# Maps Preserving Idempotence on Matrix Spaces

Authors: Yuqiu Sheng; Hanyu Zhang
Journal: Journal of Mathematics
 (2015)

---

## Abstract

SupposeF is an arbitrary field. Let |
F
| be the number of the elements of F. Let M
n
(
F
) be the space of all n
×
n matrices over F, let S
n
(
F
) be the subset of M
n
(
F
) consisting of all symmetric matrices, and let T
n
(
F
) be the subset of M
n
(
F
) consisting of all upper-triangular matrices. Let V
∈
{
S
n
(
F
)
,
M
n
(
F
)
,
T
n
(
F
)
}; a map Φ
:
V
→
V is said to preserve idempotence if A
-
λ
B is idempotent if and only if Φ
(
A
)
-
λ
Φ
(
B
) is idempotent for any A
,
B
∈
V and λ
∈
F. In this paper, the maps preserving idempotence on S
n
(
F
), M
n
(
F
), and T
n
(
F
) were characterized in case  |
F
|
=
3.


---

## Body

## 1. Introduction

SupposeF is an arbitrary field and C is the field of all complex numbers. Let |
F
| be the number of the elements of F. Let M
n
(
F
) be the space of all n
×
n matrices over F, let S
n
(
F
) be the subset of M
n
(
F
) consisting of all symmetric matrices, and let T
n
(
F
) be the subset of M
n
(
F
) consisting of all upper-triangular matrices. For a linear subspace V of M
n
(
F
), we denote by I
V the subset of V consisting of all idempotence. A map Φ
:
V
→
V is said to preserve idempotence if A
-
λ
B is idempotent if and only if Φ
(
A
)
-
λ
Φ
(
B
) is idempotent for any A
,
B
∈
V and λ
∈
F. Denote by Φ
n
(
F
) the set of all maps Φ from S
n
(
F
) to M
n
(
F
) such that (1)
A
-
λ
B
∈
I
S
n
F
⟺
Φ
A
-
λ
Φ
B
∈
I
M
n
F
,
∀
A
,
B
∈
S
n
F
,
λ
∈
F
.Denote by Ψ
n
(
F
) the set of all maps Ψ from T
n
(
F
) to M
n
(
F
) such that (2)
A
-
λ
B
∈
I
T
n
F
⟺
Φ
A
-
λ
Φ
B
∈
I
M
n
F
,
∀
A
,
B
∈
T
n
F
,
λ
∈
F
.For a matrix A
∈
M
n
(
F
), we denote by A
t the transpose of A and by A
-
1 the inverse of A. For any positive integer k
≤
n, we denote ∏
k
(
F
)
=
{
X
⊕
O
n
-
k
∣
X
∈
M
k
(
F
)
}, S
k
=
{
X
⊕
b
⊕
O
n
-
k
∣
X
∈
M
k
(
F
)
,
b
∈
F
}, where ⊕ denotes the usual direct sum of matrices. In particular, ∏
1
(
F
)
=
{
a
⊕
O
n
-
1
∣
a
∈
F
}, and ∏
n
(
F
)
=
M
n
(
F
). Denote by F the number of elements of F. Denote F
∗
=
F
∖
{
0
}, (
n
)
=
{
1,2
,
…
,
n
}. Denote by E
i
j the matrix with 1 in the (i
,
j)th entry and 0 elsewhere. Denote I
s
=
E
11
+
⋯
+
E
s
s. (Note: the dimension of E
i
j can be known from the content.)The problem studied in this paper belongs to preserver problems on spaces of matrices. In the recent years, some researchers have studied preserver problems on spaces of matrices [1–4]. Šemrl [5] characterized the bijective and continuous maps preserving idempotence on M
n
(
C
) when n
≥
3. Liu and Zhao [6] improved the result of Šemrl by relaxing the bijectivity assumption to the surjectivity and also omitting the continuous assumption and the restriction on n
≥
3. Zhang [7] investigated the maps on M
n
(
F
) preserving idempotence without the surjectivity assumption in the higher dimensions when F is any field of characteristic not 2. But the homogeneous of Lemma 4 in his paper was not proved, and its original proof in [2] needs a nonzero scalar which is not equal to 1 and -
1. In [8] and [9] the authors characterized the maps preserving idempotence on S
n
(
F
) and T
n
(
F
) in case F is any field of characteristic not 2 and |
F
|
>
3, respectively. Thus, the case of |
F
|
=
3 is not proved on M
n
(
F
)
,
T
n
(
F
), and S
n
(
F
). Clearly, it is more complex to characterize maps preserving idempotence without being homogeneous. In addition, Lemma 9 of [7], Lemma 2.2 of [8], and Lemma 2 of [9] played an important role, but their original proof is no longer applicable in the case of |
F
|
=
3. This paper will use a new method to prove these conclusions. The purpose of this paper is to prove the following theorems.Theorem 1.
SupposeF is a field with |
F
|
=
3 and n
≥
4; then Φ is a map preserving idempotence on spaces of S
n
(
F
) if and only if there exists an invertible matrix P
∈
M
n
(
F
) such that Φ
(
A
)
=
P
A
P
-
1 for every A
∈
S
n
(
F
), where P
t
P
=
a
I
n for some nonzero scalar a of F.Theorem 2.
SupposeF is a field with |
F
|
=
3 and n
≥
3. Then Ψ is a map preserving idempotence on spaces of T
n
(
F
) if and only if there exists an invertible matrix P
∈
T
n
(
F
) such that either Φ
(
A
)
=
P
A
P
-
1 for every A
∈
T
n
(
F
) or Ψ
(
A
)
=
P
J
A
t
J
P
-
1 for every A
∈
T
n
(
F
), where J
=
∑
i
=
1
n
E
i
,
n
+
1
-
i.Theorem 3.
SupposeF is a field with |
F
|
=
3; then Φ is a map preserving idempotence on spaces of M
n
(
F
) if and only if there exists an invertible matrix P
∈
M
n
(
F
) such that Φ
(
A
)
=
P
A
P
-
1 for every A
∈
M
n
(
F
) or Φ
(
A
)
=
P
A
t
P
-
1 for every A
∈
M
n
(
F
).


## 2. Some Lemmas

In order to prove the theorems, we require the following lemmas.Lemma 4.
SupposeF is a field and |
F
|
=
3. If Φ is a map preserving idempotence on M
n
(
F
), or Φ
∈
Ψ
n
(
F
), or Φ
∈
Φ
n
(
F
), then(i)
Ais idempotent if and only if Φ
(
A
) is idempotent for any A
∈
M
n
(
F
), or Φ
∈
Ψ
n
(
F
);
(ii)
Φ is injective.Proof.
The proof of it is omitted, since it is very similar to that in [6].Lemma 5.
SupposeF is a field and |
F
|
=
3. If Φ is a map preserving idempotence on M
n
(
F
), or Φ
∈
Ψ
n
(
F
), or Φ
∈
Φ
n
(
F
), then there exists an invertible matrix P such that Φ
(
E
i
i
)
=
P
-
1
E
i
i
P for every i
∈
(
n
).Proof.
The proof of it is omitted, since it is very similar to Lemmas2 and 3 of [7], respectively.Lemma 6.
SupposeF is a field and |
F
|
=
3. If Φ is a map preserving idempotence on M
n
(
F
), or Φ
∈
Φ
n
(
F
), or ϕ
∈
Ψ
n
(
F
), and Φ
(
E
i
i
)
=
E
i
i, then Φ
(
c
E
i
i
)
=
c
E
i
i for every c of F.Proof.
SinceO
,
O
+
E
i
i
∈
I
M
n
(
F
), ∀
i
∈
(
n
), then by Lemma 5 we have Φ
(
O
), Φ
(
O
)
+
E
i
i
∈
I
M
n
(
F
), ∀
i
∈
(
n
). Thus Φ
(
O
)
=
O.
For anyi
∈
(
n
), since O
-
(
-
E
i
i
)
,
(
-
E
i
i
)
+
E
i
i
,
E
j
j
-
(
-
E
i
i
)
∈
I
M
n
(
F
), ∀
j
∈
(
n
)
∖
{
i
}, then O
-
Φ
(
-
E
i
i
)
,
Φ
(
-
E
i
i
)
+
E
i
i
,
E
j
j
-
Φ
(
-
E
i
i
)
∈
I
M
n
(
F
), ∀
j
∈
(
n
)
∖
{
i
}. In view of the injectivity of Φ and by direct calculation we can conclude that Φ
(
-
E
i
i
)
=
-
E
i
i.Lemma 7.
SupposeF is a field and |
F
|
=
3, 1
≤
s
≤
n
-
1, and Φ
∈
Φ
n
(
F
). If Φ
(
A
⊕
O
n
-
s
)
=
A
⊕
O
n
-
s for every A
∈
S
s
(
F
) and Φ
c
E
i
i
=
c
E
i
i, ∀
c
∈
F, i
=
r
+
1
,
…
,
n, then Φ
(
A
⊕
a
⊕
O
n
-
s
-
1
)
=
A
⊕
a
⊕
O
n
-
s
-
1 for every A
∈
S
s
(
F
), a
∈
F.Proof.
LetX
=
A
⊕
1
⊕
O
n
-
s
-
1, Y
1
=
-
A
⊕
O
n
-
s, and Y
2
=
(
-
A
+
I
s
)
⊕
O
n
-
s; then Φ
(
Y
i
)
=
Y
i, i
=
1,2. Since X
+
Y
i
∈
I
S
n
(
F
), i
=
1,2, we can conclude that Φ
(
X
)
+
Y
i
∈
I
M
n
(
F
), i
=
1,2. Let Φ
(
X
)
+
Y
1
=
M; then M
,
Φ
(
Y
2
)
+
Φ
(
Y
1
)
+
M
∈
I
M
n
(
F
). By direct calculation we can conclude that M
=
O
s
⊕
D, where D
∈
I
M
n
-
s
(
F
); that is,(3)
Φ
X
=
A
⊕
D
.Let X
1
i
=
(
A
+
E
i
i
)
⊕
1
⊕
O
n
-
s
-
1, X
2
i
=
(
A
-
E
i
i
)
⊕
1
⊕
O
n
-
s
-
1. By (3) we have(4)
Φ
X
1
i
=
A
+
E
i
i
⊕
D
1
i
,
Φ
X
2
i
=
A
+
E
i
i
⊕
D
2
i
,where D
1
i
,
D
2
i
∈
I
M
n
-
s
F, i
=
1
,
…
,
s. Since X
1
i
-
X
,
X
2
i
-
X
1
i
,
X
-
X
2
i
∈
I
S
n
(
F
), i
=
1
,
…
,
s, by (4) we have D
1
i
-
D
,
D
2
i
-
D
1
i
,
D
-
D
2
i
∈
I
M
n
(
F
). By direct calculation we can conclude that(5)
D
=
D
1
i
=
D
2
i
.Let H
i
j
=
E
i
j
+
E
j
i, Y
i
j
=
(
A
+
H
i
j
)
⊕
1
⊕
O
n
-
s
-
1, Z
i
j
=
(
A
+
H
i
j
-
E
i
i
-
E
j
j
)
⊕
1
⊕
O
n
-
s
-
1, W
i
j
=
A
+
E
i
i
+
E
j
j
⊕
1
⊕
O
n
-
s
-
1, ∀
1
≤
i, j
≤
s, and i
≠
j; then by (3) and (5) we have(6)
Φ
Y
i
j
=
A
+
H
i
j
⊕
G
i
j
,
Φ
Z
i
j
=
A
+
H
i
j
-
E
i
i
-
E
j
j
⊕
G
i
j
,
Φ
W
i
j
=
A
+
E
i
i
+
E
j
j
⊕
D
.Since Y
i
j
-
W
i
j, W
i
j
-
Z
i
j
∈
I
S
n
(
F
), then Φ
(
X
1
i
)
-
Φ
(
Y
i
j
), Φ
(
Z
i
j
)
-
Φ
(
X
1
i
)
∈
I
S
n
(
F
). By (6) we have D
-
G
i
j, G
i
j
-
D
∈
I
M
n
-
s
(
F
); this means that D
=
G
i
j; that is, ϕ
(
(
A
+
D
i
j
)
⊕
1
⊕
O
n
-
s
-
1
)
=
(
A
+
D
i
j
)
⊕
D. Similarly we have ϕ
(
(
A
-
D
i
j
)
⊕
1
⊕
O
n
-
s
-
1
)
=
(
A
-
D
i
j
)
⊕
D. Thus D is independent of the choice of A. Since Φ
(
E
s
+
1
,
s
+
1
)
=
E
s
+
1
,
s
+
1, then Φ
(
A
⊕
1
⊕
O
n
-
s
-
1
)
=
A
⊕
1
⊕
O
n
-
s
-
1. Similarly we can conclude that Φ
(
A
⊕
(
-
1
)
⊕
O
n
-
s
-
1
)
=
A
⊕
(
-
1
)
⊕
O
n
-
s
-
1.Lemma 8.
SupposeF is a field and |
F
|
=
3, 1
≤
s
≤
n
-
1, and Φ is a map preserving idempotence on M
n
(
F
) or T
n
(
F
). If Φ
(
A
⊕
O
n
-
s
)
=
A
⊕
O
n
-
s for every A
∈
M
s
(
F
) and Φ
c
E
i
i
=
c
E
i
i, ∀
c
∈
F, i
=
r
+
1
,
…
,
n, then Φ
(
A
⊕
a
⊕
O
n
-
s
-
1
)
=
A
⊕
a
⊕
O
n
-
s
-
1 for every A
∈
M
s
(
F
)
,
a
∈
F.Proof.
UsingI
S
n
(
F
) by I
M
n
(
F
) or I
T
n
(
F
), Y
i
j by (
A
+
E
i
j
)
⊕
1
⊕
O
n
-
s
-
1, Z
i
j by (
A
+
E
i
j
+
E
i
i
)
⊕
1
⊕
O
n
-
s
-
1, and W
i
j by (
A
-
E
i
i
)
⊕
1
⊕
O
n
-
s
-
1 in the proof of Lemma 7, we can prove Lemma 8.


## 3. The Proof of Theorems1 and 2

By Lemmas4–8, we can prove Theorems 1 and 2 in similar way as in [8] and [9], respectively.


## 4. The Proof of Theorem3

The proof of Theorem3 is equivalent to the proof of the following four propositions.Proposition 9.
SupposeF is a field and |
F
|
=
3. If Φ is a map preserving idempotence on spaces of M
n
(
F
), then there exists an invertible matrix T
1
∈
M
n
(
F
) such that Φ
(
Z
)
=
T
1
Z
T
1
-
1 for every Z
∈
∏
1
(
F
).Proof.
In view of Lemma6, it is obvious.Proposition 10.
SupposeF is a field and |
F
|
=
3. If Φ is a map preserving idempotence on spaces of M
n
(
F
) and Φ
(
Z
)
=
Z for every Z
∈
∏
1
(
F
), then there exists an invertible matrix T
2
∈
M
n
(
F
) such that Φ
(
A
)
=
T
2
-
1
A
T
2 for every A
∈
∏
2
(
F
) or Φ
(
A
)
=
T
2
-
1
A
t
T
2 for every A
∈
∏
2
(
F
).Proof.
In view of Theorem1, we can prove that there exists an invertible matrix T
2
∈
M
n
(
F
) such that Φ
(
A
⊕
O
n
-
2
)
=
T
2
-
1
(
A
⊕
O
n
-
2
)
T
2 for every A
∈
S
2
(
F
). For any a
,
b
,
x
,
y
∈
F which satisfy x
≠
y, let Y
=
a
x
y
b
⊕
O
n
-
2, Y
1
=
-
a
+
1
-
x
-
x
-
b
⊕
O
n
-
2, Y
2
=
a
x
x
b
+
1
⊕
O
n
-
2, Y
3
=
a
+
1
x
x
b
⊕
O
n
-
2; then Φ
(
Y
i
)
=
Y
i, i
=
1,2
,
3. Since Y
+
Y
1, -
Y
+
Y
2, -
Y
+
Y
3
∈
I
M
n
(
F
), then Φ
(
Y
)
+
Y
1, -
Φ
(
Y
)
+
Y
2, -
Φ
(
Y
)
+
Y
3
∈
I
M
n
(
F
). Let Φ
(
Y
)
+
Y
1
=
M; then M
,
Y
2
+
Y
1
-
M
,
Y
3
+
Y
1
-
M
∈
I
M
n
(
F
). By direct calculation we can conclude that M
=
1
f
(
a
,
b
,
x
,
y
)
g
(
a
,
b
,
x
,
y
)
0
⊕
O
n
-
2, where f and g are maps from F
×
F
×
F
×
F to F, and f
(
a
,
b
,
x
,
y
)
g
(
a
,
b
,
x
,
y
)
=
0; that is,(7)
Φ
Y
=
a
f
a
,
b
,
x
,
y
+
x
g
a
,
b
,
x
,
y
+
x
b
⊕
O
n
-
2
.
Case  1. If f
(
a
,
b
,
x
,
y
)
=
0, in view of the injectivity of Φ we can conclude that g
(
a
,
b
,
x
,
y
)
≠
0; that is, (8)
Φ
Y
=
a
x
g
a
,
b
,
x
,
y
+
x
b
⊕
O
n
-
2
.Since(9)
a
x
g
a
,
b
,
x
,
y
+
x
b
⊕
O
n
-
2
+
-
a
+
1
-
g
a
,
b
,
x
,
y
-
x
-
g
a
,
b
,
x
,
y
-
x
-
b
⊕
O
n
-
2
∈
I
M
n
F
,then(10)
a
x
y
b
⊕
O
n
-
s
-
1
+
-
a
+
1
-
g
a
,
b
,
x
,
y
-
x
-
g
a
,
b
,
x
,
y
-
x
-
b
⊕
O
n
-
2
∈
I
M
n
F
.Thus g
(
a
,
b
,
x
,
y
)
+
x
=
y; that is, (11)
Φ
Y
=
a
x
y
b
⊕
O
n
-
2
.
For anyu
∈
F
∗, since (12)
u
-
1
a
+
u
x
y
b
⊕
O
n
-
2
-
a
x
y
b
⊕
O
n
-
2
∈
I
M
n
F
,then(13)
u
-
1
a
+
u
f
a
+
u
,
b
,
x
,
y
+
x
g
a
+
u
,
b
,
x
,
y
+
x
b
⊕
O
n
-
2
-
a
x
y
b
⊕
O
n
-
2
∈
I
M
n
F
.So f
(
a
+
u
,
b
,
x
,
y
)
=
0, g
(
a
+
u
,
b
,
x
,
y
)
≠
0. In a similar way as above, we can conclude that (14)
Φ
a
+
u
x
y
b
⊕
O
n
-
2
=
a
+
u
x
y
b
⊕
O
n
-
2
.Similarly, we have (15)
Φ
a
x
y
b
+
u
⊕
O
n
-
2
=
a
x
y
b
+
u
⊕
O
n
-
2
.Let c
=
y
-
x; then we have (16)
Φ
a
x
+
c
y
b
⊕
O
n
-
2
=
Φ
a
y
y
b
⊕
O
n
-
2
=
a
y
y
b
⊕
O
n
-
2
=
a
x
+
c
y
b
⊕
O
n
-
2
.For another nonzero scalar v of F, because v
=
c
+
c, (17)
Φ
a
x
+
v
y
b
⊕
O
n
-
2
=
Φ
a
x
+
c
+
c
y
b
⊕
O
n
-
2
=
a
x
+
c
+
c
y
b
⊕
O
n
-
2
=
a
x
+
v
y
b
⊕
O
n
-
2
.Similarly, we have (18)
Φ
a
x
y
+
v
b
⊕
O
n
-
2
=
a
x
y
+
v
b
⊕
O
n
-
2
.Thus, for any a
,
x
,
y
,
b, (19)
Φ
a
x
y
b
=
a
x
y
b
⊕
O
n
-
2
.
Case  2. If g
(
a
,
b
,
x
,
y
)
=
0, then in a similar way as Case 1 we can prove that (20)
Φ
a
x
y
b
⊕
O
n
-
2
=
a
x
y
b
⊕
O
n
-
2
,
∀
a
,
b
,
x
,
y
∈
F
.Proposition 11.
Suppose2
≤
s
≤
n
-
1 and F is a field and |
F
|
=
3. If Φ is a map preserving idempotence on spaces of M
n
(
F
) and Φ
(
Z
)
=
Z for every Z
∈
∏
s
(
F
), then there exists an invertible matrix T
s
∈
M
n
(
F
) such that Φ
(
A
)
=
T
s
-
1
A
T
s for every A
∈
∏
s
+
1
(
F
).Proof.
Step  1. There exists an invertible matrix T
∈
M
n
(
F
) such that (21)
Φ
A
x
1
e
1
O
a
⊕
O
n
-
s
-
1
=
T
A
x
1
e
1
O
a
⊕
O
n
-
s
-
1
T
-
1
.Let Z
=
A
x
1
e
1
O
a
⊕
O
n
-
s
-
1, Z
1
=
-
A
+
E
11
O
O
-
a
⊕
O
n
-
s
-
1, Z
2
=
A
O
O
-
a
+
1
⊕
O
n
-
s
-
1, Z
3
=
-
A
+
E
11
+
E
12
O
O
-
a
+
1
⊕
O
n
-
s
-
1; then Φ
(
Z
i
)
=
Z
i, i
=
1,2
,
3
. Since Z
+
Z
1, Z
+
Z
3, -
Z
+
Z
2
∈
I
M
n
(
F
), then Φ
(
Z
)
+
Z
1, Φ
(
Z
)
+
Z
3, -
Φ
(
Z
)
+
Z
2
∈
I
M
n
(
F
). Let Φ
(
Z
)
+
Z
1
=
M; thus M
,
Z
3
-
Z
1
+
M
,
Z
2
+
Z
1
-
M
∈
I
M
n
(
F
). This implies that M
=
E
11
f
1
(
A
,
a
,
x
1
)
e
1
O
0
⊕
O
n
-
s
-
1, where f
1 is a map from M
s
(
F
)
×
F
×
F
∗ to F
∗; that is, (22)
Φ
Z
=
A
f
1
A
,
a
,
x
1
e
1
O
a
⊕
O
n
-
s
-
1
.
For everyj
∈
{
2
,
…
,
s
}, let W
j
=
A
+
E
j
j
x
1
e
1
O
a
⊕
O
n
-
s
-
1, Q
j
=
A
+
E
j
j
x
1
e
1
O
a
⊕
O
n
-
s
-
1. Since W
j
-
Z, Z
-
Q
j
∈
I
M
n
(
F
), then Φ
(
W
j
)
-
Φ
(
Z
), Φ
(
Z
)
-
Φ
(
Q
j
)
∈
I
M
n
(
F
). By direct calculation we can conclude that(23)
f
1
A
±
E
j
j
,
a
,
x
1
=
f
1
A
,
a
,
x
1
.Similarly, for any u
∈
F
∗, 1
≤
i
≤
s which satisfy that i
≠
j, by (
Z
+
E
j
j
±
E
i
j
)
-
Z, (
Z
+
E
j
j
±
E
j
i
)
-
Z, (
Z
-
E
11
+
E
12
+
E
21
-
E
22
)
-
Z, Z
-
(
Z
+
E
11
+
E
12
+
E
21
+
E
22
), u
-
1
{
(
Z
+
u
I
s
+
1
)
-
Z
}
∈
I
M
n
(
F
), we have (24)
f
1
A
+
E
j
j
±
E
i
j
,
a
,
x
1
=
f
1
A
+
E
22
±
E
12
±
E
i
j
,
a
,
x
1
=
f
1
A
-
E
11
+
E
12
+
E
21
-
E
22
,
a
,
x
1
=
f
1
A
+
E
11
+
E
12
+
E
21
+
E
22
,
a
,
x
1
=
f
1
A
+
u
I
s
,
a
+
u
,
x
1
=
f
1
A
,
a
,
x
1
.This, together with (23), implies that f
1
(
A
,
a
,
x
1
) is independent of the choice of A and a. Without loss of generality we also denote f
1
(
A
,
a
,
x
1
) by f
1
(
x
1
). Let T
=
1
0
0
f
1
-
1
(
x
1
)
⊕
O
n
-
2; then similarity transformation by matrix T preserves the above results, and T
-
1
Φ
(
Z
)
T
=
X
. In view of the injectivity of Φ we can conclude that (25)
T
-
1
Φ
A
x
1
e
1
O
a
⊕
O
n
-
s
-
1
T
=
A
x
1
e
1
O
a
⊕
O
n
-
s
-
1
.
Step  2. If(26)
Φ
A
x
O
a
⊕
O
n
-
s
-
1
=
A
x
O
a
,
∀
A
∈
M
s
F
,
a
∈
F
,
x
=
x
1
,
…
,
x
r
-
1
,
0
,
…
,
0
t
,where 2
≤
r
≤
s, then (27)
Φ
A
x
+
x
r
e
r
O
a
⊕
O
n
-
s
-
1
=
A
x
+
x
r
e
r
O
a
⊕
O
n
-
s
-
1
,
∀
A
∈
M
s
F
,
a
,
x
1
,
…
,
x
r
∈
F
.
For anyx
r
∈
F
∗, let X
=
A
x
+
x
r
e
r
O
a
⊕
O
n
-
s
-
1, X
1
=
-
A
+
E
r
r
-
x
O
-
a
⊕
O
n
-
s
-
1, X
2
=
A
x
O
a
+
1
⊕
O
n
-
s
-
1, X
3
=
-
A
+
E
r
r
+
E
r
-
1
,
r
-
x
O
-
a
+
1
⊕
O
n
-
s
-
1. By Lemma 7 we have Φ
(
X
i
)
=
X
i, i
=
1,2
,
3
. In a similar way as Step 1 we can prove that (28)
Φ
X
=
A
x
+
f
r
x
,
x
r
e
r
O
a
⊕
O
n
-
s
-
1
,where f
r is a map from F
′
×
F to F
∗, F
′
=
{
(
x
1
,
…
,
x
r
-
1
,
0
,
…
,
0
)
t
∣
x
i
∈
F
,
i
=
1
,
…
,
r
-
1
}. By (29)
A
+
E
11
x
+
x
r
e
r
+
c
1
e
1
O
a
⊕
O
n
-
s
-
1
-
X
∈
I
M
n
F
,we can prove that f
r
(
x
,
x
r
) is independent of the choice of x
1. In view of (30)
Φ
E
r
r
+
x
r
-
1
E
1
r
x
-
x
1
-
1
e
1
+
x
r
e
r
O
0
⊕
O
n
-
s
-
1
∈
I
M
n
F
,we can conclude that f
r
(
A
,
a
,
x
,
x
r
)
=
x
r.
Step  3. If Φ
(
A
x
O
a
⊕
O
n
-
s
-
1
)
=
A
x
O
a, ∀
A
∈
M
s
(
F
), a
∈
F, x
∈
F
s, then (31)
Φ
A
x
y
1
e
1
t
a
⊕
O
n
-
s
-
1
=
A
x
y
1
e
1
t
a
⊕
O
n
-
s
-
1
,
∀
A
∈
M
s
F
,
a
,
y
1
∈
F
,
x
∈
F
s
.
For anyy
1
∈
F
∗, let K
=
A
x
y
1
e
1
t
a
⊕
O
n
-
s
-
1, K
1
=
-
A
+
E
11
-
x
O
-
a
⊕
O
n
-
s
-
1, K
2
=
A
x
O
a
+
1
⊕
O
n
-
s
-
1, K
3
=
-
A
-
x
O
-
a
+
1
⊕
O
n
-
s
-
1. By Step 2 we have Φ
(
K
i
)
=
K
i, i
=
1,2
,
3. In a similar way as Step 1 we can prove that (32)
Φ
K
=
A
x
+
h
1
x
,
y
1
e
1
l
1
x
,
y
1
e
1
t
a
⊕
O
n
-
s
-
1
,where h
1 and l
1 are maps from F
s
×
F
∗ to F, and h
1
(
x
,
y
1
)
l
1
(
x
,
y
1
)
=
0. By (33)
A
+
E
j
j
x
+
c
j
e
j
y
1
e
1
t
a
⊕
O
n
-
s
-
1
-
X
∈
I
M
n
F
,
2
≤
j
≤
s
,
A
+
E
22
+
c
1
E
12
x
+
c
1
e
1
y
1
e
1
t
a
⊕
O
n
-
s
-
1
-
A
x
-
e
2
y
1
e
1
t
a
⊕
O
n
-
s
-
1
∈
I
M
n
F
,we can prove that h
1
(
x
,
y
1
) and l
1
(
x
,
y
1
) are independent of the choice of x. In view of Φ
E
11
+
E
21
O
y
1
e
1
t
0
⊕
O
n
-
s
-
1, Φ
-
E
11
y
1
e
1
y
1
e
1
t
-
1
⊕
O
n
-
s
-
1
∈
I
M
n
(
F
), we can conclude that h
1
(
x
,
y
1
)
=
0, l
1
(
x
,
y
1
)
=
y
1.
Step  4. If (34)
Φ
A
x
y
a
⊕
O
n
-
s
-
1
=
A
x
y
a
,
∀
A
∈
M
s
F
,
a
∈
F
,
y
=
y
1
,
…
,
y
r
-
1
,
0
,
…
,
0
t
∈
F
s
,where x
∈
F
s and 2
≤
r
≤
s, then (35)
Φ
A
x
y
+
y
r
e
r
t
a
⊕
O
n
-
s
-
1
=
A
x
y
+
y
r
e
r
t
a
⊕
O
n
-
s
-
1
,
∀
A
∈
M
s
F
,
a
,
y
1
,
…
,
y
r
∈
F
,
x
∈
F
s
.
For anyy
r
∈
F
∗, let H
=
A
x
y
+
y
r
e
r
t
a
⊕
O
n
-
s
-
1, H
1
=
-
A
+
E
r
r
-
x
-
y
-
a
⊕
O
n
-
s
-
1, H
2
=
A
x
y
a
+
1
⊕
O
n
-
s
-
1, H
3
=
-
A
-
x
-
y
-
a
+
1
⊕
O
n
-
s
-
1. Then Φ
(
H
i
)
=
H
i, i
=
1,2
,
3. In a similar way as Step 1 we can prove that (36)
Φ
H
=
A
x
+
h
r
y
,
y
r
e
r
y
+
l
r
y
,
y
r
e
r
t
a
⊕
O
n
-
s
-
1
,where h
r and l
r are maps from F
′
×
F
∗ to F, F
′
=
{
(
y
1
,
…
,
y
r
-
1
,
0
,
…
,
0
)
∣
y
i
∈
F
,
i
=
1
,
…
,
r
-
1
}, and h
r
(
y
,
y
r
)
l
r
(
y
,
y
r
)
=
0. By (37)
A
+
E
i
i
x
y
+
y
r
e
r
t
+
u
i
e
i
t
a
⊕
O
n
-
s
-
1
-
X
∈
I
M
n
F
,
1
≤
i
≤
r
-
1
,we can prove that h
r
(
y
,
y
r
) and l
r
(
y
,
y
r
) are independent of the choice of y. In view of Φ
E
r
r
+
E
1
r
O
y
r
e
r
t
0
⊕
O
n
-
s
-
1, Φ
-
E
r
r
y
r
e
r
y
r
e
r
t
-
1
⊕
O
n
-
s
-
1
∈
I
M
n
(
F
), we can conclude that h
r
(
y
,
y
r
)
=
0, l
r
(
y
,
y
r
)
=
y
r.Proposition 12.
Suppose2
≤
s
≤
n
-
1, and F is a field and |
F
|
=
3. If Φ is a map preserving idempotence on spaces of M
n
(
F
) and Φ
(
Z
)
=
Z
t for every Z
∈
∏
s
(
F
), then there exists an invertible matrix T
s
∈
M
n
(
F
) such that Φ
(
A
)
=
T
s
-
1
A
t
T
s for every A
∈
∏
s
+
1
(
F
).Proof.
In a similar way as the proof of Proposition11 we can complete its proof.Remark 13.
The following example states that Theorem1 is wrong in case n
=
3.Example 14.
SupposeF is a field and |
F
|
=
3. Denote a map Ψ
:
S
3
(
F
)
→
S
3
(
F
) by (38)
Ψ
a
11
a
12
a
13
a
12
a
22
a
23
a
13
a
23
a
33
=
a
11
a
12
a
13
a
12
a
22
-
a
23
a
13
-
a
23
a
33for any a
i
j
∈
F, 1
≤
i
≤
j
≤
3. It is easy to prove that A
=
0
⊕
A
′ or a
23
=
0 for every A
=
[
a
i
j
]
∈
I
S
3
(
F
), where A
′
∈
I
S
2
(
F
). Clearly, Ψ is a map preserving idempotence, but Ψ is not the form described in Theorem 1.



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*Source: 428203-2015-09-09.xml*