# Induced Maps on Matrices over Fields

Authors: Li Yang; Xuezhi Ben; Ming Zhang; Chongguang Cao
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Suppose that𝔽 is a field and m
,
n
≥
3 are integers. Denote by M
m
n
(
𝔽
) the set of all m
×
n matrices over 𝔽 and by M
n
(
𝔽
) the set M
n
n
(
𝔽
). Let f
i
j (i
∈
[
1
,
m
]
,
j
∈
[
1
,
n
]) be functions on 𝔽, where [
1
,
n
] stands for the set {
1
,
…
,
n
}. We say that a map f
:
M
m
n
(
𝔽
)
→
M
m
n
(
𝔽
) is induced by {
f
i
j
} if f is defined by f
:
[
a
i
j
]
↦
[
f
i
j
(
a
i
j
)
]. We say that a map f on M
n
(
𝔽
) preserves similarity if A
~
B
⇒
f
(
A
)
~
f
(
B
), where A
~
B represents that A and B are similar. A map f on M
n
(
𝔽
) preserving inverses of matrices means f
(
A
)
f
(
A
-
1
)
=
I
n for every invertible A
∈
M
n
(
𝔽
). In this paper, we characterize induced maps preserving similarity and inverses of matrices, respectively.


---

## Body

## 1. Introduction

Suppose that𝔽 is a field and m
,
n
≥
3 are integers. Denote by M
m
n
(
𝔽
) the set of all m
×
n matrices over 𝔽 and by M
n
(
𝔽
) the set M
n
n
(
𝔽
). Let f
i
j (i
∈
[
1
,
m
], j
∈
[
1
,
n
]) be functions on 𝔽, where [
1
,
n
] stands for the set {
1
,
…
,
n
}. We say that map f
:
M
m
n
(
𝔽
)
→
M
m
n
(
𝔽
) is induced by {
f
i
j
} if f is defined by
(1)
f
:
A
=
[
a
i
j
]
⟼
[
f
i
j
(
a
i
j
)
]
,
for
every
A
∈
M
m
n
(
𝔽
)
.
It is easy to see that induced map may not be linear or additive.Example 1.
Letℝ be real field, f
11
(
x
)
=
sin
⁡
x, f
12
(
x
)
=
exp
⁡
⁡
(
x
), f
21
(
x
)
=
tan
⁡
(
x
), and f
22
(
x
)
=
x
2
+
1, then f
:
M
2
(
ℝ
)
→
M
2
(
ℝ
) induced by {
f
i
j
} is
(2)
f
(
[
a
b
c
d
]
)
=
[
sin
⁡
a
exp
⁡
⁡
b
tan
⁡
c
d
2
+
1
]
,
∀
[
a
b
c
d
]
∈
M
2
(
ℝ
)
.Example 2.
The transposition mapX
↦
X
t is not an induced map on M
n
(
𝔽
).Example 3.
LetP
∈
M
n
(
𝔽
) be a matrix; then f
:
X
↦
P
X is an induced map on M
n
(
𝔽
) if and only if P is diagonal.Iff
i
j is independent of the choices of i and j (i.e., f
i
j
≡
φ, for every i
∈
[
1
,
m
] and j
∈
[
1
,
n
]), then f is said to be induced by the function φ, and denote by f
(
A
)
=
A
φ
(
=
[
φ
(
a
i
j
)
]
). Denote by rank
⁡
A the rank of matrix A. We say that an induced map f preserves rank-1 if rank
⁡
f
(
A
)
=
1 whenever rank
⁡
A
=
1.Preserver problem is a hot area in matrix and operator algebra; there are many results about this area. Kalinowski [1] showed that an induced map f
(
Θ
)
=
Θ
φ, where φ is a monotonic and continuous function of real field R such that f
(
0
)
=
0, preserves ranks of matrices if and only if it is linear. Furthermore, in [2], Kalinowski generalized the results in [1] by removing any restrictions on the map φ. In [3], Liu and Zhang characterized the general form of all maps f induced by f
i
j and preserving rank-1 matrices over a field. In particular, nonlinear maps preserving similarity were studied by Du et al. [4]. One can see [5–15] and their references for some background on preserver problems.We say that a mapf on M
n
(
𝔽
) preserves similarity if
(3)
A
~
B
⟹
f
(
A
)
~
f
(
B
)
∀
A
,
B
∈
M
n
(
𝔽
)
,
where A
~
B represents that A and B are similar. A map f on M
n
(
𝔽
) preserving inverses of matrices means f
(
A
)
f
(
A
-
1
)
=
I
n for every invertible A
∈
M
n
(
𝔽
). In this paper, we describe the forms of induced map preserving similarity and inverses of matrices, respectively.We end this section by introducing some notations which will be used in the following sections. Letdiag
⁡
(
a
1
,
a
2
,
…
,
a
n
) be the diagonal matrix of order n. E
i
j is the matrix with 1 in the (
i
,
j
)th entry and 0 elsewhere and I
n is the identity matrix of order n. Denote by ⊕ the usual direct sum of matrices.


## 2. Induced Map Preserving Similarity of Matrices

In this section, we use the form of induced rank-1 preserver to describe forms of induced similarity preservers. Firstly, we need the following theorem from [3].Lemma 4 (see [3, Corollary  1]).
Suppose that𝔽 is any field and n
≥
2 are integers. Suppose that f on M
n
(
𝔽
) is induced by {
f
i
j
} such that f
(
0
)
=
0. Then f preserves rank-1 if and only if there exist invertible and diagonal P
,
Q
∈
M
n
(
𝔽
) and a multiplicative map δ on 𝔽 satifying δ
(
x
)
=
0
⇔
x
=
0 such that
(4)
f
(
A
)
=
P
A
δ
Q
,
∀
A
∈
M
n
(
𝔽
)
.Lemma 5.
Suppose that𝔽 is any field, and n is an integer with n
≥
2. If A
∈
M
n
(
𝔽
) satisfies A
2
=
0 and rank
⁡
A
=
r, then there exists an invertible matrix P
∈
M
n
(
𝔽
) such that
(5)
P
-
1
A
P
=
[
0
I
r
0
0
]
⊕
0
.Proof.
It is easy to see that there exists an invertible matrixP
1
∈
M
n
(
𝔽
) such that
(6)
A
=
P
1
[
A
1
A
2
0
0
]
P
1
-
1
,
where A
1
∈
M
r
(
𝔽
) and A
2
∈
M
r
(
n
-
r
)
(
𝔽
) satisfy rank
⁡
[
A
1
A
2
]
=
r. From A
2
=
0, we have
(7)
A
1
[
A
1
A
2
]
=
0
,
so that A
1
=
0. Thus, rank
⁡
A
2
=
r, which implies that
(8)
A
2
Q
=
[
I
r
0
]
,
for some invertible Q
∈
M
n
-
r
(
𝔽
). Let P
=
P
1
(
I
r
⊕
Q
); then (6) turns into
(9)
P
-
1
A
P
=
[
0
I
r
0
0
]
⊕
0
.
This completes the proof.Theorem 6.
Let𝔽 be a field and let m
,
n
≥
3 be positive integers. Suppose that f is a map on M
n
(
𝔽
) induced by {
f
i
j
} such that f
(
0
)
=
0. Then fpreserves similarity if and only if there exist an invertible and diagonal P
∈
M
n
(
𝔽
), q
∈
𝔽, and an injective endomorphism δ of 𝔽 such that
(10)
f
(
A
)
=
q
P
A
δ
P
-
1
,
∀
A
∈
M
n
(
𝔽
)
.Proof.
The sufficiency is obvious. We will prove the necessary part by the following four steps.
Step  1. If there exists some i
≠
j and a
≠
0 such that f
i
j
(
a
)
=
0, then f
=
0.
Proof of Step  1. For any b
≠
0, k
≠
l, since rank
⁡
b
E
l
k
=
1 and (
b
E
l
k
)
2
=
0, by Lemma 5, we have
(11)
b
E
l
k
~
E
12
~
a
E
i
j
.
Since fpreserves similarity, we derive
(12)
f
k
l
(
b
)
E
k
l
~
f
i
j
(
a
)
E
i
j
=
0
,
and thus,
(13)
f
k
l
(
b
)
=
0
,
∀
k
≠
l
∈
[
1
,
n
]
,
b
∈
𝔽
.
Because of rank(
b
E
k
k
-
b
E
l
l
-
b
E
k
l
+
b
E
l
k
)
=
1 and (
b
E
k
k
-
b
E
l
l
-
b
E
k
l
+
b
E
l
k
)
2
=
0, by Lemma 5, we have
(14)
(
b
E
k
k
-
b
E
l
l
-
b
E
k
l
+
b
E
l
k
)
~
E
12
~
a
E
i
j
.
Using f preserves similarity and (13), one can obtain that
(15)
f
k
k
(
b
)
E
k
k
+
f
l
l
(
-
b
)
E
l
l
=
0
,
hence,
(16)
f
k
k
(
b
)
=
0
,
∀
k
∈
[
1
,
n
]
,
b
∈
𝔽
.
It follows from (13) and (16) that f
i
j
=
0, that is, f
=
0.
Step  2. If there exist some i and a
≠
0 such that f
i
i
(
a
)
=
0, then f
=
0.
Proof of Step  2. For j
≠
i, it follows from a
E
i
i
~
a
E
j
j that f
i
i
(
a
)
E
i
i
~
f
j
j
(
a
)
E
j
j. Thus,
(17)
f
j
j
(
a
)
=
0
,
∀
j
∈
[
1
,
n
]
,
a
∈
𝔽
.
Because of
(18)
(
a
E
i
i
-
a
E
j
j
-
a
E
i
j
+
a
E
j
i
)
~
E
12
,
one can obtain by using (17) that (
f
i
j
(
a
)
E
i
j
+
f
j
i
(
-
a
)
E
i
j
)
~
f
12
(
1
)
E
12, and hence,
(19)
rank
⁡
(
f
i
j
(
a
)
E
i
j
+
f
j
i
(
-
a
)
E
i
j
)
=
rank
⁡
f
12
(
1
)
E
12
≤
1
.
Thus, f
i
j
(
a
)
=
0 or f
j
i
(
-
a
)
=
0. We complete the proof of this step by using the result of Step  1.
Step  3. If f
≠
0, then f preserves rank-1.
Proof of Step  3. For any rank-1 matrix A we have
(20)
A
~
[
a
1
a
2
⋯
a
n
0
0
⋯
0
⋮
⋮
⋮
⋮
0
0
⋯
0
]
,
and hence,
(21)
f
(
A
)
~
[
f
11
(
a
1
)
f
12
(
a
2
)
⋯
f
1
n
(
a
n
)
0
0
⋯
0
⋮
⋮
⋮
⋮
0
0
⋯
0
]
.
Thus, rank
⁡
f
(
A
)
≤
1; it follows from f
≠
0 that rank
⁡
f
(
A
)
=
1.
Step  4. If f
≠
0, then there exist an invertible and diagonal P
∈
M
n
(
𝔽
), 0
≠
q
∈
𝔽, and an injective endomorphism δ of 𝔽 such that
(22)
f
(
A
)
=
q
P
A
δ
P
-
1
,
∀
A
∈
M
n
(
𝔽
)
.
Proof of Step  4. Since f
≠
0, by Step  3 and Lemma 4, there exist invertible and diagonal P
,
Q
∈
M
n
(
𝔽
) and a multiplicative map δ on 𝔽 satisfying δ
(
x
)
=
0
⇔
x
=
0 such that
(23)
f
(
A
)
=
P
A
δ
(
Q
P
)
P
-
1
,
∀
A
∈
M
n
(
𝔽
)
.
Let
(24)
A
=
[
1
1
x
+
y
0
1
x
1
0
y
]
⊕
0
,
B
=
[
1
1
-
x
-
y
x
+
y
0
1
-
x
x
1
1
-
x
-
y
x
+
y
]
⊕
0
.
It is easy to see that B
=
(
I
n
+
E
32
)
A
(
I
n
+
E
32
)
-
1. Since f preserves similarity, we have that f
(
A
) and f
(
B
) are similar; further, rank
⁡
f
(
A
)
=
rank
⁡
f
(
B
). It follows from (23) that rank
⁡
f
(
B
)
≤
2, thus,
(25)
rank
⁡
[
δ
(
1
)
δ
(
1
)
δ
(
x
+
y
)
0
δ
(
1
)
δ
(
x
)
δ
(
1
)
0
δ
(
y
)
]
⊕
0
≤
2
.
This implies δ
(
x
+
y
)
=
δ
(
x
)
+
δ
(
y
), hence, δ is an injective endomorphism of 𝔽.
SetQ
P
=
diag
⁡
(
q
1
,
…
,
q
n
). Since E
i
i
~
E
j
j, one obtains by using (23) that
(26)
q
i
E
i
i
~
q
j
E
j
j
.
Thus, q
i
=
q
j. Letting q
=
q
1, then Q
P
=
q
I
n and f
(
A
)
=
q
P
A
δ
P
-
1.
This completes the proof of Theorem.


## 3. Induced Map Preserving Inverses of Matrices

Theorem 7.
Let𝔽 be a field and let m
,
n
≥
3 be positive integers. Suppose that f is a map on M
n
(
𝔽
) induced by {
f
i
j
} such that f
(
0
)
=
0. Then fpreserves inverses of matrices if and only if there exist an invertible and diagonal P
∈
M
n
(
𝔽
), c
∈
{
-
1,1
}, and an injective endomorphism φ of 𝔽 such that
(27)
f
(
A
)
=
c
P
A
φ
P
-
1
,
∀
A
∈
M
n
(
𝔽
)
.Proof.
The sufficiency is obvious. We will prove the necessary part. For anyi
≠
j
∈
[
1
,
n
], a
∈
𝔽, and b
∈
𝔽
*, since
(28)
[
E
i
i
-
a
E
i
j
+
E
j
i
+
(
b
-
1
-
a
)
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
]
-
1
=
(
1
-
a
b
)
E
i
i
+
a
b
E
i
j
-
b
E
j
i
+
b
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
,
by f preserving inverses of matrices, we have
(29)
[
f
i
i
(
1
)
f
i
j
(
-
a
)
f
j
i
(
1
)
f
j
j
(
b
-
1
-
a
)
]
[
f
i
i
(
1
-
a
b
)
f
i
j
(
a
b
)
f
j
i
(
-
b
)
f
j
j
(
b
)
]
=
I
2
,
so that
(30)
f
i
i
(
1
)
f
i
j
(
a
b
)
+
f
i
j
(
-
a
)
f
j
j
(
b
)
=
0
,
(31)
f
i
i
(
1
)
f
i
i
(
1
-
a
b
)
+
f
i
j
(
-
a
)
f
j
i
(
-
b
)
=
1
.
Let b
=
1; then (30) turns into
(32)
f
i
i
(
1
)
f
i
j
(
a
)
+
f
i
j
(
-
a
)
f
j
j
(
1
)
=
0
.
Replacing a by a
b, then the above turns into
(33)
f
i
i
(
1
)
f
i
j
(
a
b
)
+
f
i
j
(
-
a
b
)
f
j
j
(
1
)
=
0
.
It follows from (30) and (33) that
(34)
f
i
j
(
-
a
b
)
f
j
j
(
1
)
=
f
i
j
(
-
a
)
f
j
j
(
b
)
.
Replacing -
a by a, then the above turns into
(35)
f
i
j
(
a
b
)
f
j
j
(
1
)
=
f
i
j
(
a
)
f
j
j
(
b
)
.
By[
a
E
i
i
+
Σ
k
≠
i
E
k
k
]
-
1
=
a
-
1
E
i
i
+
Σ
k
≠
i
E
k
k, we have
(36)
f
i
i
(
a
)
f
i
i
(
a
-
1
)
=
1
.
In particular,
(37)
f
i
i
(
1
)
2
=
1
.
From[
a
E
i
j
+
a
E
j
i
+
Σ
k
≠
i
,
j
E
k
k
]
-
1
=
a
-
1
E
i
j
+
a
-
1
E
j
i
+
Σ
k
≠
i
,
j
E
k
k, we have
(38)
f
i
j
(
a
)
f
j
i
(
a
-
1
)
=
1
.
In particular,
(39)
f
i
j
(
1
)
f
j
i
(
1
)
=
1
.
Multiplyingf
i
j
(
-
b
-
1
) by (31), we obtain by using (38) that
(40)
f
i
i
(
1
)
f
i
i
(
1
-
a
b
)
f
i
j
(
-
b
-
1
)
+
f
i
j
(
-
a
)
=
f
i
j
(
-
b
-
1
)
.
It follows from (35) and (38) that
(41)
f
i
i
(
1
)
f
i
j
(
(
1
-
a
b
)
(
-
b
-
1
)
)
=
f
i
i
(
1
-
a
b
)
f
i
j
(
-
b
-
1
)
.
This, together with (40), implies that
(42)
f
i
i
(
1
)
f
i
i
(
1
)
f
i
j
(
-
b
-
1
+
a
)
+
f
i
j
(
-
a
)
=
f
i
j
(
-
b
-
1
)
.
Hence, it follows from f
i
i
(
1
)
2
=
1 that
(43)
f
i
j
(
-
b
-
1
+
a
)
+
f
i
j
(
-
a
)
=
f
i
j
(
-
b
-
1
)
.
Let x
=
-
b
-
1 and y
=
-
a; we have
(44)
f
i
j
(
x
-
y
)
=
f
i
j
(
x
)
-
f
i
j
(
y
)
,
∀
x
∈
𝔽
*
,
y
∈
𝔽
.
From(45)
[
E
i
i
+
E
i
j
-
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
]
-
1
=
E
i
i
+
E
i
j
-
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
we have
(46)
[
f
i
i
(
1
)
f
i
j
(
1
)
0
f
j
j
(
-
1
)
]
[
f
i
i
(
1
)
f
i
j
(
1
)
0
f
j
j
(
-
1
)
]
=
I
2
,
so that
(47)
f
i
i
(
1
)
f
i
j
(
1
)
=
-
f
i
j
(
1
)
f
j
j
(
-
1
)
.
This, together with (39), implies
(48)
f
i
i
(
1
)
=
-
f
j
j
(
-
1
)
.
Similarly, f
k
k
(
1
)
=
-
f
j
j
(
-
1
); hence,
(49)
f
i
i
(
1
)
=
f
k
k
(
1
)
,
∀
i
,
k
∈
[
1
,
n
]
.
It follows from(50)
[
E
i
i
+
x
E
i
j
+
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
]
-
1
=
E
i
i
-
x
E
i
j
+
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
,
∀
x
∈
𝔽
that
(51)
[
f
i
i
(
1
)
f
i
j
(
x
)
0
f
j
j
(
1
)
]
[
f
i
i
(
1
)
f
i
j
(
-
x
)
0
f
j
j
(
1
)
]
=
I
2
.
Hence,
(52)
f
i
i
(
1
)
f
i
j
(
-
x
)
=
-
f
i
j
(
x
)
f
j
j
(
1
)
.
This, together with (39) and (49), implies
(53)
f
i
j
(
-
x
)
=
-
f
i
j
(
x
)
,
∀
x
∈
𝔽
.
Since(54)
[
a
E
i
i
+
E
i
j
-
E
j
i
+
Σ
k
≠
i
,
j
E
k
k
]
-
1
=
-
E
i
j
+
E
j
i
+
a
E
j
j
+
Σ
k
≠
i
,
j
E
k
k
,
we have
(55)
[
f
i
i
(
a
)
f
i
j
(
1
)
f
j
i
(
-
1
)
0
]
[
0
f
i
j
(
-
1
)
f
j
i
(
1
)
f
j
j
(
a
)
]
=
I
2
,
so that
(56)
f
i
i
(
a
)
f
i
j
(
-
1
)
=
-
f
i
j
(
1
)
f
j
j
(
a
)
.
Hence,
(57)
f
i
i
(
a
)
=
f
j
j
(
a
)
.
For distincti
,
j
,
k
∈
[
1
,
n
] and x
,
y
∈
𝔽, since
(58)
[
E
i
i
+
x
E
i
j
+
x
E
i
k
+
E
j
j
+
E
j
k
+
E
k
k
+
Σ
l
≠
i
,
j
,
k
E
l
l
]
-
1
=
E
i
i
-
x
E
i
j
+
E
j
j
-
E
j
k
+
E
k
k
+
Σ
l
≠
i
,
j
,
k
E
l
l
,
we have
(59)
[
f
i
i
(
1
)
f
i
j
(
x
)
f
i
k
(
x
y
)
0
f
j
j
(
1
)
f
j
k
(
y
)
0
0
f
k
k
(
1
)
]
[
f
i
i
(
1
)
f
i
j
(
-
x
)
0
0
f
j
j
(
1
)
f
j
k
(
-
y
)
0
0
f
k
k
(
1
)
]


=
I
3
,
so that
(60)
f
i
j
(
x
)
f
j
k
(
-
y
)
+
f
i
k
(
x
y
)
f
k
k
(
1
)
=
0
.
This, together with (53), implies that
(61)
f
i
j
(
x
)
=
f
i
k
(
x
)
f
k
k
(
1
)
f
j
k
(
1
)
-
1
.
It follows from (35) and (61) that
(62)
f
i
j
(
x
)
=
f
i
k
(
x
)
f
k
k
(
1
)
f
j
k
(
1
)
-
1
=
f
i
k
(
x
)
f
11
(
1
)
f
j
k
(
1
)
-
1
=
(
f
k
k
(
1
)
-
1
f
i
1
(
1
)
f
11
(
x
)
)
×
f
11
(
1
)
(
f
k
k
(
1
)
-
1
f
j
1
(
1
)
f
11
(
1
)
)
-
1
=
f
i
1
(
1
)
f
11
(
x
)
f
j
1
(
1
)
-
1
,
∀
i
≠
j
∈
[
1
,
n
]
.
Using this, together with (57), we obtain
(63)
f
(
A
)
=
P
[
f
11
(
a
i
j
)
]
P
-
1
,
where P
=
diag
⁡
(
f
11
(
1
)
,
…
,
f
n
1
(
1
)
). Let c
=
f
11
(
1
)
∈
{
-
1,1
}; φ
(
x
)
=
f
11
(
1
)
f
11
(
x
) and then
(64)
f
(
A
)
=
c
P
[
φ
(
a
i
j
)
]
P
-
1
=
c
P
A
φ
P
-
1
.
Let ϕ
(
A
)
=
[
φ
(
a
i
j
)
]; since f preserves inverses of matrices, one can see that ϕ also preserves inverses of matrices. By φ
(
1
)
=
f
11
(
1
)
2
=
1, we obtain by using similar method to (35), (37), (44), and (53) that for any a
,
b
∈
𝔽
(65)
φ
(
a
b
)
=
φ
(
1
)
φ
(
a
b
)
=
φ
(
a
)
φ
(
b
)
,
φ
(
a
+
b
)
=
φ
(
a
)
+
φ
(
b
)
.
This completes the proof.



---

*Source: 596756-2014-02-26.xml*