Kyungpook Mathematical Journal 2018; 58(4): 589-597  
Preservers of Gershgorin Set of Jordan Product of Matrices
Manoj Joshi*, Kota Nagalakshmi Rajeshwari and Kilambi Santaram, Sandeep Kanodia
Department of Mathematics, Maharaja Ranjit Singh College of Professional Sciences, Indore (M. P.) 452001, India, e-mail : manojrjoshimrsc@yahoo.com, School of Mathematics, Devi Ahilya University, Indore (M. P.) 452001, India, e-mail : knr_k@yahoo.co.in and drksantaram@gmail.com, Department of Mathematics, Sri Aurobindo Institute of Technology, Indore (M. P.) 453555, India, e-mail : sandeepkanodia11@gmail.com
*Corresponding Author.
Received: July 24, 2017; Revised: January 29, 2018; Accepted: March 9, 2018; Published online: December 23, 2018.
© Kyungpook Mathematical Journal. All rights reserved.

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Abstract

For A, BM2(ℂ), let the Jordan product be AB + BA and G(A) the eigenvalue inclusion set, the Gershgorin set of A. Characterization is obtained for maps φ : M2(ℂ) → M2(ℂ) satisfying G[φ(A)φ(B)+φ(B)φ(A)]=G(AB+BA)

for all matrices A and B. In fact, it is shown that such a map has the form φ(A) = ±(PD)A(PD)−1, where P is a permutation matrix and D is a unitary diagonal matrix in M2(ℂ).

Keywords: Eigenvalue, Inclusion sets, Jordan Product, Preservers, Gershgorin Set.
1. Introduction

Eigenvalues of matrices play central role in linear algebra and its applications. When the order of the matrix is high, there is no efficient way to compute eigenvalues unless the matrix is of very special type. At times knowing the location of eigenvalues will be sufficient. Hence eigenvalue inclusion sets (i.e. sets containing eigenvalues) are of interest to the researchers.

The important types of eigenvalue inclusion sets of matrix A are Gershgorin set G(A), Ostrowski set Oɛ(A), ɛ ∈ [0, 1] and Brauer’s set C(A) [3, 4]. It is known that O1(A) = G(A).

The study of maps on matrix spaces which preserve a particular property, the so called preservers has been an area of active research for the past several years.

Let Mn denote the set of n × n matrices over ℂ, where ℂ is the set of complex numbers and for AMn, S(A) be an eigenvalue inclusion set mentioned above.

In [2] authors have characterized maps φ : MnMn satisfying

S(φ(A)-φ(B))=S(A-B)

and in [1] maps φ : MnMn satisfying

S(φ(A)φ(B))=S(AB)

have been characterized. Among other things, authors proved (see [1] theorem (2.1)):

A mapping φ : MnMn satisfies Oɛ[φ(A)φ(B)] = Oɛ(AB) for all A, BMn if and only if there exist c = ±1, a permutation matrix P, and an invertible diagonal matrix D, where D is unitary matrix unless (n,ɛ)=(2,12), such that φ(A) = c(DP)A(DP)−1 for all AMn.

Characterization of maps φ : MnMn that satisfy

S(φ(A)φ(B))=S(AB),

where ∘ is a binary operation on matrices such as the Jordan product AB + BA, the lie product ABBA, is still open.

The aim of this paper is to characterize the map φ : M2M2 which preserves the Gershgorin set of Jordan product in the sense that

G[(φ(A)φ(B)+φ(B)φ(A)]=G(AB+BA).

It is noteworthy that these maps admit the same characterization as in theorem stated above. The material of this paper has been organized in to two sections one on preliminaries and the other on the main result.

2. Preliminaries

The following result from matrix theory [3], (theorem 3.2.4.2) will be needed in the sequel.

Proposition 2.1

Suppose AMn has n distinct eigenvalues and BMn satisfies AB = BA. Then there is a complex polynomial p (z) of degree at most n − 1 such that B = p (A).

Definition 2.2. (Jordan product of matrices)

If A and B are two matrices of order n, then Jordan product of these matrices is defined as AB + BA.

Definition 2.3. (Gershgorin Set)

Given a matrix A = [aij ] ∈ Mn, define Rk=Rk(A)=j=1,jknakj. The Gershgorin set of A is defined by

G(A)=k=1nGk(A),

where Gk (A) = {z ∈ ℂ: |zakk| ≤ Rk}. The set Gk (A) is called a Gershgorin disk of A.

It is well known that the Gershgorin set contains all the eigenvalues of A [3, 4].

Remarks 2.4

A is the zero matrix if and only if G(A) = {0}.

A is a diagonal matrix if and only if G(A) is the set of diagonal entries of A.

For AMn, G(A) = G(PAPt), where P is a permutation matrix of order n and Pt is the transpose of the matrix P.

3. Main Result

Theorem 3.1

A mapping φ : M2M2satisfies

G[φ(A)φ(B)+φ(B)φ(A)]=G(AB+BA)

for all A, BM2 if and only if there exist c = ±1, a permutation matrix P and an unitary diagonal matrix D such that φ(A) = c (PD)A(PD)−1for all AM2.

The following propositions will be used in our proof.

Proposition 3.2

Let DM2be an invertible unitary diagonal matrix, PM2be a permutation matrix. Then for all AM2,

G[(PD)A(PD)-1]=G(A).
Proof

Let AM2, D = diag {α1, α2}. If D is an invertible diagonal unitary matrix then |αi| = 1 for i = 1, 2. If P = I, then

(PD)A(PD)-1=(D)A(D)-1=[a11α2-1α1a12α1-1α2a21a22]G[(PD)A(PD)-1]=G(A)   as αi=1   for   i=1,2.

If P=P12=[0110], then

(PD)A(PD)-1=[a22α2-1α1a21α1-1α2a12a11]G(PD)A(PD)-1=G(A).

Proposition 3.3

If a mapping φ : M2M2 satisfies

G[φ(A)φ(B)+φ(B)φ(A)]=G(AB+BA)

for all A, BM2, then there exist permutation matrix P and cij ∈ ℂ with |cij| = 1 and cijcji = 1 such that φ(Eij) = cijPEijPt for i, j ∈ {1, 2}.

Proof

Suppose i = j = 1 and A = B = E11, then

G(2φ(E11)2)=G(2E112)={2,0}φ(E11)2   must be E112or E222.

That is φ(E11)2 is a diagonal matrix with diagonal entries 0 and 1. Since φ (E11)2 has distinct eigenvalues and commutes with φ (E11), by proposition (2.1), φ (E11) is a polynomial in φ (E11)2. It follows φ (E11) is a diagonal matrix with diagonal entries 0 and 1 or −1 as φ(E11)2=E112or E222. Then φ(E11) = c11E11 or φ(E11) = c11E22, where c11 = ±1.

If φ(E11)=c11E11,then the permutation matrix P=Isatisfiesφ(E11)=c11PE11Pt.

Without loss of generality we may assume that c11 = 1 otherwise replace φ by −φ. So we assume φ(E11) = E11. Similarly φ(E22) = c22E11 or c22E22, where c22 = ±1. Suppose φ(E22) = c22E11. If A = E11, B = E22

G[φ(E11)φ(E22)+φ(E22)φ(E11)]=G(E11E22+E22E11)={0}.

But,

G[φ(E11)φ(E22)+φ(E22)φ(E11)]=G(c22E11E11+c22E11E11)={2c22,0}.

As c22 = ±1, this is a contradiction. Hence φ(E22) = c22E22, where c22 = ±1.

Suppose ij and φ(E21) = [bij ] ∈ M2. Then

G(2φ(E21)2)=G(2E212)={0}.

But,

G(2φ(E21)2)=G(2[b11b12b21b22]2)=G(2[b112+b12b21b11b12+b12b22b21b11+b22b21b21b12+b222])

Therefore,

{b112+b12b21=0,b11b12+b12b22=0,b21b11+b22b21=0,b21b12+b222=0

Taking A = E11, B = E21, we have

G[φ(E11)φ(E21)+φ(E21)φ(E11)]=G(E11E21+E21E11)=G(E21)

Note that G(E21) is a disk with center 0 and radius 1. Hence

G[φ(E11)φ(E21)+φ(E21)φ(E11)]=G([2b11b12b210])

is a disk with centre 0 and radius 1. It follows that |2b11| < 1 and |b21| ≤ 1. Two cases arise (i) |b21| = 1 or (ii) |b21| < 1, in which case b11 = 0 and |b12| = 1.

Case (i): If |b21| = 1, take A = E22, B = E21. Then

G[φ(E22)φ(E21)+φ(E21)φ(E22)]=G(E22E21+E21E22)=G(E21)

which is a disk with centre 0 and radius 1. But

G[φ(E22)φ(E21)+φ(E21)φ(E22)]=G([0c22b12c22b212c22b22])

As |b21| = 1 and c22 = ±1, b22 = 0. Otherwise this disk will not be identical with the above disk.

Now from the set of equations (1) we have b11 = 0 and b12 = 0 and hence φ(E21) = b21E21, where |b21| = 1

φ(E21)=c21PE21Pt,where c21=b21and P=I.

Case (ii): If |b21| < 1, in which case b11 = 0 and |b12| = 1, from the set of equations (1) we have b21 = 0 and b22 = 0 and hence φ(E21) = b12E12, where |b12| = 1.

This implies φ(E21) = c21PE21Pt, where c21 = b12 and P = P12.

Similarly, φ(E12) = c12PE12Pt.

Thus φ(Eij) = cijPEijPt. If PI2, then we replace the map φ by PtφP and get φ(Eij) = cijEij.

Claim :cijcji = 1

If i = j, then cijcji = 1. If ij, take A = E12 and B = E21, then

G[φ(E12)φ(E21)+φ(E21)φ(E12)]=G(E12E21+E21E12)=G(E11+E22)={1}

But,

G[φ(E12)φ(E21)+φ(E21)φ(E12)]=G(c12E12c21E21+c21E21c12E12)=G(c12c21E11+c21c12E22)={c12c21}.

Therefore, c12c21 = 1.

Proof of theorem 3.1

Sufficiency

For A, BM2 let φ(A) = c(PD)A(PD)−1, φ(B) = c(PD)B(PD)−1, where c = ±1, P is permutation matrix and D is a unitary diagonal matrix. Then

φ(A)φ(B)+φ(B)φ(A)=(PD)(AB+BA)(PD)-1(as c2=1)G[φ(A)φ(B)+φ(B)φ(A)]=G[(PD)(AB+BA)(PD)-1]=G(AB+BA)(from proposition (2.2))
Necessity

Suppose φ : M2M2 satisfies

G[(φ(A)φ(B)+φ(B)φ(A))]=G(AB+BA)

Let A = [aij] in M2. First we claim φ(A) = [cijaij], with |cij | = 1. Proof of this claim is divided into four steps. Let φ(A) = [dij ]

Step 1: To show that dij = cijaij for ij with |cij | = 1.

Step 2:φ(I) = I, where I is the identify matrix of order two.

Step 3: φ[100-1]=[100-1].

Step 4: Using the results of steps 1, 2 and 3 we prove dij = aij for i = j.

In view of proposition 3.3, we have φ(Eij) = cijEij, where |cij | = 1, c22 = ±1 and c11 = 1.

Step 1: dij = cijaij for ij.

Now, take B = E21, then

G[φ(A)φ(E21)+φ(E21)φ(A)]=G(AE21+E21A)G([c21d120c21(d11+d22)c21d12])=G([a120a11+a22a12])

Since both the sides are single disks comparing the centers we get,

c21d12=a12d12=c12a12(as c21-1=c12)

Similarly, d21 = c21a21. Hence dij = cijaij for ij.

Step 2: φ(I) = I

Let φ(I) = [bij ]. Taking A = E11, B = I, we have

G[φ(E11)φ(I)+φ(I)φ(E11)]=G(E11I+IE11)G[E11[bij]+[bij]E11]=G(2E11)G([2b11b12b210])={2,0}b11=1and b12=b21=0φ(I)=([100b22])

Now, take A = E22, B = I. Then

G[φ(E22)φ(I)+φ(I)φ(E22)]=G(E22I+IE22)G([0002c22b22])=G(2E22)={2,0}c22b22=1b22=c22as c22=±1.

Therefore φ(I)=[100c22].

Taking A = E12, B = I, we get

G[φ(E12)φ(I)+φ(I)φ(E12)]=G(E12I+IE12)G([0c12(1+c22)00])=G(2E12)

But, G(2E12) is a disk with center 0 and radius 2. This gives

c12(1+c22)=2(1+c22)=2as c12=1c22=1as c22=±1

Thus φ(I) = I.

Step 3 : φ[100-1]=[100-1]

Take A = I and B=[100-1]. Then

G[φ(I)φ(B)+φ(B)φ(I)]=G(IB+BI)G[2φ(B)]=G(2B)={2,-2}φ(B)is a diagonal matrix with diagonal entries 1and -1φ(B)=[100-1]or [-1001]

If φ(B)=[-1001], taking A = E11, we get

G[φ(E11)φ(B)+φ(B)φ(E11)]=G(E11I+IE11)G([-2000])=G([2000]){-2,0}={2,0}

Which is not possible and our supposition is wrong. Therefore

φ[100-1]=[100-1].

Step 4: d11 = a11and d22 = a22. For A = [aij] and B=[100-1], we have

G[(φ(A)φ(B)+φ(B)φ(A)]=G(AB+BA)G[2d1100-2d22]=G[2a1100-2a22]

If a11 = −a22 then the right hand side is one degenerated disk i.e. single point and therefore

a11=-a22d11=-d22a11=d11and a22=d22.

If a11 ≠ −a22 then we get two degenerated disks on the right hand side and there are two cases

Case (i)a11 = d11 and a22 = d22, then there is nothing to prove.

Case (ii)a11 = −d22 and a22 = −d11.

Now, we claim this case does not arise. We have either |a12| = |a21| or |a12| > |a21| or |a12| < |a21|. Taking B = E11, we have

G[(φ(A)φ(E11)+φ(E11)φ(A)]=G(AE11+E11A)HenceG([2d11d12d210])=G([2a11a12a210])

If they represent two disks then comparison of centers gives a11 = d11 and then from case (i) a22 = d22. Suppose they represent a single disk.

If |a12| = |a21| then |d12| = |d21|.

By comparison of centers, we get a11 = d11 = 0 and then from case (i) a22 = d22.

If |a12| > |a21|, then from step (i) |d12| > |d21|.

By comparison of centers, we get a11 = d11 and hence a22 = d22 from case (i).

If |a12| < |a21|, then |d12| < |d21|.

Taking B = E22 we have

G[φ(A)φ(E22)+φ(E22)φ(B)]=G(AE22+E22A)G([0c22d12c22d212c22d22])=G([0a12a212a22]).

As |a12| < |a21|, |c22d12| < |c22d21|, then by comparison of centers we have a22 = d22 and hence a11 = d11 from case (i).

Therefore case (ii) does not arise.

Thus dij = cijaij with |cij | = 1, cijcji = 1 and cij = 1 for i = j. Thus

φ(A)=[dij]=([d11d12d21d22])=[a11c12a12c21a21a22].

Let D=[100c21], then D-1=[100c12]. Hence

φ(A)=DAD-1.

Recall that in proposition 3.3 we obtained φ(Eij) = cijPEijPt. When PI we replaced the map φ by PtφP and assumed φ(Eij) = cijEij and the rest of the proof was carried out under this assumption. Hence

φ(A)=c(PD)A(PD)-1

and this completes the proof.

Acknowledgements

Authors thank the referees for their suggestions which have improved the presentation of the paper.

References
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  3. Horn, RA, and Johnson, CR (1985). Matrix Analysis. Cambridge: Cambridge University Press
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  4. Varga, RS (2004). Gershgorin and his circles. Berlin: Springer-Verlag
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