# The Diagonally Dominant Degree and Disc Separation for the Schur Complement of Ostrowski Matrix

Authors: Jianxing Zhao; Feng Wang; Yaotang Li
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

By applying the properties of Schur complement and some inequality techniques, some new estimates of diagonally and doubly diagonally dominant degree of the Schur complement of Ostrowski matrix are obtained, which improve the main results of Liu and Zhang (2005) and Liu et al. (2012). As an application, we present new inclusion regions for eigenvalues of the Schur complement of Ostrowski matrix. In addition, a new upper bound for the infinity norm on the inverse of the Schur complement of Ostrowski matrix is given. Finally, we give numerical examples to illustrate the theory results.


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## Body

## 1. Introduction

Letℂn×n denote the set of all n×n complex matrices, N={1,2,…,n}, and A=(aij)∈ℂn×n(n≥2). Denote
(1)Ri(A)=∑j≠i‍|aij|.We know thatA is called a strictly diagonally dominant matrix if
(2)|aii|>Ri(A),∀i∈N.A is called a generalized Ostrowski matrix if
(3)|aii||ajj|≥Ri(A)Rj(A),∀i,j∈N,i≠j.A is called Ostrowski matrix if all strict inequalities in (3) hold (see [1]).S
D
n and OSn(GOSn) will be used to denote the sets of all n×n strictly diagonally dominant matrices and the sets of all n×n (generalized) Ostrowski matrices, respectively.As shown in [2], for all i∈N, we call |aii|-Ri(A) and |aii||ajj|-Ri(A)Rj(A) the ith diagonally and doubly diagonally dominant degree of A, respectively.The infinity norm ofA is defined as
(4)∥A∥∞=max1≤i≤n{Ri(A)+|aii|}.Forβ⊆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 with row indices in β and column indices in γ. In particular, A(β,β) is abbreviated to A(β). Assuming that β={i1,i2,…,ik}⊂N, β¯=N/β={j1,j2,…,jl} and the elements of β and β¯ are both conventionally arranged in an increasing order. For 1≤t≤l, we denote
(5)At=A(β∪{jt}).
If A(β) is nonsingular,
(6)Aβ=AA(β)=A(β¯)-A(β¯,β)[A(β)]-1A(β,β¯)
is called the Schur complement of A with respect to A(β).The comparison matrix ofA, μ(A)=(αij), is defined by
(7)αij={|aij|,ifi=j,-|aij|,ifi≠j.
A matrix A=(aij)∈ℂn×n is called an M-matrix if there exist a nonnegative matrix B and a number s>ρ(B) such that A=sI-B, where ρ(B) is the spectral radius of B. We know 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]). Hn and Mn will denote the set of all n×n  H-matrices and the set of all n×n  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. A lot of work has been done on it (see [2, 4–15]). It is well known that the Schur complements of SDn and OSn are SDn and OSn, respectively. These properties have been used for the derivation of matrix inequalities in matrix analysis and for the convergence of iterations in numerical analysis (see [16–19]). Meanwhile, estimating the upper bound for the infinity norm of the inverse of the Schur complement is of great significance. We know that the upper bound of ∥A-1∥∞ plays an important role in some iterations for large scale nonhomogeneous system of linear equation Ax=b (see [20]).The paper is organized as follows. In Section2, we give several new estimates of diagonally and doubly diagonally dominant degree on the Schur complement of matrices. In Section 3, new inclusion regions for eigenvalues of the Schur complement are obtained. A new upper bound of ∥(A/β)-1∥∞ is given in Section 4. In Section 5, we present numerical examples to illustrate the theory results.


## 2. The Diagonally Dominant Degree for the Schur Complement

In this section, we give several new estimates of diagonally and doubly diagonally dominant degree on the Schur complement ofOSn.Lemma 1 (see [3]).
IfA∈Hn, then [μ(A)]-1≥|A-1|.Lemma 2 (see [3]).
IfA∈SDn or A∈OSn, then A∈Hn; that is, μ(A)∈Mn.Lemma 3 (see [6]).
IfA∈SDn or A∈OSn and β⊆N, then the Schur complement of A is in SD|β¯| or  OS|β¯|, where β¯=N-β is the complement of β in N and |β¯| is the cardinality of β¯.Lemma 4 (see [12]).
LetA∈SDn, β={i1,i2,…,ik}⊂N, β¯={j1,j2,…,jl}, and k+l=n. For any jt∈β¯, denote
(8)Bjt≡(x-|ajti1|⋯-|ajtik|-∑v=1l‍|ai1jv|⋮μ(A(β))-∑v=1l‍|aikjv|),x>0.
Then Bjt∈GOSk+1 if and only if
(9)x≥max1≤w≤kRiw(A)|aiwiw|∑v=1k‍|ajtiv|.
When the strict inequality in (9) holds, Bjt∈Mk+1, and thus detBjt>0. If the equality in (9) occurs, then detBjt≥0.Lemma 5.
LetA=(aij)∈OSn and β={i1,i2,…,ik} with an index id(1≤d≤k) satisfying |aidid|≤Rid(A), |aidid|>∑iu∈β/{id}|aidiu|, β¯={j1,j2,…,jl}, 1≤k<n, and A/β=(ats′). Then, for all 1≤t≤l,
(10)|att′|-Rt(Aβ)≥|ajtjt|-Rjt(A)+|aidid|-Pid(A)|aidid|∑v=1k‍|ajtiv|≥|ajtjt|-Pid(A)|aidid|Rjt(A)>0,
where
(11)h=max{maxi∈N/{id}|aiid||aii|-∑j∈N/{i,id}|aij|,|aidid|Rid(A)},Pid(A)=hRid(A).Proof.
From Lemmas2 and 3, we know that A(β)∈Hk and μ(A(β))∈Mk. Further, by Lemma 1, we have
(12)[μ(A(β))]-1≥|[A(β)]-1|.
Thus, for any 1≤t≤l,
(13)|att′|-Rt(Aβ)=|att′|-∑s=1,≠tl‍|ats′|=|ajtjt-(ajti1,…,ajtik)[A(β)]-1(ai1jt⋮aikjt)|-∑s≠tl‍|ajtjs-(ajti1,…,ajtik)[A(β)]-1(ai1js⋮aikjs)|≥|ajtjt|-Rjt(A)+|aidid|-Pid(A)|aidid|∑v=1k‍|ajtiv|+Pid(A)|aidid|∑v=1k‍|ajtiv|-∑s=1l‍(|ajti1|,…,|ajtik|)[μ(A(β))]-1(|ai1js|⋮|aikjs|).
Further,
(14)|att′|-Rt(Aβ)≥|ajtjt|-Rjt(A)+|aidid|-Pid(A)|aidid|∑v=1k‍|ajtiv|+1det[μ(A(β))]×det(Pid(A)|aidid|∑v=1k‍|ajtiv|-|ajti1|⋯-|ajtik|-∑s=1l‍|ai1js|⋮μ(A(β))-∑s=1l‍|aikjs|)=def.|ajtjt|-Rjt(A)+|aidid|-Pid(A)|aidid|∑v=1k‍|ajtiv|+1det[μ(A(β))]detB.
By Lemma 4, we can prove that detB≥0. Thus, inequality (10) holds.Remark 6.
Note that(15)Pid(A)|aidid|≤Rid(A)|aidid|.
This shows that Lemma 5 improves Theorem 2 of [12].Theorem 7.
LetA=(aij)∈OSn, β={i1,i2,…,ik}⊂N, β¯=N/β={j1,j2,…,jl}, 1≤k<n, and A/β=(ats′).(a)
If there exists anid∈β(1≤d≤k) such that |aidid|≤Rid(A), then, for all 1≤s,  t≤l,  t≠s,
(16)|att′||ass′|-Rt(Aβ)Rs(Aβ)≥[|ajtjt|-maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βPiv(A)|aiviv|Rjs(A)],(17)|att′||ass′|+Rt(Aβ)Rs(Aβ)≤[|ajtjt|+maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|+maxiv∈βPiv(A)|aiviv|Rjs(A)],where(18)Pi(A)=h∑j∈N/{i,id}‍|aij|+|aiid|(i≠id),andPid(A) and h are such as in Lemma 5.(b)
If|aidid|>Rid(A) for any id∈β(1≤d≤k), then, for all 1≤s,  t≤l,  t≠s,
(19)|att′||ass′|-Rt(Aβ)Rs(Aβ)≥[|ajtjt|-maxu∈N/{jt}Ru(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)],(20)|att′||ass′|+Rt(Aβ)Rs(Aβ)≤[|ajtjt|+maxu∈N/{jt}Ru(A)|auu|Rjt(A)]×[|ajsjs|+maxiv∈βQiv(A)|aiviv|Rjs(A)],where(21)η=max{max1≤ω≤k∑v=1l|aiωjv||aiωiω|-∑t≠ωk|aiωit|,max1≤ω≤k1≤v≤l|aiωjv|∑v=1l|aiωjv|},Qiω(A)=η∑t≠ωk‍|aiωit|+∑v=1l‍|aiωjv|,1≤ω≤k,and if there exists some1≤ω≤k such that ∑v=1l|aiωjv|=0, one denotes η=1.Proof.
(
a
) If there exists an id∈β such that |aidid|≤Rid(A), then, for all jt∈β¯,
(22)maxu∈N/{jt}Pu(A)|auu|=maxiv∈βPiv(A)|aiviv|=Pid(A)|aidid|.
By Lemma 5, for all 1≤t≤l,
(23)|att′|-Rt(Aβ)≥|ajtjt|-Pid(A)|aidid|Rjt(A)>0.
Thus, for all 1≤t,  s≤l,  t≠s,
(24)[|att′|-Rt(Aβ)][|ass′|-Rs(Aβ)]>0.
From Lemma 3, A/β is in OS|β¯|; that is, for all 1≤t,  s≤l,  t≠s,
(25)|att′||ass′|-Rt(Aβ)Rs(Aβ)>0.
Further, for all 1≤t,  s≤l,  t≠s,
(26)|att′||ass′|-Rt(Aβ)Rs(Aβ)≥[|att′|-Rt(Aβ)]×[|ass′|-Rs(Aβ)]≥[|ajtjt|-maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βPiv(A)|aiviv|Rjs(A)].
Therefore, inequality (16) holds. Similarly, we can prove inequality (17).
(
b
)  If |aidid|>Rid(A) for any id∈β(1≤d≤k), then, from Lemmas 1 and 2, for all 1≤t,  s≤l,  t≠s,
(27)|att′||ass′|-Rt(Aβ)Rs(Aβ)=|ajsjs-(ajsi1,…,ajsik)[A(β)]-1(ai1js⋮aikjs)|×|ajtjt-(ajti1,…,ajtik)[A(β)]-1(ai1jt⋮aikjt)|-[∑v≠sl‍|(ai1jv⋮aikjv)ajsjv-(ajsi1,…,ajsik)×[A(β)]-1(ai1jv⋮aikjv)|]×[∑u≠tl‍|(ai1ju⋮aikju)ajtju-(ajti1,…,ajtik)×[A(β)]-1(ai1ju⋮aikju)|]≥[(|ai1js|⋮|aikjs|)|ajsjs|-(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1js|⋮|aikjs|)]×[(|ai1jt|⋮|aikjt|)|ajtjt|-(|ajti1|,…,|ajtik|)×[μ(A(β))]-1(|ai1jt|⋮|aikjt|)]-{∑v≠sl‍[(|ai1jv|⋮|aikjv|)|ajsjv|+(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1jv|⋮|aikjv|)]}×{∑u≠tl‍[(|ai1ju|⋮|aikju|)|ajtju|+(|ajti1|,…,|ajtik|)×[μ(A(β))]-1(|ai1ju|⋮|aikju|)]}=def.ξ=[(|ai1js|⋮|aikjs|)|ajsjs|-(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1js|⋮|aikjs|)]×[(|ai1jt|⋮|aikjt|)|ajtjt|-maxu∈N/{jt}Ru(A)|auu|Rjt(A)+maxu∈N/{jt}Ru(A)|auu|Rjt(A)-(|ajti1|,…,|ajtik|)[μ(A(β))]-1(|ai1jt|⋮|aikjt|)]-{∑v≠sl‍[(|ai1jv|⋮|aikjv|)|ajsjv|-(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1jv|⋮|aikjv|)]}×{∑u≠tl‍[(|ai1ju|⋮|aikju|)|ajtju|-(|ajti1|,…,|ajtik|)×[μ(A(β))]-1(|ai1ju|⋮|aikju|)]}.
Therefore,(28)ξ=[(|ai1js|⋮|aikjs|)|ajsjs|-(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1js|⋮|aikjs|)]×[|ajtjt|-maxu∈N/{jt}Ru(A)|auu|Rjt(A)]+[(|ai1js|⋮|aikjs|)|ajsjs|-(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1js|⋮|aikjs|)]×[(|ai1jt|⋮|aikjt|)maxu∈N/{jt}Ru(A)|auu|Rjt(A)-(|ajti1|,…,|ajtik|)[μ(A(β))]-1(|ai1jt|⋮|aikjt|)]-{∑v≠sl‍[(|ai1jv|⋮|aikjv|)|ajsjv|-(|ajsi1|,…,|ajsik|)×[μ(A(β))]-1(|ai1jv|⋮|aikjv|)]}×{(|ai1ju|⋮|aikju|)∑u≠tl‍[(|ai1ju|⋮|aikju|)|ajtju|-(|ajti1|,…,|ajtik|)×[μ(A(β))]-1(|ai1ju|⋮|aikju|)]}.
Further,(29)|ajsjs|-(|ajsi1|,…,|ajtik|)[μ(A(β))]-1(|ai1js|⋮|aikjs|)=|ajsjs|-maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|+maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|-(η|ajsi1|,…,η|ajsik|)[ημ(A(β))]-1(|ai1js|⋮|aikjs|)=|ajsjs|-maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|+1det[ημ(A(β))]×det(maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|-η|ajsi1|⋯-η|ajsik|-|ai1js|⋮ημ(A(β))-|aikjs|)=def.|ajsjs|-maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|+1det[ημ(A(β))]detB1.
InB1, for all p=1,2,3,…,k,
(30)η|aipip|maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|≥η|aipip|Qip(A)|aipip|∑v=1k‍|ajsiv|=ηQip(A)∑v=1k‍|ajsiv|=η(η∑v≠pk‍|aipiv|+∑v=1l‍|aipjv|)∑v=1k‍|ajsiv|≥(η∑v≠pk‍|aipiv|+|aipjs|)∑v=1k‍|ηajsiv|.
And for all p,q=1,2,3,…,k, p≠q,
(31)η|aipip|η|aiqiq|>ηRip(A)ηRiq(A)=(η∑v≠pk‍|aipiv|+η∑v=1l‍|aipjv|)(η∑v≠qk‍|aiqiv|+η∑v=1l‍|aiqjv|)≥(η∑v≠pk‍|aipiv|+|aipjs|)(η∑v≠qk‍|aiqiv|+|aiqjs|).
Hence, by (30) and (31), we have B1∈GOSk+1 and so detB1≥0. Further, by (29), we obtain
(32)|ajsjs|-(|ajsi1|,…,|ajtik|)[μ(A(β))]-1(|ai1js|⋮|aikjs|)≥|ajsjs|-maxiv∈βQiv(A)|aiviv|∑v=1k‍|ajsiv|≥|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A).
By (28) and a similar method as the proof of Theorem 2.1 in [2], we can prove ξ>0. Therefore, by (29) and (32), we obtain inequality (19). Similarly, we can prove inequality (20).Remark 8.
Note that(33)0≤h,η≤1.
This shows that Theorem 7 improves Theorem 2.1 of [2].


## 3. Eigenvalue Inclusion Regions of the Schur Complement

In this section, we present new inclusion regions for eigenvalues of the Schur complement ofOSn.Lemma 9 (Brauer Ovals theorem).
LetA=(aij)∈ℂn×n. Then the eigenvalues of A are in the union of the following sets:
(34)Uij={z∈C∥z-aii∥z-ajj∣≤Ri(A)Rj(A)},∀i,j=N,i≠j.Theorem 10.
LetA=(aij)∈OSn, β={i1,i2,…,ik}⊂N, β¯=N/β={j1,j2,…,jl}, 1≤k<n, and A/β=(ats′), and let λ be eigenvalue of A/β.(a)
If there exists anid∈β(1≤d≤k) such that |aidid|≤Rid(A), then there exist 1≤t,  s≤l, t≠s, such that
(35)|λ-det(At)detA(β)||λ-det(As)detA(β)|≤2[|ajtjt|maxiv∈βPiv(A)|aiviv|Rjs|ajsjs|maxu∈N/{jt}Pu(A)|auu|Rjt(A)+|ajtjt|maxiv∈βPiv(A)|aiviv|Rjs(A)],(36)|λ-det(At)detA(β)||λ-det(As)detA(β)|≤[|ajtjt|+maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|+maxiv∈βPiv(A)|aiviv|Rjs(A)],wherePid(A) is such as in Lemma 5 and Piv(A)(v≠d) is such as in Theorem 7.(b)
If|aidid|>Rid(A) for any id∈β(1≤d≤k), then there exist 1≤t,  s≤l, t≠s, such that
(37)|λ-det(At)detA(β)||λ-det(As)detA(β)|≤2[|ajsjs|maxu∈N/{jt}Ru(A)|auu|Rjt(A)+|ajtjt|maxiv∈βQiv(A)|aiviv|Rjs(A)],(38)|λ-det(At)detA(β)||λ-det(As)detA(β)|≤[|ajtjt|+maxu∈N/{jt}Ru(A)|auu|Rjt(A)]×[|ajsjs|+maxiv∈βQiv(A)|aiviv|Rjs(A)],whereQiv(A) is such as in Theorem 7.Proof.
By Lemma9, we know that there exist 1≤t,  s≤l, t≠s, such that
(39)|λ-att′||λ-ass′|≤Rt(Aβ)Rs(Aβ).(a)
If there existsid∈β satisfying |aidid|≤Rid(A), by (16), we have
(40)Rt(Aβ)Rs(Aβ)≤|att′||ass′|-[|ajtjt|-maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βPiv(A)|aiviv|Rjs(A)]=|ajsjs-(ajsi1,…,ajsik)[A(β)]-1(ai1js⋮aikjs)|×|ajtjt-(ajti1,…,ajtik)[A(β)]-1(ai1jt⋮aikjt)|-[|ajtjt|-maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βPiv(A)|aiviv|Rjs(A)]≤[|ajtjt|+maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|+maxiv∈βPiv(A)|aiviv|Rjs(A)]-[|ajtjt|-maxu∈N/{jt}Pu(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βPiv(A)|aiviv|Rjs(A)]=2[|ajsjs|maxu∈N/{jt}Pu(A)|auu|Rjt(A)+|ajtjt|maxiv∈βPiv(A)|aiviv|Rjs(A)].On the other hand, for all1≤t≤l,
(41)|λ-att′|=|λ-ajtjt+(ajti1,…,ajtik)[A(β)]-1(ai1jt⋮aikjt)|=|λ-det(AtA(β))|=|λ-det(At)detA(β)|.Therefore, by (39), (40), and (41), we obtain inequality (35). With a Similar method, we can prove inequality (36).(b)
If|aidid|>Rid(A) for any id∈β(1≤d≤k), then by (19), (32), and a similar method as the part (a), we obtain inequality (37). Similarly, we can prove inequality (38).


## 4. Upper Bound for the Infinity Norm on the Inverse of the Schur Complement

In this section, we present a new upper bound of∥(A/β)-1∥∞.Lemma 11 (see [2]).
LetA=(aij)∈OSn and M=(mij)∈ℂn×n. Then,
(42)∥A-1M∥∞≤max1≤i,j≤ni≠j|ajj|∑v=1n|miv|+Ri(A)∑v=1n|mjv||aii||ajj|-Ri(A)Rj(A).Theorem 12.
LetA=(aij)∈SDn, M=(mij)∈ℂl×l, β={i1,i2,…,ik}⊂N, β¯=N/β={j1,j2,…,jl}, 1≤k<n, and A/β=(ats′). Then,
(43)∥(Aβ)-1M∥∞≤max1≤t,s≤lt≠s((|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)))-1(|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)))-1Δjtjs×((|ajtjt|-maxu∈N/{jt}Ru(A)|auu|Rjt(A))×(|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)))-1),(44)∥(Aβ)-1∥∞≤max1≤t,s≤lt≠s((|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)))-1(maxiv∈βQiv(A)|aiviv|(Rjt(A)+Rjs(A))|ajtjt|+Rjs(A)+maxiv∈βQiv(A)|aiviv|(Rjt(A)+Rjs(A)))×((|ajtjt|-maxu∈N/{jt}Ru(A)|auu|Rjt(A))×(|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)))-1),
where
(45)Δjtjs=(|ajtjt|+maxiv∈βQiv(A)|aiviv|Rjt(A))∑v=1l‍|msv|+(Rjs(A)+maxiv∈βQiv(A)|aiviv|Rjs(A))∑v=1l‍|mtv|,
and Qiv(A) is such as in Theorem 7.Proof.
By Lemma11, we have
(46)∥(Aβ)-1M∥∞≤max1≤t,s≤lt≠s|att′|∑v=1l|msv|+Rs(A/β)∑v=1l|mtv||att′||ass′|-Rt(A/β)Rs(A/β).
Similar to (29), we obtain
(47)|att′|=|ajtjt-(ajti1,…,ajtik)[A(β)]-1(ai1jt⋮aikjt)|≤|ajtjt|+(|ajti1|,…,|ajtik|)[μ(A(β))]-1(|ai1jt|⋮|aikjt|)≤|ajtjt|+maxiv∈βQiv(A)|aiviv|Rjt(A).
Thus, by Theorem 1 of [12], we have
(48)Rs(Aβ)≤|ass′|-|ajsjs|+Rjs(A)≤maxiv∈βQiv(A)|aiviv|Rjs(A)+Rjs(A).
Since A∈SDn, then A∈OSn. Thus, by Theorem 7, we have
(49)|att′||ass′|-Rt(Aβ)Rs(Aβ)≥[|ajtjt|-maxu∈N/{jt}Ru(A)|auu|Rjt(A)]×[|ajsjs|-maxiv∈βQiv(A)|aiviv|Rjs(A)].
Further, by (46), (47), (48), and (49), we obtain inequality (43).
LetM=I=diag(1,1,…,1); we can prove inequality (44).


## 5. Numerical Examples

In this section, we present several numerical examples to illustrate the theory results.Example 1 (see Example 2 in [2]).
Let(50)A=(1.30.20.30.40.50.220.40.50.10.30.420.10.20.40.50.130.30.50.10.20.33),β={1,2}.
By Theorem10, the eigenvalues of A/β are in the set
(51)Γ1={λ∥λ-1.87∥λ-2.78∣≤11.20}∪{λ∥λ-1.87∥λ-2.81∣≤10.40}∪{λ∥λ-2.78∥λ-2.81∣≤14.40}.
From Theorem 3.1 of [2], the eigenvalues of A/β are in the set
(52)Γ1′={λ∥λ-1.87∥λ-2.78∣≤12.06}∪{λ∥λ-1.87∥λ-2.81∣≤11.20}∪{λ∥λ-2.78∥λ-2.81∣≤15.51}.
Evidently, Γ1⊂Γ1′, and we use Figure 1 to show this fact. And the eigenvalues of A/β are denoted by “+” in Figure 1.Figure 1
The red dotted line and black dashed line denote the corresponding discsΓ1 and Γ1′, respectively.Example 2.
Let(53)A=(1.60.10.50.20.20.31.50.20.20.10.20.21.80.30.40.50.30.51.00.20.50.20.20.31.9),β={2,4}.
By Theorem10, the eigenvalues of A/β are in the set
(54)Γ2={λ∥λ-1.49∥λ-1.64∣≤7.12}∪{λ∥λ-1.49∥λ-1.84∣≤7.64}∪{λ∥λ-1.64∥λ-1.84∣≤8.50}.
From Theorem 3.1 of [2], the eigenvalues of A/β are in the set
(55)Γ2′={λ∥λ-1.49∥λ-1.64∣≤10.68}∪{λ∥λ-1.49∥λ-1.84∣≤11.46}∪{λ∥λ-1.64∥λ-1.84∣≤12.75}.
Evidently, Γ2⊂Γ2′, and we use Figure 2 to show this fact. And the eigenvalues of A/β are denoted by “+” in Figure 2.Figure 2
The red dotted line and black dashed line denote the corresponding discsΓ2 and Γ2′, respectively.Example 3.
Let(56)A=(1.50.20.30.20.10.20.21.20.10.30.10.40.60.21.60.10.20.10.50.20.11.80.30.20.20.10.20.31.30.10.10.20.10.31.22.5),β={1,3,5}.
By Theorem10, the eigenvalues of A/β are in the set
(57)Γ3={λ∥λ-1.16∥λ-1.68∣≤4.79}∪{λ∥λ-1.16∥λ-2.42∣≤6.78}∪{λ∥λ-1.68∥λ-2.42∣≤9.85}.
From Theorem 3.1 of [2], the eigenvalues of A/β are in the set
(58)Γ3′={λ∥λ-1.16∥λ-1.68∣≤5.35}∪{λ∥λ-1.16∥λ-2.42∣≤7.60}∪{λ∥λ-1.68∥λ-2.42∣≤11.09}.
Evidently, Γ3⊂Γ3′, and we use Figure 3 to show this fact. And the eigenvalues of A/β are denoted by “+” in Figure 3.
Meanwhile, by Theorem12,
(59)∥(Aβ)-1∥∞≤12.22.
From Theorem 4.2 of [2],
(60)∥(Aβ)-1∥∞≤20.60.Figure 3
The red dotted line and black dashed line denote the corresponding discsΓ3 and Γ3′, respectively.Remark 13.
Numerical examples show that the new eigenvalue inclusion set is tighter than that in Theorem3.1 of [2] and the new upper bound of ∥(A/β)-1∥∞ is sharper than that in Theorem 4.2 of [2].



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*Source: 973152-2013-10-02.xml*