# Fine Spectra of Symmetric Toeplitz Operators

Authors: Muhammed Altun
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The fine spectra of 2-banded and 3-banded infinite Toeplitz matrices were examined by several authors. The fine spectra ofn-banded triangular Toeplitz matrices and tridiagonal symmetric matrices were computed in the following papers: Altun, “On the fine spectra of triangular toeplitz operators” (2011) and Altun, “Fine spectra of tridiagonal symmetric matrices” (2011). Here, we generalize those results to the (2n+1)-banded symmetric Toeplitz matrix operators for arbitrary positive integer n.


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## Body

## 1. Introduction and Preliminaries

The spectrum of an operator over a Banach space is partitioned into three parts, which are the point spectrum, the continuous spectrum, and the residual spectrum. Some other parts also arise by examining the surjectivity of the operator and continuity of the inverse operator. Such subparts of the spectrum are called the fine spectra of the operator.The spectra and fine spectra of linear operators defined by some particular limitation matrices over some sequence spaces were 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 ℓp. 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. Akhmedov and Başar [7, 8] have determined the fine spectrum of the Cesàro operator over the sequence spaces c0, ℓ∞ and ℓp. Altun and Karakaya [9] computed the fine spectra of Lacunary matrices over c0 and c. 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 [11] computed the fine spectra of triangular Toeplitz matrices over c0 and c.The fine spectrum of the difference operatorΔ over c0 and c was studied by Altay and Başar [12]. Recently, the fine spectra of Δ over ℓp and bvp are studied by Akhmedov and Başar [13, 14], where bvp is the space of p-bounded variation sequences, introduced by Başar and Altay [15] with 1≤p<∞. The fine spectrum with respect to the Goldberg's classification of the operator B(r,s,t) over ℓp and bvp with 1<p<∞ has recently been studied by Furkan et al. [16]. Quite recently, Akhmedov and El-Shabrawy [17] have obtained the fine spectrum of the generalized difference operator Δa,b, defined as a double band matrix with the convergent sequences a~=(ak) and b~=(bk) having certain properties, over c. In 2010, Srivastava and Kumar [18] have determined the spectra and the fine spectra of the generalized difference operator Δν on ℓ1, where Δν is defined by (Δν)nn=νn and (Δν)n+1,n=-νn for all n∈ℕ, under certain conditions on the sequence ν=(νn) and they have just generalized these results by the generalized difference operator Δuv defined by Δuvx=(unxn+vn-1xn-1)n∈ℕ (see [19]).In this work, our purpose is to determine the spectra of the operator, for which the corresponding matrix is a(2n+1)-banded symmetric Toeplitz matrix, over the sequence spaces c0,c,ℓ1 and ℓ∞. We will also give the fine spectra results for the spaces c0 and c.LetX 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,
(1.1)R(T)={y∈Y:y=Tx;x∈X}.
By B(X), we denote the set of all bounded linear operators on X into itself. If X is any Banach space and T∈B(X) then the adjointT* 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
(1.2)Tλ=T-λI,
where λ is a complex number and I is the identity operator on 𝒟(T). If Tλ has an inverse, which is linear, we denote it by Tλ-1, that is,
(1.3)Tλ-1=(T-λI)-1,
and call it the resolvent operator of Tλ. If λ=0, we will simply write T-1. Many properties of Tλ and Tλ-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 Tλ-1 exists. Boundedness of Tλ-1 is another property that will be essential. We will also ask for what λ’s the domain of Tλ-1 is dense in X. For our investigation of T, Tλ and Tλ-1, we need some basic concepts in spectral theory which are given as follows (see [20, pages 370-371]).LetX≠{θ} be a complex normed space and T:𝒟(T)→X be a linear operator with domain 𝒟(T)⊂X. A regular valueλ of T is a complex number such that(R1)
Tλ-1 exists,(R2)
Tλ-1 is bounded,(R3)
Tλ-1 is defined on a set which is dense in X.Theresolvent setρ(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 spectrumσp(T) is the set such that Tλ-1 does not exist. A λ∈σp(T) is called an eigenvalue of T. The continuous spectrumσc(T) is the set such that Tλ-1 exists and satisfies (R3) but not (R2). The residual spectrumσr(T) is the set such that Tλ-1 exists but does not satisfy (R3).From Goldberg [21], if T∈B(X), X a Banach space, then there are three possibilities for ℛ(T), the range of T:(I)
ℛ(T)=X,(II)
ℛ(T)¯=X, but ℛ(T)≠X,(III)
ℛ(T)¯≠X,and three possibilities for T-1:(1)
T-1 exists and is continuous,(2)
T-1 exists, but is discontinuous,(3)
T-1 does not exist.If these possibilities are combined in all possible ways, nine different states are created. These are labelled as I1, I2, I3, II1, II2, II3, III1, III2, and III3. If λ is a complex number such that Tλ∈I1 or Tλ∈II1, then λ is in the resolvent set ρ(T,X) of T, the set of all regular values of T on X. The other classification gives rise to the fine spectrum of T. For example, we will write λ∈III1σ(T,X) if T satisfies III and 1.A triangle is a lower triangular matrix with all of the principal diagonal elements nonzero. We will writeℓ∞, c and c0 for the spaces of all bounded, convergent, and null sequences, respectively. By ℓp, we denote the space of all p-absolutely summable sequences, where 1≤p<∞. Let μ and γ be two sequence spaces and A=(ank) be an infinite matrix of real or complex numbers ank, 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=(xk)∈μ the sequence Ax={(Ax)n}, the A-transform of x, is in γ, where
(1.4)(Ax)n=∑k‍ankxk(n∈N).
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.Let an(n+1)-tuple t=(t0,t1,…,tn)∈ℂn+1 be given. A symmetric infinite Toeplitz matrix is a (2n+1)-band matrix of the form
(1.5)S=S(t)=[t0t1⋯⋯tn00⋯t1t0t1⋯⋯tn0⋯⋮t1t0t1⋯⋯tn⋯⋮⋮t1t0t1⋯⋯⋯tn⋮⋮t1t0t1⋯⋯0tn⋮⋮t1t0t1⋯00tn⋮⋮t1t0⋯⋮⋮⋮⋮⋮⋮⋮⋱].
The spectral results are clear when S is a multiple of the identity matrix, so for the sequel we will have n≥1 and tn≠0.LetR be the right shift operator:
(1.6)R=[00000⋯10000⋯01000⋯⋮⋮⋮⋮⋮⋱],
and L be the left shift operator:
(1.7)L=Rt=R-1.
Let F(z)=tn[zn+z-n]+tn-1[zn-1+z-(n-1)]+⋯+t1[z+z-1]+t0=P(z)/zn, where P is the palindromic polynomial P(z)=tnz2n+tn-1z2n-1+⋯+t0zn+t1zn-1+t2zn-2+⋯+tn. Then, we can see that S=F(R) and we will call F and P as the function and polynomial associated to the operator S, respectively. We also have
(1.8)S=LnP(R).
The roots of P(z) are nonzero and symmetric, that is, if α is a root, α-1 is also a root. Let α1,α2,…,αn,α1-1,α2-1,…,αn-1 be the roots of P(z) such that |αk|≤1 for k=1,2,…,n. Then
(1.9)S=tnLn(R-α1I)(R-α2I)⋯(R-αnI)(R-α1-1I)(R-α2-1I)⋯(R-αn-1I).
Now, by induction, we can see that
(1.10)Ln(R-α1I)(R-α2I)⋯(R-αnI)=(I-α1L)(I-α2L)⋯(I-αnL).
Let D be the unit disc {z∈ℂ:|z|≤1} and ∂D be the unit circle {z∈ℂ:|z|=1}. We have the following two lemmas as a consequence of the corresponding results in [22] and [23], respectively.Lemma 1.1.
(
I
-
α
L
)
∈
(
c
0
,
c
0
) is onto if and only if α is not on the unit circle.Lemma 1.2.
(
R
-
α
I
)
∈
(
c
0
,
c
0
) is onto if and only if α is outside the unit disc.Theorem 1.3.
S
∈
(
c
0
,
c
0
) is onto if and only if P has no root on the unit circle.Proof.
SupposeP has a root on the unit circle. Let α1,α2,…,αn,α1-1,α2-1,…,αn-1 be the roots of P(z) such that |αk|≤1 for k=1,2,…,n. We have
(1.11)S=tn(I-α1L)(I-α2L)⋯(I-αnL)(R-α1-1I)(R-α2-1I)⋯(R-αn-1I).
Since the matrix operators (I-α1L),(I-α2L),…,(I-αnL) commute with each other, without loss of generality, we can suppose α1 is a root on the unit circle. Clearly, all the operators (I-α1L),(I-α2L),…,(I-αnL),(R-α1-1I),(R-α2-1I),…,(R-αn-1I) are in (c0,c0). But, by Lemma 1.1 the operator (I-α1L) is not onto. So, S cannot be onto.
Suppose, now,P has no root on the unit circle. That means |αk|<1 for k=1,2,…,n. Then all the operators (I-α1L),(I-α2L),…,(I-αnL),(R-α1-1I),(R-α2-1I),…,(R-αn-1I) are onto by Lemma 1.1 and Lemma 1.2. Hence, S=tn(I-α1L)(I-α2L)⋯(I-αnL)(R-α1-1I)(R-α2-1I)⋯(R-αn-1I) is onto.Theorem 1.4 (cf. [24]).
LetT be an operator with the associated matrix A=(ank).(i)
T∈B(c) if and only if
(1.12)‖A‖:=sup⁡n∑k=1∞‍|ank|<∞,(1.13)ak:=lim⁡n→∞ankexistsforeachk,(1.14)a:=lim⁡n→∞∑k=1∞‍ankexists.(ii)
T∈B(c0) if and only if (1.12) and (1.13) with ak=0 for each k.(iii)
T∈B(ℓ∞) if and only if (1.12). In these cases, the operator norm of T is
(1.15)‖T‖(l∞:l∞)=‖T‖(c:c)=‖T‖(c0:c0)=‖A‖.(iv)
T∈B(ℓ1) if and only if
(1.16)‖At‖=sup⁡k∑n=1∞‍|ank|<∞.In this case, the operator norm of T is ∥T∥(ℓ1:ℓ1)=∥At∥.Corollary 1.5.
S
(
t
)
∈
B
(
μ
) for μ∈{c0,c,ℓ1,ℓ∞} and
(1.17)‖S(t)‖(μ,μ)=|t0|+2(|t1|+|t2|+⋯+|tn|).Theorem 1.6.
LetX be a Banach space and T∈B(X). Then λ∈ℂ is in the spectrum σ(T,X) if and only if T-λI is not bijective.Proof.
SupposeT-λI is not bijective. Then T-λI is not 1-1 or not onto. If it is not 1-1, then λ∈σp(T,X)⊂σ(T,X). Suppose now T-λI is 1-1. Then it is not onto and by Lemma 7.2-3 of [20], λ cannot be in ρ(T,X). Hence, λ∈σ(T,X).
Now, supposeT-λI is bijective. Then by the open mapping theorem (T-λI)-1 is continuous. Hence, λ is not in the spectrum σ(T,X).Corollary 1.7.
LetX be a Banach space and T∈B(X). Then λ∈ρ(T,X) if and only if Tλ is bijective.


## 2. The Spectra and Fine Spectra

Lemma 2.1 (Lemma 3.4 of [11]).
Letz1,z2,…,zr be distinct complex numbers with |zi|=1 for 1≤i≤r. Let 0≠x=(xk) be a sequence satisfying
(2.1)xk=(α1,0+α1,1k+⋯+α1,m1-1km1-1)z1k+(α2,0+α2,1k+⋯+α2,m2-1km2-1)z2k+⋯+(αr,0+αr,1k+⋯+αr,mr-1kmr-1)zrk,
for k=0,1,2,…, where αi,j are constants forming the polynomials Pi(k)=αi,0+αi,1k+⋯+αi,mi-1kmi-1≠0 for 1≤i≤r and 0≤j≤mi-1. Then x∉c0.Lemma 2.2.
Letz1,z2,…,zr be distinct complex numbers. Let 0≠x=(xk)∈c be a sequence satisfying
(2.2)xk=(α1,0+α1,1k+⋯+α1,m1-1km1-1)z1k+(α2,0+α2,1k+⋯+α2,m2-1km2-1)z2k+⋯+(αr,0+αr,1k+⋯+αr,mr-1kmr-1)zrk,
for k=0,1,2,…, where αi,j are constants forming the polynomials Pi(k)=αi,0+αi,1k+⋯+αi,mi-1kmi-1≠0 for 1≤i≤r and 0≤j≤mi-1. Then |zi|≤1 for 1≤i≤r, and the existence of a t≤r with |zt|=1 implies zt=1 and Pt is a constant.Proof.
Let|z1|≥|z2|≥⋯≥|zr|. To prove |zi|≤1 for all i≤r, suppose it is not true. Then let a:=|z1|>1. Let s≤r be the largest positive integer with |z1|=|z2|=⋯=|zs|. Then (xk/ak)∈c0. Let
(2.3)uk=(α1,0+α1,1k+⋯+α1,m1-1km1-1)(z1a)k+(α2,0+α2,1k+⋯+α2,m2-1km2-1)(z2a)k+⋯+(αs,0+αs,1k+⋯+αs,ms-1kms-1)(zsa)k.
We have 0=(xk/ak-uk) for s=r. If s<r, we have
(2.4)xkak-uk=(αs+1,0+αs+1,1k+⋯+αs+1,ms+1-1kms+1-1)(zs+1a)k+(αs+2,0+αs+2,1k+⋯+αs+2,ms+2-1kms+2-1)(zs+2a)k+⋯+(αr,0+αr,1k+⋯+αr,mr-1kmr-1)(zra)k.
Since |zj/a|<1 for s+1≤j≤r, we have (xk/ak-uk)∈c0. Then (uk)∈c0 but this contradicts with Lemma 2.1. Hence, we have |zi|≤1 for 1≤i≤r.
Now, let us prove the second part. Suppose, there exist a positive integerq≤r such that |zi|=1 for all i≤q. For any i, Pi is constant means mi=1. Suppose m=max{m1,…,mq}>1. Without loss of generality let m1≥m2≥⋯≥mq. Let q0≤q be the largest integer satisfying m1=m2=⋯=mq0=m. Then (xk/km-1)∈c0. This means, since |zi| have modulus less than 1, (vk)∈c0, where
(2.5)vk=α1,m-1z1k+α2,m-1z2k+⋯+αq0,m-1zq0k.
But, this again contradicts with Lemma 2.1. Hence, we have m1=m2=⋯=mq=1. Now, we have (wk)∈c, where
(2.6)wk=α1,0z1k+α2,0z2k+⋯+αq,0zqk.
Suppose, one of the elements in {z1,z2,…,zq} is equal to 1, say z1=1. Then, (wk+1-wk)∈c0, where
(2.7)wk+1-wk=α2,0(z2-1)z2k+⋯+αq,0(zq-1)zqk,
and this again contradicts with Lemma 2.1. Hence, q≤1 and 1 is the unique candidate for zt with modulus 1.Theorem 2.3.
σ
p
(
S
,
μ
)
=
∅ for μ∈{ℓ1,c0,c}.Proof.
Sinceℓ1⊂c0⊂c, it is enough to show that σp(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:
(2.8)t0x0+t1x1+t2x2+⋯+tnxn=λx0t1x0+t0x1+t1x2+⋯+tnxn+1=λx1t2x0+t1x1+t0x2+⋯+tnxn+2=λx2⋮tnx0+tn-1x1+tn-2x2+⋯+tnx2n=λxntnx1+tn-1x2+tn-2x3+⋯+tnx2n+1=λxn+1tnx2+tn-1x3+tn-2x4+⋯+tnx2n+2=λxn+2⋮
Since tn≠0 we can write this system of equations in the form:
(2.9)d0x0+d1x1+d2x2+⋯+dnxn=0d1x0+d0x1+d1x2+⋯+dnxn+1=0d2x0+d1x1+d0x2+⋯+dnxn+2=0⋮dnx0+dn-1x1+dn-2x2+⋯+dnx2n=0dnx1+dn-1x2+dn-2x3+⋯+dnx2n+1=0dnx2+dn-1x3+dn-2x4+⋯+dnx2n+2=0⋮
where d0=(t0-λ)/tn, dn=1 and dj=tj/tn for 1≤j≤n-1. This system of equations, by change of variables un+k=xk for k=0,1,2,…, is equivalent to system of equations
(2.10)dnuk+dn-1uk+1+⋯+d0un+k+d1un+k+1+⋯+dnu2n+k=0,k=0,1,2,…,
with the initial conditions u0=u1=⋯=un-1=0.
We see that this is a2n-th order linear homogenous difference equation with the corresponding characteristic polynomial
(2.11)P(z)=z2n+dn-1z2n-1+⋯+d1zn+1+d0zn+d1zn-1+⋯+dn-1z+1.
Suppose P(z) has r distinct roots z1,z2,…,zr with multiplicities m1,m2,…,mr. Then, any solution (uk) of the system of equations satisfies
(2.12)uk=(α1,0+α1,1k+⋯+α1,m1-1km1-1)z1k+(α2,0+α2,1k+⋯+α2,m2-1km2-1)z2k+⋯+(αr,0+αr,1k+⋯+αr,mr-1kmr-1)zrk.
Observe that if ζ is a root of P(z), then 1/ζ is also a root. There are two cases.
Case 1 (1 is not a root of P(z)). Since (uk)∈c, by Lemma 2.2 we can write
(2.13)uk=(α1,0+α1,1k+⋯+α1,m1-1km1-1)z1k+(α2,0+α2,1k+⋯+α2,m2-1km2-1)z2k+⋯+(αq,0+αq,1k+⋯+αq,mq-1kmq-1)zqk,
where |zi|<1 for 1≤i≤q. By the symmetry of the roots we have m1+m2+m3+⋯+mq=n. Now, using the initial conditions u0=u1=⋯=un-1=0 we have
(2.14)Vα=0,
where α=[α1,0,α1,1,…,α1,m1-1,α2,0,α2,1,…,α2,m2-1,…,αq,0,αq,1,…,αq,mq-1]t and V is the generalized Vandermonde matrix(2.15)[10⋯010⋯0⋯⋯10⋯0z1z1⋯z1z2z2⋯z2⋯⋯zqzq⋯zqz122z12⋯2m1-1z12z222z22⋯2m2-1z22⋯⋯zq22zq2⋯2mq-1zq2z133z13⋯3m1-1z13z233z23⋯3m2-1z23⋯⋯zq33zq3⋯3mq-1zq3⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮z1n-1A⋯Bz2n-1C⋯D⋯⋯zqn-1E⋯F],
where 𝔄 denotes (n-1)z1n-1,  𝔅 denotes (n-1)m1-1z1n-1,  ℭ denotes (n-1)z2n-1,  𝔇 denotes (n-1)m2-1z2n-1,  𝔈 denotes (n-1)zqn-1, and 𝔉 denotes (n-1)mq-1zqn-1.
The determinant of the matrixV was explicitly given in [25, 26]:
(2.16)det⁡V=[∏i=1q‍(∏j=0mi-1‍j!)zi(mi2)][∏1≤i<j≤q‍(zj-zi)mimj].
An inductive proof of this formula is given by Chen and Li [27]. Since zero is not a root of our polynomial P, we have detV≠0; hence, we conclude α=0, which means the sequences (uk)=0 and (xk)=0. Hence, there is no eigenvalue in this case.
Case 2 (1 is a root of P(z)). Since (uk)∈c, by Lemma 2.2 we can write
(2.17)uk=α1,01k+(α2,0+α2,1k+⋯+α2,m2-1km2-1)z2k+⋯+(αq,0+αq,1k+⋯+αq,mq-1kmq-1)zqk,
where |zi|<1 for 2≤i≤q. By the symmetry of the roots we have p:=1+m2+m3+⋯+mk≤n. Now, using the initial conditions u0=u1=⋯=un-1=0 we have
(2.18)Wα=0,
where α=[α1,0,α2,0,α2,1,…,α2,m2-1,…,αq,0,αq,1,…,αq,mq-1]t and W is an n×p submatrix of a generalized n×n Vandermonde matrix. Since the determinant of generalized Vandermonde matrix with nonzero roots is not zero, we have that the columns of W are linearly independent. So again we can conclude that α=0, which again will mean that there is no eigenvalue.IfT: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 At of the matrix A. It should be noted that the dual space c0* of c0 is isometrically isomorphic to the Banach space ℓ1 and the dual space ℓ1* of ℓ1 is isometrically isomorphic to the Banach space ℓ∞.Lemma 2.4 (see [21, page 59]).
T has a dense range if and only if T* is one to one.Corollary 2.5.
IfT∈(μ:μ) then σr(T,μ)=σp(T*,μ*)∖σp(T,μ).Theorem 2.6.
σ
r
(
S
,
c
0
)
=
∅.Proof.
σ
p
(
S
,
ℓ
1
)
=
∅ by Theorem 2.3. Now using Corollary 2.5 we have σr(S,c0)=σp(S*,c0*)∖σp(S,c0)=σp(S,ℓ1)∖σp(S,c0)=∅.IfT: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
(2.19)[χ0bAt],
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:c→c, the matrix S* is of the form
(2.20)[2(t1+t2+⋯+tn)+t000S]=[F(1)00S].Theorem 2.7.
σ
r
(
S
,
c
)
=
{
t
0
+
2
(
t
1
+
t
2
+
⋯
+
t
n
)
}
=
{
F
(
1
)
}.Proof.
Letx=(x0,x1,…)∈ℂ⊕ℓ1 be an eigenvector of S* corresponding to the eigenvalue λ. Then we have (2[t1+t2+⋯+tn]+t0)x0=λx0 and Sx′=λx′ where x′=(x1,x2,…). By Theorem 2.3  x′=(0,0,…). Then x0≠0. So λ=2[t1+t2+⋯+tn]+t0 is the only value that satisfies (2[t1+t2+⋯+tn]+t0)x0=λx0. Hence, σp(S*,c*)={2[t1+t2+⋯+tn]+t0}. Then σr(S,c)=σp(S*,c*)∖σp(S,c)={2[t1+t2+⋯+tn]+t0}.We will writeF(1) instead of {F(1)} for the sequel.Lemma 2.8.
σ
(
S
,
ℓ
1
)
=
σ
(
S
,
c
0
)
=
σ
(
S
,
c
)
=
σ
(
S
,
ℓ
∞
).Proof.
We will use the fact that the spectrum of a bounded operator over a Banach space is equal to the spectrum of the adjoint operator. The adjoint operator is the transpose of the matrix forc0 and ℓ1. So σ(S,c0)=σ(S*,c0*)=σ(S,ℓ1)=σ(S*,ℓ1*)=σ(S,ℓ∞). We know by Cartlidge [28] that 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).Theorem 2.9.
σ
(
S
,
μ
)
=
F
(
∂
D
) for μ∈{ℓ1,c0,c,ℓ∞}.Proof.
Let us first considerS as an operator on c0. By Theorems 1.6 and 2.3  λ∈σ(S,c0) if and only if S-λI is not onto over c0. By Theorem 1.3  S-λI is not onto over c0 if and only if the polynomial P(z)-λzn has a root on the unit circle. P(z)-λzn has a root on the unit circle if and only if λ=P(z)/zn=F(z) for some z∈∂D. We have λ=F(z) for some z∈∂D if and only if λ∈F(∂D). Hence, σ(S,c0)=F(∂D). Finally, we apply Lemma 2.8.Corollary 2.10.
S
∈
(
c
,
c
) is onto if and only if P has no root on the unit circle.The spectrumσ is the disjoint union of σp, σr and σc, so we have the following theorem as a consequence of Theorems 2.3, 2.6, 2.7, and 2.9.Theorem 2.11.
σ
c
(
S
,
c
0
)
=
F
(
∂
D
) and σc(S,c)=F(∂D)∖F(1).As a result of Theorems1.3, 2.3, 2.6, 2.7, and 2.9 and Corollary 2.10, we have the following.Theorem 2.12.
F
(
∂
D
)
=
I
I
2
σ
(
S
,
c
0
), F(∂D)∖F(1)=II2σ(S,c) and F(1)=III2σ(S,c).


## 3. Some Applications

Now, let us give an application of Theorem2.9. Consider the system of equations
(3.1)yk=t0xk+∑j=1n‍tj(xk+j+xk-j)k=0,1,2,…,
where xk=0 for negative k.Theorem 3.1.
LetP(z)=tnz2n+tn-1z2n-1+⋯+t0zn+t1zn-1+t2zn-2+⋯+tn, where t0,t1,…,tn are complex numbers such that the complex sequences x=(xn) and y=(yn) are solutions of system (3.1). Then the following are equivalent:(i)
boundedness of(yn) always implies a unique bounded solution (xn),(ii)
convergence of(yn) always implies a unique convergent solution (xn),(iii)
yn→0 always implies a unique solution (xn) with xn→0,(iv)
∑‍|yn|<∞ always implies a unique solution (xn) with ∑‍|xn|<∞,(v)
P has no root on the unit circle ∂D.Proof.
The system of equations (3.1) holds, so we have Sx=y. Then P is the polynomial associated to S. Let F be the function associated to S. Let us prove only (i)⇔(v) and omit the proofs of (ii)⇔(v), (iii)⇔(v), (iv)⇔(v) since they are similarly proved. Suppose boundedness of (yn) always implies a unique bounded solution (xn). Then the operator S-0I=S∈(ℓ∞,ℓ∞) is bijective. So, λ=0 is not in the spectrum σ(S,ℓ∞) by Theorem 1.6, which means 0∉F(∂D) and 0∉P(∂D).
For the reverse implication, supposeP(z) has no root on the unit circle ∂D. Then F(z) has no zero on the unit circle. So, λ=0 is in the resolvent set ρ(S,ℓ∞). Now, by Theorem 1.6, S=S-0I is bijective on ℓ∞, which means that the boundedness of (yn) implies a bounded unique solution (xn).Example 3.2.
We can see that(3.2)F(∂D)={t0+2t1cos⁡θ+2t2cos⁡2θ+⋯+2tncos⁡nθ:θ∈[0,π]}.
When n=1, S is a tridiagonal matrix, that is,
(3.3)S=[qr0000⋯rqr000⋯0rqr00⋯00rqr0⋯⋮⋮⋮⋮⋮⋮⋱],
then σ(S,μ)=F(∂D) for μ∈{ℓ1,c0,c,ℓ∞}, where F=q+r(z+z-1). Therefore,
(3.4)σ(S,μ)={q+2rcos⁡θ:θ∈[0,π]}=[q-2r,q+2r],
which is one of the main results of [29]. [q-2r,q+2r] is the closed line segment in the complex plane with endpoints q-2r and q+2r.Example 3.3.
Whenn=2, S is a pentadiagonal matrix, that is,
(3.5)S=[qrs0000⋯rqrs000⋯srqrs00⋯0srqrs0⋯00srqrs⋯⋮⋮⋮⋮⋮⋮⋮⋱],
then F=q+r(z+z-1)+s(z2+z-2) and
(3.6)σ(S,μ)={q+2rcos⁡θ+2scos⁡2θ:θ∈[0,π]}.
So the spectrum is a line segment if r is a real multiple of s. It can be proved that, the spectrum is a closed connected part of a parabola if r is not a real multiple of s. For example, if q=r=1 and s=i (the complex number i) we have
(3.7)σ(S,μ)={1+2cos⁡θ+2icos⁡2θ:θ∈[0,π]}={(x,y)∈R2:y=x2-2x-1,x∈[-1,3]}.



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*Source: 932785-2012-07-19.xml*