Math. Appl. 6 (2017), 143{150 DOI: 10.13164/ma.2017.09 TRACES OF HADAMARD AND KRONECKER PRODUCTS OF MATRICES PANKAJ KUMAR DAS and LALIT K. VASHISHT Abstract. We present some inequality/equality for traces of Hadamard product and Kronecker product of matrices. Some numerical examples are given to support the results. 1. Introduction The Hadamard (or Schur) and Kronecker products are widely studied and ap- plied in matrix theory, statistics, system theory and other areas [2]. It was Schur who initially studied algebraic and analytic properties of Hadamard product. In 1990, Horn [1] presents a widespread information focusing on the Hadamard prod- uct. Magnus and Neudecker [4] give some basic results and statistical applications involving Hadamard or Kronecker products. For basics on these two matrix prod- ucts, one may refer to [3,5{7]. In [6], the authors investigated traces of Hadamard and Kronecker products of matrices and obtained some inequalities for traces of products of matrices. In this note, we present some inequality for traces of Hadamard product, Kronecker product and mixed type product of matrices. We recall the basic definitions and notations to make our presentation self- + contained. The set of all positive real numbers is denoted by R . By Mn we denote the family of n-by-n matrices over the real field R. For A = [aij] 2 Mn, the Pn scalar i=1 aii is called the trace of A and is usually denoted by tr(A) or traceA. The Hadamard product of two matrices A = [aij] and B = [bij] of identical size is just their element-wise product which is given by A ◦ B = [aijbij]m×n: Let A = [aij] be an m-by-n matrix and let B = [bij] be a p-by-q matrix. The Kronecker product of A and B is defined as 2 3 a11B a12B : : : a1nB 6 a21B a22B : : : a2nB 7 A ⊗ B = 6 7 6 . .. 7 4 . 5 am1B am2B : : : amnB MSC (2010): primary 15A45; secondary 15A24. Keywords: trace, Hadamard product, Kronecker product. 143 144 P. K. DAS and L. K. VASHISHT Lemma 1.1. [6] Let A 2 Mn and B 2 Mm. Then, tr(A ⊗ B) = tr(B ⊗ A) = tr(A)tr(B): Theorem 1.2. [6] If A = [aij];B = [bij] 2 Mn, then Pn−1 Pn (1) tr(A ⊗ B) = n tr(A ◦ B) − i=1 j=i+1(aii − ajj)(bii − bjj): A+B A+B (2) tr(A ◦ B) ≤ tr 2 ◦ 2 : 2. Main results We start with an inequality involving the trace of m times Hadamard product of a matrix and the trace of the given matrix . Proposition 2.1. Let A 2 Mn with positive real diagonal. Then, tr A ◦ A ◦ A ◦ · · · ◦ A ≤ (tr(A))m: | {z } m + Proof. Let A = [aij ] 2 Mn and let aii 2 R . Then, by the definitions of Hadamard product and trace of matrices, we have n m m X m X m tr A ◦ A ◦ A ◦ · · · ◦ A = a ≤ aii = (tr(A)) : | {z } ii m i=1 i=1 The proposition is proved. Proposition 2.2. If A = [aij];B = [bij] 2 Mn, then n tr (A ◦ A ◦ · · · ◦ A) ◦ (B ◦ B ◦ · · · ◦ B) | {z } | {z } n n n n n−1 n X n X n X X n n n n = aii bii − (aii − ajj)(bii − bjj): i=1 i=1 i=1 j=i+1 Proof. Since n n X X tr (A ◦ A ◦ · · · ◦ A) ⊗ (B ◦ B ◦ · · · ◦ B) = an bn ; | {z } | {z } ii ii n n i=1 i=1 applying Theorem 1 in [6], we get n tr (A ◦ A ◦ · · · ◦ A) ◦ (B ◦ B ◦ · · · ◦ B) | {z } | {z } n n n n n−1 n X n X n X X n n n n = aii bii − (aii − ajj)(bii − bjj): i=1 i=1 i=1 j=i+1 The following proposition provides a relation between the trace of a matrix and Kronecker product of the given matrix. TRACES OF HADAMARD AND KRONECKER PRODUCTS OF MATRICES 145 Proposition 2.3. For any A 2 Mn, we have tr A ⊗ A ⊗ A ⊗ · · · ⊗ A = (tr(A))m: | {z } m Proof. We prove the proposition by induction. By Lemma 1.1, the equation is true for n = 2. Assume that it is true n = k, i.e., tr A ⊗ A ⊗ A ⊗ · · · ⊗ A = (tr(A))k: (2.1) | {z } k Again using Lemma 1.1, we compute tr A ⊗ A ⊗ A ⊗ · · · ⊗ A = tr A ⊗ A ⊗ A ⊗ · · · ⊗ A tr(A) | {z } | {z } k+1 k = (tr(A))ktr(A) (by (2.1)) = (tr(A))k+1: The proposition is proved. 3 1 Example 2.4. Let A = . 4 1 Then, 227 9 9 3 9 3 3 13 636 9 12 3 12 3 4 17 6 7 636 12 9 3 12 4 3 17 6 7 9 1 648 12 12 3 16 4 4 17 A ◦ A = ;A ⊗ A ⊗ A = 6 7 16 1 636 12 12 4 9 3 3 17 6 7 648 12 16 4 12 3 4 17 6 7 448 16 12 4 12 4 3 15 64 16 16 4 16 4 4 1 and tr(A) = 4. Now we have tr(A ◦ A) = 10 ≤ 42 = (tr(A))3. So, Proposition 2.1 is true. Also, tr(A ⊗ A ⊗ A) = 64 = (tr(A))3: This verifies Proposition 2.2. Proposition 2.5. For any A 2 Mn, we have tr (A + At) ◦ (A + At) = 4 tr(A ◦ A): Proof. Let A = [aij] 2 Mn. Then n n t t X X 2 tr (A + A ) ◦ (A + A ) = (aii + aii)(aii + aii) = 4 aii = 4 tr(A ◦ A): i=1 i=1 The proposition is proved. Proposition 2.6. For any A; B 2 Mn, we have tr (A + B) ◦ (A − B) = tr(A ◦ A) − tr(B ◦ B): 146 P. K. DAS and L. K. VASHISHT Proof. Let A = [aij];B = [bij] 2 Mn be arbitrary. We compute n X tr (A + B) ◦ (A − B) = (aii + bii)(aii − bii) i=1 n X 2 2 = (aii) − (bii) i=1 = tr(A ◦ A) − tr(B ◦ B): Proposition 2.7. For any A; B 2 Mn, we have tr (A + B) ◦ (A + B) = tr(A ◦ A) + 2tr(A ◦ B) + tr(B ◦ B): Proof. Similar to proof of Proposition 2.6. The following proposition gives the relationship between trace of a matrix ob- tained as Kronecker product of a finite sum of matrices and the traces of the matrices. We prove the result for two matrices. Proposition 2.8. Let A; B 2 Mn. Then tr (A + B) ⊗ (A + B) ⊗ (A + B) ⊗ · · · ⊗ (A + B) | {z } k k k−i i X k = tr(A) tr(B) : i i=1 Proof. Using Proposition 2:3, we compute tr (A + B) ⊗ (A + B) ⊗ (A + B) ⊗ · · · ⊗ (A + B) | {z } k k k = tr(A + B) = tr(A) + tr(B) k k−i i X k = tr(A) tr(B) : i i=1 The result is proved. Remark 2.9. Using Proposition 2.1, for any A; B 2 Mn with positive diagonal, we can show that tr (A + B) ◦ (A + B) ◦ (A + B) ◦ · · · ◦ (A + B) | {z } n n n−i i X n ≤ tr(A) tr(B) : i i=1 TRACES OF HADAMARD AND KRONECKER PRODUCTS OF MATRICES 147 The next proposition gives a trace inequality for the Hadamard product of matrix sums. Proposition 2.10. If A; B 2 Mn, then 1 (tr(A ◦ B))2 ≤ tr (A + B) ◦ (A + B) ◦ (A + B) ◦ (A + B) : 16 Proof. Using Theorem 1.2 and Proposition 2.3, we compute A + B A + B A + B A + B (tr(A ◦ B))2 ≤ tr ◦ tr ◦ 2 2 2 2 A + B A + B A + B A + B = tr ◦ ◦ ◦ 2 2 2 2 1 = tr (A + B) ◦ (A + B) ◦ (A + B) ◦ (A + B) : 16 The proposition is proved. The following theorem provides a relation between the trace of matrices gener- ated by Kronecker and Hadamard product of matrices. Theorem 2.11. If A; B; C; D 2 Mn, then (1) (A ⊗ B) ◦ (C ⊗ D) = (A ◦ C) ⊗ (B ◦ D): Pn−1 Pn (2) tr(A⊗(B ◦C)) = n tr(A◦B ◦C)− i=1 j=i+1(aii −ajj)(biicii −bjjcjj): Proof. Let us write A = [aij] and C = [cij]. Then, we compute (A ⊗ B) ◦ (C ⊗ D) 2 3 2 3 a11B a12B : : : a1nB c11D c12D : : : c1nD 6 a21B a22B : : : a2nB 7 6 c21D c22D : : : c2nD 7 = 6 7 ◦ 6 7 6 . .. 7 6 . .. 7 4 . 5 4 . 5 an1B an2B : : : annB cn1D cn2D : : : cnnD 2 3 a11c11B ◦ D a12c12B ◦ D : : : a1nc1nB ◦ D 6 a21c21B ◦ D a22c22B ◦ D : : : a2nc2nB ◦ D 7 = 6 7 = (A ◦ C) ⊗ (B ◦ D): 6 . .. 7 4 . 5 an1cn1B ◦ D an2cn2B ◦ D : : : anncnnB ◦ D Hence, (1) is proved. Let us write B = [bij]. Then, by Theorem 1:2, we have n−1 n X X tr(A ⊗ (B ◦ C)) = n tr(A ◦ (B ◦ C)) − (aii − ajj)(biicii − bjjcjj) i=1 j=i+1 n−1 n X X = n tr(A ◦ B ◦ C) − (aii − ajj)(biicii − bjjcjj): i=1 j=i+1 Thus, (2) is proved. 148 P. K. DAS and L. K. VASHISHT Corollary 2.12. If A; B; C; D 2 Mn, then tr (A ⊗ B) ◦ (C ⊗ D) = n tr(A ◦ B ◦ C ◦ D) n−1 n X X − (aiicii − ajjcjj)(biidii − bjjdjj): i=1 j=i+1 Indeed, by Theorem 2.11, we can write tr (A ⊗ B) ◦ (C ⊗ D) = tr (A ◦ C) ⊗ (B ◦ D) : Thus, by Theorem 1:2, we have tr (A ⊗ B) ◦ (C ⊗ D) = n tr(A ◦ C ◦ B ◦ D) n−1 n X X − (aiicii − ajjcjj)(biidii − bjjdjj): i=1 j=i+1 Remark 2.13.