Mohammed Al-Dolat , Khaldoun Al-Zoubi , Mohammed Ali and Feras Bani-Ahmad

General numerical radius inequalities for matrices of operators

De Gruyter | Published online: March 1, 2016

Abstract

Let AiB(H), (i = 1, 2, ..., n), and T = [ 0 0 A 1 A 2 0 0 A n 0 0 ] . In this paper, we present some upper bounds and lower bounds for w(T). At the end of this paper we drive a new bound for the zeros of polynomials.

MSC 2010: 47A05; 47A10; 47A12

1 Introduction

Let B(H) be the C*– algebra of all bounded linear operators on a complex Hilbert space H with inner product 〈.,.〉. The numerical range of TB(H); denoted by W(T), is the subset of complex numbers given by

W ( T ) = { T x , x : x H , | | x | | = 1 } .

The numerical radius of T, denoted by w(T), is given by

w ( T ) = { T x , x : x H , | | x | | = 1 } .

It is well-known that w(.) defines a norm on B(H), which is equivalent to the usual operator norm | | T | | = sup | | x | | = 1 | | T x | | . In fact for TB(H), we have

1 2 | | T | | w ( T ) | | T | | .

Several numerical radius inequalities that provide alternative lower and upper bounds for w(.) have received much attention from many authors. We refer the readers to [1, 2] for the history and significance, and [37] for recent developments this area. Kittaneh [5] proved that for TB(H).

w ( T ) 1 2 ( | | T | | + | | T 2 | | 1 2 ) .

So it is clear that if T2 = 0; then

(1) w ( T ) = 1 2 | | T | | . 

Also it is known that w(.) is weakly unitarily invariant, that is

(2) w ( U T U * ) = w ( T ) ,
for every unitary UB( H).

A fundamental inequality for the numerical radius is the power inequality, which says that for TB(H), we have

w ( T n ) w n ( T ) ,
for n = 1, 2, 3, ... (see, e.g. [ 1, p. 118]).

Although some open problems related to the numerical radius inequalities for bound linear operators still remain open, the investigation to establish numerical radius inequalities for several bound linear operators has been started. For example, if T1, T2B(H), it is known [1] that

w ( T 1 T 2 ) 4 w ( T 1 ) w ( T 2 ) .

Moreover, in the case that T1, T2 = T2T1,

w ( T 1 T 2 ) 2 w ( T 1 ) w ( T 2 ) .

However, the sharp inequality

w ( T 1 T 2 ) w ( T 1 ) w ( T 2 )
still has not been reached. A useful result in this direction, which can be found in [ 8], says that for T 1, T 2B( H),
w ( T 1 T 2 ± T 2 T 1 * ) 2 | | T 1 | | w ( T 2 ) .

If T1, T2B(H), and T1 is positive operator, Kittaneh in [9] showed that

w ( T 1 T 2 T 2 T 1 ) 1 2 T 1 ( T 2 + T 2 2 1 2 ) .

Recently, the authors of [10] applied a different approach to obtain a new numerical radius inequality for commutators of Hilbert space operators. They showed that for T1, T2, T3, T4B(H),

(3) w ( T 1 T 3 T 2 * ± T 2 T 4 T 1 * ) 2 | | T 1 | | | | T 2 | | w ( [ 0 T 3 T 4 0 ] ) .

The following numerical radius inequality for certain 2 × 2 operator matrices is obtained in [11],

(4) w ( [ 0 X Y 0 ] ) | | X | | + | | Y | | 2 ,
where X, YB( H). Another results in the direction can be found in [ 12].

The purpose of this work is to present new numerical radius inequalities for n × n operator matrices. Also we deduce (3) and (4) as special cases. At the end of this paper, we give some new bounds for the zeros of any monic polynomial with complex coefficients.

2. Numerical radius inequalities for the n × n operator matrix

The aim of this section is to establish new numerical radius inequalities for matrices of operators and to generalize some known inequalities. In order to do this, we need the following well-known lemma.

Lemma 2.1

([3]). LetX1, X2, ..., XnB(H). Then

w ( [ X 1 0 X 2 0 X n ] ) = max { w ( X 1 ) , w ( X 2 ) , ... , w ( X n ) }

Let us use this lemma to generalize the inequality (4).

Theorem 2.2.

Let AiB(H), i = 1, 2,..., n and T = [ 0 A 1 A 2 A n 0 ]

Then if n is even,

w ( T ) 1 2 i = 1 n | | A i | |
and if n is odd,
w ( T ) w ( A n + 1 2 ) + 1 2 i = 1 n | | A i | | .

Proof.

Let X 1 = [ 0 A 1 0 0 0 ] , X 2 = [ 0 0 A 2 0 0 0 ] , ... , X n = [ 0 0 0 A n 0 ] .

Then if n is an even number we have X i 2 = 0 for all i = 1, 2,..., n and so

w ( T ) = w ( i = 1 n X i ) i = 1 n w ( X i ) = 1 2 i = 1 n | | X i | | .

On the other hand, if n is an odd number, then following the same manner used above we achieve that

w ( T ) = w ( i = 1 n X i ) w ( X n + 1 2 ) + i n + 1 2 n w ( X i ) = w ( A n + 1 2 ) + 1 2 i n + 1 2 n | | A i | | .

Applying Theorem 2.2 with n = 2, A1 = X, A2 = Y we reach the inequality (4). Let us use (2) to prove the following theorem.

Theorem 2.3.

Let AiB(H), (i = 1, 2, ..., n) where n ≥ 3 and 1, α, α2,..., αn − 1are the roots of unity and T = [ 0 A 1 A 2 A n 0 ] . Then

w ( T ) = w ( [ 0 α n 1 A n α 2 n 3 A n 1 α 3 n 5 A n 2 α ( n 1 ) 2 A 1 0 ] ) = w ( [ 0 α A n α 3 A n 1 α 5 A n 2 α 2 n 1 A 1 0 ] ) = w ( [ 0 0 α 3 A n 1 0 0 α 5 A n 2 0 α 2 n 3 A 2 0 0 0 α n 1 A 1 0 0 α n + 1 A n 0 ] )

Proof.

Let U 1 = [ 0 α 2 n 2 I α 2 n 3 I α n 1 I 0 ] , U 2 = [ 0 I α I α 2 I α n 1 I 0 ] and [ 0 0 I 0 α I α 2 I 0 α n 2 I 0 0 0 α n 1 I ]

Then it is easy to show that U1, U2 and U3 are unitary operators so by (2) we have w ( T ) = w ( U 1 T U 1 * ) = w ( U 2 T U 2 * ) = w ( U 3 T U 3 * ) , which completes the proof. □

As a direct consequence of Theorem 2.3 we obtain the following corollary

Corollary 2.4.

LetX1, X2, X3B(H) and 1; α, α2be the roots of x3 = 1. Then

w ( [ X 2 α 2 X 1 α X 3 α X 3 X 2 α 2 X 1 α 2 X 1 α X 3 X 3 ] ) 3 w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] )

Proof.

For T = [ X 2 α 2 X 1 α X 3 α X 3 X 2 α 2 X 1 α 2 X 1 α X 3 X 2 ] , we have

w ( T ) ( [ 0 0 α X 3 0 X 2 0 α 2 X 1 0 0 ] ) + w ( [ X 2 0 0 0 0 α 2 X 1 0 α X 3 0 ] ) + w ( [ 0 α 2 X 1 0 α X 3 0 0 0 0 X 2 ] ) = 2 w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] ) + w ( [ 0 α 2 X 1 0 α X 3 0 0 0 0 X 2 ] ) = 2 w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] ) + w ( [ 0 0 α 2 X 1 0 X 2 0 α X 3 0 0 ] ) = 2 w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] ) + w ( [ 0 0 α 2 X 3 * 0 X 2 * 0 α X 1 * 0 0 ] ) = 2 w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] ) + w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] * ) = 3 w ( [ 0 0 X 3 0 X 2 0 X 1 0 0 ] ) .

Another relation for the numerical radius of the operator [ 0 A 1 A 2 A n 0 ] is as follows.

Theorem 2.5.

LetXiB(H), (i = 1, 2, ..., n) where n ≥ 3, T = [ 0 A 1 A 2 A n 0 ] , T 1 = [ 0 A 2 A 3 A n 1 0 ] and T 1 t = [ 0 A n 1 A n 2 A 2 0 ] Then

w ( T ) = w ( [ 0 0 A 1 0 T 1 0 A n 0 0 ] ) = w ( [ 0 0 A n 0 T 1 t 0 A 1 0 0 ] ) = w ( [ T 1 0 0 0 0 A n 0 A 1 0 ] ) = w ( [ T 1 t 0 0 0 0 A 1 0 A n 0 ] ) = w ( [ 0 A n 0 A 1 0 0 0 0 T 1 ] ) = w ( [ 0 A 1 0 A n 0 0 0 0 T 1 t ] ) .

An application of Theorem 2.5 yields

Corollary 2.6.

LetX1, X2, X3B(H). Then

w ( [ X 1 X 2 X 3 X 3 X 1 X 2 X 2 X 3 X 1 ] ) 3 w ( [ 0 0 X 3 0 X 1 0 X 2 0 0 ] ) .

Using a straightforward technique we derive the following lemma.

Lemma 2.7

LetAi, XiB(H), i = 1, 2, ..., n. Then

w ( i = 1 n A i X i A i * ) i = 1 n | | A i | | 2 w ( X i )

Proof.

Assume that xH is a unit vector. Then

| i = 1 n A i X i A i * x , x | = | i = 1 n X i A i * x , A i * x | i = 1 n | X i A i * x , A i * x | i = 1 n | | A i * x | | 2 w ( X i ) i = 1 n | | A i | | 2 w ( X i ) .

We attain our theorem by taking the supremum over all unit vectors xH.

Our next result can be stated as follows.

Theorem 2.8.

Let Ai, Bi, Xi, YiB(H), i = 1, 2, ..., n. Then

w ( i = 1 n A i X i B n i + 1 * B i Y i A n i + 1 * ) 2 ( ( i = 1 n | | A i | | 2 ) ( i = 1 n | | B i | | 2 ) ) w ( T ) ,

where T = [ 0 X 1 X n Y 1 Y n 0 ]

Proof.

Assume that C = [ A 1 A 2 A n B 1 B 2 B n 0 0 ] and Z = [ 0 X 1 X n Y 1 Y n 0 ]

Applying Lemma 2.1 and Lemma 2.7 we get

w ( i = 1 n A i X i B n i + 1 * + B i Y i A n i + 1 * ) = w ( [ i = 1 n A i X i B n i + 1 * + B i Y i A n i + 1 * 0 0 0 ] ) = w ( C Z C * ) | | C | | 2 w ( Z ) = | | i = 1 n A i A i * + B i B i * | | w ( Z ) ( i = 1 n | | A i A i * + B i B i * | | ) w ( Z ) ( i = 1 n | | A i | | 2 + | | B i | | 2 ) w ( Z ) .

Now, if we replace Ai and Bi by t Ai and 1 t B i , t > 0, respectively, then we have

w ( i = 1 n A i X i B n i + 1 * + B i Y i A n i + 1 * ) w ( Z ) i = 1 n ( t 4 | | A i | | 2 + | | B i | | 2 t 2 ) .

Also, since t > 0 min ( i = 1 n t 4 | | A i | | + | | B i | | t 2 ) = 2 i = 1 n | | A i | | 2 i = 1 n | | B i | | 2 so we have

w ( i = 1 n A i X i B n i + 1 * + B i Y i A n i + 1 * ) 2 w ( Z ) ( i = 1 n | | A i | | 2 i = 1 n | | B i | | 2 ) .

Finally, replace Yi by −Yi to get

w ( i = 1 n A i X i B n i + 1 * + B i Y i A n i + 1 * ) 2 w ( Z ) ( i = 1 n | | A i | | 2 i = 1 n | | B i | | 2 ) .

As an application of Theorem 2.8, we obtain the following result.

Corollary 2.9

IfTiB(H), i = 1, 2,..., n, then

w ( [ 0 T 1 T 2 T n 0 ] ) 1 n w ( i = 1 n T i )

Proof.

LetXi = Yi = Ti and Ai = Bi = I. □

Based on the inequality (1) and Lemma 2.1, an upper bound for the numerical radius of the general n × n operator matrix can be derived.

Theorem 2.10

LetAijB(H) where 1 ≤ i, jn. Then

w ( [ A 11 A 12 A 1 n A 21 A 22 A 2 n A n 1 A n 2 A n n ] ) max { w ( A i i ) : 1 i n } + 1 2 i , j = 1 , i j n | | A i j | | .

Proof.

For 1 ≤ x, yn where xy. Define the operator matrix Txy = [tij] where

t i j = { 0 otherwise . A i j if i = x , j = y

Then T x y 2 = 0 so

w ( [ A 11 A 12 A 1 n A 21 A 22 A 2 n A n 1 A n 2 A n n ] ) = w ( [ A 11 0 A 22 0 A n n ] + x , y = 1 , x y n T x y ) w ( [ A 11 0 A 22 0 A n n ] ) + x , y = 1 , x y n w ( T x y ) = max { w ( A i i ) : 1 i n } + 1 2 i , j = 1 , i j n | | A i j | |

An application of Theorem 2.10 and Corollary 2.9 yields

Corollary 2.11

LetA, B, C, D = ∈ B(H). Then

max { max { w ( A ) , w ( D ) } , 1 2 w ( A + D ) } w ( [ A B C D ] ) max { w ( A ) , w ( D ) } + | | B | | + | | C | | 2

3 A bound for the zeros of polynomials

Let p(z) = zn + anzn-1 + ... + a2z + a1 be a monic polynomial of degree n ≥ 3 with complex coefficients a1, a2, ..., an. Then the Frobenius companion matrix of p is the matrix

C ( p ) = [ a n a n 1 a 2 a 1 1 0 0 0 1 0 0 1 0 ] .

It is well-known that the zero of p are exactly the eigenvalues of C(p) (see, e.g., [13, 14]). Since the spectral radius of a matrix is dominated by its numerical radius, it follows that if z is any zero of p, then

(5) | z | w ( C ( p ) )

by the inequality (5) and Theorem 2.10 we can derive a new bound for the zeros of p.

Theorem 3.1

If z is any zero ofp(z), then

| z | max { w ( A ) , cos ( π n + 1 ) } + 1 2 ( 1 + ( i = 1 n 1 | a i | 2 ) 1 2 ) .

Proof.

Partition C(p) as

C ( p ) = [ A B C D ] ,

where A = [ a n a n 1 1 0 ] , B = [ a n 2 a 2 a 1 0 0 ] , C = [ 0 1 0 0 0 0 ] and D = [ 0 0 0 1 0 0 1 0 0 0 1 0 ]

Then we have

| z | ω ( C ( p ) ) = ω ( [ A B C D ] ) max { ω ( A ) , cos ( π n + 1 ) } + 1 2 ( 1 + ( | a i | 2 i = 1 n 1 ) 1 2 ) .

It should be mentioned here that other bounds for the zeros of p(z) can be obtained by considering different partitions of C(p). Related bounds for the zeros of p that are based on the inequality (5) and various estimates of w(C(p)) can be found in [13, 14], and references therein.

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Received: 2015-9-2
Accepted: 2016-2-8
Published Online: 2016-3-1
Published in Print: 2016-1-1

© 2016 Al-Dolat et al., published by De Gruyter Open.

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