Volume 2011, Issue 1 161209
Research Article
Open Access

Fine Spectra of Tridiagonal Symmetric Matrices

Muhammed Altun

Corresponding Author

Muhammed Altun

Faculty of Arts and Sciences, Adıyaman University, 02040 Adıyaman, Turkey adiyaman.edu.tr

Search for more papers by this author
First published: 18 May 2011
Citations: 7
Academic Editor: Martin Bohner

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, 1, and .

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 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, , and 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, 1, and . 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 : XY be a bounded linear operator. By (T), we denote the range of T, that is,
(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 TB(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 ϕX* and xX. 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)
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 , that is
(1.3)
and call it the resolvent operator of Tλ. If λ = 0, we will simply write T−1. Many properties of Tλ and 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 exists. Boundedness of is another property that will be essential. We shall also ask for what λs the domain of is dense in X. For our investigation of T, Tλ, and , 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) exists,

  • (R2) is bounded, and

  • (R3) 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) = ρ(T) in the complex plane is called the spectrum of T. Furthermore, the spectrum σ(T) is partitioned into three disjoint sets as follows: the point spectrum   σp(T) is the set such that does not exist. A λσp(T) is called an eigenvalue of T. The continuous spectrum σc(T) is the set such that exists and satisfies (R3) but not (R2). The residual spectrum σr(T) is the set such that exists but does not satisfy (R3).

A triangle is a lower triangular matrix with all of the principal diagonal elements nonzero. We shall write , c, and c0 for the spaces of all bounded, convergent, and null sequences, respectively. And by p, 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. Then, we say that A defines a matrix mapping from μ into γ, and we denote it by writing A : μγ, if for every sequence x = (xk) ∈ μ the sequence Ax = {(Ax) n}, the A-transform of x, is in γ, where
(1.4)
By (μ : γ), we denote the class of all matrices A such that A : μγ. Thus, A ∈ (μ : γ) if and only if the series on the right side of (1.4) converges for each n and every xμ, and we have Ax = {(Ax) n} nγ for all xμ.
A tridiagonal symmetric infinite matrix is of the form
(1.5)
where q, r. 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 = (ank).

  • (i)

    TB(c) if and only if

    (1.6)
    (1.7)
    (1.8)

  • (ii)

    TB(c0) if and only if (1.6) and (1.7) with ak = 0 for each k.

  • (iii)

    TB() if and only if (1.6).

In these cases, the operator norm of T is
(1.9)
  • (iv)

    TB(1) if and only if

    (1.10)
    In this case, the operator norm of T is .

Corollary 1.2. Let μ ∈ {c0, c, 1, }. S(q, r) : μμ 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 ∈ {1, c0, c}.

Proof. Since 1c0c, 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)
If x0 = 0, then xk = 0 for all k. Hence x0 ≠ 0. Without loss of generality we can suppose x0 = 1. Then x1 = (λq)/r and the system of equations turn into the linear homogeneous recurrence relation
(2.2)
where p = (qλ)/r. The characteristic polynomial of the recurrence relation is
(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

(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 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 roots α1 ≠ ±1 and α2 ≠ ±1 with α1α2 = 1. Let |α1| ≥ 1 ≥ |α2|. The solution of the recurrence relation is of the form

(2.5)
Using the first two terms and the fact that p = −(α1 + α2), we get A = α1/(α1α2) and B = α2/(α2α1). So we have
(2.6)
If |α1 | > 1>|α2|, then
(2.7)

So lim n | xn | = , which means (xn) ∉ c. Now, if |α1 | = |α2 | = 1, then there exists θ ∈ (0, π) such that α1 = eiθ and α2 = eiθ. 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 for , we get to the following.

Theorem 2.2. σp(S, ) = (q − 2r, q + 2r).

Theorem 2.3. Let p = (qλ)/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 , where

(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)Sλx = y we get to the system of equations:

(2.9)
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)
where
(2.11)
Let T = (tnk). We can see that by using Theorem 1.1, TB(c0). So is the resolvent operator of (1/r)Sλ and is continuous.

If T : μμ (μ is 1 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 of c0 is isometrically isomorphic to the Banach space 1 and the dual space , of 1 is isometrically isomorphic to the Banach space .

Corollary 2.4. σ(S, μ)⊂[q − 2r, q + 2r] for μ ∈ {1, c0, c, }.

Proof. . And by Cartlidge [14], if a matrix operator A is bounded on c, then σ(A, c) = σ(A, ). Hence we have σ(S, c0) = σ(S, 1) = σ(S, ) = 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)
So, if λσ(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)
for some θ ∈ [0,2π). Then (qλ)/r = p = −(α1 + α2) = −(eiθ + eiθ) = −2cos θ. Hence λ = q + 2rcos θ, 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 ∈ {1, c0, c, }.

Proof. By Theorem 2.2 and Corollary 2.4  (q − 2r, q + 2r) ⊂ σ(S, )⊂[q − 2r, q + 2r]. Since the spectrum of a bounded linear operator over a complex Banach space is closed, we have σ(S, ) = [q − 2r, q + 2r]. And from the proof of Corollary 2.4 we have σ(S, 1) = σ(S, c0) = σ(S, c) = σ(S, ).

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 σr(T, μ) = σp(T*, μ*)∖σp(T, μ).

Theorem 3.3. σr(S, c0) = .

Proof. σp(S, 1) = by Theorem 2.1. Now using Corollary 3.2, we have σp(S, 1)∖σp(S, c0) = .

Theorem 3.4. σr(S, 1) = (q − 2r, q + 2r).

Proof. Similarly as in the proof of the previous theorem, we have σp(S, )∖σp(S, 1) = (q − 2r, q + 2r).

If T : cc is a bounded matrix operator represented by the matrix A, then T* : c*c* acting on 1 has a matrix representation of the form
(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. For Sλ : cc, the matrix is of the form
(3.2)

Theorem 3.5. σr(S, c) = {q + 2r}.

Proof. Let x = (x0, x1, …) ∈ 1 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 spectrum σ is the disjoint union of σp, σr, and σc, we can find σc over the spaces 1, c0, and c. So we have the following.

Theorem 3.6. For the operator S, one has the following:

(3.3)

4. The Resolvent Operator

The following theorem is a generalization of Theorem 2.3.

Theorem 4.1. Let μ ∈ {c0, c, 1, }. 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)
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, 1, }. 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 = (snk) defined by

(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, 1, }. So, the matrix S−1 is the representation of the resolvent operator also for the spaces in {c, 1, }.

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 = (tnk) defined by

(4.3)
is another left inverse of S. Then λS−1 + (1 − λ)T is also a left inverse of S for any λ, 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, 1, } we have

(4.4)

Proof. Since S−1 is a symmetric matrix, the supremum of the 1 norms of the rows is equal to the supremum of the 1 norms of the columns. So, according to Theorem 1.1, what we need is to calculate the supremum of the 1 norms of the rows of S−1. Denote the nth row S−1 by for n = 0,1, …. Now, let us fix the row n and calculate the 1 norm for this row. Let . By using Theorem 4.1, we have

(4.5)
Hence
(4.6)
On the other hand
(4.7)
Then
(4.8)

Acknowledgment

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

      The full text of this article hosted at iucr.org is unavailable due to technical difficulties.