Study on Birkhoff orthogonality and symmetry of matrix operators

Yueyue Wei; Donghai Ji; Li Tang

doi:10.1515/math-2022-0591

Article Open Access

Study on Birkhoff orthogonality and symmetry of matrix operators

Yueyue Wei , Donghai Ji and Li Tang

Published/Copyright: July 10, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

We focus on the problem of generalized orthogonality of matrix operators in operator spaces. Especially, on ℬ ( l 1 n , l p n ) ( 1 ≤ p ≤ ∞ ) , we characterize Birkhoff orthogonal elements of a certain class of matrix operators and point out the conditions for matrix operators which satisfy the Bhatia-Šemrl property. Furthermore, we give some conclusions which are related to the Bhatia-Šemrl property. In a certain class of matrix operator space, such as ℬ ( l ∞ n ) , the properties of the left and right symmetry are discussed. Moreover, the equivalence condition for the left symmetry of Birkhoff orthogonality of matrix operators on ℬ ( l p n ) ( 1 < p < ∞ ) is obtained.

Keywords: Birkhoff orthogonality; matrix operator; Bhatia-Šemrl property; left symmetry; right symmetry

MSC 2010: 47A30; 47A99; 46E15

1 Introduction

The study of orthogonality has been a hot topic of research, and various results have been achieved, such as in [1,2]. In [3–5], with more and more scholars’ research on orthogonality related knowledge, many generalized orthogonality concepts with different properties have been introduced one after another in general normed spaces, among which Birkhoff orthogonality is more widely used, then the results related to orthogonality in inner product spaces have been extended to general normed spaces. Furthermore, these generalized orthogonality concepts are consistent in the inner product space.

The idea of operator theory arose in the late 19th century. The operator theory has been developed by leaps and bounds through the joint efforts of a number of mathematicians in [6–8], and the most typical operator is the matrix operator. The study of generalized orthogonality of matrices is usually divided into two aspects, one is studying it in space with matrix norm, the other is to see matrix as an element in operator space with operator norm. This article focuses on the latter study.

In this article, we only concern this problem in finite dimensional space. Let X , Y denote real normed spaces, S X = { x ∈ X : ‖ x ‖ = 1 } be the unit sphere of X . ℬ ( X , Y ) denote the set of all linear operators from X to Y . ℬ ( X , X ) = ℬ ( X ) . For an operator T ∈ ℬ ( X ) , M T = { x ∈ S X : ‖ T x ‖ = ‖ T ‖ } .

The concept of Birkhoff orthogonality was introduced by Birkhoff and defined as follows.

Definition 1.1

[5] ∀ x , y ∈ X , x is said to be orthogonal to y in the sense of Birkhoff-James, written as x ⊥ B y if

‖ x + λ y ‖ ≥ ‖ x ‖

for all λ ∈ R . Claim y as a Birkhoff orthogonal element of x .

Let A , T ∈ ℬ ( X , Y ) . After Birkhoff orthogonality was introduced, some mathematicians have studied the relationship between A ⊥ B T and A x ⊥ B T x and have obtained some important results in [9,10]. If there exists x ∈ M T such that T x ⊥ B A x , then T ⊥ B A . When H is finite dimensional Hilbert space, the converse for ℬ ( H ) was proved by Bhatia and Šemrl [11], i.e. T , A ∈ ℬ ( H ) , T ⊥ B A ⇔ ∃ x ∈ M T such that T x ⊥ A x . Later, it is generalized by Benítez et al. to inner product space [12]. Motivated by the above results, Kim defined the Bhatia-Šemrl property as follows.

Definition 1.2

[13] An operator T ∈ ℬ ( X , Y ) is said to have the Bhatia-Šemrl property, if for each A with T ⊥ B A , there exists x ∈ M T such that T x ⊥ B A x .

Li and Schneider gave an example of ∃ S , T ∈ ℬ ( X ) such that S ⊥ B T , but there does not exist x ∈ M S , such that S x ⊥ B T x in [14], then it is worth studying whether matrix operators can satisfy the Bhatia-Šemrl property.

On ℬ ( l ∞ n ) , Paul and Sain have given a characterization of Birkhoff orthogonal elements of a certain class of matrix operators [15], and then have obtained conditions for matrix operators which satisfy the Bhatia-Šemrl property. On ℬ ( X , Y ) , where X , Y be normed linear spaces. Sain et al. continued to explore the validity of the Bhatia-Šemrl Property in [16]. However, in recent years, there are few theories related to matrix operators. This article focuses on the characterization of Birkhoff orthogonal elements of a certain class of matrix operators on ℬ ( X , Y ) ( X ≠ Y ) , then obtains the conditions for matrix operators which satisfy the Bhatia-Šemrl property. Also, when T ⊥ B S is not necessarily hold, the sufficient conditions for T x ⊥ B S x and S x ∈ T x + ( S x ∈ T x − ) to hold respectively are obtained.

Meanwhile, we also study the symmetry of the matrix operator. It is well known that the orthogonality of an inner product space is symmetric, i.e. x ⊥ y ⇔ y ⊥ x . But the Birkhoff orthogonality on a general normed space are generally not symmetric, and we know that the Birkhoff orthogonality on a normed space with dimension not less than 3 is symmetric if and only if the space is an inner product space. Also, in the operator space formed by the general normed space, the Birkhoff orthogonality of the matrix operator is not symmetric either. For example, the operators T = 2 0 0 2 and A = 1 0 0 0 on ℬ ( l ∞ n ) . We have T ⊥ B A , but A ⊥̸ B T .

On ℬ ( H ) , ℬ ( l p 2 ) ( p ≥ 2 , p ≠ ∞ ) and ℬ ( X ) , Sain et al. studied the symmetry, mainly on left symmetry, of Birkhoff orthogonality of linear operators in [17–19]. In [20,21], Ghosh et al. gave equivalence conditions for the left and right symmetry of Birkhoff orthogonality of matrix operators on l 1 n and l ∞ n , respectively, and gave us a specific characterization of matrix operators with the left or right symmetry. However, there is no more perfect conclusion on matrix operators. Therefore, the study of Birkhoff orthogonality of matrix operators still needs to be deepened.

In this article, on ℬ ( X ) and ℬ ( l ∞ n ) , where X is strictly convex, we indicate the properties of the matrix operators with right symmetry. Also, on ℬ ( l p n ) ( 1 < p < ∞ ) , an equivalence condition for left symmetry of Birkhoff orthogonality of matrix operators is obtained. These results complement to the aforementioned study.

2 Preliminaries

We briefly review the preliminary knowledge related to Birkhoff orthogonality and symmetry of the linear operator.

Definition 2.1

[18] x ∈ X is called left symmetric point if x ⊥ B y implies that y ⊥ B x for all y ∈ X . Similarly, x ∈ X is called right symmetric point if y ⊥ B x implies that x ⊥ B y for all y ∈ X . If x is both a left and a right symmetric point, then x is said to be a symmetric point.

Definition 2.2

[18] ∀ x , y ∈ X , we say that y ∈ x + if ‖ x + λ y ‖ ≥ ‖ x ‖ for all λ ≥ 0 . Similarly, we say that y ∈ x − if ‖ x + λ y ‖ ≥ ‖ x ‖ for all λ ≤ 0 .

The lemmas used in this article are as follows.

Lemma 2.3

[22] T ∈ ℬ ( X , Y ) be nonzero and x ∈ M T . Then for any y ∈ X , T x ⊥ B T y ⇒ x ⊥ B y .

Lemma 2.4

[23] Let T ∈ ℬ ( X ) attain its norm at only ± x 0 ∈ S X . Then for any A ∈ ℬ ( X ) , T ⊥ B A ⇔ T x 0 ⊥ B A x 0 .

Lemma 2.5

[20] Extreme points (or their scalar multiples) of the closed unit ball are the only right symmetric points of l 1 n .

3 Characterization of Birkhoff orthogonal elements of a certain class of matrix operators

The characterizations of Birkhoff orthogonal elements of matrix operators on ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) and ℬ ( l 1 n , l ∞ n ) are as follows.

Theorem 3.1

Let S = ( a i j ) n × n ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) . ∃ j 0 ∈ { 1 , 2 , … , n } such that a i j 0 ≠ 0 for all i ∈ { 1 , 2 , … , n } and for any j ∈ { 1 , 2 , … , n } − { j 0 } have

∣ a 1 j 0 ∣ > ∣ a 1 j ∣ , ∣ a 2 j 0 ∣ > ∣ a 2 j ∣ , … , ∣ a n j 0 ∣ > ∣ a n j ∣ .

Then ∀ T = ( b i j ) n × n ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) , S ⊥ B T if and only if b 1 j 0 = ⋯ = b n j 0 = 0 .

Proof

Since S = ( a i j ) n × n , T = ( b i j ) n × n ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) , then

‖ S ‖ = max ‖ x ‖ 1 = 1 ‖ S x ‖ p = sup j ∑ i = 1 n ∣ a i j ∣ p 1 p .

Let ∀ T ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) , S ⊥ B T . Claim: b 1 j 0 = ⋯ = b n j 0 = 0 .

Without loss of generality, we assume that j 0 = 1 , i.e. for any j ∈ { 2 , 3 , … , n } have

∣ a 11 ∣ + ∣ a 21 ∣ + ⋯ + ∣ a n 1 ∣ > ∣ a 1 j ∣ + ∣ a 2 j ∣ + ⋯ + ∣ a n j ∣ .

Suppose there exists i 1 , i 2 , … , i k for all 1 ≤ k ≤ n such that b i 1 1 , b i 2 1 , … , b i k 1 are not zero. We choose λ such that

(3.1) ∣ λ ∣ < min ∣ a 11 ∣ − max 2 ≤ j ≤ n ∣ a 1 j ∣ max 1 ≤ i ≤ n ∑ j = 2 n ∣ b i j ∣ , … , ∣ a n 1 ∣ − max 2 ≤ j ≤ n ∣ a n j ∣ max 1 ≤ i ≤ n ∑ j = 2 n ∣ b i j ∣ .

Therefore, for any λ satisfying (3.1) and any j ∈ { 2 , 3 , … , n } , we have

∑ i = 1 n ∣ a i j + λ b i j ∣ p p ≤ ∑ i = 1 n ( ∣ a i j ∣ + ∣ λ ∣ ∣ b i j ∣ ) p p < ∑ i = 1 n ∣ a i 1 ∣ p p = ‖ S ‖ .

Assume that λ satisfies (3.1), then choose λ such that

(3.2) ∣ λ ∣ < min 1 ≤ t ≤ k ∣ a i t 1 ∣ max 1 ≤ t ≤ k ∣ b i t 1 ∣ .

Note that b i 1 1 , b i 2 1 , … , b i k 1 are not zero. Accordingly, they can be divided into the following two cases:

(1) Suppose b i 1 1 , b i 2 1 , … , b i k 1 have the same sign. Define a linear operator S satisfies sgn a i 1 1 = sgn a i 2 1 = ⋯ = sgn a i k 1 such that

(3.3) b i 1 1 ( sgn a i 1 1 ) , b i 2 1 ( sgn a i 2 1 ) , … , b i k 1 ( sgn a i k 1 )

have the same sign, where

sgn ( a i j ) = + 1 , a i j > 0 = − 1 , a i j < 0 = 0 , a i j = 0 .

When λ satisfies (3.2) and the sign of λ is opposite to (3.3), we have

(3.4) ∣ a 11 + λ b 11 ∣ p + ⋯ + ∣ a n 1 + λ b n 1 ∣ p p = ∑ t = 1 k ∣ ∣ a i t 1 ∣ + λ b i t 1 ( sgn a i t 1 ) ∣ p + ∑ t = k + 1 n ∣ a i t 1 ∣ p p < ∣ a 11 ∣ p + ⋯ + ∣ a n 1 ∣ p p = ‖ S ‖ .

This contradicts with S ⊥ B T for all T ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) .

(2) Suppose b i 1 1 , b i 2 1 , … , b i k 1 signs are not all the same. Without loss of generality, we assume that

sgn b i 1 1 = ⋯ = sgn b i m 1 = 1 , sgn b i m + 1 1 = ⋯ = sgn b i k 1 = − 1 ,

where 1 ≤ m < k ≤ n and m , k are integer numbers. Define S satisfies

sgn a i 1 1 = ⋯ = sgn a i m 1 = 1 , sgn a i m + 1 1 = ⋯ = sgn a i k 1 = − 1 .

Then (3.3) have the same sign. When λ satisfies (3.2) and the sign of λ is opposite to (3.3), then (3.4) also holds. This contradicts with S ⊥ B T for all S ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) . Combining (1) with (2), we obtain b 1 j 0 = ⋯ = b n j 0 = 0 .

Conversely, let b 1 j 0 = ⋯ = b n j 0 = 0 , then for any T ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) and any λ ∈ R , we have

‖ S + λ T ‖ ≥ ∣ a 1 j 0 + λ b 1 j 0 ∣ p + ⋯ + ∣ a n j 0 + λ b n j 0 ∣ p p = ∣ a 1 j 0 ∣ p + ⋯ + ∣ a n j 0 ∣ p p = ‖ S ‖ .

Hence, S ⊥ B T for all T ∈ ℬ ( l 1 n , l p n ) ( 1 ≤ p < ∞ ) . □

Theorem 3.2

Let S = ( a i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) , ∃ i 0 , j 0 ∈ { 1 , 2 , … , n } such that ∣ a i 0 j 0 ∣ > ∣ a i j ∣ and a i 0 j 0 ≠ 0 for all i , j ∈ { 1 , 2 , … , n } and a i j ≠ a i 0 j 0 . Then ∀ T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) , S ⊥ B T if and only if b i 0 j 0 = 0 .

Proof

Since S , T ∈ ℬ ( l 1 n , l ∞ n ) , then

‖ S ‖ = max ‖ x ‖ 1 = 1 ‖ S x ‖ ∞ = sup i , j ∣ a i j ∣ .

Let b i 0 j 0 = 0 . Claim: S ⊥ B T for all T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) .

Since ‖ S + λ T ‖ ≥ ∣ a i 0 j 0 + λ b i 0 j 0 ∣ = ‖ S ‖ , then S ⊥ B T for all T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) .

Conversely, let ∀ T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) , S ⊥ B T . Claim: b i 0 j 0 = 0 .

Suppose b i 0 j 0 ≠ 0 . As a i 0 j 0 ≠ 0 , without loss of generality, we can assume that λ and b i 0 j 0 ( sgn a i 0 j 0 ) have the opposite sign. We choose λ such that

(3.5) ∣ λ ∣ < ∣ a i 0 j 0 ∣ − max i , j ∣ a i j ∣ 2 max i j ∣ b i j ∣

where i , j ∈ { 1 , 2 , … , n } and a i j ≠ a i 0 j 0 . Therefore, for any λ satisfying (3.5), we have

∣ a i j + λ b i j ∣ ≤ ∣ ∣ a i j ∣ + ∣ λ ∣ ∣ b i j ∣ ∣ < ∣ a i 0 j 0 ∣ = ‖ S ‖

where i , j ∈ { 1 , 2 , … , n } and a i j ≠ a i 0 j 0 . Assume that λ satisfies (3.5), then choose λ such that

∣ λ ∣ < 1 2 min 1 ≤ i , j ≤ n , b i j ≠ 0 ∣ a i j ∣ ∣ b i j ∣ .

Then ∣ λ b i 0 j 0 ( sgn a i 0 j 0 ) ∣ < 1 2 ∣ a i 0 j 0 ∣ .

Since λ and b i 0 j 0 ( sgn a i 0 j 0 ) have the opposite sign, then

∣ a i 0 j 0 + λ b i 0 j 0 ∣ = ∣ ∣ a i 0 j 0 ∣ + λ b i 0 j 0 ( sgn a i 0 j 0 ) ∣ < ∣ a i 0 j 0 ∣ = ‖ S ‖ .

This contradicts with ∀ T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) , S ⊥ B T . Therefore, we have b i 0 j 0 = 0 . □

The following are two examples of Theorems 3.1 and 3.2.

Example 3.3

Let S , T ∈ ℬ ( l 1 3 , l 2 3 ) . Then

‖ S ‖ = max ‖ x ‖ 1 = 1 ‖ S x ‖ 2 = sup j ∑ i = 1 n ∣ a i j ∣ 2 1 2 .

Define S = ( a i j ) 3 × 3 , T = ( b i j ) 3 × 3 . For any j ∈ { 2 , 3 } , we have

∣ a 11 ∣ > ∣ a 1 j ∣ , ∣ a 21 ∣ > ∣ a 2 j ∣ , ∣ a 31 ∣ > ∣ a 3 j ∣ .

Then for S ⊥ B T for all T ∈ ℬ ( l 1 3 , l 2 3 ) , we must have b 11 = b 21 = b 31 = 0 . We may choose

T = 0 b 12 b 13 0 b 22 b 23 0 b 32 b 33 .

But if we choose

T = 4 b 12 b 13 − 1 b 22 b 23 − 2 b 32 b 33 ,

then there exists

S = 4 3 − 2 − 2 1 0 − 2 − 1 1

where S satisfies ∣ a 11 ∣ > ∣ a 1 j ∣ , ∣ a 21 ∣ > ∣ a 2 j ∣ , ∣ a 31 ∣ > ∣ a 3 j ∣ for all j ∈ { 2 , 3 } , but S ⊥̸ B T .

Example 3.4

Let S , T ∈ ℬ ( l 1 3 , l ∞ 3 ) . Then

‖ S ‖ = max ‖ x ‖ 1 = 1 ‖ S x ‖ ∞ = sup i , j ∣ a i j ∣ .

Define S = ( a i j ) 3 × 3 , ∃ i 0 = 2 , j 0 = 3 such that ∣ a 23 ∣ > ∣ a i j ∣ for all i , j ∈ { 1 , 2 , 3 } and a 23 ≠ a i j . Then for S ⊥ B T for all T ∈ ℬ ( l 1 3 , l ∞ 3 ) , we must have b 23 ( sgn a 23 ) = 0 . We may choose

T = b 11 b 12 b 13 b 21 b 22 0 b 31 b 32 b 33 .

But if we choose

T = b 11 b 12 b 13 b 21 b 22 6 b 31 b 32 b 33 ,

then there exists

S = 1 − 2 3 4 5 − 6 4 0 − 3 ,

where S satisfies ∣ a 23 ∣ > ∣ a i j ∣ for all i , j ∈ { 1 , 2 , 3 } and a 23 ≠ a i j . but S ⊥̸ B T .

4 Bhatia-Šemrl property

By Theorems 3.1 and 3.2, we have that for any T , if S ⊥ B T , then there exists x ∈ M S such that S x ⊥ B T x .

Theorem 4.1

Let S = ( a i j ) n × n , T = ( b i j ) n × n ∈ ℬ ( l 1 n , l p n ) and i 0 , j 0 ∈ { 1 , 2 , … , n } .

(1) When 1 ≤ p < ∞ , ∃ S such that ∣ a i j 0 ∣ ≠ 0 for all i ∈ { 1 , 2 , … , n } and for any j ∈ { 1 , 2 , … , n } − { j 0 } , we have

∣ a 1 j 0 ∣ > ∣ a 1 j ∣ , ∣ a 2 j 0 ∣ > ∣ a 2 j ∣ , … , ∣ a n j 0 ∣ > ∣ a n j ∣ .

If ∀ T ∈ ℬ ( l 1 n , l p n ) , S ⊥ B T , then ∃ x ∈ M S such that S x ⊥ B T x .

(2) When p = ∞ , ∃ S such that ∣ a i 0 j 0 ∣ > ∣ a i j ∣ and a i 0 j 0 ≠ 0 for all i , j ∈ { 1 , 2 , … , n } and a i j ≠ a i 0 , j 0 . If ∀ T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) , S ⊥ B T , then ∃ x ∈ M S such that S x ⊥ B T x .

Proof

(1) Without loss of generality, we assume that j 0 = 1 , i.e. for any j ∈ { 2 , 3 , … , n } , we have

∣ a 11 ∣ > ∣ a 1 j ∣ , ∣ a 21 ∣ > ∣ a 2 j ∣ , … , ∣ a n 1 ∣ > ∣ a n j ∣ .

We choose x 0 = 1 0 ⋯ 0 T . Obviously, there exists x 0 ∈ R n with ‖ x 0 ‖ 1 = 1 such that

‖ S ‖ = ‖ S x 0 ‖ p = ∣ a 11 ∣ p + ∣ a 21 ∣ p + ⋯ + ∣ a n 1 ∣ p p .

Then x 0 ∈ M S . From Theorem 3.1, we have b 11 = b 21 = ⋯ = b n 1 = 0 . Therefore, for any λ ∈ R , we have

‖ ( S + λ T ) x 0 ‖ p = ∣ a 11 + λ b 11 ∣ p + ⋯ + ∣ a n 1 + λ b n 1 ∣ p p = ∣ a 11 ∣ p + ⋯ + ∣ a n 1 ∣ p p = ‖ S x 0 ‖ p .

Hence, ∃ x 0 ∈ M S such that S x 0 ⊥ B T x 0 .

(2) We choose x 1 = 0 ⋯ 1 ⋯ 0 T , where 1 is in row j 0 . Obviously, there exists x 1 ∈ R n with ‖ x 1 ‖ 1 = 1 such that ‖ S ‖ = ‖ S x 1 ‖ ∞ = ∣ a i 0 j 0 ∣ , then x 1 ∈ M S . From Theorem 3.2.2, we have b i 0 j 0 = 0 such that

‖ ( S + λ T ) x 1 ‖ ∞ ≥ ∣ a i 0 j 0 ∣ = ‖ S x 1 ‖ ∞

for all λ ∈ R . Hence, ∃ x 1 ∈ M S such that S x 1 ⊥ B T x 1 .□

Let x 0 ∈ X be a vector. In the following, we give the connections between matrix operators S and T when S x 0 ⊥ B T x 0 . First, we introduce a family of matrix set W k = { Y i ∣ i ≤ k } , where Y i = { x ∈ R n × 1 ∣ there are i rows in x that row elements are 1 i or − 1 i , the other rows are all zero } .

Theorem 4.2

Let S = ( a i j ) n × n , T = ( b i j ) n × n ∈ ℬ ( l 1 n , l p n ) . Suppose j 1 , … , j k ∈ { 1 , 2 , … , n } such that, for any i ∈ { 1 , 2 , … , n } , a i j 1 + ⋯ + a i j k and b i j 1 + ⋯ + b i j k have the same sign (opposite sign). Then there exists x 0 ∈ Y k such that T x 0 ∈ S x 0 + ( T x 0 ∈ S x 0 − ) .

Proof

Let the elements of the j 1 , … , j k rows of x 0 be all − 1 k or 1 k , and the other rows be all zero, where x 0 ∈ Y k ⊂ W k .

We have

‖ S x 0 ‖ p = 1 ∣ k ∣ ∑ i = 1 n ∣ a i j 1 + ⋯ + a i j k ∣ p p .

Accordingly, for any λ ≥ 0 , we have

(4.1) ‖ ( S + λ T ) x 0 ‖ p = 1 ∣ k ∣ ∑ i = 1 n ∑ t = 1 k ( a i j t ) + λ ∑ t = 1 k b i j t p p ≥ 1 ∣ k ∣ ∑ i = 1 n ∑ t = 1 k ( a i j t ) p p = ‖ S x 0 ‖ p .

Hence, there exists x 0 ∈ Y k such that T x 0 ∈ S x 0 + .

Similarly, for any λ ≤ 0 , we can also obtain (4.1). Hence, there exists x 0 ∈ Y k such that T x 0 ∈ S x 0 − .□

Theorem 4.3

Let S = ( a i j ) n × n , T = ( b i j ) n × n ∈ ℬ ( l 1 n , l ∞ n ) . Suppose j 1 , … , j k ∈ { 1 , 2 , … , n } and i 0 ∈ { 1 , 2 , … , n } such that b i 0 j 1 = ⋯ = b i 0 j k = 0 and ∣ a i 0 j 1 ∣ = ⋯ = ∣ a i 0 j k ∣ ≥ ∣ a i j ∣ for all i , j ∈ { 1 , 2 , … , n } and a i 0 j t ≠ 0 for all t ∈ { 1 , 2 , … , k } . Then for any i ≤ k , there exists x i ∈ W i such that S x i ⊥ B T x i .

Proof

When k = 1 , there exists j 1 ∈ { 1 , 2 , … , n } and i 0 ∈ { 1 , 2 , … , n } such that b i 0 j 1 = 0 , a i 0 j 1 ≠ 0 and ∣ a i 0 j 1 ∣ ≥ ∣ a i j ∣ for all i , j ∈ { 1 , 2 , … , n } . Let x 1 = 0 ⋯ sgn a i 0 j 1 ⋯ 0 T ∈ W 1 , where sgn a i 0 j 1 is in row j 1 such that ‖ S x 1 ‖ ∞ = ∣ a i 0 j 1 ∣ .

Since, for any λ ∈ R , we have

‖ ( S + λ T ) x 1 ‖ ∞ ≥ ∣ a i 0 j 1 ∣ + λ b i 0 j 1 ( sgn a i 0 j 1 ) = ∣ a i 0 j 1 ∣ = ‖ S x 1 ‖ ∞ .

Hence, there exists x 1 ∈ W 1 such that S x 1 ⊥ B T x 1 .

Assume that the aforementioned conclusion holds when k = n − 1 , i.e. there exists j 1 , … , j n − 1 ∈ { 1 , 2 , … , n } and exists i 0 ∈ { 1 , 2 , … , n } such that b i 0 j 1 = ⋯ = b i 0 j n − 1 = 0 and ∣ a i 0 j 1 ∣ = ∣ a i 0 j 2 ∣ = ⋯ = ∣ a i 0 j n − 1 ∣ ≥ ∣ a i j ∣ for all i , j ∈ { 1 , 2 , … , n } and a i 0 j t ≠ 0 for all t ∈ { 1 , 2 , … , n − 1 } . Then, for any i ≤ n − 1 , there exists x i ∈ W i such that S x i ⊥ B T x i .

At this time, when k = n , we only need to prove that there exists x n ∈ Y n such that S x n ⊥ B T x n . Let x n = sgn a i 0 1 n sgn a i 0 2 n ⋯ ⋯ sgn a i 0 n n T ∈ Y n . For any λ ∈ R , we have

‖ ( S + λ T ) x n ‖ ∞ ≥ ∑ j = 1 n [ ( a i 0 j + λ b i 0 j ) sgn a i 0 j ] n = ∑ j = 1 n ∣ a i 0 j ∣ + λ ∑ j = 1 n b i 0 j sgn a i 0 j n = ∣ a i 0 1 ∣ + ⋯ + ∣ a i 0 n ∣ n = ‖ S x n ‖ ∞ .

Then S x n ⊥ B T x n . Therefore, for any i ≤ k , there exists x n ∈ W i such that S x n ⊥ B T x n .□

5 Symmetry of matrix operators in operator space

Let X be strictly convex space. We give a characterization of matrix operators whom satisfy the right and left symmetry in the sense of Birkhoff orthogonality on ℬ ( X ) and ℬ ( l ∞ n ) , respectively. Also, the equivalence condition for the left symmetry of matrix operators on ℬ ( l p n ) ( 1 < p < ∞ ) is obtained.

Theorem 5.1

Let X be an n-dimensional strictly convex space. x 0 ∈ X be a left symmetric point. T ∈ ℬ ( X ) be such that M T = { ± x 0 } and ‖ I + T ‖ = 1 + ‖ T ‖ . Then one of the following statement is true:

ker T = 0 ,
T is not a right symmetric point in ℬ ( X ) .

Proof

Note that X is a finite dimensional space. Then, there exists y 0 ∈ S X such that ‖ I + T ‖ = ‖ ( I + T ) y 0 ‖ .

Since ‖ I + T ‖ = 1 + ‖ T ‖ , we have

(5.1) 1 + ‖ T ‖ = ‖ I + T ‖ = ‖ ( I + T ) y 0 ‖ = ‖ y 0 + T y 0 ‖ ≤ ‖ y 0 ‖ + ‖ T y 0 ‖ ≤ 1 + ‖ T ‖ .

Since X is strictly convex. If y 0 and T y 0 are linearly independent, then

‖ y 0 + T y 0 ‖ < ‖ y 0 ‖ + ‖ T y 0 ‖ .

This contradicts with (5.1). Therefore, y 0 and T y 0 are linearly dependent. Let T y 0 = k 0 y 0 , where k 0 ∈ R . From (5.1), we have

‖ y 0 + T y 0 ‖ = ‖ y 0 ‖ + ‖ T y 0 ‖ = 1 + ‖ T ‖ .

Since

1 + ‖ T ‖ = ‖ y 0 + T y 0 ‖ = ‖ y 0 + k 0 y 0 ‖ = ‖ ( 1 + k 0 ) y 0 ‖ = 1 + k 0 .

Then k 0 = ‖ T ‖ . Also note

1 + ‖ T ‖ = ‖ y 0 ‖ + ‖ T y 0 ‖ .

Then ‖ T y 0 ‖ = ‖ T ‖ and thus k 0 = ‖ T y 0 ‖ = ‖ T ‖ . As M T = { ± x 0 } , then M T = { ± x 0 } = { ± y 0 } , i.e. y 0 = x 0 . Since k 0 = ‖ T x 0 ‖ = ‖ T ‖ and T y 0 = k 0 y 0 , then T x 0 = ‖ T ‖ x 0 . Let ker T ≠ 0 . Then there exists u 0 ∈ X with u 0 ≠ 0 such that T u 0 = 0 , where ‖ u 0 ‖ = 1 . Then T x 0 ⊥ B T u 0 . From Lemma 2.3, we have x 0 ⊥ B u 0 . Since x 0 is a left symmetric point, then u 0 ⊥ B x 0 . Let { u 0 , x 0 , y 1 , … , y n − 2 } be a basis of X such that u 0 ⊥ B span { x 0 , y 1 , … , y n − 2 } . Define a linear operator A ∈ ℬ ( X ) as follows:

A u 0 = u 0 , A x 0 = 1 2 x 0 , A y i = 1 2 y i ,

for all i ∈ { 1 , 2 , … , n − 2 } . Now using the strict convexity of X we obtain M A = { ± u 0 } . From Lemma 2.4.4, we have A ⊥ B T by noting A u 0 ⊥ B T u 0 . Also since T x 0 = ‖ T ‖ x 0 ⊥̸ B 1 2 x 0 = A x 0 , i.e. T x 0 ⊥̸ B A x 0 , then from Lemma 2.4.4, we have T ⊥̸ B A . Therefore, T is not right symmetric point in ℬ ( X ) . Since the original proposition and the contraposition are equally true and false; therefore, if T is a right symmetric point, then we have ker T = 0 .□

Theorem 5.2

Let T = ( t i j ) n × n ∈ ℬ ( l ∞ n ) and T be right symmetric. Then ∑ j = 1 n ∣ t 1 j ∣ = ⋯ = ∑ j = 1 n ∣ t n j ∣ = ‖ T ‖ ∞ and all row vectors of T are scalar multiples of extreme points on l ∞ n .

Proof

Without loss of generality, we assume that ‖ T ‖ ∞ = n . Let T be right symmetric.

Claim:

∑ j = 1 n ∣ t 1 j ∣ = ⋯ = ∑ j = 1 n ∣ t n j ∣ = ‖ T ‖ ∞ .

Let there exist i = 1 such that ‖ T ‖ ∞ ≠ ∑ j = 1 n ∣ t 1 j ∣ . Let t 0 = ( t 11 , t 12 , … , t 1 n ) . Define a linear operator A ∈ ℬ ( l ∞ n ) as follows:

A = W 1 W 2 ⋯ W n − t 21 − t 22 ⋯ − t 2 n ⋮ ⋮ ⋮ − t n 1 − t n 2 ⋯ − t n n .

Let W 0 = ( W 1 , … , W n ) with ‖ W 0 ‖ ∞ = n such that W 0 ⊥ B t 0 . Then ‖ A ‖ ∞ = ‖ W 0 ‖ ∞ .

Now, for any λ ∈ R , we have

‖ A + λ T ‖ ∞ ≥ ∣ W 1 + λ t 11 ∣ + ⋯ + ∣ W n + λ t 1 n ∣ = ‖ W 0 + λ t 0 ‖ ∞ ≥ ‖ W 0 ‖ ∞ = ‖ A ‖ ∞ .

Therefore, A ⊥ B T .

We choose λ such that 0 < λ < ‖ T ‖ ∞ − ∑ j = 1 n ∣ t 1 j ∣ n . Since

‖ T + λ A ‖ ∞ = max ∑ j = 1 n ∣ t 1 j + λ W j ∣ , max 2 ≤ i ≤ n ∑ j = 1 n ∣ ( 1 − λ ) t i j ∣ ,

then we have

∑ j = 1 n ∣ t 1 j + λ W j ∣ = ‖ t 0 + λ W 0 ‖ ∞ ≤ ‖ t 0 ‖ ∞ + ∣ λ ∣ ‖ W 0 ‖ ∞ < ‖ T ‖ ∞

and for any i = { 2 , 3 , … , n } , we have

∣ ( 1 − λ ) t i 1 ∣ + ⋯ + ∣ ( 1 − λ ) t i n ∣ = ( 1 − λ ) ( ∣ t i 1 ∣ + ⋯ + ∣ t i n ∣ ) < ∣ t i 1 ∣ + ⋯ + ∣ t i n ∣ ≤ ‖ T ‖ ∞ .

So T ⊥̸ B A . This contradicts that T is right symmetric. Therefore, ∑ j = 1 n ∣ t 1 j ∣ = ⋯ = ∑ j = 1 n ∣ t n j ∣ = ‖ T ‖ ∞ .

Next claim: All row vectors of T are scalar multiples of extreme points on l ∞ n .

We only need to prove that all row vectors of T are right symmetric points. Suppose the vector t 0 = ( t 11 , t 12 , … , t 1 n ) in the first row of T is not a right symmetric point. Then, there exists x 0 = ( x 1 , x 2 , … , x n ) ∈ R n such that x 0 ⊥ B t 0 but t 0 ⊥̸ B x 0 and ‖ x 0 ‖ ∞ > ‖ t 0 ‖ ∞ . For the two vectors t 0 , x 0 either ‖ t 0 + λ x 0 ‖ ∞ ≥ ‖ t 0 ‖ ∞ = ‖ T ‖ ∞ for all λ ≥ 0 , or ‖ t 0 + λ x 0 ‖ ∞ ≥ ‖ t 0 ‖ ∞ = ‖ T ‖ ∞ for all λ ≤ 0 . Without any loss of generality, we can assume that ‖ t 0 + λ x 0 ‖ ∞ ≥ ‖ t 0 ‖ ∞ = ‖ T ‖ ∞ for all λ ≥ 0 . Since t 0 ⊥̸ B x 0 . Then there exists η ∈ ( − 1 , 0 ) such that ‖ t 0 + η x 0 ‖ ∞ < ‖ t 0 ‖ ∞ = ‖ T ‖ ∞ . Define a linear operator A ∈ ℬ ( l ∞ n ) as follows:

A = x 1 x 2 ⋯ x n t 21 t 22 ⋯ t 2 n ⋮ ⋮ ⋮ t n 1 t n 2 ⋯ t n n

such that ‖ A ‖ ∞ = ‖ x 0 ‖ ∞ . As x 0 ⊥ B t 0 . For any λ ∈ R , we have

‖ A + λ T ‖ ∞ ≥ ∣ x 1 + λ t 11 ∣ + ∣ x 2 + λ t 12 ∣ + ⋯ + ∣ x n + λ t 1 n ∣ = ‖ x 0 + λ t 0 ‖ ∞ ≥ ‖ x 0 ‖ ∞ = ‖ A ‖ ∞ .

Then A ⊥ B T . Since

‖ T + η A ‖ ∞ = max ∑ j = 1 n ∣ t 1 j + η x j ∣ , max 2 ≤ i ≤ n ∑ j = 1 n ∣ ( 1 + η ) t i j ∣ , ∣ t 11 + η x 1 ∣ + ⋯ + ∣ t 1 n + η x n ∣ = ‖ t 0 + η x 0 ‖ ∞ < ‖ t 0 ‖ ∞ = ‖ T ‖ ∞ ,

and for any i ∈ { 2 , 3 , … , n } , we have

∣ ( 1 + η ) t i 1 ∣ + ⋯ + ∣ ( 1 + η ) t i n ∣ = ( 1 + η ) ( ∣ t i 1 ∣ + ⋯ + ∣ t i n ∣ ) < ∣ t i 1 ∣ + ⋯ + ∣ t i n ∣ ≤ ‖ T ‖ ∞ .

So T ⊥̸ B A . This contradicts that T is right symmetric. Therefore, all row vectors of T are right symmetric points. Hence, by Lemma 2.5, all row vectors of T are scalar multiples of extreme points on l ∞ n .□

Theorem 5.3

Let T = ( t i j ) n × n ∈ ℬ ( l p n ) ( 1 < p < ∞ ) and M T = { e i 0 } where e i 0 is a unit vector. Then T is left symmetric if and only if T e i 0 is a left symmetric point and T e i = 0 for all i ∈ { 1 , 2 , … , n } − { i 0 } .

Proof

Let T e i 0 be left symmetric point. The image of e i under T is all zero for all i ∈ { 1 , 2 , … , n } − { i 0 } . Claim: T is left symmetric.

Without loss of generality, we may assume that M T = { ± e 1 } . Let an operator A = ( a i j ) n × n ∈ ℬ ( l p n ) satisfy T ⊥ B A , then from Lemma 2.4, we have T e 1 ⊥ B A e 1 . Set T e 1 is a left symmetric point. Then A e 1 ⊥ B T e 1 , i.e. ‖ ( A + λ T ) e 1 ‖ p ≥ ‖ A e 1 ‖ p for all λ ∈ R , and thus for any λ ∈ R , we have

(5.2) ∑ i = 1 n ∣ a i 1 + λ t i 1 ∣ p ≥ ∑ i = 1 n ∣ a i 1 ∣ p .

Then by (5.2) and T e i = 0 for all i ∈ { 2 , 3 , … , n } . For any λ ∈ R , we have

‖ A + λ T ‖ p = ∑ i = 1 n ∣ a i 1 + λ t i 1 ∣ p + ∑ i = 1 n ∣ a i 2 ∣ p + ⋯ + ∑ i = 1 n ∣ a i n ∣ p p ≥ ∑ i = 1 n ∣ a i 1 ∣ p + ∑ i = 1 n ∣ a i 2 ∣ p + ⋯ + ∑ i = 1 n ∣ a i n ∣ p p = ‖ A ‖ p .

Hence A ⊥ B T . From the arbitrariness of A , for any A ∈ ℬ ( l p n ) have T ⊥ B A ⇒ A ⊥ B T . Therefore, T is left symmetric.

Conversely, let T be left symmetric, and M T = { ± e i 0 } , Claim: T e i = 0 for all i ∈ { 1 , 2 , … , n } − { i 0 } .

Suppose there exists j 0 ≠ i 0 such that T e j 0 ≠ 0 . Define a linear operator A on ℬ ( l p n ) as A e j 0 = T e j 0 , A e i = 0 , ∀ i ≠ j 0 . It is easy to verify that M A = { ± e j 0 } . Since M T = { ± e i 0 } , and T e i 0 ⊥ B A e i 0 . Then, by Lemma 2.4.4, we have T ⊥ B A . Since A e j 0 ⊥̸ B T e j 0 and M A = { ± e j 0 } . Hence, A ⊥̸ B T . This contradicts that T is left symmetric. Therefore, T e i = 0 for all i ∈ { 1 , 2 , … , n } − { i 0 } .

Next claim: T e i 0 is a left symmetric point.

Suppose T e i 0 is not a left symmetric point. Then there exists W 0 ∈ R n such that T e i 0 ⊥ B W 0 but W 0 ⊥̸ B T e i 0 . Define an operator A on ℬ ( l p n ) with A e i 0 = W 0 , A e i = 0 , ∀ i ≠ i 0 . Obviously, M A = { ± e i 0 } . Since T e i 0 ⊥ B A e i 0 and M T = { ± e i 0 } . Then by Lemma 2.4, we have T ⊥ B A . As A e i 0 ⊥̸ B T e i 0 , thus, A ⊥̸ B T . This contradicts that T is left symmetric. Therefore, T e i 0 is a left symmetric point.□

6 Conclusion

On ℬ ( l 1 n , l p n ) ( 1 ≤ p ≤ ∞ ) , we characterized Birkhoff orthogonality elements of matrix operators and obtained the conditions that matrix operators satisfy the Bhatia-Šemrl property based on the matrix theory and generalized orthogonality theory. Hence, we obtain the relations between the orthogonality of vectors on l p n ( 1 ≤ p ≤ ∞ ) and the orthogonality of operators on ℬ ( l 1 n , l p n ) ( 1 ≤ p ≤ ∞ ) both in the sense of Birkhoff-James. In addition, we further discussed the left and right symmetry of matrix operators on some classical operator spaces, such as ℬ ( l p n ) ( 1 < p < ∞ ) .

Afterward, we will further study the Bhatia-Šemrl property and symmetry of matrix operators, and try to weaken the conditions of matrix operators of the results be obtained in the sense of Birkhoff-James.

Acknowledgement

The authors would like to thank the referee for valuable comments and suggestions on this article.

Conflict of interest: The authors state no conflict of interest.

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Received: 2022-12-06

Revised: 2023-05-15

Accepted: 2023-05-17

Published Online: 2023-07-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2022-0591

Keywords for this article

Birkhoff orthogonality; matrix operator; Bhatia-Šemrl property; left symmetry; right symmetry

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