# Fine Spectra of Tridiagonal Symmetric Matrices

Authors: Muhammed Altun
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## 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 spacesc0, c, ℓ1, and ℓ∞.


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## Body

## 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 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. 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 ℓp. 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 spacesc0, 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.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 let 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 shall 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 [12, pages 370-371]).LetX≠{θ} be a complex normed space, and let 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, and(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).A triangle is a lower triangular matrix with all of the principal diagonal elements nonzero. We shall writeℓ∞, c, and c0 for the spaces of all bounded, convergent, and null sequences, respectively. And 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.A tridiagonal symmetric infinite matrix is of the form(1.5)S=S(q,r)=[qr0000⋯rqr000⋯0rqr00⋯00rqr0⋯⋮⋮⋮⋮⋮⋮⋱],
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]).
LetT be an operator with the associated matrix A=(ank).(i)
T∈B(c) if and only if
(1.6)‖A‖∶=sup⁡n∑k=1∞|ank|<∞,(1.7)ak:=lim⁡n→∞ankexists for eachk,(1.8)a:=lim⁡n→∞∑k=1∞ankexists.(ii)
T∈B(c0) if and only if (1.6) and (1.7) with ak=0 for each k.(iii)
T∈B(ℓ∞) if and only if (1.6).In these cases, the operator norm of T is
(1.9)‖T‖(l∞:l∞)=‖T‖(c:c)=‖T‖(c0:c0)=‖A‖.(iv)
T∈B(ℓ1) if and only if
(1.10)‖At‖=sup⁡k∑n=1∞|ank|<∞.
In this case, the operator norm of T is ∥T∥(ℓ1:ℓ1)=∥At∥.Corollary 1.2.
Letμ∈{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.
σp(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:
(2.1)qx0+rx1=λx0rx0+qx1+rx2=λx1rx1+qx2+rx3=λx2⋮.
If x0=0, then xk=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
(2.2)xn+pxn-1+xn-2=0forn≥2,
where p=(q-λ)/r. The characteristic polynomial of the recurrence relation is
(2.3)x2+px+1=0.
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
(2.4)xn=(A+Bn)(α)n=A+Bn,
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 xn=n+1. This means (xn)∉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 xn=(n+1)(-1)n. So, there is no eigenvalue in this case.Case 3 (p≠±2).
Then the characteristic polynomial has two distinct rootsα1≠±1 and α2≠±1 with α1α2=1. Let |α1|≥1≥|α2|. The solution of the recurrence relation is of the form
(2.5)xn=A(α1)n+B(α2)n.
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
(2.6)xn=α1n+1-α2n+1α1-α2.
If |α1|>1>|α2|, then
(2.7)|xn|≥1|α1-α2|(|α1|n+1-|α2|n+1).So limn|xn|=∞, which means (xn)∉c. Now, if |α1|=|α2|=1, then there exists θ∈(0,π) such that α1=eiθ and α2=e-iθ. So, xn=[sin(n+1)θ]/sinθ. Again we have (xn)∉c. Hence there is no eigenvalue also in this case.Repeating all the steps in the proof of this theorem forℓ∞, we get to the following.Theorem 2.2.
σp(S,ℓ∞)=(q-2r,q+2r).Theorem 2.3.
Letp=(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 Sλ-1=(snk), where
(2.8)snk=1r(α12-1)⋅{α1n-k+1-α1n+k+3ifn≥kα1-n+k+1-α1n+k+3ifn<k.
Moreover, this operator is continuous and the domain of the operator is the whole space c0.Proof.
Letα1 and α2 be as it is stated in the theorem. From (1/r)Sλx=y we get to the system of equations:
(2.9)px0+x1=y0x0+px1+x2=y1x1+px2+x3=y2⋮.
This is a nonhomogenous linear recurrence relation. Using the fact that (xn),(yn)∈c0, for (2.9) we can reach a solution with generating functions. This solution can be given by
(2.10)xn=1α12-1∑k=0∞tnkyk,
where
(2.11)tnk={α1n+1-k-α1n+3+kifn≥kα1k+1-n-α1k+3+nifn<k.
Let T=(tnk). We can see that by using Theorem 1.1, T∈B(c0). So (1/α12-1)T is the resolvent operator of (1/r)Sλ and is continuous.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∞.Corollary 2.4.
σ(S,μ)⊂[q-2r,q+2r] for μ∈{ℓ1,c0,c,ℓ∞}.Proof.
σ(S,c0)=σ(S*,c0*)=σ(S,ℓ1)=σ(S*,ℓ1*)=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 Sλ which is continuous and the whole space c0 is the domain if the roots of the polynomial P(x)=x2+px+1 satisfy
(2.12)|α2|>1>|α1|.
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
(2.13)α1=1α2=eiθ
for some θ∈[0,2π). Then (q-λ)/r=p=-(α1+α2)=-(eiθ+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 Theorem2.2 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.
IfT∈(μ:μ) then σr(T,μ)=σp(T*,μ*)∖σp(T,μ).Theorem 3.3.
σr(S,c0)=∅.Proof.
σp(S,ℓ1)=∅ by Theorem 2.1. Now using Corollary 3.2, we have σr(S,c0)=σp(S*,c0*)∖σp(S,c0)=σp(S,ℓ1)∖σp(S,c0)=∅.Theorem 3.4.
σr(S,ℓ1)=(q-2r,q+2r).Proof.
Similarly as in the proof of the previous theorem, we haveσr(S,ℓ1)=σp(S*,ℓ1*)∖σp(S,ℓ1)=σp(S,ℓ∞)∖σp(S,ℓ1)=(q-2r,q+2r).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 (3.1)[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 Sλ:c→c, the matrix Sλ* is of the form (3.2)[2r+q-λ00Sλ].Theorem 3.5.
σr(S,c)={q+2r}.Proof.
Letx=(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 σp(S*,c*)={2r+q}. Then σr(S,c)=σp(S*,c*)∖σp(S,c)={2r+q}.Now, since the spectrumσ is the disjoint union of σp, σr, and σc, we can find σc over the spaces ℓ1, c0, and c. So we have the following.Theorem 3.6.
For the operatorS, one has the following:
(3.3)σc(S,l1)={q-2r,q+2r},σc(S,c0)=[q-2r,q+2r],σc(S,c)=[q-2r,q+2r).


## 4. The Resolvent Operator

The following theorem is a generalization of Theorem2.3.Theorem 4.1.
Letμ∈{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 (snk) defined by
(4.1)snk=1r(α12-1)⋅{α1n-k+1-α1n+k+3ifn≥kα1-n+k+1-α1n+k+3ifn<k,
where α1 is the root of the polynomial P(x)=rx2+qx+r with |α1|<1.Proof.
Letμ 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, suppose0∉[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=(snk) defined by
(4.2)snk=1r(α12-1)⋅{α1n-k+1-α1n+k+3ifn≥kα1-n+k+1-α1n+k+3ifn<k,
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 matrixA 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=(tnk) defined by
(4.3)tnk=1r(α12-1)⋅{α1-n+k+1-α1n-k+1ifn<k0ifn≥k
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.
Let0∉[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
(4.4)‖S-1‖(μ:μ)=|α1|+|α1|2|r||1-α12|(1-|α1|).Proof.
SinceS-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 Sn-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-α12)|. By using Theorem 4.1, we have
(4.5)‖Sn-1‖l1=ρ(∑k=0n|α1n-k+1-α1n+k+3|+∑k=n+1∞|α1-n+k+1-α1n+k+3|)≤ρ(∑k=0n(|α1|n-k+1+|α1|n+k+3)+∑k=n+1∞(|α1|-n+k+1+|α1|n+k+3))=ρ(∑k=0n|α1|n-k+1+∑k=n+1∞|α1|-n+k+1+∑k=0∞|α1|n+k+3)=ρ(∑k=1n+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-α12|(1-|α1|).
Hence
(4.6)‖S-1‖(μ:μ)=sup⁡n‖Sn-1‖l1≤|α1|+|α1|2|r||1-α12|(1-|α1|).
On the other hand
(4.7)‖Sn-1‖l1=ρ(∑k=0n|α1n-k+1-α1n+k+3|+∑k=n+1∞|α1-n+k+1-α1n+k+3|)≥ρ(∑k=0n(|α1|n-k+1-|α1|n+k+3)+∑k=n+1∞(|α1|-n+k+1-|α1|n+k+3))=ρ(∑k=0n|α1|n-k+1+∑k=n+1∞|α1|-n+k+1-∑k=0∞|α1|n+k+3)=ρ(∑k=1n+1|α1|k+∑k=2∞|α1|k-∑k=n+3∞|α1|k).
Then
(4.8)‖S-1‖(μ:μ)=sup⁡n‖Sn-1‖l1≥lim⁡n‖Sn-1‖l1=ρ(∑k=1∞|α1|k+∑k=2∞|α1|k)=|α1|+|α1|2|r||1-α12|(1-|α1|).



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