More results on singular value inequalities of matrices

Linear Algebra and its Applications 416 (2006) 724–729 www.elsevier.com/locate/laa

More results on singular value inequalities of matrices Yunxing Tao ∗ School of Mathematics, Beijing Normal University, Beijing 100875, China Division of Mathematics, Department of Fundamental Sciences, The Academy of Armoured Forces Engineering, PLA, Beijing 100072, China Received 22 November 2005; accepted 18 December 2005 Available online 23 February 2006 Submitted by H. Schneider

Abstract The arithmetic–geometric mean inequality for singular values due to Bhatia and Kittaneh says that 2sj (AB ∗ )
sj (A∗ A + B ∗ B),

j = 1, 2, . . .

for any matrices A, B. We give a new equivalent form and some relevant generalizations of this inequality. In particular, we show that
1 3  3 1 sj A 4 B 4 + A 4 B 4
sj (A + B), j = 1, . . . , n for any n×n positive semidefinite matrices A, B, which proves a special case of Zhan’s conjecture posed in 2000. © 2006 Elsevier Inc. All rights reserved. AMS classification: 15A60; 15A42; 15A18 Keywords: Singular value; Arithmetic–geometric mean; Positive semidefinite matrix

∗ Present address: Division of Mathematics, Department of Fundamental Sciences, The Academy of Armoured Forces Engineering, PLA, Beijing 100072, China. E-mail address: [email protected]

0024-3795/$ - see front matter ( 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2005.12.017

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1. Introduction In conventional symbols, let Mm,n denote the spaces of m × n complex matrices and write Mn ≡ Mn,n which has the identity matrix In . Given any A ∈ Mm,n , let A∗ denote the conjugate 1 transpose of A, we define |A| ≡ (A∗ A) 2 , denote by sj (A) the j th largest singular value of A, j = 1, . . . , min{m, n}. Note that the singular values of the matrix A ∈ Mnare the  eigenvalues of A 0 |A|. In addition, the direct sum A ⊕ B denotes the block diagonal matrix . 0 B For Hermitian matrices A, B ∈ Mn , we write A  B to mean A − B is positive semidefinite, particularly, A  0 indicates that A is positive semidefinite. Likewise, if A is positive definite, we write A > 0. If A ∈ Mn is Hermitian, we always denote its eigenvalues in decreasing order by λ1 (A)  λ2 (A)  · · ·  λn (A). Let λ(A) and s(A) denote the n-vectors whose coordinates are the eigenvalues and singular values of A arranged in decreasing order, respectively. Weyl’s monotonicity principle [6] says A  B implies λj (A)  λj (B), j = 1, . . . , n. Let A ∈ Mn be a Hermitian matrix with eigenvalues λ1 (A)  · · ·  λp (A)  0 > λp+1 (A)  · · ·  λn (A). Choose a unitary matrix U ∈ Mn such that A = U DU ∗ with the diagonal matrix D = diag(λ1 (A), . . . , λn (A)). Let D + = diag(λ1 (A), . . . , λp (A), 0, . . . , 0) and D − = diag(0, . . . , 0, −λp+1 (A), . . . , −λn (A)). Let A+ = U D + U ∗ , A− = U D − U ∗ . Then A+ and A− are positive semidefinite and A = A+ − A− . This is called the Jordan decomposition of A (see [6, p. 99]). The well-known arithmetic–geometric mean inequalities for singular values due to Bhatia and Kittaneh [7] (see also [6]) says that 2sj (AB ∗ )
sj (A∗ A + B ∗ B),

j = 1, 2, . . . , n

(1)

for any A, B ∈ Mn . On the other hand, Zhan has proved in [3] sj (A − B)
sj (A ⊕ B),

j = 1, 2, . . . , n

(2)

for positive semidefinite A, B ∈ Mn . It is pointed out in [1, p. 37] that the two inequalities (1) and (2) are equivalent. We will give a new equivalent form of the two inequalities:   M K 2sj (K)
sj , j = 1, . . . , r K∗ N   M K for any positive semidefinite block matrix , where M ∈ Mm , N ∈ Mn , r ≡ min{m, n}. K∗ N Some related generalizations of these inequalities are discussed. 2. Main results We first give the following result with two different illustrations, then we point out that it is equivalent to inequalities (1) and (2). 

M Theorem 1. Given any positive semidefinite block matrix K∗ r ≡ min{m, n}. We have   M K 2sj (K)
sj , j = 1, . . . , r. K∗ N

 K , where M ∈ Mm , N ∈ Mn , N

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 0 K Proof 1. Let Q ≡ . We employ the arguments similar to the ones in [2] as follows: K∗ 0         M K 0 0 M K Im M −K Im = − 2Q, = 0
N K∗ N 0 −In K ∗ N 0 −In −K ∗   M K . By Weyl’s monotonicity principle, see [6], thus 2Q
K∗ N     0 K M K 2λj = 2λj (Q)
λj , j = 1, . . . , m + n. K∗ 0 K∗ N m+n−2r

   Note that λ(Q) = (s1 (K), . . . , sr (K), 0, . . . , 0, −sr (K), . . . , −s1 (K))T . Therefore, we obtain the desired result   M K , j = 1, . . . , r.  2sj (K)
sj K∗ N  I Proof 2. Let U = m 0  M UP U∗ = −K ∗

   0 M K , P = . Then, −In K∗ N    1 0 K −K , Q≡ = (P − U P U ∗ ). N K∗ 0 2

Applying the Jordan decomposition of P and Lemma IX.4.1 in [6, p. 262], we obtain 1 λj (P ), 2 1 1 λj (Q− )
λj (U P U ∗ ) = λj (P ), 2 2 λj (Q+ )


j = 1, . . . , r.

m+n−2r

   Note that λ(Q) = (s1 (K), . . . , sr (K), 0, . . . , 0, −sr (K), . . . , −s1 (K))T such that λj (Q+ ) = λj (Q− ) = sj (K), j = 1, . . . , r. Hence, we have   M K 2sj (K)
sj , j = 1, . . . , r.  K∗ N Theorem 2. The following statements are equivalent: (i) Let A, B ∈ Mn be positive semidefinite. Then sj (A − B)
sj (A ⊕ B),

j = 1, 2, . . . , n.

(ii) For any X, Y ∈ Mn , 2sj (XY ∗ )
sj (X ∗ X + Y ∗ Y ),

j = 1, . . . , n.  M (iii) Given any positive semidefinite block matrix K∗   M K , j = 1, . . . , n. 2sj (K)
sj K∗ N

 K , where M, N ∈ Mn . We have N

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Proof. (i) ⇒ (ii). We in [1, p. 37] with minor modifications. For any  cite  the demonstration   X X X, Y ∈ Mn , let C = ,D= . According to (i), Y −Y   ∗   0 YX 0 XY ∗ 2sj = sj (CC ∗ − DD ∗ ) = 2s j 0 0 XY ∗ Y X∗     ∗ 0 0 CC ∗ C C
sj = s j 0 DD ∗ 0 D∗D  ∗  0 X X + Y ∗Y = sj . 0 X∗ X + Y ∗ Y ∗ ∗ Thus, we have 2sj (XY∗ )
sj (X  X + Y Y ), j = 1, . . . , n. M K (ii) ⇒ (iii). Since  0, there must exist S, T ∈ M2n,n such that K∗ N     ∗ M K S S S∗T ∗ (S, T ) (S, T ) = =  0. T ∗S T ∗T K∗ N

The inequality (ii) obviously holds also for rectangular matrices X, Y . By this version of (ii) we have  ∗    S S S∗T M K 2sj (K) = 2sj (S ∗ T )
sj (SS ∗ + T T ∗ ) = sj = s , j T ∗S T ∗T K∗ N j = 1, . . . , n. (iii) ⇒ (i). For any positive semidefinite matrices A, B ∈ Mn , note the following unitary similarity transform:   A+B A−B

    1 I −I 1 I I A 0 2 2 =  0. √ √ A−B A+B I 0 B 2 −I I 2 I 2 2 Now the desired result follows from (iii) A+B A−B

2 A−B 2

sj (A − B)
sj

2 A+B 2



= sj

A 0

 0 , B

j = 1, . . . , n.



Theorem 3. Let A, B ∈ Mn be positive semidefinite and m be a positive integer. Then,
1  1 2sj A 2 (A + B)m−1 B 2
sj ((A + B)m ), j = 1, . . . , n,
1 3  3 1 sj A 4 B 4 + A 4 B 4
sj (A + B), j = 1, . . . , n.

1 A2 0 Proof. Let X = . Then, 1 B2 0   (A + B)m 0 ∗ m , (X X) = 0 0

∗ m

∗

(XX ) = X(X X)

m−1

1

1

A 2 (A + B)m−1 A 2 X = 1 1 B 2 (A + B)m−1 A 2 ∗

1 1 A 2 (A + B)m−1 B 2 . 1 1 B 2 (A + B)m−1 B 2

(3) (4)

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It follows from Theorem 2(iii) that
1  1 2sj A 2 (A + B)m−1 B 2
sj ((XX ∗ )m ) = sj ((X ∗ X)m ) = sj ((A + B)m ), j = 1, . . . , n. When m = 2, then
1 3  3 1 2sj A 2 B 2 + A 2 B 2
sj ((A + B)2 ),

j = 1, . . . , n.

(5)

Since 0
A2 + B 2 − AB − BA = (A − B)2 , we derive (A + B)2
2(A2 + B 2 ). Hence,
1 3  3 1 2sj A 2 B 2 + A 2 B 2
sj ((A + B)2 )
2sj (A2 + B 2 ). 1

1

Now let A 2 , B 2 substitute A, B, respectively. Then,
1 3  3 1 sj A 4 B 4 + A 4 B 4
sj (A + B), j = 1, . . . , n.



Remark. In Theorem 3, inequalities (3) generalize inequalities (5) which are proved by Bhatia and Kittaneh in [8]. In [4] Zhan posed the following conjecture (see also [1, p. 30]): Let A, B ∈ Mn be positive semidefinite and 0
r
1. Then, sj (Ar B 1−r + A1−r B r )
sj (A + B),

j = 1, . . . , n.

Inequalities (4) indicate that the case of r = 41 of the conjecture is true.   A 0 In the proof of inequalities (3), let X = and apply Theorem 2(iii), similarly we derive B 0 the following result. Corollary 4. Let A, B ∈ Mn be any two matrices, m be a positive integer. Then, 2sj (A(|A|2 + |B|2 )m−1 B ∗ )
sj ((|A|2 + |B|2 )m ),

j = 1, . . . , n.

The conclusion (i) in Lemma 5 below was proved in [5, Theorem 2.2], the other conclusion (ii) can be verified similarly. Lemma 5. Let A, B ∈ Mn be positive semidefinite matrices and let f, g be nonnegative functions on [0, ∞) which are continuous and satisfy the relation f (t)g(t) = t for all t ∈ [0, ∞). Then for j = 1, . . . , n (i) 2sj (A + B)  2  f (A) + g 2 (A) + |g(B)f (A) + f (B)g(A)| 0
sj , 0 f 2 (B) + g 2 (B) + |g(A)f (B) + f (A)g(B)|  (ii) 2sj

A g(B)g(A)

   f (A)f (B) |f (A)|2 + |g(A)|2 + |g(B)|2 + |f (B)|2 0
sj . B 0 0

Using this we get the following result.

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Theorem 6. Let A, B ∈ Mn be positive semidefinite. Then, 1 1



1 1 1 1 2B2 1 1 A2 B 2 A2 B 2 A A 2 2 sj 1 1 1 1 = 2sj (A B )
sj 1 1 A2 B 2 A2 B 2 B 2 A2 B

1 1 A + |B 2 A 2 | 0
sj (A + B)
sj 1 1 0 B + |A 2 B 2 | for j = 1, . . . , n. Proof. For the equality identities, it is sufficient to verify that        1 I I A B 1 I −I A+B 0 = . √ √ I 0 A−B 2 −I I B A 2 I 1 1     A B A+B 0 A2 B 2 That is to say, and are unitarily similar. Likewise, 1 1 B A 0 A−B A2 B 2  1 1  2 2 0 are unitarily similar. Applying Theorem 2(iii) and note that and 2A B 0 0

1 1 1 1 1 1 A A2 B 2 = [A 2 , B 2 ]∗ [A 2 , B 2 ]  0, 1 1 2 2 B A B

1

1

A2 B 2 1 1 A2 B 2



we derive the first inequalities. For the second inequalities, use Lemma 5(ii) to the nonnegative 1 functions f (t) = g(t) = t 2 . Similarly, we obtain the last inequalities from Lemma 5(i) with 1 f (t) = g(t) = t 2 . 

Acknowledgements I am grateful to Professor X. Zhan for his encouragement and the referee for careful reading and helpful suggestions. I also thank Professor Xiuping Zhang for three-year-long instruction and fostering as my supervisor. References [1] X. Zhan, Matrix Inequalities, LNM1790, Springer-Verlag, Berlin, 2002. [2] X. Zhan, On some matrix inequalities, Linear Algebra Appl. 376 (2004) 299–303. [3] X. Zhan, Singular values of differences of positive semidefinite matrices, SIAM J. Matrix Anal. Appl. 22 (3) (2000) 819–823. [4] X. Zhan, Some research problems on the Hadamard product and singular values of matrices, Linear and Multilinear Algebra 47 (2000) 191–194. [5] O. Hirzallah, Inequalities for sums and products of operators, Linear Algebra Appl. 407 (2005) 32–42. [6] R. Bhatia, Matrix Analysis, GTM169, Springer-Verlag, New York, 1997. [7] R. Bhatia, F. Kittaneh, On the singular values of a product of operators, SIAM J. Matrix Anal. Appl. 11 (1990) 272–277. [8] R. Bhatia, F. Kittaneh, Notes on matrix arithmetic–geometric mean inequalities, Linear Algebra Appl. 308 (2000) 203–211.