http://scma.maragheh.ac.ir
ON MULTIPLICATIVE (STRONG) LINEAR PRESERVERS OF MAJORIZATIONS
MOHAMMAD ALI HADIAN NADOSHAN1∗ AND ALI ARMANDNEJAD2
Abstract. In this paper, we study some kinds of majorizations on
Mn and their linear or strong linear preservers. Also, we find the structure of linear or strong linear preservers which are multiplica-tive, i.e. linear or strong linear preservers like Φ with the property Φ(AB) = Φ(A)Φ(B) for everyA, B∈Mn.
1. Introduction
An n×n nonnegative real matrixA is doubly stochastic if all of its row and column sums are one. For X, Y ∈ Mn, it is said that X is
majorized by Y (written as X ≺Y) if there exists a doubly stochastic matrix Dsuch that X =DY. If we omit the non-negativity condition, we reach to the set of all generalized n×ndoubly stochastic matrices, denoted byGDn. By considering some other kinds of stochastic matrices
such as row stochastic, upper triangular row stochastic, tridiagonal row stochastic and even stochastic matrices, we reach to various kinds of majorizations. LetT :Mn→Mn be a linear operator. For a relation∼
onMn, it is said thatT is a linear preserver of∼, ifT X ∼T Y whenever X ∼Y. Similarly, it is said that T is a strong linear preserver of ∼, if
X∼Y ⇔ T X ∼T Y.
Also, it is said that T is a multiplicative linear preserver of ∼, if in addition T(XY) = T(X)T(Y), for all X, Y ∈ Mn. In this paper, we
characterize all multiplicative linear or strong linear preservers of some kinds of majorizations. It is well-known that nonzero linear and multi-plicative operators on Mn are inner automorphisms. We use this fact in
2010Mathematics Subject Classification. 15A04, 15A21.
Key words and phrases. Doubly stochastic matrix, Linear preserver, Multiplicative map .
Received: 24 December 2015, Accepted: 27 January 2016.
∗Corresponding author.
some proofs but we prefer to prove other theorems without using this. Assume Pn be the set of alln×npermutation matrices and Eij be an n×nmatrix such that its ijth entry is 1, and all other entries are 0.
2. Multiplicative linear preservers of majorizations
In this section, we find multiplicative linear preservers of some kinds of majorizations. LetX, Y ∈Mn. Xis said to be multivariately majorized
byY (written asX≺m Y), ifX=DY for somen×ndoubly stochastic
matrixDand called directionally majorized byY (written asX ≺dY),
if for everya∈Rnthere exists a doubly stochastic matrix D
a such that Xa=DaY a. In [6] and [7], the authors found the linear preservers and
strong linear preservers of multivariate and directional majorizations as follows.
In this paper, the letter J indicates the matrix with all entries equal to 1, andtr(x) uses for the summation of coordinates of a vectorx.
Theorem 2.1 ([7], Theorem 1). Let T : Mn,m → Mn,m be a linear
function. Then, T preserves multivariate majorizations if and only if T
preserves directionally majorizations if and only if one of the following holds:
(a) There exist A1, . . . , Am∈Mn,m such that
T X =
m
∑
i=1
(tr(xi))Ai,
where xi is the i’th column of X.
(b) There exist P ∈ Pn and R, S∈Mm such that T X =P XR+J XS.
Theorem 2.2 ([6], Theorem 2.4). Let T : Mn,m → Mn,m be a linear
function. Then, T strongly preserves multivariate majorizations if and only ifT strongly preserves directional majorizations if and only if there exist P ∈ Pn and R, S∈Mm such thatT X =P XR+J XS andR(R+ nS) is invertible.
Let T : Mn → Mn has the form T X = P XR+J XS, for some P ∈ Pn. Then,T is a multiplicative linear preserver if and only ifQT Qt
is a multiplicative linear preserver for every Q ∈ Pn. So, in the proof
of the following theorem without loss of generality, we can assume that P =I.
Theorem 2.3. Let T : Mn → Mn be a linear function. Then, T is
a multiplicative linear preserver of a multivariate or a directional ma-jorization if and only if T has one of the following forms:
(ii) There exists a permutation P ∈ Pn such that
T X =P XP−1,
for allX ∈Mn.
(iii) There exists a permutation P ∈ Pn and a scalar γ ̸=−n such
that
T X = (γP +J)X(γP+J)−1,
for allX ∈Mn.
Proof. It is easy to prove that each of the conditions (i), (ii) or (iii) are sufficient for T to be a multiplicative linear preserver of a multivariate majorization. So, we only prove the necessity of the conditions. It is clear that EijEkl = 0 if and only if j ̸=k. Let T be of the form (a) in
Theorem 2.1. Then,
T X =
n
∑
i=1
(tr(xi))Ai,
where A1, . . . , An∈Mn. Let 1≤k≤n, and put X=Y =Ekk. Then
AkAk=T(X)T(Y)
=T(X) =Ak.
So, A2k = Ak. On the other hand, set X = Ekk and Y = Ejk(j ̸= k).
Then,
AkAk=T(X)T(Y)
=T(XY) =T(0) = 0.
Therefore, Ak= 0 and hence T = 0.
Now, suppose that T has the form (b) in Theorem 2.1. Then,
T(Eij) =EijR+J EijS
=
0 · · · 0 ..
. · · · ... 0 · · · 0 rj1 · · · rjn
0 · · · 0 ..
. · · · ... 0 · · · 0
+
sj1 · · · sjn
..
. ... ... sj1 · · · sjn
where rj1, . . . , rjnoccur in the i’th row. Since k̸=j,
0 =T(0) =T(EijEkl)
=T(Eij)T(Ekl)
=rjk
0 · · · 0 ..
. · · · ... 0 · · · 0 rl1 · · · rln
0 · · · 0 ..
. · · · ... 0 · · · 0
+ ( n ∑ m=1 rjm )
0 · · · 0 ..
. · · · ... 0 · · · 0 sl1 · · · sln
0 · · · 0 ..
. · · · ... 0 · · · 0
+sjk
rl1 · · · rln
..
. · · · ... rl1 · · · rln
+ ( n ∑ m=1 sjm )
sl1 · · · sln
..
. · · · ... sl1 · · · sln
,
where rl1, . . . , rln and sl1, . . . , sln in the first and second matrices occur
in the i’th row. So,
(2.1) sjk
[
rl1 . . . rln
] + ( n ∑ m=1 sjm ) [
sl1 . . . sln
] = 0,
and
(2.2) rjk[
rl1 . . . rln
] + ( n ∑ m=1 rjm ) [
sl1 . . . sln
] = 0.
We consider two cases.
Case 1: Let sjk = 0 for all j, k(1≤j ̸=k≤n). Put l = j in (2.1).
Then sjj = 0 and henceS = 0. Again, put l=j in (2.2). Thenrjk = 0
for all j̸=kand hence
R =
r11 0 0
0 . .. 0 0 0 rnn
.
Since T is multiplicative, we haveR=T(I) =T(I.I) =R2, and hence
rii= 0 or rii= 1.
If R̸= 0 andR ̸=I, without loss of generality, we may assume that
R =
0 0 0 0 1 0 0 0 L
where L is a diagonal matrix whose diagonal entries are either 0 or 1. Set
X=
1 1 0 0 1 0 0 0 0
∈Mn.
An easy calculation shows that T( X2)
̸=T(X)2 which is a contradic-tion. So, R=I orR= 0, and hence T =I orT = 0.
Case 2: Let sjk ̸= 0 for some j, k (1≤j̸=k≤n). Now, if n
∑
m=1
sjm= 0,
then, R= 0. If
n
∑
m=1
sjm̸= 0,
then, R=γS and T has the form T(X) = (γI+J)XS. Since k̸=j in (2.1) is arbitrary, we conclude that allsjk (k̸=j) are the same for each j, sayβj. Also, we denotesjj byαj.
Now, let j = k. Then, EijEkl =Eil, and T(Eij)T(Ekl) = T(Eil). If
we write this in the matrix form, we have
γ2sjj
0 · · · 0 ..
. · · · ... 0 · · · 0 sl1 · · · sln
0 · · · 0 ..
. · · · ... 0 · · · 0
+γ ( n ∑ m=1 sjm )
0 · · · 0 ..
. · · · ... 0 · · · 0 sl1 · · · sln
0 · · · 0 ..
. · · · ... 0 · · · 0
+γsjj
sl1 · · · sln
..
. · · · ... sl1 · · · sln
+ ( n ∑ m=1 sjm )
sl1 · · · sln
..
. · · · ... sl1 · · · sln
=γ
0 · · · 0 ..
. · · · ... 0 · · · 0 sl1 · · · sln
0 · · · 0 ..
. · · · ... 0 · · · 0
+
sl1 · · · sln
..
. · · · ... sl1 · · · sln
where sl1, . . . , sln occurs in the i’th row. So,
{
(γαj+∑nm=1sjm−1)
[
sl1 · · · sln
] = 0, (γβj+∑nm=1sjm)
[
sl1 · · · sln
] = 0.
If S= 0, we are done. Assume S ̸= 0. Then,
{
γαj+αj+ (n−1)βj−1 = 0, γβj+αj+ (n−1)βj = 0.
Ifγ =−norγ = 0, there is no solution. Otherwise, we haveαj = nγ(+nγ−+γ1)
and βj = γ(−n+1γ), for all j(1≤j≤n). Now, we have
1
γ(n+γ)
γ+ 1 1 · · · 1
1 . .. 1 . . . .
.
. 1 . .. 1 1 · · · 1 γ+ 1
n+γ−1 −1 · · · −1 −1 . .. −1
. . . .
.
. −1 . .. −1 −1 · · · −1 n+γ−1
=I.
This shows that S−1 =γI+J and the proof is complete. □
In the following section, we consider multiplicative strong linear pre-servers of some kinds of majorizations.
3. Multiplicative strong linear preservers of majorizations
Similar to the linear preservers of multivariate and directional ma-jorizations, some results hold for the strong linear preservers of multi-variate and directional majorizations. Now, we state a theorem that is useful in sequel.
Theorem 3.1 ([8], Theorem 1.1). Let F be an arbitrary field and φ : Mn(F)→Mn(F) a bijective linear map satisfying
φ(AB) =φ(A)φ(B),
whereA, B∈Mn(F).Then, there exists an invertible matrixT ∈Mn(F)
such that
φ(A) =T AT−1,
for every A∈Mn(F).
A nonnegative matrix A∈Mn,m is row stochastic if the sum of each
row of it is 1. Denote the set of all row stochastic matrices by Rn. For A, B ∈Mn,m, it is said thatAis a matrix majorized from left(resp. from
right) by B ifA=RB(resp. A=BR) for some rowstochastic matrices R ∈ Mn(resp. R ∈ Mm), and denoted by ≺l(resp. ≺r). In [6], the
Theorem 3.2 ([6], Theorem 5.2). A linear operator T : Mn → Mn
strongly preserves the matrix majorization ≺l if and only if there exist
a permutation matrix P and an invertible matrix L in Mn such that T X =P XLfor all X∈Mn. Moreover, ifT I =I, then L=Pt.
The structure of multiplicative strong linear preservers of a matrix majorization from left is as follows. The case of a matrix majorization from right is similar.
Theorem 3.3. A linear operator T : Mn → Mn is a multiplicative
strong linear preserver of the matrix majorization ≺l if and only if there
exists a permutation matrix P such that T X =P XPt for all X∈M n.
Proof. Since I2 = I, T(I)2 = T(I). Hence T(I)(T(I)−I) = 0. But
T(I) = P L is invertible. So, T(I) = I, and hence by Theorem 3.2, T X =P XPtfor allX∈Mn, wich showsT is multiplicative as well. □
Now, we find the structure of multiplicative strong linear preservers of lw-column majorizations. In fact, we show that the structure of mul-tiplicative strong linear preservers of a majorization and mulmul-tiplicative strong linear preservers of rw-row majorization on Mn are the same.
For x, y ∈ Mn,1, it is said that x is lw-majorized by y, if x = Ry
for some R ∈ Rn. Let A, B ∈ Mn,m. It is said that that A is
lw-column majorized by B if every column of A is lw-majorizwd by the corresponding column ofB.
Theorem 3.4 ([2], Theorem 3.10). Let T :Mn,m → Mn,m be a linear
function. Then,T strongly preserves lw-column majorization if and only if there exist b1, . . . , bm ∈ ∪mi=1Span{ei}, P1, . . . Pm ∈ Pn such that B := [b1| · · · |bm] is invertible and T X = [P1Xb1|. . .|PmXbm].
The following theorem characterizes all multiplicative linear preservers of lw-column majorization on Mn.
Theorem 3.5. Let T : Mn → Mn be a linear operator. Then, T is a
multiplicative strong linear preserver of lw-column majorization if and only if T X =P XPt for some permutationsP.
Proof. By Theorem 3.4, there exist permutation matrices P1, . . . , Pn ∈
Pn and vectorsb1, . . . , bn∈∪ni=1span{ei} such that B := [b1| · · · |bn] is
invertible and
T X = [P1Xb1| · · · |PnXbn].
By Theorem 3.1, T has also the form T X =CXC−1. Now, suppose
[P1Xb1| · · · |PnXbn] = [CXD1| · · · |CXDn],
where Di is the ith column of C−1. So, PkXbk = CXDk. Since B is
invertible, it may be assumed that (bk)i
(Pk)11 · · · (Pk)1n
..
. . .. ... (Pk)n1 · · · (Pk)nn
0 · · · 0 0 0 · · · 0 ..
. ... ... ... ... ... ... 0 · · · 0 0 0 · · · 0 0 · · · 0 1 0 · · · 0 0 · · · 0 0 0 · · · 0
..
. ... ... ... ... ... ... 0 · · · 0 0 0 · · · 0
(bk)1
.. . (bk)n
=
C11 · · · C1n
..
. . .. ... Cn1 · · · Cnn
0 · · · 0 0 0 · · · 0 ..
. ... ... ... ... ... ... 0 · · · 0 0 0 · · · 0 0 · · · 0 1 0 · · · 0 0 · · · 0 0 0 · · · 0
..
. ... ... ... ... ... ... 0 · · · 0 0 0 · · · 0
(dk)1
.. . (dk)n
.
Since Pk andC are invertible, then (bk)i= 0 if and only if (dk)i= 0.
Now, we have
(bk)ik
(Pk)1j
.. . (Pk)nj
= (dk)ik
(C)1j .. . (C)nj
.
Since j and k variate independently, we have Pk = γC and Dk = γbk.
PutP :=γC. Then,
T X = [P1Xb1| · · · |PnXbn] =P XPt.
□
4. Multiplicative linear preservers of generalized
majorizations
For x, y ∈ Rn, it is said that x is gs-majorized by y (denoted by y≻gsx), if there exists a generalized doubly stochastic matrix D such
thatx=Dy, see [3]. ForA, B∈Mn,m, it is said thatB is gd-majorized
byA(written asA≻gdB), ifAx≻gsBx, for allx∈Rm. Actually,A≻gdB
if and only if, for everyx∈Rm, there exists a g-doubly stochastic matrix Dx such that Bx= Dx(Ax). The following theorem gives the possible
structures of all linear functions from Mn,m toMn,k that preserve≻gd.
Theorem 4.1 ([3], Theorem 1.3). Let T : Mn,m → Mn,k be a linear
(i) There existA1, . . . , Am ∈Mn,k such that
T(X) =
m
∑
j=1
tr(xj)Aj,
where xj is the jth column of X.
(ii) There exist R, S∈Mm,k and an invertible matrix D∈GDn such
that
T(X) =DXR+J XS.
(iii) There exist S ∈ Mm,k, a ∈ Rm, r1, . . . , rk ∈ R and invertible
matrices D1, . . . , Dk∈GDn such that
T(X) = [r1D1Xa| · · · |rkDkXa] +J XS.
Now, we characterize the multiplicative linear preservers of gd-majorization on Mn.
Lemma 4.2. Let S ∈ Mn. The linear function T :Mn → Mn defined
by T X =J XS is multiplicative if and only if T = 0.
Proof. It is obvious that for alli, j, k ∈ {1, . . . , n},
T(Eij) =
sj1 · · · sjn
..
. ...
sj1 · · · sjn
=T(Ekj).
Then, for k̸=j,
0 =T(0) =T(EijEkl)
=T(Eij)T(Ekl)
=T(Eij)T(Ejl)
=T(EijEjl)
=T(Eil).
So,T = 0. □
Theorem 4.3. Let T : Mn → Mn be a linear function. Then, T is
a multiplicative linear preserver of gd-majorization if and only if T has one of the following forms:
(i) T X = 0 for all X∈Mn.
(ii) There exists an invertible matrixD∈GDn such that
T X =DXD−1,
(iii) There exist an invertible matrixD∈GDn and a scalar γ ̸=−n
such that
T X = (γD+J)X(γD+J)−1,
for allX ∈Mn.
Proof. Let T :Mn → Mn be a linear function that preserves≻gd. If T
is in the form of (i) or (ii) in Theorem 4.1, the proof is similar to that of Theorem 2.3. If T is in the form of (iii), we show thatT = 0.
First, suppose n= 2. If r1 = 0, we can choose D1 =D2 and T is in
the form of (ii) in Theorem 4.1. So, without loss of generality, we can assume that
T(X) = [Xa|rDXa] +J XS.
Now, let
a= [
α β
]
, D= [
d 1−d 1−d d
] ,
and for convenience,t=r(2d−1).
SupposeX = [
1 0
−1 0 ]
and Y = [
0 1
0 −1 ]
. So,
T(X) =α [
1 t
−1 −t ]
,
and
T(Y) =β [
1 t
−1 −t ]
.
Since X2 = X, T(X)2 = T(X) and hence α2(1−t) = α. So, α = 0 or α = 1−t1 . Note that if t = 1 then α = 0. On the other hand, since Y2 = −Y, β = 0 or β = −1−t1 , and if t = 1, then β = 0 and T = 0, by Lemma 4.2. So, assume that t̸= 1. Since XY =Y, αβ(1−t) =β, and if α = 0, we conclude that β = 0. Similarly, since Y X = −X, we conclude that if β = 0, then α = 0. So, we have α = β = 0 or α=−β = 1−t1 .
An easy calculation shows that
T(E11) =
[
α+s11 rαd+s12
s11 rα(1−d) +s12
] ,
T(E12) = [
−α+s21 −rαd+s22 s21 −rα(1−d) +s22
] ,
T(E22) =
[
s21 −rα(1−d) +s22
−α+s21 −rαd+s22 ]
The solution of the equations
{
T(E12)T(E11) = 0, T(E11)T(E22) = 0,
is
s11=−1+αt,
s12=−α(tr+1+rd(1t −t)),
s21= 1+αtt,
s22= αr(11+−dt+dt).
Note that α=−s11(1 +t). Assume that t̸=−1. (If t=−1, α= 0,
and we are done.)
CalculatingT(E11)T(E21), shows that
(α+s11) (rαd+s12+s11) = 0.
Now, if α+s11 = 0, and α ̸= 0, we have t = 0. Since D is invertible,
d̸= 12. So,r= 0 and we are done. Ifrαd+s12+s11= 0, then t= 1 or t=−1, which is a contradiction.
Now, suppose that n≥3. Let
T(X) = [r1D1Xa|. . .|rnDnXa] +J XS,
for some S ∈ Mn, a ∈ Rn, r1, . . . , rn ∈ R and invertible matrices D1, . . . , Dn∈ GDn. Without loss of generality, assume that r1, a1 ̸= 0,
where a1 is the first coordinate ofa. Set
X=
1 0 · · · 0 −1 0 · · ·
0 0 · · · ... ..
. ... . .. 0 0 · · · 0
.
a21 r1 n ∑ m=1 rm (
(Dm)1,1−(Dm)1,2
) (
(D1)m,1−(D1)m,2
) · · · .. . · · · r1 n ∑ m=1 rm (
(Dm)n,1−(Dm)n,2
) (
(D1)m,1−(D1)m,2
) · · ·
· · · rn n
∑
m=1
rm
(
(Dm)1,1−(Dm)1,2
) (
(Dn)m,1−(Dn)m,2
)
· · · ...
· · · rn n
∑
m=1
rm
(
(Dm)n,1−(Dm)n,2
) (
(Dn)m,1−(Dn)m,2
)
=a1
r1 (
(D1)1,1−(D1)1,2
)
· · · rn
(
(Dn)1,1−(Dn)1,2
)
..
. ...
r1
(
(D1)n,1−(D1)n,2
)
· · · rn
(
(Dn)n,1−(Dn)n,2
) .
The first column of the matrix in the right hand side of the above equation cannot be equal to zero. Since otherwise, the first and the second columns of D1 are equal and the matrix cannot be invertible. Letβ = (D1)i,1−(D1)i,2, for someisuch that (D1)i,1−(D1)i,2 ̸= 0 and
α be the corresponding entry in the right hand side matrix in the above equation. It is clear that α̸= 0. So, a21r1α=a1r1β.
If we put
X=
0 1 0 · · · 0
0 −1 0
0 0 0 ...
..
. . ..
0 · · · 0
,
we get to the equation X2 = −X and hence T(X)2 = −T(X), which gives the equation a22r1α =−a2r1β. Again, without loss of generality,
we may assume that a2 ̸= 0. If we shift the column vector
1 −1 0 .. . 0
So, amust have the form
a1
−a1
0 .. . 0
.
Now, set
X=
0 0 0 · · · 0 0 1 0 · · ·
0 −1 0 · · · ... 0 0 0 · · ·
..
. ... ... . ..
0 · · · 0
.
Then, T(X)̸= 0, and we can obtain the correspondingα and β. Using these α and β, and setting
X=
0 0 · · · 0
1 0 ...
−1 0 · · · 0
0 0
..
. ... . .. ... 0 0 · · · 0
,
we get to X2 = 0 and T(X)2 = 0. This implies that a1 orr1 = 0 and we reach to a contradiction. So, T(X) has the formT(X) =J XS, and
hence by Lemma 4.2, we have T = 0. □
5. Multiplicative strong linear preservers of generalized
majorizations
The definitions, theorems and proofs of multiplicative strong linear preservers of gs and gd-majorization are similar to multivariate and di-rectional majorizations, see [3].
Now, we consider gut-majorizations.
An n×m matrixR is called g-row stochastic if the sum of all entries in each row ofR is 1. Denote the set of alln×nupper triangular g-row stochastic matrices by Rgutn . By E, we mean a matrix whose entries in
the last column are 1 and the other entries are 0. For A, B ∈ Mn,m,
it is said that A is gut-majorized by B, and written as A ≺gut B, if
there exists R ∈ Rgutn such that A =RB. In [4] the authors found the
Theorem 5.1 ([4], Theorem 1.3). Let T : Mn,m → Mn,m be a linear
function. Then, T strongly preserves ≺gut if and only if T X =AXR+ EXS for some R, S ∈ Mm and invertible matrix A ∈ Rgutn such that R(R+S) is invertible.
LetT :Mn→Mnbe a strong linear preserver of gut-majorization. It
is obvious thatT is multiplicative if and only ifQT Q−1 is multiplicative
for every Q∈ Rgutn .
Theorem 5.2. Let T : Mn → Mn be a multiplicative strong linear
preserver of gut-majorization. Then, there exists Q ∈ Rgutn such that T X =QXQ−1 for every X∈M
n.
Proof. Without loss of generality, we can suppose that T X = XR+ EXS. By Theorem 3.1, there exists an invertible matrixD∈Mn such
that
T X =XR+EXS
=DXD−1.
If X is an arbitrary n×n matrix withnth row equal to zero, we have XR=DXD−1. Suppose that
D−1=
a1n
A1 ...
an−1n an1 · · · ann−1 ann
,
RD=
b1n
B1 ...
bn−1n bn1 · · · bnn−1 bnn
,
and
X = [
X1 0
0 0
] .
We have A1X1B1 =X1. Since X1 ∈ Mn−1 is arbitrary, A1 =B1 =
In−1. If we take X1 =In−1, then,
b1n
In−1 ...
bn−1n an1 · · · ann−1 ∗
= [
In−1 0
0 0
and hence an1 =· · ·=ann−1=b1n=· · ·=bn−1n= 0. So,
D−1 =
a1n
In−1 ...
an−1n
0 ann
,
RD= [
In−1 0
bn1 · · · bnn−1 bnn
] ,
and
D=
−a1n
ann
In−1 ...
−an−1n
ann
0 a1
nn
.
PutX =E1nin the equationXRD=DX. Sincebn1 =· · ·=bnn−1 = 0
and bnn = 1, then R=D−1.
Now,
XR+EXS=DXD−1,
and
XRD+EXSD =DX.
PuttingX =En1, . . . , Enn in this equation, we get
−a1n ann
=· · ·=−an−1n ann
= 1−ann ann
,
(SD)11=· · ·= (SD)nn, and
(SD)ij = 0, i̸=j. So,
SD =−a1n ann
I ⇒ S=−a1n ann
D−1,
D−1 =
α
In−1 ...
α 0 1 +α
,
and
D=
− α 1+α
In−1 ...
− α 1+α
0 1+1α
where α=a1n=· · ·=an−1n. Therefore,
T(X) =XD−1− α
1 +αEXD −1
= (
I− α 1 +αE
)
XD−1.
Hence T X =DXD−1, and the proof is complete. □
Acknowledgments. The authors would like to thank the referee(s) for useful comments and remarks.
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1 Department of Mathematics, Vali-e-Asr University of Rafsanjan, Zip
Code: 7718897111, Rafsanjan, Iran. E-mail address: ma.hadiann@gmail.com
2 Department of Mathematics, Vali-e-Asr University of Rafsanjan, Zip
