simply regular if there exists some S ∈ B(X) 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 R(T) and N(T) are complemented subspaces of X. Denote by R T and Q T the complementary subspaces with N(T) and R(T), respectively.
For given A ∈ B(X) and B ∈ B(Y), define where C ∈ B(Y, X) is an unknown element. e spectrum and its various subdivisions of M C are considered in many papers such as [1][2][3][4][5][6][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][6][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.

Preliminaries
is 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]). Let T ∈ B(X, Y) be a semi-Fredholm operator and S ∈ B(X, Y) be a compact operator. en, T + S is semi-Fredholm and ind(T + S) � ind(T). e following lemma is obvious, so its proof will be omitted.

Lemma 2.
Assume that L ∈ B(Y) and R ∈ B(X) have bounded inverses defined on Y and X, respectively. en, R(LT) and R(TR) (and hence R(LTR)) have the same closedness as R(T), where T ∈ B(X, Y).
Lemma 3 (see [9]). For T ∈ B(X), we have Note that the last two inclusions are strict in general, and due to the closed range theorem, equality holds here precisely when R(T) is closed. e sets of all upper and lower semi-Fredholm operators are defined by respectively. It is obvious that Lemma 4 (see [8]). Let T ∈ B(X, Y) with closed range. en, α(T ′ ) � β(T), β(T ′ ) � α(T), and Lemma 5 (see [8]). Let T ∈ B(X, Y) and S ∈ B(Y, Z).
Proof. Let B be right Fredholm. If assertion (ii) holds, then taking C � 0 gives the conclusion. We now assume that assertion (i) is true. ere are two possible cases depending on the dimension of Y 1 .
(i) Case 1: dim Y 1 < ∞. In this case, is obviously a right Weyl operator. By Lemma 1, we see that M C is a right Weyl operator for any since B is right Fredholm. en, X 2 and Y 2 have the decompositions Consequently, B can be written as Define operator C by As an operator from , M C has the following matrix form: Conversely, assume that there exists C ∈ B(X 2 , Y 1 ) such that M C is right Weyl. Obviously, B is right Fredholm.
In view of the space decompositions in (10), as an operator from Evidently, B 1 ∈ G(P B , R(B)) and en, If α(B) < ∞, then C 2 is of finite rank, and hence, from (17) and Lemmas 1 and 2, it follows that Taking S � C 2 yields assertion (i).
From eorem 1, the following perturbation result of right Weyl spectrum is obvious. where If having certain special properties for the given diagonal entries, one can further analyze the right Weylness of M C on the basis of eorem 1.
Proof. Assume that there exists C ∈ B(X 2 , Y 1 ) such that M C is right Weyl. From the assumption and the right Fredholmness of A, it follows that β(A) < ∞ and β(B) < ∞, and hence if α(B) � ∞, as required in assertion (i) of Weyl, i.e., assertion (ii) holds. By virtue of eorem 1, the desired result is obvious.
Proof. From the proof of Corollary 2, it is sufficient to notice Dually, we have the following left Weyl description. □

left Weyl if and only if A is left Fredholm, and one of the following statements is fulfilled:
Proof. Let A be left Fredholm. Similar to the proof of eorem 2, it suffices to consider the case that assertion (i) is true under the condition dimX 2 � ∞. Obviously, R(A) is complemented in Y 1 and N(A) is complemented in X 1 , and hence en, Define operator C by At this point, M C has the following matrix form: (23) en, with the aid of the operators A 1 ∈ G(P A , R(A)) and we get If β(A) < ∞, then C 2 is of finite rank, which implies that the converse is not true in general (see, e.g., [10] for more details). where In [[6], eorem 3.6], the author described the Weylness of M C on a Banach space. Using the row operator and the column operator in eorems 1 and 2, we will give a different statement.
Theorem 3. Let A ∈ B(X 1 , Y 1 ) and B ∈ B(X 2 , Y 2 ) be given operators. en, there exists C ∈ B(X 2 , Y 1 ) such that M C is Weyl if and only if A is left Fredholm, B is right Fredholm, and one of the following statements is fulfilled: Proof. Assume that A 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, α(A) < ∞, β(B) < ∞, and hence decompositions (10) and (20) are still satisfied. en, the operators S and J in assertion (i) can be written as where respectively. Hence, Taking we claim that M C is Weyl. Indeed, as an operator from where A 1 ∈ G(P A , R(A)) and B 1 ∈ G(P B , R(B)). en, Journal of Mathematics 5 are invertible operators such that From Lemmas 5 and 6, we infer that M C has the same semi-Fredholmness and index as the operator matrix on the right hand side of (36). us, R(M C ) is clearly closed, and and hence ind(M C ) � 0 by (32).
is shows that M C is Weyl, as claimed.
Conversely, assume that there exists C ∈ B(X 2 , Y 1 ) such that M C is Weyl. Obviously, A is left Fredholm, and B is a right Fredholm. en, decompositions (10) where A 1 ∈ G(P A , R(A)) and B 1 ∈ G(P B , R(B)). en, and If either α(B) < ∞ or β(A) < ∞ holds, then C 4 is of finite rank, which together with (39) implies that M 0 � A 0 0 B is Weyl, i.e., assertion (ii) is valid. We now assume that α(B) � ∞ � β(A). From (39), we equivalently have that EM C F is Weyl, which implies that is proves assertion (i). More where Ω 3 is the collection of all λ ∈ C for which there is not any pair (S, J) of operators such that A − λI S is right Obviously, A 1 ∈ B(X 1 , P(A)) is right invertible and write A −1 1r for some right inverse of A 1 . Since the operator S 2 ∈ B (N(B), Q A ) is right Fredholm. Hence, with β(S 2 ) < ∞ and erefore, operator S is the desired one satisfying assertion (i) of eorem 1, and hence we get the conclusion that there exists C ∈ B(X 2 , R(B)). en, there exists C ∈ B(X 2 , Y 1 ) such that M C is left Weyl if the following statements are fulfilled: proof. Assume that assertions (i), (ii), and (iii) hold. If taking C � 0 proves the theorem. Now let dimQ A � ∞. If α(B) < ∞, again, ind(M 0 ) � ind(A) + ind(B) ≤ 0, and C � 0 is the desired operator. It remains to consider the case of α(B) � ∞. In this case, the decompositions in the first equality of (17) and (20) are still valid, and we further have Obviously, B 1 ∈ B(P B , Y 2 ) is left invertible and write B −1 1l for some left inverse of B 1 . Since

Journal of Mathematics
the operator J 2 ∈ B(N(B), Q A ) is left Fredholm. Hence, with α(J 2 ) < ∞ and erefore, taking operator J above will give assertion (i) of eorem 2, and hence this completes the proof.

Some Extensions and Applications on a Hilbert Space
In the Hilbert space setting, we have the following results as corollaries of eorems 1, 2, and 3. In what follows, H, K and H i , K i (i � 1, 2) are always complex separable Hilbert spaces; denote by H ⊕ K the orthogonal sum of H and K, and write n(T) � dim N(T) and d(T) � dim R(T) ⊥ , as usual, for an operator T between Hilbert spaces.
proof. Comparing eorem 1 with this corollary, we have to show that, under the condition n(B) � ∞, there naturally exists an operator S ∈ B(N(B), K 1 ) such that the row operator A S is right Fredholm with ind A S � ∞ > d(B). In fact, assume first that dimH 1 � ∞. If dimK 1 � ∞, then every unitary operators from N(B) to K 1 can be chosen as the desired S. If, however, dimK 1 < ∞, then any right invertible operator meets our choice. We now let dimH 1 < ∞. At this point, decomposing N(B) into the orthogonal sum of two infinite dimensional closed subspaces will reduce to the case of dimH 1 � ∞. e following is a dual result of Corollary 10, and it gets proved directly by Corollary 10. □ Corollary 11. Let A ∈ B(H 1 , K 1 ) and B ∈ B(H 2 , K 2 ) be given operators. en, there exists C ∈ B(H 2 , K 1 ) such that M C is left Weyl if and only if A ∈ B(H 2 , K 2 ) is left Fredholm, and one of the following holds: is right Fredholm, and one of the following holds: To complete the proof, it suffices to show that there exists where M i and N i (i � 1, 2) are infinite dimensional closed subspaces of N(B) and R(A) ⊥ , respectively. Pick an arbitrary bounded operator Δ from N 2 into M 2 such that ind Δ � d(B) − n(A). Take where U is some unitary operator from N 1 to M 1 , and S i and J i (i � 1, 2) can be chosen as certain bounded operators between the corresponding spaces. Evidently, A S is right and . e proof is completed.
Let A ∈ B(H). We denote by H C the operator on H ⊕ H of the form where C ∈ B(H) 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] proof. Note that, in the underlying Hilbert space, A is left Fredholm if and only if −A * is right Fredholm, and d(−A * ) � n(A) < ∞. So, the necessity is obvious, and we now prove the sufficiency.
If n(−A * ) � ∞, since N(−A * ) � R(A) ⊥ , we see that proof. In the proofs of Corollaries 16 and 17, we actually choose the same self-adjoint operator C such that H C is Weyl. erefore, the present corollary is a direct consequence of them.

Data Availability
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Conflicts of Interest
e authors declare that they have no conflicts of interest.