Linear Algebra and its Applications 310 (000 3 7 www.elsevier.com/locate/laa On an operator inequality Jaspal Singh Aujla Department of Applied Mathematics, Regional Engineering College, Jalandhar 1011, Punjab, India Received 10 October 1999; accepted 5 January 000 Submitted by R. Bhatia Abstract Jensen s operator inequality characterizes operator convex functions of two variables (F. Hansen, Proc. Amer. Math. Soc. 15 (1997 093 10. We give a simplified proof of this theorem formulated for matrices. 000 Elsevier Science Inc. All rights reserved. AMS classification: 7A30; 7B15; 15A60 Keywords: Hermitian matrix; Tensor product; Operator convex function 1. Introduction Let M n denote the set of n n complex matrices. For A = (a ij M m and B M n, their tensor product A B is an element of M mn given by a 11 B... a 1m B..... (1 a m1 B... a mmb However, if A11 A A = 1 B11 B and B = 1 A 1 A B 1 B are block matrices, then it is more convenient to represent tensor product A B as the block matrix A 11 B 11 A 11 B 1 A 1 B 11 A 1 B 1 A 11 B 1 A 11 B A 1 B 1 A 1 B A 1 B 11 A 1 B 1 A B 11 A B 1. ( A 1 B 1 A 1 B A B 1 A B 00-3795/00/$ - see front matter 000 Elsevier Science Inc. All rights reserved. PII:S00-3795(0000036-7
J.S. Aujla / Linear Algebra and its Applications 310 (000 3 7 The definition according to ( is unitarily equivalent to the definition according to (1 and therefore will be used throughout this paper. Let I, J be intervals in R containing 0. By S n we denote the set of all hermitian matrices in M n, whereas S n (I denotes the set of all hermitian matrices in M n, whose spectrum is contained in I. We shall let Π n denote the set of projections in M n.letf : I J R be a function of two variables defined on the product of two intervals. Let A S m (I and B S n (J have spectral resolutions A = i λ ip i and B = j µ j Q j.thenf(a,bis defined as f(a,b= i,j f(λ i,µ j P i Q j. When we use the representation of tensor product of block matrices, then f(diag(a, B, diag(c, D = diag(f (A, C, f (A, D, f (B, C, f (B, D for A, B S m (I and C, D S n (J. The function f is called operator convex, if f(λa+ (1 λb, λc + (1 λd λf (A, C + (1 λf (B, D for all A, B S m (I and C, D S n (J ; m, n N and 0 λ 1. Hansen [7] characterized the operator convex functions in terms of Jensen s operator inequality. Here we shall give a simplified proof of this theorem. We shall follow the argument used in [6].. Main results Let A S n. We denote by A f(k the principal submatrix of A consisting of first k rows and columns and by A l(k, the principal submatrix of A consisting of the last k rows and columns, whereas A f(k,l(l and A l(k,f(l, respectively, denote the submatrix of A consisting of first k rows and last l columns and the submatrix of A consisting of the last k rows and first l columns. By I k, we denote the k k identity matrix. Theorem.1. Let f be a real valued continuous function of two variables defined on I J.Then diag(f (P AP, QBQ, f ((I m PA(I m P,(I n QB(I n Q ( P Qf (A, BP Q (I m P (I n Qf (A, BP Q P Qf (A, B(I m P (I n Q (I m P (I n Qf (A, B(I m P (I n Q (3 if and only if
J.S. Aujla / Linear Algebra and its Applications 310 (000 3 7 5 f(s,0 0, f(0,t 0 ( for all (s, t I J, and diag(f (A f(k,b f(l, f (A l(n k,b l(n l ( f(a,b f(kl f(a,b l((m k(n l,f (kl f(a,b f (kl,l((m k(n l f(a,b l((m k(n l for all A S m (I, B S n (J, P Π m and Q Π n., (5 Proof. Suppose (3 holds. Take P = diag(i k, 0 and Q = diag(i l, 0. Then (3 implies ( and (5. On the other hand, since any P Π m is unitarily similar to the projection diag(i k, 0 and any Q Π n is unitarily similar to the projection diag(i l, 0 and for any unitaries U M m and V M n f(u AU, V BV = (U V f (A, B(U V, inequalities ( and (5 imply (3. Theorem.. Let f be a continuous function on I J.Thenf is operator convex and f(s,0 0,f(0,t 0 for all (s, t I J if and only if (3 holds for arbitrary natural numbers m and n. Proof. Suppose (3 holds. Then by Theorem.1, f(s,0 0,f(0,t 0forall (s, t I J and (5 holds. Since f is continuous it is sufficient to prove that f is mid-point operator convex. Let A, B S m (I and C, D S n (J.Put X = U diag(a, BU, Y = V diag(c, DV where U = 1 Im I m I m I m Then it follows that X f(m = X l(m = A + B and and V = 1 In I n. I n I n, Y f(n = Y l(n = C + D, ( A + B f(x f(m,y f(n = f(x l(m,y l(n = f, C + D f(x,y = (U V f(diag(a, B, diag(c, D(U V, f(x,y f(mn = f(a,c+ f(a,d+ f(b,c+ f(b,d
6 J.S. Aujla / Linear Algebra and its Applications 310 (000 3 7 = f(x,y l(mn, f(x,y f (mn,l(mn = f(a,c f(a,d f(b,c+ f(b,d It now follows from (5 that ( ( f A+B, C+D 0 0 f ( A+B I mn = f(x,y l(mn,f (mn., C+D ( f(a,c+f(a,d+f(b,c+f(b,d f(a,c f(a,d f(b,c+f(b,d I mn f(a,c f(a,d f(b,c+f(b,d f(a,c+f(a,d+f(b,c+f(b,d Multiplying the above inequality on the left and on the right by Imn I mn, we obtain ( A + B f, C + D f(a,c+ f(b,d. On the other hand, suppose f satisfies conditions ( and is operator convex. Let A S m (I, C S n (J. Choose U = diag(i k, I m k, B = A, X = U AU, V = diag(i l, I n l, D = C, Y = V CV. Then ( B + X f, D + Y = f(diag(a f(k,a l(m k, diag(c f(l,c l(n l = diag(f (A f(k,c f(l, f (A f(k,c l(n 1, f(a l(m k,c f(l, f (A l(m k,c l(n l,. and f(b,d+ f(x,y = f(a,c+ (U V f (A, C(U V f(a,c f(kl f(a,c f (kl,l((m k(n l = * *. f(a,c l((m k(n l,f (kl f(a,c l((m k(n l Now the result follows from operator convexity of f and Theorem.1.
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