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

Yao-tang Li; Feng Wang

doi:10.1186/1029-242X-2013-431

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

Yao-tang Li¹Email author and
Feng Wang¹

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

matrix Schur complement diagonally dominant degree eigenvalue distribution

1 Introduction

Let

C^{n \times n}

denote the set of all

n \times n

complex matrices,

N = {1, 2, \dots, n}

and

A = (a_{i j}) \in C^{n \times n}

(

n \geq 2

). Denote

\begin{matrix} R_{i} (A) = \sum_{j \neq i} | a_{i j} |, C_{i} (A) = \sum_{j \neq i} | a_{j i} |, i \in N, \\ N_{r} (A) = {i : | a_{i i} | > R_{i} (A), i \in N}, N_{c} (A) = {i : | a_{i i} | > C_{i} (A), i \in N} . \end{matrix}

We call A a strictly diagonally dominant matrix (abbreviated

S D_{n}

) if

| a_{i i} | > R_{i} (A), \forall i \in 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), \forall i, j \in N, i \neq j .

As in [2], for all $i \in N$ and $α \in [0, 1]$ , 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 \subseteq N

, denote by

| β |

the cardinality of β and

\bar{β} = N / β

. If

β, γ \subseteq 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 (\bar{β}) - A (\bar{β}, β) {[A (β)]}^{- 1} A (β, \bar{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} = {\begin{matrix} | a_{i j} |, & if i = j, \\ - | a_{i j} |, & if i \neq j . \end{matrix}

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 [4–8]). Meanwhile, studying the locations of eigenvalues of the Schur complement of matrices is of great significance as shown in [2, 3, 9–14]. 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 $A \in H_{n}$ , then ${[μ (A)]}^{- 1} \geq | A^{- 1} |$ .

Lemma 2 [5]

If A is an $S D_{n}$ or A is an $O S_{n}$ , then $A \in H_{n}$ , i.e., $μ (A) \in M_{n}$ .

Lemma 3 [3]

If A is an $S D_{n}$ or A is an $O S_{n}$ and $β \subseteq N$ , then the Schur complement of A is an $S D_{| \bar{β} |}$ or an $O S_{| \bar{β} |}$ , where $\bar{β} = N - β$ is the complement of β in N, and $| \bar{β} |$ is the cardinality of $\bar{β}$ .

Lemma 4 [2]

Let

a > b

,

c > b

,

b > 0

and

0 \leq α \leq 1

. Then

a^{α} c^{1 - α} \geq {(a - b)}^{α} {(c - b)}^{1 - α} + b .

Theorem 1 Let

A = (a_{i j}) \in C^{n \times n}

,

β = {i_{1}, i_{2}, \dots, i_{k}} \subseteq N_{r} (A) \neq ϕ

,

\bar{β} = {j_{1}, j_{2}, \dots, j_{l}}

,

k < n

, and let

A / β = (a_{t s}^{'})

. Then for all

1 \leq t \leq l

,

| a_{t t}^{'} | - R_{t} (A / β) \geq | a_{j_{t} j_{t}} | - R_{j_{t}} (A) + δ_{j_{t}} \geq | a_{j_{t} j_{t}} | - R_{j_{t}} (A),

(1)

and

| a_{t t}^{'} | + R_{t} (A / β) \leq | a_{j_{t} j_{t}} | + R_{j_{t}} (A) - δ_{j_{t}} \leq | a_{j_{t} j_{t}} | + R_{j_{t}} (A),

(2)

where

\begin{matrix} δ_{j_{t}} = min_{1 \leq u \leq k} \frac{| a_{i_{u} i_{u}} | - P_{i_{u}} (A)}{| a_{i_{u} i_{u}} |} \sum_{v = 1}^{k} | a_{j_{t} i_{v}} |, r = max_{1 \leq u \leq k} \frac{\sum_{v = 1}^{l} | a_{i_{u} j_{v}} |}{| a_{i_{u} i_{u}} | - \sum_{v \neq u}^{k} | a_{i_{u} i_{v}} |}, \\ P_{i_{u}} (A) = r \sum_{v \neq u}^{k} | a_{i_{u} i_{v}} | + \sum_{v = 1}^{l} | a_{i_{u} j_{v}} |, i_{u} \in β, 1 \leq u \leq k . \end{matrix}

Proof Since

β \subseteq N_{r} (A) \neq ϕ

, then

A (β) \in H_{k}

,

μ (A (β)) \in M_{k}

. From Lemma 1 and Lemma 2, we have

{[μ (A (β))]}^{- 1} \geq | {[A (β)]}^{- 1} | .

Thus, for

1 \leq t \leq l

,

\begin{aligned} | a_{t t}^{'} | - R_{t} (A / β) \\ = | a_{t t}^{'} | - \sum_{s \neq t}^{l} | a_{t s}^{'} | \\ \geq | a_{j_{t} j_{t}} | - \sum_{s \neq t}^{l} | a_{j_{t} j_{s}} | - \sum_{s = 1}^{l} (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{s}} | \\ ⋮ \\ | a_{i_{k} j_{s}} | \end{array}) \\ = | a_{j_{t} j_{t}} | - R_{j_{t}} (A) + \sum_{v = 1}^{k} | a_{j_{t} i_{v}} | + δ_{j_{t}} - δ_{j_{t}} - \sum_{s = 1}^{l} (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{s}} | \\ ⋮ \\ | a_{i_{k} j_{s}} | \end{array}) \\ = | a_{j_{t} j_{t}} | - R_{j_{t}} (A) + δ_{j_{t}} \\ + \frac{1}{det [μ (A (β))]} det (\begin{array}{c} \sum_{v = 1}^{k} | a_{j_{t} i_{v}} | - δ_{j_{t}} & - | a_{j_{t} i_{1}} | & \dots & - | a_{j_{t} i_{k}} | \\ - \sum_{s = 1}^{l} | a_{i_{1} j_{s}} | \\ ⋮ & μ (A (β)) \\ - \sum_{s = 1}^{l} | a_{i_{k} j_{s}} | \end{array}) \\ \overset{def .}{=} | a_{j_{t} j_{t}} | - R_{j_{t}} (A) + δ_{j_{t}} + \frac{1}{det [μ (A (β))]} det B . \end{aligned}

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

Remark 1 Note that

\frac{P_{i_{u}} (A)}{| a_{i_{u} i_{u}} |} \leq r \leq \frac{R_{i_{u}} (A)}{| a_{i_{u} i_{u}} |}, 1 \leq u \leq k .

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

Theorem 2 Let

A = (a_{i j}) \in C^{n \times n}

,

β = {i_{1}, i_{2}, \dots, i_{k}} \subseteq N_{r} (A) \cap N_{c} (A) \neq ϕ

,

\bar{β} = {j_{1}, j_{2}, \dots, j_{l}}

,

k < n

, and let

A / β = (a_{t s}^{'})

. Then for all

1 \leq t \leq l

,

0 \leq α \leq 1

,

| a_{t t}^{'} | - {(R_{t} (A / β))}^{α} {(C_{t} (A / β))}^{1 - α} \geq | 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} \leq | a_{j_{t} j_{t}} | + {(R_{j_{t}} (A) - δ_{t})}^{α} {(C_{j_{t}} (A) - δ_{t}^{T})}^{1 - α},

(4)

where

\begin{matrix} δ_{t} = min_{1 \leq ω \leq k} \frac{| a_{i_{ω} i_{ω}} | - P_{i_{ω}} (A)}{| a_{i_{ω} i_{ω}} |} \sum_{v = 1}^{k} | a_{j_{t} i_{v}} |, η = max_{1 \leq ω \leq k} \frac{\sum_{v = 1}^{l} | a_{i_{ω} j_{v}} |}{| a_{i_{ω} i_{ω}} | - \sum_{v \neq ω}^{k} | a_{i_{ω} i_{v}} |}, \\ δ_{t}^{T} = min_{1 \leq ω \leq k} \frac{| a_{i_{ω} i_{ω}} | - Q_{i_{ω}} (A)}{| a_{i_{ω} i_{ω}} |} \sum_{v = 1}^{k} | a_{i_{v} j_{t}} |, ξ = max_{1 \leq ω \leq k} \frac{\sum_{v = 1}^{l} | a_{j_{v} i_{ω}} |}{| a_{i_{ω} i_{ω}} | - \sum_{v \neq ω}^{k} | a_{i_{v} i_{ω}} |}, \\ P_{i_{ω}} (A) = η \sum_{v \neq ω}^{k} | a_{i_{ω} i_{v}} | + \sum_{v = 1}^{l} | a_{i_{ω} j_{v}} |, Q_{i_{ω}} (A) = ξ \sum_{v \neq ω}^{k} | a_{i_{v} i_{ω}} | + \sum_{v = 1}^{l} | a_{j_{v} i_{ω}} | . \end{matrix}

Proof Since

β \subseteq N_{r} (A) \neq ϕ

, then

A (β) \in H_{k}

,

μ (A (β)) \in M_{k}

. From Lemma 1 and Lemma 2, we have

{[μ (A (β))]}^{- 1} \geq | {[A (β)]}^{- 1} | .

Thus, for all

1 \leq t \leq l

,

0 \leq α \leq 1

,

\begin{aligned} | a_{t t}^{'} | - {(R_{t} (A / β))}^{α} {(C_{t} (A / β))}^{1 - α} \\ \geq | a_{j_{t} j_{t}} | - (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{t}} | \\ ⋮ \\ | a_{i_{k} j_{t}} | \end{array}) \\ {- (\sum_{s \neq t}^{l} [| a_{j_{t} j_{s}} | + (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{s}} | \\ ⋮ \\ | a_{i_{k} j_{s}} | \end{array})])}^{α} \\ {\times (\sum_{s \neq t}^{l} [| a_{j_{s} j_{t}} | + (| a_{j_{s} i_{1}} |, \dots, | a_{j_{s} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{t}} | \\ ⋮ \\ | a_{i_{k} j_{t}} | \end{array})])}^{1 - α} . \end{aligned}

Let

ζ = (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{t}} | \\ ⋮ \\ | a_{i_{k} j_{t}} | \end{array}) .

By the proof of Theorem 1, we have

\sum_{s \neq t}^{l} [| a_{j_{t} j_{s}} | + (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{s}} | \\ ⋮ \\ | a_{i_{k} j_{s}} | \end{array})] \leq R_{j_{t}} (A) - δ_{t} - ζ .

Similarly,

\sum_{s \neq t}^{l} [| a_{j_{s} j_{t}} | + (| a_{j_{s} i_{1}} |, \dots, | a_{j_{s} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{t}} | \\ ⋮ \\ | a_{i_{k} j_{t}} | \end{array})] \leq C_{j_{t}} (A) - δ_{t}^{T} - ζ .

By Lemma 4, we have

\begin{aligned} | a_{t t}^{'} | - {(R_{t} (A / β))}^{α} {(C_{t} (A / β))}^{1 - α} \\ \geq | a_{j_{t} j_{t}} | - ζ - {(R_{j_{t}} (A) - δ_{t} - ζ)}^{α} {(C_{j_{t}} (A) - δ_{t}^{T} - ζ)}^{1 - α} \\ \geq | 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} . \end{aligned}

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

Remark 2 Note that

\begin{matrix} δ_{t} = min_{1 \leq ω \leq k} \frac{| a_{i_{ω} i_{ω}} | - P_{i_{ω}} (A)}{| a_{i_{ω} i_{ω}} |} \sum_{v = 1}^{k} | a_{j_{t} i_{v}} | \geq min_{1 \leq ω \leq k} \frac{| a_{i_{ω} i_{ω}} | - R_{i_{ω}} (A)}{| a_{i_{ω} i_{ω}} |} \sum_{v = 1}^{k} | a_{j_{t} i_{v}} |, \\ δ_{t}^{T} = min_{1 \leq ω \leq k} \frac{| a_{i_{ω} i_{ω}} | - Q_{i_{ω}} (A)}{| a_{i_{ω} i_{ω}} |} \sum_{v = 1}^{k} | a_{i_{v} j_{t}} | \geq min_{1 \leq ω \leq k} \frac{| a_{i_{ω} i_{ω}} | - C_{i_{ω}} (A)}{| a_{i_{ω} i_{ω}} |} \sum_{v = 1}^{k} | a_{i_{v} j_{t}} | . \end{matrix}

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}) \in C^{n \times n}

,

β = {i_{1}, i_{2}, \dots, i_{k}} \subseteq N_{r} (A) \cap N_{c} (A) \neq ϕ

,

\bar{β} = {j_{1}, j_{2}, \dots, j_{l}}

,

k < n

and

A / β = (a_{t s}^{'})

. Then for all

1 \leq t \leq l

,

0 \leq α \leq 1

,

\begin{aligned} | a_{t t}^{'} | - α R_{t} (A / β) - (1 - α) C_{t} (A / β) \\ \geq | a_{j_{t} j_{t}} | - α R_{j_{t}} (A) - (1 - α) C_{j_{t}} (A) + α δ_{t} + (1 - α) δ_{t}^{T} \\ \geq | a_{j_{t} j_{t}} | - α R_{j_{t}} (A) - (1 - α) C_{j_{t}} (A), \end{aligned}

and

\begin{aligned} | a_{t t}^{'} | + α R_{t} (A / β) + (1 - α) C_{t} (A / β) \\ \leq | a_{j_{t} j_{t}} | + α R_{j_{t}} (A) + (1 - α) C_{j_{t}} (A) - α δ_{t} - (1 - α) δ_{t}^{T} \\ \leq | a_{j_{t} j_{t}} | + α R_{j_{t}} (A) + (1 - α) C_{j_{t}} (A) . \end{aligned}

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}) \in C^{n \times n}

and

0 \leq α \leq 1

. Then for any eigenvalue λ of A, there exists

1 \leq t \leq n

such that

| λ - a_{t t} | \leq {(R_{t} (A))}^{α} {(C_{t} (A))}^{1 - α} .

Theorem 4 Let

A = (a_{i j}) \in C^{n \times n}

,

β = {i_{1}, i_{2}, \dots, i_{k}} \subseteq N_{r} (A) \neq ϕ

,

\bar{β} = {j_{1}, j_{2}, \dots, j_{l}}

, and let

A / β = (a_{t s}^{'})

. Then for any eigenvalue λ of

A / β

, there exists

1 \leq t \leq l

such that

| λ - a_{j_{t} j_{t}} | \leq R_{j_{t}} (A) - δ_{j_{t}} \leq 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 \leq t \leq l

such that

| λ - a_{t t}^{'} | \leq R_{t} (A / β)

. Hence

\begin{array}{rcl} 0 & \geq & | λ - a_{t t}^{'} | - R_{t} (A / β) \\ = & | λ - a_{j_{t} j_{t}} + (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) | \\ - \sum_{s = 1, \neq t}^{l} | a_{j_{t} j_{s}} - (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{s}} \\ ⋮ \\ a_{i_{k} j_{s}} \end{array}) | \\ \geq & | λ - a_{j_{t} j_{t}} | - \sum_{s = 1, \neq t}^{l} | a_{j_{t} j_{s}} | - \sum_{s = 1}^{l} (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{s}} | \\ ⋮ \\ | a_{i_{k} j_{s}} | \end{array}) \\ = & | λ - a_{j_{t} j_{t}} | - R_{j_{t}} (A) + \sum_{v = 1}^{k} | a_{j_{t} i_{v}} | + δ_{j_{t}} - δ_{j_{t}} \\ - \sum_{s = 1}^{l} (| a_{j_{t} i_{1}} |, \dots, | a_{j_{t} i_{k}} |) {[μ (A (β))]}^{- 1} (\begin{array}{c} | a_{i_{1} j_{s}} | \\ ⋮ \\ | a_{i_{k} j_{s}} | \end{array}) \\ \geq & | λ - a_{j_{t} j_{t}} | - R_{j_{t}} (A) + δ_{j_{t}}, \end{array}

that is

| λ - a_{j_{t} j_{t}} | \leq R_{j_{t}} (A) - δ_{j_{t}} \leq R_{j_{t}} (A) .

Thus, Inequation (5) holds. □

Theorem 5 Let

A = (a_{i j}) \in C^{n \times n}

,

β = {i_{1}, i_{2}, \dots, i_{k}} \subseteq N_{r} (A) \cap N_{c} (A) \neq ϕ

,

\bar{β} = {j_{1}, j_{2}, \dots, j_{l}}

,

k < n

, and let

A / β = (a_{t s}^{'})

. Then for any

0 \leq α \leq 1

and every eigenvalue λ of

A / β

, there exists

1 \leq t \leq l

such that

| λ - a_{j_{t} j_{t}} | \leq {(R_{j_{t}} (A) - δ_{t})}^{α} {(C_{j_{t}} (A) - δ_{t}^{T})}^{1 - α} \leq {(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 \leq t \leq l

such that

| λ - a_{t t}^{'} | \leq {(R_{t} (A / β))}^{α} {(C_{t} (A / β))}^{1 - α}, 0 \leq α \leq 1 .

Thus,

\begin{array}{rcl} 0 & \geq & | 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}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) | \\ {- [\sum_{s = 1, \neq t}^{l} | a_{j_{t} j_{s}} + (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{s}} \\ ⋮ \\ a_{i_{k} j_{s}} \end{array}) |]}^{α} \\ {\times [\sum_{s = 1, \neq t}^{l} | a_{j_{s} j_{t}} + (a_{j_{s} i_{1}}, \dots, a_{j_{s} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) |]}^{1 - α} \\ \geq & | λ - a_{j_{t} j_{t}} | - | (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) | \\ {- (\sum_{s = 1, \neq t}^{l} [| a_{j_{t} j_{s}} | + | (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{s}} \\ ⋮ \\ a_{i_{k} j_{s}} \end{array}) |])}^{α} \\ {\times (\sum_{s = 1, \neq t}^{l} [| a_{j_{s} j_{t}} | + | (a_{j_{s} i_{1}}, \dots, a_{j_{s} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) |])}^{1 - α} . \end{array}

Similarly to the proof of Theorem 2, we can prove

\begin{aligned} | (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) | \\ {+ (\sum_{s = 1, \neq t}^{l} [| a_{j_{t} j_{s}} | + | (a_{j_{t} i_{1}}, \dots, a_{j_{t} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{s}} \\ ⋮ \\ a_{i_{k} j_{s}} \end{array}) |])}^{α} \\ {\times (\sum_{s = 1, \neq t}^{l} [| a_{j_{s} j_{t}} | + | (a_{j_{s} i_{1}}, \dots, a_{j_{s} i_{k}}) {[A (β)]}^{- 1} (\begin{array}{c} a_{i_{1} j_{t}} \\ ⋮ \\ a_{i_{k} j_{t}} \end{array}) |])}^{1 - α} \\ \leq {(R_{j_{t}} (A) - δ_{t})}^{α} {(C_{j_{t}} (A) - δ_{t}^{T})}^{1 - α} . \end{aligned}

Therefore, we have

0 \geq | λ - a_{t t}^{'} | - {(R_{t} (A / β))}^{α} {(C_{t} (A / β))}^{1 - α} \geq | 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 = (\begin{array}{c} 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 \end{array}), β = {1, 3} .

By calculation with Matlab 7.1, we have that

\begin{matrix} 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 . \end{matrix}

Since

β \in N_{r} (A)

, by Theorem 4, any eigenvalue λ of

A / β

satisfies

λ \in {λ : | λ - 15 | \leq 9.2727} \cup {λ : | λ - 8 | \leq 9.5455} \cup {λ : | λ - 9 | \leq 8.1818} = Γ_{1} .

From Theorem 3 of [2], any eigenvalue λ of

A / β

satisfies

λ \in {λ : | λ - 15 | \leq 10.1250} \cup {λ : | λ - 8 | \leq 11.2500} \cup {λ : | λ - 9 | \leq 9.3750} = Γ_{1}^{'} .

Evidently,

Γ_{1} \subset 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 \in N_{r} (A) \cap N_{c} (A)

, by taking

α = 0.5

in Theorem 5, any eigenvalue λ of

A / β

satisfies

λ \in {λ : | λ - 15 | \leq 8.4968} \cup {λ : | λ - 8 | \leq 9.6648} \cup {λ : | λ - 9 | \leq 9.3002} = Γ_{2} .

From Theorem 4 of [2], any eigenvalue λ of

A / β

satisfies

λ \in {λ : | λ - 15 | \leq 8.8939} \cup {λ : | λ - 8 | \leq 10.5067} \cup {λ : | λ - 9 | \leq 9.9804} = Γ_{2}^{'} .

Evidently,

Γ_{2} \subset 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

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