International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
PROPERTIES OF FUZZY INNER PRODUCT SPACES Asit Dey1 and Madhumangal Pal2 1
Department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Midnapore-721102, India 2 Department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Midnapore-721102, India
ABSTRACT In this paper, natural inner product structure for the space of fuzzy n−tuples is introduced. Also we have introduced the ortho vector, stochastic fuzzy vectors, ortho- stochastic fuzzy vectors, ortho-stochastic fuzzy matrices and the concept of orthogonal complement of fuzzy vector subspace of a fuzzy vector space.
KEYWORDS Ortho vector, Stochastic vector, Ortho-Stochastic vector, Orthogonal Complement, Ortho-stochastic matrix, Reflection.
1. INTRODUCTION A fuzzy vector space Vn is the set of all n − tuples ( x1 , x2 , , xn ) of a fuzzy algebra ℑ = [0,1] . The elements of Vn possess a natural linear space-like structure, where the elements are called fuzzy vectors. Also, we define an operation on Vn which is analogous to an inner product and then we can define a norm and orthogonality relation on Vn . The concept of fuzzy inner product has been introduced by several authors like Ahmed and Hamouly [1], Kohli and Kumar [5], Biswas [4], etc. Also the notion of fuzzy norm on a linear space was introduced by Katsaras [9]. Later on many other mathematicians like Felbin [3], Cheng and Mordeso [10], Bag and Samanta [6], etc, have given different definitions of fuzzy normed spaces. In recent past lots of work have been done in the topic of fuzzy functional analysis, but only a few works have been done on fuzzy inner product spaces. Biswas [4] tried to give a meaningful definition of fuzzy inner product space and associated fuzzy norm functions. Later on, a modification of the definition given by Biswas [4] was done by Kohli and Kumar [5] and they also introduced the notion of fuzzy co-inner product space. But still there is no useful definition of fuzzy inner product available to work with in the sense that not much development has yet been found in fuzzy inner product (or fuzzy Hilbert) space with these definitions. The concept of ortho-vector, stochastic vector, stochastic subspace, isometry, reflection in Boolean inner product space was introduced by Gudder and Latremoliere [2]. In this paper, we modify these concepts on a fuzzy inner product spaces. We have also developed the concept of isometry, isomorphism for fuzzy inner product space. Also, some basic properties and results DOI : 10.5121/ijfls.2014.4203
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International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
such as orthogonality, the orthogonal complement and the ortho-stochastic subspace are given and established.
2. PRELIMINARIES In this article, for any a ∈ ℑ , we denote a c its complement. For any two elements a, b ∈ ℑ , we denote the infimum of a and b by a ∧ b or by ab and the supremum of a and b by a ∨ b or by a + b . If a = ( a1 , a2 , , an ) and b = (b1 , b2 , , bn ) are in Vn and c ∈ ℑ ; then
a + b = (a1 ∨ b1 , a2 ∨ b2 , , an ∨ bn ) and c a = (c ∧ a1 , c ∧ a2 , , c ∧ an ) . Throughout this paper we use the following definitions. Definition 1 (Ortho-vector) A fuzzy vector x = ( x1 , x2 , xn ) is said to be an ortho-vector if
xi x j = 0 for i , j ∈ 1,2,3, , n and i ≠ j . This implies that at most one coordinate of x is non-zero. Example 1 The vector x = (0.5,0,0,,0) is an ortho-vector. The null vector 0 = (0,0,0,,0) is an ortho vector. Definition 2 (Stochastic-vector) A fuzzy vector x = ( x1 , x2 , , xn ) is said to be a stochasticn
vector if
∨x
i
=1.
i =1
Example 2 x = (0.5,0.1,0.9,0.5,0.2,1) is a stochastic vector in V6 becau
max{0.5,0.1,0 .9,0.5,0.2 ,1} = 1 . Definition 3 (Inner Product) Let x = ( x1 , x2 ,....., xn ) , and y = ( y1 , y 2 , , y n ) in Vn . Then the inner product of x and y is denoted by 〈 x, y 〉 and is defined by 〈 x, y〉 =
n
∨ (x ∧ y ) . i
i
i =1
Definition 4 (Norm) Let x = ( x1 , x2 , , xn ) in Vn . Then the norm of x is denoted by x and is defined by x = 〈 x, x〉 .The vector x is said to be a unit vector if x = 1 . If we replace the scalar sums and products by the supremum and infimum in ℑ , then the usual properties of the Euclidean inner product are satisfied by fuzzy inner product. Let x , y , z ∈ Vn and ∈ ℑ ; then (i) 〈 x + y , z 〉 = 〈 x, z 〉 + 〈 y , z 〉 , (ii) 〈 x, y 〉 = 〈 y , x〉 , (iii) 〈 x, y 〉 = 〈 x, y 〉 = 〈 x, y 〉 , (iv) 〈 x, x〉 = 0 if and only if x = (0,0,0,,0) = 0 .
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International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Definition 5 (Orthogonal Vector) Two fuzzy vectors x and y in Vn are said to be orthogonal if
〈 x, y 〉 = 0 . In this case, we shall write x ⊥ y .
Example 3 Let x = (0,0.1,0,0.2,0.5) and x = (0.4,0,0.3,0,0) in V5 . Here 〈 x, y 〉 = 0 . Thus, x⊥ y. Now, we introduced a definition of orthogonal and orthonormal set of fuzzy vectors. Definition 6 (Orthogonal and Orthonormal Set) Let E ⊆ Vn . Then E is said to be an orthogonal set if 〈 x, y 〉 = 0 for all x , y ∈ E with x ≠ y . Also, E ⊆ Vn is said to be an orthonormal set if E is orthogonal i.e, 〈 x, y 〉 = 0 for all
x , y ∈ E with x ≠ y and x = 1 for all x ∈ E . Theorem 1 Let x , y ,∈ Vn and c ∈ ℑ . Then (i) cx = c x . (ii) x + y = x ∨ y . (iii) 〈 x, y 〉 ≤ x ∧ y . Proof. (i) cx = 〈cx, cx〉 = c〈 x, cx〉 = c〈cx, x〉 = (c ∧ c)〈 x, x〉 = c〈 x, x〉 = c x for all x ∈Vn . (ii) x + y = (iii) 〈 x, y 〉 =
n
n
n
i =1 n
i =1 n
i =1
∨( xi ∨ yi ) = (∨xi ) ∨ (∨ yi ) = x ∨ y . ∨ ( xi ∧ yi ) ≤ i =1
n
n
i =1
j =1
∨ ( xi ∧ y j ) = (∨xi ) ∧ (∨ y j ) = x ∧ y .
i , j =1
3. PROPERTIES OF FUZZY ORTHO-STOCHASTIC VECTORS Based on these definitions some results are established. Now we discuss the conditions for which two fuzzy vectors to be ortho-stochastic. Theorem 2 If x and y are two stochastic vector in Vn , then 〈 x, y 〉 = 1 if x = y , but the converse need not be true. Proof. Let x = y . Thus, xi = yi for i ∈ {1,2,..., n} . Since x and y are stochastic, so, xi = 1 and yi = 1 for some i ∈ {1,2, , n} . Thus, n
〈 x, y〉 = ∨ ( xi ∧ yi ) = 1 . i =1
But, if we take x = (0.5,0.2,1) and y = (0.3,0.5,1 ) , then x and y both are stochastic vector with 〈 x, y 〉 = 1 and x ≠ y . Thus, 〈 x, y 〉 = 1 need not implies x = y . 25
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Corollary 1 If x and y are two ortho-stochastic vector in Vn , then 〈 x, y 〉 = 1 if and only if
x = y. Proof. Let x , y be two ortho-stochastic vectors in Vn such that 〈 x, y 〉 = 1 . Taking x = ( x1 , x2 , , xn ) and y = ( y1 , y 2 , , y n ) in Vn . Since x, y are two ortho- vectors, thus, at most one coordinate of x, y is non-zero. Now, 〈 x, y 〉 = 1 n
⇒ ∨ ( xi ∧ yi ) = 1 i =1
⇒ xi ∧ yi = 1 for some i ∈ {1,2, , n} . ⇒ xi = 1, yi = 1 for some i ∈ {1,2, , n} . ⇒ x = y.
4. LINEAR COMBINATION OF VECTORS AND BASIS FOR FUZZY INNER PRODUCT SPACES In this part of this paper, we introduced the concept of dimension to fuzzy inner product spaces. It is based on the notion of basis. Definition 7. (Linear combination of vectors) Let E be a subset of Vn . A vector x ∈Vn is said to be linear combination of vectors of E if there exist a finite subset {x1 , x 2 , , x m } of E and
y1 , y2 , , ym ∈ ℑ such that x = y1 x1 + y2 x 2 + + ym x m . Let E ⊆ Vn . The set of all linear combination of vectors of E is called the linear span of E and is denoted by L (E ) . A subset E of Vn is said to be a generating subset of Vn if all vectors of Vn are linear combination of vectors of E . A subset E of Vn is said to be free if with
i = 1, 2, , m ; j = 1, 2, , k ,
m
k
i =1
j =1
∑bi a i = ∑d j c j for any bi , d j ∈ ℑ − {0} and a, c ∈ E then,
m = k , {b1 , b2 , , bm } = {d1 , d 2 , , d k }
and
{a1 , a 2 , , a m } = {c1 , c 2 , , c k } . Thus, a set E is free when a linear combination of elements in E has unique non-zero coefficients and associated vectors of E . Definition 8. (Basis) A subset E of Vn is said to be a basis of Vn if E is generating and free i.e. every element of Vn can be written as a unique linear combination of elements of E with non-zero coefficient. 26
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Now we observe that a basis must be made of unit vectors. Theorem 3. Let E be a basis of Vn . If x ∈ E then x = 1 Proof. If possible let, 0 = (0,0,0,,0) ∈ E . Then 1 = (1,1,1, ,1) ∈ Vn and 1 = 11 = 1 + 0 . Thus, 1 can be written as two distinct linear combination of elements of E . This is a contradiction because E is free. Thus, 0 ∉ E .
x ≠ 1 . Then x = 1x and x = x x because
If possible let, x ∈ E with
n
x = ∨xi , i =1
n
n
n
n
i =1
i =1
i =1
i =1
x x = (∨xi ) x = ((∨xi ) ∧ x1 , (∨xi ) ∧ x2 ,, (∨xi ) ∧ xn ) = ( x1 , x2 ,, xn ) = x . In this case, x can be written as two distinct linear combinations of elements of E with non-zero coefficient. This is a contradiction, because E is a basis. So, our assumption x ∈ E with x ≠ 1 is wrong. Hence, x = 1 . The second observation is: Theorem 4. Let E be an orthonormal set in Vn . Then E is free. Proof. Let x ∈Vn with x =
m
k
i =1
i =1
∑bi a i = ∑d i c i where a1 a 2 ,, a m ; c1 , c 2 ,, c k ∈ E and
b1 , b2 , , bm ; d1 , d 2 , , d k ∈ ℑ − {0} . Now, d i = 〈 x, c i 〉 for i = 1,2, , k . If c j ∉ {a1 , a 2 , , a m } for some j ∈ {1,2, , k } , then, d j = 〈 c j , x〉 = 〈 c j ,
m
∑b a 〉 = 0 , which i
i
i =1
is a contradiction because d j ∈ ℑ − {0} . Thus, c j ∈ {a1 , a 2 , , a m } for j = 1,2, , k
⇒ {c1 , c 2 , , c k } ⊆ {a1 , a 2 , , a m } Also, bi = 〈 x, a i 〉 for i = 1,2, , m . If a j ∉ {c1 , c 2 , , c k } for some j ∉ {1,2, , k } , k
then
a j = 〈 a j , x〉 = d j = 〈 c j , x〉 = 〈 a j , ∑d i c i 〉 = 0 , which is a contradiction because i =1
a j ∉ ℑ − {0}
.
Thus, a j ∈ {c1 , c 2 , , c k }
⇒ {a1 , a 2 , , a m } ⊆ {c1 , c 2 , , c k } . Therefore, {a1 , a 2 , , a m } = {c1 , c 2 , , c k } Thus, m = k . 27
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Now, bi = 〈 x, a i 〉 = 〈 x, c j 〉 = d j for all i = 1,2, , m and j ∈ {1,2, , k } . Hence, E is free. Corollary 2 A subset E of Vn is an orthonormal basis of Vn if it is orthonormal generating subset of Vn . We illustrate this result by an example: Example 4 Let 1 = (1.0,0.5,0,0) ∈V4 , 2 = (0,0,0.5,1.0) ∈V4 and L{1 , 2 } be the linear span of the vectors 1 , 2 .
Here 1 = 1 and 2 = 1 with 〈1 , 2 〉 = 0 . Thus, the set E = {1 , 2 } is an orthogonal generating set. Hence, E = {1 , 2 } is an orthonormal basis for L{1 , 2 } .
4.1. CONSTRUCTION OF ORTHONORMAL BASIS FROM ORTHOSTOCHASTIC VECTOR Let x = ( x1 , x2 , , xn ) ∈ Vn be an ortho-stochastic vector. The canonical basis or standard basis of Vn is defined as the basis {e1 , e 2 , e n } with
e1 = (1,0, ,0), e 2 = (0,1, ,0), , e n = (0,0, ,1) . Let us construct i = ( xi , xi +1 , , xn , x1 , x2 , , xi −1 ) for all i ∈ {1,2, , n} . Then i = 〈 i , i 〉 =
n
n
i =1
i =1
∨( xi xi ) = ∨xi = 1 , because x is stochastic.
Also for i ≠ j , we have 〈 i , j 〉 =
n
∨ xx i
j
= 0 because x is ortho-vector.
i , j =1
Therefore { 1 , 2 , , n } is an orthonormal subset of Vn . Again,
e1
=
x1 1 + x2 2 + + x1 n
e 2 = x2 1 + x3 2 + + x1 n e n = xn 1 + x1 2 + + xn −1 n Therefore, L{e1 , e 2 , , e n } ⊆ L{ 1 , 2 , , n } ⇒ Vn ⊆ L{ 1 , 2 , , n } . Therefore, Vn = L{ 1 , 2 , , n } . Thus, { 1 , 2 , , n } is an orthonormal generating subset of Vn . Hence, { 1 , 2 , , n } is an orthonormal basis of Vn . Theorem 5 Let a1 , a 2 , , a m ∈ Vn and W = L{a1 , a 2 , , a m } . If some a i is a linear combination of a1 , a 2 , , a i −1 , a i +1 , , a m ; then W = L{a1 , a 2 , , a i −1 , a i +1 , , a m } Proof. Let E = {a1 , a 2 , , a i −1 , a i +1 , , a m } . 28
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
c1 , c2 , , ci −1 , ci +1 , , cm ∈ ℑ such that a i = c1 a1 + c2 a 2 + + ci −1 a i −1 + ci +1 a i +1 + + cm a m . Since {a1 , a 2 , , a i −1 , a i +1 , , a m } ⊆ E , so, L{a1 , a 2 , , a i −1 , a i +1 , , a m } ⊆ L( E ) = W . Let
there
exists
Let x ∈ W . Then there exists d1 , d 2 , , d m ∈ ℑ such that
x = d1 a1 + d 2 a 2 + + d i a i + + d m a m . = d1 a1 + d 2 a 2 + + d i (c1 a1 + c2 a 2 + + ci −1 a i −1 + ci +1 a i +1 + + cm a m ) + + d m a m m
=
∑ (c d
j =1, j ≠ i
j
i
+ d j )a j ∈ L{a1 , a 2 ,, a i −1 , a i +1 ,, a m }
Thus, W ⊆ L{a1 , a 2 , , a i −1 , a i +1 , , a m } Hence, W = L{a1 , a 2 , , a m } = L{a1 , a 2 , , a i −1 , a i +1 , , a m } .
5. PROPERTIES OF LINEAR TRANSFORMATION AND ISOMETRY In this part of the paper, we introduced the concepts and properties of linear transformation, subspaces generated by orthonormal set, ortho-stochastic subspace and isometry. The special interest to us will be isometry and its properties. Definition 9 (Linear transformation) Let Vm and Vn be two fuzzy vector spaces over the fuzzy algebra ℑ . A mapping T : Vn Vm is called a linear transformation when for all a ∈ ℑ ,
b, c ∈ Vn we have T (ab + c) = aT (b) + T (c) . Here T (0) = 0 because
T (0) = T (00) = 0T (0) = 0 . We call T an operator on Vn when T is linear from Vn to Vn . An operator T on Vn is said to be invertible if there exist an operator S such that S T = T S = I , where I : Vn Vn is the linear operator I ( x) = x for all x ∈Vn . Definition 10 (Subspace) A subset U ⊆ Vn is said to be subspace of Vn if U is generated by an m
∑y x
orthonormal set A = {x1 , x 2 , , x m } , i.e., U = L ( A) = {
i
i
: y1 , y 2 , , y m ∈ ℑ} .
i =1
A subspace may-not contain any ortho-stochastic orthonormal basis, for example if a ∈ ℑ with a ≠ 0 , then U = {b (1, a ) : b ∈ ℑ} is a subspace of V2 with basis (1, a ) . Here U does not contain any ortho-stochastic vector. Definition 11 (Ortho-stochastic subspace) A subset U ⊆ Vn is said to be ortho-stochastic subspace of Vn if it has an ortho-stochastic orthonormal basis. 29
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Example 5 Let us consider the fuzzy vector space V3 . Let
U = {x = ( x1 , x2 , x3 ) ∈ V3 : x3 = 0, x1 , x2 ∈ ℑ} . Then U is a subspace of V3 with orthonormal basis A = {(1,0,0), (0,1,0)} , which is ortho-stochastic. Thus, U is a ortho-stochastic subspace of V3 . Definition 12 (Isometry) A linear map T : U V between two subspaces U and V of Vn and
Vm respectively, is called an isometry when for all x, y ∈ U , we have 〈T ( x ), T ( y )〉 = 〈 x, y 〉 . We are now ready to show the first observation, which is the core observation for isometry. Theorem 6 Let T : U V be an isometry, where U ⊆ Vn and V ⊆ Vm . Then T is injective. Proof. Let T : U V be an isometry, where U ⊆ Vn and V ⊆ Vm . Also, let x, y ∈ U with T ( x ) = T ( y ) and A = { 1 , 2 , , k } be an orthonormal basis of U . Now, xi = 〈 x, i 〉
= 〈T ( x), T ( i )〉 , because T is an isometry. = 〈T ( y ), T ( i )〉 , because T ( x ) = T ( y ) . = 〈 y , i 〉 because T is an isometry.
= yi for i = 1,2, , k . Therefore, x = y . Hence, T is injective. The main result for this part is stated below. Theorem 7 Let U ⊆ Vn , V ⊆ Vm be two subspaces and T : U V be a linear map. Then the followings are equivalent. (i) For any orthonormal set A = { 1 , 2 , , k } of U , the set {T ( 1 ), T ( 2 ), , T ( k )} is an orthonormal set of V . (ii) There exists an
A = { 1 , 2 , , k } of U such that {T ( 1 ), T ( 2 ), , T ( k )} is an orthonormal set of V . (iii) T is an isometry. Proof. (i) ⇒ (ii) Since A = { 1 , 2 , , k } is the orthonormal basis of U , so, A is an orthonormal set of U . Thus, by (i) {T ( 1 ), T ( 2 ), , T ( k )} is an orthonormal set of V . (ii) ⇒ (iii) Let A = { 1 , 2 , , k } be a basis of U such that {T ( 1 ), T ( 2 ), , T ( k )} is an orthonormal set of V . orthonormal
basis
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International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Let x, y ∈ U . Then x = Now, 〈T ( x ), T ( y )〉 = 〈
k
k
∑xi i , y = ∑ yi i where xi , yi ∈ ℑ for i = 1,2,...., k .
i =1 k
i =1
k
∑x T ( ), ∑y T ( i
i =1
i
j
j
j =1
)〉 =
k
∨xy i
j
〈T ( i ), T ( j )〉
i , j =1
k
= ∨ xi yi 〈T ( i ), T ( i )〉 , because T ( 2 ),, T ( k )} is an orthonormal set of i =1
V. k
= ∨ xi yi i =1
= 〈 x, y 〉 . Thus, T is an isometry. (iii) ⇒ (i) Let T be an isometry. Then 〈T ( x ), T ( y )〉 = 〈 x, y 〉 for all x, y ∈ U .
1 if 0 if
Now, 〈T ( i ), T ( j )〉 = 〈 i , j 〉 =
i= j i≠ j
Hence, {T ( 1 ), T ( 2 ), , T ( k )} is an orthonormal set.
5.1 ISOMORPHISM AND ISOMORPHIC SUBSPACES In the following, we study further into the isomorphism and isomorphic subspaces in order to obtain the results that an isomorphism maps orthonormal bases to orthonormal bases and any two orthonormal bases of a subspace in a fuzzy vector space have the same cardinality. Definition 13 (Isomorphism) Let U ⊆ Vn , V ⊆ Vm be two subspaces and T : U V be a linear map. Then T is said to be an isomorphism if T is a surjective isometry. In this case, U and V are called isomorphic sub spaces. It is clear that the composition of two isomorphism is also an isomorphism and the inverse of an isomorphism is an isomorphism. Theorem 8 Let U ⊆ Vn , V ⊆ Vm be two subspaces and T : U V be an isomorphism. Then T maps orthonormal bases to orthonormal bases. Proof. Since T : U V be an isomorphism. Thus, T is a surjective isometry. Let {x1 , x 2 , , x t } be an orthonormal basis of U . Then {T ( x1 ), T ( x 2 ), , T ( x t )} ⊆ V . Thus, L{T ( x1 ), T ( x 2 ), , T ( x t )} ⊆ V . Let y be an arbitrary element of V . Since T is surjective, so, there exists x ∈U such that T ( x ) = y . 31
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Since x ∈U and {x1 , x 2 , , x t } is an orthonormal basis of U . Thus, there exists c1 , c2 , , ct ∈ ℑ such that x =
t
∑c x . i
i
i =1
Now, y = T ( x ) = T (
t
t
i =1
i =1
∑ci x i ) = ∑ciT ( x i ) ∈ L{T ( x1 ), T ( x 2 ), , T ( x t )}
Therefore, V = {T ( x1 ), T ( x 2 ), , T ( x t )} . Thus, {T ( x1 ), T ( x 2 ), , T ( x t )} is a generating subset of V . Since T is an isometry and {x1 , x 2 , , x t } is an orthonormal basis. Thus, {T ( x1 ), T ( x 2 ), , T ( x t )} is an orthonormal generating subset of V . Hence, {T ( x1 ), T ( x 2 ), , T ( x t )} is an orthonormal basis of V . Theorem 9 Any two orthonormal bases of a subspace U of Vn have the same cardinality. Proof. Let W1 = {x1 , x 2 , , x s } and W2 = { y1 , y 2 ,, y t } be two orthonormal bases of U . Then there exist isomorphism T1 : U Vs and T2 : U Vt . Thus, T1 T2−1 : Vt Vs is an isomorphism. Therefore, Vt and Vs are two isomorphic subspaces. Hence, t = s .
6. ORTHOGONAL COMPLEMENT OF FUZZY VECTORS AND ITS PROPERTIES Definition 14 (Orthogonal complement) Let Vn be a fuzzy vector space and x ∈Vn . Then the ⊥
⊥
orthogonal complement of x is denoted by x and is defined by x = { y ∈ Vn : 〈 x, y 〉 = 0} . Let U be a subspace of Vn . Then the orthogonal complement of U is denoted by U ⊥ and is ⊥ defined by U = { y ∈Vn : 〈 x, y〉 = 0 for all x ∈ U } .
Example 6 Let us consider the fuzzy vector space V2 and x = (0.5,0) . Then ⊥
x = { y = ( y1 , y2 ) ∈ V2 : 〈 x, y〉 = 0} . Now, 〈 x, y 〉 = 0 2
⇒ ∨ xi yi = 0 i =1
⇒ max{0.5 y1 ,0 y2 } = 0 ⇒ 0.5 y1 = 0 and y2 is arbitrary. ⇒ y1 = 0 and y2 is arbitrary. ⊥
Therefore, x = { y = (0, y2 ) ∈ V2 : y2 ∈ ℑ}
. 32
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Example 7 Let us consider V3 and U = {c x : x = (1,0,0),c ∈ ℑ} . ⊥ Then U = { y = ( y1 , y2 , y3 ) ∈V3 : 〈c x, y〉 = 0} .
Now, 〈 c x, y 〉 = 0
⇒ 〈 x, y 〉 = 0 3
⇒ ∨ xi yi = 0 i =1
⇒ max{1 y1 ,0 y2 ,0 y3 } = 0
⇒ 1y1 = 0 and y2 , y3 ∈ ℑ are arbitrary. ⇒ y1 = 0 and y2 , y3 are arbitrary. ⊥ Therefore, U = { y = (0, y2 , y3 ) ∈V3 : y2 , y3 ∈ ℑ} .
Theorem 10 Let M ⊆ Vn and M ⊥ be the orthogonal complements of M . Then the following holds. ⊥⊥ (i) M ⊥ is a subspace of Vn , M ⊆ M and M ∩ M ⊥ = {0} .
(ii) Let N ⊆ Vn with M ⊆ N . Then N ⊥ ⊆ M ⊥ . (iii) {0}⊥ = Vn and Vn⊥ = {0} and M ⊥ = M ⊥⊥⊥ . (iv) Let M , N ⊆ Vn . Then ( M + N ) ⊥ = M ⊥ ∩ N ⊥ , where M + N = { x + y : x ∈ M , y ∈ N } . Proof. (i) Let a, b ∈ M ⊥ and , ∈ ℑ . Then for any c ∈ M , we have
〈 a + b, c〉 = 〈 a, c〉 ∨ 〈b, c〉 = 0 ∨ 0 = 0 . This shows that a + b ∈ M ⊥ . Thus, M ⊥ is a subspace of Vn . Again, a ∈ M . Then 〈 a, b〉 = 0 for all b ∈ M ⊥ .
⇒ a ∈ ( M ⊥ ) ⊥ = M ⊥⊥ . ⊥⊥ Therefore, M ⊆ M
Again, a ∈ M ∩ M ⊥
⇒ a ∈ M and a ∈ M ⊥ ⇒ 〈 a, a〉 = 0 ⇒a=0. Hence M ∩ M ⊥ = {0} . (ii) Let a ∈ N ⊥ 33
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
⇒ 〈 a, b〉 = 0 for all b∈ N ⇒ 〈 a, b〉 = 0 for all b ∈ M ⊆ N ⇒ a∈M ⊥. ⊥ ⊥ Therefore, N ⊆ M . (iii) {0}⊥ = {a ∈ Vn : 〈 0, a 〉 = 0} = Vn because 〈0, a〉 = 0 for all a ∈Vn . Also, if a ≠ 0 then 〈 a, a〉 ≠ 0 . In other words, a non-zero element of Vn cannot be orthogonal to the entire space Vn . Hence, Vn⊥ = {0} . Let a ∈ M . Then 〈 a, b〉 = 0 for all b ∈ M ⊥ .
⇒ a ∈ ( M ⊥ ) ⊥ = M ⊥⊥ . ⊥⊥ Thus, M ⊆ M
Changing M by M ⊥ ,we get
M ⊥ ⊆ (( M ⊥ ) ⊥ ) ⊥ = M ⊥⊥⊥ ⊥⊥ Again M ⊆ M
⇒ ( M ⊥⊥ ) ⊥ ⊆ M ⊥ , by (ii).
⇒ M ⊥⊥⊥ ⊆ M ⊥ . Hence M ⊥ = M ⊥⊥⊥ . (iv) Let a ∈ ( M + N ) ⊥ . Then 〈 a, b〉 = 0 for all b ∈ M + N . Let b = b1 + b2 , where b1 ∈ M and b2 ∈ N . Now, 〈 a, b〉 = 0
⇒ 〈 a, b1 + b2 〉 = 0 ⇒ 〈 a, b1 〉 ∨ 〈 a, b2 〉 = 0 ⇒ 〈 a, b1 〉 = 0 and 〈 a, b2 〉 = 0 for all b1 ∈ M and for all b2 ∈ N ⇒ a ∈ M ⊥ and a ∈ N ⊥ ⇒ a∈M ⊥ ∩ N⊥ Therefore, ( M + N ) ⊥ ⊆ M ⊥ ∩ N ⊥ . Conversely, let a ∈ M ⊥ ∩ N ⊥
⇒ a ∈ M ⊥ and a ∈ N ⊥ ⇒ 〈 a, b〉 = 0 for all b ∈ M and 〈 a, c〉 = 0 for all c ∈ N Let d ∈ M + N ⇒ d = b + c , where b ∈ M and c ∈ N . 34
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Now, 〈 a, d 〉
= 〈 a, b + c〉 = 〈 a, b〉 ∨ 〈 a, c〉 = 0 ∨ 0 = 0 for all d ∈ M + N . This shows that a ∈ ( M + N ) ⊥ . Therefore, M ⊥ ∩ N ⊥ ⊆ ( M + N ) ⊥ . Hence ( M + N ) ⊥ = M ⊥ ∩ N ⊥ .
7. ORTHO-STOCHASTIC MATRIX AND REFLECTION Definition 15 Let A = [ aij ]n×n be a matrix on Vn . Then the transpose of A is denoted by A∗ and ∗ is defined by A = [a ji ]n×n .
Definition 16 (Ortho-stochastic matrix) Let A = [ aij ]n×n be a matrix on Vn . Then the matrix A is said to be ortho-stochastic matrix if A∗ A ≥ I and AA∗ ≤ I . A matrix A is orthogonal if A∗ A = AA∗ = I . i.e., A and A∗ both are ortho-stochastic matrix. Lemma 1 A matrix A is symmetric ortho-stochastic if and only if A is an orthogonal matrix of order 2 . Proof. Let A be symmetric and ortho-stochastic. Then, A∗ A ≥ I , AA∗ ≤ I and A = A∗ ⇒ AA ≥ I and AA ≤ I ⇒ A 2 ≥ I and A 2 ≤ I ⇒ A2 = I . Thus, AA∗ = A2 = I = A∗ A , i.e., A is an orthogonal matrix of order 2 . Conversely, let A be an orthogonal matrix of order 2 . Then AA∗ = A∗ A = I and A2 = AA = AA = I . This shows that A is an invertible matrix with A−1 = A∗ = A . Hence, A is symmetric and ortho-stochastic. Definition 17 (Reflection) A matrix A is said to be a reflection if it is symmetric and orthostochastic, i.e, an orthogonal matrix of order 2 . Important results: The sum and product of reflections need not be a reflection. As for example,
1 0 0 0 1 0 let A = 0 0 1 and B = 1 0 0 . Then both A, B are reflection but the sum 0 1 0 0 0 1 35
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
1 1 0 1 1 1 2 A + B = 1 0 1 is not a reflection, because ( A + B) = 1 1 1 ≠ I . 0 1 1 1 1 1 0 1 0 0 0 1 ∗ Also, AB = 0 0 1 is not a reflection, because ( AB) = 1 0 0 ≠ AB . 1 0 0 0 1 0 m Definition 18 (Joint trace) Let A1 , A2 , , Ak be matrices on Vn with Am = [aij ]n×n for
m = 1,2, , k . Then the joint trace of A1 , A2 , , Ak is denoted by tr ( A1 , A2 , , Ak ) and is n
defined by tr ( A1 , A2 , , Ak ) =
∨ a a .....a 1 2 ii ii
k ii
. In particular, the trace of A = [ aij ]n×n is given by
i =1
n
tr ( A) = ∨ aii . i =1
A vector b ∈Vn is said to be an invariant vector for A if Ab = b . Also, the vector b ∈Vn is said to be common invariant vector of A1 , A2 , , Ak if Ai b = b for i = 1,2, , k . Lemma 2 Let A, B be two matrices on Vn . Then (i) tr ( AB ) = tr ( BA ) . (ii) tr ( BAB ∗ ) = tr ( A) if B is an orthogonal matrix. Proof. Let A = [ aij ]n×n and B = [bij ]n×n . n
Now, tr ( AB) =
n
∨ (∨ aik bki ) i =1 i =1
n
n
= ∨ ∨ aik bki i =1 i =1 n n
= ∨ ∨ bki aik = tr ( BA) . i =1 i =1
Also, tr ( BAB ∗ ) = tr ( B( AB ∗ ))
= tr (( AB ∗ ) B) because tr ( AB ) = tr ( BA) = tr ( AB ∗ B) = tr ( A( BB ∗ )) = tr ( AI ) , because B is an orthogonal matrix, so, BB ∗ = B ∗ B = I . = tr ( A) . 36
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Theorem 11 An ortho-stochastic matrix A has an invariant ortho-stochastic vector if and only if tr ( A) = 1 . Proof. Let A = [ aij ]n×n be a matrix on Vn . Thus, aij ∈ ℑ = [0,1] for i, j = 1,2, , n . Therefore, aij ≤ 1 for i, j = 1,2, , n .
⇒ aii ≤ 1 for i = 1,2, , n . n
⇒ ∨ aii ≤ 1 i =1
⇒ tr ( A) ≤ 1 . Let b = (b1 , b2 , , bn ) ∈ Vn be an ortho-stochastic invariant vector of A Then, Ab = b and
bi b j = 0 for i ≠ j ; i, j ∈ {1,2, , n} and
n
∨b
i
= 1.
i =1
Now, Ab = b n
⇒ ∨ aij b j = bi j =1
Multiplying both sides by bi and since bi b j = 0 for i ≠ j . Thus, aii bi = bi
⇒ aii ≥ bi n
n
i =1
i =1
⇒ ∨ aii ≥ ∨ bi = 1 ⇒ tr ( A) ≥ 1 . Therefore, tr ( A) = 1 . Conversely, let tr ( A) = 1 . n
Thus,
∨a
ii
=1.
i =1
Therefore, there exist a ortho-stochastic vector b = (b1 , b2 , , bn ) ∈ Vn such that b j ≤ aij . Since A is ortho-stochastic, so, aij a jj = 0 for i ≠ j . Now, b j ≤ a jj
⇒ aij b j ≤ aij a jj ⇒ aij b j = 0 for i ≠ j . n
Also, ( Ab) i =
∨a b ij
j
= aii bi = bi .
j =1
Thus, Ab = b . Hence, b is an invariant ortho-stochastic vector for A . We record the following observation as well: 37
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014
Corollary 3 If A is ortho-stochastic matrix and B is orthogonal matrix on Vn . Then A has an invariant ortho-stochastic vector if and only if BAB ∗ has an invariant ortho-stochastic vector. Proof. A has an invariant ortho-stochastic vector
⇔ tr ( A) = 1
⇔ tr ( BAB ∗ ) = 1 , by lemma 2 . ⇔ BAB ∗ has an invariant ortho-stochastic vector.
8. CONCLUSIONS We can use the concept of ortho-stochastic vector in Operator Theory. Also, the concepts of orthogonal complement of fuzzy vector can use to calculate the spectrum of an operator and to prove spectral theorem for compact self-adjoint operators. The results for isometry and isomorphic subspaces can use to developed the concept of partial isometry, orthogonal projection and square root of an non-negative operator. The concept presented in this paper are not limited to a specific application Thus, the results of our paper are paving the way to numerous possibilities for future research.
ACKNOWLEDGEMENTS Financial support offered by Council of Scientific and Industrial Research, New Delhi, India (Sanction no. 09/599(0054)/2013-EMR-I) is thankfully acknowledged. Also, the authors are very grateful and would like to express their sincere thanks to the anonymous referees and Editor-inChief Robert Burduk for their valuable comments.
REFERENCES [1] A.M.El-Ahmed and H.M.El-Hamouly, Fuzzy inner-product spaces, Fuzzy Sets and Systems, 44, 309326, 1991. [2] S.Gudder and F. Latremoliere, Boolean inner-product spaces and Boolean matrices, Linear Algebra Appl., 431, 274-296, 2009. [3] C.Felbin, Finite dimensional fuzzy normed linear space, Fuzzy Sets and Systems, 48, 239-248, 1992 [4] R.Biswas, Fuzzy inner product spaces and fuzzy norm functions, Information Sciences, 53, 185-190, 1991. [5] J.K.Kohli and Rajesh Kumar, On fuzzy inner-product spaces, Fuzzy Sets and Systems, 53, 227-232, 1993. [6] T.Bag and S.K.Samanta, Finite dimensional fuzzy normed linear space, J. Fuzzy Math., 11, 687-706, 2003. [7] T.Bag and S.K.Samanta, Finite bounded linear operator, Fuzzy Sets and Systems, 15, 513-547, 2005. [8] J.Z.Xiao and X.H.Zhu, Fuzzy normed spaces of operators and its completeness, Fuzzy Sets and Systems, 133, 135-146, 2003. [9] A.K.Katsaras, Fuzzy topological vector space-II, Fuzzy Sets and Systems, 12, 143-154, 1984. [10] S.C.Cheng and J.N.Mordeson, Fuzzy linear operators and fuzzy normed linear spaces, Bull. Cal. Math. Soc., 86, 429-436, 1994. [11] M. Goudarzi, S. M. Vaezpour and R. Saadati, On the intuitionistic fuzzy inner-product spaces, Chaos, Solitons and Fractals, 41, 1105-1112, 2009. [12] S.Nanda, Fuzzy fields and fuzzy linear spaces, Fuzzy Sets and Systems, 19, 89-94, 1986. 38
International Journal of Fuzzy Logic Systems (IJFLS) Vol.4, No.2, April 2014 [13] M.Yoeli, A note on a generalization of Boolean matrix theory, Amer. Math. Monthly, 68, 552-557, 1961. [14] S.Gudder, Quantum Markov chain, J. Math. Phys., 49(7), 2008.
About the Authors Asit Dey received his Bachelor of Science degree with honours in Mathematics in 2007 from Vidyasagar University, Midnapore, West Bengal, India and Master of Science degree in Mathematics in 2010 from IIT Kanpur, India. He joined the department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Midnapore, West Bengal, India as a full time Research Fellow since 2012. His research interest in fuzzy mathematics.
Dr. Madhumangal Pal is a Professor of Applied Mathematics, Vidyasagar University, India. He has received Gold and Silver medals from Vidyasagar University for rank first and second in M.Sc. and B.Sc. examinations respectively. Also he received, jointly with Prof. G.P.Bhattacherjee, “Computer Division Medal” from Institute of Engineers (India) in 1996 for best research work. He received Bharat Jyoti Award from International Friend Ship Society, New Delhi in 2012. Prof. Pal has successfully guided 16 research scholars for Ph.D. degrees and has published more than 140 articles in international and national journals, 31 articles in edited book and in conference proceedings.His specializations include Computational Graph Theory, Genetic Algorithms and Parallel Algorithms, Fuzzy Correlation & Regression, Fuzzy Game Theory, Fuzzy Matrices, Fuzzy Algebra.He is the Editor-in-Chief of “Journal of Physical Sciences” and “Annals of Pure and Applied Mathematics”, and member of the editorial Boards of several journals. Prof. Pal is the author of the eight books published from India and Oxford, UK.He organized several national seminars/ conferences/ workshop. Also, visited China, Malaysia, Thailand and Bangladesh to participated, delivered invited talks and chaired in national and international seminars/conferences/ refresher course, etc.
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