# Linear Maps on Upper Triangular Matrices Spaces Preserving Idempotent Tensor Products

Authors: Li Yang; Wei Zhang; Jinli Xu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Supposem
,
n
≥
2 are positive integers. Let 𝒯
n be the space of all n
×
n complex upper triangular matrices, and let ϕ be an injective linear map on 𝒯
m
⊗
𝒯
n. Then ϕ
(
A
⊗
B
) is an idempotent matrix in 𝒯
m
⊗
𝒯
n whenever A
⊗
B is an idempotent matrix in 𝒯
m
⊗
𝒯
n if and only if there exists an invertible matrix P
∈
𝒯
m
⊗
𝒯
n such that ϕ
(
A
⊗
B
)
=
P
(
ξ
1
(
A
)
⊗
ξ
2
(
B
)
)
P
-
1
,
∀
A
∈
𝒯
m
,
B
∈
𝒯
n, or when m
=
n, ϕ
(
A
⊗
B
)
=
P
(
ξ
1
(
B
)
⊗
ξ
2
(
A
)
)
P
-
1
,
∀
A
∈
𝒯
m
,
B
∈
𝒯
m, where ξ
1
(
[
a
i
j
]
)
=
[
a
i
j
] or ξ
1
(
[
a
i
j
]
)
=
[
a
m
-
i
+
1
,
m
-
j
+
1
] and ξ
2
(
[
b
i
j
]
)
=
[
b
i
j
] or ξ
2
(
[
b
i
j
]
)
=
[
b
n
-
i
+
1
,
n
-
j
+
1
]
.


---

## Body

## 1. Introduction

Supposem
,
n
≥
2 are positive integers. Let ℳ
n be the space of all n
×
n complex matrices, and let 𝒯
n be all upper triangular in ℳ
n. For A
∈
ℳ
m, B
∈
ℳ
n, we denote by A
⊗
B their tensor product (a.k.a. Kronecker product).Linear preserver problem is a hot area in matrix and operator theory; there are many results about this area (see [1–14]). Specially, the idempotence preservers and the rank one preservers play an important role (see [1, 2]); therefore, it is meaningful to study the idempotence preservers. Chan et al. [3] first characterize linear transformations on ℳ
n preserving idempotent matrices. Šemrl [4] applying projective geometry gives the form of transformations on rank-1 idempotents. Tang et al.  [5] investigate injective linear idempotence preservers on 𝒯
n.In quantum information science, quantum states of a system withn physical states are represented as density matrices, that is, positive semidefinite matrices with trace one. If A
∈
ℳ
m and B
∈
ℳ
n are two quantum states in two quantum systems, then A
⊗
B describes a joint state in bipartite system ℳ
m
⊗
ℳ
n. Recently, many researchers consider the problem combining linear preserver problem with quantum information science. They determine the structure of linear maps on ℳ
m
⊗
ℳ
n by using information only about the images of matrices possessing tensor product form. One can see [15–18] and their references for some background on linear preserver problems on tensor spaces arising in quantum information science.Inspired by the above, the purpose of this paper is to study injective linear mapsϕ on 𝒯
m
⊗
𝒯
n satisfying ϕ
(
A
⊗
B
)  is an idempotent matrix whenever A
⊗
B is an idempotent matrix in 𝒯
m
⊗
𝒯
n. If we remove the assumption that map is injective, then ϕ may have various forms as follows.Example 1.
A
⊗
B
↦
a
11
I
m
⊗
B is a linear idempotent preserver on 𝒯
m
⊗
𝒯
n.Example 2.
A
⊗
B
↦
a
11
(
E
11
(
m
)
+
E
22
(
m
)
)
⊗
(
b
11
E
11
(
n
)
+
b
12
E
12
(
n
)
+
b
22
E
22
(
n
)
) is a linear idempotent preserver on 𝒯
m
⊗
𝒯
n.We end this section by introducing some notations which will be used in the following sections. LetC be the complex field, I
k the k
×
k identity matrix, 0 the zero matrix whose order is omitted in different matrices just for simplicity, and X
t (resp., rank
⁡
X) the transpose (resp., rank) of X. E
i
j
(
n
), ∀
i
,
j
∈
[
1
,
n
] stands for the n
×
n matrix with 1 at the (
i
,
j
)th entry and 0 otherwise. Denote by J
k (also J) the matrix E
1
,
k
(
k
)
+
E
2
,
k
-
1
(
k
)
+
⋯
+
E
k
,
1
(
k
). Clearly, if A
=
[
a
i
j
]
∈
𝒯
n, then J
A
t
J
=
[
a
n
-
j
+
1
,
n
-
i
+
1
]
∈
𝒯
n. For positive integers n
1 and n
2 with n
1
<
n
2, let [
n
1
,
n
2
] be the set of all integers between n
1 and n
2. For any (
i
,
j
)
∈
[
1
,
m
]
×
[
1
,
n
], we define ρ by
(1)
ρ
(
i
,
j
)
=
(
i
-
1
)
n
+
j
.
For any k
∈
[
1
,
m
n
], we define σ
(
k
)
∈
[
1
,
m
] and τ
(
k
)
∈
[
1
,
n
] such that k
=
(
σ
(
k
)
-
1
)
n
+
τ
(
k
) (it is easy to see that σ and τ are well defined). It is easy to see that
(2)
E
r
s
(
m
)
⊗
E
u
v
(
n
)
=
E
(
r
-
1
)
n
+
u
,
(
s
-
1
)
n
+
v
(
m
n
)
=
E
ρ
(
r
,
u
)
,
ρ
(
s
,
v
)
(
m
n
)
,
(3)
E
i
j
(
m
n
)
=
E
σ
(
i
)
σ
(
j
)
(
m
)
⊗
E
τ
(
i
)
τ
(
j
)
(
n
)
.
We define a partial ordering of [
1
,
m
]
×
[
1
,
n
] by (
a
,
b
)
≤
(
c
,
d
) if and only if a
≤
c and b
≤
d. We say that (
a
,
b
) and (
c
,
d
) are comparable, if (
a
,
b
)
≤
(
c
,
d
) or (
c
,
d
)
≤
(
a
,
b
).


## 2. Preliminary Results

We need the form of injective linear idempotence preserver on𝒯
n, which was obtained in [5].Lemma 3 (see [5, Theorem 1]).
Letψ be an injective linear map on 𝒯
n. Then ψ
(
X
) is an idempotent matrix in 𝒯
n whenever X is an idempotent matrix in 𝒯
n if and only if there exists an invertible matrix P
∈
𝒯
n such that
(4)
ψ
(
X
)
=
P
ξ
(
X
)
P
-
1
,
∀
X
∈
𝒯
n
,
where ξ
(
X
)
=
X or ξ
(
X
)
=
J
X
t
J.It is clear that𝒯
m
⊗
𝒯
n
⊊
𝒯
m
n. For example, E
23
(
4
)
∈
𝒯
4 and
(5)
E
23
(
4
)
∉
𝒯
2
⊗
𝒯
2
=
{
[
a
11
a
12
a
13
a
14
0
a
22
0
a
24
0
0
a
33
a
34
0
0
0
a
44
]
:
a
i
j
∈
C
}
.
It is easy to see that (
σ
(
2
)
,
τ
(
2
)
)
=
(
1,2
) and (
σ
(
3
)
,
τ
(
3
)
)
=
(
2,1
). In fact, we can point out the positions of elements which are in 𝒯
m
n
∖
𝒯
m
⊗
𝒯
n.Lemma 4.
λ
E
i
j
(
m
n
)
∈
𝒯
m
⊗
𝒯
n if and only if (
σ
(
i
)
,
τ
(
i
)
)
≤
(
σ
(
j
)
,
τ
(
j
)
) or λ
=
0.Proof.
It is a direct corollary of (3).The next Lemma describes the partial ordering we defined in Section1, which is useful to prove our main Theorem.Lemma 5 (see [19, Theorem 1]).
Letm
,
n
≥
2, and let A be a matrix with m rows and n columns containing all elements of [
1
,
m
]
×
[
1
,
n
]. If every two elements of A in the same row and column are comparable, respectively, then there exist permutation matrices U
∈
ℳ
m and V
∈
ℳ
n such that
(
I
)
U
A
V
=
[
(
1,1
)
(
1,2
)
⋯
(
1
,
n
-
1
)
(
1
,
n
)
(
2,1
)
(
2,2
)
⋯
(
2
,
n
-
1
)
(
2
,
n
)
⋯
⋯
⋯
⋯
⋯
(
m
-
1,1
)
(
m
-
1,2
)
⋯
(
m
-
1
,
n
-
1
)
(
m
-
1
,
n
)
(
m
,
1
)
(
m
,
2
)
⋯
(
m
,
n
-
1
)
(
m
,
n
)
]
or when m
≥
3 or n
≥
3
(
II
)
U
A
V
=
[
(
m
,
n
)
(
1,2
)
⋯
(
1
,
n
-
1
)
(
1
,
n
)
(
2,1
)
(
2,2
)
⋯
(
2
,
n
-
1
)
(
2
,
n
)
⋯
⋯
⋯
⋯
⋯
(
m
-
1,1
)
(
m
-
1,2
)
⋯
(
m
-
1
,
n
-
1
)
(
m
-
1
,
n
)
(
m
,
1
)
(
m
,
2
)
⋯
(
m
,
n
-
1
)
(
1,1
)
]
or when m
=
n
(
III
)
U
A
V
=
[
(
1,1
)
(
2,1
)
⋯
(
m
-
1,1
)
(
m
,
1
)
(
1,2
)
(
2,2
)
⋯
(
m
-
1,2
)
(
m
,
2
)
⋯
⋯
⋯
⋯
⋯
(
1
,
m
-
1
)
(
2
,
m
-
1
)
⋯
(
m
-
1
,
m
-
1
)
(
m
,
m
-
1
)
(
1
,
m
)
(
2
,
m
)
⋯
(
m
-
1
,
m
)
(
m
,
m
)
]
or when m
=
n
≥
3
(
IV
)
U
A
V
=
[
(
m
,
m
)
(
2,1
)
⋯
(
m
-
1,1
)
(
m
,
1
)
(
1,2
)
(
2,2
)
⋯
(
m
-
1,2
)
(
m
,
2
)
⋯
⋯
⋯
⋯
⋯
(
1
,
m
-
1
)
(
2
,
m
-
1
)
⋯
(
m
-
1
,
m
-
1
)
(
m
,
m
-
1
)
(
1
,
m
)
(
2
,
m
)
⋯
(
m
-
1
,
m
)
(
1,1
)
]
.The following lemmas would make the proof of the main theorem more concise.Lemma 6.
SupposeX
∈
ℳ
n is an idempotent matrix such that 0
s
⊕
I
r
⊕
0
n
-
r
-
s
-
X also is an idempotent matrix. Then there exists an idempotent matrix X
r
∈
ℳ
r such that X
=
0
s
⊕
X
r
⊕
0
n
-
r
-
s.Proof.
ByX
2
=
X and (
0
s
⊕
I
r
⊕
0
n
-
r
-
s
-
X
)
2
=
0
s
⊕
I
r
⊕
0
n
-
r
-
s
-
X, we have
(6)
2
X
=
(
0
s
⊕
I
r
⊕
0
n
-
r
-
s
)
X
+
X
(
0
s
⊕
I
r
⊕
0
n
-
r
-
s
)
.
Set
(7)
X
=
[
X
11
X
12
X
13
X
21
X
22
X
23
X
31
X
32
X
33
]
,
where X
11
∈
ℳ
s
,
X
22
∈
ℳ
r. Then (6) implies
(8)
2
[
X
11
X
12
X
13
X
21
X
22
X
23
X
31
X
32
X
33
]
=
[
0
X
12
0
X
21
2
X
22
X
23
0
X
32
0
]
.
Hence, X
=
0
s
⊕
X
22
⊕
0
n
-
r
-
s; therefore, the lemma holds.Lemma 7.
LetX
∈
𝒯
m
⊗
𝒯
n be an idempotent matrix such that E
i
i
(
m
)
⊗
I
n
-
X also is an idempotent matrix. Then there exists an idempotent Y
∈
𝒯
n such that X
=
E
i
i
(
m
)
⊗
Y.Proof.
The proof is similar to that of Lemma6.Lemma 8.
Supposer
,
s
∈
[
1
,
n
-
1
]. If for any λ
∈
C, I
r
⊕
0
n
-
r
+
λ
X and 0
r
⊕
I
s
⊕
0
n
-
r
-
s
+
λ
X are idempotent in 𝒯
n, then
(9)
X
=
[
0
r
X
1
0
0
s
]
⊕
0
n
-
r
-
s
.Proof.
It follows fromI
r
⊕
0
n
-
r
+
λ
X, ∀
λ
∈
C is idempotent that
(10)
(
I
r
⊕
0
n
-
r
)
X
+
X
(
I
r
⊕
0
n
-
r
)
=
X
.
Let
(11)
X
=
[
X
r
X
1
X
2
0
X
s
X
3
0
0
X
n
-
r
-
s
]
,
where X
r
∈
𝒯
r, X
s
∈
𝒯
s, then (10) implies
(12)
[
2
X
r
X
1
X
2
0
0
0
0
0
0
]
=
[
X
r
X
1
X
2
0
X
s
X
3
0
0
X
n
-
r
-
s
]
.
Hence, X
r
=
0, X
s
=
0, X
n
-
r
-
s
=
0, X
3
=
0. Similarly, from 0
r
⊕
I
s
⊕
0
n
-
r
-
s
+
λ
X being idempotent, we have X
2
=
0.Lemma 9 (see [6, Page 62, Exercise 1]).
SupposeA
1
,
…
,
A
k
∈
ℳ
n are idempotent matrices such that, for any i
≠
j
∈
[
1
,
k
], A
i
+
A
j is idempotent. Let r
i
=
rank
⁡
A
i. Then there exists an invertible matrix P
∈
ℳ
n such that
(13)
A
i
=
P
diag
⁡
(
0
,
…
,
0,1
,
…
,
1,0
,
…
,
0
)
P
-
1
,
where diag
⁡
(
0
,
…
,
0,1
,
…
,
1,0
,
…
,
0
) is the diagonal matrix in which all diagonal entries are zero except those in the (
r
1
+
⋯
+
r
i
-
1
+
1
)st to the (
r
1
+
⋯
+
r
i
)th rows.Similar to Lemma9, we have the following.Lemma 10.
LetA
1
,
…
,
A
m
n
∈
𝒯
m
⊗
𝒯
n be idempotent matrices of rank-1 such that for any i
≠
j
∈
[
1
,
m
n
], A
i
+
A
j is idempotent. Then there exist a permutation π on [
1
,
m
n
] and an invertible matrix P
∈
𝒯
m
⊗
𝒯
n such that
(14)
A
i
=
P
E
π
(
i
)
π
(
i
)
(
m
n
)
P
-
1
,
i
=
1
,
…
,
m
n
.Proof.
ByA
i
∈
𝒯
m
⊗
𝒯
n being an idempotent matrix of rank-1, we can assume
(15)
A
i
=
E
π
(
i
)
π
(
i
)
(
m
n
)
+
B
i
,
i
=
1
,
…
,
m
n
,
where B
i
∈
𝒯
m
⊗
𝒯
n with zero diagonal entries. It follows from A
i
+
A
j, ∀
i
≠
j being is idempotent that π
(
i
)
≠
π
(
j
), ∀
i
≠
j. Hence, π is a permutation on [
1
,
m
n
]. By A
π
-
1
(
1
)
2
=
A
π
-
1
(
1
), we can see A
π
-
1
(
1
)
=
E
11
(
m
n
)
+
Σ
k
=
2
(
m
n
)
λ
1
k
E
1
k
(
m
n
). Let P
1
=
I
m
n
-
Σ
k
=
2
(
m
n
)
λ
1
k
E
1
k
(
m
n
)
∈
𝒯
m
⊗
𝒯
n; then P
1
-
1
=
I
m
n
+
Σ
k
=
2
(
m
n
)
λ
1
k
E
1
k
(
m
n
) and
(16)
A
π
-
1
(
1
)
=
P
1
E
11
(
m
n
)
P
1
-
1
.
By A
π
-
1
(
1
)
+
A
π
-
1
(
2
) being idempotent, we obtain P
1
-
1
A
π
-
1
(
2
)
P
1
=
E
22
(
m
n
)
+
Σ
k
=
3
(
m
n
)
λ
2
k
E
2
k
(
m
n
). Let P
2
=
I
m
n
-
Σ
k
=
3
(
m
n
)
λ
2
k
E
2
k
(
m
n
); then
(17)
A
π
-
1
(
1
)
=
P
2
P
1
E
11
(
m
n
)
P
1
-
1
P
2
-
1
,
A
π
-
1
(
2
)
=
P
2
P
1
E
22
(
m
n
)
P
1
-
1
P
2
-
1
.
Continuing to do this, we can find  P
3
,
…
,
P
m
n. Let  P
=
P
m
n
P
m
n
-
1
⋯
P
2
P
1
∈
𝒯
m
⊗
𝒯
n; then
(18)
A
π
-
1
(
i
)
=
P
E
i
i
(
m
n
)
P
-
1
,
i
=
1
,
…
,
m
n
.
This completes the proof.


## 3. The Main Result

The main result of this paper is as follows.Theorem 11.
Supposem
,
n
≥
2 are positive integers and ϕ is an injective linear map on 𝒯
m
⊗
𝒯
n. Then ϕ
(
A
⊗
B
) is an idempotent matrix in 𝒯
m
⊗
𝒯
n whenever A
⊗
B is an idempotent matrix in 𝒯
m
⊗
𝒯
n if and only if there exists an invertible matrix P
∈
𝒯
m
⊗
𝒯
n such that
(
i
)
ϕ
(
A
⊗
B
)
=
P
(
ξ
1
(
A
)
⊗
ξ
2
(
B
)
)
P
-
1
,
∀
A
∈
𝒯
m
,
B
∈
𝒯
n
,
or when m
=
n
(
ii
)
ϕ
(
A
⊗
B
)
=
P
(
ξ
1
(
B
)
⊗
ξ
2
(
A
)
)
P
-
1
,
∀
A
∈
𝒯
m
,
B
∈
𝒯
m
,
where, for i
=
1,2, ξ
i
(
X
)
=
X
or
ξ
i
(
X
)
=
J
X
t
J.Proof.
The sufficiency is obvious. We will prove the necessity by the following six steps.
Step  1. There exist a permutation π on [
1
,
m
n
] and an invertible matrix P
∈
𝒯
m
⊗
𝒯
n such that
(19)
ϕ
(
E
k
k
(
m
n
)
)
=
P
E
π
(
k
)
π
(
k
)
(
m
n
)
P
-
1
,
∀
k
∈
[
1
,
m
n
]
.
Proof of Step  1. By Lemma 10, we only need to prove that
(20)
rank
⁡
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
=
1
,
∀
i
∈
[
1
,
m
]
,
j
∈
[
1
,
n
]
.
And for any (
i
,
j
)
≠
(
u
,
v
)
∈
[
1
,
m
]
×
[
1
,
n
],
(21)
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
+
ϕ
(
E
u
u
(
m
)
⊗
E
v
v
(
n
)
)
is
an
idempotent
matrix
.
It follows fromE
11
(
m
)
⊗
I
n
,
…
,
E
m
m
(
m
)
⊗
I
n and (
E
i
i
(
m
)
+
E
j
j
(
m
)
)
⊗
I
n
,
∀
i
≠
j
∈
[
1
,
m
] are idempotent matrices in 𝒯
m
⊗
𝒯
n that ϕ
(
E
11
(
m
)
⊗
I
n
)
,
…
,
ϕ
(
E
m
m
(
m
)
⊗
I
n
) and ϕ
(
E
i
i
(
m
)
⊗
I
n
)
+
ϕ
(
E
j
j
(
m
)
⊗
I
n
), ∀
i
≠
j
∈
[
1
,
m
] are idempotent matrices. We obtain by using Lemma 9 that there exists an invertible matrix P
1
∈
ℳ
m
n such that
(22)
ϕ
(
E
i
i
(
m
)
⊗
I
n
)
=
P
1
[
0
s
i
0
0
0
I
r
i
0
0
0
0
m
n
-
s
-
r
i
]
P
1
-
1
,
where r
i
=
rank
⁡
ϕ
(
E
i
i
(
m
)
⊗
I
n
) and s
i
=
r
1
+
⋯
+
r
i
-
1. For any j
∈
[
1
,
n
], it follows from E
i
i
(
m
)
⊗
E
j
j
(
n
) and E
i
i
(
m
)
⊗
(
I
n
-
E
j
j
(
n
)
) being idempotent matrices in 𝒯
m
⊗
𝒯
n that ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
) and ϕ
(
E
i
i
(
m
)
⊗
I
n
)
-
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
) are idempotent matrices; we obtain by (22) and Lemma 6 that
(23)
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
=
P
1
[
0
s
i
0
0
0
X
j
0
0
0
0
m
n
-
s
i
-
r
i
]
P
1
-
1
,
ffffffffffffffffffffffffff
∀
j
∈
[
1
,
n
]
,
where X
j
∈
ℳ
r
i is an idempotent matrix. For any j
≠
l
∈
[
1
,
n
], E
i
i
(
m
)
⊗
(
E
j
j
(
n
)
+
E
l
l
(
n
)
) is an idempotent matrix in 𝒯
m
⊗
𝒯
n; we have ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
+
ϕ
(
E
i
i
(
m
)
⊗
E
l
l
(
n
)
) is an idempotent matrix. It follows from (23) that X
j
+
X
l is an idempotent matrix. If r
i
<
n, by Lemma 9, we can obtain that there exists some j
0
∈
[
1
,
n
] such that X
j
0
=
0. This, together with (23), implies ϕ
(
E
i
i
(
m
)
⊗
E
j
0
j
0
(
n
)
)
=
0, which is a contradiction to the fact that ϕ is injective. Hence, r
i
=
n, ∀
i
∈
[
1
,
n
]. By Lemma 9, there exists an invertible matrix Q
i
∈
ℳ
n such that X
j
=
Q
i
E
j
j
(
n
)
Q
i
-
1
,
∀
j
∈
[
1
,
n
]. Let Q
=
diag
⁡
(
Q
1
,
…
,
Q
m
); it follows from (23) that
(24)
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
=
P
1
Q
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
Q
-
1
P
1
-
1
,
ffffffff
∀
i
∈
[
1
,
m
]
,
k
∈
[
1
,
n
]
.
Hence, (20) and (21) hold. This completes the proof of Step 1.
By Step  1, we may assume that for anyi
∈
[
1
,
m
], j
∈
[
1
,
n
],
(25)
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
=
E
π
(
ρ
(
i
,
j
)
)
π
(
ρ
(
i
,
j
)
)
(
m
n
)
.
From this, together with (3), we can also write
(26)
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
=
E
σ
(
π
(
ρ
(
i
,
j
)
)
)
σ
(
π
(
ρ
(
i
,
j
)
)
)
(
m
)
⊗
E
τ
(
π
(
ρ
(
i
,
j
)
)
)
τ
(
π
(
ρ
(
i
,
j
)
)
)
(
n
)
.
Step   2. (i) For any i
∈
[
1
,
m
], j
1
,
j
2
∈
[
1
,
n
], (
σ
(
π
(
ρ
(
i
,
j
1
)
)
)
,
τ
(
π
(
ρ
(
i
,
j
1
)
)
)
) and (
σ
(
π
(
ρ
(
i
,
j
2
)
)
)
,
τ
(
π
(
ρ
(
i
,
j
2
)
)
)
) are comparable.
(ii) For anyi
1
,
i
2
∈
[
1
,
m
], j
∈
[
1
,
n
], (
σ
(
π
(
ρ
(
i
1
,
j
)
)
)
,
τ
(
π
(
ρ
(
i
1
,
j
)
)
)
) and (
σ
(
π
(
ρ
(
i
2
,
j
)
)
)
,
τ
(
π
(
ρ
(
i
2
,
j
)
)
)
) are comparable.
Proof of Step  2. (i) Suppose there exist some i
0
∈
[
1
,
m
] and j
1
<
j
2
∈
[
1
,
n
] such that (
σ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
,
τ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
) and (
σ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
,
τ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
) are not comparable. Without loss of generality, we may assume that
(27)
σ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
<
σ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
,
τ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
>
τ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
.
It follows that
(28)
π
(
ρ
(
i
0
,
j
1
)
)
=
(
σ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
-
1
)
n
+
τ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
≤
(
σ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
-
1
)
n
+
n
≤
(
σ
(
π
(
ρ
(
i
0
,
j
1
)
)
)
)
n
≤
(
σ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
-
1
)
n
<
(
σ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
-
1
)
n
+
τ
(
π
(
ρ
(
i
0
,
j
2
)
)
)
=
π
(
ρ
(
i
0
,
j
2
)
)
.
For anyx
∈
C, by E
i
0
i
0
(
m
)
⊗
(
E
j
1
j
1
(
n
)
+
x
E
j
1
j
2
(
n
)
) and E
i
0
i
0
(
m
)
⊗
(
E
j
2
j
2
(
n
)
+
x
E
j
1
j
2
(
n
)
) being idempotent matrices in 𝒯
m
⊗
𝒯
n, we obtain by (25) that
(29)
E
π
(
ρ
(
i
0
,
j
1
)
)
π
(
ρ
(
i
0
,
j
1
)
)
(
m
n
)
+
x
ϕ
(
E
i
0
i
0
(
m
)
⊗
E
j
1
j
2
(
n
)
)
,
E
π
(
ρ
(
i
0
,
j
2
)
)
π
(
ρ
(
i
0
,
j
2
)
)
(
m
n
)
+
x
ϕ
(
E
i
0
i
0
(
m
)
⊗
E
j
1
j
2
(
n
)
)
are idempotent matrices in 𝒯
m
⊗
𝒯
n; hence, by Lemma 8
(30)
ϕ
(
E
i
0
i
0
(
m
)
⊗
E
j
1
j
2
(
n
)
)
=
λ
E
π
(
ρ
(
i
0
,
j
1
)
)
π
(
ρ
(
i
0
,
j
2
)
)
(
m
n
)
.
From (27), by Lemma 4, we obtain that λ
=
0, which is a contradiction to the fact that ϕ is injective. Using a similar method, we may prove (ii) holds. This completes the proof of Step 2.
Note. It is easy to see that ρ is a bijective map from [
1
,
m
]
×
[
1
,
n
] to [
1
,
m
n
] with ρ
-
1
:
k
↦
(
σ
(
k
)
,
τ
(
k
)
) from [
1
,
m
n
] to [
1
,
m
]
×
[
1
,
n
]. This, together with π, is a permutation on [
1
,
m
n
]; we obtain that
(31)
{
(
σ
(
π
ρ
(
i
,
j
)
)
,
τ
(
π
ρ
(
i
,
j
)
)
)
:
i
∈
[
1
,
m
]
,
j
∈
[
1
,
n
]
}
=
[
1
,
m
]
×
[
1
,
n
]
.
Let a
i
j
=
(
σ
(
π
ρ
(
i
,
j
)
)
,
τ
(
π
ρ
(
i
,
j
)
)
); then [
a
i
j
] forms an m
×
n matrix containing all elements of [
1
,
m
]
×
[
1
,
n
]. Step 2 implies that every two elements of [
a
i
j
] in the same row and column are comparable, respectively. Thus, applying Lemma 5 to [
a
i
j
], we conclude that one of (
I
)–(
IV
) holds. If (
I
) holds, then, for any but fixed i
∈
[
1
,
m
], all (
σ
(
π
ρ
(
i
,
j
)
)
,
τ
(
π
ρ
(
i
,
j
)
)
), j
=
1
,
…
,
n, are in the same row; that is, σ
(
π
(
ρ
(
i
,
j
)
)
)
=
σ
(
π
(
ρ
(
i
,
1
)
)
), ∀
j
∈
[
1
,
n
] and {
τ
(
π
ρ
(
i
,
j
)
)
:
j
=
1
,
…
,
n
}
=
[
1
,
n
]; hence, it follows from (26) that
(32)
ϕ
(
E
i
i
(
m
)
⊗
I
n
)
=
E
σ
(
π
(
ρ
(
i
,
1
)
)
)
σ
(
π
(
ρ
(
i
,
1
)
)
)
(
m
)
⊗
I
n
.
Similarly, if (
III
) holds, then m
=
n and, for any but fixed j
∈
[
1
,
m
], all (
σ
(
π
ρ
(
i
,
j
)
)
,
τ
(
π
ρ
(
i
,
j
)
)
), i
=
1
,
…
,
m, are in the same row; that is, σ
(
π
ρ
(
i
,
j
)
)
=
σ
(
π
ρ
(
1
,
j
)
), ∀
i
∈
[
1
,
m
] and {
τ
(
π
ρ
(
i
,
j
)
)
:
i
=
1
,
…
,
m
}
=
[
1
,
m
]; hence, it follows from (26) that
(33)
ϕ
(
I
m
⊗
E
j
j
(
m
)
)
=
E
σ
(
π
(
ρ
(
1
,
j
)
)
)
σ
(
π
(
ρ
(
1
,
j
)
)
)
(
m
)
⊗
I
m
.
We claim that(
II
) and (
IV
) do not hold. Indeed, if (
II
) holds, for convenience, we assume m
≥
3 and we first consider the special case U
=
I
m, V
=
I
n in (
II
) (one can use a similar method to prove the case of n
≥
3). Thus, by (26), we have
(34)
ϕ
(
E
11
(
m
)
⊗
E
11
(
m
)
)
=
E
m
m
(
m
)
⊗
E
n
n
(
n
)
,
ϕ
(
E
m
m
(
m
)
⊗
E
n
n
(
n
)
)
=
E
11
(
m
)
⊗
E
11
(
m
)
,
ϕ
(
E
i
i
(
m
)
⊗
E
j
j
(
n
)
)
=
E
i
i
(
m
)
⊗
E
j
j
(
n
)
,
∀
(
i
,
j
)
≠
(
1,1
)
,
(
m
,
n
)
.
Since, for anyx
∈
C, (
E
11
(
m
)
+
x
E
12
(
m
)
)
⊗
E
11
(
n
), E
11
(
m
)
⊗
(
E
11
(
n
)
+
x
E
12
(
n
)
) and (
E
22
(
m
)
+
x
E
12
(
m
)
)
⊗
E
11
(
n
), E
11
(
m
)
⊗
(
E
22
(
n
)
+
x
E
12
(
n
)
) are idempotent matrices in 𝒯
m
⊗
𝒯
n, we obtain by (34) that
(35)
E
m
m
(
m
)
⊗
E
n
n
(
n
)
+
x
ϕ
(
E
12
(
m
)
⊗
E
11
(
n
)
)
,
E
m
m
(
m
)
⊗
E
n
n
(
n
)
+
x
ϕ
(
E
11
(
m
)
⊗
E
12
(
n
)
)
,
E
22
(
m
)
⊗
E
11
(
n
)
+
x
ϕ
(
E
12
(
m
)
⊗
E
11
(
n
)
)
,
E
11
(
m
)
⊗
E
22
(
n
)
+
x
ϕ
(
E
11
(
m
)
⊗
E
12
(
n
)
)
are idempotent matrices in 𝒯
m
⊗
𝒯
n. This, together with Lemma 8, implies
(36)
ϕ
(
E
12
(
m
)
⊗
E
11
(
n
)
)
=
λ
E
(
n
+
1
)
,
m
n
(
m
n
)
≠
0
,
ϕ
(
E
11
(
m
)
⊗
E
12
(
n
)
)
=
μ
E
2
,
m
n
(
m
n
)
≠
0
.
For any x
∈
C, (
E
11
(
m
)
+
x
E
12
(
m
)
)
⊗
(
E
11
(
n
)
+
E
12
(
n
)
) is an idempotent matrix in 𝒯
m
⊗
𝒯
n. Thus, by (34) and (36)
(37)
ϕ
(
E
12
(
m
)
⊗
E
12
(
n
)
)
=
Σ
k
β
k
E
k
,
m
n
(
m
n
)
.
Similarly, since for any x
∈
C, (
E
22
(
m
)
+
x
E
12
(
m
)
)
⊗
E
22
(
n
), (
E
11
(
m
)
+
x
E
12
(
m
)
)
⊗
E
22
(
n
) and E
22
(
m
)
⊗
(
E
22
(
n
)
+
x
E
12
(
n
)
), E
22
(
m
)
⊗
(
E
11
(
n
)
+
x
E
12
(
n
)
) are idempotent matrices in 𝒯
m
⊗
𝒯
n, we have
(38)
ϕ
(
E
12
(
m
)
⊗
E
22
(
n
)
)
=
λ
′
E
2
,
(
n
+
2
)
(
m
n
)
≠
0
,
ϕ
(
E
22
(
m
)
⊗
E
12
(
n
)
)
=
μ
′
E
(
n
+
1
)
,
(
n
+
2
)
(
m
n
)
≠
0
.
For any x
∈
C, (
E
22
(
m
)
+
E
12
(
m
)
)
⊗
(
E
22
(
n
)
+
E
12
(
n
)
) is an idempotent matrix in 𝒯
m
⊗
𝒯
n; we obtain
(39)
ϕ
(
E
12
(
m
)
⊗
E
12
(
n
)
)
=
Σ
k
β
k
′
E
k
,
(
n
+
2
)
(
m
n
)
.
It follows from (37) and (39) that ϕ
(
E
12
(
m
)
⊗
E
12
(
n
)
)
=
0, which is a contradiction to that ϕ is injective.
For general case, by(
II
), we can choose a permutation p
1
,
…
,
p
m of [
1
,
m
] and a permutation q
1
,
…
,
q
n of [
1
,
n
] such that
(40)
(
σ
(
π
(
ρ
(
p
1
,
q
1
)
)
)
,
τ
(
π
(
ρ
(
p
1
,
q
1
)
)
)
)
=
(
m
,
n
)
,
(
σ
(
π
(
ρ
(
p
m
,
q
n
)
)
)
,
τ
(
π
(
ρ
(
p
m
,
q
n
)
)
)
)
=
(
1,1
)
,
(
σ
(
π
(
ρ
(
p
i
,
q
j
)
)
)
,
τ
(
π
(
ρ
(
p
i
,
q
j
)
)
)
)
=
(
i
,
j
)
,
fffffffffffffffffffffffffffffff
∀
(
i
,
j
)
≠
(
1,1
)
,
(
m
,
n
)
.
From this, together with (26), we obtain
(41)
ϕ
(
E
p
1
p
1
(
m
)
⊗
E
q
1
q
1
(
m
)
)
=
E
m
m
(
m
)
⊗
E
n
n
(
n
)
,
ϕ
(
E
p
m
p
m
(
m
)
⊗
E
q
n
q
n
(
n
)
)
=
E
11
(
m
)
⊗
E
11
(
m
)
,
ϕ
(
E
p
i
p
i
(
m
)
⊗
E
q
j
q
j
(
n
)
)
=
E
i
i
(
m
)
⊗
E
j
j
(
n
)
,
∀
(
i
,
j
)
≠
(
1,1
)
,
(
m
,
n
)
.
Using a similar method as the above, we can drive a contradiction. Similarly, we may prove (
IV
) does not hold.
If (32) holds, we may assume
(42)
ϕ
(
E
i
i
(
m
)
⊗
I
n
)
=
E
g
(
i
)
g
(
i
)
(
m
)
⊗
I
n
,
where g is a permutation on [
1
,
m
]. If (33) holds, we may assume
(43)
ϕ
(
I
m
⊗
E
i
i
(
m
)
)
=
E
g
(
i
)
g
(
i
)
(
m
)
⊗
I
m
.
We next assume (42) to prove (
i
) of theorem holds and one can use similar methods to prove (
ii
) of theorem if (43) holds.
Step  3. There exists an invertible matrix P
∈
𝒯
m
⊗
𝒯
n such that
(44)
ϕ
(
E
i
i
(
m
)
⊗
X
)
=
P
(
E
g
(
i
)
g
(
i
)
(
m
)
⊗
ξ
i
(
X
)
)
P
-
1
,
fffffffffff
∀
i
∈
[
1
,
m
]
,
X
∈
𝒯
n
,
where, for i
∈
[
1
,
m
], ξ
i
(
X
)
=
X or ξ
i
(
X
)
=
J
X
t
J.
Proof of Step  3. For any idempotent matrix A
∈
𝒯
n, since E
i
i
(
m
)
⊗
A and E
i
i
(
m
)
⊗
(
I
n
-
A
) are idempotent matrices in 𝒯
m
⊗
𝒯
n, we obtain by (42) that
(45)
ϕ
(
E
i
i
(
m
)
⊗
A
)
,
E
g
(
i
)
g
(
i
)
(
m
)
⊗
I
n
-
ϕ
(
E
i
i
(
m
)
⊗
A
)
are idempotent matrices in 𝒯
m
⊗
𝒯
n. By Lemma 7, we have
(46)
ϕ
(
E
i
i
(
m
)
⊗
A
)
=
E
g
(
i
)
g
(
i
)
(
m
)
⊗
ψ
i
(
A
)
,
where ψ
i
(
A
)
∈
𝒯
n is an idempotent matrix. By the arbitrariness of A, we can expand ψ
i to be a linear map on 𝒯
n. Hence
(47)
ϕ
(
E
i
i
(
m
)
⊗
X
)
=
E
g
(
i
)
g
(
i
)
(
m
)
⊗
ψ
i
(
X
)
,
∀
X
∈
𝒯
n
.
It is easy to see that ψ
i is injective and preserving idempotents. Thus, by Lemma 3, there exists an invertible P
g
(
i
)
∈
𝒯
n such that ψ
i
(
X
)
=
P
g
(
i
)
ξ
i
(
X
)
P
g
(
i
)
-
1, where ξ
i
(
X
)
=
X or ξ
i
(
X
)
=
J
X
t
J. Let P
=
diag
⁡
(
P
1
,
…
,
P
m
)
∈
𝒯
m
⊗
𝒯
n; we complete the proof of this step.
By Step 3, we may assume(48)
ϕ
(
E
i
i
(
m
)
⊗
X
)
=
E
g
(
i
)
g
(
i
)
(
m
)
⊗
ξ
i
(
X
)
,
∀
i
∈
[
1
,
m
]
,
X
∈
𝒯
n
.
Step  4. g
(
i
)
=
i or g
(
i
)
=
m
-
i
+
1.
Proof of Step  4. If m
=
2, then this claim is clear. For m
≥
3, we prove that if i
<
j
<
k, then g
(
i
)
<
g
(
j
)
<
g
(
k
) or g
(
i
)
>
g
(
j
)
>
g
(
k
). Otherwise, we assume g
(
i
)
<
g
(
k
)
<
g
(
j
) (other cases can be proven by using similar methods). Since for any x
∈
C, (
E
i
i
(
m
)
+
x
E
i
j
(
m
)
)
⊗
I
n and (
E
j
j
(
m
)
+
x
E
i
j
(
m
)
)
⊗
I
n are idempotent matrices in 𝒯
m
⊗
𝒯
n, we have by using (48) that E
g
(
i
)
g
(
i
)
(
m
)
⊗
I
n
+
x
ϕ
(
E
i
j
(
m
)
⊗
I
n
) and E
g
(
j
)
g
(
j
)
(
m
)
⊗
I
n
+
x
ϕ
(
E
i
j
(
m
)
⊗
I
n
) are idempotent matrices in T
m
⊗
T
n. This, together with Lemma 8, implies
(49)
ϕ
(
E
i
j
(
m
)
⊗
I
n
)
=
E
g
(
i
)
g
(
j
)
(
m
)
⊗
A
for
some
A
≠
0
∈
𝒯
n
.
Similarly,
(50)
ϕ
(
E
i
k
(
m
)
⊗
I
n
)
=
E
g
(
i
)
g
(
k
)
(
m
)
⊗
B
for
some
B
≠
0
∈
𝒯
n
,
ϕ
(
E
j
k
(
m
)
⊗
I
n
)
=
E
g
(
k
)
g
(
j
)
(
m
)
⊗
C
for
some
C
≠
0
∈
𝒯
n
.
By (
E
j
j
(
m
)
+
E
i
j
(
m
)
+
E
i
k
(
m
)
+
E
j
k
(
m
)
)
⊗
I
n being an idempotent matrix in 𝒯
m
⊗
𝒯
n, we obtain by using (48), (49), and (50) that
(51)
E
g
(
j
)
g
(
j
)
(
m
)
⊗
I
n
+
E
g
(
i
)
g
(
j
)
(
m
)
⊗
A
+
E
g
(
i
)
g
(
k
)
(
m
)
⊗
B
+
E
g
(
k
)
g
(
j
)
(
m
)
⊗
C
is an idempotent matrix in 𝒯
m
⊗
𝒯
n; that is,
(52)
[
0
A
B
0
0
C
0
0
I
n
]
2
=
[
0
A
B
0
0
C
0
0
I
n
]
.
This implies A
=
0, which is a contradiction. Hence, we complete the proof of Step  4.
Step  5.   For any i
<
j, ξ
i
=
ξ
j,and there exists λ
i
j
≠
0 such that
(53)
ϕ
(
E
i
j
(
m
)
⊗
X
)
=
λ
i
j
E
g
(
i
)
g
(
j
)
(
m
)
⊗
ξ
i
(
X
)
,
∀
i
≠
j
,
X
∈
𝒯
n
.
Proof of Step  5. We prove the case of g
(
i
)
=
i (one can use a similar method to prove the case of g
(
i
)
=
m
-
i
+
1). Hence,
(54)
ϕ
(
E
i
i
(
m
)
⊗
E
k
k
(
n
)
)
=
E
i
i
(
m
)
⊗
E
k
k
(
n
)
,
∀
i
∈
[
1
,
m
]
,
k
∈
[
1
,
n
]
.
Without loss of generality, we may assume i
=
1, j
=
2. Since for any x
∈
C, (
E
11
(
m
)
+
x
E
12
(
m
)
)
⊗
E
k
k
(
n
) and (
E
22
(
m
)
+
x
E
12
(
m
)
)
⊗
E
k
k
(
n
) are idempotent matrices in 𝒯
m
⊗
𝒯
n, we obtain by (54) that E
11
(
m
)
⊗
E
k
k
(
n
)
+
x
ϕ
(
E
12
(
m
)
⊗
E
k
k
(
n
)
) and E
22
(
m
)
⊗
E
k
k
(
n
)
+
x
ϕ
(
E
12
(
m
)
⊗
E
k
k
(
n
)
) are idempotent matrices in T
m
⊗
T
n. This, together with Lemma 8, implies ϕ
(
E
12
(
m
)
⊗
E
k
k
(
n
)
)
=
λ
k
E
12
(
m
)
⊗
E
k
k
(
n
), where λ
k
≠
0. Let Λ
=
diag
⁡
(
λ
1
,
…
,
λ
m
); then
(55)
ϕ
(
E
12
(
m
)
⊗
I
n
)
=
[
0
Λ
0
0
]
⊕
0
.
Let Q
=
[
I
m
-
Λ
0
I
m
]
⊕
I
(
m
-
2
)
n
∈
𝒯
m
⊗
𝒯
n; by (55), one can obtain
(56)
ϕ
(
(
E
11
(
m
)
+
E
12
(
m
)
)
⊗
I
n
)
=
[
I
n
Λ
0
0
]
⊕
0
=
Q
(
E
11
(
m
)
⊗
I
n
)
Q
-
1
,
ϕ
(
(
E
22
(
m
)
-
E
12
(
m
)
)
⊗
I
n
)
=
[
0
-
Λ
0
I
n
]
⊕
0
=
Q
(
E
22
(
m
)
⊗
I
n
)
Q
-
1
.
Let F
1
=
E
11
(
m
)
+
E
12
(
m
), F
2
=
E
22
(
m
)
-
E
12
(
m
). Then, (56) turn into
(57)
ϕ
(
F
i
⊗
I
n
)
=
Q
(
E
i
i
(
m
)
⊗
I
n
)
Q
-
1
,
i
=
1,2
.
By (57), using a similar method to Step 3, one can obtain
(58)
ϕ
(
F
i
⊗
X
)
=
Q
(
E
i
i
(
m
)
⊗
ζ
i
(
X
)
)
Q
-
1
,
∀
X
∈
𝒯
n
,
i
=
1,2
,
where ζ
i
(
X
)
=
X or ζ
i
(
X
)
=
J
X
t
J. Hence
(59)
ϕ
(
F
1
⊗
X
)
=
[
I
m
-
Λ
0
I
m
]
[
ζ
1
(
X
)
0
0
0
]
[
I
m
Λ
0
I
m
]
=
[
ζ
1
(
X
)
ζ
1
(
X
)
Λ
0
0
]
,
∀
X
∈
𝒯
n
,
ϕ
(
F
2
⊗
X
)
=
[
I
m
-
Λ
0
I
m
]
[
0
0
0
ζ
2
(
X
)
]
[
I
m
Λ
0
I
m
]
=
[
0
-
Λ
ζ
2
(
X
)
0
ζ
2
(
X
)
]
,
∀
X
∈
𝒯
n
.
Thus,
(60)
ϕ
(
E
12
(
m
)
⊗
X
)
=
ϕ
(
F
1
⊗
X
)
-
ϕ
(
E
11
(
m
)
⊗
X
)
=
[
ζ
1
(
X
)
-
ξ
1
(
X
)
ζ
1
(
X
)
Λ
0
0
]
,
ϕ
(
E
12
(
m
)
⊗
X
)
=
ϕ
(
E
22
(
m
)
⊗
X
)
-
ϕ
(
F
2
⊗
X
)
=
[
0
Λ
ζ
2
(
X
)
0
ξ
2
(
X
)
-
ζ
2
(
X
)
]
.
This implies
(61)
[
ζ
1
(
X
)
-
ξ
1
(
X
)
ζ
1
(
X
)
Λ
0
0
]
=
[
0
Λ
ζ
2
(
X
)
0
ξ
2
(
X
)
-
ζ
2
(
X
)
]
.
Fromζ
1
(
X
)
Λ
=
Λ
ζ
2
(
X
), ∀
X
∈
𝒯
n, one can easily see that Λ
=
λ
12
I
n
≠
0 and ζ
1
(
X
)
=
ζ
2
(
X
); thus ξ
1
(
X
)
=
ξ
2
(
X
). This completes the proof of Step  5.
By Step 5, we may assumeξ
=
ξ
i.
Step  6. For m
≥
3 and i
<
j
<
k, we have λ
i
j
λ
j
k
=
λ
i
k.
Proof of Step  6. From (
E
j
j
(
m
)
+
E
i
j
(
m
)
+
E
j
k
(
m
)
+
E
i
k
(
m
)
)
⊗
I
n is an idempotent matrix in 𝒯
m
⊗
𝒯
n, we have
(62)
(
E
g
(
j
)
g
(
j
)
(
m
)
+
λ
i
j
E
g
(
i
)
g
(
j
)
(
m
)
+
λ
j
k
E
g
(
j
)
g
(
k
)
(
m
)
+
λ
i
k
E
g
(
i
)
g
(
k
)
(
m
)
)
⊗
I
n
is an idempotent matrix in 𝒯
m
⊗
𝒯
n. It follows from g
(
i
)
=
i or g
(
i
)
=
m
-
i
+
1 that λ
i
j
λ
j
k
=
λ
i
k.
Wheng
(
i
)
=
i, let P
=
diag
⁡
(
λ
12
,
…
,
λ
1
m
); then
(63)
ϕ
(
E
i
j
(
m
)
⊗
X
)
=
P
E
i
j
(
m
)
P
-
1
⊗
ξ
(
X
)
,
∀
i
≤
j
,
X
∈
𝒯
n
.
Hence,
(64)
ϕ
(
A
⊗
X
)
=
(
P
⊗
I
n
)
(
A
⊗
ξ
(
X
)
)
(
P
⊗
I
n
)
-
1
,
FFFFFFFFFFFFFFFFFFFFF
∀
A
∈
𝒯
m
,
X
∈
𝒯
n
.
Wheng
(
i
)
=
m
-
i
+
1, let P
=
diag
⁡
(
λ
1
m
,
…
,
λ
12
); then
(65)
ϕ
(
E
i
j
(
m
)
⊗
X
)
=
P
E
n
-
i
+
1
,
n
-
j
+
1
(
m
)
P
-
1
⊗
ξ
(
X
)
,
∀
i
≤
j
,
X
∈
𝒯
n
.
Thus
(66)
ϕ
(
A
⊗
X
)
=
(
P
⊗
I
n
)
(
(
J
A
T
J
)
⊗
ξ
(
X
)
)
(
P
⊗
I
n
)
-
1
,
FFFFFFFFFFFFFFFFFFFFFFFFF
∀
A
∈
𝒯
m
,
X
∈
𝒯
n
.
This completes the proof of the theorem.



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*Source: 148321-2014-01-28.xml*