Fine Spectra of Tridiagonal Symmetric Matrices

Abstract

The fine spectra of upper and lower triangular banded matrices were examined by several authors. Here we determine the fine spectra of tridiagonal symmetric infinite matrices and also give the explicit form of the resolvent operator for the sequence spaces 𝑐₀, 𝑐, ℓ₁, and ℓ_∞.

1. Introduction

The spectrum of an operator is a generalization of the notion of eigenvalues for matrices. The spectrum over a Banach space is partitioned into three parts, which are the point spectrum, the continuous spectrum, and the residual spectrum. The calculation of these three parts of the spectrum of an operator is called the fine spectrum of the operator.

The spectrum and fine spectrum of linear operators defined by some particular limitation matrices over some sequence spaces was studied by several authors. We introduce the knowledge in the existing literature concerning the spectrum and the fine spectrum. Wenger [1] examined the fine spectrum of the integer power of the Cesàro operator over 𝑐 and, Rhoades [2] generalized this result to the weighted mean methods. Reade [3] worked on the spectrum of the Cesàro operator over the sequence space 𝑐₀. Gonzáles [4] studied the fine spectrum of the Cesàro operator over the sequence space ℓ_𝑝. Okutoyi [5] computed the spectrum of the Cesàro operator over the sequence space 𝑏𝑣. Recently, Rhoades and Yildirim [6] examined the fine spectrum of factorable matrices over 𝑐₀ and 𝑐. Coşkun [7] studied the spectrum and fine spectrum for the p-Cesàro operator acting over the space 𝑐₀. Akhmedov and Başar [8, 9] have determined the fine spectrum of the Cesàro operator over the sequence spaces 𝑐₀, ℓ_∞, and ℓ_𝑝. In a recent paper, Furkan, et al. [10] determined the fine spectrum of 𝐵(𝑟,𝑠,𝑡) over the sequence spaces 𝑐₀ and 𝑐, where 𝐵(𝑟,𝑠,𝑡) is a lower triangular triple-band matrix. Later, Altun and Karakaya [11] computed the fine spectra for Lacunary matrices over 𝑐₀ and 𝑐.

In this work, our purpose is to determine the fine spectra of the operator, for which the corresponding matrix is a tridiagonal symmetric matrix, over the sequence spaces 𝑐₀, 𝑐, ℓ₁, and ℓ_∞. Also we will give the explicit form of the resolvent for this operator and compute the norm of the resolvent operator when it exists and is continuous.

Let 𝑋 and 𝑌 be Banach spaces and 𝑇∶𝑋→𝑌 be a bounded linear operator. By ℛ(𝑇), we denote the range of 𝑇, that is, ℛ(𝑇)={𝑦∈𝑌∶𝑦=𝑇𝑥;𝑥∈𝑋}.(1.1) By 𝐵(𝑋), we denote the set of all bounded linear operators on 𝑋 into itself. If 𝑋 is any Banach space, and let 𝑇∈𝐵(𝑋) then the adjoint 𝑇^∗ of 𝑇 is a bounded linear operator on the dual 𝑋^∗ of 𝑋 defined by (𝑇^∗𝜙)(𝑥)=𝜙(𝑇𝑥) for all 𝜙∈𝑋^∗ and 𝑥∈𝑋. Let 𝑋≠{𝜃} be a complex normed space and 𝑇∶𝒟(𝑇)→𝑋 be a linear operator with domain 𝒟(𝑇)⊂𝑋. With 𝑇, we associate the operator 𝑇_𝜆=𝑇−𝜆𝐼,(1.2) where 𝜆 is a complex number and 𝐼 is the identity operator on 𝒟(𝑇). If 𝑇_𝜆 has an inverse, which is linear, we denote it by 𝑇^𝜆−1, that is 𝑇^𝜆−1=(𝑇−𝜆𝐼)⁻¹(1.3) and call it the resolvent operator of 𝑇_𝜆. If 𝜆=0, we will simply write 𝑇⁻¹. Many properties of 𝑇_𝜆 and 𝑇^𝜆−1 depend on 𝜆, and spectral theory is concerned with those properties. For instance, we will be interested in the set of all 𝜆 in the complex plane such that 𝑇^𝜆−1 exists. Boundedness of 𝑇^𝜆−1 is another property that will be essential. We shall also ask for what 𝜆^s the domain of 𝑇^𝜆−1 is dense in 𝑋. For our investigation of 𝑇, 𝑇_𝜆, and 𝑇^𝜆−1, we need some basic concepts in spectral theory which are given as follows (see [12, pages 370-371]).

Let 𝑋≠{𝜃} be a complex normed space, and let 𝑇∶𝒟(𝑇)→𝑋 be a linear operator with domain 𝒟(𝑇)⊂𝑋. A regular value 𝜆 of 𝑇 is a complex number such that (R1)𝑇^𝜆−1 exists,(R2)𝑇^𝜆−1 is bounded, and(R3)𝑇^𝜆−1 is defined on a set which is dense in 𝑋.

The resolvent set 𝜌(𝑇) of 𝑇 is the set of all regular values 𝜆 of 𝑇. Its complement 𝜎(𝑇)=ℂ⧵𝜌(𝑇) in the complex plane ℂ is called the spectrum of 𝑇. Furthermore, the spectrum 𝜎(𝑇) is partitioned into three disjoint sets as follows: the point spectrum 𝜎_𝑝(𝑇) is the set such that 𝑇^𝜆−1 does not exist. A 𝜆∈𝜎_𝑝(𝑇) is called an eigenvalue of 𝑇. The continuous spectrum 𝜎_𝑐(𝑇) is the set such that 𝑇^𝜆−1 exists and satisfies (R3) but not (R2). The residual spectrum 𝜎_𝑟(𝑇) is the set such that 𝑇^𝜆−1 exists but does not satisfy (R3).

A triangle is a lower triangular matrix with all of the principal diagonal elements nonzero. We shall write ℓ_∞, 𝑐, and 𝑐₀ for the spaces of all bounded, convergent, and null sequences, respectively. And by ℓ_𝑝, we denote the space of all 𝑝-absolutely summable sequences, where 1≤𝑝<∞. Let 𝜇 and 𝛾 be two sequence spaces and 𝐴=(𝑎_𝑛𝑘) be an infinite matrix of real or complex numbers 𝑎_𝑛𝑘, where 𝑛,𝑘∈ℕ. Then, we say that 𝐴 defines a matrix mapping from 𝜇 into 𝛾, and we denote it by writing 𝐴∶𝜇→𝛾, if for every sequence 𝑥=(𝑥_𝑘)∈𝜇 the sequence 𝐴𝑥={(𝐴𝑥)_𝑛}, the 𝐴-transform of 𝑥, is in 𝛾, where(𝐴𝑥)_𝑛=_𝑘𝑎_𝑛𝑘𝑥_𝑘(𝑛∈ℕ).(1.4) By (𝜇∶𝛾), we denote the class of all matrices 𝐴 such that 𝐴∶𝜇→𝛾. Thus, 𝐴∈(𝜇∶𝛾) if and only if the series on the right side of (1.4) converges for each 𝑛∈ℕ and every 𝑥∈𝜇, and we have 𝐴𝑥={(𝐴𝑥)_𝑛}_𝑛∈ℕ∈𝛾 for all 𝑥∈𝜇.

A tridiagonal symmetric infinite matrix is of the form 𝑆=𝑆(𝑞,𝑟)=⎡⎢⎢⎢⎢⎢⎣⋮00𝑞𝑟⋮0𝑞𝑟𝑟⋮0𝑞𝑟𝑟⋮00𝑞𝑟⋮000𝑟⋮0000⋱⋯⋯⋯⋯⎤⎥⎥⎥⎥⎥⎦,(1.5) where 𝑞,𝑟∈ℂ. The spectral results are clear when 𝑟=0, so for the sequel we will have 𝑟≠0.

Theorem 1.1 (cf. [13]). Let 𝑇 be an operator with the associated matrix 𝐴=(𝑎_𝑛𝑘). (i)𝑇∈𝐵(𝑐) if and only if ‖𝐴‖∶=sup_𝑛∞_𝑘=1𝑎||𝑎_𝑛𝑘||<∞,(1.6)_𝑘∶=𝑎∶l=im_𝑛→∞li𝑎m_{𝑛𝑘∞𝑛→∞}existsforeach𝑘,(1.7)_𝑘=1𝑎_𝑛𝑘exists.(1.8)(ii)𝑇∈𝐵(𝑐₀) if and only if (1.6) and (1.7) with 𝑎_𝑘=0 for each 𝑘.(iii)𝑇∈𝐵(ℓ_∞) if and only if (1.6). In these cases, the operator norm of 𝑇 is ‖𝑇‖_{(ℓ∞∶ℓ∞)}=‖𝑇‖_{(𝑐∶𝑐)}=‖𝑇‖_{(𝑐0∶𝑐0)}=‖𝐴‖.(1.9)(iv)𝑇∈𝐵(ℓ₁) if and only if ‖‖𝐴^𝑡‖‖=sup_𝑘∞_𝑛=1||𝑎_𝑛𝑘||<∞.(1.10) In this case, the operator norm of 𝑇 is ‖𝑇‖_{(ℓ1∶ℓ1)}=‖𝐴^𝑡‖.

Corollary 1.2. Let 𝜇∈{𝑐₀,𝑐,ℓ₁,ℓ_∞}. 𝑆(𝑞,𝑟)∶𝜇→𝜇 is a bounded linear operator and ‖𝑆(𝑞,𝑟)‖_{(𝜇∶𝜇)}=|𝑞|+2|𝑟|.

2. The Spectra and Point Spectra

Theorem 2.1. 𝜎_𝑝(𝑆,𝜇)=∅ for 𝜇∈{ℓ₁,𝑐₀,𝑐}.

Proof. Since ℓ₁⊂𝑐₀⊂𝑐, it is enough to show that 𝜎_𝑝(𝑆,𝑐)=∅. Let 𝜆 be an eigenvalue of the operator 𝑆. An eigenvector 𝑥=(𝑥₀,𝑥₁,…)∈𝑐 corresponding to this eigenvalue satisfies the linear system of equations: 𝑞𝑥₀+𝑟𝑥₁=𝜆𝑥₀𝑟𝑥₀+𝑞𝑥₁+𝑟𝑥₂=𝜆𝑥₁𝑟𝑥₁+𝑞𝑥₂+𝑟𝑥₃=𝜆𝑥₂⋮.(2.1) If 𝑥₀=0, then 𝑥_𝑘=0 for all 𝑘∈ℕ. Hence 𝑥₀≠0. Without loss of generality we can suppose 𝑥₀=1. Then 𝑥₁=(𝜆−𝑞)/𝑟 and the system of equations turn into the linear homogeneous recurrence relation 𝑥_𝑛+𝑝𝑥_𝑛−1+𝑥_𝑛−2=0for𝑛≥2,(2.2) where 𝑝=(𝑞−𝜆)/𝑟. The characteristic polynomial of the recurrence relation is 𝑥²+𝑝𝑥+1=0.(2.3) There are three cases here.Case 1 (𝑝=−2). Then characteristic polynomial has only one root: 𝛼=1. Hence, the solution of the recurrence relation is of the form 𝑥_𝑛=(𝐴+𝐵𝑛)(𝛼)^𝑛=𝐴+𝐵𝑛,(2.4) where 𝐴 and 𝐵 are constants which can be determined by the first two terms 𝑥₀ and 𝑥₁. 1=𝑥₀=𝐴+𝐵0, so we have 𝐴=1. And −𝑝=𝑥₁=𝐴+𝐵1, so we have 𝐵=1. Then 𝑥_𝑛=𝑛+1. This means (𝑥_𝑛)∉𝑐. So, we conclude that there is no eigenvalue in this case.Case 2 (𝑝=2). Then characteristic polynomial has only one root: 𝛼=−1. The solution of the recurrence relation, found as in Case 1, is 𝑥_𝑛=(𝑛+1)(−1)^𝑛. So, there is no eigenvalue in this case.Case 3 (𝑝≠±2). Then the characteristic polynomial has two distinct roots 𝛼₁≠±1 and 𝛼₂≠±1 with 𝛼₁𝛼₂=1. Let |𝛼₁|≥1≥|𝛼₂|. The solution of the recurrence relation is of the form 𝑥_𝑛=𝐴𝛼₁^𝑛+𝐵𝛼₂^𝑛.(2.5) Using the first two terms and the fact that 𝑝=−(𝛼₁+𝛼₂), we get 𝐴=𝛼₁/(𝛼₁−𝛼₂) and 𝐵=𝛼₂/(𝛼₂−𝛼₁). So we have 𝑥_𝑛=𝛼^1𝑛+1−𝛼^2𝑛+1𝛼₁−𝛼₂.(2.6) If |𝛼₁|>1>|𝛼₂|, then ||𝑥_𝑛||≥1||𝛼₁−𝛼₂||||𝛼¹||^𝑛+1−||𝛼₂||^𝑛+1.(2.7)So lim_𝑛|𝑥_𝑛|=∞, which means (𝑥_𝑛)∉𝑐. Now, if |𝛼₁|=|𝛼₂|=1, then there exists 𝜃∈(0,𝜋) such that 𝛼₁=𝑒^𝑖𝜃 and 𝛼₂=𝑒^−𝑖𝜃. So, 𝑥_𝑛=[sin(𝑛+1)𝜃]/sin𝜃. Again we have (𝑥_𝑛)∉𝑐. Hence there is no eigenvalue also in this case.

Repeating all the steps in the proof of this theorem for ℓ_∞, we get to the following.

Theorem 2.2. 𝜎_𝑝(𝑆,ℓ_∞)=(𝑞−2𝑟,𝑞+2𝑟).

Theorem 2.3. Let 𝑝=(𝑞−𝜆)/𝑟. Let 𝛼₁ and 𝛼₂ be the roots of the polynomial 𝑃(𝑥)=𝑥²+𝑝𝑥+1, with |𝛼₂|>1>|𝛼₁|. Then the resolvent operator over 𝑐₀ is 𝑆^𝜆−1=(𝑠_𝑛𝑘), where 𝑠_𝑛𝑘=1𝑟𝛼²¹−1⋅𝛼^{1𝑛−𝑘+1}𝛼−𝛼^1𝑛+𝑘+3if𝑛≥𝑘_{1−𝑛+𝑘+1}−𝛼^1𝑛+𝑘+3if𝑛<𝑘.(2.8) Moreover, this operator is continuous and the domain of the operator is the whole space 𝑐₀.

Proof. Let 𝛼₁ and 𝛼₂ be as it is stated in the theorem. From (1/𝑟)𝑆_𝜆𝑥=𝑦 we get to the system of equations: 𝑝𝑥₀+𝑥₁=𝑦₀𝑥₀+𝑝𝑥₁+𝑥₂=𝑦₁𝑥₁+𝑝𝑥₂+𝑥₃=𝑦₂⋮.(2.9) This is a nonhomogenous linear recurrence relation. Using the fact that (𝑥_𝑛),(𝑦_𝑛)∈𝑐₀, for (2.9) we can reach a solution with generating functions. This solution can be given by 𝑥_𝑛=1𝛼²¹−1^∞_𝑘=0𝑡_𝑛𝑘𝑦_𝑘,(2.10) where 𝑡_𝑛𝑘=𝛼^{1𝑛+1−𝑘}𝛼−𝛼^1𝑛+3+𝑘if𝑛≥𝑘_{1𝑘+1−𝑛}−𝛼^1𝑘+3+𝑛if𝑛<𝑘.(2.11) Let 𝑇=(𝑡_𝑛𝑘). We can see that by using Theorem 1.1, 𝑇∈𝐵(𝑐₀). So (1/𝛼²¹−1)𝑇 is the resolvent operator of (1/𝑟)𝑆_𝜆 and is continuous.

If 𝑇∶𝜇→𝜇 (𝜇 is ℓ₁ or 𝑐₀) is a bounded linear operator represented by the matrix 𝐴, then it is known that the adjoint operator 𝑇^∗∶𝜇^∗→𝜇^∗ is defined by the transpose 𝐴^𝑡 of the matrix 𝐴. It should be noted that the dual space 𝑐^∗0 of 𝑐₀ is isometrically isomorphic to the Banach space ℓ₁ and the dual space ℓ^∗1, of ℓ₁ is isometrically isomorphic to the Banach space ℓ_∞.

Corollary 2.4. 𝜎(𝑆,𝜇)⊂[𝑞−2𝑟,𝑞+2𝑟] for 𝜇∈{ℓ₁,𝑐₀,𝑐,ℓ_∞}.

Proof. 𝜎(𝑆,𝑐₀)=𝜎(𝑆^∗,𝑐^∗0)=𝜎(𝑆,ℓ₁)=𝜎(𝑆^∗,ℓ^∗1)=𝜎(𝑆,ℓ_∞). And by Cartlidge [14], if a matrix operator 𝐴 is bounded on 𝑐, then 𝜎(𝐴,𝑐)=𝜎(𝐴,ℓ_∞). Hence we have 𝜎(𝑆,𝑐₀)=𝜎(𝑆,ℓ₁)=𝜎(𝑆,ℓ_∞)=𝜎(𝑆,𝑐). What remains is to show that 𝜎(𝑆,𝑐₀)⊂[𝑞−2𝑟,𝑞+2𝑟]. By Theorem 2.3, there exists a resolvent operator of 𝑆_𝜆 which is continuous and the whole space 𝑐₀ is the domain if the roots of the polynomial 𝑃(𝑥)=𝑥²+𝑝𝑥+1 satisfy ||𝛼₂||>1>||𝛼₁||.(2.12) So, if 𝜆∈𝜎(𝑆,𝑐₀) then (2.12) is not satisfied. Since 𝛼₁𝛼₂=1, (2.12) is not satisfied means, the roots can be only of the form 𝛼₁=1𝛼₂=𝑒^𝑖𝜃(2.13) for some 𝜃∈[0,2𝜋). Then (𝑞−𝜆)/𝑟=𝑝=−(𝛼₁+𝛼₂)=−(𝑒^𝑖𝜃+𝑒^−𝑖𝜃)=−2cos𝜃. Hence 𝜆=𝑞+2𝑟cos𝜃, which means 𝜆 can be only on the line segment [𝑞−2𝑟,𝑞+2𝑟].

Theorem 2.5. 𝜎(𝑆,𝜇)=[𝑞−2𝑟,𝑞+2𝑟] for 𝜇∈{ℓ₁,𝑐₀,𝑐,ℓ_∞}.

Proof. By Theorem 2.2 and Corollary 2.4(𝑞−2𝑟,𝑞+2𝑟)⊂𝜎(𝑆,ℓ_∞)⊂[𝑞−2𝑟,𝑞+2𝑟]. Since the spectrum of a bounded linear operator over a complex Banach space is closed, we have 𝜎(𝑆,ℓ_∞)=[𝑞−2𝑟,𝑞+2𝑟]. And from the proof of Corollary 2.4 we have 𝜎(𝑆,ℓ₁)=𝜎(𝑆,𝑐₀)=𝜎(𝑆,𝑐)=𝜎(𝑆,ℓ_∞).

3. The Continuous Spectra and Residual Spectra

Lemma 3.1 (see [15, page 59]). 𝑇 has a dense range if and only if 𝑇^∗ is one to one.

Corollary 3.2. If 𝑇∈(𝜇∶𝜇) then 𝜎_𝑟(𝑇,𝜇)=𝜎_𝑝(𝑇^∗,𝜇^∗)⧵𝜎_𝑝(𝑇,𝜇).

Theorem 3.3. 𝜎_𝑟(𝑆,𝑐₀)=∅.

Proof. 𝜎_𝑝(𝑆,ℓ₁)=∅ by Theorem 2.1. Now using Corollary 3.2, we have 𝜎_𝑟(𝑆,𝑐₀)=𝜎_𝑝(𝑆^∗,𝑐^∗0)⧵𝜎_𝑝(𝑆,𝑐₀)=𝜎_𝑝(𝑆,ℓ₁)⧵𝜎_𝑝(𝑆,𝑐₀)=∅.

Theorem 3.4. 𝜎_𝑟(𝑆,ℓ₁)=(𝑞−2𝑟,𝑞+2𝑟).

Proof. Similarly as in the proof of the previous theorem, we have 𝜎_𝑟(𝑆,ℓ₁)=𝜎_𝑝(𝑆^∗,ℓ^∗1)⧵𝜎_𝑝(𝑆,ℓ₁)=𝜎_𝑝(𝑆,ℓ_∞)⧵𝜎_𝑝(𝑆,ℓ₁)=(𝑞−2𝑟,𝑞+2𝑟).

If 𝑇∶𝑐→𝑐 is a bounded matrix operator represented by the matrix 𝐴, then 𝑇^∗∶𝑐^∗→𝑐^∗ acting on ℂ⊕ℓ₁ has a matrix representation of the form 𝜒𝑏𝐴0^𝑡,(3.1) where 𝜒 is the limit of the sequence of row sums of 𝐴 minus the sum of the limits of the columns of 𝐴, and 𝑏 is the column vector whose 𝑘th entry is the limit of the 𝑘th column of 𝐴 for each 𝑘∈ℕ. For 𝑆_𝜆∶𝑐→𝑐, the matrix 𝑆^∗𝜆 is of the form 2𝑟+0𝑞−𝜆𝑆0_𝜆.(3.2)

Theorem 3.5. 𝜎_𝑟(𝑆,𝑐)={𝑞+2𝑟}.

Proof. Let 𝑥=(𝑥₀,𝑥₁,…)∈ℂ⊕ℓ₁ be an eigenvector of 𝑆^∗ corresponding to the eigenvalue 𝜆. Then we have (2𝑟+𝑞)𝑥₀=𝜆𝑥₀ and 𝑆𝑥′=𝜆𝑥′ where 𝑥′=(𝑥₁,𝑥₂,…). By Theorem 2.1,𝑥′=(0,0,…). Then 𝑥₀≠0. And 𝜆=2𝑟+𝑞 is the only value that satisfies (2𝑟+𝑞)𝑥₀=𝜆𝑥₀. Hence 𝜎_𝑝(𝑆^∗,𝑐^∗)={2𝑟+𝑞}. Then 𝜎_𝑟(𝑆,𝑐)=𝜎_𝑝(𝑆^∗,𝑐^∗)⧵𝜎_𝑝(𝑆,𝑐)={2𝑟+𝑞}.

Now, since the spectrum 𝜎 is the disjoint union of 𝜎_𝑝, 𝜎_𝑟, and 𝜎_𝑐, we can find 𝜎_𝑐 over the spaces ℓ₁, 𝑐₀, and 𝑐. So we have the following.

Theorem 3.6. For the operator 𝑆, one has the following: 𝜎𝜎_𝑐𝑆,ℓ₁={𝑞−2𝑟,𝑞+2𝑟},_𝑐𝜎𝑆,𝑐₀=[𝑞−2𝑟,𝑞+2𝑟_𝑐],(𝑆,𝑐)=[𝑞−2𝑟,𝑞+2𝑟).(3.3)

4. The Resolvent Operator

The following theorem is a generalization of Theorem 2.3.

Theorem 4.1. Let 𝜇∈{𝑐₀,𝑐,ℓ₁,ℓ_∞}. The resolvent operator 𝑆⁻¹ over 𝜇 exists and is continuous, and the domain of 𝑆⁻¹ is the whole space 𝜇 if and only if 0∉[𝑞−2𝑟,𝑞+2𝑟]. In this case, 𝑆⁻¹ has a matrix representation (𝑠_𝑛𝑘) defined by 𝑠_𝑛𝑘=1𝑟𝛼²¹−1⋅𝛼^{1𝑛−𝑘+1}𝛼−𝛼^1𝑛+𝑘+3if𝑛≥𝑘_{1−𝑛+𝑘+1}−𝛼^1𝑛+𝑘+3if𝑛<𝑘,(4.1) where 𝛼₁ is the root of the polynomial 𝑃(𝑥)=𝑟𝑥²+𝑞𝑥+𝑟 with |𝛼₁|<1.

Proof. Let 𝜇 be one of the sequence spaces in {𝑐₀,𝑐,ℓ₁,ℓ_∞}. Suppose 𝑆 has a continuous resolvent operator where the domain of the resolvent operator is the whole space 𝜇. Then 𝜆=0 is not in 𝜎(𝑆,𝜇)=[𝑞−2𝑟,𝑞+2𝑟]. Conversely if 0∉[𝑞−2𝑟,𝑞+2𝑟], then 𝑆 has a continuous resolvent operator, and since 𝑆 is bounded by Lemma 7.2-7.3 of [12] the domain of this resolvent operator is the whole space 𝜇.
Now, suppose 0∉[𝑞−2𝑟,𝑞+2𝑟]. Let 𝛼₁ and 𝛼₂ be the roots of the polynomial 𝑃(𝑥)=𝑟𝑥²+𝑞𝑥+𝑟 where |𝛼₁|≤|𝛼₂|. Since 0∉[𝑞−2𝑟,𝑞+2𝑟], by the proof of Corollary 2.4|𝛼₁|≠|𝛼₂|. Then |𝛼₁|<1<|𝛼₂|. So 𝑆 satisfies the conditions of Theorem 2.3. Hence the resolvent operator of 𝑆 is represented by the matrix 𝑆⁻¹=(𝑠_𝑛𝑘) defined by 𝑠_𝑛𝑘=1𝑟𝛼²¹−1⋅𝛼^{1𝑛−𝑘+1}𝛼−𝛼^1𝑛+𝑘+3if𝑛≥𝑘_{1−𝑛+𝑘+1}−𝛼^1𝑛+𝑘+3if𝑛<𝑘,(4.2) when 𝜇=𝑐₀ by that theorem. The matrix 𝑆⁻¹ is already a left inverse of the matrix 𝑆. Observe that 𝑆⁻¹ satisfies also the corresponding conditions of Theorem 1.1, which means 𝑆⁻¹∈(𝜇,𝜇) for 𝜇∈{𝑐,ℓ₁,ℓ_∞}. So, the matrix 𝑆⁻¹ is the representation of the resolvent operator also for the spaces in {𝑐,ℓ₁,ℓ_∞}.

Remark 4.2. If a matrix 𝐴 is a triangle, we can see that the resolvent (when it exists) is the unique lower triangular left hand inverse of 𝐴. In our case, 𝑆 is far away from being a triangle. The matrix 𝑆⁻¹ of this theorem is not the unique left inverse of the matrix 𝑆 for 0∉[𝑞−2𝑟,𝑞+2𝑟]. For example, the matrix 𝑇=(𝑡_𝑛𝑘) defined by 𝑡_𝑛𝑘=1𝑟𝛼²¹−1⋅𝛼^{1−𝑛+𝑘+1}0−𝛼^{1𝑛−𝑘+1}iiff𝑛𝑛<≥𝑘𝑘(4.3) is another left inverse of 𝑆. Then 𝜆𝑆⁻¹+(1−𝜆)𝑇 is also a left inverse of 𝑆 for any 𝜆∈ℂ, which means there exist infinitely many left inverses for 𝑆.

Theorem 4.3. Let 0∉[𝑞−2𝑟,𝑞+2𝑟], and, 𝛼₁ be the root of 𝑃(𝑥)=𝑟𝑥²+𝑞𝑥+𝑟 with |𝛼₁|<1. Then for 𝜇∈{𝑐₀,𝑐,ℓ₁,ℓ_∞} we have ‖‖𝑆⁻¹‖‖_{(𝜇∶𝜇)}=||𝛼¹||+||𝛼₁||²|𝑟|||1−𝛼²¹||1−||𝛼₁||.(4.4)

Proof. Since 𝑆⁻¹ is a symmetric matrix, the supremum of the ℓ₁ norms of the rows is equal to the supremum of the ℓ₁ norms of the columns. So, according to Theorem 1.1, what we need is to calculate the supremum of the ℓ₁ norms of the rows of 𝑆⁻¹. Denote the 𝑛th row 𝑆⁻¹ by 𝑆^𝑛−1 for 𝑛=0,1,…. Now, let us fix the row 𝑛 and calculate the ℓ₁ norm for this row. Let 𝜌=1/|𝑟(1−𝛼²¹)|. By using Theorem 4.1, we have ‖‖𝑆^𝑛−1‖‖_ℓ1=𝜌^𝑛_𝑘=0||𝛼^{1𝑛−𝑘+1}−𝛼^1𝑛+𝑘+3||+^∞_𝑘_=𝑛+1≤𝜌_𝑛||𝛼^{1−𝑛+𝑘+1}−𝛼^1𝑛+𝑘+3||_𝑘=0||𝛼¹||^{𝑛−𝑘+1}+||𝛼₁||^𝑛+𝑘+3+^∞_𝑘=𝑛+1||𝛼¹||^{−𝑛+𝑘+1}+||𝛼₁||^𝑛+𝑘+3=𝜌_𝑛_𝑘=0||𝛼₁||^{𝑛−𝑘+1}+^∞_𝑘_=𝑛+1||𝛼₁||^{−𝑛+𝑘+1}+^∞_𝑘=0||𝛼₁||^𝑛+𝑘+3=𝜌_𝑛+1_𝑘=1||𝛼₁||^𝑘+^∞_𝑘=2||𝛼₁||^𝑘+^∞_𝑘_=𝑛+3||𝛼₁||^𝑘=𝜌2_∞_𝑘=0||𝛼₁||^𝑘−2−||𝛼₁||−||𝛼₁||^𝑛+2≤𝜌2_∞_𝑘=0||𝛼₁||^𝑘−2−||𝛼₁||=||𝛼¹||+||𝛼₁||²|𝑟|||1−𝛼²¹||1−||𝛼₁||.(4.5) Hence ‖‖𝑆⁻¹‖‖_{(𝜇∶𝜇)}=sup_𝑛‖‖𝑆^𝑛−1‖‖_ℓ1≤||𝛼¹||+||𝛼₁||²|𝑟|||1−𝛼²¹||1−||𝛼₁||.(4.6) On the other hand ‖‖𝑆^𝑛−1‖‖_ℓ1=𝜌^𝑛_𝑘=0||𝛼^{1𝑛−𝑘+1}−𝛼^1𝑛+𝑘+3||+^∞_𝑘_=𝑛+1≥𝜌_𝑛||𝛼^{1−𝑛+𝑘+1}−𝛼^1𝑛+𝑘+3||_𝑘=0||𝛼¹||^{𝑛−𝑘+1}−||𝛼₁||^𝑛+𝑘+3+^∞_𝑘=𝑛+1||𝛼¹||^{−𝑛+𝑘+1}−||𝛼₁||^𝑛+𝑘+3=𝜌_𝑛_𝑘=0|𝛼₁|^{𝑛−𝑘+1}+^∞_𝑘_=𝑛+1||𝛼₁||^{−𝑛+𝑘+1}−^∞_𝑘=0||𝛼₁||^𝑛+𝑘+3=𝜌_𝑛+1_𝑘=1||𝛼₁||^𝑘+^∞_𝑘=2||𝛼₁||^𝑘−^∞_𝑘_=𝑛+3||𝛼₁||^𝑘.(4.7) Then ‖‖𝑆⁻¹‖‖_{(𝜇∶𝜇)}=sup_𝑛‖‖𝑆^𝑛−1‖‖_ℓ1≥lim_𝑛‖‖𝑆^𝑛−1‖‖_ℓ1=𝜌^∞_𝑘=1||𝛼₁||^𝑘+^∞_𝑘=2||𝛼₁||^𝑘=||𝛼¹||+||𝛼₁||²|𝑟|||1−𝛼²¹||1−||𝛼₁||.(4.8)

Acknowledgment

The author thanks the referees for their careful reading of the original paper and for their valuable comments.