Journal of Inequalities in Pure and Applied Mathematics

Transcription

1 Journal of Inequalities in Pure and Applied Mathematics MATRIX AND OPERATOR INEQUALITIES FOZI M DANNAN Department of Mathematics Faculty of Science Qatar University Doha - Qatar fmdannan@queduqa volume, issue 3, article 34, 001 Received 3 April, 001; accepted 16 May, 001 Communicated by: D Bainov Abstract Home Page c 000 Victoria University ISSN (electronic:

2 Abstract In this paper we prove certain inequalities involving matrices and operators on Hilbert spaces In particular inequalities involving the trace and the determinant of the product of certain positive definite matrices 000 Mathematics Subject Classification: 15A45, 47A50 Key words: Inequality, Matrix, Operator 1 Introduction 3 Matrix Inequalities 4 3 Operator Inequalities 13 References Page of 17 J Ineq Pure and Appl Math (3 Art 34, 001

3 1 Introduction Inequalities have proved to be a powerful tool in mathematics, in particular in modeling error analysis for filtering and estimation problems, in adaptive stochastic control and for investigation of quantum mechanical Hamiltonians as it has been shown by Patel and Toda [10, 11, 1] and Lieb and Thirring [5] It is the object of this paper to prove new interesting matrix and operator inequalities We refer the reader to [4, 7, 8] for the basics of matrix and operator inequalities and for a survey of many other basic and important inequalities Through out the paper if A is an n n matrix, we write tra to denote the trace of A and det A for the determinant of A If A is positive definite we write A > 0 The adjoint of A (a matrix or operator is denoted by A Page 3 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

4 Matrix Inequalities Through out this section, we work with square matrices on a finite dimensional Hilbert space Theorem 1 If A > 0 and B > 0, then (1 0 < tr (AB m < (tr (AB m for any integer m > 0 Proof The equality holds for m = 1 For m > 1, let B = I, and λ 1, λ,, λ n be the eigenvalues of A Since n λm i < ( n λ i m, then ( 0 < tr (A m < (tra m Since ( is true for any A > 0, we let D = B 1 AB 1 Then inequality ( holds for D Thus 0 < tr (D m < (trd m, from which the result follows Theorem Let A, B be positive definite matrices Then (3 0 < tr (AB m < [tr (AB s ] m s, provided that m and s are positive integers and m > s ( m Proof Clearly tr (AB m = tr A 1 BA 1 > 0 Let l1, l,, l n be the eigenvalues of A 1 BA 1 Then from Hardy s inequality [3] (l1 m + l m + + ln m 1 m < (l1 s + l s + + ln s 1 s for m > s > 0, we get [ ( tr A 1 1 m ] 1 [ ( m BA < tr A 1 1 s ] 1 s BA This implies (3 Theorem 3 Page 4 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

5 If A i > 0 and B i > 0 (i = 1,,, k, then ( ( (4 tr A i B i tr A i ( tr B i If A i B i > 0 (i = 1,,, k, then (5 ( tr ( A i B i < tr Proof Since ( 0 tr (θa i + B i = θ tr A i A i ( tr B i ( ( +θtr A i B i +tr we conclude (4 To prove (5, it suffices to prove that ( ( (6 tr A i B i < tr (A i B i Since A i B i > 0 for i = 1,,, k, then U = k A ib i > 0 Therefore the inequality tr (U < (tru for positive definite U implies (6 and the proof is complete Remark 1 The condition A i B i example shows B i > 0 in (5 is essential as the following, Page 5 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

6 Example 1 Let A = C = ( ( 4 3 4, B = ( ( , B = , It is clear that A, B, C, and D are positive definite matrices Now ( ( (AB + CD =, (AB + CD = Thus tr (AB + CD = 476 > [tr (AB + CD] = 3136 Remark R Bellman [1] proved that tr (AB tr (A B (* for positive definite matrices A and B Further he asked: Does the above inequality (* hold for higher powers? Such a question had been solved by EHLieb and WE Thirring [5],where they proved (7 tr (AB m < tr (A m B m for any positive integer m, and for A, B positive definite matrices In 1995, Changqin Xu [] proved a particular case of (7: that is when A and B are positive definite matrices Notice that (trab m and tr (A m B m are upper bounds for tr (AB m in (1 and (7 One may ask what is max{tr (A m B m, tr (AB m } The following examples show that either (trab m or tr (A m B m can be the least Page 6 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

7 Example Let A = ( ( 5, B = 1 Then tr (AB = 144 < 04 = tr (A B Example 3 Let A = ( 3 ( 1, B = 1 Then tr (A B = 5 < 36 = tr (AB Theorem 4 If 0 < A 1 B 1 and 0 < A B, then (8 0 < tr (A 1 A tr (B 1 B Proof Since 0 < A 1 B 1 and 0 < A B, it follows that 0 < A 1 A 1 A 1 A 1 B 1 A 1 and (9 0 < B 1 1 A B 1 1 B 1 1 B B 1 1 Since trace is a monotone function on the definite matrices, we get (10 0 < tr (A 1 A tr (B 1 A Page 7 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

8 and (11 0 < tr (B 1 A tr (B 1 B This implies (8 Remark 3 The conditions A 1 > 0 and A > 0 in Theorem 4 are essential even if A 1 A and B 1 B are symmetric as the following example shows Example 4 Let A 1 = A = ( ( , B =, 0 ( ( , B = 0 It is clear that A 1 < B 1 and A < B We have also ( 1 7 A 1 A =, B 1 B = 7 15 and tr (A 1 A = 8 > 7 = tr (B 1 B Theorem 5 If A > 0 and B > 0, then (1 n (det A det B m n tr (A m B m for any positive integer m ( Page 8 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

9 Proof Since A is diagonalizable, there exists an orthogonal matrix P and a diagonal matrix Λ such that Λ = P T AP So if the eigenvalues of A are λ 1, λ,, λ n, then Λ = diag (λ 1, λ,, λ n Let b 11 (m, b (m,, b nn (m denote the elements of ( P BP T m Then (13 1 n tr (Am B m = 1 n tr ( P Λ m P T B m = 1 n tr ( Λ m P T B m P = 1 n tr [ Λ m ( P T BP m] = 1 n [λm 1 b 11 (m + λ m b (m + + λ m n b nn (m] Using the arithmetic-mean geometric- mean inequality [9], we get (14 1 n tr (Am B m [λ m 1 λ m λ m n ] 1 n [b11 (m b (m b nn (m] 1 n Since det A a 11 a a nn for any positive definite matrix A, [4] we conclude that (15 det ( P T BP m b11 (m b (m b nn (m and (16 det Λ m λ m 1 λ m λ m n Page 9 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

10 Therefore from (14 it follows that 1 n tr (Am B m [det (Λ m ] 1 n [ det ( P T BP m] 1 n = [ det ( P T AP ] mn [det ( P T BP ] m n = (det A det B m n Here we used the fact that A > 0 and B > 0 The proof is complete Corollary 6 [6] Let A and X be positive definite n n- matrices such that det X = 1 Then (17 n (det A 1 n tr (AX Proof Take B = X and m = 1 in Theorem 5 Theorem 7 If A 0, B 0 and AB = BA, then (18 (m 1n det (A m + B m [det (A + B] m and (19 m 1 tr (A m + B m tr (A + B m for any positive integer m Proof To prove inequality (18, it is enough to prove ( A m + B m m A + B (0 Page 10 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

11 for any pair of commuting positive definite matrices A and B We use induction to prove (0 Clearly (0 holds true for m = Assume that (0 is true for m = k We have to prove (0 for m = k + 1 Indeed, since A k + B k A + B it follows that ( k+1 A + B (1 A + B = A + B Ak + B k = Ak+1 + B k+1 = Ak+1 + B k+1 = Ak+1 + B k+1 Ak + B k, Ak+1 + B k+1 + BAk + AB k 4 4 Ak+1 + B k+1 BA k AB k ( 4 A k B k (A B 4 Now the equality ( A k B k (A B = ( A k 1 + A k B + + AB k + B k 1 (A B for A 0, B 0 and AB = BA, implies AB 0 [8] Consequently L = A k 1 + A k B + + AB k + B k 1 0 Since L (A B = (A B L, then ( A k B k (A B 0 Therefore, from (1 we obtain ( k+1 A + B Ak+1 + B k+1 Page 11 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

12 The proof is complete Inequality (19 follows directly from (0 Remark 4 The condition AB = BA in inequality (0 is essential as the following example shows Example 5 Let A = ( ( 3, B = It is clear that A > 0, B > 0 and AB BA For m = 3 inequality (0 becomes ( 4 ( A 3 + B 3 (A + B 3 Easily we find that 4 ( A 3 + B 3 = ( (, (A + B = 45 4 and det C = 9 which implies that C < 0 Page 1 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

13 3 Operator Inequalities In this section we consider inequalities involving operators on separable Hilbert space X We start with the following simple well-known inequality Theorem 31 Let S and T be self-adjoint bounded linear operators on the Hilbert space H Then ( ST + T S S + T (31 Proof Since ( S + T ST + T S ( S + T = and since the square of the self-adjoint operator is a non-negative operator, we get ( S+T 0 The claim of the theorem now follows Now we present a similar type result as Theorem 31 but for non-self-adjoint case More precisely: Theorem 3 Let S and T be bounded linear operators on a Hilbert space X Assume S to be self-adjoint Then 1 (3 4 K + H p H Q, where (33 (34 P = 1 (ST + T S, Q = ( S + T, H P = 1 (P + P, H Q = 1 (Q + Q Page 13 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

14 and K = 1 (T + T Proof For any bounded linear operator T we have T = 1 [(T + T + (T T ] = H T + K, where H T = 1 (T + T Inequality (31 can be applied to the self-adjoint operators S and H T, so we get (35 Ux, x 0, where U = (S + H T (SH T + H T S Now we have (36 U = (S + H T S (T + T (T + T S = (S + H T 4H [ P ] = 1 S + T + (T + 1 (T T + T T 4H P = 1 ( S + T + (T 4H P K = 1 = 4 ( 8HQ 4H P 4H P K (H Q H P 14 K Page 14 of 17 Therefore the required inequality (3 follows from (36 and (35 J Ineq Pure and Appl Math (3 Art 34, 001

15 Remark 31 When both S and T are not self-adjoint operators, Theorem 3 does not hold The following example illustrates this fact Example 31 Let S and T be defined on R R by the following matrices ( ( S =, T = By computation we find that P = 1 ( 7 1 (ST + T S =, 6 14 ( ( S + T 1 8 Q = =, K = 4 7 ( ( H P =, H 3 14 Q =, 7 H Q H P 1 4 K = < 0 ( ( , Page 15 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

16 References [1] R BELLMAN, Some inequalities for positive definite matrices, in General Inequalities Proceedings, nd Internat Conf on General Inequalities, (EFBeckenbach, Ed, p 89 90, Birkhauser,1980 [] CHNG-QIN XU, Bellman s inequality, Linear Algebra and its Appl, 9 (1995, 9 14 [3] GH HARDY, JE LITTLEWOOD AND G POLYA, Inequalities, Cambridge Univ Press, Cambridge, 195 [4] RA HORN AND CR JOHNSON, Matrix Analysis, Cambridge Univ Press, 1999 [5] EH LIEB AND WE THIRRING, Inequalities for the moments of the eigenvalues of the Schrodinger Hamiltonian and their relation to Sobolev inequalities, Studies in Mathematical Physics, Essays in Honor of Valentine, (1976, Bartmann, Princeton, NJ, [6] JR MAGNUS AND H NEUDECKER, Matrix Differential calculus with Applications in Statistics and Econometrics, John Wiley & Sons, 1990 [7] M MARCUS AND H MINC, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Inc, Boston, 1964 [8] ML MEHTA, Matrix Theory, Selected topics and useful results, Les Editions de Physique, 1989 [9] DS MITRINOVIĆ, Analytic Inequalities, Springer -Verlag Berlin, Heidelberg, New York, 1970 Page 16 of 17 J Ineq Pure and Appl Math (3 Art 34, 001

17 [10] R PATEL AND M TODA, Modeling error analysis of stationary linear discrete time filters, NASA TM X-73, 5(Feb (1977 [11] R PATEL AND M TODA, Trace inequalities involving Hermitian matrices, Linear Algebra and its Appl, 3 (1979, 13 0 [1] M TODA AND R PATEL, Algorithms for adaptive stochastic control for a class of linear systems, NASA TM X-73, 40(Apr (1977 Page 17 of 17 J Ineq Pure and Appl Math (3 Art 34, 001