1、第 2 章模糊聚类分析2.1 模糊矩阵定义 1 设 R = (rij)mn, 若 0rij1, 则称 R为 模糊矩阵 . 当 rij只取 0或 1时,称 R为 布尔 (Boole)矩阵 . 当模糊方阵 R = (rij)nn的对角线上的元素 rii都为 1时,称 R为 模糊自反矩阵 .定义 2 设 A=(aij)mn,B=(bij)mn都 是模糊矩阵,相等 : A = B aij = bij;包含 : AB aijbij;并 : A B = (aij bij)mn;交 : AB = (aij bij)mn;余 : Ac = (1- aij)mn.模糊矩阵的并、交、余运算性质幂等律: A A =
2、 A, AA = A;交换律: A B = B A, AB = BA;结合律: (A B) C = A (B C),(AB)C = A(BC);吸收律: A (AB) = A, A(A B) = A; 分配律: (A B)C = (AC ) (BC);(AB) C = (A C )(B C);0-1律: A O = A, AO = O;A E = E, AE = A;还原律: (Ac)c = A;对偶律: (A B)c =AcBc, (AB)c =Ac Bc.模糊矩阵的合成运算与模糊方阵的幂设 A = (aik)ms, B = (bkj)sn, 定义模糊矩阵 A 与 B 的合成为:A B =
3、(cij)mn,其中 cij = (aik bkj) | 1ks .模糊方阵的幂定义:若 A为 n 阶方阵,定义 A2 = A A, A3 = A2 A, , Ak = Ak-1 A.合成 ( )运算的性质:性质 1: (A B) C = A (B C);性质 2: Ak Al = Ak + l, (Am)n = Amn;性质 3: A ( B C ) = ( A B ) ( A C );( B C ) A = ( B A ) ( C A );性质 4: O A = A O = O, I A=A I =A;性质 5: AB, CD A C B D.注:合成 ()运算关于 ()的分配律不成立,即
4、( AB ) C ( A C )( B C )( AB ) C ( A C )( B C )( AB ) C ( A C )( B C )模糊矩阵的转置定义 设 A = (aij)mn, 称 AT = (aijT )nm为 A的转置矩阵,其中 aijT = aji.转置运算的性质:性质 1: ( AT )T = A;性质 2: ( A B )T = AT BT,( AB )T = ATBT;性质 3: ( A B )T = BT AT; ( An )T = ( AT )n ;性质 4: ( Ac )T = ( AT )c ;性质 5: AB AT BT .证明性质 3: ( A B )T =
5、BT AT; ( An )T = ( AT )n .证明 :设 A=(aij)ms, B=(bij)sn, A B=C =(cij)mn,记 ( A B )T = (cijT )nm , AT = (aijT )sm , BT = (bijT )ns ,由转置的定义知 ,cijT = cji , aijT = aji , bijT = bji .BT AT= (bikT akjT )nm= (bki ajk)nm= (ajk bki)nm = (cji)nm = (cijT )nm= ( A B )T . 模糊矩阵的 - 截矩阵定义 7 设 A = (aij)mn,对任意的 0, 1, 称A=
6、 (aij()mn,为模糊矩阵 A的 - 截矩阵 , 其中当 aij 时, aij() =1; 当 aij 时, aij() =0.显然, A的 - 截矩阵为布尔矩阵 . 对任意的 0, 1,有性质 1: AB A B;性质 2: (A B) = A B, (AB) = AB;性质 3: ( A B ) = A B;性质 4: ( AT ) = ( A )T.下面证明性质 1: AB AB 和性质 3.性质 1的证明: AB aijbij;当 aijbij时, aij() =bij() =1;当 aij bij时, aij() =0, bij() =1;当 aijbij 时, aij() = bij() =0;综上所述 aij()bij()时 , 故 A B .