# k-Kernel Symmetric Matrices

Authors: A. R. Meenakshi; D. Jaya Shree
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

In this paper we present equivalent characterizations ofk-Kernel symmetric Matrices. Necessary and sufficient conditions are determined for a matrix to be k-Kernel Symmetric. We give some basic results of kernel symmetric matrices. It is shown that k-symmetric implies k-Kernel symmetric but the converse need not be true. We derive some basic properties of k-Kernel symmetric fuzzy matrices. We obtain k-similar and scalar product of a fuzzy matrix.


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## Body

## 1. Introduction

Throughout we deal with fuzzy matrices that is, matrices over a fuzzy algebraℱ with support [0,1] under max-min operations. For a,b∈ℱ, a+b=max{a,b}, a·b=min{a,b}, let ℱmn be the set of all m×n matrices over ℱ, in short ℱnn is denoted as ℱn. For A∈ℱn, let AT, A+, R(A), C(A), N(A), and ρ(A) denote the transpose, Moore-Penrose inverse, Row space, Column space, Null space, and rank of A, respectively. A is said to be regular if AXA=A has a solution. We denote a solution X of the equation AXA=A by A- and is called a generalized inverse, in short, g-inverse of A. However A{1} denotes the set of all g-inverses of a regular fuzzy matrix A. For a fuzzy matrix A, if A+ exists, then it coincides with AT [1, Theorem 3.16]. A fuzzy matrix A is range symmetric if R(A)=R(AT) and Kernel symmetric if N(A)=N(AT)={x:xA=0}. It is well known that for complex matrices, the concept of range symmetric and kernel symmetric is identical. For fuzzy matrix A∈ℱn, A is range symmetric, that is, R(A)=R(AT) implies N(A)=N(AT) but converse needs not be true [2, page 217]. Throughout, let k-be a fixed product of disjoint transpositions in Sn=1,2,…,n and, K be the associated permutation matrix. A matrix A=(aij)∈ℱn is k-Symmetric if aij=ak(j)k(i) for i,j=1 to n. A theory for k-hermitian matrices over the complex field is developed in [3] and the concept of k-EP matrices as a generalization of k-hermitian and EP (or) equivalently kernel symmetric matrices over the complex field is studied in [4–6]. Further, many of the basic results on k-hermitian and EP matrices are obtained for the k-EP matrices. In this paper we extend the concept of k-Kernel symmetric matrices for fuzzy matrices and characterizations of a k-Kernel symmetric matrix is obtained which includes the result found in [2] as a particular case analogous to that of the results on complex matrices found in [5].


## 2. Preliminaries

Forx=(x1,x2,…,xn)∈ℱ1×n, let us define the function κ(x)=(xk(1),xk(2),…,xk(n))T∈ℱn×1. Since K is involutory, it can be verified that the associated permutation matrix satisfy the following properties.SinceK is a permutation matrix, KKT=KTK=In and K is an involution, that is, K2=I, we have KT=K.(P.1)
K=KT, K2=I, and κ(x)=Kx For A∈ℱn ,(P.2)
N(A)=N(AK),(P.3)
ifA+ exists, then (KA)+=A+K and (AK)+=KA+(P.4)
A+ exist if and only if AT is a g-inverse of A.Definition 2.1 (see [2, page 119]).
ForA∈ℱn is kernel symmetric if N(A)=N(AT), where N(A)={x/xA=0andx∈ℱ1×n}, we will make use of the following results.Lemma 2.2 (see [2, page 125]).
ForA,B∈ℱn and P being a permutation matrix, N(A)=N(B)⇔N(PAPT)=N(PBPT)Theorem 2.3 (see [2, page 127]).
ForA∈ℱn, the following statements are equivalent:(1)
A is Kernel symmetric,(2)
PAPT is Kernel symmetric for some permutation matrix P,(3)
there exists a permutation matrixP such that PAPT=[D000] with detD>0.


## 3.k-Kernel Symmetric Matrices

Definition 3.1.
A matrixA∈ℱn is said to be k-Kernel symmetric if N(A)=N(KATK)Remark 3.2.
In particular, whenκ(i)=i for each i=1 to n, the associated permutation matrix K reduces to the identity matrix and Definition 3.1 reduces to N(A)=N(AT), that is, A is Kernel symmetric. If A is symmetric, then A is k-Kernel symmetric for all transpositions k in Sn.
Further,A is k-Symmetric implies it is k-kernel symmetric, for A=KATK automatically implies N(A)=N(KATK). However, converse needs not be true. This is, illustrated in the following example.Example 3.3.
Let(3.1)A=[000.60.5100.50.30],K=[001010100]KATK=[000.60.3100.50.50].
Therefore,A is not k-symmetric.
For thisA, N(A)={0}, since A has no zero rows and no zero columns.
N(KATK)={0}. Hence A is k-Kernel symmetric, but A is not k-symmetric.Lemma 3.4.
ForA∈ℱn, A+ exists if and only if (KA)+ exists.Proof.
By [1, Theorem 3.16], For A∈ℱmn if A+ exists then A+=AT which implies AT is a g-inverse of A. Conversely if AT is a g-inverse of A, then AATA=A⇒ATAAT=AT. Hence AT is a 2 inverse of A. Both AAT and ATA are symmetric. Hence AT=A+:
(3.2)A+exists⟺AATA=A⟺KAATA=KA⟺(KA)(KA)T(KA)=KA⟺(KA)T∈(KA){1}⟺(KA)+,exists(By,P.4).For sake of completeness we will state the characterization of k-kernel symmetric fuzzy matrices in the following. The proof directly follows from Definition 3.1 and by (P.2).Theorem 3.5.
ForA∈ℱn, the following statements are equivalent:(1)
A is k-Kernel symmetric,(2)
KA is Kernel symmetric,(3)
AK is Kernel symmetric,(4)
N(AT)=N(KA),(5)
N(A)=N((AK)T),Lemma 3.6.
LetA∈ℱn, then any two of the following conditions imply the other one, (1)
A is Kernel symmetric,(2)
A is k-Kernel symmetric,(3)
N(AT)=N((AK)T).Proof.
However, (1) and (2) ⇒ (3):
(3.3)Aisk-Kernelsymmetric⇒N(A)=N(KATK)⇒N(A)=N(KAT)(By,P.2)Hence,(1)and(2)⇒N(AT)=N(A)=N((AK)T).
Thus (3) holds.
Also (1) and (3) ⇒ (2):
(3.4)AisKernelsymmetric⇒N(A)=N(AT)Hence,(1)and(3)⇒N(A)=N((AK)T)⇒N(AK)=N((AK)T)(By,P.2)⇒AKisKernelsymmetric⇒Aisk-Kernelsymmetric(byTheorem(3.5)).
Thus (2) holds.
However, (2) and (3) ⇒ (1):
(3.5)Aisk-Kernelsymmetric⇒N(A)=N(KATK)⇒N(A)=N((AK)T)(by,P.2)Hence(2)and(3)⇒N(A)=N(AT).
Thus, (1) holds.
Hence the theorem.Toward characterizing a matrix beingk-Kernel symmetric, we first prove the following lemma.Lemma 3.7.
LetB=[D000], where D is r×r fuzzy matrix with no zero rows and no zero columns, then the following equivalent conditions hold:  (1)
B is k-Kernel symmetric,(2)
N(BT)=N((BK)T),(3)
K=[K100K2] where K1 and K2 are permutation matrices of order r and n-r, respectively,(4)
k=k1k2 where k1 is the product of disjoint transpositions on Sn={1,2,…,n} leaving (r+1,r+2,…,n) fixed and k2 is the product of disjoint transposition leaving (1,2,…,r) fixed.Proof.
SinceD has no zero rows and no zero columns N(D)=N(DT)={0}. Therefore N(B)=N(BT)≠{0} and B is Kernel symmetric.
Now we will prove the equivalence of (1),(2), and (3). B is k-Kernel symmetric ⇔N(BT)=N((BK)T) follows from By Lemma (3.6).
Choosez=[0y] with each component of y≠0 and partitioned in conformity with that of B=[D000]. Clearly, z∈N(B)=N((BT))=N((BK)T). Let us partition K as K=[K1K3K3TK2], Then
(3.6)KBT=[K1K3K3TK2][DT000]=[K1DT0K3TDT0].
Now
(3.7)z=[0y]∈N(B)=N(KBT)⇒[0y][K1DT0K3TDT0]=0⇒yK3TDT=0
Since N(DT)=0, it follows that yK3T=0.
Since each component ofy≠0 under max-min composition yK3T=0, this implies K3T=0⇒K3=0.
Therefore(3.8)K=[K100K2].
Thus, (3) holds, Conversely, if (3) holds, then
(3.9)KBT=[K1DT000],N(KBT)=N(B).
Thus (1)⇔(2)⇔(3) holds.
However,(3)⇔(4): the equivalence of (3) and (4) is clear from the definition of k.Definition 3.8.
ForA,B∈ℱn, A is k-similar to B if there exists a permutation matrix P such that A=(KPTK)BP.Theorem 3.9.
ForA∈ℱn and k=k1k2 (where k1k2 as defined in Lemma 3.7). Then the following are equivalent:(1)
A is k-Kernel symmetric of rank r,(2)
A is k-similar to a diagonal block matrix [D000] with detD>0,(3)
A=KGLGT and L∈ℱr with detL>0 and GTG=Ir.Proof.
(1)⇔(2).
By using Theorem2.3 and Lemma 3.7 the proof runs as follows. 							(3.10)Aisk-Kernelsymmetric⟺KAisKernelsymmetric:⟺PKAPT=[E000]withdetE>0forsomepermutationmatrixP(ByTheorem(2.3))⟺A=KPT[E000]P⟺A=(KPTK)K[E000]P(ByP.1)⟺A=KPTK[K100K2][E000]P⟺A=KPTK[K1E000]P⟺A=KPTK[D000]P.
Thus A is k-similar to a diagonal block matrix [D000], where D=K1E and det D>0 .
However, (2)⇔(3):
(3.11)A=KPTK[K1E000]P=K[P1TP3TP2TP4T][K100K2][D000][P1P2P3P4]=K[P1TP2T]K1D[P1P2]=KGLGT,whereG=[P1TP2T],GT=[P1P2],L=K1D∈ℱrGTG=[P1P2][P1TP2T]=P1P1T+P2P2T=Ir,L∈ℱr.
Hence the Proof.Letx,y∈ℱ1×nȦ scalar product of x and y is defined by xyT=〈x,y〉. For any subset S∈ℱ1×n,S⊥={y:〈x,y〉=0,forallx∈S}.Remark 3.10.
In particular, whenκ(i)=i,K reduces to the identity matrix, then Theorem 3.9 reduces to Theorem 2.3. For a complex matrix A, it is well known that N(A)⊥=R(A*), where N(A)⊥ is the orthogonal complement of N(A). However, this fails for a fuzzy matrix hence Cn=N(A)⊕R(A) this decomposition fails for Kernel fuzzy matrix. Here we shall prove the partial inclusion relation in the following.Theorem 3.11.
ForA∈ℱn, if N(A)≠{0}, then R(AT)⊆N(A)⊥ and R(AT)≠ℱ1×n.Proof.
Letx≠0∈N(A), since x≠0, xio≠0 for atleast one io. Suppose xi≠0 (say) then under the max-min composition xA=0 implies, the ith row of A=0, therefore, the ith column of AT=0. If x∈R(AT), then there exists y∈ℱ1×n such that yAT=x. Since ith column of AT=0, it follows that, ith component of x=0, that is, xi=0 which is a contradiction. Hence x∉R(AT) and R(AT)≠ℱ1×n.
For anyz∈R(AT), z=yAT for some y∈ℱ1×n. For any x∈N(A), xA=0 and
(3.12)〈x,z〉=xzT=x(yAT)T=xAyT=0.
Therefore, z ∈N(A)⊥, R(AT)⊆N(A)⊥.Remark 3.12.
We observe that the converse of Theorem3.11 needs not be true. That is , if R(AT)≠ℱ1×n, then N(A)≠{0} and N(A)⊥⊆R(AT) need not be true. These are illustrated in the following Examples.Example 3.13.
Let(3.13)A=[000.60.5100.50.30]
since A has no zero columns, N(A)={0}.
For thisA,R(AT)={(x,y,z):0≤x≤0.6,0≤y≤1,0≤z≤0.5}.
Therefore,R(AT)≠ℱ1×3.Example 3.14.
Let(3.14)A=[110010000].
For this A,
(3.15)N(A)={(0,0,z):z∈ℱ},N(A)⊥={(x,y,0):x,y∈ℱ},
Here, R(AT)={(x,y,0):0≤y≤x≤1}≠ℱ1×3.
Therefore, forx>y∈ℱ, (x,y,0)∈N(A)⊥ but (x,y,0)∉R(AT).
Therefore,N(A)⊥ is not contained in R(AT).



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*Source: 926217-2009-09-08.xml*