On some generalizations of the Brunn-Minkowski inequality

Linear Algebra and its Applications 586 (2020) 103–110

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Linear Algebra and its Applications www.elsevier.com/locate/laa

On some generalizations of the Brunn-Minkowski inequality Yanpeng Zheng a , Xiaoyu Jiang a,∗ , Xiaoting Chen b , Fawaz Alsaadi c a

Key Laboratory of Complex Systems and Intelligent Computing in Universities of Shandong, Linyi University, Linyi 276005, China b School of Statistics, Qufu Normal University, Qufu 273165, China c Department of Information Technology, King Abdulaziz University, Jeddah 21589, Saudi Arabia

a r t i c l e

i n f o

Article history: Received 15 August 2019 Accepted 24 October 2019 Available online 25 October 2019 Submitted by P. Semrl

a b s t r a c t Recently, Liu (2016) [9] obtained some determinantal inequalities for sector matrices. In this paper, we make considerable improvement on Liu’s results. © 2019 Elsevier Inc. All rights reserved.

MSC: 15A45 15A60 Keywords: Determinant inequality Schur complement Numerical range Sector

* Corresponding author. E-mail addresses: [email protected] (Y. Zheng), [email protected] (X. Jiang), [email protected] (X. Chen), [email protected] (F. Alsaadi). https://doi.org/10.1016/j.laa.2019.10.022 0024-3795/© 2019 Elsevier Inc. All rights reserved.

Y. Zheng et al. / Linear Algebra and its Applications 586 (2020) 103–110

104

1. Introduction Denoted by Mn the set of n × n complex matrices. The famous matrix form of the Brunn-Minkowski inequality (e.g., [4, Theorem 7.8.21]) states that if A, B ∈ Mn are positive definite, then (det(A + B))1/n ≥ (det A)1/n + (det B)1/n .

(1)

In the last 50s, Fan [1] gave a generalization of (1). He obtained the following imaginative result. Theorem 1.1. Let A, B ∈ Mn be positive definite and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. Then for each k = 1, . . . , n, 

det(A + B) det(Ak + Bk )

1  n−k

 ≥

det A det Ak

1  n−k

 +

det B det Bk

1  n−k

.

(2)

In [10], Yuan and Leng gave an extension of the inequality (2). They proved the following matrix form of the Brunn-Minkowski inequality. Theorem 1.2. [10, Theorem 1.1] Let A, B ∈ Mn be positive definite and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. If a and b are two nonnegative real numbers such that A > aIn and B > bIn , then 

1  n−k

det(A + B) − det((a + b)In−k ) det(Ak + Bk )

 ≥

1  n−k det A − det(aIn−k ) det Ak 1   n−k det B + − det(bIn−k ) . det Bk

Recently, Liu [9] obtained generalizations of Theorem 1.1 and Theorem 1.2 to a larger class of matrices, namely, matrices whose numerical range is contained in a sector. This direction of research has been carried out by a number of authors and seems fruitful; see [6–8,3,2,11]. Recall that the numerical range (also known as the field of values) of A ∈ Mn is defined by W (A) = {x∗ Ax : x ∈ C n , x∗ x = 1}. Also, we define a sector on the complex plane: for some fixed angle α ∈ [0, π/2) Sα = {z ∈ C : z > 0, |z| ≤ (z) tan α}. Clearly, if A is positive definite, then W (A) ⊂ S0 . As 0 ∈ / Sα , if W (A) ⊂ Sα , then A is necessarily nonsingular.

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The main results of [9] are the following two theorems. Theorem 1.3. [9, Theorem 3.1] Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. Then   1    1  1   det A  n−k  det B  n−k  det(A + B)  n−k n+k       n−k ≥ (cos α) + .  det Ak   det(Ak + Bk )  det Bk  Theorem 1.4. [9, Theorem 3.3] Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. If a and b are two nonnegative real numbers such that A > aIn and B > bIn , then (sec α) ≥

n+k n−k

1    n−k  det(A + B)    − (cos α)k det((a + b)In−k )  det(Ak + Bk ) 

   1  1    det A  det(aInk ) n−k  det B  det(bInk ) n−k  −  − + .  det Ak   det Bk  (cos α)n (cos α)n

Instead of extending more determinantal inequalities from positive definite matrices to sector matrices, the goal of this paper is to improve and complement the previous two theorems. 2. Preliminaries In this section, we present some lemmas that we use to facilitate the proofs in the next section. For A ∈ Mn , the conjugate transpose of A is denoted by A∗ , the real (or Hermitian) part of A is denoted by A = 12 (A + A∗ ) and the imaginary part of A is denoted by 1 A = 2i (A − A∗ ). For two Hermitian matrices A, B ∈ Mn , we write A > B to mean that A − B is positive definite. We also consider A ∈ Mn to be partitioned as 

A11 A= A21

 A12 , A22

(3)

where diagonal blocks are square matrices. If A11 is nonsingular, then the Schur complement of A11 in A is defined as A/A11 = A22 − A21 A−1 11 A12 . For our purpose, it is important to notice the basic equality det A/A11 =

det A . det A11

Lemma 2.1. [5, Proposition 2.1] Let A ∈ Mn be partitioned as in (3). If W (A) ⊂ Sα , then W (A/A11 ) ⊂ Sα .

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Lemma 2.2. [5, Lemma 2.5] Let A ∈ Mn be partitioned as in (3). If A is positive definite, then (A/A11 ) ≥ (A)/(A11 ). The next lemma is known as the Ostrowski-Taussky inequality in the literature (see [4, p. 510]). Lemma 2.3. Let A ∈ Mn with A being positive definite. Then det A ≤ | det A|. A reverse of the Ostrowski-Taussky inequality was proved by Lin [5]. Lemma 2.4. [5, Lemma 2.6] Let A ∈ Mn with W (A) ⊂ Sα . Then (sec α)n det A ≥ | det A|. The last lemma gives a fundamental relation between the Schur complement of the matrix sum and the sum of the Schur complements. Lemma 2.5. [5, Theorem 3.1] Let A, B ∈ Mn with W (A), W (B) ⊂ Sα be comformally partitioned as in (3). Then

sec2 (α) (A + B)/(A11 + B11 ) ≥ (A/A11 ) + (B/B11 ). 3. Main results The next theorem provides a considerable improvement of Theorem 1.3 as far as the coefficient is concerned. Theorem 3.1. Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. Then   1    1  1   det A  n−k  det B  n−k  det(A + B)  n−k n     ≥ (cos α) n−k  +  .  det(Ak + Bk )  det Ak  det Bk 

(4)

Proof. Let C = A + B. Clearly, W (C) ⊂ Sα . By Lemma 2.1, W (C/Ck ) ⊂ Sα , in particular, (C/Ck ) is positive definite. By Lemma 2.3, | det(C/Ck )| ≥ det (C/Ck ). Moreover, since (C/Ck ) ≥ (C)/(Ck ) by Lemma 2.2, we have

(5)

Y. Zheng et al. / Linear Algebra and its Applications 586 (2020) 103–110



det (C/Ck ) ≥ det (C)/(Ck ) .

107

(6)

Compute   1  det C  n−k 1   = |det(C/Ck )| n−k  det Ck  ≥ ≥



 =

1

n−k

det (C/Ck )

1

n−k det (C)/(Ck )

det(C) det(Ck )

1  n−k

 1 det(B) n−k ≥ + det(Bk ) 1   1   (cos α)n | det A| n−k (cos α)n | det B| n−k ≥ + det(Ak ) det(Bk )  1 1   n−k  n−k  n | det A| | det B| = (cos α) n−k + det(Ak ) det(Bk )    1  1   det A  n−k  det B  n−k n   ≥ (cos α) n−k  +  , det Ak  det Bk  

det(A) det(Ak )

1  n−k



in which the first inequality is by (5), the second inequality is by (6), the third inequality is by the Brunn-Minkowski inequality (1), the fourth inequality is by Lemma 2.4, and the last inequality is by Lemma 2.3, respectively. 2 Remark 3.2. From the proof of previous theorem, we have   1  1   det C  n−k det(C) n−k   ≥ .  det Ck  det(Ck )

(7)

However, in Liu’s proof of [9, Theorem 3.1], he did not apply the Schur complement so he only obtained a weaker result   1   1  det C  n−k det(C) n−k k   ≥ (cos α) .  det Ck  det(Ck ) Theorem 3.3. Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. Then  1 1   1   n−k   n−k  det(A + B)  n−k     det A det B     ≥ (cos α)3  +  .  det(Ak + Bk )  det Ak  det Bk 

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Proof. Let C = A + B. By Lemma 2.5, we have

det (C/Ck ) ≥ det (cos α)2 (A/Ak ) + (B/Bk ) . This gives

1

n−k

det (C/Ck )

1

n−k ≥ (cos α)2 det (A/Ak ) + (B/Bk ) .

(8)

Now compute   1  det C  n−k 1   = |det(C/Ck )| n−k  det Ck  ≥



det (C/Ck )

1

n−k

1

n−k ≥ (cos α)2 det (A/Ak ) + (B/Bk )

≥ (cos α)2



1

n−k

det (A/Ak )

1



n−k + det (B/Bk )

1

1

≥ (cos α)2 (cos α| det(A/Ak )| n−k + cos α| det(B/Bk )| n−k  1  n−k   1     det B  n−k 3  det A   = (cos α)  +  , det Ak  det Bk 



in which the first inequality is by (5), the second inequality is by (8), the third inequality is by (1), and the last inequality is by Lemma 2.4, respectively. 2 n

3 n−k . Since 0 < cos α ≤ 1 for α ∈ [0, π/2], when k ≥ 2n 3 , and we have (cos α) ≥ (cos α) That is, the inequality in Theorem 3.3 is stronger than that in Theorem 3.1 when k ≥ 2n 3 and weaker when 1 ≤ k < 2n . 3 The next theorem provides a considerable improvement of Theorem 1.4.

Theorem 3.4. Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the kth leading principal submatrix of A, B, respectively. If a and b are two nonnegative real numbers such that A > aIn and B > bIn , then

(sec α)

n n−k

1    n−k  det(A + B)   − det((a + b)In−k )   det(Ak + Bk ) 

   1  1    det A  det(aInk ) n−k  det B  det(bInk ) n−k     − − ≥  +  . det Ak  (cos α)n det Bk  (cos α)n

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Remark 3.5. Compared with Theorem 1.4, note that we have also dropped the coefficient (cos α)k of det((a + b)In−k ) on the left side of the inequality, which makes our result even stronger. Proof. Let C = A + B. By (7),    det C  det(C)    det Ck  ≥ det(Ck ) . Now by Compute 1    n−k  det C    − det((a + b)In−k )  det Ck 



1  n−k det C ≥ − det((a + b)In−k ) det Ck 1 1   n−k   n−k det A det B ≥ − det(aIn−k ) + − det(bIn−k ) det Ak det Bk   1   1 n n | det A| | det B| det(aIn−k ) n−k det(bIn−k ) n−k n−k n−k ≥ (cos α) − + (cos α) − det Ak (cos α)n det Bk (cos α)n     1   1  det A  det(aInk ) n−k  det B  det(bInk ) n−k n n  − − n−k ≥ (cos α) n−k  + (cos α) ,  det Bk  det Ak  (cos α)n (cos α)n

in which the second inequality is by Theorem 1.2, the third inequality is by Lemma 2.4, and the fourth inequality is by Lemma 2.3, respectively. 2 We call A ∈ Mn an accretive-dissipative matrix if both A and A are positive definite. Similar to the corollaries in [9], one could also present some results for accretivedissipative matrices, we omit the details. Declaration of competing interest There is no competing interest. Acknowledgement This work was supported by the National Natural Science Foundation of China (No. 11671187), Natural Science Foundation of Shandong Province (Nos. ZR2016AM14 and 2019GGX101006 ) and the Ph.D. Research Foundation of Linyi University (Nos. LYDX2018BS067 and LYDX2018BS052).

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References [1] K. Fan, Some inequalities concerning positive-definite Hermitian matrices, Proc. Camb. Philos. Soc. 51 (1958) 414–421. [2] S. Dong, L. Hou, A complement of the Hadamard-Fischer inequality, J. Intell. Fuzzy Systems 35 (2018) 4011–4015. [3] X. Jiang, Y. Zheng, X. Chen, Extending a refinement of Koteljanski˘ıs inequality, Linear Algebra Appl. 574 (2019) 252–261. [4] R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 2013. [5] M. Lin, Extension of a result of Hanynsworth and Hartfiel, Arch. Math. 1 (2015) 93–100. [6] M. Lin, Some inequalities for sector matrices, Oper. Matrices 10 (2016) 915–921. [7] D. Choi, T.-Y. Tam, P. Zhang, Extensions of Fischer’s inequality, Linear Algebra Appl. 569 (2019) 311–322. [8] X. Fu, Y. Liu, S. Liu, Extension of determinantal inequalities of positive definite matrices, J. Math. Inequal. 11 (2017) 355–359. [9] J. Liu, Generalizations of the Brunn-Minkowski inequality, Linear Algebra Appl. 508 (2016) 206–213. [10] J. Yuan, G. Leng, A generalization of the matrix form of the Brunn-Minkowski inequality, J. Aust. Math. Soc. 83 (2007) 125–134. [11] P. Zhang, Extension of Matic’s results, Linear Algebra Appl. 486 (2015) 328–334.