JFSA Journal of Function Spaces and Applications 1758-4965 0972-6802 Hindawi Publishing Corporation 820930 10.1155/2013/820930 820930 Research Article Matrix Mappings on the Domains of Invertible Matrices http://orcid.org/0000-0001-5294-1716 Altun Muhammed Okoudjou Kasso A. Faculty of Arts and Sciences Melikşah University 38280 Kayseri Turkey meliksah.edu.tr 2013 24 10 2013 2013 31 05 2013 05 09 2013 2013 Copyright © 2013 Muhammed Altun. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We focus on sequence spaces which are matrix domains of Banach sequence spaces. We show that the characterization of a random matrix operator T=(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=BTA1(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.

1. Introduction and Preliminaries

Let ω denote the set of all complex sequences. Any subspace of ω is called as a sequence space. We will write , c, and c0 for the spaces of all bounded, convergent, and null sequences, respectively. By p, we denote the space of all p absolutely summable sequences, where 1p<.

Let X and Y be Banach spaces. Then, (X,Y) is the set of all operators L:XY and (X,Y) is the set of all continuous linear operators L:XY. (X,Y) is a Banach space with the operator norm defined by L=sup{L(x)/x:0xX}  (L(X,Y)).

Let E and F be two sequence spaces and A=(ank) an infinite matrix of real or complex numbers ank, where n, k={1,2,3,}. Then, we say that A defines a matrix mapping from E into F, and we denote it by writing A:EF, 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). Thus, A(E,F) if and only if the series on the right side of (1) converges for each n and every xE, and we have Ax={(Ax)n}nF for all xE.

Let A=(ank) and B=(bnk). Suppose the sums (2)cnk=janjbjk exist for all n, k. Then, the product of A and B is defined by AB=(cnk).

If E is a subset of ω, then EA={xω:AxE} 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:EAE is bijective; that is, there exists an operator A-1 such that A-1(Ax)=x for all xEA and A(A-1y)=y for all yE. 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 spaces c0, c, p, and 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 as p, c0, c, or . 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 , and of the Nörlund matrices in . A general approach was done in  reducing the characterizations of the classes (XT,Y) for arbitrary FK spaces X with AK and Yω 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 . The gliding hump properties of matrix domains were examined in .

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 [<xref ref-type="bibr" rid="B16">12</xref>, Theorem <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M82"><mml:mn>4.2</mml:mn><mml:mo>.</mml:mo><mml:mn>8</mml:mn></mml:math></inline-formula>, page 57]).

Matrix operators between BK spaces are continuous.

An infinite matrix T=(tnk) is said to be a triangle if tnk=0 for n<k and tnn0 for n. The following is a well-known result about triangles.

Theorem 2 (see [<xref ref-type="bibr" rid="B16">12</xref>, 1.4.8, page 9]).

Every triangle T has an inverse T-1 which also is a triangle, and x=T(T-1x)=T-1(Tx) for all xω.

Let A1,A2 be two matrices and μ a sequence space. If A1(A2x)=(A1A2)x holds for all xμ, 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  xω. Moreover, we have the following result.

Theorem 3 (see [<xref ref-type="bibr" rid="B16">12</xref>, Theorem <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M107"><mml:mn>1.4</mml:mn><mml:mo>.</mml:mo><mml:mn>4</mml:mn></mml:math></inline-formula>, page 8]).

Let A1,A2 be two matrices.

If rows of A1 are in 1 and A2(,), then A1 and A2 are associative over .

If A1 is row finite, then A1 and A2 are associative over ωA2.

Corollary 4.

The set of matrices (,) are associative over .

Remark 5.

The associativity property does not hold in general (see, e.g., [12, Example 1.4.6]).

Theorem 6.

Let A be an invertible matrix over a normed sequence space E with norm ·E. Then, EA is a normed sequence space with norm zEA=AzE.

Proof.

Let α and zEA. Then, (4)αzEA=A(αz)E=αAzE=|α|AzE=|α|zEA. Secondly, for y,zEA we have (5)y+zEA=A(y+z)E=Ay+AzEAyE+AzE=yEA+zEA, so the triangle inequality holds.

Now, suppose that zEA=0. Then, AzE=0 and since ·E is a norm we have Az=θ. Since A is invertible, we have z=θ.

Theorem 7.

Let E be a Banach sequence space and A a matrix. Then, the operator A:EAE is linear and continuous.

Proof.

Let An be the operator that corresponds to the nth row of the matrix operator A; that is, Anx=k=1ankxk for all x=(xk)EA. Let x=(xk)EA and y=(yk)EA. Clearly, we have Ancx=cAnx for any c. Let x=Anx=k=1ankxk and y=Any=k=1ankyk. We have An(x+y)=k=1ank(xk+yk)=limmk=1mank(xk+yk)=limmk=1mankxk+limmk=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{AxExEA:0xEA}=sup{AxEAxE:0xEA}=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-xmEA=A(xn-xm)E=Axn-AxmE=yn-ymE. Let yE such that limyn=y in E. Let x=A-1y. Then, (8)limnxn-xEA=limnA(xn-x)E=limnyn-yE=0.

Theorem 9.

Let A and B be invertible matrices over the BK spaces E and F, respectively. Suppose that T(EA,FB). If BTA-1=BTA-1 over E, then the operator T(EA,FB) is continuous.

Proof.

Let T(EA,FB) and let · be the norm of E. Since A is invertible over E, A:EAE is bijective. Also, A is continuous by Theorem 7.

By the bounded inverse theorem A-1 is continuous. Similarly the inverse B-1 of B:FBF is continuous.

Now, we have that the matrix operator R=BTA-1(E,F) and is continuous by Theorem 1. Since A and B-1 are continuous, the operator T=B-1RA is continuous.

Theorem 10.

Let A 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 BTA-1(E,F).

Proof.

Suppose that T(EA,FB), and let z=(zk)E. Then, clearly A-1zEA. Hence, we have (TA-1)zFB and so (BTA-1)zF.

For the inverse implication, suppose that BTA-1(E,F), and let zEA. Then, AzE and so (BTA-1A)z=(BT)zF. Since B is invertible, we have TzFB.

Corollary 11.

Let E and F be sequence spaces and A and B triangles. Then, for an operator T, T(EA,FB) if and only if BTA-1(E,F).

Theorem 12.

Let E and F be two normed sequence spaces and A and B invertible matrices over E and F, respectively. Then, for an operator T, T(EA,FB) if and only if BTA-1(E,F). In this case, one has (9)T(EA,FB)=BTA-1(E,F).

Proof.

It is enough to show the following equality: (10)T(EA,FB)=supθzEATzFBzEA=supθAzE(BT)zFAzE=supθAzE(BTA-1A)zFAzE=(BTA-1)E,F.

3. Examples and Applications

Theorem 10, 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.

Let λ=(λk) be a sequence of nonzero scalars in . 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)={zω:λzE}. Let (12)Λ=[λ10000λ20000λ30000λ4]. Then, the matrix Λ is invertible with (13)Λ-1=[λ1-10000λ2-10000λ3-10000λ4-1], and E(λ)=EΛ. So, all the Theorems in Section 2 are applicable for multiplier sequence spaces (see  for detailed applications and examples).

Example 14.

Let T=(tnk) be a row-finite matrix and B the operator with matrix representation (14)B=.B is invertible and the inverse operator has the matrix representation (15)B-1=[1000-11000-11000-11].B and B-1 are triangles and so they are one to one. B-1 is the operator B(r,s) of  with r=1 and s=-1. Since both T and B-1 are row finite, we have TB-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 ) we have that T(cB,c) if and only if the following three conditions hold:

limntnk-tn(k+1) exists for each k;

limnktnk-tn(k+1) exists;

supnk|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:

limn  snk exists for each k;

limnksnk exists;

supnk|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:

limnsnk-sn(k+1) exists for each k;

limnksnk-sn(k+1) exists;

supnk|snk-sn(k+1)|<.

Theorem 15.

An infinite matrix A=(ank)(,bv) if and only if (20)supNf()k|nN(ank-a(n-1)k)|<, where f() denotes the collection of all finite subsets of .

Proof.

First, we observe that bv=1Δ, where Δ is the difference operator B-1 of Example 14. The inverse of Δ is B. By Corollary 11, A(,1Δ) if and only if ΔA(,1). Since bvc, we have A(,c) and so by the Schur Theorem (see, e.g., Theorem 2.11 of ) rows of A are in 1. By Theorem 3 part (ii), Δ and A are associative over . So, ΔA=ΔA over . We have (21)[ΔA]nk=ank-a(n-1)k, where a0k=0 for all k. Now, using Theorem 2.14 of , we have A(,bv) if and only if (22)supNf()k|nN(ank-a(n-1)k)|<.

Using Theorem 12, we have the following.

Corollary 16.

If A=(ank)(,bv), then A=(ank)(,bv) and (23)A(,bv)=ΔA(,1).

Example 17.

Let T=(tnk) be a row finite matrix and C the Cesàro operator which has a matrix representation (24)C=[10001/21/2001/31/31/301/41/41/41/4]. 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=[1000-12000-23000-34]. Since T,C, and C-1 are all row finite, the operator CTC-1 is represented by the matrix CTC-1. Now, using Corollary 11, we have that T is in (()C,()C) if and only if CTC-1 is in (,), where ()C denotes the space, introduced by Ng and Lee , of all sequences whose C-transforms are in the space .

Let snk 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 (,) if and only if (27)supnkkn|snk-sn(k+1)|< by Theorem 2.6 of . So, we have T(()C,()C) if and only if (27) holds. Similarly, we have T(cC,()C) or T((c0)C,()C) if and only if (27) holds, where (c0)C and cC denote the spaces c0~ and c~, studied by Şengönül and Başar , 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[1nk|snk|]<,limnsnknexists  for  each  k,limn1nksnkexists, where snk=a1k+a2k++ank.

Proof.

By Corollary 11, A(c,cC) if and only if CA(c,c). Since A(c,cC)(c,ω), we have that rows of A are in 1. By Theorem 3 part (ii), C and A are associative over c. So, CA=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=1akck is convergent for all Cesàro summable sequences (ck), then limkak=0.

Proof.

Suppose that limkak0. 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 knn2kn-1. Let (30)ck={kn|akn|naknif  k=kn0otherwise. Then, the sequence (ck) is Cesàro summable because for knm<kn+1 we have (31)|c1+c2++cmm||c1|+|c2|++|cm|m=|ck1|+|ck2|++|ckn|m=k1/1+k1/2++kn/nmk1/1+k1/2++kn-1/(n-1)+kn/nkn=k1kn+k22kn++kn-1(n-1)kn+1n1n2+12n2++1(n-1)n2+1n(n-1)1n2+1n2n. On the other hand, (32)k=1akck=n=1aknckn=n=1kn|akn|n>ϵn=11n, and so we get to the contradiction k=1akck= for the Cesàro summable sequence (ck). So, our assumption limkak0 is not true.

Theorem 20.

Let C be the Cesàro matrix. Then, A=(ank)(cC,cC) if and only if (33)limkkank=0        for  each  n,(34)supnkkn|snk-sn(k+1)|<,(35)limnsnk-sn(k+1)n        exists  for  each  k,(36)limn1nkk(snk-sn(k+1))        exists, where snk=a1k+a2k++ank.

Proof.

Suppose that A(cC,cC). Then, by Lemma 19, we have (33). By Corollary 11, A(cC,cC) if and only if CAC-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=1ank(kxk-(k-1)xk-1) is convergent in . 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=1k(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, CAC-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), AC-1=AC-1 over c. By Part (ii) of Theorem 3, C and AC-1 are associative over c, and so we have CAC-1=CAC-1 over c. Hence, CAC-1(c,c), and now by Corollary 11, A(cC,cC).

Corollary 21.

If A=(ank)(cC,cC), then A=(ank)(cC,cC) and (41)A(cC,cC)=CAC-1(c,c)=supnkkn|snk-sn(k+1)|. Finally, let one give an application on a matrix operator which is not a triangle.

Theorem 22.

Let (42)Γ=[γ1γ2γ3γ40γ2γ3γ400γ3γ4000γ4], where γk0 for all k. Then, a matrix A(c0Γ,c0) if and only if the following three conditions hold: (43)(ankγk)k=1        is  bounded  for  each  n,supnk=1|ankγk-an(k-1)γk-1|  <,limn(ankγk-an(k-1)γk-1)=0        for  each  k, where an0/γ0=0.

Proof.

Suppose that A(c0Γ,c0). The operator Γ:ωΓω is one to one, because for Γx=0 with x=(xk)ω we have the system of equations (44)0=j=kγjxj  for  k=1,2,3,,0=γkxk+j=k+1γjxj=γkxk+0=γkxkfor  k=1,2,3,.

Now, let us show that Γ:c0Γc0 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:c0Γc0 is onto. So, Γ:c0Γc0 is bijective. The inverse operator is then (46)Γ-1=[γ1-1-γ1-1000γ2-1-γ2-1000γ3-1-γ3-1000γ4-1]. Now, by Theorem 10, A(c0Γ,c0) if and only if AΓ-1(c0,c0).

For x=(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 Lemma 19, 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 , are (49)supnk=1|ankγk-1-an(k-1)γk-1-1|<  ,              limn(ankγk-1-an(k-1)γk-1-1)=0for  each  k. Hence, we have all conditions (43). We leave the reverse implication part to the reader.

Acknowledgment

The author thanks the referees for their valuable comments and suggestions.

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