# The Spectrum of the OperatorD(r,0,s,0,t)  over the Sequence Spaces c0 and c

Authors: Binod Chandra Tripathy; Avinoy Paul
Journal: Journal of Mathematics
 (2013)

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## Abstract

We have examined the spectra of the operatorD(r,0,s,0,t) on the sequence spaces c0 and c.


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## Body

## 1. Introduction

Spectral theory is an important branch of mathematics due to its application in other branches of science. It has been proved to be a standard tool of mathematical sciences because of its usefulness and application-oriented scope in different fields. In numerical analysis, the spectral values may determine whether a discretization of a differential equation will get the right answer or how fast a conjugate gradient iteration will converge. In aeronautics, the spectral values may determine whether the flow over a wing is laminar or turbulent. In electrical engineering, it may determine the frequency response of an amplifier or the reliability of a power system. In quantum mechanics, it may determine atomic energy levels and thus the frequency of a laser or the spectral signature of a star. In structural mechanics, it may determine whether an automobile is too noisy or whether a building will collapse in an earthquake. In ecology, the spectral values may determine whether a food web will settle into a steady equilibrium. In probability theory, they may determine the rate of convergence of a Markov process.In summability theory, different classes of matrices have been investigated. Characterization of matrix classes is found in Rath and Tripathy [1], Tripathy [2], Tripathy and Sen [3], and many others. There are particular types of summability methods like Nörlund, Riesz, Euler, and Abel. Matrix methods have been studied from different aspects recently by Altin et al. [4], Tripathy and Baruah [5], and others. Still there is a lot to be explored on spectra of some matrix operators transforming one class of sequences into another class of sequences. The spectra of the difference operator have also been investigated on some classes of sequences. Altay and Başar [6–8] studied the spectra of difference operator Δ and generalized difference operator on c0,c, and ℓp. Recently, the fine spectrum of B(r,s,t) over the sequence spaces c0,c,ℓp, and bvp has been studied by Furkan et al. [9, 10]. A detailed account of the development and initial works on spectra of some matrix classes are found in the monograph of Başar [11].Throughout the paperw,ℓ∞,c, and c0 denote the space of all bounded, convergent, and null sequences x=(xk) with complex terms, respectively, normed by ∥x∥∞=supk|xk|. The zero sequence is denoted by θ=(0,0,0,…). Kızmaz [12] defined the difference sequence spaces ℓ∞(Δ), c(Δ) and c0(Δ) as follows:
(1)Z(Δ)={x=(xk):(Δxk)∈Z},forZ∈{l∞,c,c0},
where
(2)Δx=(Δxk)=(xk-xk+1),∀k∈ℕ={0,1,2,…}.The previous spaces are Banach spaces, normed by∥x∥Δ=∥x1∥+supk∥Δxk∥.Different classes of sequence spaces using the difference operator have been introduced and investigated in the recent past by Tripathy et al. [13, 14], Tripathy and Mahanta [15], Tripathy and Sarma [16], and many others. The idea of Kizmaz [12] was applied to introduce a new type of generalized difference operator on sequence spaces by Tripathy and Esi [17].Letm∈ℕ be fixed; then Tripathy and Esi [17] have introduced the following type of difference sequence spaces:
(3)Z(Δm)={x=(xk):(Δmxk)∈Z},nbjhuyforZ∈{ℓ∞,c,c0},
where
(4)Δmx=(Δmxk)=(xk-xk+m).Letm,n≥0 be fixed integers; then Esi et al. [18] have introduced the following type of difference sequence spaces:
(5)Z(Δmn)={x=(xk)∈w:Δmnx=(Δmnxk)∈Z},sddasdasdasdsadvforZ∈{ℓ∞,c,c0},
where
(6)Δmnx=(Δmnxk)=(Δmn-1xk-Δmn-1xk+m),Δm0xk=xk∀k∈ℕ.Takingn=1, we have the sequence spaces ℓ∞(Δm), c(Δm), and c0(Δm) studied by Tripathy and Esi [17]. Taking m=1, we have the sequence spaces ℓ∞(Δn), c(Δn), and c0(Δn) studied by Et and Çolak [19]. Taking m=1 and n=1, we have the sequence spaces ℓ∞(Δ), c(Δ), and c0(Δ) studied by Kızmaz [12].


## 2. Preliminaries and Definition

LetX be a linear space. By B(X), we denote the set of all bounded linear operators on X into itself. If T∈B(X), where X is a Banach space, then the adjoint operator T* of T is a bounded linear operator on the dual X* of X defined by (T*φ)(x)=φ(Tx) for all φ∈X* and x∈X.LetT:D(T)→X be a linear operator, defined on D(T)⊂X, where D(T) denote the  domain of T and X is a complex normed linear space. For T∈B(X) we associate a complex number α with the operator (T-αI) denoted by Tα defined on the same domain D(T), where I is the identity operator. The inverse (T-αI)-1, denoted by Tα-1, is known as the resolvent operator of Tα.A regular value ofα is a complex number α of T such that (R1)
Tα-1 exists,(R2)
Tα-1 is bounde,(R3)
Tα-1 is defined on a set which is dense in X.The resolvent set ofT is the set of all such regular values α of T, denoted by ρ(T). Its complement is given by ℂ∖ρ(T) in the complex plane ℂ which is called the spectrum of T, denoted by σ(T). Thus the spectrum σ(T) consists of those values of α∈ℂ, for which Tα is not invertible.


### 2.1. Classification of Spectrum

The spectrumσ(T) is partitioned into three disjoint sets as follows.(i)
The point (discrete) spectrumσp(T) is the set such that Tα-1 does not exist. Further α∈σp(T) is called the eigenvalue of T.(ii)
The continuous spectrumσc(T) is the set such that Tα-1 exists and satisfies (R3) but not (R2); that is, Tα-1 is unbounded.(iii)
The residual spectrumσr(T) is the set such that Tα-1 exists (and may be bounded or not) but does not satisfy (R3); that is, the domain of Tα-1 is not dense in X.This is to note that in finite dimensional case, continuous spectrum coincides with the residual spectrum and are equal to the empty set and the spectrum consists of only the point spectrum.LetE and F be two sequence spaces and let A=(ank) be an infinite matrix of real or complex numbers ank, where n,k∈ℕ. Then, we say that A defines a matrix mapping from E into F, denoted by A:E→F, if for every sequence x=(xn)∈E the sequence Ax={(Ax)n} is in F where (Ax)n=∑k=0∞ankxk(n∈ℕandx∈E), provided the right hand side converges for every n∈ℕ and x∈E.Our main focus in this paper is on the matrixA=D(r,0,s,0,t), where
(7)D(r,0,s,0,t)=[r00000⋯0r0000⋯s0r000⋯0s0r00⋯t0s0r0⋯0t0s0r⋯⋮⋮⋮⋮⋮⋮⋱].We assume here and hereafter thats and t are complex parameters which do not simultaneously vanish.Lemma 1.
The matrixA=(ank) gives rise to a bounded linear operator T∈B(c) from c to itself if and only if(1)
the rows of A inℓ1 and their ℓ1 norms are bounded,(2)
the columns ofA are in c,(3)
the sequence of row sums ofA is in c.
The operator norm of T is the supremum of theℓ1 norms of the rows.Lemma 2.
The matrixA=(ank) gives rise to a bounded linear operator T∈B(c0) from c0 to itself if and only if (1)
the rows of A inℓ1 and their ℓ1 norms are bounded,(2)
the columns ofA are in c0.
The operator norm of T is the supremum of theℓ1 norms of the rows.From the previous two lemmas we have the following results.Lemma 3.
D
(
r
,
0
,
s
,
0
,
t
)
:
c
→
c is a bounded linear operator with
(8)∥D(r,0,s,0,t)∥(c,,c)=|r|+|s|+|t|.Lemma 4.
D
(
r
,
0
,
s
,
0
,
t
)
:
c
0
→
c
0 is a bounded linear operator with
(9)∥D(r,0,s,0,t)∥(c,c)=∥D(r,0,s,0,t)∥(c0,c0).In this paper, our purpose is to determine the fine spectrum of the operatorD(r,0,s,0,t) over the sequence spaces c0 and c.


## 2.1. Classification of Spectrum

The spectrumσ(T) is partitioned into three disjoint sets as follows.(i)
The point (discrete) spectrumσp(T) is the set such that Tα-1 does not exist. Further α∈σp(T) is called the eigenvalue of T.(ii)
The continuous spectrumσc(T) is the set such that Tα-1 exists and satisfies (R3) but not (R2); that is, Tα-1 is unbounded.(iii)
The residual spectrumσr(T) is the set such that Tα-1 exists (and may be bounded or not) but does not satisfy (R3); that is, the domain of Tα-1 is not dense in X.This is to note that in finite dimensional case, continuous spectrum coincides with the residual spectrum and are equal to the empty set and the spectrum consists of only the point spectrum.LetE and F be two sequence spaces and let A=(ank) be an infinite matrix of real or complex numbers ank, where n,k∈ℕ. Then, we say that A defines a matrix mapping from E into F, denoted by A:E→F, if for every sequence x=(xn)∈E the sequence Ax={(Ax)n} is in F where (Ax)n=∑k=0∞ankxk(n∈ℕandx∈E), provided the right hand side converges for every n∈ℕ and x∈E.Our main focus in this paper is on the matrixA=D(r,0,s,0,t), where
(7)D(r,0,s,0,t)=[r00000⋯0r0000⋯s0r000⋯0s0r00⋯t0s0r0⋯0t0s0r⋯⋮⋮⋮⋮⋮⋮⋱].We assume here and hereafter thats and t are complex parameters which do not simultaneously vanish.Lemma 1.
The matrixA=(ank) gives rise to a bounded linear operator T∈B(c) from c to itself if and only if(1)
the rows of A inℓ1 and their ℓ1 norms are bounded,(2)
the columns ofA are in c,(3)
the sequence of row sums ofA is in c.
The operator norm of T is the supremum of theℓ1 norms of the rows.Lemma 2.
The matrixA=(ank) gives rise to a bounded linear operator T∈B(c0) from c0 to itself if and only if (1)
the rows of A inℓ1 and their ℓ1 norms are bounded,(2)
the columns ofA are in c0.
The operator norm of T is the supremum of theℓ1 norms of the rows.From the previous two lemmas we have the following results.Lemma 3.
D
(
r
,
0
,
s
,
0
,
t
)
:
c
→
c is a bounded linear operator with
(8)∥D(r,0,s,0,t)∥(c,,c)=|r|+|s|+|t|.Lemma 4.
D
(
r
,
0
,
s
,
0
,
t
)
:
c
0
→
c
0 is a bounded linear operator with
(9)∥D(r,0,s,0,t)∥(c,c)=∥D(r,0,s,0,t)∥(c0,c0).In this paper, our purpose is to determine the fine spectrum of the operatorD(r,0,s,0,t) over the sequence spaces c0 and c.


## 3. The Fine Spectrum of the OperatorD(r,0,s,0,t) over the Sequence Space c0

Theorem 5.
Lets be a complex number such that s2=-s and define the set by
(10)S={α∈C:|2(r-α)-s+s2-4t(r-α)|≤1}.
Then
(11)σ(D(r,0,s,0,t),c0)=S.Proof.
At first we have to prove that(D(r,0,s,0,t)-αI)-1 exists and is in B(c0) for α∉S and secondly we have to show that (D(r,0,s,0,t)-αI) is not invertible for α∈S.
Without loss of generality we may assume thats2=-s. Let α∉S; then it is easy to see that α≠r and so (D(r,0,s,0,t)-αI) is triangle and has an inverse.
Let(12)(D(r,0,s,0,t)-αI)-1=[a10000⋯a2a1000⋯a3a2a100⋯a4a3a2a10⋯a5a4a3a2a1⋯⋮⋮⋮⋮⋮⋱].
We have(13)[r-α0000⋯0r-α000⋯s0r-α00⋯0s0r-α0⋯t0s0r-α⋯⋮⋮⋮⋮⋮⋱]×[a10000⋯a2a1000⋯a3a2a100⋯a4a3a2a10⋯a5a4a3a2a1⋯⋮⋮⋮⋮⋮⋱]=[10000⋯01000⋯00100⋯00010⋯00001⋯⋮⋮⋮⋮⋮⋱].
On solving the previous system of equations we get(14)a1=1r-α,a2=0,a3=-s(r-α)2,a4=0,a5=s2-(r-α)t(r-α)3,⋮
In general this sequence is obtained recursively by(15)an=-san-2+tan-4r-α,forn≥5.
It is easy to verify that(16)a2n+1=1s2-4t(r-α)×{[-s+s2-4t(r-α)2(r-α)]n+1b-[-s-s2-4t(r-α)2(r-α)]n+1}nm,.hnhnhnhnhnmforn∈Z+,a2n=0,forn∈ℕ.
On taking(17)u1=[-s+s2-4t(r-α)2(r-α)],u2=[-s-s2-4t(r-α)2(r-α)],
we have
(18)a2n+1=1s2-4t(r-α)(u1n+1-u2n+1),forn∈ℕ.
If one assumess2=s, then we will get the same sequence as in the case of s2=-s.
Ify=s2-4t(r-α), then
(19)a2n+1=1y{2(r-α)}n+1{(-s+y)n+1-(-s-y)n+1}=1{2(r-α)}n+1×[2(n+1)(-s)n+(n+1)n(n-1)6(-s)n-2y2+⋯].
If we puts2=4t(r-α), then a2n+1=(2(n+1)/-s)[-s/2(r-α)]n+1, for n∈ℕ and a2n=0, for n∈ℕ.
Thus, on simple calculation we get(an)∈ℓ1 if and only if |-s/2(r-α)|<1. Therefore α∉S implies an→0, as n→∞.
Next, we assume thats2≠4t(r-α). Since α∉S, therefore |u1|<1. Now we have to show that |u2|<1.
We have that|u1|<1 implies |1+1-(4t(r-α))/s2|<|2(r-α)/-s|.
Since|1-z|≤|1+z| for any z∈ℂ, we must have |1-1+(4t(r-α))/s2|<|2(r-α)/-s| which leads us to conclude that |u2|<1 and hence in this case also α∉S implies that an→0 as n→∞.
Now,(20)∥(D(r,0,s,0,t-αI)-1∥(c0,c0)=supn∑k=1n|ak|=∑k=1∞|ak|≤1|s2-4t(r-α)|(∑n=0∞|u1|n+1+∑n=0∞|u2|n+1)<∞.
Since|u1|<1 and |u2|<1, we have α∉σ(D(r,0,s,0,t),c0).
Thus,α∉S⇒α∉σ(D(r,0,s,0,t),c0) and hence σ(D(r,0,s,0,t),c0)⊆S.
Next we have to show thatS⊆σ(D(r,0,s,0,t),c0).
Letα∈S. If α=r, then (D(r,0,s,0,t)-αI) is represented by the matrix
(21)D(0,0,s,0,t)=[000000⋯000000⋯s00000⋯0s0000⋯t0s000⋯0t0s00⋯⋮⋮⋮⋮⋮⋮⋱].
SinceD(r,0,s,0,t)-rI=D(0,0,s,0,t) does not have a dense range, so it is not invertible.
Again ifs2=4t(r-α), then a2n+1=(2(n+1)/-s)[-s/2(r-α)]n+1 for n∈Z+ and a2n=0, for n∈ℕ.
Nowα∈S⇒|-s/2(r-α)|≥1 and hence (an)∉c0; therefore, we have (D(r,0,s,0,t)-αI)-1∉B(c0).
Next we assume thatα≠r  and s2≠4t(r-α).
Sinceα≠r, we have that (D(r,0,s,0,t)-αI) is a triangle. Further s2≠4t(r-α), so we must have |u1|>|u2|. This implies (an)∉c0 and hence (D(r,0,s,0,t)-αI)-1∉B(c0). This shows that S⊆σ(D(r,0,s,0,t),c0).
This completes the proof.Theorem 6.
One hasσp(D(r,0,s,0,t),c0)=∅.Proof.
SupposeD(r,0,s,0,t)x=αx for x≠θ=(0,0,0,…) in c0. Then by solving the system of linear equations we have
(22)rx0=αx0,sx0+rx2=αx2,tx0+sx2+rx4=αx4,⋮and(23)rx1=αx1,sx1+rx3=αx3,tx1+sx3+rx5=αx5,⋮
Ifxn0 is the first nonzero entry of the sequence x=(xn), then from the previous system of linear (22) and (23) we have txn0-4+sxn0-2+rxn0=αxn0and we obtain that α=r and from the next of either (22) or (23) we get xn0=0 which is a contradiction. This completes the proof.IfT:c0→c0 is a bounded linear operator with the matrix A, then it is known that the adjoint operator T*:c0*→c0* is defined by the transpose At of the matrix A. It should be noted that the dual space c0* of c0 is isometrically isomorphic to the Banach space ℓ1 of absolutely summable sequences normed by ∥x∥=∑|xn|.Theorem 7.
One hasσp(D(r,0,s,0,t)*,c0*)=S1, where
(24)S1={α∈C:|2(r-α)-s+s2-4t(r-α)|<1}.Proof.
SupposeD(r,0,s,0,t)*x=αx for x≠θ=(0,0,0,…) in c0*≅ℓ1; that is, considering the following system of linear equations we have
(25)rx0+sx2+tx4=αx0,rx1+sx3+tx5=αx1,rx2+sx4+tx6=αx2,rx3+sx5+tx7=αx3,⋮
Ifα=r, then we may choose x0≠0 and so x=(x0,0,0,…) is an eigenvector corresponding to α=r and hence α∈σp(D(r,0,s,0,t)*,c0*).
Next we assume thatα≠r; then from the previous system of equations we have
(26)x2n+4=a2n+1(r-α)n+1tn+1(α-r)x0+a2n+3(r-α)n+2tn+1x2,m,.lvbforn∈ℕ,(27)x2n+5=a2n+1(r-α)n+1tn+1(α-r)x1+a2n+3(r-α)n+2tn+1x3m,.lvbforn∈ℕ.
Ifα is a number such that |2(r-α)/-s+s2-4t(r-α)|<1, then we may choose x0=x1=1 and x2=x3=2(r-α)/-s+s2-4t(r-α).
We can easily verify thatu1-u2=(s2-4t(r-α))/(r-α) and u1u2=t/(r-α); using these results and combining the fact x2=x3=1/u1 with relation (26) we observe that(28)x2n+4=a2n+1(r-α)n+1tn+1(α-r)x0+a2n+3(r-α)n+2tn+1x2=(u1n+1-u2n+1)s2-4t(r-α)×(r-α)n+1tn+1(α-r)+(u1n+2-u2n+2)s2-4t(r-α)×(r-α)n+2tn+1×1u1=(r-α)s2-4t(r-α)×(r-αt)n+1×(u2n+1-u1n+1+u1n+2-u2n+2u1)=1(u1-u2)×1u1n+1u2n+1×1u1(u1u2n+1-u2n+1)=u2n+1(u1-u2)(u1-u2)u1n+2u2n+1=1u1n+2=(x2)n+2,forn∈ℕ.
Similarly we can show thatx2n+5=(x3)n+2, for n∈ℕ.
The same result may be obtained in the caseu1=u2,  that is, for the case s2=4t(r-α).
Nowx=(xk)∈ℓ1 since |x2|=|x3|<1; this implies that α∈σp(D(r,0,s,0,t),c0*) and hence S1⊆σp(D(r,0,s,0,t)*,c0*).
Conversely, letα∉S1; therefore |2(r-α)/(-s+s2-4t(r-α))|≥1; that is, |u1|≤1. We have to show that α∉σp(D(r,0,s,0,t)*,c0*); that is, to show (xk)∉ℓ1 for this it is sufficient to show that ∑n=0∞|xn|=∑n=0∞|x2n|+∑n=0∞|x2n+1| is divergent.
Now consider∑n=0∞|x2n|.
Here,(29)x2n+6x2n+4=1t×a2n+3(r-α)n+2(α-r)x0+a2n+5(r-α)n+3x2a2n+1(r-α)n+1(α-r)x0+a2n+3(r-α)n+2x2=(r-αt)×(a2n+3a2n+1)×(-x0+(a2n+5/a2n+3)x2-x0+(a2n+3/a2n+1)x2).
Thus we have ((r-α)/t)=1/u1u2.
Now we examine three cases.
Case  1 (|u2|<|u1|≤1). In this case s2≠4t(r-α) and
(30)limn→∞a2n+3a2n+1=limn→∞u1n+2-u2n+2u1n+1-u2n+1=limn→∞u1n+2(1-(u2/u1)n+2)u1n+1(1-(u2/u1)n+1)=u1.
Now, if-x0+u1x2=0, then we get x2n=(1/u1n)x0 which is not in ℓ1since |u1|≤1. Otherwise
(31)limn→∞|x2n+6x2n+4|=1|u1||u2||u1|=1|u2|>1.
Case  2 (|u2|=|u1|<1). In this case s2=4t(r-α) and using the formula
(32)a2n+1=(2(n+1)-s)[-s2(r-α)]n+1,forn∈Z+,a2n=0,forn∈ℕ.
We have(33)limn→∞a2n+3a2n+1=-s2(r-α)=u2=u1,
and so
(34)limn→∞|x2n+6x2n+4|=1|u1||u2||u1|=1|u2|>1.
Case  3 (|u2|=|u1|=1). In this case s2=4t(r-α) and so we have |(r-α)/t|=1/(|u1||u2|)=1 and |u2|=|u1|=|-s/2(r-α)|=1 and so |-s/2t|=1. Our aim is to show α∉σp(D(r,0,s,0,t)*,c0*). On the contrary we assume that α∈σp(D(r,0,s,0,t)*,c0*). This implies that θ≠x∈ℓ1.
From (26)
(35)x2n+4=a2n+1(r-α)n+1tn+1(α-r)x0+a2n+3(r-α)n+2tn+1x2,forn∈Z+=(2(n+1)-s)[-s2(r-α)]n+1×1tn+1×(2(n+2)-sr-α)n+1×(2(n+2)-sα-r)x0+(2(n+2)-s)[-s2(r-α)]n+2×1tn+1×(r-α)n+2x2=(-s2t)n+1[2(n+1)-s(α-r)x0+(n+2)x2]=(-s2t)n+1[(s2t)(n+1)x0+(n+2)x2].
Now, if we consider the series∑n=0∞|x2n+1| in place of ∑n=0∞|x2n|, one will get results parallel to all the results obtained previous just by replacing x1 in place of x0 and x3 in place of x2.
Sincelimn→∞|xn|=0, we must have from (35) x0=x2=0 and similarly we get x1=x3=0. This implies x=θ, a contradiction and so we must have α∉σp(D(r,0,s,0,t)*,c0*).
In Cases  1 and 2, by the d’Alembert ratio test we get that∑n=0∞|x2n| is divergent and similarly we get that ∑n=0∞|x2n+1| is also divergent and hence ∑n=0∞|xn|=∑n=0∞|x2n|+∑n=0∞|x2n+1| is divergent, since the sum of two absolutely divergent series is divergent. In Case  3 α∈σp(D(r,0,s,0,t)*,c0*) leads to contradiction. Thus in all the previous cases α∉σp(D(r,0,s,0,t)*,c0*) and hence   σp(D(r,0,s,0,t)*,c0*)⊆S1. This completes the proof of  theorem.Lemma 8.
T has a dense range if and only ifT* is one to one, where T* denotes the adjoint operator of T.Theorem 9.
σ
r
(
D
(
r
,
0
,
s
,
0
,
t
)
,
c
0
)
=
S
1, where S1 is defined as in Theorem 7.Proof.
One hasσp(D(r,0,s,0,t)*,c0*)=S1; then D(r,0,s,0,t)*-αI is not one-to-one for all α∈S1. Therefore by Lemma 8, D(r,0,s,0,t)-αI have a dense range for all α∈S1 and hence σr(D(r,0,s,0,t),c0)=S1.Theorem 10.
Considerσc(D(r,0,s,0,t),c0)=S2, where
(36)S2={α∈C:|2(r-α)-s+s2-4t(r-α)|=1}.Proof.
The proof immediately follows from the fact that the set of spectra is the disjoint union of the point spectrum, residual spectrum, and continuous spectrum; that is,(37)σ(D(r,0,s,0,t),c0)=σp(D(r,0,s,0,t),c0)∪σr(D(r,0,s,0,t),c0)∪σc(D(r,0,s,0,t),c0).


## 4. The Fine Spectrum of the More Generalized OperatorD(r,0,s,0,t) on the Sequence Space c

Theorem 11.
Considerσ(D(r,0,s,0,t),c)=S, where S is defined as in Theorem 5.Proof.
This is obtained in a similar way used in the proof of Theorem5.Theorem 12.
One hasσp(D(r,0,s,0,t),c)=∅.Proof.
This is obtained in a similar way that is used in the proof of Theorem6.IfT:c→c is a bounded matrix operator with the matrix A, then T*:c*→c* acting on c⊕ℓ1 has a matrix representation of the form [χ0bAt], where χ is the limit of the sequence of row sums of A minus the sum of the columns of A and b is the column vector whose kth entry is the limit of the kth column of A for each k∈ℕ. For D(r,0,s,0,t):c→c, the matrix D(r,0,s,0,t)*∈B(ℓ1) is of the form D(r,0,s,0,t)*=[r+s+t00D(r,0,s,0,t)t].Theorem 13.
Considerσp(D(r,0,s,0,t)*,c*)=S1∪{r+s+t}.Proof.
SupposeD(r,0,s,0,t)*x=αx for x≠θ=(0,0,0,…) in c0*≅ℓ1; that is, consider the following system of linear equations:
(38)(r+s+t)x0=αx0,rx1+sx3+tx5=αx1,rx2+sx4+tx6=αx2,rx3+sx5+tx7=αx3,rx4+sx6+tx8=αx4,⋮
Then, we obtain that(39)x2n+4=a2n+1(r-α)n+1tn+1(α-r)x1+a2n+3(r-α)n+2tn+1x3,forn∈Z+,x2n+5=a2n+1(r-α)n+1tn+1(α-r)x2+a2n+3(r-α)n+2tn+1x4,forn∈Z+.
Ifx0≠0, then (r+s+t)=α. So, α=(r+s+t) is an eigenvalue with the corresponding eigenvector x=(x0,0,0,…).
Ifα≠(r+s+t), then x0=0 and so using arguments similar to those in the proof of Theorem 7 one can see by (39) that x∉ℓ1. This completes the proof.Theorem 14.
Considerσr(D(r,0,s,0,t),c)=σp(D(r,0,s,0,t)*,c*).Theorem 15.
Considerσc(D(r,0,s,0,t),c)=S2∖{r+s+t}, where S2 is defined as in Theorem 10.


## 5. Particular Case

The spectrum of the operatorΔ22 over the sequence space c0 and c may be derived as the set {α∈C:|1-α|≤1}; one can directly produce the same result from the present paper since Δ22=D(1,0,-2,0,1).



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