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 produ
Trang 1Volume 2010, Article ID 201486, 8 pages
doi:10.1155/2010/201486
Research Article
On Some Matrix Trace Inequalities
Z ¨ubeyde Uluk ¨ok and Ramazan T ¨urkmen
Department of Mathematics, Science Faculty, Selc¸uk University, 42003 Konya, Turkey
Correspondence should be addressed to Z ¨ubeyde Uluk ¨ok,zulukok@selcuk.edu.tr
Received 23 December 2009; Revised 4 March 2010; Accepted 14 March 2010
Academic Editor: Martin Bohner
Copyrightq 2010 Z Uluk¨ok and R T ¨urkmen 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
LetM m,n denote the space ofm × n complex matrices and write M n ≡ M n,n The identity matrix inM n is denotedI n As usual,A∗ A T denotes the conjugate transpose of matrix
A A matrix A ∈ M n is Hermitian ifA∗ A A Hermitian matrix A is said to be positive
semidefinite or nonnegative definite, written asA ≥ 0, if
A is further called positive definite, symbolized A > 0, if the strict inequality in 1.1 holds for all nonzerox ∈ C n An equivalent condition forA ∈ M nto be positive definite is thatA is
Hermitian and all eigenvalues ofA are positive real numbers Given a positive semidefinite
matrixA and p > 0, A pdenotes the unique positive semidefinitepth power of A.
Let A and B be two Hermitian matrices of the same size If A − B is positive
semidefinite, we write
Denoteλ1A, , λ n A and s1A, , s n A eigenvalues and singular values of matrix A,
respectively SinceA is Hermitian matrix, its eigenvalues are arranged in decreasing order,
that is,λ1A ≥ λ2A ≥ · · · ≥ λ n A and if A is any matrix, its singular values are arranged
in decreasing order, that is,s1A ≥ s2A ≥ · · · ≥ s n A > 0 The trace of a square matrix A
Trang 2the sum of its main diagonal entries, or, equivalently, the sum of its eigenvalues is denoted
by trA.
LetA be any m × n matrix The Frobenius Euclidean norm of matrix A is
A F
⎡
⎣m
i1
n
j1
a ij2
⎤
⎦
1/2
It is also equal to the square root of the matrix trace ofAA∗, that is,
A F trAA∗. 1.4
A norm · on M m,n is called unitarily invariantUAV A for all A ∈ M m,nand all unitaryU ∈ M m , V ∈ M n
Given two real vectorsx x1, , x n and y y1, , y n in decreasing order, we say thatx is weakly log majorized by y, denoted x ≺ w log y, if Π k
i1 x i ≤ Πk
i1 y i , k 1, 2, , n, and
we say thatx is weakly majorized by y, denoted x ≺ w y, if k i1 x i≤ k i1 y i , k 1, 2, , n We
sayx is majorized by y denoted by x ≺ y, if
x ≺ w y, n
i1
x in
i1
As is well known,x ≺ w log y yields x ≺ w y see, e.g., 1, pages 17–19
LetA be a square complex matrix partitioned as
A A11 A12
A21 A22
whereA11is a square submatrix ofA If A11is nonsingular, we call
A11 A22− A21A−1
the Schur complement ofA11 inA see, e.g., 2, page 175 If A is a positive definite matrix, thenA11is nonsingular and
Recently, Yang 3 proved two matrix trace inequalities for positive semidefinite matricesA ∈ M nandB ∈ M n,
0≤ tr AB2n ≤ tr A2 trA2n−1
trB2n
,
0≤ tr AB2n1 ≤ tr Atr B trA2n
trB2n
,
1.9
forn 1, 2,
Trang 3Also, authors in 4 proved the matrix trace inequality for positive semidefinite matricesA and B,
trAB m≤trA2mtrB2m1/2
wherem is a positive integer.
Furthermore, one of the results given in5 is
ndet A · det B m/n ≤ trA m B m 1.11 forA 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 ∈ M n , σA ◦ B≺ w σA ◦ σB.
Lemma 2.2 see, e.g., 7 Let A, B ∈ M m,n , then
t
i1
δ i AB2m ≤t
i1
λ iA∗ABB∗m
≤t
i1
λ iA∗A m BB∗m
, 1 ≤ t ≤ n, m ∈ N.
2.1
Lemma 2.3 Cauchy-Schwarz inequality Let a1, a2, , a n and b1, b2, , b n be real numbers Then,
n
i1
a i b i
2
≤ n
i1
a2
i n
i1
b2
i
, ∀a i , b i ∈ R. 2.2
Lemma 2.4 see, e.g., 8, page 269 If A and B are poitive semidefinite matrices, then,
0≤ trAB ≤ 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
i1
s i AB ≤k
i1
s i As i B 1 ≤ k ≤ n. 2.4
Lemma 2.6 see, e.g., 10 Let F and G are positive semidefinite matrices Then,
t
i1
λ m
i FG ≤t
i1
λ i F m G m , 1 ≤ t ≤ n, 2.5
where m is a positive integer.
Trang 43 Main Results
Horn and Mathias11 show that for any unitarily invariant norm · on M n
A∗B2≤ A∗AB∗B ∀A, B ∈ M m,n ,
A ◦ B2≤ A∗AB∗B ∀A, B ∈ M n 3.1
Also, the authors in12 show that for positive semidefinite matrix A L X
X∗M
, whereX ∈
M m,n
|X| p2
for allp > 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
F ≤A∗A m
F B∗B m
where m is a positive integer In particular, if A and B are positive semidefinite matrices, then
A ◦ B m2
F ≤A2m
F
B2m
Proof From definition of Frobenius norm, we write
A ◦ B m2
F trA ◦ B m A ◦ B m∗. 3.5 Also, for anyA and B, it follows that see, e.g., 13
AA∗◦ BB∗ A ◦ B
A∗◦ B∗ I
A ◦ BA ◦ B∗≤ AA∗◦ BB∗. 3.7 Since| tr A2m | ≤ trA m A∗m ≤ trAA∗m for A ∈ M nand from inequality3.7, we write
A ◦ B m2
F tr A ◦ B m A ◦ B m∗
≤ trA ◦ BA ◦ B∗m
≤ trAA∗◦ BB∗m.
3.8
Trang 5FromLemma 2.1and Cauchy-Schwarz inequality, we write
trAm ◦ B m n
i1
λ i A m ◦ B m ≤n
i1
λ i A m λ i B m
≤
n
i1
λ2
i A mn
i1
λ2
i B m
1/2
trA2mtrB2m1/2
.
3.9
By combining inequalities3.7, 3.8, and 3.9, we arrive at
tr
AA∗◦ BB∗m
≤trAA∗AA∗mtrBB∗BB∗m1/2
≤trAA∗AA∗mtrBB∗BB∗m1/2
trAA∗2m1/2
trBB∗2m1/2
A∗A m
F B∗B m
F
3.10
Thus, the proof is completed LetA and B be positive semidefinite matrices Then
A ◦ B m2
F ≤A2m
F
B2m
wherem > 0.
Theorem 3.2 Let A i ∈ M n i 1, 2, , k be positive semidefinite matrices For positive real
numbers s, m, t
k
i1
A st/2m i 2
F
2
≤ k
i1
A sm
i 2
F
k
i1
A tm
i 2
F
Proof Let
A
⎛
⎜
⎜
⎜
⎜
A S/21 0 · · · 0
0 A s/2
2 · · · 0
.
0 0 · · · A s/2 k
⎞
⎟
⎟
⎟
⎟, B
⎛
⎜
⎜
⎜
⎜
A t/21 0 · · · 0
0 A t/2
2 · · · 0
.
0 0 · · · A t/2 k
⎞
⎟
⎟
⎟
⎟. 3.13
Trang 6We know thatA, B ≥ 0, then by using the definition of Frobenius norm, we write
A ◦ B m2
F k
i1
A st/2m i 2
F ,
A2m
F!k
i1
A sm
i 2
F , B2m
F !k
i1
A tm
i 2
F
3.14
Thus, by usingTheorem 3.1, the desired is obtained
Now, we give a trace inequality for positive semidefinite block matrices
Theorem 3.3 Let
A A11 A12
A21 A22
≥ 0, B B11 B12
B21 B22
then,
tr A 221/2
B1/2
11
$2m
tr
#
A1/2
22 B11
1/2$2m
≤ tr AB m ≤ trA m B m , 3.16
where m is an integer.
Proof Let
M X 0
Y Z
3.17
withZ A1/2
22 , Y A −1/2
22 A21, X A11− A12A−1
22A211/2 ThenA M∗M see, e.g., 14 Let
K X 0
Y Z
3.18
withZ B22− B21B−1
11B121/2,Y B21B −1/2
11 ,X B1/2
11 ThenB KK∗ see, e.g., 14 We know that
M k X k 0
∗ Z k
,
M · K
⎡
⎣ A11− A12A−1
22A21
1/2
B1/2
A −1/222 A21B1/2
11 A1/2
22 B21B11−1/2 A1/2
22
B22− B21B−1
11B12
1/2
⎤
⎦,
Trang 7M · K2m
⎡
⎢ A11− A12A−1
22A21
1/2
B1/2
11
'2m
0
∗ &A1/2
22
B22− B21B−1
11B12
1/2'2m
⎤
⎥
⎦.
3.19
By usingLemma 2.2, it follows that
tr MK2m ≤n
i1
s i MK2m
≤n
i1
s i MK2m
n
i1
s2
i MKmn
i1
λ i
M∗MKK∗m
n
i1
λ iAB m
n
i1
trAB m≤n
i1
λ iM∗M m KK∗m
n
i1
λ iA m B m
n
i1
trAm B m .
3.20
Therefore, we get
tr MK2m tr A
11− A12A−1
22A21
1/2*
B1/2
11
$2m
tr
#
A1/2
22 B22− B21B−1
11B12
1/2$2m
≤ tr AB m ≤ trA m B m .
3.21
As result, we write
tr A 221/2
B1/2
11
$2m
tr
#
A1/2
22 B11
1/2$2m
≤ tr AB m ≤ trA m B m . 3.22
Example 3.4 Let
A 4 1
1 1
> 0, B 5 2
2 1
Then trAB 25, det A 3, det B 1 From inequality 1.11, for m 1, we get
ndet A det B1/n 2√3 ∼ 3.464. 3.24
Trang 8Also, form 1, since tr + A22
1/2
B1/2
11 2 15 and tr A1/2
22 B+111/2
2 0.2, we get
tr
) +
A22
1/2
B1/2
11
*2
tr
)
A1/2
22 B+111/2*2
Thus, according to this example from3.24 and 3.25, we get
ndet A det B1/n≤ tr
) +
A22
1/2
B1/2
11
*2
tr
)
A1/2
22 B+111/2*2
≤ trAB. 3.26
Acknowledgment
This study was supported by the Coordinatorship of Selc¸uk University’s Scientific Research ProjectsBAP
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