Bound on the Minimum Eigenvalue of 𝐻-Matrices Involving Hadamard Products

Abstract

We present a new lower bound on the minimum eigenvalue of 𝐻-matrices involving Hadamard products 𝜏(𝐴^(𝛼1)₁∘⋯∘𝐴^(𝛼𝑚)_𝑚), and we show that our lower bound is larger than the lower bound ∏^𝑚_𝑘=1[𝜏(𝐴_𝑘)]^𝛼𝑘. Three examples verify our result.

1. Introduction

In [1], it is shown by Theorem  5.7.15 that if 𝐴_𝑘 are 𝑛×𝑛  𝐻-matrices for all 𝑘∈{1,…,𝑚}, and 𝛼_𝑘≥0 satisfy ∑^𝑚_𝑘=1𝛼_𝑘≥1, then 𝜏𝐴^(𝛼1)₁∘⋯∘𝐴^(𝛼𝑚)_𝑚≥^𝑚_𝑘=1𝜏𝐴_𝑘^𝛼𝑘, (1) where 𝐴^(𝛼) is defined as entrywise and any scalar definition of 𝑎^𝛼 such that |𝑎^𝛼|=|𝑎|^𝛼 is allowed. This theorem provided a beautiful result about the minimum eigenvalue of 𝐻-matrices involving Hadamard products, but sometimes this inequality could be very weak. For example, 𝐴=,100−21−3001−12001,𝐵=100−191000−81−171𝐶=10001000100 (2) are 𝐻-matrices, and 𝜏𝐴⁽²⁾∘𝐵⁽²⁾∘𝐶⁽²⁾=10000≫𝜏(𝐴)²𝜏(𝐵)²𝜏(𝐶)²=1, (3) see the details in Section 3.

A lot of works have been done on the minimum eigenvalue of 𝑀-matrices and 𝐻-matrices involving Hadamard products, see the results in [1–6].

In this paper, we present a new lower bound by including diagonal entries and prove that our bound is larger than the bound in (1).

We now introduce some notations, see [1]. The Hadamard product of 𝐴=[𝑎_𝑖𝑗]_𝑚×𝑛 and 𝐵=[𝑏_𝑖𝑗]_𝑚×𝑛 is defined by 𝐴∘𝐵≡[𝑎_𝑖𝑗𝑏_𝑖𝑗]_𝑚×𝑛. We define |𝑎^𝛼|=|𝑎|^𝛼 and 0⁰=1. Let 𝐴=[𝑎_𝑖𝑗]_𝑛×𝑛 and denote 𝑎_𝑖𝑗 by 𝐴(𝑖,𝑗) for every 𝑖,𝑗∈{1,…,𝑛}. We denote 𝐴^(𝑟)=[𝑎^𝑟_𝑖𝑗], where 𝑟≥0, and denote by 𝑍_𝑛 the class of all 𝑛×𝑛 real matrices all of whose off-diagonal entries are nonpositive. Let 𝐴∈𝑍_𝑛, then the minimum eigenvalue of 𝐴 is defined by 𝜏(𝐴)=min{Re(𝜆)∶𝜆∈𝜎(𝐴)}; 𝜏(𝐴) is a eigenvalue of 𝐴.

For two real matrices 𝐴,𝐵∈𝑍_𝑛, the Fan product of 𝐴 and 𝐵, denoted by 𝐴⋆𝐵=𝐶=[𝑐_𝑖𝑗], is defined by 𝑐_𝑖𝑗=−𝑎_𝑖𝑗𝑏_𝑖𝑗,if𝑎𝑖≠𝑗,_𝑖𝑖𝑏_𝑖𝑖,if𝑖=𝑗, (4) and then 𝐴⋆𝐵∈𝑍_𝑛. The comparison matrix 𝑀(𝐴)=[𝑚_𝑖𝑗] of a given matrix 𝐴=[𝑎_𝑖𝑗]∈ℂ^𝑛×𝑛 is defined by 𝑚_𝑖𝑗=−𝑎_𝑖𝑗,if𝑎𝑖≠𝑗,_𝑖𝑖,if𝑖=𝑗. (5) A matrix 𝐴 is an 𝐻-matrix if its comparison matrix 𝑀(𝐴) is an 𝑀-matrix. For 𝐴∈ℂ^𝑛×𝑛, we define 𝜏(𝐴)≡𝜏(𝑀(𝐴)).

Last, for 𝐴∈𝑍_𝑛, we introduce a new definition 𝐴^[𝑟]=[𝑑_𝑖𝑗], where 𝑑_𝑖𝑗=−𝑎_𝑖𝑗^𝑟,if𝑎𝑖≠𝑗,_𝑖𝑖𝑟,if𝑖=𝑗. (6)

2. Main Results

To prove the main theorem, we need several lemmas.

Lemma 1. Let 𝑥_𝑗=(𝑥_𝑗(1),…,𝑥_𝑗(𝑛))^𝑇≥0 for all 𝑗∈{1,…,𝑚}. If 𝑝_𝑗>0 and ∑^𝑚_𝑗=1(1/𝑝_𝑗)=𝑟≥1, then ^𝑛𝑚_𝑖=1𝑗=1𝑥_𝑗(𝑖)≤^𝑚_𝑗=1^𝑛_𝑖=1𝑥_𝑗(𝑖)^{𝑝𝑗1/𝑝𝑗}. (7)

Proof. When 𝑟=1, we can prove this inequality by the Hölder inequality and by induction. When 𝑟>1, apparently, ∑^𝑚_𝑗=1(1/𝑟𝑝_𝑗)=1, similarly, we can prove this inequality.

If 𝐴∈𝑍_𝑛, then 𝐴=𝛼𝐼−𝐵 for some 𝐵≥0 and some 𝛼∈ℝ. Let 𝜎(𝐵)={𝜌(𝐵),𝜆₂,…,𝜆_𝑛}. Then 𝜎(𝐴)={𝛼−𝜌(𝐵),𝛼−𝜆₂,…,𝛼−𝜆_𝑛}. If there is a 𝜆_𝑖 such that 𝛼−𝜌(𝐵)=Re(𝛼−𝜌(𝐵))>Re(𝛼−𝜆_𝑖)=𝛼−Re(𝜆_𝑖), then 𝜌(𝐵)<Re(𝜆_𝑖), but this is a contradiction. So, 𝜏(𝐴)=𝛼−𝜌(𝐵). If 𝑥≥0 is the right Perron eigenvector of 𝐵, then 𝐴𝑥=(𝛼𝐼−𝐵)𝑥=𝛼𝑥−𝐵𝑥=𝛼𝑥−𝜌(𝐵)𝑥=(𝛼−𝜌(𝐵))𝑥=𝜏(𝐴)𝑥. If 𝑦≥0 is the left Perron eigenvector of 𝐵, then 𝑦^𝑇𝐴=𝑦^𝑇(𝛼𝐼−𝐵)=𝛼𝑦^𝑇−𝑦^𝑇𝐵=𝛼𝑦^𝑇−𝜌(𝐵)𝑦^𝑇=(𝛼−𝜌(𝐵))𝑦^𝑇=𝜏(𝐴)𝑦^𝑇. So, similar to the Perron-Frobenius theorem, we have the following: if 𝐴∈𝑍_𝑛 became irreducible, then there exist positive vectors 𝑢 and 𝑣 such that 𝐴𝑢=𝜏(𝐴)𝑢, and 𝑣^𝑇𝐴=𝜏(𝐴)𝑣^𝑇, 𝑢 and 𝑣 being called right and left Perron eigenvectors of 𝐴, respectively.

Lemma 2. If 𝐴∈𝑍_𝑛 is irreducible, and 𝐴𝑧≥𝑘𝑧 for a nonnegative nonzero vector 𝑧, then 𝑘≤𝜏(𝐴).

Proof. 𝐴=𝛼𝐼−𝑃, where 𝑃≥0 is irreducible and 𝛼∈ℝ. By the Perron-Frobenius theorem, 𝑃 has a positive left Perron vector 𝑦, that is, 𝑦^𝑇𝑃=𝜌(𝑃)𝑦^𝑇. Note that 𝐴𝑧−𝑘𝑧=(𝛼−𝑘)𝑧−𝑃𝑧≥0. Hence, 𝑦^𝑇(𝐴𝑧−𝑘𝑧)=(𝛼−𝑘)𝑦^𝑇𝑧−𝜌(𝑃)𝑦^𝑇𝑧=(𝛼−𝑘−𝜌(𝑃))𝑦^𝑇𝑧≥0. Since 𝑦^𝑇𝑧>0, we have 𝑘≤𝛼−𝜌(𝑃)=𝜏(𝐴).

Lemma 3. Let 𝐴_𝑘∈𝑍_𝑛 for all 𝑘∈{1,…,𝑚}. If 𝑝_𝑘>0 and ∑^𝑚_𝑘=1(1/𝑝_𝑘)≥1, then 𝜏𝐴₁⋆⋯⋆𝐴_𝑚≥min_{1≤𝑖≤𝑛}^𝑚_𝑘=1𝐴_𝑘()−𝑖,𝑖_𝑚𝑘=1𝐴_𝑘(𝑖,𝑖)^𝑝𝑘−𝜏𝐴^[𝑝𝑘]_𝑘^1/𝑝𝑘. (8)

Proof. It is quite evident that the conclusion holds with equality for 𝑛=1.
For 𝑛≥2, we have two cases.
Case  1. 𝐴₁⋆⋯⋆𝐴_𝑚∈𝑍_𝑛 is irreducible. Then 𝐴_𝑘∈𝑍_𝑛 are irreducible for all 𝑘∈{1,…,𝑚}. Thus, 𝐴^[𝑝𝑘]_𝑘∈𝑍_𝑛 are irreducible for all 𝑘∈{1,…,𝑚}. Let 𝑢^(𝑝𝑘)_𝑘=(𝑢_𝑘(1)^𝑝𝑘,…,𝑢_𝑘(𝑛)^𝑝𝑘)^𝑇>0 be the right Perron eigenvectors of 𝐴^[𝑝𝑘]_𝑘 for all 𝑘∈{1,…,𝑚}. If 𝑢_𝑘⁽¹⁾=𝑢_𝑘=(𝑢_𝑘(1),…,𝑢_𝑘(𝑛))^𝑇>0 for all 𝑘∈{1,…,𝑚}, then for every 𝑖∈{1,…,𝑛}, we have 𝐴^[𝑝𝑘]_𝑘𝑢^(𝑝𝑘)_𝑘=𝜏𝐴^[𝑝𝑘]_𝑘𝑢^(𝑝𝑘)_𝑘,𝐴_𝑘(𝑖,𝑖)^𝑝𝑘𝑢_𝑘(𝑖)^𝑝𝑘−_𝑗≠𝑖𝐴_𝑘𝑖,𝑗^𝑝𝑘𝑢_𝑘𝑗^𝑝𝑘=𝜏𝐴^[𝑝𝑘]_𝑘𝑢_𝑘(𝑖)^𝑝𝑘,_𝑗≠𝑖𝐴_𝑘𝑖,𝑗^𝑝𝑘𝑢_𝑘𝑗^𝑝𝑘=𝐴_𝑘(𝑖,𝑖)^𝑝𝑘−𝜏𝐴^[𝑝𝑘]_𝑘𝑢_𝑘(𝑖)^𝑝𝑘≥0. (9) Let 𝐶=𝐴₁⋆⋯⋆𝐴_𝑚∈𝑍_𝑛. Let 𝑧=𝑢₁∘⋯∘𝑢_𝑚=(𝑧(1),…,𝑧(𝑛))^𝑇>0. Then 𝑧(𝑖)=∏^𝑚_𝑘=1𝑢_𝑘(𝑖) for all 𝑖∈{1,…,𝑛}. Then, for all 𝑖∈{1,…,𝑛}, we have (𝐶𝑧)_𝑖=^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)𝑧(𝑖)−_𝑗≠𝑖^𝑚_𝑘=1𝐴_𝑘𝑧𝑗=𝑖,𝑗_𝑚𝑘=1𝐴_𝑘(𝑖,𝑖)𝑧(𝑖)−^𝑚_{𝑗≠𝑖𝑘=1}𝐴_𝑘𝑖,𝑗𝑢_𝑘𝑗≥^𝑚_𝑘=1𝐴_𝑘()𝑧(𝑖)−𝑖,𝑖_{𝑚𝑘=1𝑗≠𝑖}𝐴_𝑘𝑖,𝑗^𝑝𝑘𝑢_𝑘𝑗^{𝑝𝑘1/𝑝𝑘}=^𝑚_𝑘=1𝐴_𝑘()𝑧(𝑖)−𝑖,𝑖_𝑚𝑘=1𝐴_𝑘(𝑖,𝑖)^𝑝𝑘−𝜏𝐴^[𝑝𝑘]_𝑘𝑢_𝑘(𝑖)^{𝑝𝑘1/𝑝𝑘}=^𝑚_𝑘=1𝐴_𝑘()−𝑖,𝑖_𝑚𝑘=1𝐴_𝑘(𝑖,𝑖)^𝑝𝑘−𝜏𝐴^[𝑝𝑘]_𝑘^1/𝑝𝑘𝑧(𝑖). (10) By Lemma 1, the “≥” hold. By Lemma 2, we have 𝜏𝐴₁⋆⋯⋆𝐴_𝑚≥min_{1≤𝑖≤𝑛}^𝑚_𝑘=1𝐴_𝑘()−𝑖,𝑖_𝑚𝑘=1𝐴_𝑘(𝑖,𝑖)^𝑝𝑘−𝜏𝐴^[𝑝𝑘]_𝑘^1/𝑝𝑘. (11)
Case  2. 𝐴₁⋆⋯⋆𝐴_𝑚∈𝑍_𝑛 is reducible. We denote by 𝑇 the 𝑛×𝑛 permutation matrix (𝑡_𝑖𝑗) with 𝑡₁₂=𝑡₂₃=⋯=𝑡_{𝑛−1,𝑛}=𝑡_𝑛1=1, (12) the remaining 𝑡_𝑖𝑗 zero, then both 𝐴_𝑘−𝜀𝑇∈𝑍_𝑛 are irreducible for any chosen positive real number 𝜀. Now we substitute 𝐴_𝑘−𝜀𝑇 for 𝐴_𝑘 in the previous case, and then by letting 𝜀→0, the result follows by continuity.

Lemma 4 (see [7]). Let for all 𝑖∈{1,…,𝑛}, 𝑥_𝑖≥𝑦_𝑖≥0 and 𝑝_𝑖>0 satisfying ∑^𝑛_𝑖=1(1/𝑝_𝑖)≥1. Then ^𝑛_𝑖=1𝑥_𝑖−^𝑛_𝑖=1𝑦_𝑖≥^𝑛_𝑖=1𝑥^𝑝𝑖_𝑖−𝑦^𝑝𝑖_𝑖^1/𝑝𝑖. (13)

The following is our main theorem.

Theorem 5. Let 𝛼_𝑘≥0 and ∑^𝑚_𝑘=1𝛼_𝑘≥1. If 𝐴_𝑘 is 𝑛×𝑛  𝐻-matrices for all 𝑘∈{1,…,𝑚}, then 𝜏𝐴^(𝛼1)₁∘⋯∘𝐴^(𝛼𝑚)_𝑚≥min_{1≤𝑖≤𝑛}^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)^𝛼𝑘−^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)−𝜏𝐴_𝑘^𝛼𝑘≥^𝑚_𝑘=1𝜏𝐴_𝑘^𝛼𝑘. (14)

Proof. Without loss of generality, we can assume that 𝛼_𝑘>0 for all 𝑘∈{1,2,…,𝑚}. Let 𝑝_𝑘=1/𝛼_𝑘>0, then ∑^𝑚_𝑘=1(1/𝑝_𝑘)=∑^𝑚_𝑘=1𝛼_𝑘≥1. Now we have: 𝜏𝐴^(𝛼1)₁∘⋯∘𝐴^(𝛼𝑚)_𝑚=𝜏𝑀𝐴^(𝛼1)₁∘⋯∘𝐴^(𝛼𝑚)_𝑚=𝜏𝑀𝐴^(𝛼1)₁⋆⋯⋆𝑀𝐴^(𝛼𝑚)_𝑚≥min_{1≤𝑖≤𝑛}^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)^𝛼𝑘−^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)−𝜏𝑀𝐴^(𝛼𝑘)_𝑘^{[𝑝𝑘]1/𝑝𝑘}=min_{1≤𝑖≤𝑛}^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)^𝛼𝑘−^𝑚_𝑘=1𝐴_𝑘(𝑖,𝑖)−𝜏𝐴_𝑘^𝛼𝑘≥min^𝑚_{1≤𝑖≤𝑛𝑘=1}𝐴_𝑘(𝑖,𝑖)−𝐴_𝑘(𝑖,𝑖)−𝜏𝐴_𝑘^1/𝑝𝑘=^𝑚_𝑘=1𝜏𝐴_𝑘^𝛼𝑘. (15) The first “≥” hold by Lemma 3. Because 𝐴_𝑘 is 𝐻-matrix, we have 0≤𝜏(𝐴_𝑘)≤|𝐴_𝑘(𝑖,𝑖)|, which means that |𝐴_𝑘(𝑖,𝑖)|≥|𝐴_𝑘(𝑖,𝑖)|−𝜏(𝐴_𝑘)≥0. So the second “≥” hold by |𝐴_𝑘(𝑖,𝑖)|^𝛼𝑘≥[|𝐴_𝑘(𝑖,𝑖)|−𝜏(𝐴_𝑘)]^𝛼𝑘≥0 and Lemma 4.

3. Examples

In this section, we present three examples to illustrate our improved bound.

Example 6. We take the matrices 𝐴, 𝐵, and 𝐶 in Section 1. It is easy to get 𝜏(𝐴)=𝜏(𝐵)=𝜏(𝐶)=1, so the low bound in (1) is 𝜏(𝐴)²𝜏(𝐵)²𝜏(𝐶)²=1; on the other side, we can get 𝜏(𝐴⁽²⁾∘𝐵⁽²⁾∘𝐶⁽²⁾)=10000, which is greatly larger than the lower bound 𝜏(𝐴)=𝜏(𝐵)=𝜏(𝐶), that is, 𝜏𝐴⁽²⁾∘𝐵⁽²⁾∘𝐶⁽²⁾=10000≫𝜏(𝐴)²𝜏(𝐵)²𝜏(𝐶)²=1. (16) However, our lower bound in Theorem 5 is min_1≤𝑖≤3𝑎²_𝑖𝑖𝑏²_𝑖𝑖𝑐²_𝑖𝑖−𝑎_𝑖𝑖−𝜏(𝐴)²×𝑏_𝑖𝑖−𝜏(𝐵)²𝑐_𝑖𝑖−𝜏(𝐶)²=min_1≤𝑖≤3100²×1²×1²−(100−1)²×(1−1)²(1−1)²=10000, (17) which is the exact value of 𝜏(𝐴⁽²⁾∘𝐵⁽²⁾∘𝐶⁽²⁾).

Example 7. Let 𝐴=.11−5−8−214−7−1−39,𝐵=9−8−7−213−4−1−615 (18) It is easy to see that 𝐴 and 𝐵 are 𝐻-matrices and 𝜏(𝐴)=3.9152,𝜏(𝐵)=4.1345. Thus, we have 𝜏𝐴⁽²⁾∘𝐵⁽²⁾=9799>𝜏(𝐴)²𝜏(𝐵)²=262.0319, (19) but our bound is min_1≤𝑖≤3𝑎²_𝑖𝑖𝑏²_𝑖𝑖−𝑎_𝑖𝑖−𝜏(𝐴)²𝑏_𝑖𝑖−𝜏(𝐵)²=8612.8, (20) which is close to the exact value of 9799.

Example 8. Take the 𝐻-matrices 𝐴=.15−12−4−314−2−4011,𝐵=27−14−6−420−8−10−523 (21) It is easy to see that 𝜏𝐴⁽²⁾∘𝐵⁽¹⁾=2775.5>𝜏(𝐴)²𝜏(𝐵)¹=244.0, (22) but our bound is min_1≤𝑖≤3𝑎²_𝑖𝑖𝑏¹_𝑖𝑖−𝑎_𝑖𝑖−𝜏(𝐴)²𝑏_𝑖𝑖−𝜏(𝐵)¹=2340.3. (23) Also, our bound is much closer to the exact value than the low bound in (1).

Acknowledgments

The authors would like to thank the anonymous reviewers for their helpful comments and valuable suggestions which improved the quality of this paper. This work is supported by the Cultivation Fund of the Key Scientific and Technical Innovation Project, Ministry of Education of China (no. 708040), the Leading Academic Discipline Program, 211 Project for Shanghai University of Finance and Economics (the third phase), and Graduate Research and Innovation Fund of Shanghai University of Finance and Economics (CXJJ-2012-385).