aa r X i v : . [ m a t h . F A ] O c t EXTENSION OF A RESULT OF HAYNSWORTH ANDHARTFIEL
MINGHUA LIN
Abstract.
About last 70s, Haynsworth [6] used a result of the Schurcomplement to refine a determinant inequality for positive definite matri-ces. Haynsworth’s result was improved by Hartfiel [5]. We extend theirresult to a larger class of matrices, namely, matrices whose numericalrange is contained in a sector. Our proof relies on a number of newrelations for the Schur complement of this class of matrices. Introduction
We start with the notation used in this paper. Let M n be the set ofall n × n complex matrices. For A ∈ M n , the conjugate transpose of A isdenoted by A ∗ , the real and imaginary part of A are in the sense of theCartesian decomposition and they are denoted by ℜ A = ( A + A ∗ ) and ℑ A = i ( A − A ∗ ), respectively. For two Hermitian matrices A, B ∈ M n , wewrite A ≥ B (or B ≤ A ) to mean that A − B is positive semidefinite. Wealso consider A ∈ M n to be partitioned as A = (cid:20) A A A A (cid:21) , (1.1)where diagonal blocks are square matrices. If A is nonsingular, then wepartition A − conformally as A . If A is nonsingular, then the Schur com-plement of A in A is defined by A/A = A − A A − A . The term“Schur complement” and the notation were first brought in by Haynsworth.We refer the readers to [14] for a survey of this important notion and its farreaching applications in various branches of mathematics.Recall that the numerical range (also known as the field of values) of A ∈ M n is defined by W ( A ) = { x ∗ Ax : x ∈ C n , x ∗ x = 1 } . Also, we define a sector on the complex plane S α = { z ∈ C : ℜ z > , |ℑ z | ≤ ( ℜ z ) tan α } , α ∈ [0 , π/ . Clearly, if A is positive definite, then W ( A ) ⊂ S . Mathematics Subject Classification.
Key words and phrases. determinant inequality, numerical range, sector, Schurcomplement.
For fundamentals of numerical range, see [4, 8]. As 0 / ∈ S α , if W ( A ) ⊂ S α ,then A is necessarily nonsingular.The main object we are dealing in this paper is a class of matrices whosenumerical range is contained in S α . Part of the motivation for investigatingthis class of matrices comes from the search for the optimal growth factorin Gaussian elimination; see, for example, [1, 2, 7, 10, 12].Let A, B ∈ M n be positive definite. It is well known thatdet( A + B ) ≥ det A + det B. (1.2)Haynsworth proved the following refinement of (1.2). Theorem 1.1. [6, Theorem 3]
Suppose
A, B ∈ M n are positive definite. Let A k and B k , k = 1 , . . . , n − , denote the k -th principal submatrices of A and B respectively. Then det( A + B ) ≥ n − X k =1 det B k det A k ! det A + n − X k =1 det A k det B k ! det B. (1.3)Hartfiel [5] obtained an improvement of (1.3): under the same conditionas in Theorem 1.1,det( A + B ) ≥ n − X k =1 det B k det A k ! det A + n − X k =1 det A k det B k ! det B + (2 n − n ) √ det AB. (1.4)Haynsworth’s proof of (1.3) relies on an inequality for the Schur comple-ment [6, Theorem 2]: Let
A, B ∈ M n be positive definite and be comformallypartitioned as in (1.1). Then( A + B ) / ( A + B ) ≥ A/A + B/B . (1.5)In this paper, we first extend (1.5), then as an application, we obtain ageneralization of (1.4) and so (1.3).2. Preliminaries
The larger class of matrices dealt in this paper has some nice closureproperties, just like the class of positive definite matrices. For example, thefollowing proposition says that the Schur complement is closed.
Proposition 2.1.
Let A ∈ M n be partitioned as in (1.1). If W ( A ) ⊂ S α ,then W ( A/A ) ⊂ S α as well.Proof. Clearly, if W ( A ) ⊂ S α , then W ( A ∗ ) ⊂ S α and W ( A ) ⊂ S α . Also,for any nonsingular X ∈ M n , W ( A ) = W ( XAX ∗ ). Therefore, W ( A − ) = ETERMINANT INEQUALITY 3 W ( AA − A ∗ ) = W ( A ∗ ) ⊂ S α . The desired result follows by observing that( A/A ) − = ( A − ) . (cid:3) In the remaining of this section, we present a few auxiliary results.
Lemma 2.2.
Let A ∈ M n with W ( A ) ⊂ S α . Then A can be decomposed as A = XZX ∗ for some invertible X ∈ M n and Z = diag( e iθ , . . . , e iθ n ) with | θ j | ≤ α for all j . Remark 2.3.
The decomposition appears first in [1, Lemma 1.1]. In [15],it is shown that the diagonal entries of Z are unique up to permutation. Lemma 2.4.
Let A ∈ M n with ℜ A positive definite. Then ( ℜ A ) − ≥ ℜ ( A − ) . Proof.
By [13, Lemma 2.1], ℜ ( A − ) = (cid:16) ℜ A + ( ℑ A )( ℜ A ) − ( ℑ A ) (cid:17) − . As( ℑ A )( ℜ A ) − ( ℑ A ) is positive semidefinite, ℜ ( A − ) ≤ ( ℜ A ) − follows. (cid:3) Lemma 2.5.
Let A ∈ M n be partitioned as in (1.1). If ℜ A is positivedefinite, then ℜ ( A/A ) ≥ ( ℜ A ) / ( ℜ A ) . Proof.
The notation ( ℜ A ) / ( ℜ A ) makes sense as ℜ A is the (1 ,
1) blockof ℜ A . Consider the Cartesian decomposition A = M + iN with M = ℜ A , N = ℑ A being conformally partitioned as A . Then we have the followingequality relating the Schur complements [11, Lemma 2.2], A/A = M/M + i ( N/N ) + Y ( M − − iN − ) − Y ∗ , where Y = M M − − N N − .As ℜ (cid:16) ( M − − iN − ) − (cid:17) is positive semidefinite, so is ℜ (cid:16) Y ( M − − iN − ) − Y ∗ (cid:17) .The desired result follows. (cid:3) Lemma 2.6.
Let A ∈ M n with W ( A ) ⊂ S α . Then sec n ( α ) det( ℜ A ) ≥ | det A | . Proof.
Consider the decomposition A = XZX ∗ as in Lemma 2.2. Thenafter dividing by | det X | , it suffices to show sec n ( α ) det( ℜ Z ) ≥
1. Buteach diagonal entry of the diagonal matrix sec( α ) ℜ Z is no less than one,implying the result. (cid:3) Remark 2.7.
The above inequality may be regarded as a complement ofthe Ostrowski-Taussky inequality (see [9, p. 510]). With some minor mod-ification in the proof of [15, Lemma 3.1], Zhang showed that actually theeigenvalues of sec( α ) ℜ Z weakly log majorize the singular values of A . M. LIN An extension of (1.5)
First of all, we remark that a direct extension of (1.5) is not valid. That is,assuming
A, B ∈ M n with W ( A ) , W ( B ) ⊂ S α are comformally partitionedas in (1.1), it does not hold in general that ℜ (cid:16) ( A + B ) / ( A + B ) (cid:17) ≥ ℜ ( A/A ) + ℜ ( B/B ) . (3.1)To see this, take B = A ∗ , then (3.1) contradicts Lemma 2.5.The main result of this section is a correct version of (3.1). Theorem 3.1.
Let
A, B ∈ M n with W ( A ) , W ( B ) ⊂ S α be comformallypartitioned as in (1.1). Then sec ( α ) ℜ (cid:16) ( A + B ) / ( A + B ) (cid:17) ≥ ℜ ( A/A ) + ℜ ( B/B ) . Proof.
We prove the following claim first, which may be regarded as a reversecomplement of Lemma 2.5.
Claim 1. sec ( α )( ℜ A ) / ( ℜ A ) ≥ ℜ ( A/A ). Proof of Claim 1.
We consider the decomposition A = XZX ∗ as inLemma 2.2. We further partition X as a 2-by-1 block matrix X = (cid:20) X X (cid:21) .Then A = (cid:20) X ZX ∗ X ZX ∗ X ZX ∗ X ZX ∗ (cid:21) . Let Y = ( X ∗ ) − = (cid:20) Y Y (cid:21) be comformallypartitioned as X . Then A − = (cid:20) Y Z − Y ∗ Y Z − Y ∗ Y Z − Y ∗ Y Z − Y ∗ (cid:21) . Clearly,cos ( α )( ℜ Z ) − ≤ ℜ ( Z − ) , it follows that cos ( α ) Y ( ℜ Z ) − Y ∗ ≤ ℜ ( Y Z − Y ∗ ) , i.e., cos ( α ) (cid:16) ( ℜ A ) − (cid:17) ≤ ℜ ( A − ) , or cos ( α ) (cid:16) ( ℜ A ) / ( ℜ A ) (cid:17) − ≤ ℜ (cid:16) ( A/A ) − (cid:17) . Taking the inverse on both hand sides yieldssec ( α ) (cid:16) ( ℜ A ) / ( ℜ A ) (cid:17) ≥ (cid:16) ℜ (cid:16) ( A/A ) − (cid:17)(cid:17) − ≥ ℜ ( A/A ) , in which the second inequality is by Lemma 2.4. This completes the proofof Claim 1. ETERMINANT INEQUALITY 5
To finish the proof of Theorem 3.1, we observe the following chain ofinequalities ℜ (cid:16) ( A + B ) / ( A + B ) (cid:17) ≥ ℜ ( A + B ) / ℜ ( A + B ) by Lemma 2.5 ≥ ( ℜ A ) / ( ℜ A ) + ( ℜ B ) / ( ℜ B ) by (1.5) ≥ cos ( α ) (cid:16) ℜ ( A/A ) + ℜ ( B/B ) (cid:17) by Claim 1. (cid:3) An extension of (1.4)
As an applicaton of Theorem 3.1, we present the following extension ofHaynsworth and Hartfiel’s result mentioned in the Introduction.
Theorem 4.1.
Suppose
A, B ∈ M n such that W ( A ) , W ( B ) ⊂ S α . Let A k and B k , k = 1 , . . . , n − , denote the k -th principal submatrices of A and B respectively. Then sec n − ( α ) | det( A + B ) | ≥ n − X k =1 (cid:12)(cid:12)(cid:12)(cid:12) det B k det A k (cid:12)(cid:12)(cid:12)(cid:12)! | det A | + n − X k =1 (cid:12)(cid:12)(cid:12)(cid:12) det A k det B k (cid:12)(cid:12)(cid:12)(cid:12)! | det B | + (2 n − n ) p | det AB | . Proof.
Clearly, ( A k +1 + B k +1 ) / ( A k + B k ) ∈ C , so | ( A k +1 + B k +1 ) / ( A k + B k ) | ≥ ℜ (cid:16) ( A k +1 + B k +1 ) / ( A k + B k ) (cid:17) , k = 1 , . . . , n − . Here we set A n = A, B n = B . By Proposition 2.1, W ( A k +1 /A k ) , W ( B k +1 /B k ) ⊂ S α ; then by Theorem 3.1 and Lemma 2.6,sec ( α ) ℜ (cid:16) ( A k +1 + B k +1 ) / ( A k + B k ) (cid:17) ≥ ℜ ( A k +1 /A k ) + ℜ ( B k +1 /B k ) ≥ cos( α ) (cid:16) | A k +1 /A k | + | B k +1 /B k | (cid:17) . Hence, sec ( α ) | ( A k +1 + B k +1 ) / ( A k + B k ) | ≥ | A k +1 /A k | + | B k +1 /B k | , that is, sec ( α ) (cid:12)(cid:12)(cid:12)(cid:12) det( A k +1 + B k +1 )det( A k + B k ) (cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12) det A k +1 det A k (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) det B k +1 det B k (cid:12)(cid:12)(cid:12)(cid:12) (4.1)for k = 1 , . . . , n − k from 1 to n − n − ( α ) | det( A + B ) | ≥ | A + B | n − Y k =1 (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) det A k +1 det A k (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) det B k +1 det B k (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) . M. LIN As | A + B | ≥ cos( α )( | A | + | B | ), we therefore arrive atsec n − ( α ) | det( A + B ) | ≥ ( | A | + | B | ) n − Y k =1 (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) det A k +1 det A k (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) det B k +1 det B k (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) = n Y k =1 (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) det A k det A k − (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) det B k det B k − (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) , where, by convention, det A = det B = 1.The conclusion follows by taking a k = | det A k | , b k = | det B k | , k =0 , , . . . , n , in Claim 2. Claim 2.
Let a k , b k > k = 1 , . . . , n , also let a = b = 1. Then n Y k =1 (cid:18) a k a k − + b k b k − (cid:19) ≥ a n n − X s +1 b s a s ! + b n n − X s +1 a s b s ! +(2 n − n ) p a n b n . Proof of Claim 2.
The present proof is due to O. Kuba. The orginalproof by the author, which is by induction, is considerably longer. Let N n = { , , . . . , n } , and let P ( N n ) be the set of subsets of N n . We consider specialsubsets ( B s ) ≤ s ≤ n and ( B ′ s ) ≤ s ≤ n defined by B s = { , , . . . , s } , B ′ s = { s, s + 1 , . . . , n } . Finally we define Ω = {∅} ∪ {B s : 1 ≤ s ≤ n } ∪ {B ′ s : 2 ≤ s ≤ n } andΩ ′ = P ( N n ) \ Ω. Note that | Ω ′ | = 2 n − n , and that each k ∈ N n belongs toexactly n of the subsets of Ω.With this notation, for every x , x , . . . , x n >
0, we infer that Y B∈ Ω Y k ∈B x k = n Y k =1 x nk and so Y B∈ Ω ′ Y k ∈B x k = n Y k =1 x n − − nk , moreover, n Y k =1 (1 + x k ) = X B∈P ( N n ) Y k ∈B x k = X B∈ Ω Y k ∈B x k + X B∈ Ω ′ Y k ∈B x k . But X B∈ Ω Y k ∈B x k = 1 + n X s =1 x x · · · x s + n X s =2 x s x s +1 · · · x n ETERMINANT INEQUALITY 7 and using the arithemtic mean-geometric mean inequality X B∈ Ω ′ Y k ∈B x k ≥ | Ω ′ | Y B∈ Ω ′ Y k ∈B x k ! / | Ω ′ | = (2 n − n ) n Y k =1 x n − − nk ! / (2 n − n ) = (2 n − n ) √ x x · · · x n . So we have n Y k =1 (1 + x k ) ≥ n X s =1 x x · · · x s + n X s =2 x s x s +1 · · · x n + (2 n − n ) √ x x · · · x n . Taking x k = a k − b k b k − a k , for k = 1 , . . . , n , gives n Y k =1 (cid:18) a k − b k b k − a k (cid:19) ≥ n X s =1 b s a s + b n a n n X s =2 a s − b s − + (2 n − n ) p b n /a n = 1 + n − X s =1 b s a s + b n a n n − X s =1 a s b s ! + (2 n − n ) p b n /a n . Multiplying both sides of the inequality by n Y k =1 a k a k − = a n yields the desiredinequality. This completes the proof of Claim 2. (cid:3) Apparently, Theorem 4.1 reduces to (1.4) when α = 0. A matrix A ∈ M n is accretive-dissipative if both ℜ A , ℑ A are positive definite (see [3]). Notethat if A is accretive-dissipative, then W ( e − iπ/ A ) ⊂ S π/ . Thus, we havethe following corollary. Corollary 4.2.
Suppose
A, B ∈ M n are accretive-dissipative. Let A k and B k , k = 1 , . . . , n − , denote the k -th principal submatrices of A and B respectively. Then n − | det( A + B ) | ≥ n − X k =1 (cid:12)(cid:12)(cid:12)(cid:12) det B k det A k (cid:12)(cid:12)(cid:12)(cid:12)! | det A | + n − X k =1 (cid:12)(cid:12)(cid:12)(cid:12) det A k det B k (cid:12)(cid:12)(cid:12)(cid:12)! | det B | + (2 n − n ) p | det AB | . Acknowledgments.
The author thanks P. van den Driessche and P. Zhang forsome helpful remarks. In particular, the present proof of Claim 2 is due to O.Kuba. The author is currently a PIMS Postdoctoral Fellow at the University ofVictoria. The support of PIMS is gratefully acknowledged . M. LIN
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Department of Mathematics and Statistics, University of Victoria, Vic-toria, BC, Canada, V8W 3R4.
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