Eigenvalue inclusion sets for linear response eigenvalue problems

Jun He; Yanmin Liu; Wei Lv

doi:10.1515/dema-2022-0029

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Eigenvalue inclusion sets for linear response eigenvalue problems

Jun He , Yanmin Liu and Wei Lv

Published/Copyright: August 5, 2022

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From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

In this article, some inclusion sets for eigenvalues of a matrix in the linear response eigenvalue problem (LREP) are established. It is proved that the inclusion sets are tighter than the Geršgorin-type sets. A numerical experiment shows the effectiveness of our new results.

Keywords: linear response eigenvalue problem; Geršgorin-type sets; localization set

MSC 2010: 15A18; 65F15

1 Introduction

Consider the following linear response eigenvalue problem (LREP):

(1) H z = 0 M K 0 u v = λ u v = λ z ,

where K = ( k i j ) = A − B , M = ( m i j ) = A + B ∈ R n × n are symmetric positive definite, and

A B B A

is positive definite. The LREP (1) plays an important role in computational quantum chemistry and physics [1,2], which is also called random phase approximation eigenvalue problem or the Bethe-Salpeter eigenvalue problem [3,4].

The eigenvalues of H are real and come in pairs { λ , − λ } , and we denote eigenvalues of H by

− λ n ≤ … ≤ − λ 1 ≤ λ 1 ≤ … ≤ λ n .

If both K and M are positive definite, then λ 1 > 0 . The following minimization principle for the smallest positive eigenvalue λ 1 of the LREP (1) is given by Thouless [5],

(2) λ 1 = min x , y ρ ( x , y ) , with ρ ( x , y ) = x T K x + y T M y 2 x T y ,

which is also called Thouless’ minimization principle, and ρ ( x , y ) is called Thouless’ functional. A minimization principle for the sum of the smallest eigenvalues with the positive sign and Cauchy-like interlacing inequalities for the LREP are presented in [1]. Recently, several numerical algorithms are introduced to compute the few smallest eigenvalues with the positive sign, such as 4D subspace search conjugate gradient type methods [2,6], extended locally optimal block preconditioned 4D conjugate gradient algorithm [7], and Golub-Kahan-Lanczos process [8].

In this article, we focus on the inclusion sets, which contain all eigenvalues of a matrix in the LREP. The well-known Geršgorin’s theorem defines in the complex plane a union of n disks that contains all the n eigenvalues [9]. It is an extremely useful tool in estimating eigenvalue bounds and plays an important role in the eigenvalue perturbation theory [10,11].

Theorem 1

If a matrix A = ( a i j ) ∈ C n × n , then all eigenvalues of A are contained in

Δ ( A ) ≔ ⋃ i = 1 n Δ i ( A ) ,

where

Δ i ( A ) = { λ ∈ C : ∣ λ − a i i ∣ ≤ R i ( A ) } , R i ( A ) = ∑ j = 1 , j ≠ i n ∣ a i j ∣ .

By the Geršgorins theorem, then all eigenvalues of the LREP are contained in

(3) Δ ( H ) ≔ ⋃ i = 1 n Δ i ( H ) ,

where

Δ i ( H ) = { λ ∈ C : ∣ λ ∣ ≤ max { R i ( M ) , R i ( K ) } } , R i ( M ) = ∑ j = 1 n ∣ m i j ∣ , R i ( K ) = ∑ j = 1 n ∣ k i j ∣ .

In this article, some new inclusion sets for eigenvalues of a matrix in the LREP are established, which are always better that the eigenvalue set (3). In Section 2, two new eigenvalue inclusion sets for LREPs are presented, the comparison results among Δ ( H ) , Γ ( H ) , and Ω ( H ) are also given. A numerical experiment is presented in Section 3. Finally, a conclusion is given in Section 4.

2 Eigenvalue inclusion sets for LREPs

Theorem 2

Let H = 0 M K 0 , then all eigenvalues of H are contained in

(4) Γ ( A ) ≔ Γ 1 ( H ) ⋃ Γ 2 ( H ) ,

where

Γ 1 ( H ) ≔ ⋃ i = 1 n { λ ∈ C : ∣ λ 2 − k i i m i i ∣ ≤ ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) } , Γ 2 ( H ) ≔ ⋃ i = 1 n { λ ∈ C : ∣ λ 2 − k i i m i i ∣ ≤ ∣ λ ∣ r i ( M ) + ∣ m i i ∣ r i ( K ) } ,

and

r i ( M ) = ∑ j = 1 , j ≠ i n ∣ m i j ∣ , r i ( K ) = ∑ j = 1 , j ≠ i n ∣ k i j ∣ .

Proof

Let λ be an eigenvalue of H with vector z = ( x , y ) T , x = ( x 1 , x 2 , … , x n ) T and y = ( y 1 , y 2 , … , y n ) T , then

(5) λ x = M y and λ y = K x .

Denote

∣ x p ∣ = max { ∣ x i ∣ , 1 ≤ i ≤ n } , ∣ y q ∣ = max { ∣ y i ∣ , 1 ≤ i ≤ n } .

Now, we assume that ∣ x p ∣ ≤ ∣ y q ∣ , the q th equations in (5) imply

(6) λ x q − m q q y q = ∑ j = 1 , j ≠ q n m q j y j ,

(7) λ y q − k q q x q = ∑ j = 1 , j ≠ q n k q j x j .

Solving for y q , we can obtain

(8) ( λ 2 − k q q m q q ) y q = k q q ∑ j = 1 , j ≠ q n m q j y j + λ ∑ j = 1 , j ≠ q n k q j x j .

Taking the absolute value on both sides of equation (8) and using the triangle inequality yields

(9) ∣ λ 2 − k q q m q q ∣ ∣ y q ∣ ≤ ∣ k q q ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ ∣ y j ∣ + λ ∑ j = 1 , j ≠ q n ∣ k q j ∣ ∣ x j ∣ .

Then, we can obtain

∣ λ 2 − k q q m q q ∣ ≤ ∣ k q q ∣ r q ( M ) + ∣ λ ∣ r q ( K ) .

If ∣ y q ∣ ≤ ∣ x p ∣ , similarly, considering the p th equations in (5) and solving for x p , we can obtain

(10) ( λ 2 − k p p m p p ) x p = λ ∑ j = 1 , j ≠ p n m p j y j + m p p ∑ j = 1 , j ≠ p n k p j x j .

Taking the absolute value on both sides of equation (10) and using the triangle inequality yields

(11) ∣ λ 2 − k p p m p p ∣ ∣ x p ∣ ≤ ∣ λ ∣ ∑ j = 1 , j ≠ p n ∣ m p j ∣ ∣ y j ∣ + ∣ m p p ∣ ∑ j = 1 , j ≠ p n ∣ k p j ∣ ∣ x j ∣ .

we can obtain

∣ λ 2 − k p p m p p ∣ ≤ ∣ λ ∣ r p ( M ) + ∣ m p p ∣ r p ( K ) .

This completes the proof.□

Theorem 3

Let H = 0 M K 0 , then all eigenvalues of H are contained in

Ω ( H ) ≔ Ω 1 ( H ) ⋃ Ω 2 ( H ) ⋃ Ω 3 ( H ) ⋃ Ω 4 ( H ) ,

where

Ω 1 ( H ) ≔ ⋃ i ≠ j { λ ∈ C : ∣ λ 2 − k i i m i i ∣ ∣ λ 2 − k j j m j j ∣ ≤ ( ∣ λ ∣ r i ( M ) + ∣ m i i ∣ r i ( K ) ) ( ∣ λ ∣ r j ( M ) + ∣ m j j ∣ r j ( K ) ) } , Ω 2 ( H ) ≔ ⋃ i ≠ j { λ ∈ C : ∣ λ 2 − k i i m i i ∣ ∣ λ 2 − k j j m j j ∣ ≤ ( ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ) ( ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ) } , Ω 3 ( H ) ≔ ⋃ i ≠ j { λ ∈ C : ∣ λ 2 − k i i m i i ∣ ∣ λ 2 − k j j m j j ∣ ≤ ( ∣ λ ∣ r i ( M ) + ∣ m i i ∣ r i ( K ) ) ( ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ) } ,

and

Ω 4 ( H ) ≔ ⋃ i ≠ j { λ ∈ C : ∣ λ 2 − k i i m i i ∣ ∣ λ 2 − k j j m j j ∣ ≤ ( ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ) ( ∣ λ ∣ r j ( M ) + ∣ m j j ∣ r j ( K ) ) } .

Proof

Let λ be an eigenvalue of H with vector z = ( x , y ) T , x = ( x 1 , x 2 , … , x n ) T and y = ( y 1 , y 2 , … , y n ) T , then

(12) λ x = M y and λ y = K x .

Let ω i = max { ∣ x i ∣ , ∣ y i ∣ } , q be an index such that ω q = max { ∣ ω i ∣ , 1 ≤ i ≤ n } and p be an index such that ω p = max { ∣ ω i ∣ , 1 ≤ i ≤ n , i ≠ q } . Obviously, ω q ≠ 0 .

Case I: Suppose that ω q = ∣ x q ∣ , ω p = ∣ x p ∣ , similar to the proof of Theorem 2, the q th equations in (12) imply

(13) ∣ λ 2 − k q q m q q ∣ ω q ≤ ∣ λ ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ ∣ y j ∣ + m q q ∑ j = 1 , j ≠ q n ∣ k q j ∣ ∣ x j ∣ ≤ ∣ λ ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ + ∣ m q q ∣ ∑ j = 1 , j ≠ q n ∣ k q j ∣ ω p .

Similarly, p th equations in (12) imply

(14) ∣ λ 2 − k p p m p p ∣ ω p ≤ ∣ λ ∣ ∑ j = 1 , j ≠ p n ∣ m p j ∣ + ∣ m p p ∣ ∑ j = 1 , j ≠ p n ∣ k p j ∣ ω q .

Multiplying inequalities (13) with (14), we have

∣ λ 2 − k q q m q q ∣ ∣ λ 2 − k p p m p p ∣ ≤ ( ∣ λ ∣ r q ( M ) + ∣ m q q ∣ r q ( K ) ) ( ∣ λ ∣ r p ( M ) + ∣ m p p ∣ r p ( K ) ) .

Case II: Suppose that ω q = ∣ y q ∣ , ω p = ∣ y p ∣ , similar to the proof of Theorem 2, the q th equations in (12) imply

(15) ∣ λ 2 − k q q m q q ∣ ω q ≤ ∣ k q q ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ ∣ y j ∣ + λ ∑ j = 1 , j ≠ q n ∣ k q j ∣ ∣ x j ∣ ≤ ∣ k q q ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ + λ ∑ j = 1 , j ≠ q n ∣ k q j ∣ ω p .

And the p th equations in (12) imply

(16) ∣ λ 2 − k p p m p p ∣ ω p ≤ ∣ k p p ∣ ∑ j = 1 , j ≠ p n ∣ m p j ∣ + λ ∑ j = 1 , j ≠ p n ∣ k p j ∣ ω q .

Multiplying inequalities (15) with (16), we have

∣ λ 2 − k q q m q q ∣ ∣ λ 2 − k p p m p p ∣ ≤ ( ∣ k q q ∣ r q ( ) M + ∣ λ ∣ r q ( K ) ) ( ∣ k p p r p ( M ) + ∣ λ ∣ r p ( K ) ) .

Case III: Suppose that ω q = ∣ x q ∣ , ω p = ∣ y p ∣ , considering the q th equations in (12) and solving for x q , we can obtain

(17) ∣ λ 2 − k q q m q q ∣ ω q ≤ ∣ λ ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ ∣ y j ∣ + m q q ∑ j = 1 , j ≠ q n ∣ k q j ∣ ∣ x j ∣ ≤ ∣ λ ∣ ∑ j = 1 , j ≠ q n ∣ m q j ∣ + ∣ m q q ∣ ∑ j = 1 , j ≠ q n ∣ k q j ∣ ω p .

And the p th equations in (12) imply

(18) ∣ λ 2 − k p p m p p ∣ ω p ≤ ∣ k p p ∣ ∑ j = 1 , j ≠ p n ∣ m p j ∣ + λ ∑ j = 1 , j ≠ p n ∣ k p j ∣ ω q .

Multiplying inequalities (17) with (18), we have

∣ λ 2 − k q q m q q ∣ ∣ λ 2 − k p p m p p ∣ ≤ ( ∣ λ ∣ r q ( M ) + ∣ m q q ∣ r q ( K ) ) ( ∣ k p p ∣ r p ( M ) + ∣ λ ∣ r p ( K ) ) .

Case IV: Suppose that ω q = ∣ y q ∣ , ω p = ∣ x p ∣ , similar to the proof of Case III, we have

∣ λ 2 − k q q m q q ∣ ∣ λ 2 − k p p m p p ∣ ≤ ( ∣ k q q ∣ r q ( M ) + ∣ λ ∣ r q ( K ) ) ( ∣ λ ∣ r p ( M ) + ∣ m p p ∣ r p ( K ) ) .

This completes the proof.□

Remark

By comparing the number of the basic arithmetic operations of results in Theorems 2 and 3, we see that the result in Theorem 2 requires less basic arithmetic operations compared with the result in Theorem 3. However, the result in Theorem 3 is always tighter than the result in Theorem 2.

The comparison results among Δ ( H ) , Γ ( H ) , and Ω ( H ) are given as follows.

Theorem 4

Let H = 0 M K 0 and λ be an eigenvalue of H, then

λ ∈ Ω ( H ) ⊆ Γ ( H ) ⊆ Δ ( H ) .

Proof

First, we prove that Ω ( H ) ⊆ Γ ( H ) . Let λ be any point of Ω 2 ( H ) . Then there are i , j ∈ N , i ≠ j , such that λ ∈ Ω 2 ( H ) , i.e.,

(19) ∣ λ 2 − k i i m i i ∣ ∣ λ 2 − k j j m j j ∣ ≤ ( ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ) ( ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ) .

If ( ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ) ( ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ) = 0 , then

∣ λ 2 − k i i m i i ∣ = 0 ,

∣ λ 2 − k j j m j j ∣ = 0 .

Therefore, λ ∈ Γ ( H ) . Moreover, if ( ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ) ( ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ) > 0 , then from inequality (19), we have

(20) ∣ λ 2 − k i i m i i ∣ ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ∣ λ 2 − k j j m j j ∣ ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ≤ 1 .

Hence, from inequality (20), we have that

∣ λ 2 − k i i m i i ∣ ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ≤ 1 ,

∣ λ 2 − k j j m j j ∣ ∣ k j j ∣ r j ( M ) + ∣ λ ∣ r j ( K ) ≤ 1 .

That is, λ ∈ Γ 1 ( H ) ⊆ Γ ( H ) .

Similarly, let λ be any point of Ω 1 ( H ) , Ω 3 ( H ) or Ω 4 ( H ) , we can obtain

λ ∈ Ω 1 ( H ) ⊆ Γ ( H ) , λ ∈ Ω 3 ( H ) ⊆ Γ ( H ) ,

and

λ ∈ Ω 4 ( H ) ⊆ Γ ( H ) .

Finally, we prove that Γ ( H ) ⊆ Δ ( H ) . Let λ be any point of Γ 1 ( H ) . Then there are i ∈ N , such that

∣ λ 2 − k i i m i i ∣ ≤ ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) ,

then

λ 2 ≤ ∣ k i i m i i ∣ + ∣ k i i ∣ r i ( M ) + ∣ λ ∣ r i ( K ) .

That is,

λ 2 ≤ ∣ k i i ∣ R i ( M ) + ∣ λ ∣ r i ( K ) .

If ∣ λ ∣ ≤ R i ( M ) , then

λ 2 ≤ ∣ k i i ∣ R i ( M ) + R i ( M ) r i ( K ) ≤ R i ( M ) R i ( K ) ≤ max { R i ( M ) , R i ( K ) } ,

and therefore,

∣ λ ∣ ≤ R i ( M ) R i ( K ) ≤ max { R i ( M ) , R i ( K ) } .

If R i ( M ) ≤ ∣ λ ∣ , then

λ 2 ≤ ∣ k i i ∣ ∣ λ ∣ + ∣ λ ∣ r i ( K ) ,

and therefore,

∣ λ ∣ ≤ R i ( K ) ≤ max { R i ( M ) , R i ( K ) } .

Then, Γ 1 ( H ) ⊆ Δ ( H ) .

Similarly, we can obtain Γ 2 ( H ) ⊆ Δ ( H ) .

This completes the proof.□

3 Numerical example

Example 1

Let

M = 3 2 2 4 , K = 1 1 1 2 , H = 0 M K 0 .

M , K are symmetric and positive definite, and the eigenvalues of H are

{ − 3.8008 , − 0.7442 , 0.7442 , 3.8008 } .

In this example, we only compare the results among Ω 1 ( H ) , Γ 2 ( H ) and Δ ( H ) , and the comparison results of other eigenvalue inclusion sets can be obtain similarly. The eigenvalue inclusion sets Ω 1 ( H ) , Γ 2 ( H ) , Δ ( H ) and the exact eigenvalues are drawn in Figure 1, where Δ ( H ) is represented by green boundary, Γ 2 ( H ) is represented by black boundary, and Ω 1 ( H ) is represented by blue boundary. The exact eigenvalues are plotted by red “ ∗ .” Obviously, the results Ω 1 ( H ) in Theorem 3 are tighter than the results Γ 2 ( H ) in Theorem 2, which can be obtained in the proof of Theorem 4. And it is easy to see that λ ∈ Ω 1 ( H ) ⊆ Γ 2 ( H ) ⊆ Δ ( H ) .

Figure 1

Comparisons of eigenvalue inclusion sets.

4 Conclusion

In this article, some new eigenvalue inclusion sets are given, and theoretical analysis and numerical example show that these estimates are more efficient than the Geršgorin-type inclusion sets.

Acknowledgements

The authors would like to thank the referees for their suggestions that helped improve the original manuscript in its present form.

Funding information: This work was supported by New academic talents and innovative exploration fostering project in China(Qian Ke He Pingtai Rencai [2017]5727-21), Guizhou Province Natural Science Foundation in China (Qian Jiao He KY [2020]094), Science and Technology Foundation of Guizhou Province, China (Qian Ke He Ji Chu ZK[2021]Yi Ban 014), and General project of philosophy and Social sciences planning in Guizhou Province (19GZYB11).
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors declare no conflict of interest in this article.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2022-04-15

Revised: 2022-06-15

Accepted: 2022-06-16

Published Online: 2022-08-05

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https://doi.org/10.1515/dema-2022-0029

Keywords for this article

linear response eigenvalue problem; Geršgorin-type sets; localization set

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