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International Journal of Algebra, Vol. 9, 2015, no. 8, 379 - 394 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2015.5954

Bounds for the Smallest and the Largest Eigenvalues of Hermitian Matrices1 Rachid Marsli Mathematics and Statistics Department King Fahd University of Petroleum and Minerals Dhahran, 31261, Kingdom of Saudi Arabia c 2015 Rachid Marsli. This article is distributed under the Creative ComCopyright mons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this paper the author derives a lower bound for the largest eigenvalue and an upper bound for the smallest eigenvalue of Hermitian matrices, based on Weyl’s inequalities. Some related results, consequences, applications, and examples are provided. In the fifth section, the main result is applied to some integer matrices.

Mathematics Subject Classification: 15A18 Keywords: Hermitian matrix; smallest eigenvalue; largest eigenvalue; spread

1

Introduction

In matrix theory, some of the most useful inequalities are Weyl’s inequalities, named after Hermann Weyl, and which compare the eigenvalues of the sum A1 + A2 of n × n Hermitian matrices with the sum of the eigenvalues of A1 and A2 . These inequalities are as follows, for 1 ≤ i ≤ n: λi (A1 + A2 ) ≤ λi+j (A1 ) + λn−j (A2 ) for j = 0, 1, ..., n − i ; λi (A1 + A2 ) ≥ λi−j+1 (A1 ) + λj (A2 ) for j = 1, 2, ..., i, 1

This work was partially supported by NSA grant, H98230-14-1-0152.

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where λ1 (A1 ) ≤ λ2 (A1 ) ≤ ... ≤ λn (A1 ) are the eigenvalues of A1 in nondecreasing order, and similarly for A2 and A1 + A2 . In particular we have λi (A1 ) + λ1 (A2 ) ≤ λi (A1 + A2 ) ≤ λi (A1 ) + λn (A2 ),

(1)

for 1 ≤ i ≤ n. Letting i = 1 and i = n we then have, respectively: λn (A1 + A2 ) ≤ λn (A1 ) + λn (A2 ), and λ1 (A1 + A2 ) ≥ λ1 (A1 ) + λ1 (A2 ). For t n × n Hermitian matrices A1 , A2 , ..., At , we get λn (

t X

Ak ) ≤

t X

k=1

k=1

t X

t X

(λn (Ak )),

(2)

(λ1 (Ak )).

(3)

and λ1 (

k=1

Ak ) ≥

k=1

In the following sections, based on the above inequalities, we derive some other inequalities and bounds for the eigenvalues of Hermitian matrices. We also provide related results, consequences, applications, and examples. For these purposes we use the following notation. Let S = {P1 , . . . , Pn! } be the set of all n × n permutation matrices, let w be any vector in Cn , and let A be an n × n matrix. Then we define: L(A) =

n! X

PkT APk ;

k=1

for n ≥ 2, n

n

n

n

n

n

X 1 XX 1X 1 aij , aii − aij }; γ1 (A) = min{ n i=1 j=1 n i=1 n(n − 1) i6=j X 1 XX 1X 1 γ2 (A) = max{ aij , aii − aij }; n i=1 j=1 n i=1 n(n − 1) i6=j for n = 1, γ1 (A) = γ2 (A) = a11 . As usual, ρ(A) is the spectral radius of A, kAk1 and kAk∞ are the respective induced matrix norms of A, and kwk2 is the Euclidean norm of vector w . As before, if A is Hermitian, then we use the ordering λ1 (A) ≤ λ2 (A) ≤ ... ≤ λn (A).

Bounds for the smallest and the largest eigenvalues of Hermitian matrices

2

381

The main result

Lemma 2.1. Let A = [aij ] be an n × n Hermitian matrix. Then (L(A))ii = (n − 1)!

n X

app ,

for i = 1, 2 · · · n;

p=1

and (L(A))ij = (n − 2)!

X

apq ,

for i 6= j.

p6=q

Proof. Let e1 , e2 , ..., en be the n canonical vectors in Cn . In general, for 1 ≤ i ≤ n and 1 ≤ k ≤ n, ei is the k th column in (n − 1)! elements of S. Also, for i 6= j and k 6= m, ei and ej are respectively the k th and mth columns in (n − 2)! elements of S. Let ek,i be the ith column of the permutation matrix Pk . Then  T  ek,1  eTk,2      PkT A Pk =  .  A ek,1 ek,2 · · · ek,n ,  ..  eTk,n so that (PkT A Pk )ii = eTk,i A ek,i which is equal to a11 for (n − 1)! terms of L(A)ii of the form (PkT A Pk )ii since ek,i = e1 for (n − 1)! permutation matrices, and it is equal to a2,2 for another (n − 1)! terms and so on. Hence, for 1 ≤ i ≤ n, (L(A))ii =

n! X

eTk,i A ek,i = (n − 1)!(a11 ) + (n − 1)!(a22 ) + ... + (n − 1)!(ann )

k=1

= (n − 1)!

n X

app .

p=1

(PkT A Pk )ij = eTk,i A ek,j ,

Also, for i 6= j,

which is equal to a12 for (n − 2)! terms of (L(A))ij , since ek,i and ek,j are respectively, the first and the second columns in (n − 2)! elements of S. In an analogous way, eTk,i A ek,j is equal to a1,3 for another (n − 2)! terms and so on. Hence, for i 6= j, (L(A))ij =

n! X k=1

eTk,i A ek,j = (n − 2)!(a12 ) + (n − 2)!(a13 ) + ...

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= (n − 2)!

n X

apq .

p6=q

Applying inequalities (2) and (3) to L(A), and using the fact that all the Hermitian matrices (Pk A Pk ) are similar to A, and therefore have the same eigenvalues, we obtain: Lemma 2.2. Let A be an n × n Hermitian matrix. Then λ1 (A) ≤ λ1 (L(A))/n! ≤ λn (L(A))/n! ≤ λn (A). Lemma 2.3. Let α and β be complex numbers, and let M be an n × n matrix with α on all diagonal positions and β on all off-diagonal positions. Then M has eigenvalues α−β with multiplicity n−1 and α+(n−1)β with multiplicity 1. Proof. Observe that M = (α − β)I + βJ where I is the identity matrix and J is the all 10 s matrix. Now J has eigenvalues n , 0 ... 0, so that M has eigenvalues (α − β) + nβ, ie α + (n − 1)β, with multiplicity 1 and α − β with multiplicity n − 1. We now come to the main result of the paper. Theorem 2.4. Let A be an n × n Hermitian matrix. Then λ1 (A) ≤ γ1 (A) ≤ γ2 (A) ≤ λn (A). Proof. By Lemma 2.1, L(A) has one distinct diagonal entry which is equal n X to (n − 1)! aii and one distinct off-diagonal entry which is equal to (n − i=1 X aij . Hence, by Lemma 2.3 2)! i6=j

λ1 (L(A)) = min{(n − 1)!

n X

aii − (n − 2)!

i=1

X

aij ,

(n − 1)!

n XX

aij }

i=1 j=1

i6=j

= n!γ1 (A), and λn (L(A)) = max{(n − 1)!

n X

aii − (n − 2)!

i=1

= n!γ2 (A). Then the theorem follows from Lemma (2.2).

X i6=j

aij , (n − 1)!

n XX i=1 j=1

aij }

Bounds for the smallest and the largest eigenvalues of Hermitian matrices

383

Remarks 1. Let B be an n × n Hermitian matrix with one distinct diagonal element α and one distinct off-diagonal elements β. Then, by Lemma 2.3 and the definition of γ1 (A) and γ2 (A), γ1 (B) = λ1 (B) = min{α − β , α + (n − 1)(β)}, and γ2 (B) = λ2 (B) = max{α − β , α + (n − 1)(β)}. Since γ1 (B), γ2 (B), λ1 (B), and λ2 (B) are continuous functions of the entries of B, a relatively small perturbation applied to B and leading to a Hermitian matrix C will result into small differences (γ1 (C) − λ1 (C)) and (λn (C) − γ2 (C)). As an example, let     200 100 100 201 101 102 B = 100 200 100 and C = 101 202 103 . 100 100 200 102 103 203 Then ( λ1 (B) = γ1 (B) = 100 λ3 (B) = γ2 (B) = 400

( λ1 (C) ≈ 99.4183 ≤ 100 = γ1 (C) and λ3 (C) ≈ 406.0087 ≥ 406 = γ2 (C)

2. For a real n × n matrix A = [aij ] with real eigenvalues, which is not necessary Hermitian, we have n 1 X aii ≤ λn (A). λ1 (A) ≤ n i=1

This follows from the fact that

n X i=1

aii =

n X

λi . If A is Hermitian, then

i=1

we also have n

λ1 (A) ≤ γ1 (A) ≤

1X aii ≤ γ2 (A) ≤ λn (A). n i=1

Hence, for every Hermitian matrix A, γ1 (A) and γ2 (A) give equal or trace(A) better bounds for, respectively, λ1 (A) and λn (A) than . n 3. If a matrix is not Hermitian then Theorem 2.4 may apply to it and may not. For example, Theorem 2.4 applies to   1 −1 C = −2 1

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and does not apply to 

   0 −1 0 −1 A = and B = 0 0 −2 0 4. For the eigenvalues of a given n × n Hermitian matrix A = [aij ] other than λ1 (A) and λn (A), we can derive an upper bound that is dependent on the entries of A and λn (A), and a lower bound dependent on the entries of A and λ1 (A). In fact for 1 < i < n, inequalities in (1) applied to L(A), with induction and argument similar to the proof of Lemma 2.2, imply λi (L(A)) − (n! − 1) λn (A) ≤ λi (A) ≤ λi (L(A)) − (n! − 1) λ1 (A), which leads to λn (A) − n!h1 ≤ λi (A) ≤ λ1 (A) + n!h2 , with h1 = λn (A) − γ1 (A) ; h2 = γ1 (A) − λ1 (A) if

X

aij ≥ 0,

i6=j

and h1 = λn (A) − γ2 (A) ; h2 = γ2 (A) − λ1 (A) if

X

aij ≤ 0.

i6=j

3

Consequences and applications of Theorem 2.4

The following corollary of Theorem 2.4 gives a lower bound for the spread of a Hermitian matrix in terms of its entries. Corollary 3.1. Let A = [aij ] be an n×n Hermitian matrix, with n ≥ 2. Then P | i6=j aij | . λn (A) − λ1 (A) ≥ n−1 Proof. From Theorem 2.4, we have λ1 (A) ≤ γ1 (A) ≤ γ2 (A) ≤ λn (A), which implies: P | i6=j aij | λn (A) − λ1 (A) ≥ γ2 (A) − γ1 (A) = . n−1 The next result also follows easily from Theorem 2.4, and gives necessary conditions for Hermitian matrices to be similar to each other.

Bounds for the smallest and the largest eigenvalues of Hermitian matrices

385

Corollary 3.2. Let A and C be two Hermitian matrices that are similar to each other, and let λ1 = λ1 (A) = λ1 (C) and λn = λn (A) = λn (C). Then, in addition to the inequalities γ2 (A)−γ1 (A) ≤ λn − λ1 and γ2 (C)−γ1 (C) ≤ λn − λ1 , we also have |γ2 (C) − γ2 (A)| ≤ λn − λ1 ; |γ2 (C) − γ1 (A)| ≤ λn − λ1 ; |γ1 (C) − γ1 (A)| ≤ λn − λ1 ; |γ1 (C) − γ2 (A)| ≤ λn − λ1 . Example 3.3. Let   1 0 0 0 A = 0 3 0 0 −2



 1 −2 −4 and C = −2 2 −1 . −4 −1 3

Then λ3 (A) − λ1 (A) = 5 and γ2 (C) − γ1 (C) = 7. This implies, according to Corollary 3.2, that A and C are not similar. Corollary 3.4. Let A be a Hermitian matrix, and kAk be a matrix norm of A. Then max {|γ1 (A)| , |γ2 (A)|} ≤ ρ(A) ≤ kAk. and ρ(Ak ) ≥ max {|γ1 (A)|k , |γ2 (A)|k }. Proof. Clearly, −ρ(A) ≤ λ1 ≤ γ1 (A) ≤ γ2 (A) ≤ λn (A) ≤ ρ(A) ≤ kAk. The second part of the corollary follows from the first part and the fact that ρ(Ak ) = (ρk (A)). Theorem 3.5. Let A and B be n × n commuting Hermitian matrices. Let α and β be, respectively, eigenvalues of A and B, such that |α| = ρ(A), |β| = ρ(B), Av = αv and Bv = βv for some nonzero vector v. Then for every positive integers k and m we have    ρ(Ak B m ) ≥ max {(|γ1 (A))|k , |γ2 (A)|k } max {(|γ1 (B))|m , |γ2 (B)|m } . Proof. αβ is an eigenvalue of A B, so that ρ(A B) = αβ = ρ(A)ρ(B) and more generally ρ(Ak B m ) = ρ(A)k ρ(B)m . Then we apply Corollary 3.4. Theorem 3.6. Let N be an n × n normal matrix and let β1 , β2 · · · βn be its eigenvalues with |β1 | ≤ |β2 | ≤ · · · ≤ |βn |. Then p p |β1 | ≤ γ1 (N N ∗ ) ≤ γ2 (N N ∗ ) ≤ |βn |. In particular, if γ1 (N N ∗ ) = 0, then N is singular.

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Proof. Apply Theorem 2.4 to the matrix N N ∗ which is hermitian, and its eigenvalues are |β1 |2 , |β2 |2 , · · · , |βn |2 . Hermitian matrices are normal. Therefore we have: Corollary 3.7. Let A be Hermitian matrix. If γ1 (A2 ) = 0 then A is singular. Lemma 3.8. Let A be an n × n matrix, and let C2 (A) be the matrix with the same entries as A, except one off-diagonal entry of smallest absolute value in each row of A is replaced by 0 in C2 (A). Let λ be an eigenvalue of A with geometric multiplicity greater or equals to 2. Then n o |λ| ≤ min kC2 (A)k1 , kC2 (A)k∞ . The above result and its proof can be found in [3]. Theorem 3.9. Let A be an n × n Hermitian matrix and let C2 (A) be as in Lemma 3.8. n o If min kC2 (A)k1 , kC2 (A)k∞ < γ2 (A), then λn (A) is simple, and o n if min kC2 (A)k1 , kC2 (A)k∞ < −γ1 (A), then λ1 (A) is simple. Proof. The first part of the theorem follows from Theorem 2.4 and Lemma 2.3. In the second part, use the facts: λ1 (A) = −λn (−A), γ2 (A) = −γ1 (−A), kC2 (A)k1 = kC2 (−A)k1 and kC2 (A)k∞ = kC2 (−A)k∞ Then apply Theorem 2.4 and Lemma 2.3 to (−A). Example 3.10. Let 

 3 3 4 A = 3 2 3 , 4 3 0

kC2 (A)k1 = 7
0, and define r1 = min{

1 1 , }, −1 |γ1 (A )| γ2 (A−1 )

r2 = max{

1 1 , }. −1 |γ1 (A )| γ2 (A−1 )

and

Then Theorem 3.11 implies that min{α , |β|} ≤ r1 , and max{α , |β|} ≤ r2 . Based on the discussion above, we state the following corollary. Corollary 3.12. Let A be a nonsingular Hermitian matrix such that γ1 (A) < 0 and γ2 (A) > 0. Let rˆ1 = min



 1 1 1 1 , , and rˆ2 = max |γ1 (A)| γ2 (A) |γ1 (A)| γ2 (A)

. Then • A−1 has both positive and negative eigenvalues, • A−1 has at least one eigenvalue within the interval [−ˆ r1 , rˆ1 ], • A−1 hast at least two distinct eigenvalues within the the interval [−ˆ r2 , rˆ2 ]. Remark 3.13. Let A be a Hermitian matrix for which γ1 (A) < 0 and γ2 (A) > 0. Then A has both positive and negative eigenvalues. If we interchange the positions of some diagonal entries of A, or if we interchange the positions of some off-diagonal entries of A such that the obtained matrix B is Hermitian, then B also has negative and positive eigenvalues. This is because γ1 (A) = γ1 (B) and γ2 (A) = γ2 (B).

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Example 3.14. Consider the Hermitian matrix     0 3 2 λ1 (A) ≈ −6, 4058 1  with A = 3 −5 λ2 (A) ≈ −2.3351   2 1 −1 λ3 (A) ≈ 2.7409 Since A has both positive and negative eigenvalues, then according to Remark 3.13, matrices such as     0 3 1 −5 3 2 2  and C =  3 −1 1 B = 3 −5 1 2 −1 2 1 0 has both positive and negative eigenvalues.

4

Some other results related to positive definite matrices

An argument similar to the proof of Theorem 3.11 yields the following. Theorem 4.1. Let A be a positive definite or a negative definite matrix. Then 1 1 ≥ γ2 (A−1 ) ≥ γ1 (A−1 ) ≥ . λ1 (A) λn (A) Example 4.2. The following example illustrates the case of a positive definite 1 gives a better upper-bound for λ1 (A) than γ1 (A). matrix where γ2 (A−1 )     2 0 1 3 1 −2 1 3 −2 If A = 0 2 1 , then A−1 =  1 4 1 1 2 −2 −2 4 and λ1 (A) ≈ 0.5858 ≤

1 ≈ 0.9231 ≤ γ1 (A) ≈ 1.3333. γ2 (A−1 )

Theorem 4.3. Let A be a Hermitian matrix. If γ1 (A) γ2 (A) ≤ 0, then A is neither positive definite nor negative definite. This is also the case if A is nonsingular and γ1 (A−1 ) γ2 (A−1 ) ≤ 0. In other words, for A to be positive definite, both γ1 (A) and γ2 (A) need to be positive, and for A to be negative definite, both γ1 (A) and γ2 (A) need to be negative.

Bounds for the smallest and the largest eigenvalues of Hermitian matrices

389

Proof. The first part follows directly from Theorem 2.4, and the second follows from Theorem 4.1

Next, we give lower bound for the condition number of positive definite and negative definite matrices. Theorem 4.4. Let A and B be, respectively, positive definite and negative definite. Then λn (A) γ2 (A) ≥ > 0 and λ1 (A) γ1 (A)

5

λ1 (B) γ1 (B) ≥ > 0. λ2 (B) γ2 (B)

Application of Theorem 2.4 to some integer matrices

We first give a result that follows directly from Theorem 2.4. Corollary 5.1. Let {An } be a sequence of n × n Hermitian matrices. If

lim γ2 (An ) = ∞,

n→∞

and if

5.1

lim λn (An ) = ∞,

then

n→∞

lim γ1 (An ) = −∞,

then

n→∞

lim λ1 (An ) = −∞.

n→∞

Toeplitz matrices

Consider the set of Toeplitz matrices of the form: 

1

     Tn =      

2

2 1

3 2 ..

. 3 2 .. . . . . . . . .. .. . . n ... ...

... .. . .. . .. . .. . 3

 ... n ..  .  . . ..  . . ,  .. . 3   .. . 2 2

1

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Rachid Marsli

where n ranges over all positive integers greater or equal to 2. Then n−2

n

n−1 n X n }| { z }| { z X z }| { (Tn )ij = (1 + · · · + 1) + 2(2 + · · · + 2 + 2 (3 + · · · + 3) + · · · + 2n i=1 j=1

= n + 2(2(n − 1) + 2(3(n − 2) + · · · + 2(n(n − (n − 1)))   = n + 2 [2n + 3n + · · · + n(n)] − [2 + 3 × 2 + · · · + n × (n − 1)] n−1  X n(n + 1) − 1)] − k(k + 1) = n + 2 [n( 2 k=1



=

n3 + 3 n2 − n . 3

Hence γ2 (Tn ) =

n2 + 3n − 1 , 3

and n X

γ1 (Tn ) =

i=1

X

(Tn )ii −

n X n X (Tn )ij

(Tn )ij

i6=j

n n(n − 1) Now by applying Theorem 2.4, we obtain:

=

i=1 j=1

n(n − 1)

= −

n+1 . 3

Theorem 5.2. Let Tn be defined as above with n ≥ 2. Then Tn is neither positive definite nor positive semidefinite, and we have n+1 λ1 (Tn ) ≤ − so that lim λ1 (Tn ) = −∞, n→∞ 3 and n2 + n − 1 λn (Tn ) ≥ so that lim λn (Tn ) = ∞. n→∞ 3

5.2

Hankel matrices

Consider the set of Hankel matrices Hn of the form:   1 2 3 ... ... n   . . . 2 .. .. 3 .. n+1    ..  . . . . . . . . 3 . . . . .     ..  . . . . Hn =  .. . . . . . . . . .  .  .  . . . . . . . . . . . . . . 2n − 2     n n + 1 . . . . . . 2n − 2 2n − 1

Bounds for the smallest and the largest eigenvalues of Hermitian matrices

391

where n ranges over all positive integers greater or equal to 2. Then n X

(Hn )ii = n2 ;

i=1

and n X n X (Hn )ij = 1 + 2 × 2 + · · · + n × n + (n + 1) × (n − 1) + i=1 j=1

(n + 2) × (n − 1) + · · · + 2n − 1 = 1 + 4 + · · · + n2 + (n2 − 1) + (n2 − 4) + · · · + (n2 − (n − 1)2 ) = n2 + n2 + · · · + n2 = n3 Hence γ2 (Hn ) = n2

and γ1 (Hn ) = −1.

Therefore, by applying Theorem 2.4, we obtain: Theorem 5.3. Let Hn be defined as above, with n ≥ 2. Then Hn is neither positive definite nor semi-positive definite, and we have λ1 (Hn ) ≤ −1 and λn (Hn ) ≥ n2 .

5.3

GCD matrices

Let Bn = [bij ] be the n × n symmetric matrix such that bij = gcd(i, j) = the greatest common divisor of i and j. Since bii = i, n X n X

γ2 (Bn ) =

n X

bij

i=1 j=1

n

=

i=1

X

bij

n

X

bii +

bij

i6=j

n

2 n+1 = + 2

X

gcd(i, j)

i