International Journal of Mathematics and Mathematical Sciences

Volume 2011 (2011), Article ID 161209, 10 pages

http://dx.doi.org/10.1155/2011/161209

## Fine Spectra of Tridiagonal Symmetric Matrices

Faculty of Arts and Sciences, Adıyaman University, 02040 Adıyaman, Turkey

Received 14 December 2010; Accepted 10 March 2011

Academic Editor: Martin Bohner

Copyright © 2011 Muhammed Altun. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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 , , , 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 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 . Gonzáles [4] studied the fine spectrum of the Cesàro operator over the sequence space . Okutoyi [5] computed the spectrum of the Cesàro operator over the sequence space . Recently, Rhoades and Yildirim [6] examined the fine spectrum of factorable matrices over and . Coşkun [7] studied the spectrum and fine spectrum for the p-Cesàro operator acting over the space . Akhmedov and Başar [8, 9] have determined the fine spectrum of the Cesàro operator over the sequence spaces , , and . In a recent paper, Furkan, et al. [10] determined the fine spectrum of over the sequence spaces and , where is a lower triangular triple-band matrix. Later, Altun and Karakaya [11] computed the fine spectra for Lacunary matrices over and .

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 , , , 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 and be Banach spaces and be a bounded linear operator. By , we denote the range of , that is,
By , we denote the set of all bounded linear operators on into itself. If is any Banach space, and let then the *adjoint * of is a bounded linear operator on the dual of defined by for all and . Let be a complex normed space and be a linear operator with domain . With , we associate the operator
where is a complex number and is the identity operator on . If has an inverse, which is linear, we denote it by , that is
and call it the *resolvent operator* of . If , we will simply write . Many properties of and depend on , and spectral theory is concerned with those properties. For instance, we will be interested in the set of all in the complex plane such that exists. Boundedness of is another property that will be essential. We shall also ask for what s the domain of is dense in . For our investigation of , , and , we need some basic concepts in spectral theory which are given as follows (see [12, pages 370-371]).

Let be a complex normed space, and let be a linear operator with domain . A *regular value * of is a complex number such that exists, is bounded, and is defined on a set which is dense in .

The *resolvent set * of is the set of all regular values of . Its complement in the complex plane is called the *spectrum* of . Furthermore, the spectrum is partitioned into three disjoint sets as follows: the *point spectrum * is the set such that does not exist. A is called an *eigenvalue* of . The *continuous spectrum * is the set such that exists and satisfies (R3) but not (R2). The *residual spectrum * 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 , , and for the spaces of all bounded, convergent, and null sequences, respectively. And by , we denote the space of all -absolutely summable sequences, where . Let and be two sequence spaces and be an infinite matrix of real or complex numbers , where . Then, we say that defines a matrix mapping from into , and we denote it by writing , if for every sequence the sequence , the -transform of , is in , where By , we denote the class of all matrices such that . Thus, if and only if the series on the right side of (1.4) converges for each and every , and we have for all .

A tridiagonal symmetric infinite matrix is of the form where . The spectral results are clear when , so for the sequel we will have .

Theorem 1.1 (cf. [13]). *Let be an operator with the associated matrix . *(i)* if and only if
*(ii)* if and only if (1.6) and (1.7) with for each .*(iii)* if and only if (1.6). **In these cases, the operator norm of is
*(iv)* if and only if
In this case, the operator norm of is .*

Corollary 1.2. *Let . is a bounded linear operator and .*

#### 2. The Spectra and Point Spectra

Theorem 2.1. * for .*

*Proof. *Since , it is enough to show that . Let be an eigenvalue of the operator . An eigenvector corresponding to this eigenvalue satisfies the linear system of equations:
If , then for all . Hence . Without loss of generality we can suppose . Then and the system of equations turn into the linear homogeneous recurrence relation
where . The characteristic polynomial of the recurrence relation is
There are three cases here.*Case 1 (). *Then characteristic polynomial has only one root: . Hence, the solution of the recurrence relation is of the form
where and are constants which can be determined by the first two terms and . , so we have . And , so we have . Then . This means . So, we conclude that there is no eigenvalue in this case.*Case 2 (). *Then characteristic polynomial has only one root: . The solution of the recurrence relation, found as in Case 1, is . So, there is no eigenvalue in this case.*Case 3 (). *Then the characteristic polynomial has two distinct roots and with . Let . The solution of the recurrence relation is of the form
Using the first two terms and the fact that , we get and . So we have
If , then
So , which means . Now, if , then there exists such that and . So, . Again we have . 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. *.*

Theorem 2.3. *Let . Let and be the roots of the polynomial , with . Then the resolvent operator over is , where
**
Moreover, this operator is continuous and the domain of the operator is the whole space .*

*Proof. *Let and be as it is stated in the theorem. From we get to the system of equations:
This is a nonhomogenous linear recurrence relation. Using the fact that , for (2.9) we can reach a solution with generating functions. This solution can be given by
where
Let . We can see that by using Theorem 1.1, . So is the resolvent operator of and is continuous.

If ( is or ) is a bounded linear operator represented by the matrix , then it is known that the adjoint operator is defined by the transpose of the matrix . It should be noted that the dual space of is isometrically isomorphic to the Banach space and the dual space , of is isometrically isomorphic to the Banach space .

Corollary 2.4. * for .*

*Proof. *. And by Cartlidge [14], if a matrix operator is bounded on , then . Hence we have . What remains is to show that . By Theorem 2.3, there exists a resolvent operator of which is continuous and the whole space is the domain if the roots of the polynomial satisfy
So, if then (2.12) is not satisfied. Since , (2.12) is not satisfied means, the roots can be only of the form
for some . Then . Hence , which means can be only on the line segment .

Theorem 2.5. * for .*

*Proof. *By Theorem 2.2 and Corollary 2.4. Since the spectrum of a bounded linear operator over a complex Banach space is closed, we have . And from the proof of Corollary 2.4 we have .

#### 3. The Continuous Spectra and Residual Spectra

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

Corollary 3.2. *If then .*

Theorem 3.3. *. *

*Proof. * by Theorem 2.1. Now using Corollary 3.2, we have .

Theorem 3.4. *. *

*Proof. *Similarly as in the proof of the previous theorem, we have .

If is a bounded matrix operator represented by the matrix , then acting on has a matrix representation of the form where is the limit of the sequence of row sums of minus the sum of the limits of the columns of , and is the column vector whose th entry is the limit of the th column of for each . For , the matrix is of the form

Theorem 3.5. *. *

*Proof. *Let be an eigenvector of corresponding to the eigenvalue . Then we have and where . By Theorem 2.1,. Then . And is the only value that satisfies . Hence . Then .

Now, since the spectrum is the disjoint union of , , and , we can find over the spaces , , and . So we have the following.

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

#### 4. The Resolvent Operator

The following theorem is a generalization of Theorem 2.3.

Theorem 4.1. *Let . The resolvent operator over exists and is continuous, and the domain of is the whole space if and only if . In this case, has a matrix representation defined by
**
where is the root of the polynomial with .*

*Proof. *Let be one of the sequence spaces in . Suppose has a continuous resolvent operator where the domain of the resolvent operator is the whole space . Then is not in . Conversely if , then has a continuous resolvent operator, and since is bounded by Lemma 7.2-7.3 of [12] the domain of this resolvent operator is the whole space .

Now, suppose . Let and be the roots of the polynomial where . Since , by the proof of Corollary 2.4. Then . So satisfies the conditions of Theorem 2.3. Hence the resolvent operator of is represented by the matrix defined by
when by that theorem. The matrix is already a left inverse of the matrix . Observe that satisfies also the corresponding conditions of Theorem 1.1, which means for . So, the matrix is the representation of the resolvent operator also for the spaces in .

*Remark 4.2. *If a matrix is a triangle, we can see that the resolvent (when it exists) is the unique lower triangular left hand inverse of . In our case, is far away from being a triangle. The matrix of this theorem is not the unique left inverse of the matrix for . For example, the matrix defined by
is another left inverse of . Then is also a left inverse of for any , which means there exist infinitely many left inverses for .

Theorem 4.3. *Let , and, be the root of with . Then for we have
*

*Proof. *Since is a symmetric matrix, the supremum of the norms of the rows is equal to the supremum of the norms of the columns. So, according to Theorem 1.1, what we need is to calculate the supremum of the norms of the rows of . Denote the th row by for . Now, let us fix the row and calculate the norm for this row. Let . By using Theorem 4.1, we have
Hence
On the other hand
Then

#### Acknowledgment

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

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