THE STRUCTURE OF LINEAR PRESERVERS OF LEFT MATRIX MAJORIZATION ONRP ∗
FATEMEH KHALOOEI† AND ABBAS SALEMI†
Abstract. For vectors X, Y ∈Rn,Y is said to be left matrix majorized byX (Y ≺ℓ X) if for some row stochastic matrixR, Y =RX. A linear operator T:Rp→Rnis said to be a linear preserver of≺ℓifY ≺ℓXonRpimplies thatT Y ≺ℓT XonRn. The linear operatorsT:Rp→Rn
(n < p(p−1)) which preserve≺ℓhave been characterized. In this paper, linear operatorsT:Rp→Rn
which preserve≺ℓare characterized without any condition onnandp.
Key words. Row stochastic matrix, Doubly stochastic matrix, Matrix majorization, Weak matrix majorization, Left (right) multivariate majorization, Linear preserver.
AMS subject classifications.15A04, 15A21, 15A51.
1. Introduction. LetMnm be the algebra of all n×m real matrices. A ma-trix R = [rij] ∈Mnm is called a row stochastic (resp., row substochastic) matrix if rij ≥0 and Σm
k=1rik = 1 (resp., ≤1) for alli, j. For A, B in Mnm, A is said to be left matrix majorizedbyB(A≺ℓB),ifA=RBfor somen×nrow stochastic matrix R. These notions were introduced in [11]. If A≺ℓ B ≺ℓ A, we write A ∼ℓ B. Let T:Rp→Rn be a linear operator.T is said to be a linear preserver of≺
ℓifY ≺ℓX on
Rpimplies thatT Y ≺ℓT X onRn. For more information about types of majorization see [1], [5] and [10]; for their preservers see [2]-[4], [6] and [9].
We shall use the following conventions throughout the paper: Let T :Rp →Rn be a nonzero linear operator and let [T] = [tij] denote the matrix representation ofT with respect to the standard bases{e1, e2, . . . , ep}ofRpand{f1, f2, . . . , fn}ofRn.If
p= 1, then all linear operators onR1are preservers of≺ℓ. Thus, we assumep≥2.Let
Ai be mi×pmatrices, i= 1, . . . , k. We use the notation [A1/A2/ . . . /Ak] to denote
the corresponding (m1+m2+. . .+mk)×pmatrix. We let e= (1,1, . . . ,1)t∈Rp,
and denote
a: = max{maxT(e1), . . . ,maxT(ep)},
b: = min{minT(e1), . . . ,minT(ep)}.
(1.1)
∗Received by the editors November 18, 2008. Accepted for publication February 3, 2009. Handling
Editor: Stephen J. Kirkland.
†Department of Mathematics, Shahid Bahonar University of Kerman, Kerman, Iran
Theorem 1.1. ([9, Theorem 2.2])LetT:Rp→Rn be a nonzero linear preserver
of ≺ℓ and suppose p ≥ 2. Then p ≤ n, b ≤ 0 ≤ a and for each i ∈ {1, . . . , p}, a= maxT(ei)andb= minT(ei). In particular, every column of[T]contains at least one entry equal toaand at least one entry equal to b.
Definition 1.2. LetT:Rp→Rn be a linear operator. We denote byPi (resp., Ni) the sum of the nonnegative (resp., non positive) entries in the ithrow of [T]. If all the entries in theith row are positive (resp., negative), we defineNi = 0 (resp., Pi= 0).
We know that T is a linear preserver of ≺ℓ if and only if αT is also a linear preserver of≺ℓfor some nonzero real numberα. Without loss of generality we make the following assumption.
Assumption 1.3. Let T:Rp →
Rn be a nonzero linear preserver of ≺ℓ . Let a
andb be as in (1.1). We assume that0≤ −b≤1 =a.
Definition 1.4. Let P be the permutation matrix such that P(ei) = ei+1,
1≤i≤p−1, P(ep) =e1.LetI denote thep×pidentity matrix, and letr, s∈Rbe
such thatrs <0. Define thep(p−1)×pmatrixPp(r, s) = [P1/P2/ . . . /Pp−1], where
Pj =rI+sPj, for all j = 1,2, . . . , p−1. It is clear that up to a row permutation, the matricesPp(r, s) andPp(s, r) are equal. Also definePp(r,0) :=rI,Pp(0, s) :=sI andPp(0,0) as a zero row.
The structure of all linear operatorsT:Mnm→Mnmpreserving matrix majoriza-tions was considered in [6, 7, 8]. Also the linear operators T from Rp to Rn that preserve the left matrix majorization ≺ℓ were characterized in [9] forn < p(p−1). In the present paper, we will characterize all linear preservers of≺ℓ mapping Rp to
Rn without any additional conditions.
2. Left matrix majorization. In this section we obtain a key condition that is necessary forT :Rp→Rn to be a linear preserver of≺ℓ. We first need the following.
Lemma 2.1. Let T : Rp →
Rn be a linear operator such that minT(Y) ≤
minT(X)for allX ≺ℓY.Then T is a preserver of ≺ℓ.
Proof. LetX ≺ℓY.It is enough to show that maxT(X)≤maxT(Y). SinceX≺ℓ Y,−X ≺ℓ −Y,and hence minT(−Y)≤minT(−X).This means that maxT(X)≤ maxT(Y).Then T is a preserver of≺ℓ.
Remark 2.2. LetT :Rp→
we define
Ik ={i: 1≤i≤n, tik = 1}, Jk={j: 1≤j≤n, tjk=b}.
Next we state the key theorem of this paper.
Theorem 2.3. Let T :Rp→Rn be a linear preserver of ≺
ℓ and let aand b be
as in Assumption 1.3. Then there exist 0≤α≤1 andb≤β ≤0 such that Pp(1, β)
andPp(α, b)are submatrices of[T], wherePp(r, s)is as in Definition 1.4.
Proof. Let 1≤k≤pbe a fixed number and let Ik andJk be as in Remark 2.2. SinceT is a linear preserver of≺ℓ,it follows that Ik andJk are nonempty sets. Also ek+el ≺ℓ ek, l =k. Thus, the other entries in the ith row, i ∈ Ik (resp., jth row, j∈Jk) are non positive (resp., nonnegative). Hence, til ≤0, tjl ≥0, l=k, i∈Ik, andj∈Jk. Letβi
k=
l=ktil≤0, i∈Ik and α j k =
l=ktjl ≥0, j∈Jk. Set βk := min{βi
k, i∈Ik}, αk := max{α j
k, j∈Jk}. (2.1)
DefineXk=−(N+ 1)ek+e. ChooseN0 large enough such that for allN ≥N0 and
1≤i≤n,
minT(Xk) =−N+βk ≤ −N tik+ l=k
til≤ −N b+αk= maxT(Xk). (2.2)
We know thatXk ∼ℓ Xr =−(N+ 1)er+e, 1 ≤r≤pand T is a linear preserver of ≺ℓ. Hence by (2.2), α := αk = αr and β := βk = βr,1 ≤ r ≤ p. Also, Xk ∼ℓ
−N ei+ej, i=j. F or eachN ≥N0, there exists 1≤h≤nsuch that−N thi+thj=
minT(−N ei +ej) = minT(Xk) = −N +β and for each 1 ≤ i ≤ p, 1 ≤ j ≤ p and N ≥ N0, there exists 1 ≤ h ≤n such that −N(1−thi) = thj −β. It follows
that thi = 1, thj =β. Hence Pp(1, β) is a submatrix of [T]. Similarly, there exists N1, such that for each N ≥ N1 there exists 1 ≤ h ≤ n so that −N thi +thj =
maxT(−N ei+ej) = maxT(Xk) =−N b+αand−N(b−thi) =thj−α.Thus,thi =b and thj = α. Since 1 ≤ i = j ≤ p was arbitrary, Pp(b, α) is a submatrix of [T]. Therefore,Pp(1, β) andPp(b, α) are submatrices of [T].
Remark 2.4. Let T :Rp →Rn andT:Rp →Rm be two linear operators such that [T] = [T1/T2/ . . . /Tn] and let [T] = [T1/T2/ . . . /Tm] be the matrix representation
of these operators with respect to the standard basis. LetR(T) ={T1, T2, . . . , Tn}
be the set of all rows of [T]. If R(T) = R(T), thenT preserves≺ℓ if and only if T preserves≺ℓ.
Proof. Without loss of generality, letβ ≤0 ≤αand let X = (x1, . . . , xp)t, Y =
(y1, . . . , yp)t∈Rp such thatX ≺ℓ Y.Thenym= minY ≤xi ≤maxY =yM, for all 1≤i≤p.It is easy to check thatαym+βyM ≤αxi+βxj,for alli=j∈ {1, . . . , p}, which implies minT Y ≤minT X.Hence by Lemma 2.1,T X ≺ℓT Y.
3. Left matrix majorization on R2 . LetT:R2 → Rn be a linear operator and leta, b, be as in Assumption 1.3. We consider the squareS= [b,1]×[b,1] inR2.
Definition 3.1. LetT :R2→Rnbe a linear operator and let [T] = [T
1/ . . . /Tn],
whereTi= (ti1, ti2),1≤i≤n. Define
∆ := Conv ({(ti1, ti2),(ti2, ti1),1≤i≤n})⊆R2.
Also, letC(T) denote the set of all corners of ∆.
Lemma 3.2. LetT :R2→Rn be a linear preserver of≺
ℓand[T] = [T1/ . . . /Tn], whereTj = (tj1, tj2), 1≤j ≤n. If for some 1≤i≤n, ti1ti2>0, then Ti∈/ C(T), whereC(T)is as in Definition 3.1.
Proof. Assume that, if possible, there exists 1≤i≤n such thatTi ∈C(T) and ti1ti2>0. By Remark 2.4 we can assume that [T] has no identical rows. Without loss
of generality, we assume that there exist 1 ≤i≤n and real numbers m≤M such thatti1>0, ti2>0 andmti1+M ti2< mtj1+M tj2, j=i. Chooseε >0 small enough
so thatmti1+ (M+ε)ti2< mtj1+ (M+ε)tj2, j=i. Since (m, M)t≺ℓ(m, M+ε)t, T(m, M)t≺
ℓ T(m, M +ε)t. But min(T(m, M +ε)t) =mti1+ (M +ε)ti2 > mti1+
M ti2= min(T(m, M)t), a contradiction.
Next we shall characterize all linear operatorsT :R2→Rn which preserve≺ℓ.
Theorem 3.3. Let T : R2 → Rn be a linear operator. Then T is a linear
preserver of ≺ℓ if and only if P2(x, y) is a submatrix of [T] and xy ≤ 0 for all
(x, y)∈C(T).
Proof. Let T be a linear preserver of≺ℓwith 0≤ −b≤1 =a. Let (x, y)∈C(T), then by Lemma 3.2,xy≤0.Without loss of generality, letTi= (ti1, ti2)∈C(T) and
ti1ti2≤0. By Remark 2.4, we assume that [T] has no identical rows. Then there exist
real numbersm, M ∈Rsuch thatmti1+M ti2< mtj1+M tj2, j=i. Chooseε0>0
small enough so that (m−ε)ti1+(M+ε)ti2<(m−ε)tj1+(M+ε)tj2, j=i,0< ε≤ε0.
Since (M +ε, m−ε)t∼
ℓ (m−ε, M +ε)t, T(M +ε, m−ε)t ∼ℓ T(m−ε, M +ε)t. Hence, for all 0 < ε ≤ ε0, there exist 1 ≤k ≤ n such that Tk = (tk1, tk2) ∈ C(T)
and (m−ε)ti1+ (M +ε)ti2 = minT(m−ε, M +ε)t = minT(M +ε, m−ε)t =
(M+ε)tk1+ (m−ε)tk2. Sincek∈ {1,2, . . . , n}is a finite set, there existsksuch that
tk1=ti2 andtk2=ti1. Therefore,P2(ti1, ti2) is a submatrix of [T].
xy≤0.Define the linear operatorTonR2such that [T] = [P
2(x1, y1)/· · ·/P2(xr, yr)],
where (xi, yi) ∈ C(T),1 ≤ i ≤ r. By elementary convex analysis, we know that maxT(X) = maxT(X ) and minT(X) = minT(X) for all X ∈ R2. Hence it is enough to show thatT is a linear preserver of≺ℓ.By Lemma 2.5, eachP2(xi, yi) is
a linear preserver of≺ℓ. Thus,Tis a linear preserver of≺ℓ.
4. Left matrix majorization on Rp. In this section we shall characterize all linear operatorsT :Rp→Rn which preserve≺ℓ.We shall prove several lemmas and prove the main theorem of this paper.
Definition 4.1. LetT :Rp→Rnbe a linear operator and let [T] = [T
1/ . . . /Tn].
Define
Ω := Conv({Ti= (ti1, . . . , tip),1≤i≤n})⊆Rp.
Also, letC(T) be the set of all corners of Ω.
Lemma 4.2. LetT :Rp →Rn be a linear preserver of≺
ℓand[T] = [T1/ . . . /Tn], where Ti = (ti1, ti2, . . . , tip), 1 ≤ i ≤ n. Suppose there exists 1 ≤ i ≤ n such that
tij >0,∀1 ≤j ≤p, or tij < 0,∀1 ≤ j ≤ p. Then Ti ∈/ C(T), where C(T) is as in Definition 4.1.
Proof. Assume that, if possible, there exists 1 ≤ i ≤ n such that Ti ∈ C(T) and tij > 0, for all 1 ≤ j ≤ p, or tij < 0, for all 1 ≤ j ≤ p. By Remark 2.4, without loss of generality, we can assume that [T] has no identical rows and there exists 1 ≤ i ≤ n such that tij > 0, for all 1 ≤ j ≤ p. Since Ti ∈ C(T), there exists X = (x1, . . . , xp)t such thatx1ti1+x2ti2+· · ·+xptip < x1tj1+x2tj2+· · ·+
xptjp, j = i. Let xk = max{xi,1 ≤ i ≤ p}. Choose ε > 0 small enough so that x1ti1+· · ·+ (xk+ε)tik+· · ·+xptip < x1tj1+· · ·+ (xk+ε)tjk+· · ·+xptjp, j=i.
DefineX = (x1, . . . , xk+ε, . . . , xp)t.Sincetik >0, hence minT(X) =x1ti1+x2ti2+
· · ·+xptip < x1ti1+· · ·+ (xk +ε)tik +· · ·+xptip = minT(X). But X ≺ℓ X, a contradiction.
LetT :Rp→Rnbe a linear operator. Without loss of generality, we assume that [T] = [Tp/Tn/T],where all entries ofTp(resp.,Tn) are positive (resp., negative) and each row ofThas nonnegative and non positive entries.
Corollary 4.3. Let T and T be as above. Then T preserves ≺ℓ if and only if
C(T) =C(T) andT preserves≺ℓ,where C(T) is as in Definition 4.1.
Definition 4.4. LetT :Rp→Rn be a linear operator. Define ∆ = Conv({(Pi, Ni),(Ni, Pi) : 1≤i≤n}),
where Pi, Ni be as in (1.2). Let E(T) ={(Pi, Ni) : (Pi, Ni) is a corner of ∆}.Let 1≤i≤n,define [i] ={j: 1≤j ≤n, Pi=Pj and Ni= Nj}.
Lemma 4.5. Let T:Rp →
Rn be a linear preserver of≺ℓ and let C(T), E(T) be
as in Definitions 4.1, 4.4, respectively. If (Pr, Nr)∈E(T) for some1≤r≤n, then there existsk∈[r]such that Tk∈C(T).
Proof. Suppose there exist 1 ≤ r ≤ n such that (Pr, Nr) ∈ E(T). Then there existsm≤M such that
Prm+NrM < Pjm+NjM, j /∈[r]. (4.1)
LetX ∈Rp such that min(X) =m and max(X) =M. Then there exists 1≤k≤n such that minT X =pl=1tklxl. Hence
Prm+NrM ≤Pkm+NkM ≤
p
l=1
tklxl= minT(X). (4.2)
DefineY ∈Rp byyl=m, iftrl>0 andyl=M,iftrl≤0.ObviouslyY ≺ℓX.Since T preserves≺ℓ, T Y ≺ℓT X which implies that
Pkm+NkM ≤
p
l=1
tklxl= minT X≤minT Y ≤Prm+NrM. (4.3)
Now, by (4.2) and (4.3), we havePrm+NrM =Pkm+NkM. Thus by (4.1),k∈[r] and minT X=pl=1tklxl. HenceTk∈C(T) for somek∈[r].
Next we state the main result in this paper.
Theorem 4.6. Let T andE(T)be as in Definition 4.4. ThenT preserves ≺ℓ if and only ifPp(α, β) is a submatrix of[T] for all(α, β)∈E(T).
Proof. Let T be a preserver of ≺ℓ and let (Pr, Nr) ∈ E(T). Then there exists m≤M such thatPrm+NrM < Pjm+NjM, j /∈[r]. Chooseε0 small enough so
that for all 0< ε < ε0,
Pr(m−ε) +Nr(M+ε)< Pj(m−ε) +Nj(M +ε), j /∈[r], Ifj∈[r], thenPj =Pr andNj=Nr.Thus
Let 0< ε < ε0,be fixed and let Xε= (xε1, . . . , xpε)t∈Rp with minXε=m−ε and maxXε=M +ε.As in the proof of Lemma 4.5, there existsk∈[r] such that
Pr(m−ε) +Nr(M +ε) = minT(Xε) = p
l=1
tklxεl.
Fixi=j ∈ {1, . . . , p} and define Yε= (yε
1, . . . , yεp)t∈Rp such that yiε=m−ε, yε
j =M+εandy ε
l =γl, m−ε < γl< M+ε, l=i, j.SinceX ε∼
ℓYε,T Xε∼ℓT Yε, there exists q∈ [r] such thattqi(m−ε) +tqj(M +ε) +l=i,jγltql=Pr(m−ε) + Nr(M+ε).Since 0< ε < ε0andm−ε≤γl≤M+ε, l=r, sare arbitrary, it is easy
to show that there existss∈[r] such thattsi=Pr andtsj=Nr andtsl = 0, l=i, j. Therefore [T] hasPp(Pr, Nr) as a submatrix.
Conversely, Let E(T) = {(Pi1, Ni1), . . . ,(Pis, Nis)}.Then up to a row permuta-tion [T] = [Pp(Pi1, Ni1)/ . . . /Pp(Pis, Nis)/Q].
LetTbe the operator on Rpsuch that [T] = [ Pp(Pi1, Ni1)/ . . . /Pp(Pik, Nik)]. LetTi∈Qand suppose there existsX ∈Rp such that
minT(X) = p
l=1
tilxl≤
p
l=1
tjlxl,1≤j≤n.
Obviously,Pim+NiM ≤pl=1tilxl≤
p
l=1tjlxl,1≤j≤n, wherem= minX and
M = maxX.We know that (Pi, Ni)∈∆ and ∆ is convex. Hence there is 1≤k≤n such that (Pk, Nk) ∈ E(T) and Pkm+NkM ≤ Pim+NiM. As in the proof of Lemma 4.5, minT X = Pkm+NkM. Then minT X ≤ minT X. But we know that minT(X) ≤ minT X and thus minT X = minT X. Similarly, maxT X = maxT X. Therefore,T is a preserver of ≺ℓif and only if T preserves≺ℓ.By Lemma 2.5 each
Pp(Pil, Nil) is a preserver of ≺ℓ,1 ≤ l ≤ k. Hence T is a preserver of ≺ℓ and the theorem is proved.
Next we state necessary conditions for T :Rp →Rn to be a linear preserver of
≺ℓ. We use the notation of Theorem 2.3 in the following corollary.
Corollary 4.7. Let T : Rp →
Rn be a linear operator and let a and b be as given in (1.1). If the following conditions hold, thenT is a linear preserver of ≺ℓ.
• [T]has [Pp(a,0)/Pp(0, b)/Pp(a, b)]as a submatrix.
• 0≤Pi≤aandb≤Ni≤0, 1≤i≤n,
wherePi andNi,1≤i≤n are as in Definition 1.2.
Proof. It is clear thatE(T) ={(a,0),(0, b),(a, b)}.Since [T] hasPp(a,0),Pp(0, b) andPp(a, b) as submatrices, it follows by Theorem 4.6 thatT is a linear preserver of
LetT :Rp→Rn be a linear preserver of≺ℓ,and let [T] = [T1|T2|. . .|Tp],where Ti is theith column of [T]. For i= j ∈ {1, . . . , p} define Tij : R2 →Rn such that [Tij] = [Ti|Tj].
Lemma 4.8. Let T : Rp → Rn be a linear preserver of ≺ℓ, and let Tij be as
above. Then Tij is a linear preserver of ≺
ℓ for alli=j∈ {1, . . . , p}.
Proof. Let i = j ∈ {1, . . . , p} and let x = (x1, x2)t, y = (y1, y2)t ∈ R2 such
that x≺ℓ y. Define X, Y ∈ Rp such that Xi =x1, Xj = x2, Yi =y1, Yj =y2 and
Xk =Yk= 0,for allk=i, j.It is obvious thatX ≺ℓY in Rp and henceT X≺ℓT Y inRn.ButTijx=x
1Ti+x2Tj =T X ≺ℓT Y =y1Ti+y2Tj=Tijy.Therefore,Tij
is a linear preserver of≺ℓ.
The following example shows that the converse of Lemma 4.8 is not necessarily true.
Example 4.9. Assume [T] = [P3(1,−0.5)/ 0.25 0.25 0.25]. Consider X =
(−1,−1,−1)t and Y = (−1,−1,−0.75)t, we know that X ≺
ℓ Y and minT X < minT Y. Thus T is not a linear preserver of ≺ℓ . However, by Corollary 4.7, for all i=j∈ {1,2,3}, Tij preserves≺
ℓ.
5. Additional results. In this section we give short proofs of some Theorems from [6, 9].
Theorem 5.1. [6]Let T:R2→R2 be a linear operator. Then T preserves ≺
ℓ if
and only if T has the formT(X) = (aI+bP)X for allX ∈R2, whereP is the2×2
permutation matrix not equal toI, andab≤0.
Proof. LetT be a preserver of≺ℓ. By Assumption 1.3, a= 1.By Theorem 2.3, there exist 0≤α≤1 and b≤β ≤0 such thatP(1, β) and P(b, α) are submatrices of [T]. Since [T] is a 2×2 matrix, β =b and α = 1. Therefore, [T] =
1 b b 1
and henceT(X) = (I+bP)X,for allX ∈R2.Conversely, up to a row permutation, [T] =P2(1, b) and by Lemma 2.5,T preserves≺ℓ.
Theorem 5.2. [6] Let p ≥ 3. Then T:Rp → Rp is a linear preserver of left
matrix majorization if and only if T is of the form X →aP X for some a∈R and some permutation matrixP.
Theorem 5.3. ([9, Theorem 3.1]) For a linear preserver T of Rp to Rn the
following assertions hold.
(a)If n <2pandp≥3, thenT is nonnegative.
(b) If T is nonnegative, then there exists an n×n permutation matrix Q such that[T] =Q[I/W], where W is a (possibly vacuous)(n−p)×pmatrix of one of the following forms (i),(ii)or (iii):
(i)W is row stochastic;
(ii)W is row substochastic and has a zero row;
(iii)W = [(cI)/B], where0< c <1andBis an (n−2p)×prow substochastic matrix with row sums at least c.
(c)Let Qbe ann×npermutation matrix, and letW be an(n−p)×pmatrix of the form(i),(ii), or (iii)in part (b). Then the operator X →Q[X/(W X)]from Rp
intoRn is a nonnegative linear preserver of ≺ℓ.
Proof.
(a) Assume that, if possible, b < 0. By Theorem 2.3 n ≥ p(p−1). Since p ≥
3, n≥2p,a contradiction.
(b) SinceT is nonnegative,Ni= 0,1≤i≤n,and 0≤Pi ≤1. By Theorem 2.3, [T] hasPp(1,0) as its submatrix and therefore up to a row permutation [T] = [I/W]. Letc= min{Pi,1≤i≤n}.ThenE(T) ={(1,0),(c,0)}.By Theorem 4.6,Pp(c,0) is a submatrix of [T]. Ifc= 1 then (i) holds; if c= 0 then (ii) holds and if 0< c <1, then (iii) holds.
(c) Let [T] = [I/W], where W is an (n−p)×p matrix of the form (i), (ii), or (iii) in part (b). Then E(T) = {(1,0),(c,0)}. By Theorem 4.6, T is a nonnegative linear preserver of≺ℓ.
Theorem 5.4. ([9, Theorem 4.5]) AssumeT : Rp →
Rn is a linear preserver of ≺ℓ, b < 0 and 2p ≤ n < p(p−1). Let Pi (resp., Ni) denote the sum of the
positive (resp., negative) entries of theithrow of[T]. Then, up to a row permutation, [T] = [I/bI/B] andmin(Ni+bPi) =b,(i= 1,2, . . . , n).
Acknowledgment. This research has been supported by the Mahani Mathemat-ical Research Center and the Linear Algebra and Optimization Center of Excellence of the Shahid Bahonar University of Kerman.
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