# Matrix Mappings on the Domains of Invertible Matrices

Authors: Muhammed Altun
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We focus on sequence spaces which are matrix domains of Banach sequence spaces. We show that the characterization of a random matrix operatorT=(tnk)∈(EA,FB), where EA and FB are matrix domains with invertible matrices A and B, can be reduced to the characterization of the operator S=B∘T∘A−1∈(E,F). As an application, the necessary and sufficient conditions for the matrix operators between invertible matrix domains of the classical sequence spaces and norms of these operators are given.


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## Body

## 1. Introduction and Preliminaries

Letω denote the set of all complex sequences. Any subspace of ω is called as a sequence space. We will write ℓ∞, c, and c0 for the spaces of all bounded, convergent, and null sequences, respectively. By ℓp, we denote the space of all p absolutely summable sequences, where 1≤p<∞.LetX and Y be Banach spaces. Then, (X,Y) is the set of all operators L:X→Y and ℬ(X,Y) is the set of all continuous linear operators L:X→Y. ℬ(X,Y) is a Banach space with the operator norm defined by ∥L∥=sup{∥L(x)∥/∥x∥:0≠x∈X}(L∈ℬ(X,Y)).LetE and F be two sequence spaces and A=(ank) an infinite matrix of real or complex numbers ank, where n, k∈ℕ={1,2,3,…}. Then, we say that A defines a matrix mapping from E into F, and we denote it by writing A:E→F, if for every sequence x=(xk)∈E the sequence Ax={(Ax)n}, the A-transform of x, is in F, where
(1)(Ax)n=∑k‍ankxk(n∈ℕ).
Thus, A∈(E,F) if and only if the series on the right side of (1) converges for each n∈ℕ and every x∈E, and we have Ax={(Ax)n}n∈ℕ∈F for all x∈E.LetA=(ank) and B=(bnk). Suppose the sums
(2)cnk=∑j‍anjbjk
exist for all n, k∈ℕ. Then, the product of A and B is defined by AB=(cnk).IfE is a subset of ω, then EA={x∈ω:Ax∈E} is the matrix domain of A in E. We will say that a matrix A is invertible over a sequence space  E if the operator A:EA→E is bijective; that is, there exists an operator A-1 such that A-1(Ax)=x for all x∈EA and A(A-1y)=y for all y∈E. We will say A is invertible if A is invertible over ω.A BK space is a Banach sequence space with continuous coordinates. The sequence spacesc0, c, ℓp, and ℓ∞ are the well-known examples of BK spaces.Several authors studied matrix mappings on sequence spaces that are matrix domains of the difference operator or of the matrices of some classical methods of summability in spaces such asℓp, c0, c, or ℓ∞. For instance, some matrix domains of the difference operator were studied in [1, 2], of the Cesàro matrices in [3, 4], of the Euler matrices in [5–7], and of the Nörlund matrices in [8]. A general approach was done in [9] reducing the characterizations of the classes (XT,Y) for arbitrary FK spaces X with AK and Y⊂ω to those of the classes (X,Y) and (X,c), where T is a triangle. Compact operators on matrix domains of triangles were examined in [10]. The gliding hump properties of matrix domains were examined in [11].In this work, our aim is to give some general results for matrix mappings between sequence spaces, which are matrix domains of invertible matrices of sequence spaces. Also, we give some applications of the results.Theorem 1 (see [12, Theorem 4.2.8, page 57]).
Matrix operators between BK spaces are continuous.An infinite matrixT=(tnk) is said to be a triangle if tnk=0 for n<k and tnn≠0 for n∈ℕ. The following is a well-known result about triangles.Theorem 2 (see [12, 1.4.8, page 9]).
Every triangleT has an inverse T-1 which also is a triangle, and x=T(T-1x)=T-1(Tx) for all x∈ω.LetA1,A2 be two matrices and μ a sequence space. If A1(A2x)=(A1A2)x holds for all x∈μ, then we will say A1 and A2 are associative over the space  μ. If A1 and A2 are associative over the space ω, we will shortly say A1 and A2 are associative. For row finite matrices, we do not generally have an inverse. But we have associativity; that is, for any two row finite matrices R1 and R2 we have
(3)R1(R2x)=(R1R2)x∀x∈ω.
Moreover, we have the following result.Theorem 3 (see [12, Theorem 1.4.4, page 8]).
LetA1,A2 be two matrices.(i)
If rows ofA1 are in ℓ1 and A2∈(ℓ∞,ℓ∞), then A1 and A2 are associative over ℓ∞.(ii)
IfA1 is row finite, then A1 and A2 are associative over ωA2.Corollary 4.
The set of matrices(ℓ∞,ℓ∞) are associative over ℓ∞.Remark 5.
The associativity property does not hold in general (see, e.g., [12, Example 1.4.6]).Theorem 6.
LetA be an invertible matrix over a normed sequence space E with norm ∥·∥E. Then, EA is a normed sequence space with norm ∥z∥EA=∥Az∥E.Proof.
Letα∈ℂ and z∈EA. Then,
(4)∥αz∥EA=∥A(αz)∥E=∥αAz∥E=|α|∥Az∥E=|α|∥z∥EA.
Secondly, for y,z∈EA we have
(5)∥y+z∥EA=∥A(y+z)∥E=∥Ay+Az∥E≤∥Ay∥E+∥Az∥E=∥y∥EA+∥z∥EA,
so the triangle inequality holds.
Now, suppose that∥z∥EA=0. Then, ∥Az∥E=0 and since ∥·∥E is a norm we have Az=θ. Since A is invertible, we have z=θ.Theorem 7.
LetE be a Banach sequence space and A a matrix. Then, the operator A:EA→E is linear and continuous.Proof.
LetAn be the operator that corresponds to the nth row of the matrix operator A; that is, Anx=∑k=1∞‍ankxk for all x=(xk)∈EA. Let x=(xk)∈EA and y=(yk)∈EA. Clearly, we have Ancx=cAnx for any c∈ℂ. Let x′=Anx=∑k=1∞‍ankxk and y′=Any=∑k=1∞‍ankyk. We have An(x+y)=∑k=1∞‍ank(xk+yk)=limm∑k=1m‍ank(xk+yk)=limm∑k=1m‍ankxk+limm∑k=1m‍ankyk=x′+y′=Anx+Any.   That means the operator An is linear for arbitrary n, which implies the linearity of A.
A is continuous since
(6)∥A∥=sup{∥Ax∥E∥x∥EA:0≠x∈EA}=sup{∥Ax∥E∥Ax∥E:0≠x∈EA}=1.


## 2. Main Results

Theorem 8.
If(E,∥·∥) is a Banach sequence space and A is an invertible matrix over E, then (EA,∥·∥A) is a Banach space.Proof.
Let(E,∥·∥) be a Banach sequence space. Then, (EA,∥·∥A) is a normed space by the previous theorem. Now, let (xn) be a Cauchy sequence in EA. Let yn=Axn. Then (yn) is a sequence in E and is Cauchy in E since
(7)∥xn-xm∥EA=∥A(xn-xm)∥E=∥Axn-Axm∥E=∥yn-ym∥E.
Let y∈E such that limyn=y in E. Let x=A-1y. Then,
(8)limn→∞∥xn-x∥EA=limn→∞∥A(xn-x)∥E=limn→∞∥yn-y∥E=0.Theorem 9.
LetA and B be invertible matrices over the BK spaces E and F, respectively. Suppose that T∈(EA,FB). If B∘T∘A-1=BTA-1 over E, then the operator T∈(EA,FB) is continuous.Proof.
LetT∈(EA,FB) and let ∥·∥ be the norm of E. Since A is invertible over E, A:EA→E is bijective. Also, A is continuous by Theorem 7.
By the bounded inverse theoremA-1 is continuous. Similarly the inverse B-1 of B:FB→F is continuous.
Now, we have that the matrix operatorR=B∘T∘A-1∈(E,F) and is continuous by Theorem 1. Since A and B-1 are continuous, the operator T=B-1∘R∘A is continuous.Theorem 10.
LetA and B be invertible matrices over the sequence spaces E and F, respectively. Then, for an operator T, T∈(EA,FB) if and only if B∘T∘A-1∈(E,F).Proof.
Suppose thatT∈(EA,FB), and let z=(zk)∈E. Then, clearly A-1z∈EA. Hence, we have (T∘A-1)z∈FB and so (B∘T∘A-1)z∈F.
For the inverse implication, suppose thatB∘T∘A-1∈(E,F), and let z∈EA. Then, Az∈E and so (B∘T∘A-1∘A)z=(B∘T)z∈F. Since B is invertible, we have Tz∈FB.Corollary 11.
LetE and F be sequence spaces and A and B triangles. Then, for an operator T, T∈(EA,FB) if and only if B∘T∘A-1∈(E,F).Theorem 12.
LetE and F be two normed sequence spaces and A and B invertible matrices over E and F, respectively. Then, for an operator T, T∈ℬ(EA,FB) if and only if B∘T∘A-1∈ℬ(E,F). In this case, one has
(9)∥T∥(EA,FB)=∥B∘T∘A-1∥(E,F).Proof.
It is enough to show the following equality:(10)∥T∥(EA,FB)=supθ≠z∈EA∥Tz∥FB∥z∥EA=supθ≠Az∈E∥(B∘T)z∥F∥Az∥E=supθ≠Az∈E∥(B∘T∘A-1∘A)z∥F∥Az∥E=∥(B∘T∘A-1)∥E,F.


## 3. Examples and Applications

Theorem10, Corollary 11, and Theorem 12 have many applications, especially in the subject of characterization of matrices which act as operators between certain sequence spaces. We just give a taste by the following examples and theorems.Example 13.
Letλ=(λk) be a sequence of nonzero scalars in ℂ. For any z=(zk)∈ω, let λz=(λkzk). Then, for a sequence space E, the multiplier sequence space E(λ), associated with the multiplier sequence λ, is defined as
(11)E(λ)={z∈ω:λz∈E}.
Let
(12)Λ=[λ1000⋯0λ200⋯00λ30⋯000λ4⋯⋮⋮⋮⋮⋱].
Then, the matrix Λ is invertible with
(13)Λ-1=[λ1-1000⋯0λ2-100⋯00λ3-10⋯000λ4-1⋯⋮⋮⋮⋮⋱],
and E(λ)=EΛ. So, all the Theorems in Section 2 are applicable for multiplier sequence spaces (see [13] for detailed applications and examples).Example 14.
LetT=(tnk) be a row-finite matrix and B the operator with matrix representation
(14)B=[1000⋯1100⋯1110⋯1111⋯⋮⋮⋮⋮⋱].B is invertible and the inverse operator has the matrix representation
(15)B-1=[1000⋯-1100⋯0-110⋯00-11⋯⋮⋮⋮⋮⋱].B and B-1 are triangles and so they are one to one. B-1 is the operator B(r,s) of [14] with r=1 and s=-1. Since both T and B-1 are row finite, we have T∘B-1=TB-1. Now, using Corollary 11 we have that T is in (cB,c) if and only if TB-1 is in (c,c):(16)[TB-1]nk=tnk-tn(k+1),
so by the Kojima-Schur Theorem (see, e.g., Theorem 2.7 of [15]) we have that T∈(cB,c) if and only if the following three conditions hold:(i)
limntnk-tn(k+1) exists for each k;(ii)
limn∑k‍tnk-tn(k+1) exists;(iii)
supn∑k‍|tnk-tn(k+1)|<∞.
Now, let us examine(c,cB). T is in (c,cB) if and only if BT is in (c,c). We have
(17)[BT]nk=snk,
where snk is defined as the sum of the first n terms of the kth column of the matrix T; that is,
(18)snk=t1k+t2k+⋯+tnk.
Now, by using the Kojima-Schur Theorem, we have, T∈(c,cB) if and only if the following three conditions hold:(i)
limnsnk exists for each k;(ii)
limn∑k‍snk exists;(iii)
supn∑k‍|snk|<∞.Now, let us examine (cB,cB). T is in (cB,cB) if and only if BTB-1 is in (c,c). We have
(19)[BTB-1]nk=snk-sn(k+1).
Now, by using the Kojima-Schur Theorem we have that T∈(cB,cB) if and only if the following three conditions hold:(i)
limnsnk-sn(k+1) exists for each k;(ii)
limn∑k‍snk-sn(k+1) exists;(iii)
supn∑k‍|snk-sn(k+1)|<∞.Theorem 15.
An infinite matrixA=(ank)∈(ℓ∞,bv) if and only if
(20)supN∈f(ℕ)∑k‍|∑n∈N‍(ank-a(n-1)k)|<∞,
where f(ℕ) denotes the collection of all finite subsets of ℕ.Proof.
First, we observe thatbv=ℓ1Δ, where Δ is the difference operator B-1 of Example 14. The inverse of Δ is B. By Corollary 11, A∈(ℓ∞,ℓ1Δ) if and only if Δ∘A∈(ℓ∞,ℓ1). Since bv⊂c, we have A∈(ℓ∞,c) and so by the Schur Theorem (see, e.g., Theorem 2.11 of [15]) rows of A are in ℓ1. By Theorem 3 part (ii), Δ and A are associative over ℓ∞. So, Δ∘A=ΔA over ℓ∞. We have
(21)[ΔA]nk=ank-a(n-1)k,
where a0k=0 for all k∈ℕ. Now, using Theorem 2.14 of [15], we have A∈(ℓ∞,bv) if and only if
(22)supN∈f(ℕ)∑k‍|∑n∈N‍(ank-a(n-1)k)|<∞.Using Theorem12, we have the following.Corollary 16.
IfA=(ank)∈(ℓ∞,bv), then A=(ank)∈ℬ(ℓ∞,bv) and
(23)∥A∥(ℓ∞,bv)=∥ΔA∥(ℓ∞,ℓ1).Example 17.
LetT=(tnk) be a row finite matrix and C the Cesàro operator which has a matrix representation
(24)C=[1000⋯1/21/200⋯1/31/31/30⋯1/41/41/41/4⋯⋮⋮⋮⋮⋱].
Then, cC is the well-known space of Cesàro summable sequences. This is a sequence space which includes the space of convergent sequences c. C is invertible and the inverse operator has the matrix representation
(25)C-1=[1000⋯-1200⋯0-230⋯00-34⋯⋮⋮⋮⋮⋱].
Since T,C, and C-1 are all row finite, the operator C∘T∘C-1 is represented by the matrix CTC-1. Now, using Corollary 11, we have that T is in ((ℓ∞)C,(ℓ∞)C) if and only if CTC-1 is in (ℓ∞,ℓ∞), where (ℓ∞)C denotes the space, introduced by Ng and Lee [3], of all sequences whose C-transforms are in the space ℓ∞.
Letsnk be defined as in the previous example. Then, after some calculations, we have
(26)[CTC-1]nk=kn(snk-sn(k+1)).
Then, CTC-1 is in (ℓ∞,ℓ∞) if and only if
(27)supn∑k‍kn|snk-sn(k+1)|<∞
by Theorem 2.6 of [15]. So, we have T∈((ℓ∞)C,(ℓ∞)C) if and only if (27) holds. Similarly, we have T∈(cC,(ℓ∞)C) or T∈((c0)C,(ℓ∞)C) if and only if (27) holds, where (c0)C and cC denote the spaces c0~ and c~, studied by Şengönül and Başar [4], of all sequences whose C-transforms are in the spaces c0 and c, respectively.Theorem 18.
Let C be the Cesàro matrix. Then,A=(ank)∈(c,cC) if and only if
(28)supn[1n∑k‍|snk|]<∞,limnsnknexistsforeachk,limn1n∑k‍snkexists,
where snk=a1k+a2k+⋯+ank.Proof.
By Corollary11, A∈(c,cC) if and only if C∘A∈(c,c). Since A∈(c,cC)⊂(c,ω), we have that rows of A are in ℓ1. By Theorem 3 part (ii), C and A are associative over c. So, C∘A=CA over c. We have
(29)[CA]nk=snkn.
Now, using Kojima-Schur Theorem, we have A∈(c,cC) if and only if conditions (28) hold.Lemma 19.
If∑k=1∞‍akck is convergent for all Cesàro summable sequences (ck), then limkak=0.Proof.
Suppose thatlimkak≠0. Then, there exist ϵ>0 and a subsequence (akn) of (ak) such that kn|akn|>ϵ. Without loss of generality, we can choose this subsequence (akn) such that kn≥n2kn-1. Let
(30)ck={kn|akn|naknifk=kn0otherwise.
Then, the sequence (ck) is Cesàro summable because for kn≤m<kn+1 we have
(31)|c1+c2+⋯+cmm|≤|c1|+|c2|+⋯+|cm|m=|ck1|+|ck2|+⋯+|ckn|m=k1/1+k1/2+⋯+kn/nm≤k1/1+k1/2+⋯+kn-1/(n-1)+kn/nkn=k1kn+k22kn+⋯+kn-1(n-1)kn+1n≤1n2+12n2+⋯+1(n-1)n2+1n≤(n-1)1n2+1n≤2n.
On the other hand,
(32)∑k=1∞‍akck=∑n=1∞‍aknckn=∑n=1∞‍kn|akn|n>ϵ∑n=1∞‍1n,
and so we get to the contradiction ∑k=1∞‍akck=∞ for the Cesàro summable sequence (ck). So, our assumption limkak≠0 is not true.Theorem 20.
Let C be the Cesàro matrix. Then,A=(ank)∈(cC,cC) if and only if
(33)limkkank=0foreachn,(34)supn∑k‍kn|snk-sn(k+1)|<∞,(35)limnsnk-sn(k+1)nexistsforeachk,(36)limn1n∑k‍k(snk-sn(k+1))exists,
where snk=a1k+a2k+⋯+ank.Proof.
Suppose thatA∈(cC,cC). Then, by Lemma 19, we have (33). By Corollary 11, A∈(cC,cC) if and only if C∘A∘C-1∈(c,c). First let us show that A and C-1 are associative over c. Let x=(xk)∈c. We have that
(37)(A(C-1x))n=∑k=1∞‍ank(kxk-(k-1)xk-1)
is convergent in ℂ. Let Sm be the mth partial sum of this series. Then,
(38)Sm=∑k=1m-1‍k(ank-an(k+1))xk+manmxm;limmmanm=0, so
(39)(A(C-1x))n=limmSm=∑k=1∞‍k(ank-an(k+1))xk=((AC-1)x)n;AC-1∈(c,ω) and C is row finite. Then, by Theorem 3 part (ii), C and AC-1 are associative over c. So, C∘A∘C-1=CAC-1 over c. We have
(40)[CAC-1]nk=kn(snk-sn(k+1)).
Now, using Kojima-Schur Theorem, A∈(cC,cC) implies conditions (34)–(36).
For the reverse implication, suppose that conditions (33)–(36) hold. Then, by the Kojima-Schur Theorem, conditions (34)–(36) imply CAC-1∈(c,c). By (33), A∘C-1=AC-1 over c. By Part (ii) of Theorem 3, C and AC-1 are associative over c, and so we have C∘A∘C-1=CAC-1 over c. Hence, C∘A∘C-1∈(c,c), and now by Corollary 11, A∈(cC,cC).Corollary 21.
IfA=(ank)∈(cC,cC), then A=(ank)∈ℬ(cC,cC) and
(41)∥A∥(cC,cC)=∥CAC-1∥(c,c)=supn∑k‍kn|snk-sn(k+1)|.
Finally, let one give an application on a matrix operator which is not a triangle.Theorem 22.
Let(42)Γ=[γ1γ2γ3γ4⋯0γ2γ3γ4⋯00γ3γ4⋯000γ4⋯⋮⋮⋮⋮⋱],
where γk≠0 for all k∈ℕ. Then, a matrix A∈(c0Γ,c0) if and only if the following three conditions hold:
(43)(ankγk)k=1∞isboundedforeachn,supn∑k=1∞‍|ankγk-an(k-1)γk-1|<∞,limn(ankγk-an(k-1)γk-1)=0foreachk,
where an0/γ0=0.Proof.
Suppose thatA∈(c0Γ,c0). The operator Γ:ωΓ→ω is one to one, because for Γx=0 with x=(xk)∈ω we have the system of equations
(44)0=∑j=k∞‍γjxjfork=1,2,3,…,0=γkxk+∑j=k+1∞‍γjxj=γkxk+0=γkxkfork=1,2,3,….
Now, let us show thatΓ:c0Γ→c0 is onto. Let y=(yk)∈c0. Then Γx=y for x=(xk) with xk=γk-1(yk-yk+1) since
(45)(Γx)n=γnxn+γn+1xn+1+⋯=(yn-yn+1)+(yn+1-yn+2)+⋯=yn.
Hence, Γ:c0Γ→c0 is onto. So, Γ:c0Γ→c0 is bijective. The inverse operator is then
(46)Γ-1=[γ1-1-γ1-100⋯0γ2-1-γ2-10⋯00γ3-1-γ3-1⋯000γ4-1⋯⋮⋮⋮⋮⋱].
Now, by Theorem 10, A∈(c0Γ,c0) if and only if A∘Γ-1∈(c0,c0).
Forx=(xk)∈c0 we have Γ-1x=(γ1-1[x1-x2],γ2-1[x2-x3],…), and then for A=(ank) we have (A(Γ-1x))n=∑k‍ankγk-1(xk-xk+1). Let Sm be the mth partial sum of this series. Then,
(47)Sm=∑k=1m‍(ankγk-1-an(k-1)γk-1-1)xk-anmγm-1xm+1,
where an0=0.
By a similar proof to the proof of Lemma19, we can see that the sequence (anmγm-1)m=1∞ is bounded for each n, when A∈(c0Γ,ω). Then,
(48)(A(Γ-1x))n=limmSm=∑k=1∞‍(ankγk-1-an(k-1)γk-1-1)xk=((AΓ-1)x)n.
The conditions for AΓ-1 to be in (c0,c0), by Example 5A page 129 of [12], are
(49)supn∑k=1∞‍|ankγk-1-an(k-1)γk-1-1|<∞,limn(ankγk-1-an(k-1)γk-1-1)=0foreachk.
Hence, we have all conditions (43). We leave the reverse implication part to the reader.



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