#### Abstract

The fine spectra of 2-banded and 3-banded infinite Toeplitz matrices were examined by several authors. The fine spectra of *n*-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 ()-banded symmetric Toeplitz matrix operators for arbitrary positive integer .

#### 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 , 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 . Akhmedov and Başar [7, 8] have determined the fine spectrum of the Cesàro operator over the sequence spaces , and . Altun and Karakaya [9] computed the fine spectra of Lacunary matrices over and . Furkan et al. [10] determined the fine spectrum of over the sequence spaces and , where is a lower triangular triple-band matrix. Later, Altun [11] computed the fine spectra of triangular Toeplitz matrices over and .

The fine spectrum of the difference operator over and was studied by Altay and Başar [12]. Recently, the fine spectra of over and are studied by Akhmedov and Başar [13, 14], where is the space of -bounded variation sequences, introduced by Başar and Altay [15] with . The fine spectrum with respect to the Goldberg's classification of the operator over and with 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 , defined as a double band matrix with the convergent sequences and having certain properties, over . In 2010, Srivastava and Kumar [18] have determined the spectra and the fine spectra of the generalized difference operator on , where is defined by and for all , under certain conditions on the sequence and they have just generalized these results by the generalized difference operator defined by (see [19]).

In this work, our purpose is to determine the spectra of the operator, for which the corresponding matrix is a -banded symmetric Toeplitz matrix, over the sequence spaces and . We will also give the fine spectra results for the spaces and .

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 X is any Banach space and 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 will 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 [20, pages 370-371]).

Let be a complex normed space and be a linear operator with domain . A *regular value * of is a complex number such that(R1) exists, (R2) is bounded, (R3) 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).

From Goldberg [21], if , a Banach space, then there are three possibilities for , the range of :(I),(II), but ,(III),and three possibilities for :(1) exists and is continuous,(2) exists, but is discontinuous,(3) does not exist.If these possibilities are combined in all possible ways, nine different states are created. These are labelled as I_{1}, I_{2}, I_{3}, II_{1}, II_{2}, II_{3}, III_{1}, III_{2}, and III_{3}. If is a complex number such that or , then is in the resolvent set of , the set of all regular values of on . The other classification gives rise to the fine spectrum of . For example, we will write if satisfies III and .

A triangle is a lower triangular matrix with all of the principal diagonal elements nonzero. We will write , and for the spaces of all bounded, convergent, and null sequences, respectively. 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 .

Let an -tuple be given. A symmetric infinite Toeplitz matrix is a -band matrix of the form The spectral results are clear when is a multiple of the identity matrix, so for the sequel we will have and .

Let be the right shift operator: and be the left shift operator: Let , where is the palindromic polynomial . Then, we can see that and we will call and as the function and polynomial associated to the operator , respectively. We also have The roots of are nonzero and symmetric, that is, if is a root, is also a root. Let be the roots of such that for . Then Now, by induction, we can see that Let be the unit disc and be the unit circle . We have the following two lemmas as a consequence of the corresponding results in [22] and [23], respectively.

Lemma 1.1. * is onto if and only if is not on the unit circle.*

Lemma 1.2. * is onto if and only if is outside the unit disc.*

Theorem 1.3. * is onto if and only if has no root on the unit circle.*

*Proof. *Suppose has a root on the unit circle. Let be the roots of such that for . We have
Since the matrix operators commute with each other, without loss of generality, we can suppose is a root on the unit circle. Clearly, all the operators are in . But, by Lemma 1.1 the operator is not onto. So, cannot be onto.

Suppose, now, has no root on the unit circle. That means for . Then all the operators are onto by Lemma 1.1 and Lemma 1.2. Hence, is onto.

Theorem 1.4 (cf. [24]). *Let be an operator with the associated matrix .*(i)* if and only if
*(ii)* if and only if (1.12) and (1.13) with for each .*(iii)* if and only if (1.12). In these cases, the operator norm of is
*(iv)* if and only if
**In this case, the operator norm of is .*

Corollary 1.5. * for and
*

Theorem 1.6. *Let be a Banach space and . Then is in the spectrum if and only if is not bijective.*

* Proof. *Suppose is not bijective. Then is not 1-1 or not onto. If it is not 1-1, then . Suppose now is 1-1. Then it is not onto and by Lemma 7.2-3 of [20], cannot be in . Hence, .

Now, suppose is bijective. Then by the open mapping theorem is continuous. Hence, is not in the spectrum .

Corollary 1.7. *Let be a Banach space and . Then if and only if is bijective.*

#### 2. The Spectra and Fine Spectra

Lemma 2.1 (Lemma 3.4 of [11]). *Let be distinct complex numbers with for . Let be a sequence satisfying
**
for , where are constants forming the polynomials for and . Then .*

Lemma 2.2. *Let be distinct complex numbers. Let be a sequence satisfying
**
for , where are constants forming the polynomials for and . Then for , and the existence of a with implies and is a constant.*

*Proof. *Let . To prove for all , suppose it is not true. Then let . Let be the largest positive integer with . Then . Let
We have for . If , we have
Since for , we have . Then but this contradicts with Lemma 2.1. Hence, we have for .

Now, let us prove the second part. Suppose, there exist a positive integer such that for all . For any , is constant means . Suppose . Without loss of generality let . Let be the largest integer satisfying . Then . This means, since have modulus less than 1, , where
But, this again contradicts with Lemma 2.1. Hence, we have . Now, we have , where
Suppose, one of the elements in is equal to 1, say . Then, , where
and this again contradicts with Lemma 2.1. Hence, and 1 is the unique candidate for with modulus 1.

Theorem 2.3. * 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:
Since we can write this system of equations in the form:
where , and for . This system of equations, by change of variables for , is equivalent to system of equations
with the initial conditions .

We see that this is a -th order linear homogenous difference equation with the corresponding characteristic polynomial
Suppose has distinct roots with multiplicities . Then, any solution of the system of equations satisfies
Observe that if is a root of , then is also a root. There are two cases.*Case 1* ( is not a root of ). Since , by Lemma 2.2 we can write
where for . By the symmetry of the roots we have . Now, using the initial conditions we have
where and is the generalized Vandermonde matrix
where denotes , denotes , denotes , denotes , denotes , and denotes .

The determinant of the matrix was explicitly given in [25, 26]:
An inductive proof of this formula is given by Chen and Li [27]. Since zero is not a root of our polynomial , we have ; hence, we conclude , which means the sequences and . Hence, there is no eigenvalue in this case.*Case 2* ( is a root of ). Since , by Lemma 2.2 we can write
where for . By the symmetry of the roots we have . Now, using the initial conditions we have
where and is an submatrix of a generalized Vandermonde matrix. Since the determinant of generalized Vandermonde matrix with nonzero roots is not zero, we have that the columns of are linearly independent. So again we can conclude that , which again will mean that there is no eigenvalue.

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 .

Lemma 2.4 (see [21, page 59]). * has a dense range if and only if is one to one.*

Corollary 2.5. *If then .*

Theorem 2.6. *.*

*Proof. * by Theorem 2.3. Now using Corollary 2.5 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 2.7. *.*

*Proof. *Let be an eigenvector of corresponding to the eigenvalue . Then we have and where . By Theorem 2.3 . Then . So is the only value that satisfies . Hence, . Then .

We will write instead of for the sequel.

Lemma 2.8. *. *

*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 for and . So . We know by Cartlidge [28] that if a matrix operator is bounded on , then . Hence, we have .

Theorem 2.9. * for .*

*Proof. *Let us first consider as an operator on . By Theorems 1.6 and 2.3 if and only if is not onto over . By Theorem 1.3 is not onto over if and only if the polynomial has a root on the unit circle. has a root on the unit circle if and only if for some . We have for some if and only if . Hence, . Finally, we apply Lemma 2.8.

Corollary 2.10. * is onto if and only if has no root on the unit circle.*

The spectrum is the disjoint union of , and , so we have the following theorem as a consequence of Theorems 2.3, 2.6, 2.7, and 2.9.

Theorem 2.11. * and .*

As a result of Theorems 1.3, 2.3, 2.6, 2.7, and 2.9 and Corollary 2.10, we have the following.

Theorem 2.12. *, and .*

#### 3. Some Applications

Now, let us give an application of Theorem 2.9. Consider the system of equations where for negative .

Theorem 3.1. *Let , where are complex numbers such that the complex sequences and are solutions of system (3.1). Then the following are equivalent:*(i)*boundedness of always implies a unique bounded solution ,*(ii)*convergence of always implies a unique convergent solution ,*(iii)* always implies a unique solution with ,*(iv)* always implies a unique solution with ,*(v)* has no root on the unit circle .*

*Proof. *The system of equations (3.1) holds, so we have . Then is the polynomial associated to . Let be the function associated to . 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 always implies a unique bounded solution . Then the operator is bijective. So, is not in the spectrum by Theorem 1.6, which means and .

For the reverse implication, suppose has no root on the unit circle . Then has no zero on the unit circle. So, is in the resolvent set . Now, by Theorem 1.6, is bijective on , which means that the boundedness of implies a bounded unique solution .

*Example 3.2. *We can see that
When , is a tridiagonal matrix, that is,
then for , where . Therefore,
which is one of the main results of [29]. is the closed line segment in the complex plane with endpoints and .

*Example 3.3. *When , is a pentadiagonal matrix, that is,
then and
So the spectrum is a line segment if is a real multiple of . It can be proved that, the spectrum is a closed connected part of a parabola if is not a real multiple of . For example, if and (the complex number ) we have