Selçuk J. Appl. Math. Selçuk Journal of Vol. 14. No. 1. pp. 31-36, 2013 Applied Mathematics
On the Traces of Hadamard and Kronecker Products of Matrices Necati Taskara1; Ibrahim H.Gumus2
1Department of Mathematics, Science Faculty, Selcuk University, Campus, 42075,
Konya, Turkiye
e-mail:ntaskara@ selcuk.edu.tr
2Department of Mathematics, Science and Art Faculty, Ad¬yaman University,
Ad¬ya-man, Turkiye
e-mail:igumus@ adiyam an.edu.tr
Received Date: July 28, 2011 Accepted Date: March 10, 2013
Abstract. In this paper we investigated traces of Hadamard and Kronecker products of matrices and obtained some inequalities for traces of products of matrices.
Key words: Trace; hadamard product; kronecker product. AMS Classi…cation: 15A45; 15A24.
1. Introduction
The initial studies about algebraic and analytic properties of Hadamard product are done by Schur in 1911. Because of the originality of this study, hadamard product is called Schur product. Horn gave a widespread information about Hadamard product in 1990 [1]. Marcus and Moyls [2] dwelled upon Hadamard product to be a principal submatrix of Kronecker product, also Visick [3] showed that A B = PnT(A B)Pn, where A and B are arbitrary n n matrices and P is an n2 n selection matrix such that PTP = I. Also, some studies, related with traces and matrix products, can be seen in [4-10].
Let A = (aij) be n n matrix. Then the trace of A, denoted by tr(A) is the sum of the major diagonal elements of A , that is
tr(A) = n X i=1 aii: 1Corresponding Author
The Hadamard product of two m n matrices A = (aij) and B = (bij) is de…ned as
A B = (aijbij)m n:
If A = (aij) is an m n matrix and B = (bij) is a p q matrix, the Kronecker product A B is the mp nq matrix
A B = 0 B B B @ a11B a12B a1nB a21B a21B a2nB .. . ... . .. ... am1B am2B amnB 1 C C C A: Lemma 1. For 8i 2 N; 8ai 0, (1) 1 n n X i=1 ai ! n Y i=1 ai !1=n :
Lemma 2. Let A and B be arbitrary n n matrices. Then
(2) tr(A B) = tr(A):tr(B) .
Proposition 1. For 8a; b 2 R;we have
(a b)2 0 a2 2ab + b2 0 : Then
(3) a2+ b2 2ab .
2. Main Results
In this section, we have a trace equality related with Hadamard product and Kronecker product. Also, we obtain some inequalities for traces of products of matrices.
The following theorem give the relation between trace of Hadamard product and trace of Kronecker product.
Theorem 1. Let A = (aij) and B = (bij) be n n matrices. Then
(4) tr(A B) = n:tr(A B) n 1X i=1 n X j=i+1 (aii ajj)(bii bjj) .
Proof. By Lemma 2, we have (5) tr(A):tr(B) = n:tr(A B) n 1X i=1 n X j=i+1 (aii ajj)(bii bjj) .
Let use the principle of mathematical induction on n. For n = 2, 2 X i=1 aii 2 X i=1 bii = 2: 2 X i=1 aii:bii 2 X i=1 2 X j=2 (aii ajj)(bii bjj); where tr(A) = 2 X i=1 aii; tr(B) = 2 X i=1 bii; tr(A B) = 2 X i=1 aiibii . Let us rewrite the right hand side of above equation as follows;
2 X i=1 aii 2 X i=1 bii = 2: 2 X i=1 aii:bii a11b11+ a11b22+ a22b11 a22b22 = 2: 2 X i=1 aii:bii a11b11 a22b22+ a11b22+ a22b11 = a11b11+ a22b22+ a11b22+ a22b11 = (a11+ a22)(b11+ b22) = 2 X i=1 aii 2 X i=1 bii .
Then it is clear that the result is hold for n = 2.
Assume that it is true for all positive integers n = k. That is,
(6) k X i=1 aii k X i=1 bii= k: k X i=1 aii:bii k 1 X i=1 k X j=2 (aii ajj)(bii bjj):
We have to show that it is true for n = k + 1. That is,
(7) k+1 X i=1 aii k+1 X i=1 bii= (k + 1) k+1 X :i=1 aii:bii k X i=1 k+1 X j=2 (aii ajj)(bii bjj) . If we add k:a(k+1)(k+1)b(k+1)(k+1)+ k+1 X i=1 aii:bii k X i=1 (aii a(k+1)(k+1))(bii b(k+1)(k+1))
term to the right and the left sides of equation (6), then we obtain equation (7). Therefore, the result is true for every n 2.
In the following theorem, for arbitrary square matrices A and B; we obtain the trace inequality for Hadamard product of matrix sums.
Theorem 2. Let A = (aij) and B = (bij) be n n matrices. Then tr(A B) tr (A + B
2 ) ( A + B
2 ) .
Proof. From tr(A B) = n P i=1 aii:bii, we have tr (A + B 2 ) ( A + B 2 ) = n X i=1 aii+ bii 2 aii+ bii 2 = 1 4 n X :i=1 (aii+ bii)(aii+ bii) = 1 4 n X :i=1 a2ii+ 2aiibii+ b2ii = n X i=1 aiibii 2 + 1 4 n X :i=1 a2ii+ b2ii .
Then from equation (3) we obtain
tr (A + B 2 ) ( A + B 2 ) = n X i=1 aiibii 2 + 1 4 n X :i=1 a2ii+ b2ii n X i=1 aiibii 2 + 1 4 n X :i=1 2aiibii = n X i=1 aiibii 2 + n X :i=1 aiibii 2 = n X :i=1 aiibii= tr(A B):
By considering above theorem, we give a generalization for trace inequality of Hadamard product.
Theorem 3. Let A(j) be n n matrices having positive real numbers as diagonal elements for every j 2 N. Then
tr m j=1 A(j) tr 0 @m m X j=1 A(j) m 1 A .
Proof. We write (8) tr m j=1 A(j) =Pni=1a(1)ii a(2)ii a(m)ii =Pni=1 m Q j=1 a(j)ii . Also we can write
tr 0 @m m X j=1 A(j) m 1 A = tr m A(1)+ A(2)+ A(m) m = tr A (1)+ A(2)+ A(m) m A(1)+ A(2)+ A(m) m = n X i=1 a(1)ii + a(2)ii + + a(m)ii m !m :
Thus from Lemma 1 and equation (8) we have
tr 0 @m m X j=1 A(j) m 1 A = n X i=1 a(1)ii + a(2)ii + + a(m)ii m !m n X i=1 m Y j=1 a(j)ii = tr m j=1 A(j) . Example 1. Let A = 3 1 4 1 and B = 4 1 3 2 be matrices. Then we compute A B, A B, tr(A B), tr(A B) as follows;
A B = 12 1 12 2 ; A B = 0 B B @ 12 3 4 1 9 6 3 2 16 4 4 1 12 8 3 2 1 C C A , tr(A B) = 10 ; tr(A B) = 12 By applying Theorem 1, we have
2:tr(A B) 1 X i=1 2 X j=i+1 (aii ajj)(bii bjj) = 2:10 (a11 a22)(b11 b22) = 20 8 = 12 = tr(A B)
Therefore, the theorem is true. Example 2. Let A = 3 1 4 1 and B = 4 1 3 2 be matrices. Then tr(A B) = 10; tr A + B 2 A + B 2 = 53 4 . By applying Theorem 2, we have
10 53 4 . Therefore, the theorem is true .
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