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α-Admissible mapping in C-algebra-valued b-metric spaces and fixed point theorems

  • In the present paper, for a unital C-algebra A, we introduce a version of αA-admissible on C-algebra-valued b-metric space, we proved some Banach and common fixed point theorems using αA-admissible. Also, we give some non-trivial examples and an application to illustrate our results.

    Citation: Saleh Omran, Ibtisam Masmali. α-Admissible mapping in C-algebra-valued b-metric spaces and fixed point theorems[J]. AIMS Mathematics, 2021, 6(9): 10192-10206. doi: 10.3934/math.2021590

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  • In the present paper, for a unital C-algebra A, we introduce a version of αA-admissible on C-algebra-valued b-metric space, we proved some Banach and common fixed point theorems using αA-admissible. Also, we give some non-trivial examples and an application to illustrate our results.



    In 2014, Ma et al [18] introduced the concept of C-algebra-valued metric spaces by replacing the range of R with an unital C-algebra. Later in 2015, Ma et al [19] introduced the nation of C-algebra-valued metric spaces as a generalization of C-algebra-valued metric space. They proved some Banach fixed point theorems. Several research are obtained some results in Banach and common fixed point theorems in C-algebra-valued metric spaces (see [2,3,7,10,13,14,25,26,31,34]. The notion of C-algebra-valued partial metric space and C-algebra-valued partial b-metric spaces are introduced in [8,22] and proved fixed point results as analogous of Banach contraction principle.

    In [27] introduced the study of fixed point for the α-admissibility of mappings and generalized several known results of metric spaces see also [28]. Later on, many authors proved α-admissible mappings theorems with various contraction condition see [1,5,9,12,17,20,21,30,32,33,35,36]. The aim of this paper is generalizing some results of metric spaces and C-algebra b-valued metric spaces.

    We start with some definition and results about C-algebra b-valued metric spaces. Suppose that A is a unital C-algebra with a unit I. Set Ah={xA:x=x}. An element xA is a positive element, if x=x and σ(x)R+ is the spectrum of x. We define a partial ordering on Ah as xy if 0Ayx, where 0A means the zero element in A and we let A+ denote the {xA:x0A} and |x|=(xx)12.

    On the other hand, [27] introduced the study of fixed point for the α-admissibility of mappings and generalized several know results of metric spaces.

    Throughout this paper, we use the concept of α-admissibility of mappings defined on C-algebra b-valued metric spaces and we defined the generalized Lipschitz contractions on such spaces. The aim of this paper is generalizing some results of metric space and C-algebra b-valued metric spaces.

    Lemma 1.1. Suppose that A is a unital C-algebra with unit IA. The following are holds.

    (1) If aA, with a<12, then 1a is invertible and a(1a)1<1.

    (2) For any xA and a,bA+, such that ab, we have xax and xbx are positive element and xaxxbx.

    (3) If 0Aab then ab.

    (4) If a,bA+ and ab=ba, then a.b0A.

    (5) Let A denote the set {aA:ab=babA} and let aA, if b,cA with bc0A and 1a(A)+ is an invertible element, then (IAa)1b(IAa)1c.

    We refer [24] for more Calgebra details.

    Definition 1.2. Let X be a non-empty set and bIA, bA, suppose the mapping

    dA:X×XA, satisfies:

    (1) dA(x,y)0A for all x,yX and dA(x,y)=0Ax=y.

    (2) dA(x,y)=dA(y,x) for all x,yX.

    (3) dA(x,z)b[dA(x,y)+dA(y,z)] for all x,y,zX, where 0A is zero-element in A and IA is the unit element in A. Then dA is called a C-algebra valued b-metric on X and (X,A,dA) is called C-algebra- valued b-metric space.

    Example 1.3. Let X be a Banach space, dA:X×XA given by dA(x,y)=xypa, for all x,yX, aA+, a0 and p>1.

    Its easy to variety that (X,A,dA) is a C-algebra -valued b-metric space.

    Using the inequality (a+b)p2p(ap+bp) for all a,b0, p>1, we have

    xzp2p(xyp+yzp)

    for x,y,zX, which implies that

    dA(x,z)2p(dA(x,y)+dA(y,z))

    In the next we give a counter example, show that in general, a C-algebra valued b-metric space in not necessary a C-algebra valued metric space.

    Example 1.4. Let X=R and A=M2(R). Define

    dA(x,y)=(|xy|200k|xy|2)

    x,yR, k>0, it is clear that (X,A,dA) is a C-algebra valued b-metric space by using the same argument in example 1.3 when p=2

    Now, dA(0,1)=(100k.1), dA(1,2)=(100k.1), dA(0,2)=(400k.4).

    Its obvious that dA(0,2)dA(0,1)+dA(1,2).

    So (X,A,dA) is not a C-algebra valued metric space

    Definition 1.5. Let (X,A,dA) be a C-algebra- valued b-metric space, xX, and {xn}n=1 be a sequence in X, then

    (i) {xn}n=1 convergent to x whenever, for every cA with c0A, there is a natural number NN such that

    dA(xn,x)c,

    for all n>N. We denote this by limnxn=x or xnx as n+.

    (ii) {xn}n=1 is said to be a Cauchy sequence whenever, for every cA with c0A, there is a natural number NN such that

    dA(xn,xm)c,

    for all n,m>N.

    Lemma 1.6. (i) {xn}n=1 is a convergence in X. If for any element ϵ>0 there is NN such that for all n>N, d(xn,x)ϵ.

    (ii) {xn}n=1 is a Cauchy sequence in X, for any ϵ>0 there NN such that

    dA(xn,xm)ϵ, for all n,m>N. We say that (X,A,dA) is a complete C-algebra- valued b-metric space if every Cauchy sequence is convergent with respect to A.

    Example 1.7. Let X=R and A=Mn(R) the set of all n×n-matrices with entries in R. Define

    dA(a,b)=(λ1|aiibii|p00λn|annbnn|p)

    where a=(aij)ni,j=1, b=(bij)ni,j=1 are two n×n-matrices, aij,bijR for all i,j=1,...,n,λi0 for i=1,...,n are positive real numbers.

    One can define a partial ordering on (Mn(R)) on Mn(R) as following aMn(R)b if and only if aijbiji,j=1,...,n. And an element aMn(R)0 is positive in Mn(R) if and only if aij0 for all i,j=1,...,n. (X,Mn(R),dMn(R)) is C-algebra- valued b-metric space.

    One can prove that

    dA(a,c)Mn(R)2p(dA(a,b)+dA(b,c)),

    for all a,b,cMn(R).

    We need only to use the following inequality in R

    |xz|p2p(|xy|p+|yz|p,

    where b=2pIMn(R)IMn(R) p1, where IMn(R)is the unit element in Mn(R).

    Remark 1.8. In the above example the inequality |xz|p|xy|p+|yz|p it is impossible for x>y>z. Then the (X,A,d) is not a C-algebra valued metric space.

    It is useful to discuss the relation between C-algebra valued metric spaces and lattices-valued metric spaces. To classify C-algebra-valued-metric spaces and its relation with lattices, we have to discuss the concept of quantale which introduced by Mulvey [23]. A quantale Q is a complete lattice together with an associative multiplication :Q×QQ such that a(ibi)=i(abi), and i(ai)b=i(aib) for all ai,bi,a,bQ, iI, I is an index set the quantale is said to be unit, if it has a unital 'e' satisfy ae=ea for all aQ. And Q is called an involuative quantale with relation :QQ satisfy (a)=a, (ab)=ba and (iai)=iai for all ai,bi,aQ.

    The top element of Q is denoted by 1 and the bottom element denoted by 0. A typical example of quantale is given by End(S), the set of all sublattices of Endomorphisms of the complete lattices S is a unital quantale with join calculated by point wise (ifi)(x)=ifi(x) and multiplication as composition (fg)(x)=(fy)(x).

    And it is unit identity is IdS. An element aQ is said to be right-sided if a1a, for all aQ, denote by R(Q) the set of all right-sided elements. Similarly, an element aQ is said to be left- side if 1aa for all aQ, L(Q) denote the set of left-sided elements.if aQ is right-sided elements and left-sided elements it is said to be 2-sided elements and the set of 2-sided-elements denoted by I(Q). Any two sided-elements a is distributive in the sense that aibi=i(abi).

    A quantale is commutative if it is commutative under the multiplication. If the quntale is commutative then QI(Q). If A is a C-algebra and R(A) is the lattice of all closed right ideals of A, then R(A) is a quantale and the multiplication of closed right ideals obtained by taking the topological closure of the usual product of ideals, simply, IJ=¯IJ for any two ideals I,JR(A).

    By Gelfand duality theorem [11]. Any commutative C-algebra is isomorphic to the set of all continuous functions of the compact Hausdorf topological space. So, in this case R(A) is isomorphic to the lattice O(ˆA) of all open sublattices of ˆA, where ˆA is the topological space determined by A, the spectrum of A. Therefore, commutative C-algebra classify by commutative quantales as given in [6]. A is a commutative C-algebra if and only if R(A) is commutative quantale.

    On the other hand 'Sherman [29]' show that if Asa is the space of self-.adjoint elements of a C-algebra A with the canonical order given by ab if and only if ba0 is positive. Then Asa is a lattice ordered if and only if A is commutative. Therefore, the C-algebra valued metric space in commutative case coinside with the commutative quantale -valued-metric space, with a suitable metric For a non-commutative C-algebra A with unit, by MaxA is meant. The set of all subspace of A together with the multiplication defined by MN=¯MN to be the closure of product liner subspace, for each M,NMaxA, and the join defined by iMi=¯iMi, and the involution M={a:aM} and the unit of MaxA is given by the identity. So, MaxA is defined A unital involutive quantale. In the case the non-commutative C-algebra is classify by MaxA, following [15,16]. If A and B are two unital C-algebras. Then A and B are isomorphic if and only if MaxA and MaxB are isomorphic as a unital involutive quantale. So, C-algebra valued metric spaces are classify by the unital involutive quantale-valued metric space.

    In 2012 Samet et al [27], introduced the concept of α-admissible mapping as follows.

    Definition 2.1. Let T:XX be self map and α:X×X[0,+). Then T is called α-admissible if for all x,yX with α(x,y)1 implies α(Tx,Ty)1.

    Next, we introduced an analogue definition of α-admissible for a unital C-algebra.

    Definition 2.2. Let X be a non-empty set and αA:X×X(A)+ be a function, we say that the self map T is αA - admissible if (x,y)X×X, αA(x,y)IA αA(Tx,Ty)IA, where IA the unity of A.

    Definition 2.3. Let (X,A,d) be a complete C-algebra- valued b-metric space, the mapping T:XX is said to be generalised Lipschitz condition if there exist aA such that a<1 and

    dA(Tx,Ty)adA(x,y)a, (2.1)

    for all x,yX with αA(x,y)IA.

    Example 2.4. Let X=R and A=Mn(R) as given in example (1.7), define T:XX, by Tx=x2, and αMn(R):X×XMn(R)+, given by αMn(R)(x,y)=IMn(R) and αMn(R)(Tx,Ty)=αMn(R)(x2,y2)=IMn(R) thus T is αMn(R)admissible, where Mn(R)+ is the set of all positive elements

    αMn(R)(x,y)dMn(R)(Tx,Ty)Mn(R)IMn(R).(λ1|x2 y2|p00λn|x2 y2|p)Mn(R)IMn(R)(2)p.dMn(R)(x,y),

    and a=IMn(R)(2)p, a=IMn(R)(2)p, so T satisfy the generalised Lipschitz condition.

    Theorem 2.5. Let (X,A,dA) be a complete C-algebra- valued b-metric space, with bIA, bA, ba2<1 suppose that T:XX, be a generalised Lipschitz contraction satisfies the following conditions:

    (i) T is αA-admissible.

    (ii) There exists x0X such that αA(x0,Tx0)IA.

    (iii) T is continuous.

    Then T has a fixed point.

    Proof: let x0X such that αA(x0,Tx0)IA and define a sequence {xn}n=0X such that xn=Txn1 for all nN. If xn=xn+1 for some nN, then x=xn is a fixed point for T.

    Assume that xnxn+1 for all nN, since T is αA-admissible, we have

    αA(x0,x1)=αA(x0,Tx0)IAαA(Tx0,T2x0)=α(x1,x2)IA.

    By induction we get

    αA(xn,xn+1)IA. (2.2)

    Since T is generalised Lipschitz condition, then

    dA(xn,xn+1)=dA(Txn1,Txn)adA(xn1,xn)a(a)2dA(xn2,xn1)a2...(a)ndA(xo,x1)an(a)nd0an.

    Denote that d0:=dA(x0,x1) in A, notice that in C-algebra, if a,bA+ and 0Aab, then for any xA both xax and xbx are positive elements and

    0Axaxxbx.

    Now, for m1, p1 it following that

    dA(xm,xm+p)b[dA(xm,xm+1)+dA(xm+1,xm+p)]bdA(xm,xm+1)+b2dA(xm+1,xm+2)+...+bp1dA(xm+p2,xm+p1)+bp1dA(xm+p1,xm+p)b((a)md0am)+b2((a)m+1d0am+1)+...+bp1((a)m+p2d0am+p2)+bp1((a)m+p1d0am+p1)=p1k=1bk((a)m+k1d0am+k1)+bp1(a)m+p1d0am+p1=p1k=1bk((a)m+k1d120d120am+k1)+bp1(a)m+p1d0am+p1=p1k=1((a)m+k1bk2d120)(d120bk2am+k1)+(bp12(a)m+p1d120d120bp12(a)m+p1)=p1k=1(d120bk2am+k1)(d120bk2am+k1)+(d120bp12am+p1)(d120bp12am+p1)=p1k=1|d120bk2am+k1|2+|d120bp12am+p1|2p1k=1d120bk2am+k12.IA+d120bp12am+p12.IAd1202p1k=1a2(m+k1)bk.IA+d1202bp1a2(m+p1).IA=d0[ba2m(1(ba2)p11ba2)].IA+d0bp1.a2(m+p1).IA=d0[ba2m((ba2)p11ba21)].IA+d0bp1.a2(m+p1)0A,

    with the condition ba2<1 and at m+.

    It implies that {xn}n=0 is a Cauchy sequence. By completeness of X that exists xX such that xnx as n+.

    Since T is continuous and the lim is unique it follows xn+1=TxnTx as n+ such that x=limnxn+1=limnTxn=Tx, so, Tx=x is a fixed point for T.

    Now, we replace the assumption of continuoity of T in the above theorem by another condition.

    Theorem 2.6. Let (X,A,dA) be a complete C-algebra- valued b-metric space, with bIA.

    Let T:XX be generalized Lipschitz condition as in (2.5) and the following conditions are satisfies:

    (i) T is αA-admissible.

    (ii) There exists x0X such that αA(x0,Tx0)IA.

    (iii) If {xn}n=0 is a sequence in X such that αA(xn,xn+1IA for all nN and xnxX, as n+, then αA(xn,x)IA for all nN. Then T has a fixed point in X.

    Proof: From theorem 2.5, we Know that {xn}n=0 is a Couchy sequence in (X,A,dA), then there exists xX such that xnx as n+.

    On the other hand from equation(3.1) and by hypothesis (iii), we have dA(xn,x)IA, for all nN, since T is generalized Lipschitz Contraction using 2.2 we get

    dA(x,Tx)b[dA(x,xn+1)+dA(xn+1,Tx)=b[dA(x,xn+1)+dA(Txn,Tx)b[dA(x,xn+1)+a(d(xn,x)a]0Aasn+.
    dA(x,Tx)=0ATx=x.

    To prove the uniqueness of the fixed point of generalized Lipschitz mapping we have to consider the following property.

    (H): For all x,yX, there exists zX such that dA(x,z)IA and dA(y,z)IA.

    Theorem 2.7. Adding condition (H) to the hypothesis of theorem (2.5) we obtain the uniqueness of the fixed point of T.

    Proof: Suppose that x and y are two fixed points of T from (H), there exists zX such that

    αA(x,z)IAandαA(y,z)IA. (2.3)

    Since T is αA-admissible, from (2.2) we have

    αA(x,Tnz))IAandαA(y,Tnz))IA. (2.4)

    Since T is generalized Lipschitz contraction, so by using (2.4), we have

    dA(x,Tnz)=dA(Tx,T(Tn1z))adA(x,Tn1z)a...(a)ndA(x,z)aforallnNdA(x,Tnz)a2ndA(x,z).

    Since ba2<1, a<1, we have a2n0A as n+ and dA(x,Tnz)0A, Tnz=x as n+.

    Similarly we get Tnz=y as n+, there for by uniqueness of the limit, we obtain x=y. This complete the proof.

    Now, we give a common fixed point theorems for two mappings satisfy a common αA-admissible.

    Definition 3.1. let (T,S):XX be a continuous self mappings on X. αA:X×XA+. (T,S) are said to be common αA-admissible if for any x0X,

    αA(x0,y)IAαA(Tx0,Sy)IAαA(T2x0,S2y)IA.

    Theorem 3.2. Let (X,A,dA) be complete C-algebra- valued b-metric space and T,S:XX, such that

    αA(x,y)dA(Tx,Sy)adA(x,y)a, (3.1)

    and a<1, b.a2<1 and the following conditions are satisfies:

    (i) (T,S) are common αA-admissible.

    (ii) The exists x0X such that

    αA(x0,y)IAαA(Tx0,Sy)IA.

    (iii) T and S are continuous and have a common fixed point in X.

    Proof: Let x0X and construct a sequence {xn}X such that Tx2n=x2n+1, Sx2n+1=x2n+2 form (3.1), we get

    αA(x0,x1)=αA(Tx0,Sx1)IAαA(T2x0,S2x1)IAαA(x2,x3)IA,

    by induction, we have αA(x2n,x2n+1)IA, for all nN.

    dA(x2n+1,x2n+2)=dA(Tx2n,Sx2n+1)αA(x2n,x2n+1)dA(Tx2n,Sx2n+1)adA(x2n,x2n+1)a,

    by induction, we obtain

    dA(x2n+1,x2n+2)(a)2n+1dA(x0,x1)a2n+1.

    Similarly,

    dA(x2n,x2n+1)(a)2ndA(x0,x1)a2n.

    Now, we can obtain for any nN

    dA(xn,xn+1)(a)ndA(x0,x1)an.

    Then for pN, p1, m1, and applying the triangle inequality, we have

    dA(xm,xm+p)b[dA(xm,xm+1)+b2dA(xm+1,xm+2)+......+bp2dA(xm+p2,xm+p1)+bp1dA(xm+p1,xm+p)p1k=1bk((a)m+k1d0am+k1)+bp1(a)m+p1d0am+p1,

    by similar calculation as theorem (2.5), we get

    dA(xm,xm+p)d0[ba2m(1(ba2)p11ba2)]IA+d0bp1a2(m+p1)IA0A,

    as n+, where IA is the unitary in A, d0:=dA(x0,x1),b(A+).

    So, {xn} is a Cauchy sequence in X.

    The completion of (X,A,dA) implies that there exists xX such that limnxn=x

    Now, we using triangle inequality and (3.1), we set

    dA(x,Sx)b[dA(x,x2n+1)+dA(x2n+1,Sx)]b[dA(x,x2n+1)+dA(Tx2n,Sx)]andαA(x2n,x)IA,wegetdA(x,Sx)b[dA(x,x2n+1)+adA(x2n,x)a]dA(x,Sx)bdA(x,x2n+1)+ba2dA(x2n,x)dA(x,Sx)dA(x,x2n+1)(b+ba2).

    Since a<1, we have a contradiction dA(x,Sx)=0ASx=x, similarly, we get Tx=x, so, S and T have a common fixed point.

    In the following, we will show that the uniquely of common fixed point in X, for that assume that is another fixed point yX such that Ty=y=y.

    Since x satisfy property H, and (T,S) are αA -admissible, we have

    dA(x,Snz)=dA(Tx,Snz)adA(Tx,Sn1z)a...(a)ndA(x,z)andA(x,Snz)a2ndA(x,z)0asn+.

    So, dA(x,Snz)=0A this implies that Snz=x.

    Similarly, we get Snz=y Thus x is a unique common fixed point.

    Theorem 3.3. Let (X,A,dA) be a complete C-algebra- valued b-metric space, suppose that two mappings T,S:XX, satisfy

    α(x,y)dA(Tx,Ty)adA(Sx,Sy)aforanyx,yX, (3.2)

    where aA, with ba2<1 and a<1.

    If R(T)R(S) and R(S) is complete in X, T and S are weakly compatible, such that the following holds

    (i) (T,S) are common αA-admissible.

    (ii) There is x0X such that αA(x0,y)IAαA(Tx0,Sy)IA.

    (iii) T and S are continuous.

    (iv) X has a property (H), they T and S have a unique common fixed point in X.

    Proof: Let x0X, choose x1X, such that Sx1=Sx0, which can be done since R(T)R(S). Let x0X such that Sx2=Tx1.

    Repeating the process, we have a sequence {Sxn}n=1 in X satisfying Sxn=Txn1.

    Then, since (T,S) are αA-admissible, we get

    αA(Sx1,Sx2)=αA(Tx0,Tx1)IAαA(T2x0,T2x1)IAαA(Sx2,Sx3)IA...αA(Sxn,Sxn+1)IA.

    Now,

    dA(Sxn,Sxn+1)=dA(Txn1,Txn)adA(Sxn1,Sxn)a...(a)ndA(Sx0,Sx1)an.

    For m1, p1.

    dA(Sxm,Sxm+p)bdA(Sxm,Sxm+1)+b2dA(Sxm+1,Sxm+2)+...++bp1dA(Sxm+p2,Sxm+p+1)+bp1dA(Sxm+p1,Sxm+p)p1k=1bk(a)m+k1d0(a)m+k1+...+bp1(a)m+p1d0(a)m+p1.

    Using similar calculation as in theorem 2.5, we get

    dA(Sxm,Sxm+p)d0[ba2m(ba2)p11]ba21IA+d0bp1a2(m+p+1)IA0Aasm+.

    Where d0=dA(Sx0,Sx1).

    So, {Sxn}n=1 is a Cauchy sequence in R(S) and is complete in X, there exists xX such that limn+Sxn=Sx.

    Also,

    dA(Sxn,Tx)=dA(Txn1,Tx)adA(Sxn,x)a0A,asn+.

    So, SxnTx as n+. Hens Sxn=Tx=Sx, so x is coincidence common fixed point in X.

    Morovere of y is another common fixed point such that Ty=Sy=y, so

    dA(Sx,Sy)=dA(Tx,Ty)adA(Sx,Sy)a
    dA(Sx,Sy)a2dA(Sx,Sy).

    Since a<1, so we yet dA(Sx,Sy)=0ASx=Sy.

    So S, T have coincidence fixed point is unique Sx=Tx=x.

    Since {Sxn}n=1 is a sequence in X, convergent to Sx and Sy respectively,

    Sx=limn+Sxn=Tx, since the lim is unique, so Tx=Sx=x, so S and T have a common fixed point in X.\\ Since X has a property (H) and (S,T) are αA-admissible, we get

    dA(x,Tnx)=dA(Tx1,Tnz)=dA(Tx,Tn1z)adA(Sx,S(Tn1zn))a...(a)ndA(Sx,Sz)andA(x,Tnz)a2ndA(Sx,Sz)0asn+.dA(x,Tnz)=0ATnz=x.

    Similarly Tnz=y, so x=y and this complete the proof.

    We introduce a non-trivial example satisfy the theorem 2.5.

    Example 4.1. Let X=[0,1], A=M2(R), p>1 and k>0 is a constant, we define dA=X×XA as dA(x,y)=(|xy|p00k|xy|p) for all x,yX. Then (X,A,dA) is C-algebra valued b-metric space. Define T:XX as Tx=x2, then

    dA(Tx,Ty)=(|x2y2|p00k|x2y2|p)=(|xy|p|x+y|p00k|xy|p|x+y|p)2p.I(|xy|p00k|xy|p)

    Define, αA:X×XA, by αA(x,y)={(x00y)if  x=y=10otherwise,

    it is clear that αA(x,y)={IM2(R)if  x=y,0otherwise,

    αA(Tx,Ty)={(x200y2)if  x=y=10otherwise,

    αA(x,y)=IM2(R)αA(Tx,Ty)=IM2(R)

    So, αA(x,y)dA(Tx,Ty)(2)p)dA(x,y)(2)p. So, it is satisfy the conditions of theorem 2.5, and then T has a fixed point 0X.

    As an application, we use the C-algebra-valued b-metric space to study the existence and uniqueness of the system of matrix equations in [4] by using theorem 2.5.

    Example 4.2. Application: Suppose that Mn(C) the set of all m×n matrices with complex entries. Mn(C) is a C-algebra with the operator norm. Let B1, B2, ..., BnMn(C) are diagonal matrices which satisfy nk=1|Bk|2<1.

    Let A=(aij)1i,jnMn(C) and C=(cij)1i,jnMn(C)+, where Mn(C)+ denote the set of all positive definite matrices "hermitian and the eigenvalues are non-negative". Then the matrix equations

    Ank=1BkABk=C, (4.1)

    has a unique solution.

    Proof: Set α=nk=1|Bk|2, clear if α=0, then the equations has a unique solution in Mn(C). Without loss of generality, suppose that α>0. For A,DMn(C) and p1, define dMn(C):Mn(C)×Mn(C)Mn(C)+ as

    dMn(C)(A,D) = diag(λ1|a11d11,...,λn|anndnn|p), λ1,...,λn>0, then (Mn(C),dMn(C)) is a C-algebra valued b-metric space and is complete since the set Mn(C) is complete (the proof is given in the example 1.7). Consider the map T=(Tii):Mn(C)Mn(C) defined by

    Tii(aij)1i,jn=nk=1Bk(aii)Bk+cii. Define αMn(C):Mn(C)×Mn(C)Mn(C)+,

    αMn(C)(A,B)=IMn(C), clear that T is αM2(C) admissible. Then

    dMn(C)(TA,TD)=diag(λ1|(nk=1Bka11Bk+c11)(nk=1Bkd11Bk+c11),...,λn|(nk=1BkannBk+cnn)(nk=1BkdnnBk+cnn)|p)=diag(λ1|(nk=1Bk(a11d11)Bk|p,...,λn|(nk=1Bk(anndnn)Bk|p)=diag(λ1(nk=1|Bk|2)p|a11d11|p,...,λn(nk=1|Bk|2)p|anndnn|p)=diag(nk=1|Bk|2)p(λ1|a11d11|p,...,λn|anndnn|p)=αpdMn(C)(A,D).

    Therefore, T satisfy the condition of theorem 2.5 and has a fixed point. So the matrix equations (4.1) has a solution on Mn(C). Moreover αMn(C) is satisfy the condition (H), so the system of matrix equations have a unique hermitian matrix solution A.

    In this paper, we define a new version of αA-admissible in the case of self map T:AA and αA-admissible in two self mappings (T,S). We prove the principal Banach fixed point theorem and two common fixed point theorems in the C-algebra- valued b-metric space, which generalized the given results in [18,19,26,27]. Moreover, we introduced an application to show that the useful of C-algebra- valued b-metric space to study the existence and unique of system matrix equations.

    The first author thanks the south valley university, Egypt, for partially supporting the study. The second author thanks the Jazan university, Saudi Arabia, for supporting the study.

    The authors of this current research declaring that this study has been done without any competing intersts.



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