General Letters in Mathematics Vol. 4, No. 1, Feb 2018, pp.13-22 e-ISSN 2519-9277, p-ISSN 2519-9269 Available online at http:// www.refaad.com On Hadamard and Kronecker Products Over Matrix of Matrices Z. Kishka1, M. Saleem 2, M. Abdalla3 and A. Elrawy 4 1;2 Dept. of Math., Faculty of Science, Sohag University, Sohag 82524, Egypt. 3;4 Dept. of Math., Faculty of Science, South Valley University, Qena 83523, Egypt. [email protected], 2 [email protected],3 [email protected],4 [email protected], Abstract. In this paper, we define and study Hadmard and Kronecker products over matrix of matrices and their basic properties. Furthermore, we establish a connection the Hadamard product of matrix of matrices and the usual matrix of matrices multiplication. In addition, we show some application of the Kronecker product. Keywords: Hadamard (Schur) product, Kronecker sum, Kronecker product, matrix of matrices. 2010 MSC No: 15A15, 15A09, 34A30, 39A10. 1 Introduction Matrices and matrix operations play an important role in almost every branch of mathematics, computer graph- ics, communication, computational mathematics, natural and social sciences and engineering. The Hadamard and Kronecker products are studied and utilized widely in matrix theory, statistics [1, 2], physics[2], system theory and other areas; see, e.g. [3, 4, 5, 6, 7, 8, 9, 10, 11]. An equality connection between the Hadamard and Kronecker products look to be firstly used by e.g. [12, 13, 14]. In partitioned matrices, the Khatri-Rao product can be seen as a generalized Hadamard product which is discussed and used by many authors e.g. [15, 16]. Also Tracy-Singh product as a generalized Kronecker product is studied in [19, 20, 21]. Finally, the approach of this paper may not be practical conventional in all situations. In the present paper, we define and study Hadamard and Kronecker product over the matrix of matrices (in a short way; MMs) which was presented newly by Kishka et al [22]. We now give a short overview of this paper. In section 2, we define and study Hadamard product over MMs and give some properties. Then we show the association between Hadamard and MMs product in section 3. Finally, in section 4, we introduce the Kronecker product and prove a number of its properties. In addition, we introduce the notation of the vector matrices (VMs)-operator from which applications can be submitted to Kronecker product. Throughout this paper, the accompanying notations are utilized: Let K be a field and Ml (K) be the set of all l × l matrices defined on K, I and O stand for the identity matrix and the zero matrix in Ml (K), respectively. We denote by Mm×n (Ml (K)) the set of all m × n MMs over Ml (K) ; the elements of Mm×n (Ml (K)) are denoted by A, B, C, D, G,:::. [22] Let A 2 Mm×n (Ml (K)) ; then it can be written as a rectangular table of elements Aij; i = 1; 2; :::; m and j = 1; 2; :::; n; as follows: 0 1 A11 ::: A1n B . .. C A = @ . A : Am1 ::: Amn 14 Z.Kishka et al. t Definition 1.1. [22] Given a m × n-MMs; A = (Aij). Its transpose is the n × m-MMs A , given by t [A ]ji = (Aij) = [A]ij; i = 1; :::; m; j = 1; :::; n; where Aij 2 Ml (K) : 2 Hadamard Product In this section, we give a definition of Hadamard product over MMs with some properties. Definition 2.1. Let A = (Aij) and B = (Bij) 2 Mm×n(Ml(K)), then A ◦ B = (AijBij); i = 1; :::; m; j = 1; :::; n; where Aij;Bij 2 Ml(K), this product is called Hadamard or Schur product over MMs. Example 2.2. Let A and B 2 M2(M2(K)) where 0 2 −3 2 3 1 B −1 1 1 3 C A = B C ; @ 2 6 1 0 A 2 4 0 1 and 0 0 1 1 0 1 B 1 1 0 1 C B = B 3 3 C ; @ 2 −3 2 0 A −1 1 0 2 then 0 −1 1 2 3 1 B 1 −2 1 3 C A ◦ B = B 3 3 C : @ −2 0 2 0 A 0 −2 0 2 Right away we might investigate some fundamentals properties of the Hadamard product over MMs. Theorem 2.3. Let A and B 2 Mm×n(Ml(K)); then A ◦ B = B ◦ A if Ml(K) is commutative ring. Proof. The proof follows directly from the fact that Aij and Bij are commutative. Let A and B be m × n-MMs where A = (Aij) and B = (Bij); then A ◦ B = (AijBij) = (BijAij) = B ◦ A: Theorem 2.4. The identity MMs under the Hadamard product is the m × n MMs with all entries equal to I, denoted Jmn. Proof. Let A = (Aij) 2 Mm×n(Ml(K)); then A ◦ Jmn = (AijIij) = (Aij) and so Jmn ◦ A = (Aij): Therefore, Jmn as defined above is indeed the identity MMs under the Hadamard product. We have denoted the Hadamard identity as Jmn to avoid confusion with the \usual" identity MMs, In: ^ Theorem 2.5. Let A = (Aij) 2 Mm×n(Ml(K)); then A has a Hadamard inverse, denoted by A if and only if ^ −1 Aij are non-singular 8i = 1; 2; :::; m; j = 1; 2; :::; n: Furthermore, A = (Aij ): On Hadamard and Kronecker products over Matrix of Matrices 15 ^ ^ −1 Proof. ()) Let A = (Aij) 2 Mm×n(Ml(K)) with Hadamard inverse A; then A ◦ A = AijAij = (Iij) = Jmn; which is only possible when all entries of A are invertible. In other words Aij are non-singular matrices 8i = 1; 2; :::; m; j = 1; 2; :::; n: (() Take any A = (Aij) 2 Mm×n(Ml(K)) such that Aij are non-singular 8i = 1; 2; :::; m; j = 1; 2; :::; n: Then there ^ ^ ^ ^ −1 −1 exists A = (Aij ) such that A ◦ A = A ◦ A = J mn; and so A has an inverse A = (Aij ): ^ We have denoted the Hadamard inverse as A to avoid confusion with the \usual" inverse MMs, A−1: Theorem 2.6. Let A, B 2 Mm×n(Ml(K)), then (A ◦ B)t = At ◦ Bt: Proof. Suppose that A, B 2 Mm×n(Ml(K)) where A = (Aij) and B = (Bij); then t t (A ◦ B) = (AijBij) = (AjiBji) = At ◦ Bt: t t t Remark 2.7. If Ml (K) is commutative ring with unity, then (A ◦ B) = B ◦ A . Theorem 2.8. The set Mn(Ml(K)) of non-singular MMs is a group under Hadamard product. Proof. Let A = (Aij), B = (Bij) and C = (Cij) 2 Mn(Ml(K)) be MMs then it is easy to see that Mn(Ml(K)) is a group. Theorem 2.9. Suppose that E 2 Ml(K) and A, B and C 2 Mn(Ml(K)); then (1) C◦ (A + B) = C ◦ A + C ◦ B. (2) E (A ◦ B) = (EA) ◦ B = A ◦ (EB) : where Ml(K) is commutative ring with unity. Proof. Part 1. [C◦ (A + B)]ij = [C]ij [A + B]ij = [C]ij [A]ij + [B]ij = [C]ij [A]ij + [C]ij [B]ij = [C ◦ A]ij + [C ◦ B]ij = [C ◦ A + C ◦ B]ij Part 2. [E (A ◦ B)]ij = E[A]ij[B]ij = [EA]ij[B]ij = [EA ◦ B]ij = [A]ijE[B]ij = [A]ij[EB]ij = [A ◦ EB]ij ; since Ml(K) is commutative ring with unity. Remark 2.10. The above part 2 is also true if we replace E by c 2 K. 16 Z.Kishka et al. 3 Connection between Hadamard and MMs products In this section, we show connection between Hadamard and MMs multiplications. Let A = (Aij) 2 Mm(Ml(K)) and consider the set S12:::m of m cyclic permutations of 12...m given by S12:::m = f123::: (m − 1) m; 23::: (m − 1) m1; :::; m12::: (m − 2) (m − 1)g: (1) th If Ai is the i column of the m × m MMs A, then we define A(s) = (As1 : As2 : ::: : Asm) ; (2) i.e., A(s) is the MMs A with permuted columns according to the permutation s = s1s2:::sm: It is easy to see that A(s) = A.I(s). Also, for B = (Bij) 2 Mm(Ml(K)); let us define 0 1 Bs11 Bs22 ··· Bsmm B Bs11 Bs22 ··· Bsmm C B = B C : (3) [S] B . .. C @ . A Bs11 Bs22 ··· Bsmm With these notation, we have the following theorem. Theorem 3.1. Let A = (Aij) and B = (Bij) 2 Mm(Ml(K)); then X X A.B = A(s) ◦ B[s] = A.I(s) ◦ B[s]; s S where A(S) and B[S] are defined as in (2) and (3) ; respectively, for a particular permutation s 2 S12:::m: Proof. Suppose that A = (Aij) and B = (Bij) 2 Mm(Ml(K)); then 0 Pm Pm 1 j=1 A1jBj1 ··· j=1 A1jBjm Pm Pm B j=1 A2jBj1 ··· j=1 A2jBjm C A.B= B C B . .. C @ . A Pm Pm j=1 AmjBj1 ··· j=1 AmjBjm 0 m m m 1 X X X = @ AijBj1 : AijBj2 : ::: : AijBjmA : j=1 j=1 j=1 Clearly, the usual product of MMs A.B decomposes uniquely as the sum of m MMs C(s); where s = s1s2:::sm 2 S123:::m: Such a decomposition can be constructed as follows: The first MMs C(12:::m) is obtained by C12:::m = (Ai1B11 : Ai2B22 : ::: : AimBmm) = A(12:::m) ◦ B[12:::m]: The second MMs is selected according to the permutation 23:::m1; i:e:; C23:::m1 = (Ai2B21 : Ai3B32 : ::: : Ai1B1m) = A(23:::m1) ◦ B[23:::m1]: Following this procedure and taking the MMs according to the complete set of m cyclic permutations of 12:::m in (1) ; we obtain the (m − 1)th MMs as C(m−1)m1:::(m−3)(m−2) = A ◦ B : ((m−1)m1:::(m−3)(m−2)) [(m−1)m1:::(m−3)(m−2)] Finally, the MMs corresponding to the ultimate permutation m1···(m − 2)(m − 1) is formed by the remaining summands as Cm1···(m−2)(m−1) = A ◦ B : (m1···(m−2)(m−1)) [m1···(m−2)(m−1)] Thus, we have A.B= A(12:::m) ◦ B[12:::m] + A(23:::m1) ◦ B[23:::m1] + ::::+ A ◦ B + A ◦ B ; ((m−1)m1:::(m−3)(m−2)) [(m−1)m1:::(m−3)(m−2)] (m1···(m−2)(m−1)) [m1···(m−2)(m−1)] which is the required result.