Open Access

The estimates of diagonally dominant degree and eigenvalues distributions for the Schur complements of matrices

Journal of Inequalities and Applications20132013:431

https://doi.org/10.1186/1029-242X-2013-431

Received: 28 May 2013

Accepted: 26 August 2013

Published: 11 September 2013

Abstract

By applying the properties of Schur complement and some inequality techniques, some new estimates of diagonally dominant degree on the Schur complement of matrices are obtained, which improve the main results of Liu (SIAM J. Matrix Anal. Appl. 27:665-674, 2005) and Liu (Linear Algebra Appl. 432:1090-1104, 2010). As an application, we present some new distribution theorems for eigenvalues of the Schur complement. Finally, we give a numerical example to illustrate the theory results.

MSC:15A45, 15A48.

Keywords

matrixSchur complementdiagonally dominant degreeeigenvalue distribution

1 Introduction

Let C n × n denote the set of all n × n complex matrices, N = { 1 , 2 , , n } and A = ( a i j ) C n × n ( n 2 ). Denote
R i ( A ) = j i | a i j | , C i ( A ) = j i | a j i | , i N , N r ( A ) = { i : | a i i | > R i ( A ) , i N } , N c ( A ) = { i : | a i i | > C i ( A ) , i N } .
We call A a strictly diagonally dominant matrix (abbreviated S D n ) if
| a i i | > R i ( A ) , i N .
A is called an Ostrowski matrix (abbreviated O S n ) (see [1]) if
| a i i | | a j j | > R i ( A ) R j ( A ) , i , j N , i j .

As in [2], for all i N and α [ 0 , 1 ] , we call | a i i | R i ( A ) , | a i i | α R i ( A ) ( 1 α ) C i ( A ) and | a i i | [ R i ( A ) ] α [ C i ( A ) ] 1 α the i th dominant degree, α-dominant degree and product α-dominant degree of A, respectively.

For β N , denote by | β | the cardinality of β and β ¯ = N / β . If β , γ N , then A ( β , γ ) is the submatrix of A lying in the rows indexed by β and the columns indicated by γ. In particular, A ( β , β ) is abbreviated to A ( β ) . If A ( β ) is nonsingular, then
A / β = A / A ( β ) = A ( β ¯ ) A ( β ¯ , β ) [ A ( β ) ] 1 A ( β , β ¯ )

is called the Schur complement of A with respect to A ( β ) .

The comparison matrix of A, μ ( A ) = ( α i j ) is defined by
α i j = { | a i j | , if  i = j , | a i j | , if  i j .

A matrix A is called an M-matrix if there exist a nonnegative matrix B and a real number s > ρ ( B ) such that A = s I B , where ρ ( B ) is the spectral radius of B. It is well known that A is an H-matrix if and only if μ ( A ) is an M-matrix, and if A is an M-matrix, then the Schur complement of A is also an M-matrix and det A > 0 (see [3]). We denote by H n and M n the sets of H-matrices and M-matrices, respectively.

The Schur complement has been proved to be a useful tool in many fields such as control theory, statistics and computational mathematics, and many works have been done on it (see [48]). Meanwhile, studying the locations of eigenvalues of the Schur complement of matrices is of great significance as shown in [2, 3, 914]. In this paper, we present some new estimates of diagonally dominant degree on the Schur complement of matrices and use them to study the distributions for the eigenvalues of the Schur complement of matrices.

The paper is organized as follows. In Section 2, we give several new estimates of the diagonally dominant degree, the α-diagonally dominant degree and product α-diagonally dominant degree on the Schur complement of matrices. In Section 3, several new distribution theorems for eigenvalues of the Schur complements are obtained. In Section 4, we present a numerical example to illustrate the theory results.

2 Estimates of diagonally dominant degree for the Schur complement

In this section, we present several new estimates for the diagonally dominant degree, the α-diagonally dominant degree and product α-diagonally dominant degree of the Schur complement of matrices.

Lemma 1 [5]

If A H n , then [ μ ( A ) ] 1 | A 1 | .

Lemma 2 [5]

If A is an S D n or A is an O S n , then A H n , i.e., μ ( A ) M n .

Lemma 3 [3]

If A is an S D n or A is an O S n and β N , then the Schur complement of A is an S D | β ¯ | or an O S | β ¯ | , where β ¯ = N β is the complement of β in N, and | β ¯ | is the cardinality of β ¯ .

Lemma 4 [2]

Let a > b , c > b , b > 0 and 0 α 1 . Then
a α c 1 α ( a b ) α ( c b ) 1 α + b .
Theorem 1 Let A = ( a i j ) C n × n , β = { i 1 , i 2 , , i k } N r ( A ) ϕ , β ¯ = { j 1 , j 2 , , j l } , k < n , and let A / β = ( a t s ) . Then for all 1 t l ,
| a t t | R t ( A / β ) | a j t j t | R j t ( A ) + δ j t | a j t j t | R j t ( A ) ,
(1)
and
| a t t | + R t ( A / β ) | a j t j t | + R j t ( A ) δ j t | a j t j t | + R j t ( A ) ,
(2)
where
δ j t = min 1 u k | a i u i u | P i u ( A ) | a i u i u | v = 1 k | a j t i v | , r = max 1 u k v = 1 l | a i u j v | | a i u i u | v u k | a i u i v | , P i u ( A ) = r v u k | a i u i v | + v = 1 l | a i u j v | , i u β , 1 u k .
Proof Since β N r ( A ) ϕ , then A ( β ) H k , μ ( A ( β ) ) M k . From Lemma 1 and Lemma 2, we have
[ μ ( A ( β ) ) ] 1 | [ A ( β ) ] 1 | .
Thus, for 1 t l ,
| a t t | R t ( A / β ) = | a t t | s t l | a t s | | a j t j t | s t l | a j t j s | s = 1 l ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j s | | a i k j s | ) = | a j t j t | R j t ( A ) + v = 1 k | a j t i v | + δ j t δ j t s = 1 l ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j s | | a i k j s | ) = | a j t j t | R j t ( A ) + δ j t + 1 det [ μ ( A ( β ) ) ] det ( v = 1 k | a j t i v | δ j t | a j t i 1 | | a j t i k | s = 1 l | a i 1 j s | μ ( A ( β ) ) s = 1 l | a i k j s | ) = def . | a j t j t | R j t ( A ) + δ j t + 1 det [ μ ( A ( β ) ) ] det B .

Similarly to the proof of Lemma 4 in [13], we can prove that det B 0 . Thus, we obtain Inequation (1). Similarly, we can prove Inequation (2). □

Remark 1 Note that
P i u ( A ) | a i u i u | r R i u ( A ) | a i u i u | , 1 u k .

This shows that Theorem 1 improves Theorem 1 of [13].

Theorem 2 Let A = ( a i j ) C n × n , β = { i 1 , i 2 , , i k } N r ( A ) N c ( A ) ϕ , β ¯ = { j 1 , j 2 , , j l } , k < n , and let A / β = ( a t s ) . Then for all 1 t l , 0 α 1 ,
| a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α | a j t j t | ( R j t ( A ) δ t ) α ( C j t ( A ) δ t T ) 1 α ,
(3)
and
| a t t | + ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α | a j t j t | + ( R j t ( A ) δ t ) α ( C j t ( A ) δ t T ) 1 α ,
(4)
where
δ t = min 1 ω k | a i ω i ω | P i ω ( A ) | a i ω i ω | v = 1 k | a j t i v | , η = max 1 ω k v = 1 l | a i ω j v | | a i ω i ω | v ω k | a i ω i v | , δ t T = min 1 ω k | a i ω i ω | Q i ω ( A ) | a i ω i ω | v = 1 k | a i v j t | , ξ = max 1 ω k v = 1 l | a j v i ω | | a i ω i ω | v ω k | a i v i ω | , P i ω ( A ) = η v ω k | a i ω i v | + v = 1 l | a i ω j v | , Q i ω ( A ) = ξ v ω k | a i v i ω | + v = 1 l | a j v i ω | .
Proof Since β N r ( A ) ϕ , then A ( β ) H k , μ ( A ( β ) ) M k . From Lemma 1 and Lemma 2, we have
[ μ ( A ( β ) ) ] 1 | [ A ( β ) ] 1 | .
Thus, for all 1 t l , 0 α 1 ,
| a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α | a j t j t | ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j t | | a i k j t | ) ( s t l [ | a j t j s | + ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j s | | a i k j s | ) ] ) α × ( s t l [ | a j s j t | + ( | a j s i 1 | , , | a j s i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j t | | a i k j t | ) ] ) 1 α .
Let
ζ = ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j t | | a i k j t | ) .
By the proof of Theorem 1, we have
s t l [ | a j t j s | + ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j s | | a i k j s | ) ] R j t ( A ) δ t ζ .
Similarly,
s t l [ | a j s j t | + ( | a j s i 1 | , , | a j s i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j t | | a i k j t | ) ] C j t ( A ) δ t T ζ .
By Lemma 4, we have
| a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α | a j t j t | ζ ( R j t ( A ) δ t ζ ) α ( C j t ( A ) δ t T ζ ) 1 α | a j t j t | ζ [ ( R j t ( A ) δ t ) α ( C j t ( A ) δ t T ) 1 α ζ ] = | a j t j t | ( R j t ( A ) δ t ) α ( C j t ( A ) δ t T ) 1 α .

Thus, we obtain Inequation (3). Similarly, we can prove Inequation (4). □

Remark 2 Note that
δ t = min 1 ω k | a i ω i ω | P i ω ( A ) | a i ω i ω | v = 1 k | a j t i v | min 1 ω k | a i ω i ω | R i ω ( A ) | a i ω i ω | v = 1 k | a j t i v | , δ t T = min 1 ω k | a i ω i ω | Q i ω ( A ) | a i ω i ω | v = 1 k | a i v j t | min 1 ω k | a i ω i ω | C i ω ( A ) | a i ω i ω | v = 1 k | a i v j t | .

This shows that Theorem 2 improves Theorem 2 of [2].

Similarly to the proof of Theorem 2, we can prove the following theorem.

Theorem 3 Let A = ( a i j ) C n × n , β = { i 1 , i 2 , , i k } N r ( A ) N c ( A ) ϕ , β ¯ = { j 1 , j 2 , , j l } , k < n and A / β = ( a t s ) . Then for all 1 t l , 0 α 1 ,
| a t t | α R t ( A / β ) ( 1 α ) C t ( A / β ) | a j t j t | α R j t ( A ) ( 1 α ) C j t ( A ) + α δ t + ( 1 α ) δ t T | a j t j t | α R j t ( A ) ( 1 α ) C j t ( A ) ,
and
| a t t | + α R t ( A / β ) + ( 1 α ) C t ( A / β ) | a j t j t | + α R j t ( A ) + ( 1 α ) C j t ( A ) α δ t ( 1 α ) δ t T | a j t j t | + α R j t ( A ) + ( 1 α ) C j t ( A ) .

3 Distribution for eigenvalues of the Schur complement

In this section, we present two new distribution theorems for eigenvalues of the Schur complement.

Lemma 5 [2]

Let A = ( a i j ) C n × n and 0 α 1 . Then for any eigenvalue λ of A, there exists 1 t n such that
| λ a t t | ( R t ( A ) ) α ( C t ( A ) ) 1 α .
Theorem 4 Let A = ( a i j ) C n × n , β = { i 1 , i 2 , , i k } N r ( A ) ϕ , β ¯ = { j 1 , j 2 , , j l } , and let A / β = ( a t s ) . Then for any eigenvalue λ of A / β , there exists 1 t l such that
| λ a j t j t | R j t ( A ) δ j t R j t ( A ) .
(5)
Proof Let λ be an eigenvalue of A / β . From the famous Gerschgorin circle theorem, we know that there exists 1 t l such that | λ a t t | R t ( A / β ) . Hence
0 | λ a t t | R t ( A / β ) = | λ a j t j t + ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | s = 1 , t l | a j t j s ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j s a i k j s ) | | λ a j t j t | s = 1 , t l | a j t j s | s = 1 l ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j s | | a i k j s | ) = | λ a j t j t | R j t ( A ) + v = 1 k | a j t i v | + δ j t δ j t s = 1 l ( | a j t i 1 | , , | a j t i k | ) [ μ ( A ( β ) ) ] 1 ( | a i 1 j s | | a i k j s | ) | λ a j t j t | R j t ( A ) + δ j t ,
that is
| λ a j t j t | R j t ( A ) δ j t R j t ( A ) .

Thus, Inequation (5) holds. □

Theorem 5 Let A = ( a i j ) C n × n , β = { i 1 , i 2 , , i k } N r ( A ) N c ( A ) ϕ , β ¯ = { j 1 , j 2 , , j l } , k < n , and let A / β = ( a t s ) . Then for any 0 α 1 and every eigenvalue λ of A / β , there exists 1 t l such that
| λ a j t j t | ( R j t ( A ) δ t ) α ( C j t ( A ) δ t T ) 1 α ( R j t ( A ) ) α ( C j t ( A ) ) 1 α .
(6)
Proof Let λ be an eigenvalue of A / β . By Lemma 5, we know that there exists 1 t l such that
| λ a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α , 0 α 1 .
Thus,
0 | λ a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α = | λ a j t j t ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | [ s = 1 , t l | a j t j s + ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j s a i k j s ) | ] α × [ s = 1 , t l | a j s j t + ( a j s i 1 , , a j s i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | ] 1 α | λ a j t j t | | ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | ( s = 1 , t l [ | a j t j s | + | ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j s a i k j s ) | ] ) α × ( s = 1 , t l [ | a j s j t | + | ( a j s i 1 , , a j s i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | ] ) 1 α .
Similarly to the proof of Theorem 2, we can prove
| ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | + ( s = 1 , t l [ | a j t j s | + | ( a j t i 1 , , a j t i k ) [ A ( β ) ] 1 ( a i 1 j s a i k j s ) | ] ) α × ( s = 1 , t l [ | a j s j t | + | ( a j s i 1 , , a j s i k ) [ A ( β ) ] 1 ( a i 1 j t a i k j t ) | ] ) 1 α ( R j t ( A ) δ t ) α ( C j t ( A ) δ t T ) 1 α .
Therefore, we have
0 | λ a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α | λ a j t j t | ( R j t ( A ) δ t ) ( C j t ( A ) δ t T ) 1 α .

That is, Inequation (6) holds. □

4 A numerical example

In this section, we present a numerical example to illustrate the theory results.

Example 1 Let
A = ( 16 1 5 2 2 3 15 2 4 3 2 2 18 1 4 5 3 5 8 2 5 2 2 3 9 ) , β = { 1 , 3 } .
By calculation with Matlab 7.1, we have that
R 1 ( A ) = 10 ; R 2 ( A ) = 12 ; R 3 ( A ) = 9 ; R 4 ( A ) = 15 ; R 5 ( A ) = 12 ; C 1 ( A ) = 15 ; C 2 ( A ) = 8 ; C 3 ( A ) = 14 ; C 4 ( A ) = 10 ; C 5 ( A ) = 11 ; δ 2 = 2.7273 ; δ 4 = 5.4545 ; δ 5 = 3.8182 ; δ 2 T = 0.2143 ; δ 4 T = 0.2143 ; δ 5 T = 0.4286 .
Since β N r ( A ) , by Theorem 4, any eigenvalue λ of A / β satisfies
λ { λ : | λ 15 | 9.2727 } { λ : | λ 8 | 9.5455 } { λ : | λ 9 | 8.1818 } = Γ 1 .
From Theorem 3 of [2], any eigenvalue λ of A / β satisfies
λ { λ : | λ 15 | 10.1250 } { λ : | λ 8 | 11.2500 } { λ : | λ 9 | 9.3750 } = Γ 1 .
Evidently, Γ 1 Γ 1 , we use Figure 1 to show the fact.
Figure 1

The blue solid line and the green dashed line denote the corresponding discs Γ 1 and Γ 1 , respectively.

Meanwhile, since β N r ( A ) N c ( A ) , by taking α = 0.5 in Theorem 5, any eigenvalue λ of A / β satisfies
λ { λ : | λ 15 | 8.4968 } { λ : | λ 8 | 9.6648 } { λ : | λ 9 | 9.3002 } = Γ 2 .
From Theorem 4 of [2], any eigenvalue λ of A / β satisfies
λ { λ : | λ 15 | 8.8939 } { λ : | λ 8 | 10.5067 } { λ : | λ 9 | 9.9804 } = Γ 2 .
Evidently, Γ 2 Γ 2 , we use Figure 2 to show the fact.
Figure 2

The blue solid line and the green dashed line denote the corresponding discs Γ 2 and Γ 2 , respectively.

Declarations

Acknowledgements

The authors are very indebted to the referees for their valuable comments and corrections, which improved the original manuscript of this paper. This work was supported by the National Natural Science Foundation of China (71161020, 11361074) and IRTSTYN.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Yunnan University

References

  1. Cvetković LJ: A new subclass of H -matrices. Appl. Math. Comput. 2009, 208: 206–210. 10.1016/j.amc.2008.11.037MathSciNetView ArticleMATHGoogle Scholar
  2. Liu JZ, Huang ZJ: The Schur complements of γ -diagonally and product γ -diagonally dominant matrix and their disc separation. Linear Algebra Appl. 2010, 432: 1090–1104. 10.1016/j.laa.2009.10.021MathSciNetView ArticleMATHGoogle Scholar
  3. Liu JZ, Li JC, Huang ZH, Kong X: Some properties on Schur complement and diagonal Schur complement of some diagonally dominant matrices. Linear Algebra Appl. 2008, 428: 1009–1030. 10.1016/j.laa.2007.09.008MathSciNetView ArticleMATHGoogle Scholar
  4. Carlson D, Markham T: Schur complements on diagonally dominant matrices. Czechoslov. Math. J. 1979, 29: 246–251.MathSciNetMATHGoogle Scholar
  5. Horn RA, Johnson CR: Topics in Matrix Analysis. Cambridge University Press, New York; 1991.View ArticleMATHGoogle Scholar
  6. Ikramov KD: Invariance of the Brauer diagonal dominance in Gaussian elimination. Vestn. Mosk. Univ., Ser. 15 Vyčisl. Mat. Kibern. 1989, 2: 91–94.MATHGoogle Scholar
  7. Smith R: Some interlacing properties of the Schur complement of a Hermitian matrix. Linear Algebra Appl. 1992, 177: 137–144.MathSciNetView ArticleMATHGoogle Scholar
  8. Zhang FZ: The Schur Complement and Its Applications. Springer, New York; 2005.View ArticleMATHGoogle Scholar
  9. Li B, Tsatsomeros M: Doubly diagonally dominant matrices. Linear Algebra Appl. 1997, 261: 221–235. 10.1016/S0024-3795(96)00406-5MathSciNetView ArticleMATHGoogle Scholar
  10. Liu JZ, Huang YQ: The Schur complements of generalized doubly diagonally dominant matrices. Linear Algebra Appl. 2004, 378: 231–244.MathSciNetView ArticleMATHGoogle Scholar
  11. Liu JZ, Huang YQ: Some properties on Schur complements of H -matrices and diagonally dominant matrices. Linear Algebra Appl. 2004, 389: 365–380.MathSciNetView ArticleMATHGoogle Scholar
  12. Liu JZ, Huang ZJ: The dominant degree and disc theorem for the Schur complement. Appl. Math. Comput. 2010, 215: 4055–4066. 10.1016/j.amc.2009.12.063MathSciNetView ArticleMATHGoogle Scholar
  13. Liu JZ, Zhang FZ: Disc separation of the Schur complements of diagonally dominant matrices and determinantal bounds. SIAM J. Matrix Anal. Appl. 2005, 27: 665–674. 10.1137/040620369MathSciNetView ArticleMATHGoogle Scholar
  14. Liu JZ: Some inequalities for singular values and eigenvalues of generalized Schur complements of products of matrices. Linear Algebra Appl. 1999, 286: 209–221. 10.1016/S0024-3795(98)10156-8MathSciNetView ArticleMATHGoogle Scholar

Copyright

© Li and Wang; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.