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\markboth{A.\,Kara{\scriptsize{\.{I}}}sa, Y.\,Asar and D.\,T.\,Tollu}{Fine spectra of
triangular triple-band matrices}
\title{Fine spectra of triangular triple-band matrices on sequence spaces $c$ and $\ell_{p}$, $(0<p<1)$}




\author[A.\,Kara{\footnotesize{\.{I}}}sa, Y.\,Asar, D.\,T.\,Tollu]{Al{\footnotesize{\.{I}}} Kara{\footnotesize{\.{I}}}sa\affil{1}\comma\corrauth, Yas{\footnotesize{\.{I}}}n Asar\affil{1} and Durhasan Turgut Tollu\affil{1}}
\address{{\affilnum{1}\ Department of Mathematics-Computer Sciences, Faculty of Science, Necmettin Erbakan University, Meram Campus 42\,090 Konya, Turkey}}
%
\emails{{\tt akaraisa@konya.edu.tr}\,\,(A.\,Kara{\footnotesize{\.{i}}}sa), {\tt yasar@konya.edu.tr} (Y.\,Asar), {\tt dttollu@konya.edu.tr} (D.\,T.\,Tollu)}
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%%%%% Begin Abstract %%%%%%%%%%%
\begin{abstract}
The purpose of this study is to determine the fine spectra of the
operator for which the corresponding upper and lower triangular
matrices $A(r,s,t)$ and $B(r,s,t)$  are on the sequence spaces $c$
and $\ell_{p}$, where $(0<p<1)$, respectively. Further, we obtain
the approximate point spectrum, defect spectrum and compression
spectrum on these spaces. Furthermore, we give the graphical
representations of the spectrum of the triangular triple-band matrix
over the sequence spaces $c$ and  $\ell_{p}$.
\end{abstract}
%%%%% end %%%%%%%%%%%

%%%%% Keywords %%%%%%%%%%%
\keywords{Spectrum of an operator, triple band matrix, spectral
mapping theorem, sequence spaces $\ell_{p}$ and $c$, Goldberg's
classification}

%%%% AMS subject classifications %%%%
\ams{47A10, 47B37}

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\maketitle

%%%% Start %%%%%%
\section{Introduction}


It is well known that matrices play an important role in operator
theory. The spectrum of an operator generalizes the notion of
eigenvalues for matrices. The spectrum of an operator over a Banach
space is partitioned into three parts, i.e., the point spectrum,
the continuous spectrum and the residual spectrum. Calculation
of these three parts of the spectrum of an operator is called
determination of the fine spectrum of the operator.

By a sequence space we understand a linear subspace of the space
$\omega=\mathbb{C}^{\mathbb{N}_1}$ of all complex sequences
containing $\phi$, which is the set of all finitely non--zero
sequences, where $\mathbb{N}_1$ denotes the set of positive
integers. We write $\ell_\infty$, $c$, $c_0$ and $bv$ for spaces
of all bounded, convergent, null and bounded variation sequences,
which are Banach spaces with the sup-norm
$\|x\|_{\infty}=\sup_{k\in\mathbb N}|x_k|$,
$x\in\{\ell_{\infty},c_{0},c\}$ and
$\|x\|_{bv}=\sum_{k=0}^{\infty}|x_k-x_{k+1}|$, while $\phi$ is not a
Banach space with respect to any norm, respectively, where
$\mathbb{N}=\{0,1,2,\dots\}$. Also, by $\ell_p$, $(0<p<\infty)$ we
denote the sequence space of all sequences associated with a
$p-$absolutely convergent series which is a Banach space with the
norm $\|x\|_p=\left(\sum_{k=0}^{\infty}|x_k|^p\right)^{1/p}$, $1\leq
p<\infty$ and $\|x\|_p=\sum_{k=0}^{\infty}|x_k|^p$, $0< p<1$.

Let $A=(a_{nk})$ be an infinite matrix of complex numbers $a_{nk}$,
where $n,k\in\mathbb{N}$, and write
\newpage
\begin{eqnarray}\label{A-trans}
(Ax)_{n}=\sum_k a_{nk}x_k~;~~ (n\in \mathbb{N},~x\in D_{00}(A)),
\end{eqnarray}
where $D_{00}(A)$ denotes the subspace of $w$ consisting of $x\in w$
for which the sum exists as a finite sum.  For simplicity in
notation, we write the summation without limits
running from $0$ to $\infty$ and we use the convention that any
term with a negative subscript is equal to naught. More generally, if
$\mu$ is a normed sequence space, we can write $D_\mu(A)$ for
$x\in w$ for which the sum in \eqref{A-trans} converges in the norm
of $\mu$. We write
\begin{eqnarray*}(\lambda:\mu)=\{A:\lambda\subseteq D_\mu(A) \}\end{eqnarray*}
for the space of those matrices sending the whole of the sequence
space $\lambda$ into $\mu$ in this sense.

We give a short survey concerning the spectrum and the fine spectrum
of linear operators defined by some particular triangle matrices
over certain sequence spaces. The fine spectrum of the Ces\`{a}ro
operator of order one on the sequence space $\ell_p$ was studied by
Gonz\`{a}lez \cite{gon}, where $1<p<\infty$. Also, weighted mean
matrices of the operator on $\ell_p$ were investigated by Cartlidge
\cite{cart}. The spectrum of the Ces\`{a}ro operator of order one on
the sequence spaces $bv_0$ and $bv$ was investigated by Okutoyi
\cite{oku1,oku}. The spectrum and the fine spectrum of the Rhally
operator on the sequence spaces $c_0$ and $c$ were examined by Y\i
ld\i r\i m \cite{yil2}. The fine spectrum of the difference operator
$\Delta$ over the sequence spaces $c_0$ and $c$ were studied by Altay
and Ba\c sar \cite{bafbb}.  The fine spectra of the difference
operator $\Delta$ over the sequence spaces $\ell_p$ and $bv_p$ was
studied by Akhmedov and Ba\c sar \cite{aafb1,aafb2}, where $bv_p$ is
the space of $p$-bounded variation sequences introduced by Ba\c
sar and Altay \cite{fbbabv} with $1\leq p<\infty$. Also, the fine
spectrum of $B(r,s,t)$ over the sequence spaces $c_0$ and $c$ was
studied by Furkan et al. \cite{Fu07}. In 2010, Srivastava and Kumar
\cite{Sr10} determined the spectra and the fine spectra of the
generalized difference operator $\Delta_{\nu}$ on $\ell_{1}$, where
$\Delta_{\nu}$ is defined by $(\Delta_{\nu})_{nn}=\nu_n$ and
$(\Delta_{\nu})_{n+1,n}=-\nu_n$ for all $n\in\mathbb N$, under
certain conditions on the sequence $\nu =(\nu_{n})$, and they
generalized these results by the generalized difference operator
$\Delta_{uv}$ defined by
$\Delta_{uv}x=(u_{n}x_{n}+v_{n-1}x_{n-1})_{n\in\mathbb N}$ for all
$n\in\mathbb N$, (see \cite{Sr11}). Karakaya and Altun determined
the fine spectra of upper triangular double-band matrices over the
sequence spaces $c_0$ and $c$  \cite{vkma}. Akhmedov and
El-Shabrawy \cite{aafb3}, and El-Shabrawy \cite{saad, aafb42}
obtained the fine spectrum of the generalized difference operator
$\Delta_{a,b}$, defined as a double band matrix with the convergent
sequences $\widetilde{a}=(a_k)$ and $\widetilde{b}=(b_k)$ having
certain properties, over the sequence spaces $c$, $\ell_{p}$,
$(1<p<\infty)$ and $c_{0}$, respectively. Dutta and Baliarsingh
\cite{p1,p4} examined the fine spectra of the generalized $rth$
difference operator $\triangle_{v}^r$ on the sequence spaces
$\ell_1$ and $c$.

Recently, Karaisa \cite{kar,kar1} and Karaisa and Ba\c sar
\cite{kar2} have determined the fine spectrum of matrix
operators with the corresponding upper and lower triangular
matrices $A(\widetilde{r},\widetilde{s})$ and
$B(\widetilde{r},\widetilde{s})$ with the convergent sequences
$\widetilde{r}=(r_k)$ and $\widetilde{s}=(s_k)$ having certain
properties, over the sequence space $\ell_{p}$ for $(1\leq
p<\infty)$ and $(1< p<\infty)$, respectively. Later, Karaisa and
Ba\c sar \cite{akfb,akfb1,akfb2} have determined the fine spectrum
of the upper triangular triple band matrix $A(r,s,t)$ over some
sequence spaces. Finally, D\"{u}ndar and  Ba\c sar have determined
the fine spectrum of the matrix operator $\triangle^{+}$ defined by
an upper triangle double band matrix acting on the sequence space
$c_{0}$ \cite{dundar}.

This paper is organized as follows: In Section 2, some notations and
fundamental definitions are given. In Section 3, the spectrum,
the point spectrum, the residual spectrum and the continuous
spectrum of the operator $A(r,s,t)$ on the sequence space $c$ have
been computed. We also give the approximate point spectrum, the
defect spectrum and the compression spectrum of the operator
$A(r,s,t)$ over the space $c$. In Section 4, we have computed the
spectrum and the fine spectrum of the lower-triangular triple band matrix
$B(r,s,t)$ over the sequence space  $\ell_{p}$, $(0<p<1)$ as well.
Also, the boundedness and the norm of the operator $B(r,s,t)$ are given. Finally, we give the graphical representation of the
spectrum of upper triangular triple-band matrices and conclude the
study.



\section{Notations and known results}

Let $X$ and $Y$ be the Banach spaces, and $T:X \rightarrow Y$  a
bounded linear operator. By $R(T)$ we denote range of $T$, i.e.,
\begin{eqnarray*}
R(T)=\{ y\in Y:y=Tx,\text{ }x\in X\}.
\end{eqnarray*}
By $B(X)$ we also denote the set of all bounded linear
operators on $X$ into itself. If $T\in B(X) $, then the adjoint $T^{\ast }$ of $T$ is a bounded linear operator
on the dual $X^{\ast }$ of $X$ defined by $\left( T^{\ast }f\right) \left(x\right) =f\left( Tx\right)$ for all $f\in X^{\ast}$ and $x\in X$.

Let $X\neq\{\theta\}$ be a complex normed space and $T :D(T
)\rightarrow X$  a linear operator with domain $D(T )\subseteq X$.
With $T$ we associate the operator $T_{\alpha}=T-\alpha I$, where
$\alpha$ is a complex number and $I$ is the identity operator on
$D(T)$. If $T_{\alpha}$ has an inverse which is linear, it is denoted
by $T_{\alpha}^{-1}$, that is, $T_{\alpha}^{-1}=(T-\alpha I)^{-1}$
and it is called the resolvent operator of $T$.

Many properties of $T_{\alpha}$ and $T_{\alpha}^{-1}$ depend on $\alpha$, and spectral theory is concerned with those properties. For instance, we are
interested in the set of all $\alpha$'s in the complex plane such that $T_{\alpha}^{-1}$ exists. The boundedness of $T_{\alpha}^{-1}$ is another property that will be essential. We shall also ask for what $\alpha$'s the domain of $T_{\alpha}^{-1}$ is dense in $X$, to name just a few aspects. For our investigation of $T$, $T_{\alpha}$ and $T_{\alpha}^{-1}$, we need some basic concepts of spectral theory, which are given as follows (see \cite[pp. 370-371]{kre}):

A regular value $\alpha$ of $T$ is a complex number such that
\begin{itemize}
\item [(R1)] $T_{\alpha}^{-1}$ exists,
\item [(R2)] $T_{\alpha}^{-1}$ is bounded,
\item [(R3)] $T_{\alpha}^{-1}$ is defined on a set which is dense in $X$.
\end{itemize}
The resolvent set $\rho(T)$ of $T$ is a set of all regular values $\alpha$ of $T$. Its complement $ \mathbb{C}\setminus\sigma(T)$ in the complex plane
$\mathbb{C}$ is called the spectrum of $T$. Furthermore, the spectrum $\sigma(T)$ is partitioned into three disjoint sets as follows: The point
spectrum $\sigma_{p}(T)$ is a set such that $T_{\alpha}^{-1}$ does not exist. $\alpha \in\sigma_{p}(T)$ is called an eigenvalue of T. The continuous spectrum $\sigma_{c}(T)$ is a set such that $T_{\alpha}^{-1}$ exists and satisfies (R3) but not (R2). The residual spectrum $\sigma_{r}(T)$ is a set such that $T_{\alpha}^{-1}$ exists but does not satisfy (R3).

In this section, before Appell et al. \cite{Ap04}, we define the
three more subdivisions of the spectrum called the approximate
point spectrum, the defect spectrum and the compression spectrum, respectively.

Given a bounded linear operator $T$ in a Banach space $X$, we call a
sequence $(x_{k})$ in $X$  a Weyl sequence for $T$ if $\left\Vert
x_{k}\right\Vert =1$ and $\left\Vert Tx_{k}\right\Vert \rightarrow
0$, as $ k\rightarrow \infty$.

From now on, we call the set
\begin{eqnarray}\label{eq14}
\sigma_{ap}(T,X):=\{\alpha\in \mathbb{C}:\text{there exists a Weyl
sequence for }T-\alpha I\}
\end{eqnarray}
the approximate point spectrum of $T$. Moreover, the subspectrum
\begin{eqnarray}\label{eq15}
\sigma_{\delta }(T,X):=\{\alpha\in \mathbb{C}:T-\alpha I\text{ is
not surjective}\}
\end{eqnarray}
is called the defect spectrum of $T$.

The two subspectra given by (\ref{eq14}) and (\ref{eq15}) form
(not necessarily disjoint) subdivisions
\begin{eqnarray*}
\sigma (T,X)=\sigma_{ap}(T,X)\cup \sigma_{\delta }(T,X)
\end{eqnarray*}
of the spectrum. There is another subspectrum,i.e.,
\begin{eqnarray*}
\sigma_{co}(T,X)=\{\lambda \in \mathbb{C}:\overline{R(T-\alpha
I)}\neq X\},
\end{eqnarray*}
which is often called the compression spectrum in the literature.
In Goldberg \cite {Go66}, if $T\in B(X) $, with $X$
 a Banach
space, then there are three possibilities for $R(T)$:
\begin{itemize}
\item [(A)] $\quad R(T)=X$,
\item [(B)] $\quad R(T)\neq\overline{R(T)}=X$,
\item [(C)] $\quad \overline{R(T)}\neq X$.,\end{itemize}
and three possibilities for $T^{-1}$:
\begin{itemize}
\item [(1)] $\quad T^{-1}$ exists and is continuous,
\item [(2)] $\quad T^{-1}$ exists but is discontinuous,
\item [(3)] $\quad T^{-1}$ does not exist.
\end{itemize}

If these possibilities are combined in all possible ways, nine
different states are created. These are labelled by: $A_{1}$,
$A_{2}$, $A_{3}$, $ B_{1} $, $B_{2}$, $B_{3}$, $C_{1}$,
$C_{2}$, and $C_{3}$. If an operator is in state $C_{2}$, for
example, then $\overline{R(T)} \neq X$ and $T^{-1}$ exists but is
discontinuous and we can write $\sigma (T,X)C_{2}$.

 By the definitions given above, we can illustrate the subdivisions
of the spectrum in the following table:
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline &  & 1 & 2 & 3 \\ \hline
&  & $T^{-1}_\alpha$ exists & $T^{-1}_\alpha$ exists & $T^{-1}_\alpha$ \\
&  & and is bounded & and is unbounded & does not exist \\
\hline\hline
&  &  &  & $\alpha \in \sigma_{p}(T,X)$ \\
A & $R(T-\alpha I)=X$ & $\alpha \in \rho (T,X)$ & -- & $\alpha
\in \sigma_{ap}(T,X)$ \\
&  &  &  &  \\ \hline\hline
&  &  & $\alpha \in \sigma_{c}(T,X)$ & $\alpha \in \sigma_{p}(T,X)$ \\
B & $\overline{R(T-\alpha I)}=X$ & $\alpha \in \rho (T,X)$ & $\alpha
\in
\sigma_{ap}(T,X)$ & $\alpha \in \sigma_{ap}(T,X)$ \\
&  &  & $\alpha \in \sigma_{\delta }(T,X)$ & $\alpha\in \sigma
_{\delta }(T,X)$ \\ \hline\hline &  & $\alpha \in \sigma_{r}(T,X)$ &
$\alpha \in \sigma_{r}(T,X)$ & $\alpha
\in \sigma_{p}(T,X)$ \\
C & $\overline{R(T-\alpha I)}\not=X$ & $\alpha\in \sigma _{\delta
}(T,X)$
& $\alpha \in \sigma_{ap}(T,X)$ & $\alpha \in \sigma_{ap}(T,X)$ \\
&  &  & $\alpha \in \sigma_{\delta }(T,X)$ & $\alpha \in \sigma
_{\delta
}(T,X)$ \\
&  & $\alpha \in \sigma_{co}(T,X)$ & $\alpha \in \sigma _{co}(T,X)$
& $ \alpha \in \sigma_{co}(T,X)$ \\ \hline
 \end{tabular}

\vspace*{0.3cm}
{\small {\sf Table 1:} {\em Subdivisions of the spectrum of a linear operator}}
\vspace*{0.3cm}\end{center}

\begin{proposition}[see \cite{Ap04}, Proposition 1.3, p.\,28] \label{pr11}
Spectra and subspectra of an operator $T\in B(X)$ and its adjoint
$T^{\ast}\in B(X^{\ast })$ are related by the following relations:
\begin{enumerate}
 \item[(a)] $\sigma(T^{\ast},X^{\ast})=\sigma (T,X)$,
  \item[(b)] $\sigma_{c}(T^{\ast},X^{\ast})\subseteq \sigma_{ap}(T,X)$,
  \item[(c)] $\sigma_{ap}(T^{\ast},X^{\ast})=\sigma_{\delta }(T,X)$,
  \item[(d)] $ \sigma_{\delta}(T^{\ast},X^{\ast})=\sigma_{ap}(T,X)$,
  \item[(e)] $\sigma_{p}(T^{\ast},X^{\ast})=\sigma_{co}(T,X)$,
  \item[(f)] $\sigma_{co}(T^{\ast},X^{\ast})\supseteq \sigma_{p}(T,X)$,
  \item[(g)] $\sigma(T,X)=\sigma_{ap}(T,X)\cup\sigma_{p}(T^{\ast},X^{\ast})=\sigma_{p}(T,X)\cup\sigma_{ap}(T^{\ast},X^{\ast})$.\end{enumerate}
\end{proposition}
Relations (c)--(f) show that the approximate point spectrum is
in a certain sense dual to the defect spectrum, and the point spectrum is
dual to the compression spectrum. Equality (g) implies, in
particular, that $\sigma (T,X)=\sigma_{ap}(T,X)$ if $X$ is a Hilbert
space and $T$ is normal. Roughly speaking, this shows that normal
(in particular, self-adjoint) operators on the Hilbert spaces are
most similar to matrices in finite dimensional spaces (see
\cite{Ap04}).

\section{Fine spectra of upper triangular triple-band matrices over the space of convergent sequences}
In this section, our main focus is on the upper triple-band matrix $A(r,s,t)$,
where
\begin{eqnarray*}
A(r,s,t)= \left[ \begin{array}{ccccc}
r & s & t &0&\ldots \\
0 & r& s&t& \ldots \\
0 & 0 & r &s& \ldots \\0&0 & 0 & r & \ldots \\
\vdots &   \vdots& \vdots& \vdots& \ddots
\end{array} \right].
\end{eqnarray*}
From now on, we assume that $s$ and $t$ are complex parameters which
do not vanish simultaneously.

Let us introduce the operator $A(r,s,t)$ from $c$ to itself by
\begin{eqnarray*}
A(r,s,t)x=(rx_{k}+sx_{k+1}+tx_{k+2})^{\infty}_{k=0} ~\textrm{ where }~x=(x_k)\in c.
\end{eqnarray*}
In this section, we examine the spectrum, the point spectrum, the
residual spectrum and the continuous spectrum of the operator
$A(r,s,t)$ on the sequence space $c$. Besides, we give the
approximate point spectrum, the defect spectrum and the compression
spectrum of the matrix operator $A(r,s,t)$  over the space $c$.


\begin{lemma}[see \cite{will}, Theorem 1.3.6, p.\,6]
The matrix $A=(a_{nk})$ gives rise to a bounded linear operator
$T\in B(c)$ from $c$ to itself if and only if
\begin{itemize}
\item [(1)] the rows of $A$ are in $\ell_{1}$ and their $\ell_{1}$ norms are
bounded,
\item [(2)] the columns of $A$  are in $c$,
\item [(3)] the sequence of row sums of $A$ is in $c$.
\end{itemize}
The operator norm of $T$ is the supremum of the $\ell_{1}$ norms of
the rows.
\end{lemma}

\begin{corollary}  $A(r,s,t):c\rightarrow c$ is a bounded linear operator and
\begin{eqnarray}\label{eq21}
\|A(r,s,t)\|_{(c:c)}=|r|+|s|+|t|.
\end{eqnarray}
\end{corollary}

\begin{lemma}[see \cite{Go66}, p. 59]\label{lem4}
$T$ has a dense range if and only if $T^{\ast}$ is one to one.
\end{lemma}

If $T:c\to c$ is a bounded matrix operator with the
matrix $A$, then $T^{\ast}:c^{\ast}\to c^{\ast}$ acting on
$\mathbb{C}\oplus \ell_{1}$ has a matrix representation of the form $\left[\begin{array}{cc}
\chi& 0 \\
b& A^{t}
\end{array}\right]$, where $\chi$ denotes the characteristic of the matrix $A$ and $b$ is the
column vector whose $k^{th}$ entry is the limit of the column of $A$
for each $k\in \mathbb{N} $. For
$A(r,s,t):c\to c$, the matrix
$A(r,s,t)^{\ast}\in B(\ell_{1})$ is of the
form \begin{eqnarray*}
A(r,s,t)^{\ast}= \left[
\begin{array}{cc}
r+s+t& 0 \\
0& A^{t}(r,s,t)
\end{array}\right].
\end{eqnarray*}

\begin{theorem}\label{thm4.1}
$A(r,s,t):c\to c $ has a dense range if and only if $\alpha \neq
r+s+t$.
\end{theorem}
\begin{proof}
First, let us show that $\sigma_{p}[A(r,s,t)^{\ast},\mathbb{C}\oplus
\ell_{1}]=\{r+s+t\}$. Suppose that $\alpha$ is an eigenvalue of the
operator $A(r,s,t)^{\ast}:\mathbb{C}\oplus \ell_{1}\to
\mathbb{C}\oplus \ell_{1}$. Then there exists $f\in\ell_{1}$
satisfying the system of equations
\begin{eqnarray}\label{aaa}\left.
\begin{array}{rcl}
(r+s+t)f_{0}&=&\alpha f_{0}\\rf_{1}&=&\alpha
f_{1}\\sf_{1}+rf_{2}&=&\alpha
f_{2}\\tf_{1}+sf_{2}+rf_{3}&=&\alpha f_{3}\\ &\vdots&
\end{array}\right\}
\end{eqnarray}
From the above system of linear equation, one can see that $\alpha =
r+s+t$ is an eigenvalue corresponding to the eigenvector
$(1,0,0,0,\ldots)$. Now, suppose that $\alpha \neq r+s+t$. Then,
$f_0=0$. Let  $f_{k}$ be the first nonzero entry of the sequence
$f$. Then by the $k$-th equation of \eqref{aaa}, $\alpha=r$. But, the
$(k+1)$-th equation $f_{k}=0$ for $s\neq 0$, which contraditcts our
assumption. Hence there is no other eigenvalue. So
$\sigma_{p}[A(r,s,t)^{\ast},\mathbb{C}\oplus \ell_{1}]=\{r+s+t\}$.
\end{proof}




\begin{lemma}[see \cite{Go66}, p.\,60]\label{lem3}
The adjoint operator $T^{\ast}$ of $T$ is onto if and only if $T$ has a bounded inverse.
\end{lemma}


Before giving the main theorem of this section, we should note the
following remark. In this paper, from now on, if $z$ is a complex
number, then by $\sqrt{z}$  we always mean the square root of $z$
with a nonnegative real part. If $Re(\sqrt{z})=0$, then $\sqrt{z}$
represents the square root of z with $Im(\sqrt{z})>0$. The same
results are obtained if  $\sqrt{z}$ represents the other square
roots.
\begin{theorem}\label{thm3.2}
Let $s$ be a complex number such that $\sqrt{s^{2}}=-s$ and define
the set $D_{1}$ by
\begin{eqnarray*}
D_{1}=\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|\leq \big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}.
\end{eqnarray*} Then,
$\sigma_{c}[A(r,s,t),c]\subseteq D_{1}.$
\end{theorem}
\begin{proof}
Let $y=(y_{k})\in \ell_{1}$. Then, by solving the equation
$A_{\alpha}(r,s,t)^{\ast}x=y$
for $x= (x_{k})$ in terms of  $y$, we obtain
\begin{eqnarray*}
x_{0}&=&\frac{ y_{0} }{r+s+t-\alpha} ,\\
x_{1}&=&\frac{ y_{1} }{r-\alpha} ,\\
x_{2}&=&\frac{ y_{2} }{r-\alpha}+\frac{- sy_{1}
}{(r-\alpha)^{2}} ,\\
x_{3}&=&\frac{ y_{3} }{r-\alpha}+\frac{
-sy_{2}}{(r-\alpha)^{2}}+\frac{( s^{2}-t(r-\alpha))y_{1}
}{(r-\alpha)^{3}}, \\
&\vdots&
\end{eqnarray*}
and if we denote $a_{1}=1/(r-\alpha)$, $a_{2}=-s/(r-\alpha)^{2}$, $a_{3}=[s^{2}-t(r-\alpha)]/(r-\alpha)^{3}$, we have
\begin{eqnarray}\label{3.4}
x_{0}&=&\frac{ y_{0} }{r+s+t-\alpha}\nonumber ,\\
x_{1}&=& a_{1}y_{1}  \nonumber,\\
x_{2}&=& a_{1}y_{2}+a_{2}y_{1}  \nonumber,\\
x_{3}&=& a_{1}y_{3}+a_{2}y_{2}+a_{3}y_{1}\nonumber ,\\
&\vdots&\nonumber\\
x_{n}&=& a_{1}y_{n}+a_{2}y_{n-1}+\cdots+
a_{n}y_{1}=\sum_{k=1}^{n}a_{n+1-k}y_{k}\,\,\textrm{ for }\,n\geq1.
\end{eqnarray}
Now, we must find $a_{n}$. We have
$y_{n}=tx_{n-2}+sx_{n-1}+(r-\alpha)x_{n}$, for $n\geq 3$ and if we
use relation (\ref{3.4}), we have
\begin{eqnarray*}
y_{n}&=& t\sum_{k=1}^{n-2}a_{n-1-k}y_{k}+s\sum_{k=1}^{n-1}a_{n-k}y_{k}+ (r-\alpha)\sum_{k=1}^{n}a_{n+1-k}y_{k}\\
&=&\displaystyle y_{1}[ta_{n-2}+sa_{n-1}+(r-\alpha)a_{n}]+y_{2}[ta_{n-3}+sa_{n-2}+(r-\alpha)a_{n-1}]\\&&\displaystyle +\cdots+y_{n-1}[sa_{1}+(r-\alpha)a_{2}]+
y_{n}a_{1}(r-\alpha)\,\,\textrm{ for all}\,n\geq3.
\end{eqnarray*}
This implies that $ta_{n-2}+sa_{n-1}+(r-\alpha)a_{n}=0$,
$ta_{n-3}+sa_{n-2}+(r-\alpha)a_{n-1}=0,\ldots$,$sa_{1}+(r-\alpha)a_{2}=0$,
$(r-\alpha)a_{1}=1$. In fact, this sequence is obtained recursively
by letting
\begin{eqnarray*}
a_{1}=\frac{1}{r-\alpha},~~a_{2}=\frac{-s}{(r-\alpha)^{2}} ~\textrm{ and }~ ta_{n-2}+sa_{n-1}+(r-\alpha)a_{n}=0~\textrm{ for all }~ n\geq 3.
\end{eqnarray*}
The characteristic equation of the recurrence relation is
$(r-\alpha)\lambda^{2}+s\lambda+t=0$. If
$\Delta=s^{2}-4t(r-\alpha)\neq 0$, then the straightforward
calculation gives that
\begin{eqnarray}\label{3.5}
a_{n}&=&\frac{\lambda_{1}^{n}-\lambda_{2}^{n}}{\sqrt{s^{2}-4t(r-\alpha)}}~\textrm{
for all }~ n\geq 1,~\lambda_{1}=
\frac{-s+\sqrt{\Delta}}{2(r-\alpha)},~~\lambda_{2}=\frac{-s-\sqrt{\Delta}}{2(r-\alpha)}.
\end{eqnarray}
By (\ref{3.4}), one can see that
\begin{eqnarray*}
|x_{n}|&\leq &\sum_{k=1}^{n}|a_{n+1-k}||y_{k}|\nonumber,
\,\,\textrm{ for all }\,\,n\in\mathbb{N}_{1},
\end{eqnarray*}
and we have
\begin{eqnarray*}
|x_{0}|+|x_{1}|+\cdots+|x_{n}|&\leq& \frac{ |y_{0}| }{|r+s+t-\alpha|}+\sum_{k=1}^{1}|a_{2-k}||y_{k}|+\sum_{k=1}^{2}|a_{3-k}||y_{k}|\\
&&+\cdots+\sum_{k=1}^{n}|a_{n+1-k}||y_{k}|\\
&=&\frac{ |y_{0}| }{|r+s+t-\alpha|}\!+\!\sum_{j=1}^{n}|a_{j}||y_{1}|\!+\!\sum_{j=1}^{n-1}|a_{j}||y_{2}|\!+\cdots+\!\sum_{j=1}^{1}|a_{j}||y_{n}|\\
&\leq&\frac{ |y_{0}| }{|r+s+t-\alpha|}+
\sum_{j=1}^{n}|a_{j}|(|y_{1}|+|y_{2}|+\cdots+|y_{n}|)
\end{eqnarray*}
for all $n\in \mathbb{N}$. By letting $n\to\infty$, we get
\begin{eqnarray*}
\|x\|_{1}\leq |\|y\|_{1}\sum_{j=1}^{\infty}|a_{j}|+\frac{ |y_{0}|
}{|r+s+t-\alpha|}.
\end{eqnarray*}
We must show that $\sum_{j=1}^{\infty}|a_{j}|<\infty$. There are two cases here:

\textbf{Case 1.} If $\Delta=s^{2}-4t(r-\alpha)\neq 0$,  relation
(\ref{3.5}) holds for all $k\in\mathbb{N}_{1}$. Now, we show that if
$|\lambda_{1}|<1,|\lambda_{2}|<1$. Assume that $|\lambda_{1}|<1$. So
we have
\begin{eqnarray*}
\left|-s+\sqrt{s^{2}-4t(r-\alpha)}\right|<\left|2(r-\alpha)\right|.
\end{eqnarray*}
Since $\sqrt{s^{2}}=-s$, one can  see that

\begin{eqnarray*}
\left|1+\sqrt{1-\frac{4t(r-\alpha)}{s^{2}}}\right|<\left|\frac{2(r-\alpha)}{-s}\right|.
\end{eqnarray*}
Since $|1-\sqrt{z}|\leq |1+\sqrt{z}|$ for any $z\in \mathbb{C}$, we
get
\begin{eqnarray*}
\left|1-\sqrt{1-\frac{4t(r-\alpha)}{s^{2}}}\right|<\left|\frac{2(r-\alpha)}{-s}\right|.
\end{eqnarray*}
It follows that $|\lambda_{2}|<1$.
 Now, for $|\lambda_{1}|<1$ we can see that
\begin{eqnarray*}
\sum_{j=1}^{\infty}|a_{j}|\leq\frac{1}{|\sqrt{\Delta}|}\left(\sum_{j=1}^{\infty}|\lambda_{1}|^{j}+\sum_{j=1}^{\infty}|\lambda_{2}|^{j}\right).
\end{eqnarray*}
$x=(x_{k})\in\ell_{1}$ since $|\lambda_{1}|<1$. Hence, $A_{\alpha}(r,s,t)^{\ast}$ is onto. By Lemma \ref{lem3}, $A_{\alpha}(r,s,t)$ has a bounded inverse.
This means that
\begin{eqnarray*}
\sigma_{c}[A(r,s,t),c]\subseteq\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|\leq
\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}=D_{1}.
\end{eqnarray*}

\textbf{Case 2.} If $\Delta=s^{2}-4t(r-\alpha)= 0,$ calculation on the
recurrence sequence gives
\begin{eqnarray}\label{aaaa}
a_{n}=\left(\frac{2n}{-s}\right)\left[\frac{-s}{2(r-\alpha)}\right]^{n}~~~~\textrm{for
all}~~ n\geq 1.
\end{eqnarray}
Now, for $|-s|<2|r-\alpha|$ we can see that
\begin{eqnarray*}
\lim_{n \to \infty}\left|\frac{a_{n}}{a_{n-1}}\right|=\left|\frac{-s}{2(r-\alpha)}\right|<1.
\end{eqnarray*}
Since $\sum_{k=1}^{\infty}|a_{k}|$ is convergent,
$x=(x_{k})\in\ell_{1}$. Hence, $A_{\alpha}(r,s,t)^{\ast}$ is onto.
By Lemma \ref{lem3}, $A_{\alpha}(r,s,t)$ has a bounded inverse. This
means that
\begin{eqnarray*}
\sigma_{c}[A(r,s,t),c]\subseteq\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|\leq
\big|{-s}\big|\right\}\subseteq D_{1}.
\end{eqnarray*}
\end{proof}

\begin{theorem}\label{thm3.4}
Define $D_{2}$ by
\begin{eqnarray*}
D_{2}=\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|<\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}.
\end{eqnarray*} Then $\sigma_{p}[A(r,s,t),c]=D_{2}\cup\{r+s+t\}$.
\end{theorem}
\begin{proof}
Let $A(r,s,t)x=\alpha x$ for $\theta\neq
x\in c$. Then, by solving the system of linear equations
\begin{eqnarray*}\left.\begin{array}{rcl}
rx_{0}+sx_{1}+tx_{2}&=&\alpha x_{0}\\[1ex]
rx_{1}+sx_{2}+tx_{3}&=&\alpha
x_{1}\\[1ex]rx_{2}+sx_{3}+tx_{4}&=&\alpha
x_{2}\\[1ex]&\vdots&\\[1ex]rx_{k-2}+sx_{k-1}+tx_{k}&=&\alpha x_{k-2}\\[1ex]
&\vdots&
\end{array}\right\}\end{eqnarray*}
we have
\begin{eqnarray}\label{3}\left.\begin{array}{rcl}\label{3}
x_{2}&=&\displaystyle\frac{-s}{t}x_{1}-\frac{r-\alpha}{t}x_{0}\\
x_{3}&=&\displaystyle\frac{s^{2}-t(r-\alpha)}{t^{2}}x_{1}+\frac{s(r-\alpha)}{t^{2}}x_{0}\\
&\vdots&\\
x_{n}&=&\displaystyle\frac{a_{n}(r-\alpha)^{n}}{t^{n-1}}x_{1}-\frac{a_{n-1}(r-\alpha)^{n}}{t^{n-1}}x_{0}
\end{array}\right\}\end{eqnarray}

for all $n\geq2$. Assume that $\alpha\in D_{2}$. Then we choose
$x_{0}=1$ and $x_{1}=2(r-\alpha)/[-s+\sqrt{s^{2}-4t(r-\alpha)}]$. We
will show that $x_{n}=x_{1}^{n}$ for all $n\geq2$. Since
$\lambda_{1}$ and $\lambda_{2}$ are roots of the characteristic
equation $(r-\alpha)\lambda^{2}+s\lambda+t=0 $, we must have
\begin{eqnarray*}
\lambda_{1}\lambda_{2}=\frac{t}{r-\alpha}~\textrm{ and }~\lambda_{1}-\lambda_{2}=\frac{\sqrt{\Delta}}{r-\alpha}
\end{eqnarray*}
combining $x_{1}=1/\lambda_{1}$ with  relation (\ref{3}) one can see that
\begin{eqnarray}
x_{n}&=&\frac{a_{n}(r-\alpha)^{n}}{t^{n-1}}x_{1}-\frac{a_{n-1}(r-\alpha)^{n}}{t^{n-1}}x_{0}\label{a1}\\
&=&\left(\frac{r-\alpha}{t}\right)^{n-1}(r-\alpha)(-a_{n-1}x_{0}+a_{n}x_{1})\label{aa1}\\
&=&\frac{1}{(\lambda_{1}\lambda_{2})^{n-1}}\frac{r-\alpha}{\sqrt{\Delta}}(-\lambda_{1}^{n-1}+\lambda_{2}^{n-1}+\lambda_{1}^{n-1}-\lambda_{2}^{n}\lambda_{1}^{-1})\nonumber\\
&=&\frac{1}{\lambda_{1}^{n-1}\lambda_{2}^{n-1}}\left(\frac{1}{\lambda_{1}-\lambda_{2}}\right)\lambda_{2}^{n-1}\left(\frac{\lambda_{1}-\lambda_{2}}{\lambda_{1}}\right)\nonumber\\
&=&\frac{1}{\lambda_{1}^{n}}\nonumber\\
&=&x_{1}^{n}.\label{a2}
\end{eqnarray}
The same result is obtained in the case $\triangle=0$ as follows;
$\Delta=0$ implies that $s^2-4t(r-\alpha)=0$. Thus,
$r-\alpha=\frac{s^{2}}{4t}$ and $\lambda_1=\lambda_2=-\frac{2t}{s}$.
By \eqref{aaaa}, we get
\begin{eqnarray}\label{yaa1}
a_{n}=\left(\frac{2n}{-s}\right)\left(\frac{-2t}{s}\right)^{n}.
\end{eqnarray}

Then, substituting \eqref{aa1} and choosing $x_0=1$ and
$x_1=-\frac{s}{2t}$, we can write
\begin{eqnarray*}
x_n &=& \left(\frac{s}{-2t}\right)^{2n-2}\frac{s^2}{4t}\left[\frac{2n-2}{s}\left(\frac{-2t}{s}\right)^{n-1}-\frac{2n}{s}\left(\frac{-2t}{s}\right)^n\left(\frac{s}{-2t}\right)\right]\\
&=& \left(\frac{s}{-2t}\right)^{2n-2}\left(\frac{-2t}{s}\right)^{n-1}\frac{s^2}{4t}\left(\frac{2n-2}{s}-\frac{2n}{s}\right)\\
&=&\left(\frac{s}{-2t}\right)^n\left(\frac{-2t}{s}\right)\frac{s^2}{4t}\left(\frac{-2}{s}\right)\\
&=& \left(\frac{s}{-2t}\right)^n\\
&=& x_1^n.
\end{eqnarray*}
 Since $|x_{1}|<1$ and $x_{1}=1$, i.e., $\alpha=r+s+t$, $x=(x_{n})\in c$.
This shows that $D_{2}\cup\{r+s+t\}\subseteq\sigma_{p}[A(r,s,t),c]$.

Now, we assume that $\alpha\notin D_{2}$, i.e, $|\lambda_{1}|\leq1$.
We must show that $\alpha\notin\sigma_{p}[A(r,s,t),c]$. In this
situation, we examine the following three cases.

\textbf{Case 1.}  $|\lambda_{2}|<|\lambda_{1}|< 1.$ In this case, we have $s^{2}\neq4t(r-\alpha)$ and  from relation (\ref{3}) we obtain  that
\begin{eqnarray*}
x_{n}&=&\frac{a_{n}(r-\alpha)^{n}}{t^{n-1}}x_{1}-\frac{a_{n-1}(r-\alpha)^{n}}{t^{n-1}}x_{0}\\
&=&\left(\frac{r-\alpha}{t}\right)^{n-1}(r-\alpha)(-a_{n-1}x_{0}+a_{n}x_{1})\\
&=&\frac{r-\alpha}{\sqrt{\Delta}(\lambda_{1}\lambda_{2})^{n-1}}(-\lambda_{1}^{n-1}x_{0}+\lambda_{2}^{n-1}x_{0}+\lambda_{1}^{n-1}x_{1}-\lambda_{2}^{n}x_{1})\\
&=&\frac{r-\alpha}{\sqrt{\Delta}}\left[\left(\frac{1}{\lambda_{1}^{n-1}}-\frac{1}{\lambda_{2}^{n-1}} \right)x_{0}+\left(\frac{\lambda_{1}}{\lambda_{2}^{n-1}}-\frac{\lambda_{2}}{\lambda_{1}^{n-1}} \right)x_{1}\right]\\
&=& \frac{r-\alpha}{\sqrt{\Delta}}\left[\frac{1}{\lambda_{1}^{n-1}}\left(x_{0}-\lambda_{2}x_{1} \right)+\frac{1}{\lambda_{2}^{n-1}}\left(-x_{0}+\lambda_{1}x_{1} \right) \right].
\end{eqnarray*}
Now, if $-x_{0}+\lambda_{1}x_{1}=0$ and $x_{0}-\lambda_{2}x_{1}=0$,  then we have $\lambda_{1}=\lambda_{2}$, which is a contradiction. Otherwise, $x=(x_{k})\notin c$.

\textbf{Case 2.} $|\lambda_{2}|=|\lambda_{1}|<1.$
In this case, we have $s^{2}=4t(r-\alpha)$ and using the formula
\begin{eqnarray}\label{aa2}
a_{n}=\left(\frac{2n}{-s}\right)\left[\frac{-s}{2(r-\alpha)}\right]^{n}~~~~\textrm{for
all}~~ n\geq 1.
\end{eqnarray}
Substituting \eqref{aa2} into \eqref{aa1}, we get the following
\begin{eqnarray*}
x_{n}=\frac{2(r-\alpha)}{s\lambda_{1}^{n-1}}\left[x_{0}(n-1)-nx_{1}\lambda_{1}\right].
\end{eqnarray*}
If $x_{0}=x_{1}=0$, then $x=\theta$, which is a contradiction.
Otherwise, $x=(x_{k})\notin c$ since $1/|\lambda_{1}|>1$.

\textbf{Case 3.} $|\lambda_{2}|=|\lambda_{1}|=1.$ In this case, we
have $s^{2}=4t(r-\alpha)$ and so we have $|-s/2t|=1$. Substituting
\eqref{yaa1} into \eqref{aa1}, we obtain the following
\begin{eqnarray*}
x_{n}=\left(\frac{-s}{2t}\right)^{n-1}\left[-(n-1)\frac{-s}{2t}x_{0}+nx_{1}\right].
\end{eqnarray*}
If $x_{0}=x_{1}=0$, then $x=\theta$, which is a contradiction.
Otherwise, $x=(x_{k})\notin c$, which means
$\alpha\notin\sigma_{p}[A(r,s,t),c]$. Thus
$\sigma_{p}[A(r,s,t),c]\subseteq D_{2}\cup\{r+s+t\}$. This completes
the proof.
\end{proof}

\begin{theorem}\label{thm3.6}
$\sigma_{r}[A(r,s,t),c]=\sigma_{p}[A^{\ast}(r,s,t),c^{\ast}]\setminus\sigma_{p}[A(r,s,t),c]=\emptyset$.
\end{theorem}
\begin{proof}
For
$\alpha\in\sigma_{p}[A^{\ast}(r,s,t),c^{\ast}]\setminus\sigma_{p}[A(r,s,t),c]$,
the operator $A(r,s,t)-\alpha I$ is one to one. Thus,
$[A(r,s,t)-\alpha I]^{-1}$ exists. On the other hand,
$A(r,s,t)^{\ast}-\alpha I$ is not one to one. Hence,
$A(r,s,t)-\alpha I$ does not have a dense range in $c$ by Lemma \ref{lem4}.
From Theorem \ref{thm3.4} and Theorem \ref{thm4.1}, we have
$\sigma_{r}[A(r,s,t),c]=\emptyset$. This completes the proof.
\end{proof}



\begin{theorem}\label{thm3.7}
 Let $s$ be a complex number such that $\sqrt{s^{2}}=-s$. Then,
\begin{eqnarray*}
\sigma[A(r,s,t),c]=D_{1}.
\end{eqnarray*}
\end{theorem}
\begin{proof}
The inclusion
\begin{eqnarray*}
\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|<
\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}\subseteq\sigma[A(r,s,t),c]
\end{eqnarray*}
holds by Theorem \ref{thm3.4}. Since the spectrum of any bounded
operator is closed \cite{kre} , we have
\begin{eqnarray}\label{1}
\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|\leq
\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}\subseteq\sigma[A(r,s,t),c].
\end{eqnarray}
Again, Theorems \ref{thm3.2}-\ref{thm3.6} give that
\begin{eqnarray}\label{2}
\sigma[A(r,s,t),c]\subseteq\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|\leq
\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}.
\end{eqnarray}
By combining (\ref{1}) and (\ref{2}), one can observe that $\sigma[A(r,s,t),c]=D_{1}$, as desired.
\end{proof}
\begin{theorem}\label{thm3.8}
$\sigma_{c}[A(r,s,t),c]=D_{3}\setminus\{r+s+t\}$, where
\begin{eqnarray*}
D_{3}=\left\{\alpha\in\mathbb{C}:2\big|{r-\alpha}\big|=\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|\right\}.
\end{eqnarray*}
\end{theorem}
\begin{proof}
Since the sets $\sigma_{c}[A(r,s,t),c]$, $\sigma_{r}[A(r,s,t),c]$
and $\sigma_{p}[A(r,s,t),c]$ are pairwise disjoint and their union
is $\sigma[A(r,s,t),c]$. Thus the proof is immediate from Theorems
\ref{thm3.4}-\ref{thm3.7}.
\end{proof}

\begin{theorem} \label{thm3.8}
If $|t|>|-s|$, $r\in\sigma[A(r,s,t),c]A_{3}$.
\end{theorem}
\begin{proof}
From Theorem \ref{thm3.4}, $r\in \sigma_{p}[A(r,s,t),c]$. Thus,
$[A(r,s,t)-\alpha I]^{-1}$ does not exist. It is sufficient to show
that $A(r,s,t)-Ir$ is onto and for given $y=(y_{k})\in c$, we have
to find $x=(x_{k})\in c$ such that $[A(r,s,t)-Ir]x=y$. Solving this
equation, we get
\begin{eqnarray}\label{ali}
x_{k}=\frac{1}{t}\sum_{i=0}^{k-2}\left(\frac{-s}{t}\right)^{k-2-i}y_{i}+\left(\frac{-s}{t}\right)^{k}x_{1}
\end{eqnarray}
for $k\geq 2$. By \eqref{ali}, $x_{k}$ satisfies
\begin{eqnarray*}
x_{k}=\frac{-s}{t}x_{k-1}+\frac{y_{k-2}}{t}
\end{eqnarray*}
for $k\geq 2$. Since$|\frac{-s}{t}|<1$ and $(y_{k-2}/t)\in c$, by
Lemma 2.1 \cite{aafb3}, $x=(x_{k})\in c$. The operator $A(r,s,t)-Ir$
is onto. Hence, $r\in\sigma[A(r,s,t),c]A_{3}$.
\end{proof}



\begin{theorem}
The following statements hold:
\begin{enumerate}
\item[(i)] $\sigma_{ap}[A(r,s,t),c]=D_{1}$,
\item[(ii)] $\sigma_{\delta}[A(r,s,t),c]=D_{1}\backslash\{r\}$,
\item[(iii)] $\sigma_{co}[A(r,s,t),c]=\{r+s+t\}$.
\end{enumerate}
\end{theorem}
\begin{proof} (i) From Table 1, $\sigma_{ap}[A(r,s,t),c]=\sigma
\left[ A( r,s,t),c\right] \backslash \sigma \left[ A(
r,s,t),c\right] C_{1}$. By Theorem \ref{thm3.6}, we get
$\sigma_{r}[A(r,s,t),c]=\sigma \left[ A( r,s,t),c\right] C_{1}\cup
\sigma \left[ A( r,s,t),c\right] C_{2}=\emptyset$. Again by Table 1,
one can see that $\sigma \left[ A(r,s,t),c\right] C_{1}=\sigma
\left[A(r,s,t),c\right] C_{2}=\emptyset$. Hence,
$\sigma_{ap}[A(r,s,t),c]=D_{1}$.

(ii) The following equality
\begin{eqnarray*}
\sigma_{\delta }[A(r,s,t),c]=\sigma [A( r,s,t),c]\backslash \sigma
\left[ A(r,s,t),c\right]A_{3}
\end{eqnarray*}can be deduced from Table 1. By using Theorems \ref{thm3.7} and \ref{thm3.8}
we conclude that
$\sigma_{\delta}[A(r,s,t),c]=D_{1}\backslash\{r\}$.

(iii)  From Table 1, we have
\begin{eqnarray*}
\sigma_{co}[A(r,s,t),c]=\sigma \left[ A( r,s,t),c\right] C_{1}\cup
\sigma \left[ A( r,s,t),c\right] C_{2}\cup \sigma \left[ A(r,s,t),
c\right] C_{3}.
\end{eqnarray*} Thus, $\sigma_{co}[A(r,s,t),c]=\{r+s+t\}$ by Theorems \ref{thm4.1} and \ref{thm3.6}.
\end{proof}



\section{Fine spectra of triangular triple-band matrices over the space of $\ell_{p}$, $(0<p<1)$ }

In this section, we determine the fine spectra of a lower triangular
triple-band matrix over the sequence spaces $\ell_{p}$, where
$(0<p<1)$.

A lower  triple-band infinite matrix is of the following form:
\begin{eqnarray*}
B(r,s,t)= \left[ \begin{array}{ccccc}
r & 0 & 0 &0&\ldots \\
s & r& 0&0& \ldots \\
t & s & r &0& \ldots \\0&t & s & r & \ldots \\
\vdots &   \vdots& \vdots& \vdots& \ddots.
\end{array} \right].
\end{eqnarray*}
Let us begin with a theorem concerning the bounded linearity of the
operator $B(r, s, t)$ acting on the sequence space  $\ell_{p}$,
$(0<p<1)$.

\begin{theorem}
The operator $B(r,s,t):\ell_{p}\to \ell_{p}$ is a bounded linear
operator and
\begin{eqnarray}\label{eq211}
\|B(r,s,t)\|_{(\ell_{p}:\ell_{p})}=
|r|^{p}+|s|^{p}+|t|^{p}.
\end{eqnarray}
\end{theorem}
\begin{proof}
The linearity of the operator $B(r,s,t)$ is clear. Now, we prove
that (\ref{eq211}) holds on the space $\ell_{p}$. Let us take
$e^{(0)}=(1,0,0,\cdots)\in\ell_p$. Then
$B(r,s,t)e^{(0)}=(r,s,t,0,\ldots)$ and observe that
\begin{eqnarray*}
\|B(r,s,t)\|_{(\ell_{p}:\ell_{p})}\geq\frac{\|{B(r,s,t)}e^{(0)}\|_{p}}{\|e^{(0)}\|_{p}}=|r|^p+|s|^p+|t|^p,
\end{eqnarray*}
which gives the following
\begin{eqnarray}\label{eq221}
\|B(r,s,t)\|_{(\ell_{p}:\ell_{p})}\geq |r|^{p}+|s|^{p}+|t|^{p}.
\end{eqnarray}
Let $x=(x_k)\in\ell_p$, where $0<p<1$.  Then, by using the triangle inequality and taking $x_{-1} = 0$, we have
\begin{eqnarray*}
\|B(r,s,t)x\|_{p} &=&\sum_{k=0}^{\infty}|rx_{k}+sx_{k-1}+tx_{k-2}|^p \\
 &\leq&\sum_{k=0}^{\infty}|rx_{k}|^p+\sum_{k=0}^{\infty}|sx_{k-1}|^p+\sum_{k=0}^{\infty}|tx_{k-2}|^p\\
 &=&|r|^{p}\sum_{k=0}^{\infty}|x_{k}|^p+|s|^{p}\sum_{k=0}^{\infty}|x_{k-1}|^p+|t|^{p}\sum_{k=0}^{\infty}|x_{k-2}|^p\\
 &=&|s|^{p}\|x\|_{p}+|r|^{p}\|x\|_{p}+|t|^{p}\|x\|_{p}\\
 &=&(|r|^{p}+|s|^{p}+|t|^{p})\|x\|_{p},
\end{eqnarray*}which gives
\begin{eqnarray}\label{eq231}
\|B(r,s,t)\|_{(\ell_{p}:\ell_{p})}\leq|r|^{p}+|s|^{p}+|t|^{p}.
\end{eqnarray}Therefore, by combining inequalities (\ref{eq221}) and (\ref{eq231}) we complete the proof.
\end{proof}


If $T:\ell_{p}\longrightarrow \ell_{p}$ is a bounded matrix operator
with the matrix $A$, then it is known that the adjoint operator
$T^{\ast}:\ell_{p}^{\ast}\longrightarrow \ell_{p}^{\ast}$  is
defined by the transpose of the matrix $A$ and the dual space
$\ell^{\ast}_{p}$ of $\ell_{p}$ is isomorphic to
$\ell_{\infty}$, where $0<p<1$.

\begin{theorem}\label{thm1a}
$\sigma[B(r,s,t),\ell_{p}]=D_{1}.$
\end{theorem}
\begin{proof}
It is known from Cartlidge \cite{cart} that if a matrix operator
$A$ is bounded on $c$, then $\sigma(A, c)=\sigma(A,\ell_{\infty})$.
So we have $\sigma[A(r,s,t),\ell_{\infty}]=D_{1}.$ By Proposition
\ref{pr11},
$\sigma[B(r,s,t),\ell_{p}]=\sigma[B^{\ast}(r,s,t),\ell^{\ast}_{p}]=\sigma[A(r,s,t),\ell_{\infty}]=D_{1}$,
which completes the proof.
\end{proof}


\begin{theorem}\label{thm2a}
$\sigma_{p}[B(r,s,t),\ell_{p}]=\emptyset$.
\end{theorem}
\begin{proof}
Consider $B(r,s,t)x=\alpha x$ with
$x\neq\theta=(0,0,0,\ldots)$ in $\ell_{p}$. Then, by
solving the system of linear equations
\begin{eqnarray*}
\begin{array}{rcl}
rx_{0}&=&\alpha x_{0}\\sx_{0}+rx_{1}&=&\alpha
x_{1}\\[1ex]tx_{0}+sx_{1}+rx_{2}&=&\alpha
x_{2}\\[1ex]tx_{1}+sx_{2}+rx_{3}&=&\alpha x_{3}\\[1ex] &\vdots&
\end{array}
\end{eqnarray*}
Let $x_{k}$ be the first nonzero entry of $x$. Then the system of
equations reduces to

\begin{eqnarray*}
\begin{array}{rcl}
rx_{k}&=&\alpha x_{k}\\[1ex]sx_{k}+rx_{k+1}&=&\alpha
x_{k+1}\\[1ex]tx_{k}+sx_{k+1}+rx_{k+2}&=&\alpha
x_{k+2}\\[1ex]tx_{k+1}+sx_{k+2}+rx_{k+3}&=&\alpha x_{k+3}\\[1ex]
&\vdots&
\end{array}
\end{eqnarray*}
From the first equation we get $\alpha=r$ and using the other
equations in the given order we get $s,t=0$, which contradicts the
fact that $s,t\neq0$.
\end{proof}

\begin{theorem}\label{thm3a}
 $\sigma_{p}[B^{\ast}(r,s,t),\ell^{\ast}_{p}]=\sigma_{p}[A(r,s,t),
\ell_{\infty}] =D_{1}$.
\end{theorem}
\begin{proof}
Assume that $\alpha\in D_{1}$. By using the methodology used in the
proof of Theorem \ref{thm3.4}, and it is easy to see that $\alpha
\in D_1$ implies $|x_1|\leq 1$, we can see that $(x_n)=(x_1)^n$,  as
in equation \eqref{a2}. Thus $x_{n}\in \ell_{\infty}$. Moreover,
assume that $\alpha \notin D_2$, which implies $|\lambda|<1$. Using
the same reasoning given in Case 1 and Case 2 in the proof of
Theorem \ref{thm3.4}, $x_n \notin \ell_{\infty}$. Therefore
$\sigma_{p}[A(r,s,t), \ell_{\infty}]=D_{1}$.
\end{proof}


\begin{theorem}\label{thm4a} $\sigma_{r}[B(r,s,t),\ell_{p}]= D_{1}$.
\end{theorem}
\begin{proof}
 We show that the operator $B(r,s,t)-\alpha I$ has an inverse and
$\overline{R[B(r,s,t)-\alpha I]}$ $\neq \ell_{p}$ for $\alpha$
satisfying
$2\big|{r-\alpha}\big|\leq\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|$.
For $\alpha\neq r$, $B(r,s,t)-\alpha I$ is a triangle so it has an
inverse. For $\alpha=r$, the operator $B(r,s,t)-\alpha I$ is one to
one  by Theorem \ref{thm2a}. So it has an inverse. By Theorem
\ref{thm3a}, the operator $[B(r,s,t)-\alpha
I)]^{\ast}=B(r,s,t)^{\ast}-\alpha I$ is not one to one for
$\alpha\in\mathbb{C}$ such that
$2\big|{r-\alpha}\big|\leq\big|{-s+\sqrt{s^{2}-4t(r-\alpha)}}\big|$.
Hence the range of the operator $B(r,s,t)-\alpha I$ is not dense in
$\ell_{p}$ by Lemma \ref{lem4}. So,
$\sigma_{r}[B(r,s,t),\ell_{p}]=D_{1}$.
\end{proof}


\begin{theorem}\label{thm5a}
$\sigma_{c}[B(r,s,t),\ell_{p}]=\emptyset$.
\end{theorem}
\begin{proof}

Since the parts $\sigma_{c}[B(r,s,t),\ell_{p}]$,
$\sigma_{r}[B(r,s,t),\ell_{p}]$ and $\sigma_{p}[B(r,s,t),\ell_{p}]$
are pairwise disjoint and their union is
$\sigma[B(r,s,t),\ell_{p}]$, the proof is immediate, from Theorems
\ref{thm1a},  \ref{thm2a} and \ref{thm4a}.
\end{proof}





\begin{theorem} \label{thm6a}If $|t|>|-s|$, $r\in \sigma[B(r,s,t),\ell_{p}]C_{1}$.
\end{theorem}
\begin{proof}
From Theorem \ref{thm4a}, $r\in \sigma_{r}[B(r,s,t),\ell_{p}]$. It
is sufficient to show that the operator $[B(r,s,t)-Ir]^{-1}$ is
continuous. By Lemma \ref{lem3}, it is enough to show that
$[B(r,s,t)-Ir]^{\ast}$ is onto and for given
$y=(y_{k})\in\ell_{p}^{\ast}=\ell_{\infty}$, we have to find
$x=(x_{k})\in\ell_{\infty}$ such that $[B(r,s,t)-Ir]^{\ast}x=y$.
Solving the system of linear equations
\begin{eqnarray*}\begin{array}{rcl}
sx_{1}+tx_{2}&=&\ y_{0}\\[1ex]sx_{2}+tx_{3}&=& y_{1}\\[1ex]sx_{3}+tx_{4}&=&
y_{2}\\[1ex]&\vdots&\\[1ex]sx_{k}+tx_{k+1}&=&y_{k-1}\\[1ex]
&\vdots&
\end{array}\end{eqnarray*}one can easily observe that
\begin{eqnarray*}
x_{k}=\frac{1}{t}\sum_{i=0}^{k-2}\left(\frac{-s}{t}\right)^{k-2-i}y_{i}+\left(\frac{-s}{t}\right)^{k}x_{1}.
\end{eqnarray*}
We can easily see that $x=(x_{k})\in\ell_{\infty}$ since
$|t|>|-s|$. This shows that $[B(r,s,t)-Ir]^{\ast}$ is onto. Hence,
$r\in\sigma[B(r,s,t),\ell_{p}]C_{1}$.
\end{proof}

\begin{theorem}
The following statements hold:
\begin{enumerate}
\item[(i)] $\sigma_{ap}[B(r,s,t),\ell_{p}]=D_{1}\backslash\{r\}$ for $|t|>|-s|$,
\item[(ii)] $\sigma_{\delta}[B(r,s,t),\ell_{p}]=D_{1}$,
\item[(iii)] $\sigma_{co}[B(r,s,t),\ell_{p}]=D_{1}$.
\end{enumerate}
\end{theorem}
\begin{proof} (i)  From Table 1, we get  $\sigma_{ap}[B(r,s,t),\ell_{p}]=\sigma
\left[ B( r,s,t),\ell_{p}\right] \backslash \sigma \left[ B(
r,s,t),\ell_{p}\right] C_{1}$. By Theorem \ref{thm6a},  one can
obtain $\sigma \left[ B(r,s,t),\ell_{p}\right] C_{1}=\{r\}$. Hence,
$\sigma_{ap}[B(r,s,t),\ell_{p}]=D_{1}\backslash\{r\}$.

(ii) The following equality
\begin{eqnarray*}
\sigma_{\delta }[B(r,s,t),\ell_{p}]=\sigma [B(
r,s,t),\ell_{p}]\backslash \sigma \left[
B(r,s,t),\ell_{p}\right]A_{3}
\end{eqnarray*} is obtained from Table 1. By Theorems \ref{thm1a} and  \ref{thm2a}
 $\sigma_{\delta}[B(r,s,t),\ell_{p}]=D_{1}$.

(iii)  From Table 1, we have
\begin{eqnarray*}
\sigma_{co}[B(r,s,t),\ell_{p}]=\sigma \left[ B(
r,s,t),\ell_{p}\right] C_{1}\cup \sigma \left[ B(
r,s,t),\ell_{p}\right] C_{2}\cup \sigma \left[
B(r,s,t),\ell_{p}\right]C_{3}=\sigma_{p}[B^{\ast}(r,s,t),\ell^{\ast}_{p}].
\end{eqnarray*}


By Theorem \ref{thm3a}, it is immediate that
$\sigma_{co}[B(r,s,t),\ell_{p}]=D_{1}$.
\end{proof}

\section{Graphical representation }

In this section, we give the graphical representations of the
spectrum of the triangular triple-band matrix over the sequence
space $c$.

If we choose $r=t=1$, $s=-2$, we get
\begin{eqnarray*}
\sigma[A(1,-2,1),c]=\left\{\alpha\in\mathbb{C}:\big|1-\sqrt{\alpha}\big|\leq1\right\}.
\end{eqnarray*}
Then, in polar coordinates, the boundary of $\sigma[A(1,-2,1),c]$ is
as follows:

Let $\alpha =\rho e^{i\theta}$. Then,
\begin{eqnarray*}
1 &= &\left\vert 1-\sqrt{\rho}e^\frac{i\theta}{2}\right\vert\\
&= &\left[ 1-\sqrt{\rho}\cos \left( \frac{\theta }{2}\right)
\right]^{2} +\left[-\sqrt{\rho}\sin \left( \frac{\theta
}{2}\right) \right]^{2} \\
&= &-2\sqrt{\rho}\cos \left( \frac{\theta }{2}\right)+\rho+1\\
\rho&= &4\cos ^{2}\left( \frac{\theta
}{2}\right),\,\,-\pi<\theta<\pi.
\end{eqnarray*}

\begin{figure}[htp]
\centering
\includegraphics[scale=0.27]{Figure1.eps}
\vspace*{0.3cm}
\caption{\it The continuous spectrum of the operator  $A(r,s,t)$ over
the space $c$} \label{figure2}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[scale=0.27]{Figure2.eps}
\vspace*{0.3cm}\caption{\it The point spectrum of the operator $A(r,s,t)$ over the
space $c$} \label{figure2}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[scale=0.28]{Figure3.eps}
\vspace*{0.3cm}\caption{\it The spectrum of the operator $B(r,s,t)$  and $A(r,s,t)$ over the spaces $\ell_{p}$ and $c$}
\label{figure3}
\end{figure}
\newpage
\section*{Conclusion}

In this  paper, we determine the spectrum, the continuous spectrum, thepoint spectrum and the residual spectrum of
the operator triple-band matrix over the sequence spaces $\ell_{p}$ and $c$ and give their graphical
representations. We also obtain a new type of subspectral classes.

Finally, we should note that in the case $t=0$, the operator
$A(r,s,t)$ defined  by an upper triangular triple-band matrix reduces
to the operator $U(r,s)$ defined by an upper triangular   double
band matrix, and in the case $r=1$, $s=-1$ and $t=0$, the operator
$A(r,s,t)$ is reduced to $\triangle^{+}$ defined by an upper
triangular difference matrix. Our results are more general and more
comprehensive than the corresponding results obtained by Karakaya
and Altun \cite{vkma}, and D\"{u}ndar and Ba\c sar \cite{dundar},
respectively.

%%%% Acknowledgment %%%%%%%%
\section*{Acknowledgement}
The authors would like to express their gratitude to the referee for
making some useful comments on the first draft of the manuscript
which improved the presentation of the paper.
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%\end{linenumbers}
\end{document}
