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Journal of Inequalities and Applications
Volume 2010, Article ID 201486, 8 pages
doi:10.1155/2010/201486
Research Article
On Some Matrix Trace Inequalities
Zübeyde Ulukök and Ramazan Türkmen
Department of Mathematics, Science Faculty, Selçuk University, 42003 Konya, Turkey
Correspondence should be addressed to Zübeyde Ulukök, [email protected]
Received 23 December 2009; Revised 4 March 2010; Accepted 14 March 2010
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 &times; 2 block matrices.
1. Introduction and Preliminaries
Let Mm,n denote the space of m &times; 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
x∗ Ax ≥ 0,
∀x ∈ Cn .
1.1
A is further called positive definite, symbolized A &gt; 0, if the strict inequality in 1.1 holds
for all nonzero x ∈ Cn . 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
matrix A and p &gt; 0, Ap 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
Denote λ1 A, . . . , λn A and s1 A, . . . , sn A eigenvalues and singular values of matrix A,
respectively. Since A is Hermitian matrix, its eigenvalues are arranged in decreasing order,
that is, λ1 A ≥ λ2 A ≥ &middot; &middot; &middot; ≥ λn A and if A is any matrix, its singular values are arranged
in decreasing order, that is, s1 A ≥ s2 A ≥ &middot; &middot; &middot; ≥ sn A &gt; 0. The trace of a square matrix A
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 &times; n matrix. The Frobenius Euclidean norm of matrix A is
⎤1/2
⎡
n m 2
aij ⎦ .
AF ⎣
1.3
i1 j1
It is also equal to the square root of the matrix trace of AA∗ , that is,
AF trAA∗ .
1.4
A norm &middot; on Mm,n is called unitarily invariant UAV A for all A ∈ Mm,n and all
unitary U ∈ Mm , V ∈ Mn .
Given two real vectors x x1 , . . . , xn and y y1 , . . . , yn in decreasing order, we say
that x is weakly log majorized by y, denoted x ≺w log y, if Πki1 xi ≤ Πki1 yi , k 1, 2, . . . , n, and
we say that x is weakly majorized by y, denoted x ≺w y, if ki1 xi ≤ ki1 yi , k 1, 2, . . . , n. We
say x is majorized by y denoted by x ≺ y, if
x ≺w y,
n
i1
xi n
yi .
i1
1.5
As is well known, x ≺w log y yields x ≺w y see, e.g., 1, pages 17–19.
Let A be a square complex matrix partitioned as
A
A11 A12
A21 A22
,
1.6
where A11 is a square submatrix of A. If A11 is nonsingular, we call
11 A22 − A21 A−1 A12
A
11
1.7
the Schur complement of A11 in A see, e.g., 2, page 175. If A is a positive definite matrix,
then A11 is nonsingular and
11 ≥ 0.
A22 ≥ A
1.8
Recently, Yang 3 proved two matrix trace inequalities for positive semidefinite
matrices A ∈ Mn and B ∈ Mn ,
n−1 n
tr B2 ,
0 ≤ tr AB2n ≤ tr A2 tr A2
n n
tr B2 ,
0 ≤ tr AB2n1 ≤ tr Atr B tr A2
for n 1, 2, . . ..
1.9
Journal of Inequalities and Applications
3
Also, authors in 4 proved the matrix trace inequality for positive semidefinite
matrices A and B,
1/2
,
tr ABm ≤ tr A2m tr B2m
1.10
where m is a positive integer.
Furthermore, one of the results given in 5 is
ndet A &middot; det Bm/n ≤ trAm Bm 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 , σA ◦ B≺w σA ◦ σB.
Lemma 2.2 see, e.g., 7. Let A, B ∈ Mm,n , then
t t
λi A∗ ABB∗ m
δi AB2m ≤
i1
i1
t
≤
λi A∗ Am BB∗ m ,
2.1
1 ≤ t ≤ n, m ∈ N.
i1
Lemma 2.3 Cauchy-Schwarz inequality. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be real numbers.
Then,
n
2
ai bi
i1
n
n
2
2
≤
ai
bi ,
i1
∀ai , bi ∈ R.
2.2
i1
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 &times; n matrices. Then,
k
si AB ≤
i1
k
si Asi B
1 ≤ k ≤ n.
2.4
i1
Lemma 2.6 see, e.g., 10. Let F and G are positive semidefinite matrices. Then,
t
i1
where m is a positive integer.
λm
i FG ≤
t
λi F m Gm ,
i1
1 ≤ t ≤ n,
2.5
4
Journal of Inequalities and Applications
3. Main Results
Horn and Mathias 11 show that for any unitarily invariant norm &middot; on Mn
A∗ B2 ≤ A∗ AB∗ B
∀A, B ∈ Mm,n ,
3.1
A ◦ B2 ≤ A∗ AB∗ B ∀A, B ∈ Mn .
Also, the authors in 12 show that for positive semidefinite matrix A Mm,n
L X
X∗ M
, where X ∈
p 2
|X| ≤ Lp Mp 3.2
for all p &gt; 0 and all unitarily invariant norms &middot; .
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 ◦ Bm 2 ≤ A∗ Am B∗ Bm ,
F
F
F
3.3
where m is a positive integer. In particular, if A and B are positive semidefinite matrices, then
2m 2m A ◦ Bm 2 ≤ B
.
A
F
F
F
3.4
Proof. From definition of Frobenius norm, we write
A ◦ Bm 2 tr A ◦ Bm A ◦ Bm∗ .
F
3.5
Also, for any A and B, it follows that see, e.g., 13
AA∗ ◦ BB∗ A ◦ B
A∗ ◦ B ∗
I
≥ 0,
A ◦ BA ◦ B∗ ≤ AA∗ ◦ BB∗ .
3.6
3.7
Since | tr A2m | ≤ trAm A∗ m ≤ trAA∗ m for A ∈ Mn and from inequality 3.7, we write
A ◦ Bm 2 tr A ◦ Bm A ◦ Bm∗
F
m ≤ tr A ◦ BA ◦ B∗
≤ tr AA∗ ◦ BB∗ m .
3.8
Journal of Inequalities and Applications
5
From Lemma 2.1 and Cauchy-Schwarz inequality, we write
trAm ◦ Bm n
n
λi Am ◦ Bm ≤
λi Am λi Bm i1
≤
i1
n
i1
λ2i Am n
i1
1/2
3.9
λ2i Bm 1/2
tr A2m tr B2m
.
By combining inequalities 3.7, 3.8, and 3.9, we arrive at
1/2
tr AA∗ ◦ BB∗ m ≤ tr AA∗ AA∗ m tr BB∗ BB∗ m
1/2
≤ tr AA∗ AA∗ m tr BB∗ BB∗ m
3.10
1/2 1/2
tr BB∗ 2m
tr AA∗ 2m
A∗ Am F B∗ Bm F .
Thus, the proof is completed. Let A and B be positive semidefinite matrices. Then
A ◦ Bm 2 ≤ A2m B2m ,
F
F
3.11
F
where m &gt; 0.
Theorem 3.2. Let Ai ∈ Mn i 1, 2, . . . , k be positive semidefinite matrices. For positive real
numbers s, m, t
k k k 2
2
2
st/2m 2
sm
tm
A A .
≤
Ai
i
i
F
F
F
i1
i1
3.12
i1
Proof. Let
⎛
⎜
⎜
⎜
⎜
A⎜
⎜
⎜
⎝
AS/2
1
0
0
⎞
⎛
⎟
⎟
0 ⎟
⎟
,
.. ⎟
..
⎟
.
. ⎟
⎠
&middot; &middot; &middot; As/2
k
⎜
⎜
⎜
⎜
B⎜
⎜
⎜
⎝
&middot;&middot;&middot;
&middot;&middot;&middot;
As/2
2
..
.
..
.
0
0
0
At/2
1
0
0
&middot;&middot;&middot;
..
.
0
0
⎞
⎟
⎟
0 ⎟
⎟
.
.. ⎟
..
⎟
. . ⎟
⎠
&middot; &middot; &middot; At/2
k
&middot;&middot;&middot;
At/2
2
..
.
0
3.13
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Journal of Inequalities and Applications
We know that A, B ≥ 0, then by using the definition of Frobenius norm, we write
k st/2m 2
A ◦ Bm 2 Ai
,
F
F
i1
! k
!
2
2m ,
A &quot; Asm
i
F
F
3.14
! k
!
2
2m .
B &quot; Atm
i
F
F
i1
i1
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
A11 A12
A21 A22
≥ 0,
B11 B12
B
≥ 0,
B21 B22
3.15
then,
tr
#
22
A
1/2
1/2
B11
\$2m
#
1/2 \$2m
tr A1/2
≤ tr ABm ≤ trAm Bm ,
B11
22
3.16
where m is an integer.
Proof. Let
M
X 0
3.17
Y Z
1/2
−1/2
−1
with Z A1/2
. Then A M∗ M see, e.g., 14. Let
22 , Y A22 A21 , X A11 − A12 A22 A21 X 0
K
3.18
Y Z
−1/2
1/2
−1
with Z B22 − B21 B11
B12 1/2 , Y B21 B11
, X B11
. Then B KK ∗ see, e.g., 14. We
know that
k
M ⎡
M&middot;K ⎣
A11 − A12 A−1
22 A21
Xk
0
∗
Zk
1/2 1/2
B11
,
0
⎤
⎦
1/2 ,
1/2
1/2
−1/2
1/2 −1
A−1/2
A
B
A
B
B
A
−
B
B
B
B
21
21
22
21
12
22
22
22
11
11
11
Journal of Inequalities and Applications
⎡&amp;
⎢
M &middot; K2m ⎣
7
A11 − A12 A−1
22 A21
1/2 1/2
B11
'2m
∗
0
⎤
⎥
&amp;
'2m ⎦.
1/2
−1
A1/2
22 B22 − B21 B11 B12
3.19
By using Lemma 2.2, it follows that
n
n
si MK2m ≤
si MK2m
tr MK2m ≤
i1
i1
n i1
s2i MK
m
n
λi M∗ MKK ∗ m
i1
n
n
n
λi ABm tr ABm ≤
λi M∗ Mm KK ∗ m
i1
i1
3.20
i1
n
n
λi Am Bm trAm Bm .
i1
i1
Therefore, we get
#)
#
\$2m
1/2 *
1/2 \$2m
1/2
1/2
−1
B
A11 − A12 A−1
A
tr
A
−
B
B
B
B
tr MK2m tr
22
21 11 12
22 21
22
11
≤ tr ABm ≤ trAm Bm .
3.21
As result, we write
tr
#
22
A
1/2
1/2
B11
\$2m
#
1/2 \$2m
1/2 tr A22 B11
≤ tr ABm ≤ trAm Bm .
3.22
Example 3.4. Let
A
4 1
1 1
&gt; 0,
B
5 2
2 1
&gt; 0.
3.23
Then tr AB 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
8
Journal of Inequalities and Applications
2
2
1/2 1/2
+
+1/2 0.2, we get
Also, for m 1, since tr A
B11 15 and tr A1/2
22
22 B11
)
)
*2
*2
1/2 1/2
1/2
+
+
tr A
B11
tr A1/2
15.2.
B
22
11
22
3.25
Thus, according to this example from 3.24 and 3.25, we get
)
)
*2
*2
1/2 1/2
1/2 + 1/2
+
ndet A det B1/n ≤ tr A
B
tr
A
≤ trAB.
B
22
11
22
11
3.26
Acknowledgment
This study was supported by the Coordinatorship of Selçuk University’s Scientific Research
Projects BAP.
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