Abstract and Applied Analysis

Volume 2013, Article ID 581683, 8 pages

http://dx.doi.org/10.1155/2013/581683

## A Note on -Potence Preservers on Matrix Spaces over Complex Field

^{1}Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China^{2}School of Mathematical Sciences, Heilongjiang University, Harbin 150080, China

Received 13 March 2013; Accepted 3 May 2013

Academic Editor: Chunrui Zhang

Copyright © 2013 Xiaofei Song et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let be the field of all complex numbers, the space of all matrices over , and the subspace of consisting of all symmetric matrices. The map satisfies that is -potent in implying that is -potent in , where then there exist an invertible matrix and with such that for every . Moreover, the inductive method used in this paper can be used to characterise similar maps from to .

#### 1. Introduction

Let be the field of all complex numbers, the space of all matrices over , the subspace of consisting of all triangular matrices, and the subspace of consisting of all symmetric matrices. For fixed integer , is called a -potent matrix if ; especially, is an idempotent matrix when . The map satisfies that is a -potent matrix in implying that is a -potent matrix in , where , is a kind of the so-called weak preservers. While replacing “implying that” with “if and only if,” is called strong preserver. Obviously, a strong preserver must be a weak preserver, while a weak preserver may not be a strong preserver.

The preserver problem in this paper is from LPPs but without linear assumption (more details about LPP in [1–3]). You and Wang characterized the strong -potence preservers from to in [4]; then Song and Cao extended the result to weak preservers from to in [5]. In [6], Wang and You characterized the strong -potence preservers from to . In this paper, the authors characterized the weak -potence preservers from to and proved the following theorem.

Theorem 1. *Suppose satisfy that is a -potent matrix in implying that is a -potent matrix in , where . Then there exist invertible and with such that for every .*

Furthermore, we can derive the following corollary from Theorem 1.

Corollary 2. *Suppose satisfy that is a -potent matrix in implying that is a -potent matrix in , where . Then there exist invertible and with such that for every , where for some nonzero . *

In fact, the proof of Theorem 1 through some adjustments is suitable for the weak -potence preserver from to , and more details can be seen in remarks.

#### 2. Notations and Lemmas

denotes the set of all -potent matrices in , while . denotes the set of all complex number satisfying , . denotes matrices in with in and elsewhere, and denotes the unit matrix in . denotes the set of integer satisfy . denotes the general linear group consisting of all invertible matrices in . denotes an arbitrary diagonal matrix in . For , , and are orthogonal if . denotes the space of all matrices over . denotes the set of all maps satisfying that is a -potent matrix in implying that is a -potent matrix in , where .

For an arbitrary matrix , we denote by the term in position of , by the matrix with the term in its position equal to , where and . Moreover, we denote by the matrix with the term in its position equal to and terms elsewhere equal to . We especially simplify it with when , and for every . Naturally, for every .

Without fixing , also denotes a matrix in with in its position, where , , and , .

At first, we need the following Lemmas 3, 4, 5, and 7, which are about -potent matrices and orthogonal matrices.

Lemma 3 (see [2]). *Suppose , , and for every ; then and are orthogonal. *

Lemma 4 ([7, Lemma 1]). *Suppose , , , are mutually orthogonal nonzero -potent matrices; then there exists such that with for every . *

Lemma 5. *Suppose , , , , with , , for arbitrary nonzero with and , . Then , , and there exist , with such that and . *

*Proof. *By the assumption of and , is idempotent. Denote this matrix by , and then we can get the following equation:

Since the matrices on both sides of satisfy the following equation:
then the following matrix is -potent by the assumption of lemma:

We denote by the following matrix:
then the following equation is obvious:
Unfolding it, we get ; that is, , where is the coefficient matrix of for every .

Let , then we calculate it and get the following equations:

It is easy to get and the following equation:
Note that the highest degree of in is ; then the highest degree of in is less or equal to for every with , and the highest degree of in is , where is the coefficient matrix of in and is the coefficient of in .

By the assumption of , we have and . Then the following equations are true:
and , where the highest degree of is and is the coefficient matrix of .

Now, we calculate the upper left part of .

When , , of which the upper left part is . Then in the upper left part of , the highest degree of is , and the coefficient matrix is .

When , if appears in the left (or right) end of an additive item of , then the upper left part of this item is . So, the upper left part of is equal to the upper left part of ; that is, the upper left part is , and the highest degree of is with as the coefficient matrix of .

By the assumption of , we have .

By , we have , , and if and only if . When , we can get by , and by , where and satisfy ; that is, by , which is equivalent to . When , .

*Remark 6. *Replacing with in Lemma 5, we have implies or , and implies or . These cases will not appear in the proof of Theorem 1, but are necessary for the weak preservers from to .

Lemma 7. *Suppose for arbitrary , with , where is a map satisfying for every . Then there exists nonzero such that for every . *

*Proof. *Since the trace of is equal to , then , or , especially, when equal to , with . Denote by , and by ; then we have or .(1)If , then , that is, ; (2)if , then . When , ; when , . But implies , or . It is a contradiction! So it is impossible that .

Hence, there exists nonzero such that for every .

We can prove the following Lemmas 8 and 9 similar as Lemmas 4 and 5 in [4].

Lemma 8 (see [4], Lemma 4). *Suppose , and are orthogonal -potent matrices; then and are orthogonal. *

Lemma 9 (see [4], Lemma 5). *Suppose ; then are homogeneous; that is, for every and every . *

Corollary 10. *Suppose , , , and for every , , . Then and are orthogonal. *

*Proof. *By the assumption and Lemma 9, we have , , . By Lemma 3, and are orthogonal.

Corollary 11. *Suppose and for arbitrary diagonal matrix . Then for every , with , , where is only decided by and . *

*Proof. *Let , , and ; then , and satisfy the assumption of Corollary 10, and and are orthogonal; that is, for some , , , and .

Since for arbitrary nonzero with , after applying , we have . By Lemma 5, , .

Let , where for every ; then is the function of , , and and denote by the value of on , , , , and .

Fix , , and and add a free variable to for some ; then becomes into a map of . Since for arbitrary and with , then by and , we can derive that . By Lemma 7, for fixed , , and ; that is, for arbitrary . Similarly, we can prove for arbitrary .

In fact, we have proved that and for arbitrary , and arbitrary ; then follows.

Since for fixed , , and with , , and arbitrary , , then implies . By Lemma 7, we can get for arbitrary and with ; that is, for arbitrary .

Until now, we have proved that for arbitrary ; that is, is only decided by and .

*Remark 12. *The proof of Corollary 11 presents the basic procedure of proof of Theorem 1. In order to decide the image of matrix , we use Corollary 10 and the images of and , which usually are diagonal matrices or some matrices with images already decided.

If is a weak preserver from to , then Corollary 11 is also true. Let , , and ; then we can prove similarly as proving , and . Since for arbitrary nonzero , then the following matrix is -potent:
Remark 6 tells us that , , or ; that is, , or , . Similarly, we can prove , , or . Since is arbitrary, we set for convenience.

If ; then implies ; that is, , which is a contradiction. Hence, we proved that it is impossible or .

If and , then implies ; that is, , which is a contradiction. Hence, we proved that and , or and .

#### 3. Proof of Theorem 1

Suppose , then we can derive Theorem 1 from Propositions 13, 14, and 16.

Proposition 13. *Suppose , with ; then if and only if . *

*Proof. *Suppose and for some , with . At first, we prove that for arbitrary . Since the equation is already true when , then we assume in the following proof.

Let , , and ; then it is easy to verify , , and satisfying the assumption of Corollary 10. So and are orthogonal. Moreover, we can derive from and . Let , then and are orthogonal -potent matrices. While implies ; then . There are two cases on .(1)If , then ; that is, ; (2)if , we can derive that from . Note that , so it is true that ; that is, . Finally, we can derive from and . At the same time, .

Anyway, for arbitrary .

Since for every nonzero with , then , and by . While the equation is equivalent to . Note that is the constant term of the equation; then by the infinite property of , and follows. Then we can derive which is a contradiction to the assumption.

Proposition 14. *Suppose for every ; then for arbitrary . *

*Proof. *The proof will be completed by induction on the following equation for arbitrary with for every :
where .

When , (10) is equivalent to for arbitrary .

At first, by the assumption, it is already true that for every .

Suppose for every with ; then by the homogeneity of , we just need to prove the following equation for with :
There are two cases on . (1) If , then there exists such that , and the following statements are true:

Note that and by the assumption; then the following statements are true:

Since , then , and follows.(2) If , then we have the following statements:

Since ; then by case 1, and follows. While , hence we get .

Anyway, we prove ; then by the induction, (10) is true for .

Suppose (10) is true for , then we prove the case on .

Let , , ; then we have and the following equation:
We will prove the following equation which is equivalent to (10) on :

For arbitrary nonzero with , the following matrix is idempotent:
where .

Note that and satisfy the following equation:

After applying on the above matrices, we have by the inductive assumption. Then because of the assumption of ; that is, (10) holds for .

Finally, we prove that for every by the induction.

*Remark 15. *If is a weak -potence preserver from to ; then Propositions 13 and 14 (replacing with for arbitrary in the proof of Proposition 14) hold since Corollary 10 is true under this assumption.

Proposition 16. *Suppose for every , then there exist and such that for every . *

*Proof. *The proof will be completed in the following 4 steps.*Step **1. *, where for every :

Since is nonzero -potent, then we can derive from Lemma 4 that there exists such that for every , where . It is obvious that the following map and for every .
Without loss of generality, we can assume .*Step **2. *, for arbitrary diagonal matrix .

The proof of this step can be seen in Step 3, Section 3 in [5].*Step **3. * for every .

Let , , and , we can derive the following equation from Step 2 and Corollary 10:
where , , , , , with .

Note that for , with . In fact, and are all the eigenvalues of this matrix. Applying on the matrix , we have .

Since is fixed, then is the finite set which contains all of eigenvalues of , and there exists such that the trace of is for infinite choices of ; that is, there exist , with such that the traces of and are all equal to ; then we have the following equation:
which is equivalent to
where , for , .

Naturally, there are infinite choices of for fixed such that the above equation is true. If is equal to some , where , and are fixed, then we can derive from the following equation:
that there are infinite choices of for constant if and only if . While and imply , which is a contradiction to , hence varies with .

Since and are all fixed numbers for fixed , then implies that there are at least two different values of for fixed and infinite choices of ; it is a contradiction. So and follows. Hence for every .*Step **4. * for every .

After the discussion in Steps 1, 2, and 3, we already have the following equation:
where , for every . Since the map , then we can assume without loss of generality.

The proof in this step will be completed by induction on the following equation for arbitrary with for every :
where with .

When , (25) is equivalent to for arbitrary diagonal matrix and , with , since is homogeneous. The proof will be completed in the following and .(1) for every .

We already derive from Corollary 11 that for every , where is only decided by .

Suppose the map satisfies the following equation for every ,
then , and for arbitrary diagonal matrix and every , and .

Without loss of generality, we can assume for every and arbitrary .(2) Suppose for every , with ; then for every , with .

At first, we have to prove that for arbitrary nonzero and .

By the assumption, we already have the following equations:

Let , , and . Then the following statements are true
where and are -potent.

Let , , and , then , , and satisfy the assumption of Corollary 10. Hence we get and are orthogonal; that is,

Similarly, we can derive the following equation from Corollary 10:

Comparing the above two equations, we have , , , and , that is, .

We will prove . For arbitrary nonzero with , let , and ; then implies ; that is, the following matrix is -potent since by the assumption
by Lemma 5, . Hence we prove .

Now we prove .

By Corollary 11, we already have .

For arbitrary nonzero with ,