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

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.

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 (n2). 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),iN.

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,jN,ij.

As in [2], for all iN and α[0,1], we call | a i i | R i (A), | a i i |α R i (A)(1α) 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 βN, denote by |β| the cardinality of β and β ¯ =N/β. If β,γN, 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=sIB, 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 detA>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 βN, then the Schur complement of A is an S D | β ¯ | or an O S | β ¯ | , where β ¯ =Nβ 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 0α1. 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 1tl,

| 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 1tl,

| 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 detB0. 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 | ,1uk.

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 1tl, 0α1,

| 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 1tl, 0α1,

| 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 1tl, 0α1,

| 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 0α1. Then for any eigenvalue λ of A, there exists 1tn 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 1tl 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 1tl 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 0α1 and every eigenvalue λ of A/β, there exists 1tl 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 1tl such that

| λ a t t | ( R t ( A / β ) ) α ( C t ( A / β ) ) 1 α ,0α1.

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
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
figure 2

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

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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.

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Li, Yt., Wang, F. The estimates of diagonally dominant degree and eigenvalues distributions for the Schur complements of matrices. J Inequal Appl 2013, 431 (2013). https://doi.org/10.1186/1029-242X-2013-431

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Keywords

  • matrix
  • Schur complement
  • diagonally dominant degree
  • eigenvalue distribution