Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID 201486, 8 pages doi:10.1155/2010/201486
Research Article On Some Matrix Trace Inequalities
Zubeyde¨ Ulukok¨ and Ramazan Turkmen¨
Department of Mathematics, Science Faculty, Selc¸uk University, 42003 Konya, Turkey
Correspondence should be addressed to Zubeyde¨ Ulukok,¨ [email protected]
Received 23 December 2009; Revised 4 March 2010; Accepted 14 March 2010
Academic Editor: Martin Bohner
Copyright q 2010 Z. Ulukok¨ and R. Turkmen.¨ This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We first present an inequality for the Frobenius norm of the Hadamard product of two any square matrices and positive semidefinite matrices. Then, we obtain a trace inequality for products of two positive semidefinite block matrices by using 2 × 2 block matrices.
1. Introduction and Preliminaries
Let Mm,n denote the space of m × n complex matrices and write Mn ≡ Mn,n. The identity ∗ T matrix in Mn is denoted In. As usual, A A denotes the conjugate transpose of matrix ∗ A. A matrix A ∈ Mn is Hermitian if A A. A Hermitian matrix A is said to be positive semidefinite or nonnegative definite, written as A ≥ 0, if
∗ n x Ax ≥ 0, ∀x ∈ C . 1.1
A is further called positive definite, symbolized A>0, if the strict inequality in 1.1 holds n for all nonzero x ∈ C . An equivalent condition for A ∈ Mn to be positive definite is that A is Hermitian and all eigenvalues of A are positive real numbers. Given a positive semidefinite p matrix A and p>0, A denotes the unique positive semidefinite pth power of A. Let A and B be two Hermitian matrices of the same size. If A − B is positive semidefinite, we write
A ≥ B or B ≤ A. 1.2
λ A ,...,λ A s A ,...,s A A Denote 1 n and 1 n eigenvalues and singular values of matrix , respectively. Since A is Hermitian matrix, its eigenvalues are arranged in decreasing order, λ A ≥ λ A ≥ ··· ≥ λ A A that is, 1 2 n and if is any matrix, its singular values are arranged s A ≥ s A ≥ ··· ≥ s A > . A in decreasing order, that is, 1 2 n 0 The trace of a square matrix 2 Journal of Inequalities and Applications
the sum of its main diagonal entries, or, equivalently, the sum of its eigenvalues is denoted by tr A. Let A be any m × n matrix. The Frobenius Euclidean norm of matrix A is ⎡ ⎤ 1/2 m n ⎣ 2⎦ AF aij . 1.3 i 1 j 1
∗ It is also equal to the square root of the matrix trace of AA , that is, ∗ AF tr AA . 1.4
Anorm·on Mm,n is called unitarily invariant UAV A for all A ∈ Mm,n and all unitary U ∈ Mm,V ∈ Mn. x x ,...,x y y ,...,y Given two real vectors 1 n and 1 n in decreasing order, we say k k x y x ≺w y Π xi ≤ Π yi,k , ,...,n that is weakly log majorized by , denoted log ,if i 1 i 1 1 2 ,and x y x ≺ y k x ≤ k y ,k , ,...,n we say that is weakly majorized by , denoted w ,if i 1 i i 1 i 1 2 .We say x is majorized by y denoted by x ≺ y,if
n n x ≺wy, xi yi. 1.5 i 1 i 1
x ≺ y x ≺ y As is well known, w log yields w see, e.g., 1, pages 17–19 . Let A be a square complex matrix partitioned as
A A 11 12 A , 1.6 A A 21 22
A A A where 11 is a square submatrix of .If 11 is nonsingular, we call
A A − A A−1A 11 22 21 11 12 1.7
A A A the Schur complement of 11 in see, e.g., 2, page 175 .If is a positive definite matrix, A then 11 is nonsingular and A ≥ A ≥ . 22 11 0 1.8
Recently, Yang 3 proved two matrix trace inequalities for positive semidefinite matrices A ∈ Mn and B ∈ Mn,
n− n n 1 0 ≤ tr AB 2 ≤ tr A 2 tr A2 tr B2 , n n 1.9 n 0 ≤ tr AB 2 1 ≤ tr A tr B tr A2 tr B2 , for n 1, 2,.... Journal of Inequalities and Applications 3
Also, authors in 4 proved the matrix trace inequality for positive semidefinite matrices A and B,
/ m m m 1 2 tr AB ≤ tr A 2 tr B 2 , 1.10 where m is a positive integer. Furthermore, one of the results given in 5 is
m/n m m n det A · det B ≤ tr A B 1.11 for A and B positive definite matrices, where m is any positive integer.
2. Lemmas
Lemma 2.1 see, e.g., 6 . For any A and B ∈ Mn,σ A ◦ B ≺wσ A ◦ σ B .
Lemma 2.2 see, e.g., 7 . Let A, B ∈ Mm,n, then
t t 2m ∗ ∗ m δi AB ≤ λi A ABB i 1 i 1 2.1 t ∗ m ∗ m ≤ λi A A BB , 1 ≤ t ≤ n, m ∈ N. i 1
Lemma 2.3 a ,a ,...,a b ,b ,...,b Cauchy-Schwarz inequality . Let 1 2 n and 1 2 n be real numbers. Then,
n 2 n n 2 2 aibi ≤ ai bi , ∀ai,bi ∈ R. 2.2 i 1 i 1 i 1
Lemma 2.4 see, e.g., 8, page 269 . If A and B are poitive semidefinite matrices, then,
0 ≤ tr AB ≤ tr A tr B. 2.3
Lemma 2.5 see, e.g., 9, page 177 . Let A and B are n × n matrices. Then,
k k si AB ≤ si A si B 1 ≤ k ≤ n . 2.4 i 1 i 1
Lemma 2.6 see, e.g., 10 . Let F and G are positive semidefinite matrices. Then,
t t m m m λi FG ≤ λi F G , 1 ≤ t ≤ n, 2.5 i 1 i 1 where m is a positive integer. 4 Journal of Inequalities and Applications
3. Main Results
Horn and Mathias 11 show that for any unitarily invariant norm ·on Mn
∗ 2 ∗ ∗ A B ≤ A AB B ∀A, B ∈ Mm,n, 3.1 2 ∗ ∗ A ◦ B ≤ A AB B ∀A, B ∈ Mn.
A LX X ∈ Also, the authors in 12 show that for positive semidefinite matrix X∗ M , where Mm,n p |X| 2 ≤ LpMp 3.2 for all p>0 and all unitarily invariant norms ·. By the following theorem, we present an inequality for Frobenius norm of the power of Hadamard product of two matrices.
Theorem 3.1. Let A and B be n-square complex matrices. Then A ◦ B m2 ≤ A∗A m B∗B m , F F F 3.3 where m is a positive integer. In particular, if A and B are positive semidefinite matrices, then A ◦ B m2 ≤ A2m B2m . F F F 3.4
Proof. From definition of Frobenius norm, we write A ◦ B m2 A ◦ B m A ◦ B m∗ . F tr 3.5
Also, for any A and B, it follows that see, e.g., 13
AA∗ ◦ BB∗ A ◦ B ≥ 0, 3.6 A∗ ◦ B∗ I
∗ ∗ ∗ A ◦ B A ◦ B ≤ AA ◦ BB . 3.7
m m ∗ m ∗ m Since | tr A2 |≤tr A A ≤ tr AA for A ∈ Mn and from inequality 3.7 , we write A ◦ B m2 A ◦ B m A ◦ B m∗ F tr ∗m ≤ tr A ◦ B A ◦ B 3.8 ∗ ∗ m ≤ tr AA ◦ BB . Journal of Inequalities and Applications 5
From Lemma 2.1 and Cauchy-Schwarz inequality, we write
n n m m m m m m tr A ◦ B λi A ◦ B ≤ λi A λi B i 1 i 1 n n 1/2 2 m 2 m 3.9 ≤ λi A λi B i 1 i 1 / m m 1 2 tr A2 tr B2 .
By combining inequalities 3.7 , 3.8 ,and 3.9 , we arrive at
∗ ∗ m ∗ ∗ m ∗ ∗ m / tr AA ◦ BB ≤ tr AA AA tr BB BB 1 2 ∗ ∗ m ∗ ∗ m / ≤ tr AA AA tr BB BB 1 2 / / 3.10 ∗ m 1 2 ∗ m 1 2 tr AA 2 tr BB 2 A∗A m B∗B m . F F
Thus, the proof is completed. Let A and B be positive semidefinite matrices. Then
A ◦ B m2 ≤ A2m B2m , F F F 3.11 where m>0.
Theorem 3.2. Let Ai ∈ Mn i 1, 2,...,k be positive semidefinite matrices. For positive real numbers s, m, t
k 2 k k s t / m2 A 2 ≤ Asm2 Atm2 . 3.12 i F i F i F i 1 i 1 i 1
Proof. Let
⎛ ⎞ ⎛ ⎞ S/ t/ A 2 0 ··· 0 A 2 0 ··· 0 ⎜ 1 ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ As/2 ··· ⎟ ⎜ At/2 ··· ⎟ ⎜ 0 2 0 ⎟ ⎜ 0 2 0 ⎟ A ⎜ ⎟,B ⎜ ⎟. ⎜ ⎟ ⎜ ⎟ 3.13 ⎜ . . .. . ⎟ ⎜ . . .. . ⎟ ⎝ . . . . ⎠ ⎝ . . . . ⎠ ··· As/2 ··· At/2 00 k 00 k 6 Journal of Inequalities and Applications
We know that A, B ≥ 0, then by using the definition of Frobenius norm, we write
k 2 A ◦ B m2 A s t /2 m , F i F i 1 ! ! 3.14 k k ! ! A2m " Asm2 , B2m " Atm2 . F i F F i F i 1 i 1
Thus, by using Theorem 3.1, the desired is obtained.
Now, we give a trace inequality for positive semidefinite block matrices.
Theorem 3.3. Let
A A B B 11 12 11 12 A ≥ 0,B ≥ 0, 3.15 A A B B 21 22 21 22 then,
# $ m # $ m 1/2 2 1/2 2 A B1/2 A1/2 B ≤ AB m ≤ AmBm , 3.16 tr 22 11 tr 22 11 tr tr where m is an integer.
Proof. Let
X 0 M 3.17 YZ
Z A1/2,Y A−1/2A ,X A − A A−1A 1/2 A M∗M with 22 22 21 11 12 22 21 . Then see, e.g., 14 .Let
X 0 K 3.18 YZ
Z B − B B−1B 1/2 Y B B−1/2 X B1/2 B KK∗ with 22 21 11 12 , 21 11 , 11 . Then see, e.g., 14 .We know that
k X 0 Mk , ∗ Zk
⎡ ⎤ − 1/2 / A − A A 1A B1 2 0 M · K ⎣ 11 12 22 21 11 ⎦, / A−1/2A B1/2 A1/2B B−1/2 A1/2 B − B B−1B 1 2 22 21 11 22 21 11 22 22 21 11 12 Journal of Inequalities and Applications 7
⎡& ' m ⎤ − 1/2 / 2 A − A A 1A B1 2 0 m ⎢ 11 12 22 21 11 ⎥ M · K 2 ⎣ & ' ⎦. / 2m ∗ A1/2 B − B B−1B 1 2 22 22 21 11 12 3.19
By using Lemma 2.2, it follows that
n n 2m 2m 2m tr MK ≤ si MK ≤ si MK i 1 i 1
n m n 2 ∗ ∗ m si MK λi M MKK i 1 i 1 3.20 n n n m m ∗ m ∗ m λi AB tr AB ≤ λi M M KK i 1 i 1 i 1 n n m m m m λi A B tr A B . i 1 i 1
Therefore, we get
# $ m # $ m 1/2 2 1/2 2 MK 2m A − A A−1A B1/2 A1/2 B − B B−1B tr tr 11 12 22 21 11 tr 22 22 21 11 12
m m m ≤ tr AB ≤ tr A B . 3.21
As result, we write
# $ m # $ m 1/2 2 1/2 2 A B1/2 A1/2 B ≤ AB m ≤ AmBm . 3.22 tr 22 11 tr 22 11 tr tr
Example 3.4. Let
41 52 A > 0,B > 0. 3.23 11 21
Then tr AB 25, det A 3, det B 1. From inequality 1.11 ,form 1, we get
√ /n ∼ n det A det B 1 2 3 3.464. 3.24 8 Journal of Inequalities and Applications
/ 2 / 2 m A 1 2B1/2 A1/2B 1 2 . Also, for 1, since tr 22 11 15 and tr 22 11 0 2, we get 2 2 1/2 1/2 A B1/2 A1/2B . . 3.25 tr 22 11 tr 22 11 15 2
Thus, according to this example from 3.24 and 3.25 ,weget 2 2 1/2 1/2 n A B 1/n ≤ A B1/2 A1/2B ≤ AB . 3.26 det det tr 22 11 tr 22 11 tr
Acknowledgment
This study was supported by the Coordinatorship of Selc¸uk University’s Scientific Research Projects BAP .
References
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