Spectrum and fine spectrum of generalised lower triangular triple band matrices over the sequence space $l_p$

TL;DR

This paper investigates the spectrum and fine spectrum of a new generalized difference operator represented by a lower triangular triple band matrix with periodic sequences over the sequence space l_p, extending previous studies on band matrices.

Contribution

It introduces a more general class of lower triangular triple band matrices with periodic sequences and analyzes their spectral properties on l_p spaces.

Findings

Spectrum and fine spectrum characterized for the new operator

Includes approximate point, defect, and compression spectra

Provides examples and discusses special cases

Abstract

The spectrum of triangular band matrices defined on the sequence spaces where the entries of each band is a constant or convergent sequence is well studied. In this article, the spectrum and fine spectrum of a new generalised difference operator defined by a lower triangular triple band matrix on the sequence space $l_p (1 \leq p < \infty)$ are obtained where the bands are considered as periodic sequences. The approximate point spectrum, defect spectrum, compression spectrum and the Goldberg classification of the spectrum are also discussed. Suitable examples are given in order to supplement the results. Several special cases of our findings are discussed which confirm that our study is more general and extensive.

Tables

Table 1. Table 1. Subdivisions of spectrum of a linear operator

		1	2	3
		$T_{\lambda }^{-1}$ exists and bounded	$T_{\lambda }^{-1}$ exists and unbounded	$T_{\lambda }^{-1}$ does not exist
A	$R\left(T_{\lambda }\right)=X$	$\lambda \in \rho (T,X)$	—
B	$\overline{R\left(T_{\lambda }\right)}=X$	$\lambda \in \rho (T,X)$
C	$\overline{R\left(T_{\lambda }\right)}\ne X$

Equations

$$B(r_{1},r_{2};s_{1},s_{2};t_{1},t_{2})=\left({\begin{array}[]{cccccc}r_{1}&0&0&0&0&\cdots\\ s_{1}&r_{2}&0&0&0&\cdots\\ t_{1}&s_{2}&r_{1}&0&0&\cdots\\ 0&t_{2}&s_{1}&r_{2}&0&\cdots\\ 0&0&t_{1}&s_{2}&r_{1}&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}}\right)$$

$$(T^{\ast }\varphi )(x)=\varphi (Tx)\text{mbox}forall\varphi \in Y^{\ast }\text{mbox}andx\in X$$

$$\sigma (T,X)=\{\lambda \in \mathbb{C}:(T-\lambda I)\text{mbox}hasnoinverseinB(X)\}.$$

$${\sigma }_{ap}(T,X)$$

$${\sigma }_{\delta }(T,X)$$

$${\sigma }_{co}(T,X)$$

$$\sigma (T,X)={\sigma }_{ap}(T,X)\cup {\sigma }_{\delta }(T,X)$$

$$\sigma (T,X)={\sigma }_{ap}(T,X)\cup {\sigma }_{co}(T,X).$$

$$(Ax{)}_n=\sum _k{a}_{nk}{x}_k,n\in \mathbb{N}$$

$$B_1$$

$$B_2$$

$$\Vert Bx{\Vert }_{l_p}$$

$$\Vert B{\Vert }_{l_p}\ge \frac{\Vert B{e}^{(k)}{\Vert }_{l_p}}{\Vert e^{(k)}{\Vert }_{l_p}}=\{\begin{array}{rlr}& (|r_1{|}^p+|s_1{|}^p+|t_1{|}^p{)}^{\frac{1}{p}},k\text{mbox}isodd& \\ & (|r_2{|}^p+|s_2{|}^p+|t_2{|}^p{)}^{\frac{1}{p}},k\text{mbox}iseven.& \end{array}$$

$$\sigma (B,l_p)=\left\{\lambda \in \mathbb{C}:\left|\frac{2(r_1-\lambda )(r_2-\lambda )}{s_1{s}_2-t_1(r_2-\lambda )-t_2(r_1-\lambda )+\sqrt{\chi }}\right|\le 1\right\}$$

$$\chi$$

$$(B-\lambda I)^{-1}=\left({\begin{array}[]{ccccc}a_{1}&0&0&0&\cdots\\ a_{2}&b_{1}&0&0&\cdots\\ a_{3}&b_{2}&a_{1}&0&\cdots\\ a_{4}&b_{3}&a_{2}&b_{1}&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots\end{array}}\right)$$

$$\begin{array}{rlr}{t}_1{a}_{2k-1}+s_2{a}_{2k}+(r_1-\lambda )a_{2k+1}& =& 0\\ t_2{a}_{2k}+s_1{a}_{2k+1}+(r_2-\lambda )a_{2k+2}& =& 0\end{array}\}$$

$$\begin{array}{rlr}{t}_2{b}_{2k-1}+s_1{b}_{2k}+(r_2-\lambda )b_{2k+1}& =& 0\\ t_1{b}_{2k}+s_2{b}_{2k+1}+(r_1-\lambda )b_{2k+2}& =& 0\end{array}\}$$

$$a_1=\frac{1}{r_1-\lambda },a_2=-\frac{s_1}{(r_1-\lambda )(r_2-\lambda )},$$

$$b_1=\frac{1}{r_2-\lambda },b_2=-\frac{s_2}{(r_1-\lambda )(r_2-\lambda )}.$$

$$a_{2k+1}$$

$$a_{2k+2}$$

$$x_{k+1}=A_1{x}_k,k\in \mathbb{N},$$

$$A_{1}=\left({\begin{array}[]{cc}-\frac{t_{1}}{r_{1}-\lambda}&-\frac{s_{2}}{r_{1}-\lambda}\\ \\ \frac{s_{1}t_{1}}{(r_{1}-\lambda)(r_{2}-\lambda)}&-\frac{t_{2}}{r_{2}-\lambda}+\frac{s_{1}s_{2}}{(r_{1}-\lambda)(r_{2}-\lambda)}\end{array}}\right),\ \ x_{k}=\left({\begin{array}[]{c}a_{2k-1}\\ a_{2k}\end{array}}\right)\ \ \mbox{for}\ \ k\in\mathbb{N}.$$

$$f(\alpha )={\alpha }^2+\left(\frac{t_1}{r_1-\lambda }+\frac{t_2}{r_2-\lambda }-\frac{s_1{s}_2}{(r_1-\lambda )(r_2-\lambda )}\right)\alpha +\frac{t_1{t}_2}{(r_1-\lambda )(r_2-\lambda )}=0.$$

$${\alpha }_1=\frac{s_1{s}_2-t_1(r_2-\lambda )-t_2(r_1-\lambda )+\sqrt{\chi }}{2(r_1-\lambda )(r_2-\lambda )},{\alpha }_2=\frac{s_1{s}_2-t_1(r_2-\lambda )-t_2(r_1-\lambda )-\sqrt{\chi }}{2(r_1-\lambda )(r_2-\lambda )}$$

$$\left({\begin{array}[]{cc}\frac{-s_{1}s_{2}+t_{2}(r_{1}-\lambda)-t_{1}(r_{2}-\lambda)-\sqrt{\chi}}{2(r_{1}-\lambda)(r_{2}-\lambda)}&-\frac{s_{2}}{r_{1}-\lambda}\\ \\ \frac{s_{1}t_{1}}{(r_{1}-\lambda)(r_{2}-\lambda)}&\frac{s_{1}s_{2}-t_{2}(r_{1}-\lambda)+t_{1}(r_{2}-\lambda)-\sqrt{\chi}}{2(r_{1}-\lambda)(r_{2}-\lambda)}\end{array}}\right)f=\left({\begin{array}[]{c}0\\ 0\end{array}}\right).$$

$$f=(f_1,f_2{)}^t={\left(1,\frac{-s_1{s}_2+t_2(r_1-\lambda )-t_1(r_2-\lambda )-\sqrt{\chi }}{2s_2(r_2-\lambda )}\right)}^t$$

$$g=(g_1,g_2{)}^t={\left(1,\frac{-s_1{s}_2+t_2(r_1-\lambda )-t_1(r_2-\lambda )+\sqrt{\chi }}{2s_2(r_2-\lambda )}\right)}^t$$

$$x_{k}=\left({\begin{array}[]{c}a_{2k-1}\\ a_{2k}\end{array}}\right)=c_{1}\left({\begin{array}[]{c}f_{1}\\ f_{2}\end{array}}\right)\alpha_{1}^{k}+c_{2}\left({\begin{array}[]{c}g_{1}\\ g_{2}\end{array}}\right)\alpha_{2}^{k},\ \ k\in\mathbb{N},$$

$$x_{1}=c_{1}\left({\begin{array}[]{c}f_{1}\\ f_{2}\end{array}}\right)\alpha_{1}+c_{2}\left({\begin{array}[]{c}g_{1}\\ g_{2}\end{array}}\right)\alpha_{2}=\left({\begin{array}[]{c}a_{1}\\ a_{2}\end{array}}\right)=\left({\begin{array}[]{c}\frac{1}{r_{1}-\lambda}\\ \\ -\frac{s_{1}}{(r_{1}-\lambda)(r_{2}-\lambda)}\end{array}}\right).$$

$$c_1=\frac{s_1{g}_1+(r_2-\lambda )g_2}{{\alpha }_1(r_1-\lambda )(r_2-\lambda )(f_1{g}_2-g_1{f}_2)},c_2=-\frac{s_1{f}_1+(r_2-\lambda )f_2}{{\alpha }_2(r_1-\lambda )(r_2-\lambda )(f_1{g}_2-g_1{f}_2)}.$$

Full text

1. Introduction

In infinite dimensional Banach spaces, the spectrum of a bounded linear operator has more complex structure than the finite dimensional case. It can be partitioned into three disjoint parts, namely point spectrum, continuous spectrum and residual spectrum and these three parts of the spectrum is termed as fine spectrum of the operator.

Several researchers have studied the spectrum and fine spectrum of various bounded linear operators defined on various sequence spaces. Altay and Başar [1], Kayaduman and Furkan [2], Akhmedov and Başar [3] obtained the spctrum and fine spectrum of the difference operator $\Delta$ over the sequence spaces $c_{0},$ $c$ and $l_{1},$ $bv$ and $bv_{p}(1\leq p<\infty)$ respectively. The fine spectrum of the generalised difference operator $B(r,s)$ is studied by Altay and Başar [4], Furkan et al. [5], Bilgiç and Furkan [6] over the sequence spaces $c_{0}$ and $c$, $l_{1}$ and $bv$, $l_{p}$ and $bv_{p}$ $(1<p<\infty)$ respectively. Furkan et al. [7, 8], Bilgiç and Furkan [9] further generalised these results to the operator $B(r,s,t).$ The fine spectrum of triangular Toeplitz operator defined on $c_{0}$ and $c$ is obtained by Altun [10]. Srivastava and Kumar [11, 12] have studied the spectrum and fine spectrm of the generalised differnece operators $\vartriangle_{v}$ and $\vartriangle_{uv}$ on $l_{1}.$ Later on, El-Shabrawy [13, 14] has studied the fine spectrum of the operator $\vartriangle_{ab}$ on the sequence space $l_{p}$ $(1<p<\infty)$ and $c_{0}$ respectively. The spectrum of the upper triangular matrix $A(\tilde{r},\tilde{s})$ defined on $l_{1}$ and $l_{p}$ $(1<p<\infty)$ is studied by Karaisa [15, 16]. In 2012, the fine spectrum of lower triangular triple band matrix $\Delta_{uvw}$ on $l_{1}$ is studied by Panigrahi and Srivastava [17] and analogusly, the upper triangular case is studied by Altundag and Abay [18]. The spectrum and fine spectrum of the of the generalised difference operator $\vartriangle^{r}_{v}$ over the sequence spaces $c_{0}$ and $l_{1}$ have been studied by Dutta and Baliarsingh (see [19, 20]). In 2017, Birbonshi and Srivastava [21] have studied various spectral properties of $n$ band triangular matrices of constant bands. Recently El-Shabrawy and Abu-Janah [22] studied the spectrum of the operators $B(r,s)$ and $\Delta_{ab}$ over the sequence space $bv_{0}$ and $h.$

It has been observed that, the entries in each band of the band matrices chosen by the earlier researchers are either constant sequences or convergent sequences. In the present paper, we have considered an operator which is represented by a triple band lower triangular matrix defined over the sequence space $l_{p}(1\leq p<\infty),$ where the bands are taken as periodic sequences of period two. The operator can be represented by the matrix

[TABLE]

where $r_{1},r_{2}$ are complex numbers, $s_{1},s_{2}$ are non-zero complex numbers and $t_{1},t_{2}$ are complex numbers such that, either $t_{1},t_{2}$ both are non-zero or both are equal to zero. For the sake of convenience we will use the notation $B$ instead of $B(r_{1},r_{2};s_{1},s_{2};t_{1},t_{2})$. Several operators whose spectrum and fine spectrum studied previously can be derived from the above operator with perticular choices of $r_{1},r_{2},$ $s_{1},s_{2}$ and $t_{1},t_{2}$. The purpose of this work is to obtain the spectrum and fine spectrum of the operator $B$ on the sequence space $l_{p}(1<p<\infty).$ The theory of system of linear difference equations plays an important role in our work. For detail study of difference equation we refer the book [23].

The rest of the paper is organised as follows. Section 2 deals with some notations and review of few concepts of operator theory. Section 3 and 4 contains our results on the spectrum and some spectral classification of the operator $B$ over the sequence space $l_{p}(1<p<\infty)$ and $l_{1}$ respectively. Some examples are provided in Section 5 in support of our results, and finally, in Section 6 we conclude our paper by mentioning some previous results which can be derived from our results.

2. Preliminaries and Notations

In this paper, the sequence space $l_{p}(1\leq p<\infty)$ represents the set of all $p$-absolutely summable sequences of real or complex numbers. All the infinite sequences and matrices are indexed by the set of natural numbers $\mathbb{N}.$ Let $T:X\rightarrow Y$ be a bounded linear operator where $X$ and $Y$ are complex Banach spaces. Then the range space of $T$ and null space of $T$ are denoted by $R(T)$ and $N(T)$ respectively. The adjoint of $T$, denoted by $T^{*}$, is a bounded linear operator $T^{*}:Y^{*}\rightarrow X^{*}$ which is defined by

[TABLE]

where $X^{*}$ and $Y^{*}$ are the dual spaces of $X$ and $Y$ respectively. The set of all bounded linear operator on $X$ onto itself is denoted by $B(X).$ For any operator $T\in B(X),$ the resolvent set of $T$ is the set of all complex numbers $\lambda$ for which the operator $T-\lambda I$ has a bounded inverse in $X$ where $I$ is the identity operator in $X.$ The resolvent set of $T$ is denoted by $\rho(T,X).$ The compliment of the resolvant set in the complex plane $\mathbb{C}$ is called the spectrum of $T$ and it is denoted by $\sigma(T,X)$, i.e.,

[TABLE]

The spectrum $\sigma(T,X)$ can be classified into three disjoint sets as follows:

The set of points $\lambda\in\mathbb{C}$ for which $N(T-\lambda I)\neq\{0\}$ is called the point spectrum of $T$ and it is denoted by $\sigma_{p}(T,X)$. The set of points $\lambda\in\mathbb{C}$ for which $N(T-\lambda I)=\{0\}$ and $\overline{R(T-\lambda I)}=X$ but ${R(T-\lambda I)}\neq X$ is called the continuous spectrum of $T.$ It is denoted by $\sigma_{c}(T,X).$ The set of points $\lambda\in\mathbb{C}$ for which $N(T-\lambda I)=\{0\}$ and $\overline{R(T-\lambda I)}\neq X$ is called the residual spectrum of $T$ and it is denoted by $\sigma_{r}(T,X)$. The three sets $\sigma_{p}(T,X),\sigma_{c}(T,X),\sigma_{r}(T,X)$ are disjoint and their union is the whole spectrum $\sigma(T,X).$

Next we mention some more subdivisions of the spectrum. For any $T\in B(X)$ the approximate point spectrum, defect spectrum, and compression spectrum of $T$ is denoted by $\sigma_{ap}(T,X),$ $\sigma_{\delta}(T,X)$ and $\sigma_{co}(T,X)$ respectively and defined as,

[TABLE]

The two sets $\sigma_{ap}(T,X)$ and $\sigma_{\delta}(T,X)$ form a subdivision (not necessarily disjoint) of spectrum, i.e.,

[TABLE]

and $\sigma_{ap}(T,X),$ $\sigma_{co}(T,X)$ also form a subdivision (not necessarily disjoint) of spectrum, i.e.,

[TABLE]

The following results are useful in this context.

Proposition 2.1**.**

[24, p. 28] The following relations hold for an operator $T\in B(X)$

1. (a)

$\sigma(T^{*},X^{*})=\sigma(T,X),$ 2. (b)

$\sigma_{ap}(T^{*},X^{*})=\sigma_{\delta}(T,X),$ 3. (c)

$\sigma_{p}(T^{*},X^{*})=\sigma_{co}(T,X),$ 4. (d)

$\sigma(T,X)=\sigma_{ap}(T,X)\cup\sigma_{p}(T^{*},X^{*})=\sigma_{p}(T,X)\cup\sigma_{ap}(T^{*},X^{*}).$

Depending upon the set $R(T_{\lambda})$ and the inverse $T_{\lambda}^{-1}$ of an operator $T\in B(X)$ where $T_{\lambda}=T-\lambda I,$ Goldberg [25] classified the spectrum and the resolvant set of $T$ which is given in the following table.

If we combine the possibilities A, B, C and 1, 2, 3 then nine different states are created. These are labelled by $A_{1}$, $A_{2}$, $A_{3}$, $B_{1}$, $B_{2}$, $B_{3},$ $C_{1}$, $C_{2}$, $C_{3}.$ For example, for any $\lambda\in\mathbb{C},$ $T-\lambda I\in A_{1}$ or $T-\lambda I\in B_{1}$ then $\lambda\in\rho(T,X).$ If the operator $T$ is in state $C_{1}$ for example, then $(T-\lambda I)^{-1}$ exists and bounded and $\overline{R(T_{\lambda})}\neq X$ and we write $\lambda\in C_{1}\sigma(T,X).$

Let $A=(a_{nk})$ be an infinite matrix of complex numbers and $\lambda$ and $\mu$ be two sequence spaces. Then, the matrix $A$ defines a matrix mapping from $\lambda$ into $\mu$ if for every sequence $x=(x_{k})\in\lambda$ the sequence $Ax=\{(Ax)_{n}\}$ is in $\mu$ where

[TABLE]

and it is denoted by $A:\lambda\rightarrow\mu.$ We denote the class of all matrices $A$ such that $A:\lambda\rightarrow\mu$ by $(\lambda,\mu).$

Lemma 2.2**.**

[25, p. 59] The bounded linear operator $T:X\rightarrow Y$ has dense range if and only if $A^{*}$ is one to one

Lemma 2.3**.**

[26, p. 126] A matrix $A=(a_{nk})$ gives rise to a bounded linear operator $T\in B(l_{1})$ from $l_{1}$ to itself if and only if the supremum of $l_{1}$ norms of the columns of A is bounded.

Lemma 2.4**.**

[26, p. 126] A matrix $A=(a_{nk})$ gives rise to a bounded linear operator $T\in B(l_{\infty})$ from $l_{\infty}$ to itself if and only if the supremum of $l_{1}$ norms of the rows of A is bounded.

Lemma 2.5**.**

[27, p. 174, Theorem 9] Let $1<p<\infty$ and suppose $A\in(l_{1},l_{1})\cap(l_{\infty},l_{\infty})$. Then $A\in(l_{p},l_{p}).$

Remark 2.6*.*

Throughout our work, if $z$ is a complex number then $\sqrt{z}$ means the square root of $z$ with non-negative real part. If $Re(\sqrt{z})=0$ then $\sqrt{z}$ means the square root of $z$ with $Im(z)\geq 0.$

3. Spectra of $B$ on $l_{p}(1<p<\infty)$

Theorem 3.1**.**

The operator $B:l_{p}\rightarrow l_{p}$ is a bounded linear operator and $B_{1}\leq\|B\|_{l_{p}}\leq B_{2}$ where

[TABLE]

Proof.

The linearity of $B$ is easy to check so we drop it. For any $(x_{k})\in l_{p}$ with $x_{-1}=x_{0}=0,$ we have

[TABLE]

This shows that $\|B\|_{l_{p}}\leq B_{2}.$ Also let $e^{(k)}$ be a sequence in $l_{p}$ whose $k$-th component is one and other components are zero. Then

[TABLE]

This proves the other part. ∎

Theorem 3.2**.**

Let $s_{1},s_{2}$ be two complex numbers such that $\sqrt{s_{1}^{2}}=s_{1}$ and $\sqrt{s_{2}^{2}}=s_{2}$ then the spectrum of $B$ on $l_{p}$ is given by

[TABLE]

where

[TABLE]

Proof.

Let $S=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\}$ and let $\lambda\notin S.$ This implies that, $\lambda\notin\{r_{1},r_{2}\}$ and $(B-\lambda I)$ has an inverse and it is of the form

[TABLE]

where

[TABLE]

and

[TABLE]

for $k\in\mathbb{N}$ with

[TABLE]

[TABLE]

The equations in (3.3) imply

[TABLE]

The above equations can be written as

[TABLE]

where

[TABLE]

In order to obtain the solution of the system of difference equations (3.5), we need to find the eigenvalues and eigenvectors of $A_{1}$. We consider two cases here. In case 1 we consider $A_{1}$ is diagonalisable and in case 2 we consider $A_{1}$ is not diagonalisable.

Case 1 : $A_{1}$ is diagonalisable .

To find the eigenvalues of $A_{1},$ let $\det(A_{1}-\alpha I)=0$. Which gives the characteristic equation

[TABLE]

The roots of the equation are

[TABLE]

where $\chi$ is mentioned in (3.2). Since $A_{1}$ is diagonalisable and not a scalar multiple of the identity matrix, so $\alpha_{1}\neq\alpha_{2}$. This implies $\chi\neq 0.$ Let f=\left({\begin{array}[]{c}f_{1}\\ f_{2}\end{array}}\right) be the eigenvector corresponding to $\alpha_{1}.$ Then $(A_{1}-\alpha_{1}I)f=0.$ This gives

[TABLE]

Then

[TABLE]

is the eigenvector corresponding to $\alpha_{1}.$ A similar calculation shows that

[TABLE]

is the eigenvector corresponding to $\alpha_{2}.$ Hence the general solution of (3.5) is given by (see [23, p. 137])

[TABLE]

where $c_{1},$ $c_{2}$ are constants which can be determined from the relation

[TABLE]

By solving the above system of equations we obtain, if $t_{1}\neq 0,$ $t_{2}\neq 0$ then $\alpha_{1}\neq 0,$ $\alpha_{2}\neq 0$ and

[TABLE]

Also if $t_{1}=t_{2}=0$ then $\alpha_{1}\neq 0,$ $\alpha_{2}=0$ and $c_{1}=\frac{r_{2}-\lambda}{s_{1}s_{2}}.$ It can be easily verified that, $c_{1}\neq 0$ for both the cases $t_{1}\neq 0,$ $t_{2}\neq 0$ and $t_{1}=t_{2}=0.$

Since $\lambda\notin S$ so $|\alpha_{1}|<1.$ Now we show that $|\alpha_{2}|<1.$ If $s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)=0$ then, $|\alpha_{2}|=|\alpha_{1}|<1.$ Let $s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)\neq 0.$ Since $|\alpha_{1}|<1,$ we have

[TABLE]

Since $|1-\sqrt{z}|\leq|1+\sqrt{z}|$ for all $z\in\mathbb{C},$ we have

[TABLE]

This proves that $|\alpha_{2}|<1.$ Hence from (3.7) it follows that, $(a_{k})\in l_{1}.$

Case 2: $A_{1}$ is not diagonalisable.

In this case we have

[TABLE]

Also if $t_{1}=t_{2}=0$ then from (3.6) it follows that one eigenvalue is zero and other is non-zero which is not possible in this case. Therefore $t_{1}\neq 0$ and $t_{2}\neq 0.$

Let h=\left({\begin{array}[]{c}h_{1}\\ h_{2}\end{array}}\right) be the corresponding eigenvector of $\alpha.$ Then $(A_{1}-\alpha I)h=0$ gives

[TABLE]

Then

[TABLE]

is the eigenvector corresponding to $\alpha.$ Let $u=(u_{1},\ u_{2})^{t}$ be the generalised eigenvector of $\alpha.$ Then

[TABLE]

The Jordan form of $A_{1}$ is given by

[TABLE]

where

[TABLE]

Hence the general solution of (3.3) can be written as (see [23, p. 145])

[TABLE]

Where \tilde{c}=\left({\begin{array}[]{c}\tilde{c_{1}}\\ \tilde{c_{2}}\end{array}}\right)\  and  J^{k}=\left({\begin{array}[]{cc}\alpha^{k}&k\alpha^{k-1}\\ 0&\alpha^{k}\end{array}}\right). Therefore the solution $x_{k}$ is of the form

[TABLE]

with

[TABLE]

Since $t_{1}\neq 0,$ $t_{2}\neq 0$ then $\alpha\neq 0$ and

[TABLE]

Since $|\alpha|=\left|\frac{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)}{2(r_{1}-\lambda)(r_{2}-\lambda)}\right|<1,$ from (3.8), we have $(a_{k})\in l_{1}.$

Now from the equations in (3.4) we have for $k\in\mathbb{N}$

[TABLE]

The above equations can be written as

[TABLE]

where

[TABLE]

The characteristic equation of $A_{2}$ is $\det(A_{2}-\alpha I)=0.$ Which gives

[TABLE]

Since the characteristic equation of the matrices $A_{1}$ and $A_{2}$ are same so the eigenvalues of $A_{2}$ are $\alpha_{1}$ and $\alpha_{2}$. Then by similar method as above it can be shown that $(b_{k})\in l_{1}.$ This implies $(B-\lambda I)^{-1}\in(l_{1},l_{1}).$ Also since $(a_{k})\in l_{1}$ and $(b_{k})\in l_{1}$, the supremum of the $l_{1}$ norms of the rows of $(B-\lambda I)^{-1}$ is finite. This proves that $(B-\lambda I)^{-1}\in(l_{\infty},l_{\infty}).$ Therefore $(B-\lambda I)^{-1}\in(l_{p},l_{p})$. Hence $\lambda\notin\sigma(B,l_{p})$ and this shows that $\sigma(B,l_{p})\subseteq S.$

Conversely, let $\lambda\in S.$ If $\lambda=r_{1}$ or $\lambda=r_{2}$ then $B-\lambda I$ does not have dense range, so it is not invertible. Hence let $\lambda\notin\{r_{1},r_{2}\}.$ Then

[TABLE]

where $(a_{k})$ and $(b_{k})$ are given in (3.3) and (3.4) respectively. If $\chi=0$ then $\left|\frac{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)}{2(r_{1}-\lambda I)(r_{2}-\lambda I)}\right|\geq 1.$ Therefore from (3.8) it follows that $(a_{k})\notin l_{p}.$ Also if $\chi\neq 0$ then from (3.7) we have $|\alpha_{1}|\geq 1$ and $c_{1}\neq 0.$ This also implies that $(a_{k})\notin l_{p}.$ Now $x=(1,0,0,\cdots)\in l_{p}$ but $(B-\lambda I)^{-1}x=(a_{1},a_{2},\cdots)\notin l_{p}.$ This shows that $(B-\lambda I)^{-1}\notin B(l_{p}).$ Hence $S\subseteq\sigma(B,l_{p}).$ ∎

Theorem 3.3**.**

The point spectrum of $B$ on $l_{p}$ is given by

[TABLE]

Proof.

Suppose $(B-\lambda I)x=0$ for $x\neq\theta$ in $l_{p}.$ This gives the following equations

[TABLE]

From the above equation it follows that, if $\lambda\notin\{r_{1},r_{2}\}$ then $x_{k}=0$ for all $k\in\mathbb{N}$. This implies $\sigma_{p}(B,l_{p})\subseteq\{r_{1},r_{2}\}.$ Now we consider two cases here.

Case 1: $r_{1}=r_{2}=r.$

Consider $(B-rI)x=0$ for $x\neq\theta$. Then we have the following equations

[TABLE]

Since $s_{1},s_{2}$ are non-zero complex numbers, from above equations it follows that $x_{k}=0$ for $k\in\mathbb{N}.$ Hence $\sigma_{p}(B,l_{p})=\emptyset$ in this case.

Case 2: $r_{1}\neq r_{2}.$

Consider $(B-r_{1}I)x=0$ for $x\neq\theta$. Then we have the following equations

[TABLE]

Here we consider two subcases.

Subcase 1: $t_{1}\neq 0,$ $t_{2}\neq 0.$

Let $s_{1}s_{2}-t_{1}(r_{2}-r_{1})\neq 0.$ Then second and third equation of (3.10) give

[TABLE]

This implies $x_{1}=0$ and $x_{2}=0.$ From fourth and fifth equation of (3.10) we have

[TABLE]

Then $x_{3}=0$ and $x_{4}=0.$ Proceeding in this way we have $x_{k}=0$ for all $k\in\mathbb{N}.$ Which is a contradiction.

Now let $s_{1}s_{2}-t_{1}(r_{2}-r_{1})=0$ where $s_{1},$ $s_{2},$ $t_{1}$ are non-zero and $r_{1}\neq r_{2}.$ Then from (3.11) we have either $(x_{1},x_{2})=(0,0)$ or $x_{1}\neq 0,$ $x_{2}\neq 0.$ If $(x_{1},x_{2})=(0,0)$ then from from (3.12) we have either $(x_{3},x_{4})=(0,0)$ or $x_{3}\neq 0,$ $x_{4}\neq 0.$ In this way if we consider every pair $(x_{1},x_{2}),(x_{3},x_{4}),\cdots$ equals to $(0,0)$ then $x_{k}=0$ for all $k\in\mathbb{N}.$ Hence let $(x_{p},x_{p+1})$ be the first pair such that $x_{p}\neq 0$ and $x_{p+1}\neq 0$ for some odd $p\in\mathbb{N}$. This implies

[TABLE]

Then the equations in (3.10) reduce to

[TABLE]

From 4th equation of (3.13) we have $x_{p+3}=-\frac{t_{1}}{s_{2}}x_{p+2}$. Substituting it to 3rd equation of (3.13) we get $t_{2}s_{2}x_{p+1}+(s_{1}s_{2}-t_{1}(r_{2}-r_{1}))x_{p+2}=0$ and this implies $t_{2}s_{2}x_{p+1}=0$ which is not possible. Therefore $r_{1}\notin\sigma_{p}(B,l_{p}).$

Sub-case 2: $(t_{1},t_{2})=(0,0).$

Here the system of equations (3.10) reduces to

[TABLE]

where $k\in\mathbb{N}.$ This implies $x_{2k}=0$ and consequently $x_{2k-1}=0$ for all $k\in\mathbb{N}.$ Therefore $r_{1}\notin\sigma_{p}(B,l_{p}).$

Similarly as above it can be proved that $r_{2}\notin\sigma_{p}(B,l_{p})$ and this proves the theorem. ∎

Theorem 3.4**.**

The point spectrum of $B^{*}$ over $l_{p}^{*}$ is given by

[TABLE]

where $\chi$ is mentioned in (3.2).

Proof.

Let $S_{1}=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|<1\right\}$ and $\lambda\in S_{1}$. Consider $(B^{*}-\lambda I)x=0$ where $x\neq\theta.$ This implies the following equations

[TABLE]

If $\lambda=r_{1}$ then $(1,0,0,\cdots)$ is an eigenvector corresponding to the eigenvalue $r_{1}$. This implies $r_{1}\in\sigma_{p}(B^{*},l_{p}^{*}).$ If $\lambda=r_{2}$ then $(1,-\frac{r_{1}-r_{2}}{s_{1}},0,0,\cdots)$ is an eigenvector corresponding to the eigenvalue $r_{2}$. This implies $r_{2}\in\sigma_{p}(B^{*},l_{p}^{*}).$ Assume $\lambda\notin\{r_{1},r_{2}\}.$ Then (3.14) can be written as

[TABLE]

for $k=1,2,3,\cdots.$ We prove the theorem in two cases.

Case 1: $t_{1}\neq 0,$ $t_{2}\neq 0.$

Equations in (3.15) implies

[TABLE]

Therefore we have the following system of difference equation

[TABLE]

where

[TABLE]

The characteristic equation of the matrix $C$ is

[TABLE]

It can be easily verified that $g(\alpha)=0$ is the reciprocal equation of $f(\alpha)=0$ in (3.6) and since $t_{1}\neq 0,$ $t_{2}\neq 0$ and $\lambda\notin\{r_{1},r_{2}\}$ then the roots of both the equations $f(\alpha)=0$ and $g(\alpha)=0$ are non zero. Then

[TABLE]

is an eigenvalue of $C$ and let $f=(f_{1},f_{2})^{t}$ be the corresponding eigenvector. Then

[TABLE]

is a solution of (3.15). Since $\lambda\in S_{1}$ so $|\alpha|<1$ and this implies $(x_{k})\in l_{p}^{*}\cong l_{q}$ where $\frac{1}{p}+\frac{1}{q}=1.$ This proves that $\lambda\in\sigma_{p}(B^{*},l_{p}^{*}).$ Hence

[TABLE]

Also $S_{1}\subseteq\sigma_{p}(B^{*},l_{p}^{*})\subseteq\sigma(B^{*},l_{p}^{*})=\sigma(B,l_{p})=S_{1}\cup S_{2}$ where

[TABLE]

Let $\lambda\in S_{2}.$ Then the eigenvalues of $C$ are

[TABLE]

where $\beta_{1}=\frac{1}{\alpha_{1}},$ $\beta_{2}=\frac{1}{\alpha_{2}}.$ Let $C$ is diagonalisable. Then the general solution of (3.15) is of the form

[TABLE]

where $(\tilde{f}_{1},\ \tilde{f}_{2})^{t}$ and $(\tilde{g}_{1},\ \tilde{g}_{2})^{t}$ are eigenvectors corresponding to $\beta_{1}$ and $\beta_{2}$ respectively and $c_{1}$ and $c_{2}$ are arbitrary constants. Since $\lambda\in S_{2}$ so $\left|\frac{1}{\alpha_{1}}\right|=1$ i.e., $|\beta_{1}|=1.$ Also from the relation $|1-\sqrt{z}|\leq|1+\sqrt{z}|$ for all $z\in\mathbb{C},$ we have $|\alpha_{1}|=1$ implies $|\alpha_{2}|\leq 1.$ Therefore $\left|\frac{1}{\alpha_{2}}\right|=|\beta_{2}|\geq 1.$ Using these relations in (3.17) we get $x_{k}\nrightarrow 0$ as $k\rightarrow\infty.$ Hence $(x_{k})\notin l_{p}^{*}.$ Also let $C$ is not diagonalisable. Then $\chi=0$ and the eigenvalue of $C$ is

[TABLE]

Since $\chi=0,$

[TABLE]

This implies $|\beta|=1$ since $\lambda\in S_{2}.$ Then similar as Theorem 3.2, the general solution of (3.15) is of the form

[TABLE]

where $(\tilde{h}_{1},\ \tilde{h}_{2})^{t}$ and $(\tilde{u}_{1},\ \tilde{u}_{2})^{t}$ are the eigenvector and generalised eigenvector of $\beta$ respectively and $\tilde{c_{1}},\tilde{c_{2}}$ are arbitrary constants. Since $|\beta|=1,$ from (3.18) it follows that, $x_{k}\nrightarrow 0$ as $k\rightarrow\infty.$ Hence $(x_{k})\notin l_{p}^{*}.$ Therefore $\lambda\notin\sigma_{p}(B^{*},l_{p}^{*}).$ This proves that $\sigma_{p}(B^{*},l_{p}^{*})=S_{1}$.

Case 2: $t_{1}=t_{2}=0.$

In this case we have

[TABLE]

Then (3.15) reduces to

[TABLE]

for $k=1,2,3,\cdots$. If $x_{1}=0$ then the above equations imply $x_{k}=0$ for all $k\in\mathbb{N}.$ Hence let $x_{1}\neq 0.$ Then from (3.19) we have

[TABLE]

for $k=1,2,3,\cdots.$ Therefore $(x_{k})\in l_{p}^{*}\cong l_{q}$ if and only if $\left|\frac{(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}}\right|<1.$ This proves the required result.

∎

Theorem 3.5**.**

The residual spectrum and continuous spectrum of $B$ over $l_{p}$ are given by

1. (i)

$\sigma_{r}(B,l_{p})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|<1\right\},$ 2. (ii)

$\sigma_{c}(B,l_{p})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|=1\right\}.$

Where $\chi$ is mentioned in (3.2).

Proof.

From Lemma 2.2 it can be easily derived that,

[TABLE]

Hence the results for residual spectrum follows from Theorem 3.3 and Theorem 3.4. Also since the spectrum is a disjoint union of point spectrum, residual spectrum and continuous spectrum, the other result follows immediately. ∎

Theorem 3.6**.**

The operator $B$ satisfies the following relations,

1. (a)

$A_{3}\sigma(B,l_{p})=B_{3}\sigma(B,l_{p})=C_{3}\sigma(B,l_{p})=\emptyset,$ 2. (b)

$B_{2}\sigma(B,l_{p})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|=1\right\},$ 3. (c)

$C_{2}\sigma(B,l_{p})\supseteq\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|<1\right\}\setminus\{r_{1},r_{2}\},$ 4. (d)

$C_{1}\sigma(B,l_{p})\subseteq\{r_{1},r_{2}\}.$

Proof.

From Table 1 we have the following relations

[TABLE]

The result in $(a)$ follows from the above relations and Theorem 3.3. Also since $\sigma_{c}(B,l_{p})=B_{2}\sigma(B,l_{p}),$ the result in $(b)$ follows from Theorem 3.5. Again from the proof of Theorem 3.2 it follows that, for any $\alpha\in\sigma_{r}(B,l_{p})\setminus\{r_{1},r_{2}\}$ the operator $(B-\alpha I)^{-1}\notin B(l_{p})$. This implies that $\sigma_{r}(B,l_{p})\setminus\{r_{1},r_{2}\}\subseteq C_{2}\sigma(B,l_{p})$ and $C_{1}\sigma(B,l_{p})\subseteq\{r_{1},r_{2}\}.$ ∎

Theorem 3.7**.**

The operator $B$ satisfies the following relations

1. (a)

$\sigma_{ap}(B,l_{p})\supseteq\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\}\setminus\{r_{1},r_{2}\},$ 2. (b)

$\sigma_{ap}(B^{*},l_{p}^{*})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\},$ 3. (c)

$\sigma_{\delta}(B,l_{p})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\},$ 4. (d)

$\sigma_{co}(B,l_{p})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|<1\right\}.$

Proof.

From Table 1 we have $\sigma_{ap}(B,l_{p})=\sigma(B,l_{p})\setminus C_{1}\sigma(B,l_{p})$ and the result in $(a)$ follows from Theorem 3.6. The results in $(b),$ $(c)$ and $(d)$ follow from the relations $(d),$ $(b)$ and $(c)$ in Proposition 2.1 respectively. ∎

4. Spectra of $B$ on $l_{1}$

In this section we give the results on the fine spectrum of the operator $B$ on the sequence space $l_{1}.$ First we mention the result on the boundedness of $B.$

Theorem 4.1**.**

The operator $B:l_{1}\rightarrow l_{1}$ is a bounded linear operator and

[TABLE]

Proof.

From Lemma 2.3 it follows that $B$ is a bounded linear operator from $l_{1}$ to itself. For any $(x_{k})\in l_{1}$ we have

[TABLE]

This implies $\|B\|_{l_{1}}\leq\max\{|r_{1}|+|s_{1}|+|t_{1}|,\ |r_{2}|+|s_{2}|+|t_{2}|\}.$ Also let $e^{(k)}$ be a sequence in $l_{1}$ whose $k$-th component is one and other components are zero. Then similarly as Theorem 3.1 it can be derived that

[TABLE]

This proves the result. ∎

Since the spectrum and point spectrum of $B$ on $l_{1}$ can be derived using the similar arguments used in the case of $l_{p},$ we omit the proof and give the statements only.

Theorem 4.2**.**

Let $s_{1},s_{2}$ be two complex numbers such that $\sqrt{s_{1}^{2}}=s_{1}$ and $\sqrt{s_{2}^{2}}=s_{2}$ then the spectrum of $B$ on $l_{1}$ is given by

[TABLE]

where $\chi$ is mentioned in (3.2).

Theorem 4.3**.**

The point spectrum of $B$ on $l_{1}$ is given by

[TABLE]

Theorem 4.4**.**

The point spectrum of $B^{*}$ over $l_{1}^{*}$ is given by

[TABLE]

where $\chi$ is mentioned in (3.2).

Proof.

Proceeding similarly upto the equation (3.16) of Theorem 3.4, it can be proved that, $\sigma(B,l_{1})\subseteq\sigma_{p}(B^{*},l_{1}^{*})$. To prove the converse part we have

[TABLE]

∎

Theorem 4.5**.**

The residual spectrum and continuous spectrum of $B$ over $l_{1}$ are given by

1. (i)

$\sigma_{r}(B,l_{1})=\sigma(B,l_{1})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\}$, 2. (ii)

$\sigma_{c}(B,l_{1})=\emptyset,$

Where $\chi$ is mentioned in Theorem 3.2.

Proof.

Proofs are similar as Theorem 3.5. ∎

Theorem 4.6**.**

The operator $B$ satisfies the following relations

1. (a)

$A_{3}\sigma(B,l_{1})=B_{3}\sigma(B,l_{1})=C_{3}\sigma(B,l_{1})=\emptyset,$ 2. (b)

$B_{2}\sigma(B,l_{1})=\sigma_{c}(B,l_{1})=\emptyset,$ 3. (c)

$C_{2}\sigma(B,l_{1})\supseteq\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\}\setminus\{r_{1},r_{2}\},$ 4. (d)

$C_{1}\sigma(B,l_{1})\subseteq\{r_{1},r_{2}\}.$

Proof.

Proofs are similar as Theorem 3.6. ∎

Theorem 4.7**.**

The operator $B$ satisfies the following relations

1. (a)

$\sigma_{ap}(B,l_{1})\supseteq\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\}\setminus\{r_{1},r_{2}\},$ 2. (b)

$\sigma_{ap}(B^{*},l_{1}^{*})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\},$ 3. (c)

$\sigma_{\delta}(B,l_{1})=\sigma_{co}(B,l_{1})=\left\{\lambda\in\mathbb{C}:\left|\frac{2(r_{1}-\lambda)(r_{2}-\lambda)}{s_{1}s_{2}-t_{1}(r_{2}-\lambda)-t_{2}(r_{1}-\lambda)+\sqrt{\chi}}\right|\leq 1\right\}.$

Proof.

Since $\sigma_{ap}(B,l_{1})=\sigma(B,l_{1})\setminus C_{1}\sigma(B,l_{1}),$ the result in $(a)$ follows from Theorem 4.6. The other results follow from Proposition 2.1. ∎

5. Example

Here some examples are given.

1. (i)

Let $B_{1}$ be an operator of the form (1.1) where, $r_{1}=1,r_{2}=i,$ $s_{1}=2,s_{2}=1,$ and $t_{1}=-i,t_{2}=1.$ Then the spectrum of the operator $B_{1}$ is given by

[TABLE] 2. (ii)

Let $B_{2}$ be an operator of the form (1.1) where, $r_{1}=i,r_{2}=2,$ $s_{1}=1+i,s_{2}=1$ and $t_{1}=t_{2}=0$ then the spectrum of the operator $B_{2}$ is given by the set

[TABLE]

6. Conclusion

Many results on the spectrum and fine spectrum of difference operators obtained by earlier researchers can be derived from our results. Here we mention some of them.

1. (i)

If $r_{1}=r_{2}=r,$ $s_{1}=s_{2}=s$ and $t_{1}=t_{2}=t$ then the operator $B$ reduces to the operator $B(r,s,t)$ and some of the results given in [9, 8] follow from our results. 2. (ii)

If $r_{1}=r_{2}=r,$ $s_{1}=s_{2}=s$ and $t_{1}=t_{2}=0$ then the operator $B$ reduces to the operator $B(r,s).$ Therefore some of the results given in [5, 6] follow from our results. 3. (iii)

If $r_{1}=r_{2}=1,$ $s_{1}=s_{2}=-1$ and $t_{1}=t_{2}=0$ then the operator $B$ reduces to the difference operator $\Delta$ and the corresponding results given by [2] follows from our reults. 4. (iv)

If $r_{1}=r_{2}=s,$ $s_{1}=s_{2}=1-s$ and $t_{1}=t_{2}=0$ where $s$ is real number and $s\neq 0,1$ we get the Zweier matrix $Z^{s}$ and some of the results in [28] follows from our results.