Research ArticleInduced Maps on Matrices over Fields

Li Yang1 Xuezhi Ben2 Ming Zhang2 and Chongguang Cao2

1 Department of Foundation Harbin Finance University Harbin 150030 China2 School of Mathematical Science Heilongjiang University Harbin 150080 China

Correspondence should be addressed to Chongguang Cao caochongguang163com

Received 21 October 2013 Revised 23 December 2013 Accepted 2 January 2014 Published 26 February 2014

Academic Editor Chi-Keung Ng

Copyright copy 2014 Li Yang et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Suppose that F is a field and119898 119899 ge 3 are integers Denote by119872119898119899(F) the set of all119898times119899matrices over F and by119872

119899(F) the set119872

119899119899(F)

Let119891119894119895(119894 isin [1119898] 119895 isin [1 119899]) be functions on F where [1 119899] stands for the set 1 119899We say that amap119891 119872

119898119899(F) rarr 119872

119898119899(F) is

induced by 119891119894119895 if 119891 is defined by119891 [119886

119894119895] 997891rarr [119891

119894119895(119886119894119895)] We say that a map 119891 on119872

119899(F) preserves similarity if119860 sim 119861 rArr 119891(119860) sim 119891(119861)

where 119860 sim 119861 represents that 119860 and 119861 are similar A map 119891 on119872119899(F) preserving inverses of matrices means 119891(119860)119891(119860minus1) = 119868

119899for

every invertible119860 isin 119872119899(F) In this paper we characterize inducedmaps preserving similarity and inverses ofmatrices respectively

1 Introduction

Suppose that F is a field and 119898 119899 ge 3 are integers Denoteby119872119898119899(F) the set of all119898 times 119899matrices over F and by119872

119899(F)

the set119872119899119899(F) Let 119891

119894119895(119894 isin [1119898] 119895 isin [1 119899]) be functions on

F where [1 119899] stands for the set 1 119899 We say that map119891 119872

119898119899(F) rarr 119872

119898119899(F) is induced by 119891

119894119895 if 119891 is defined by

119891 119860 = [119886119894119895] 997891997888rarr [119891

119894119895(119886119894119895)] for every 119860 isin 119872

119898119899(F) (1)

It is easy to see that inducedmapmaynot be linear or additive

Example 1 Let R be real field 11989111(119909) = sin119909 119891

12(119909) =

exp(119909) 11989121(119909) = tan(119909) and 119891

22(119909) = 119909

2

+ 1 then 119891

1198722(R) rarr 119872

2(R) induced by 119891

119894119895 is

119891([

119886 119887

119888 119889]) = [

sin 119886 exp 119887tan 119888 1198892 + 1] forall [

119886 119887

119888 119889] isin 119872

2(R) (2)

Example 2 The transpositionmap119883 997891rarr 119883119905 is not an induced

map on119872119899(F)

Example 3 Let 119875 isin 119872119899(F) be a matrix then 119891 119883 997891rarr 119875119883 is

an induced map on119872119899(F) if and only if 119875 is diagonal

If 119891119894119895is independent of the choices of 119894 and 119895 (ie 119891

119894119895equiv 120593

for every 119894 isin [1119898] and 119895 isin [1 119899]) then 119891 is said to be

induced by the function 120593 and denote by 119891(119860) = 119860120593

(=

[120593(119886119894119895)]) Denote by rank119860 the rank of matrix 119860 We say

that an induced map 119891 preserves rank-1 if rank119891(119860) = 1

whenever rank119860 = 1Preserver problem is a hot area in matrix and operator

algebra there are many results about this area Kalinowski[1] showed that an induced map 119891(Θ) = Θ

120593 where 120593 isa monotonic and continuous function of real field 119877 suchthat 119891(0) = 0 preserves ranks of matrices if and only ifit is linear Furthermore in [2] Kalinowski generalized theresults in [1] by removing any restrictions on themap120593 In [3]Liu and Zhang characterized the general form of all maps 119891induced by 119891

119894119895and preserving rank-1 matrices over a field In

particular nonlinearmaps preserving similarity were studiedby Du et al [4] One can see [5ndash15] and their references forsome background on preserver problems

We say that a map 119891 on119872119899(F) preserves similarity if

119860 sim 119861 997904rArr 119891 (119860) sim 119891 (119861) forall119860 119861 isin 119872119899(F) (3)

where119860 sim 119861 represents that119860 and 119861 are similar A map 119891 on119872119899(F) preserving inverses of matrices means 119891(119860)119891(119860minus1) =

119868119899for every invertible 119860 isin 119872

119899(F) In this paper we describe

the forms of induced map preserving similarity and inversesof matrices respectively

We end this section by introducing some notations whichwill be used in the following sections Let diag(119886

1 1198862 119886

119899)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 596756 5 pageshttpdxdoiorg1011552014596756

2 Abstract and Applied Analysis

be the diagonal matrix of order 119899 119864119894119895is the matrix with 1 in

the (119894 119895)th entry and 0 elsewhere and 119868119899is the identity matrix

of order 119899 Denote by oplus the usual direct sum of matrices

2 Induced Map PreservingSimilarity of Matrices

In this section we use the form of induced rank-1 preserverto describe forms of induced similarity preservers Firstly weneed the following theorem from [3]

Lemma 4 (see [3 Corollary 1]) Suppose that F is any fieldand 119899 ge 2 are integers Suppose that 119891 on 119872

119899(F) is induced

by 119891119894119895 such that 119891(0) = 0 Then 119891 preserves rank-1 if and

only if there exist invertible and diagonal 119875119876 isin 119872119899(F) and a

multiplicative map 120575 on F satifying 120575(119909) = 0 hArr 119909 = 0 suchthat

119891 (119860) = 119875119860120575

119876 forall119860 isin 119872119899(F) (4)

Lemma 5 Suppose that F is any field and 119899 is an integer with119899 ge 2 If 119860 isin 119872

119899(F) satisfies 1198602 = 0 and rank119860 = 119903 then

there exists an invertible matrix 119875 isin 119872119899(F) such that

119875minus1

119860119875 = [

0 119868119903

0 0] oplus 0 (5)

Proof It is easy to see that there exists an invertible matrix1198751isin 119872119899(F) such that

119860 = 1198751[

11986011198602

0 0]119875minus1

1 (6)

where 1198601

isin 119872119903(F) and 119860

2isin 119872

119903(119899minus119903)(F) satisfy

rank [11986011198602] = 119903 From 1198602 = 0 we have

1198601[11986011198602] = 0 (7)

so that 1198601= 0 Thus rank119860

2= 119903 which implies that

1198602119876 = [119868

1199030] (8)

for some invertible119876 isin 119872119899minus119903(F) Let 119875 = 119875

1(119868119903oplus119876) then (6)

turns into

119875minus1

119860119875 = [

0 119868119903

0 0] oplus 0 (9)

This completes the proof

Theorem 6 Let F be a field and let 119898 119899 ge 3 be positiveintegers Suppose that 119891 is a map on 119872

119899(F) induced by 119891

119894119895

such that 119891(0) = 0 Then 119891 preserves similarity if and only ifthere exist an invertible and diagonal 119875 isin 119872

119899(F) 119902 isin F and

an injective endomorphism 120575 of F such that

119891 (119860) = 119902119875119860120575

119875minus1

forall119860 isin 119872119899(F) (10)

Proof The sufficiency is obviousWe will prove the necessarypart by the following four steps

Step 1 If there exists some 119894 = 119895 and 119886 = 0 such that 119891119894119895(119886) = 0

then 119891 = 0

Proof of Step 1 For any 119887 = 0 119896 = 119897 since rank 119887119864119897119896= 1 and

(119887119864119897119896)2

= 0 by Lemma 5 we have

119887119864119897119896sim 11986412sim 119886119864119894119895 (11)

Since 119891 preserves similarity we derive

119891119896119897(119887) 119864119896119897sim 119891119894119895(119886) 119864119894119895= 0 (12)

and thus

119891119896119897(119887) = 0 forall119896 = 119897 isin [1 119899] 119887 isin F (13)

Because of rank (119887119864119896119896minus119887119864119897119897minus119887119864119896119897+119887119864119897119896) = 1 and (119887119864

119896119896minus

119887119864119897119897minus 119887119864119896119897+ 119887119864119897119896)2

= 0 by Lemma 5 we have

(119887119864119896119896minus 119887119864119897119897minus 119887119864119896119897+ 119887119864119897119896) sim 11986412sim 119886119864119894119895 (14)

Using 119891 preserves similarity and (13) one can obtain that

119891119896119896(119887) 119864119896119896+ 119891119897119897(minus119887) 119864

119897119897= 0 (15)

hence

119891119896119896(119887) = 0 forall119896 isin [1 119899] 119887 isin F (16)

It follows from (13) and (16) that 119891119894119895= 0 that is 119891 = 0

Step 2 If there exist some 119894 and 119886 = 0 such that119891119894119894(119886) = 0 then

119891 = 0

Proof of Step 2 For 119895 = 119894 it follows from 119886119864119894119894sim 119886119864

119895119895that

119891119894119894(119886)119864119894119894sim 119891119895119895(119886)119864119895119895 Thus

119891119895119895(119886) = 0 forall119895 isin [1 119899] 119886 isin F (17)

Because of

(119886119864119894119894minus 119886119864119895119895minus 119886119864119894119895+ 119886119864119895119894) sim 11986412 (18)

one can obtain by using (17) that (119891119894119895(119886)119864119894119895+ 119891119895119894(minus119886)119864

119894119895) sim

11989112(1)11986412 and hence

rank (119891119894119895(119886) 119864119894119895+ 119891119895119894(minus119886) 119864

119894119895) = rank119891

12(1) 11986412le 1 (19)

Thus 119891119894119895(119886) = 0 or 119891

119895119894(minus119886) = 0 We complete the proof of this

step by using the result of Step 1

Step 3 If 119891 = 0 then 119891 preserves rank-1

Proof of Step 3 For any rank-1 matrix 119860 we have

119860 sim

[

[

[

[

[

11988611198862sdot sdot sdot 119886119899

0 0 sdot sdot sdot 0

0 0 sdot sdot sdot 0

]

]

]

]

]

(20)

and hence

119891 (119860) sim

[

[

[

[

[

11989111(1198861) 11989112(1198862) sdot sdot sdot 119891

1119899(119886119899)

0 0 sdot sdot sdot 0

0 0 sdot sdot sdot 0

]

]

]

]

]

(21)

Thus rank119891(119860) le 1 it follows from119891 = 0 that rank119891(119860) = 1

Abstract and Applied Analysis 3

Step 4 If119891 = 0 then there exist an invertible and diagonal119875 isin119872119899(F) 0 = 119902 isin F and an injective endomorphism 120575 of F such

that

119891 (119860) = 119902119875119860120575

119875minus1

forall119860 isin 119872119899(F) (22)

Proof of Step 4 Since 119891 = 0 by Step 3 and Lemma 4 thereexist invertible and diagonal 119875119876 isin 119872

119899(F) and a multiplica-

tive map 120575 on F satisfying 120575(119909) = 0 hArr 119909 = 0 such that

119891 (119860) = 119875119860120575

(119876119875) 119875minus1

forall119860 isin 119872119899(F) (23)

Let

119860 =[

[

1 1 119909 + 119910

0 1 119909

1 0 119910

]

]

oplus 0

119861 =[

[

1 1 minus 119909 minus 119910 119909 + 119910

0 1 minus 119909 119909

1 1 minus 119909 minus 119910 119909 + 119910

]

]

oplus 0

(24)

It is easy to see that 119861 = (119868119899+ 11986432)119860(119868119899+ 11986432)minus1 Since 119891

preserves similarity we have that 119891(119860) and 119891(119861) are similarfurther rank119891(119860) = rank119891(119861) It follows from (23) thatrank119891(119861) le 2 thus

rank[

[

120575 (1) 120575 (1) 120575 (119909 + 119910)

0 120575 (1) 120575 (119909)

120575 (1) 0 120575 (119910)

]

]

oplus 0 le 2 (25)

This implies 120575(119909 + 119910) = 120575(119909) + 120575(119910) hence 120575 is an injectiveendomorphism of F

Set 119876119875 = diag(1199021 119902

119899) Since 119864

119894119894sim 119864119895119895 one obtains by

using (23) that

119902119894119864119894119894sim 119902119895119864119895119895 (26)

Thus 119902119894= 119902119895 Letting 119902 = 119902

1 then 119876119875 = 119902119868

119899and 119891(119860) =

119902119875119860120575

119875minus1

This completes the proof of Theorem

3 Induced Map PreservingInverses of Matrices

Theorem 7 Let F be a field and let 119898 119899 ge 3 be positiveintegers Suppose that 119891 is a map on 119872

119899(F) induced by 119891

119894119895

such that 119891(0) = 0 Then 119891 preserves inverses of matrices ifand only if there exist an invertible and diagonal 119875 isin 119872

119899(F)

119888 isin minus1 1 and an injective endomorphism 120593 of F such that

119891 (119860) = 119888119875119860120593

119875minus1

forall119860 isin 119872119899(F) (27)

Proof The sufficiency is obviousWe will prove the necessarypart For any 119894 = 119895 isin [1 119899] 119886 isin F and 119887 isin Flowast since

[119864119894119894minus 119886119864119894119895+ 119864119895119894+ (119887minus1

minus 119886) 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= (1 minus 119886119887) 119864119894119894+ 119886119887119864

119894119895minus 119887119864119895119894+ 119887119864119895119895+ Σ119896 = 119894119895

119864119896119896

(28)

by 119891 preserving inverses of matrices we have

[

119891119894119894(1) 119891

119894119895(minus119886)

119891119895119894(1) 119891

119895119895(119887minus1

minus 119886)

] [

119891119894119894(1 minus 119886119887) 119891

119894119895(119886119887)

119891119895119894(minus119887) 119891

119895119895(119887)] = 1198682 (29)

so that

119891119894119894(1) 119891119894119895(119886119887) + 119891

119894119895(minus119886) 119891

119895119895(119887) = 0 (30)

119891119894119894(1) 119891119894119894(1 minus 119886119887) + 119891

119894119895(minus119886) 119891

119895119894(minus119887) = 1 (31)

Let 119887 = 1 then (30) turns into

119891119894119894(1) 119891119894119895(119886) + 119891

119894119895(minus119886) 119891

119895119895(1) = 0 (32)

Replacing 119886 by 119886119887 then the above turns into

119891119894119894(1) 119891119894119895(119886119887) + 119891

119894119895(minus119886119887) 119891

119895119895(1) = 0 (33)

It follows from (30) and (33) that

119891119894119895(minus119886119887) 119891

119895119895(1) = 119891

119894119895(minus119886) 119891

119895119895(119887) (34)

Replacing minus119886 by 119886 then the above turns into

119891119894119895(119886119887) 119891

119895119895(1) = 119891

119894119895(119886) 119891119895119895(119887) (35)

By [119886119864119894119894+ Σ119896 = 119894119864119896119896]minus1

= 119886minus1

119864119894119894+ Σ119896 = 119894119864119896119896 we have

119891119894119894(119886) 119891119894119894(119886minus1

) = 1 (36)

In particular

119891119894119894(1)2

= 1 (37)

From [119886119864119894119895+ 119886119864119895119894+ Σ119896 = 119894119895

119864119896119896]minus1

= 119886minus1

119864119894119895+ 119886minus1

119864119895119894+

Σ119896 = 119894119895

119864119896119896 we have

119891119894119895(119886) 119891119895119894(119886minus1

) = 1 (38)

In particular

119891119894119895(1) 119891119895119894(1) = 1 (39)

Multiplying119891119894119895(minus119887minus1

) by (31) we obtain by using (38) that

119891119894119894(1) 119891119894119894(1 minus 119886119887) 119891

119894119895(minus119887minus1

) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (40)

It follows from (35) and (38) that

119891119894119894(1) 119891119894119895((1 minus 119886119887) (minus119887

minus1

)) = 119891119894119894(1 minus 119886119887) 119891

119894119895(minus119887minus1

) (41)

This together with (40) implies that

119891119894119894(1) 119891119894119894(1) 119891119894119895(minus119887minus1

+ 119886) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (42)

Hence it follows from 119891119894119894(1)2

= 1 that

119891119894119895(minus119887minus1

+ 119886) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (43)

4 Abstract and Applied Analysis

Let 119909 = minus119887minus1 and 119910 = minus119886 we have

119891119894119895(119909 minus 119910) = 119891

119894119895(119909) minus 119891

119894119895(119910) forall119909 isin F

lowast

119910 isin F (44)

From

[119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896

(45)

we have

[

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] [

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] = 1198682 (46)

so that

119891119894119894(1) 119891119894119895(1) = minus119891

119894119895(1) 119891119895119895(minus1) (47)

This together with (39) implies

119891119894119894(1) = minus119891

119895119895(minus1) (48)

Similarly 119891119896119896(1) = minus119891

119895119895(minus1) hence

119891119894119894(1) = 119891

119896119896(1) forall119894 119896 isin [1 119899] (49)

It follows from

[119864119894119894+ 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896 forall119909 isin F

(50)

that

[

119891119894119894(1) 119891

119894119895(119909)

0 119891119895119895(1)] [

119891119894119894(1) 119891

119894119895(minus119909)

0 119891119895119895(1)] = 1198682 (51)

Hence

119891119894119894(1) 119891119894119895(minus119909) = minus119891

119894119895(119909) 119891119895119895(1) (52)

This together with (39) and (49) implies

119891119894119895(minus119909) = minus119891

119894119895(119909) forall119909 isin F (53)

Since

[119886119864119894119894+ 119864119894119895minus 119864119895119894+ Σ119896 = 119894119895

119864119896119896]

minus1

= minus119864119894119895+ 119864119895119894+ 119886119864119895119895+ Σ119896 = 119894119895

119864119896119896

(54)

we have

[

119891119894119894(119886) 119891

119894119895(1)

119891119895119894(minus1) 0

] [

0 119891119894119895(minus1)

119891119895119894(1) 119891

119895119895(119886)] = 1198682 (55)

so that

119891119894119894(119886) 119891119894119895(minus1) = minus119891

119894119895(1) 119891119895119895(119886) (56)

Hence

119891119894119894(119886) = 119891

119895119895(119886) (57)

For distinct 119894 119895 119896 isin [1 119899] and 119909 119910 isin F since

[119864119894119894+ 119909119864119894119895+ 119909119864119894119896+ 119864119895119895+ 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895minus 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897

(58)

we have

[

[

119891119894119894(1) 119891

119894119895(119909) 119891

119894119896(119909119910)

0 119891119895119895(1) 119891

119895119896(119910)

0 0 119891119896119896(1)

]

]

[

[

119891119894119894(1) 119891

119894119895(minus119909) 0

0 119891119895119895(1) 119891

119895119896(minus119910)

0 0 119891119896119896(1)

]

]

= 1198683

(59)

so that

119891119894119895(119909) 119891119895119896(minus119910) + 119891

119894119896(119909119910) 119891

119896119896(1) = 0 (60)

This together with (53) implies that

119891119894119895(119909) = 119891

119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

(61)

It follows from (35) and (61) that

119891119894119895(119909)

= 119891119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

= 119891119894119896(119909) 11989111(1) 119891119895119896(1)minus1

= (119891119896119896(1)minus1

1198911198941(1) 11989111(119909))

times 11989111(1) (119891

119896119896(1)minus1

1198911198951(1) 11989111(1))

minus1

= 1198911198941(1) 11989111(119909) 1198911198951(1)minus1

forall119894 = 119895 isin [1 119899]

(62)

Using this together with (57) we obtain

119891 (119860) = 119875 [11989111(119886119894119895)] 119875minus1

(63)

where 119875 = diag(11989111(1) 119891

1198991(1)) Let 119888 = 119891

11(1) isin minus1 1

120593(119909) = 11989111(1)11989111(119909) and then

119891 (119860) = 119888119875 [120593 (119886119894119895)] 119875minus1

= 119888119875119860120593

119875minus1

(64)

Let 120601(119860) = [120593(119886119894119895)] since 119891 preserves inverses of matrices

one can see that 120601 also preserves inverses of matrices By120593(1) = 119891

11(1)2

= 1 we obtain by using similar method to(35) (37) (44) and (53) that for any 119886 119887 isin F

120593 (119886119887) = 120593 (1) 120593 (119886119887) = 120593 (119886) 120593 (119887)

120593 (119886 + 119887) = 120593 (119886) + 120593 (119887)

(65)

This completes the proof

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 5

Acknowledgments

The authors show great thanks to the referee for hishercareful reading of the paper and valuable comments whichgreatly improved the readability of the paper Li Yang issupported by Vocational education institute in HeilongjiangProvince ldquo12th five-year development planrdquo The GuidingFunction and Practice Research of Mathematical modelingin Advanced Mathematics Teaching of New Rise FinancialInstitutionsrsquo (Grant no GG0666) Chongguang Cao is sup-ported by National Natural Science Foundation Grants ofChina (Grant no 11371109)

References

[1] J Kalinowski ldquoOn rank equivalence and rank preservingoperatorsrdquo Novi Sad Journal of Mathematics vol 32 no 1 pp133ndash139 2002

[2] J Kalinowski ldquoOn functions preserving rank of matricesrdquoMathematical Notes A Publication of the University of Miskolcvol 4 no 1 pp 35ndash37 2003

[3] S-W Liu and G-D Zhang ldquoMaps preserving rank 1 matricesover fieldsrdquo Journal ofNatural Science ofHeilongjiangUniversityvol 23 no 1 pp 138ndash140 2006

[4] S Du J Hou and Z Bai ldquoNonlinearmaps preserving similarityon 119861(119867)rdquo Linear Algebra and its Applications vol 422 no 2-3pp 506ndash516 2007

[5] T Oikhberg and A M Peralta ldquoAutomatic continuity oforthogonality preservers on a non-commutative 119871119901(120591) spacerdquoJournal of Functional Analysis vol 264 no 8 pp 1848ndash18722013

[6] M A Chebotar W-F Ke P-H Lee and N-C Wong ldquoMap-pings preserving zero productsrdquo Studia Mathematica vol 155no 1 pp 77ndash94 2003

[7] C-W Leung C-K Ng and N-C Wong ldquoLinear orthogonalitypreservers of Hilbert Clowast-modules over Clowast-algebras with realrank zerordquo Proceedings of the American Mathematical Societyvol 140 no 9 pp 3151ndash3160 2012

[8] I V Konnov and J C Yao ldquoOn the generalized vector vari-ational inequality problemrdquo Journal of Mathematical Analysisand Applications vol 206 no 1 pp 42ndash58 1997

[9] D Huo B Zheng and H Liu ldquoCharacterizations of nonlinearLie derivations of 119861(119883)rdquo Abstract and Applied Analysis vol2013 Article ID 245452 7 pages 2013

[10] H Yao and B Zheng ldquoZero triple product determined matrixalgebrasrdquo Journal of Applied Mathematics vol 2012 Article ID925092 18 pages 2012

[11] J Xu B Zheng and H Yao ldquoLinear transformations betweenmultipartite quantum systems that map the set of tensorproduct of idempotent matrices into idempotent matrix setrdquoJournal of Function Spaces and Applications vol 2013 ArticleID 182569 7 pages 2013

[12] J Xu B Zheng and L Yang ldquoThe automorphism group of theLie ring of real skew-symmetric matricesrdquo Abstract and AppliedAnalysis vol 2013 Article ID 638230 8 pages 2013

[13] X Song C Cao and B Zheng ldquoA note on 119896-potence preserverson matrix spaces over complex fieldrdquo Abstract and AppliedAnalysis vol 2013 Article ID 581683 8 pages 2013

[14] B Zheng J Xu and A Fosner ldquoLinear maps preserving rankof tensor products of matricesrdquo Linear and Multilinear Algebra2014

[15] L Yang W Zhang and J L Xu ldquoLinear maps on upper trian-gular matrices spaces preserving idempotent tensor productsrdquoAbstract and Applied Analysis vol 2014 Article ID 148321 8pages 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

be the diagonal matrix of order 119899 119864119894119895is the matrix with 1 in

the (119894 119895)th entry and 0 elsewhere and 119868119899is the identity matrix

of order 119899 Denote by oplus the usual direct sum of matrices

2 Induced Map PreservingSimilarity of Matrices

In this section we use the form of induced rank-1 preserverto describe forms of induced similarity preservers Firstly weneed the following theorem from [3]

Lemma 4 (see [3 Corollary 1]) Suppose that F is any fieldand 119899 ge 2 are integers Suppose that 119891 on 119872

119899(F) is induced

by 119891119894119895 such that 119891(0) = 0 Then 119891 preserves rank-1 if and

only if there exist invertible and diagonal 119875119876 isin 119872119899(F) and a

multiplicative map 120575 on F satifying 120575(119909) = 0 hArr 119909 = 0 suchthat

119891 (119860) = 119875119860120575

119876 forall119860 isin 119872119899(F) (4)

Lemma 5 Suppose that F is any field and 119899 is an integer with119899 ge 2 If 119860 isin 119872

119899(F) satisfies 1198602 = 0 and rank119860 = 119903 then

there exists an invertible matrix 119875 isin 119872119899(F) such that

119875minus1

119860119875 = [

0 119868119903

0 0] oplus 0 (5)

Proof It is easy to see that there exists an invertible matrix1198751isin 119872119899(F) such that

119860 = 1198751[

11986011198602

0 0]119875minus1

1 (6)

where 1198601

isin 119872119903(F) and 119860

2isin 119872

119903(119899minus119903)(F) satisfy

rank [11986011198602] = 119903 From 1198602 = 0 we have

1198601[11986011198602] = 0 (7)

so that 1198601= 0 Thus rank119860

2= 119903 which implies that

1198602119876 = [119868

1199030] (8)

for some invertible119876 isin 119872119899minus119903(F) Let 119875 = 119875

1(119868119903oplus119876) then (6)

turns into

119875minus1

119860119875 = [

0 119868119903

0 0] oplus 0 (9)

This completes the proof

Theorem 6 Let F be a field and let 119898 119899 ge 3 be positiveintegers Suppose that 119891 is a map on 119872

119899(F) induced by 119891

119894119895

such that 119891(0) = 0 Then 119891 preserves similarity if and only ifthere exist an invertible and diagonal 119875 isin 119872

119899(F) 119902 isin F and

an injective endomorphism 120575 of F such that

119891 (119860) = 119902119875119860120575

119875minus1

forall119860 isin 119872119899(F) (10)

Proof The sufficiency is obviousWe will prove the necessarypart by the following four steps

Step 1 If there exists some 119894 = 119895 and 119886 = 0 such that 119891119894119895(119886) = 0

then 119891 = 0

Proof of Step 1 For any 119887 = 0 119896 = 119897 since rank 119887119864119897119896= 1 and

(119887119864119897119896)2

= 0 by Lemma 5 we have

119887119864119897119896sim 11986412sim 119886119864119894119895 (11)

Since 119891 preserves similarity we derive

119891119896119897(119887) 119864119896119897sim 119891119894119895(119886) 119864119894119895= 0 (12)

and thus

119891119896119897(119887) = 0 forall119896 = 119897 isin [1 119899] 119887 isin F (13)

Because of rank (119887119864119896119896minus119887119864119897119897minus119887119864119896119897+119887119864119897119896) = 1 and (119887119864

119896119896minus

119887119864119897119897minus 119887119864119896119897+ 119887119864119897119896)2

= 0 by Lemma 5 we have

(119887119864119896119896minus 119887119864119897119897minus 119887119864119896119897+ 119887119864119897119896) sim 11986412sim 119886119864119894119895 (14)

Using 119891 preserves similarity and (13) one can obtain that

119891119896119896(119887) 119864119896119896+ 119891119897119897(minus119887) 119864

119897119897= 0 (15)

hence

119891119896119896(119887) = 0 forall119896 isin [1 119899] 119887 isin F (16)

It follows from (13) and (16) that 119891119894119895= 0 that is 119891 = 0

Step 2 If there exist some 119894 and 119886 = 0 such that119891119894119894(119886) = 0 then

119891 = 0

Proof of Step 2 For 119895 = 119894 it follows from 119886119864119894119894sim 119886119864

119895119895that

119891119894119894(119886)119864119894119894sim 119891119895119895(119886)119864119895119895 Thus

119891119895119895(119886) = 0 forall119895 isin [1 119899] 119886 isin F (17)

Because of

(119886119864119894119894minus 119886119864119895119895minus 119886119864119894119895+ 119886119864119895119894) sim 11986412 (18)

one can obtain by using (17) that (119891119894119895(119886)119864119894119895+ 119891119895119894(minus119886)119864

119894119895) sim

11989112(1)11986412 and hence

rank (119891119894119895(119886) 119864119894119895+ 119891119895119894(minus119886) 119864

119894119895) = rank119891

12(1) 11986412le 1 (19)

Thus 119891119894119895(119886) = 0 or 119891

119895119894(minus119886) = 0 We complete the proof of this

step by using the result of Step 1

Step 3 If 119891 = 0 then 119891 preserves rank-1

Proof of Step 3 For any rank-1 matrix 119860 we have

119860 sim

[

[

[

[

[

11988611198862sdot sdot sdot 119886119899

0 0 sdot sdot sdot 0

0 0 sdot sdot sdot 0

]

]

]

]

]

(20)

and hence

119891 (119860) sim

[

[

[

[

[

11989111(1198861) 11989112(1198862) sdot sdot sdot 119891

1119899(119886119899)

0 0 sdot sdot sdot 0

0 0 sdot sdot sdot 0

]

]

]

]

]

(21)

Thus rank119891(119860) le 1 it follows from119891 = 0 that rank119891(119860) = 1

Abstract and Applied Analysis 3

Step 4 If119891 = 0 then there exist an invertible and diagonal119875 isin119872119899(F) 0 = 119902 isin F and an injective endomorphism 120575 of F such

that

119891 (119860) = 119902119875119860120575

119875minus1

forall119860 isin 119872119899(F) (22)

Proof of Step 4 Since 119891 = 0 by Step 3 and Lemma 4 thereexist invertible and diagonal 119875119876 isin 119872

119899(F) and a multiplica-

tive map 120575 on F satisfying 120575(119909) = 0 hArr 119909 = 0 such that

119891 (119860) = 119875119860120575

(119876119875) 119875minus1

forall119860 isin 119872119899(F) (23)

Let

119860 =[

[

1 1 119909 + 119910

0 1 119909

1 0 119910

]

]

oplus 0

119861 =[

[

1 1 minus 119909 minus 119910 119909 + 119910

0 1 minus 119909 119909

1 1 minus 119909 minus 119910 119909 + 119910

]

]

oplus 0

(24)

It is easy to see that 119861 = (119868119899+ 11986432)119860(119868119899+ 11986432)minus1 Since 119891

preserves similarity we have that 119891(119860) and 119891(119861) are similarfurther rank119891(119860) = rank119891(119861) It follows from (23) thatrank119891(119861) le 2 thus

rank[

[

120575 (1) 120575 (1) 120575 (119909 + 119910)

0 120575 (1) 120575 (119909)

120575 (1) 0 120575 (119910)

]

]

oplus 0 le 2 (25)

This implies 120575(119909 + 119910) = 120575(119909) + 120575(119910) hence 120575 is an injectiveendomorphism of F

Set 119876119875 = diag(1199021 119902

119899) Since 119864

119894119894sim 119864119895119895 one obtains by

using (23) that

119902119894119864119894119894sim 119902119895119864119895119895 (26)

Thus 119902119894= 119902119895 Letting 119902 = 119902

1 then 119876119875 = 119902119868

119899and 119891(119860) =

119902119875119860120575

119875minus1

This completes the proof of Theorem

3 Induced Map PreservingInverses of Matrices

Theorem 7 Let F be a field and let 119898 119899 ge 3 be positiveintegers Suppose that 119891 is a map on 119872

119899(F) induced by 119891

119894119895

such that 119891(0) = 0 Then 119891 preserves inverses of matrices ifand only if there exist an invertible and diagonal 119875 isin 119872

119899(F)

119888 isin minus1 1 and an injective endomorphism 120593 of F such that

119891 (119860) = 119888119875119860120593

119875minus1

forall119860 isin 119872119899(F) (27)

Proof The sufficiency is obviousWe will prove the necessarypart For any 119894 = 119895 isin [1 119899] 119886 isin F and 119887 isin Flowast since

[119864119894119894minus 119886119864119894119895+ 119864119895119894+ (119887minus1

minus 119886) 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= (1 minus 119886119887) 119864119894119894+ 119886119887119864

119894119895minus 119887119864119895119894+ 119887119864119895119895+ Σ119896 = 119894119895

119864119896119896

(28)

by 119891 preserving inverses of matrices we have

[

119891119894119894(1) 119891

119894119895(minus119886)

119891119895119894(1) 119891

119895119895(119887minus1

minus 119886)

] [

119891119894119894(1 minus 119886119887) 119891

119894119895(119886119887)

119891119895119894(minus119887) 119891

119895119895(119887)] = 1198682 (29)

so that

119891119894119894(1) 119891119894119895(119886119887) + 119891

119894119895(minus119886) 119891

119895119895(119887) = 0 (30)

119891119894119894(1) 119891119894119894(1 minus 119886119887) + 119891

119894119895(minus119886) 119891

119895119894(minus119887) = 1 (31)

Let 119887 = 1 then (30) turns into

119891119894119894(1) 119891119894119895(119886) + 119891

119894119895(minus119886) 119891

119895119895(1) = 0 (32)

Replacing 119886 by 119886119887 then the above turns into

119891119894119894(1) 119891119894119895(119886119887) + 119891

119894119895(minus119886119887) 119891

119895119895(1) = 0 (33)

It follows from (30) and (33) that

119891119894119895(minus119886119887) 119891

119895119895(1) = 119891

119894119895(minus119886) 119891

119895119895(119887) (34)

Replacing minus119886 by 119886 then the above turns into

119891119894119895(119886119887) 119891

119895119895(1) = 119891

119894119895(119886) 119891119895119895(119887) (35)

By [119886119864119894119894+ Σ119896 = 119894119864119896119896]minus1

= 119886minus1

119864119894119894+ Σ119896 = 119894119864119896119896 we have

119891119894119894(119886) 119891119894119894(119886minus1

) = 1 (36)

In particular

119891119894119894(1)2

= 1 (37)

From [119886119864119894119895+ 119886119864119895119894+ Σ119896 = 119894119895

119864119896119896]minus1

= 119886minus1

119864119894119895+ 119886minus1

119864119895119894+

Σ119896 = 119894119895

119864119896119896 we have

119891119894119895(119886) 119891119895119894(119886minus1

) = 1 (38)

In particular

119891119894119895(1) 119891119895119894(1) = 1 (39)

Multiplying119891119894119895(minus119887minus1

) by (31) we obtain by using (38) that

119891119894119894(1) 119891119894119894(1 minus 119886119887) 119891

119894119895(minus119887minus1

) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (40)

It follows from (35) and (38) that

119891119894119894(1) 119891119894119895((1 minus 119886119887) (minus119887

minus1

)) = 119891119894119894(1 minus 119886119887) 119891

119894119895(minus119887minus1

) (41)

This together with (40) implies that

119891119894119894(1) 119891119894119894(1) 119891119894119895(minus119887minus1

+ 119886) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (42)

Hence it follows from 119891119894119894(1)2

= 1 that

119891119894119895(minus119887minus1

+ 119886) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (43)

4 Abstract and Applied Analysis

Let 119909 = minus119887minus1 and 119910 = minus119886 we have

119891119894119895(119909 minus 119910) = 119891

119894119895(119909) minus 119891

119894119895(119910) forall119909 isin F

lowast

119910 isin F (44)

From

[119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896

(45)

we have

[

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] [

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] = 1198682 (46)

so that

119891119894119894(1) 119891119894119895(1) = minus119891

119894119895(1) 119891119895119895(minus1) (47)

This together with (39) implies

119891119894119894(1) = minus119891

119895119895(minus1) (48)

Similarly 119891119896119896(1) = minus119891

119895119895(minus1) hence

119891119894119894(1) = 119891

119896119896(1) forall119894 119896 isin [1 119899] (49)

It follows from

[119864119894119894+ 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896 forall119909 isin F

(50)

that

[

119891119894119894(1) 119891

119894119895(119909)

0 119891119895119895(1)] [

119891119894119894(1) 119891

119894119895(minus119909)

0 119891119895119895(1)] = 1198682 (51)

Hence

119891119894119894(1) 119891119894119895(minus119909) = minus119891

119894119895(119909) 119891119895119895(1) (52)

This together with (39) and (49) implies

119891119894119895(minus119909) = minus119891

119894119895(119909) forall119909 isin F (53)

Since

[119886119864119894119894+ 119864119894119895minus 119864119895119894+ Σ119896 = 119894119895

119864119896119896]

minus1

= minus119864119894119895+ 119864119895119894+ 119886119864119895119895+ Σ119896 = 119894119895

119864119896119896

(54)

we have

[

119891119894119894(119886) 119891

119894119895(1)

119891119895119894(minus1) 0

] [

0 119891119894119895(minus1)

119891119895119894(1) 119891

119895119895(119886)] = 1198682 (55)

so that

119891119894119894(119886) 119891119894119895(minus1) = minus119891

119894119895(1) 119891119895119895(119886) (56)

Hence

119891119894119894(119886) = 119891

119895119895(119886) (57)

For distinct 119894 119895 119896 isin [1 119899] and 119909 119910 isin F since

[119864119894119894+ 119909119864119894119895+ 119909119864119894119896+ 119864119895119895+ 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895minus 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897

(58)

we have

[

[

119891119894119894(1) 119891

119894119895(119909) 119891

119894119896(119909119910)

0 119891119895119895(1) 119891

119895119896(119910)

0 0 119891119896119896(1)

]

]

[

[

119891119894119894(1) 119891

119894119895(minus119909) 0

0 119891119895119895(1) 119891

119895119896(minus119910)

0 0 119891119896119896(1)

]

]

= 1198683

(59)

so that

119891119894119895(119909) 119891119895119896(minus119910) + 119891

119894119896(119909119910) 119891

119896119896(1) = 0 (60)

This together with (53) implies that

119891119894119895(119909) = 119891

119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

(61)

It follows from (35) and (61) that

119891119894119895(119909)

= 119891119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

= 119891119894119896(119909) 11989111(1) 119891119895119896(1)minus1

= (119891119896119896(1)minus1

1198911198941(1) 11989111(119909))

times 11989111(1) (119891

119896119896(1)minus1

1198911198951(1) 11989111(1))

minus1

= 1198911198941(1) 11989111(119909) 1198911198951(1)minus1

forall119894 = 119895 isin [1 119899]

(62)

Using this together with (57) we obtain

119891 (119860) = 119875 [11989111(119886119894119895)] 119875minus1

(63)

where 119875 = diag(11989111(1) 119891

1198991(1)) Let 119888 = 119891

11(1) isin minus1 1

120593(119909) = 11989111(1)11989111(119909) and then

119891 (119860) = 119888119875 [120593 (119886119894119895)] 119875minus1

= 119888119875119860120593

119875minus1

(64)

Let 120601(119860) = [120593(119886119894119895)] since 119891 preserves inverses of matrices

one can see that 120601 also preserves inverses of matrices By120593(1) = 119891

11(1)2

= 1 we obtain by using similar method to(35) (37) (44) and (53) that for any 119886 119887 isin F

120593 (119886119887) = 120593 (1) 120593 (119886119887) = 120593 (119886) 120593 (119887)

120593 (119886 + 119887) = 120593 (119886) + 120593 (119887)

(65)

This completes the proof

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 5

Acknowledgments

The authors show great thanks to the referee for hishercareful reading of the paper and valuable comments whichgreatly improved the readability of the paper Li Yang issupported by Vocational education institute in HeilongjiangProvince ldquo12th five-year development planrdquo The GuidingFunction and Practice Research of Mathematical modelingin Advanced Mathematics Teaching of New Rise FinancialInstitutionsrsquo (Grant no GG0666) Chongguang Cao is sup-ported by National Natural Science Foundation Grants ofChina (Grant no 11371109)

References

[1] J Kalinowski ldquoOn rank equivalence and rank preservingoperatorsrdquo Novi Sad Journal of Mathematics vol 32 no 1 pp133ndash139 2002

[2] J Kalinowski ldquoOn functions preserving rank of matricesrdquoMathematical Notes A Publication of the University of Miskolcvol 4 no 1 pp 35ndash37 2003

[3] S-W Liu and G-D Zhang ldquoMaps preserving rank 1 matricesover fieldsrdquo Journal ofNatural Science ofHeilongjiangUniversityvol 23 no 1 pp 138ndash140 2006

[4] S Du J Hou and Z Bai ldquoNonlinearmaps preserving similarityon 119861(119867)rdquo Linear Algebra and its Applications vol 422 no 2-3pp 506ndash516 2007

[5] T Oikhberg and A M Peralta ldquoAutomatic continuity oforthogonality preservers on a non-commutative 119871119901(120591) spacerdquoJournal of Functional Analysis vol 264 no 8 pp 1848ndash18722013

[6] M A Chebotar W-F Ke P-H Lee and N-C Wong ldquoMap-pings preserving zero productsrdquo Studia Mathematica vol 155no 1 pp 77ndash94 2003

[7] C-W Leung C-K Ng and N-C Wong ldquoLinear orthogonalitypreservers of Hilbert Clowast-modules over Clowast-algebras with realrank zerordquo Proceedings of the American Mathematical Societyvol 140 no 9 pp 3151ndash3160 2012

[8] I V Konnov and J C Yao ldquoOn the generalized vector vari-ational inequality problemrdquo Journal of Mathematical Analysisand Applications vol 206 no 1 pp 42ndash58 1997

[9] D Huo B Zheng and H Liu ldquoCharacterizations of nonlinearLie derivations of 119861(119883)rdquo Abstract and Applied Analysis vol2013 Article ID 245452 7 pages 2013

[10] H Yao and B Zheng ldquoZero triple product determined matrixalgebrasrdquo Journal of Applied Mathematics vol 2012 Article ID925092 18 pages 2012

[11] J Xu B Zheng and H Yao ldquoLinear transformations betweenmultipartite quantum systems that map the set of tensorproduct of idempotent matrices into idempotent matrix setrdquoJournal of Function Spaces and Applications vol 2013 ArticleID 182569 7 pages 2013

[12] J Xu B Zheng and L Yang ldquoThe automorphism group of theLie ring of real skew-symmetric matricesrdquo Abstract and AppliedAnalysis vol 2013 Article ID 638230 8 pages 2013

[13] X Song C Cao and B Zheng ldquoA note on 119896-potence preserverson matrix spaces over complex fieldrdquo Abstract and AppliedAnalysis vol 2013 Article ID 581683 8 pages 2013

[14] B Zheng J Xu and A Fosner ldquoLinear maps preserving rankof tensor products of matricesrdquo Linear and Multilinear Algebra2014

[15] L Yang W Zhang and J L Xu ldquoLinear maps on upper trian-gular matrices spaces preserving idempotent tensor productsrdquoAbstract and Applied Analysis vol 2014 Article ID 148321 8pages 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

Step 4 If119891 = 0 then there exist an invertible and diagonal119875 isin119872119899(F) 0 = 119902 isin F and an injective endomorphism 120575 of F such

that

119891 (119860) = 119902119875119860120575

119875minus1

forall119860 isin 119872119899(F) (22)

Proof of Step 4 Since 119891 = 0 by Step 3 and Lemma 4 thereexist invertible and diagonal 119875119876 isin 119872

119899(F) and a multiplica-

tive map 120575 on F satisfying 120575(119909) = 0 hArr 119909 = 0 such that

119891 (119860) = 119875119860120575

(119876119875) 119875minus1

forall119860 isin 119872119899(F) (23)

Let

119860 =[

[

1 1 119909 + 119910

0 1 119909

1 0 119910

]

]

oplus 0

119861 =[

[

1 1 minus 119909 minus 119910 119909 + 119910

0 1 minus 119909 119909

1 1 minus 119909 minus 119910 119909 + 119910

]

]

oplus 0

(24)

It is easy to see that 119861 = (119868119899+ 11986432)119860(119868119899+ 11986432)minus1 Since 119891

preserves similarity we have that 119891(119860) and 119891(119861) are similarfurther rank119891(119860) = rank119891(119861) It follows from (23) thatrank119891(119861) le 2 thus

rank[

[

120575 (1) 120575 (1) 120575 (119909 + 119910)

0 120575 (1) 120575 (119909)

120575 (1) 0 120575 (119910)

]

]

oplus 0 le 2 (25)

This implies 120575(119909 + 119910) = 120575(119909) + 120575(119910) hence 120575 is an injectiveendomorphism of F

Set 119876119875 = diag(1199021 119902

119899) Since 119864

119894119894sim 119864119895119895 one obtains by

using (23) that

119902119894119864119894119894sim 119902119895119864119895119895 (26)

Thus 119902119894= 119902119895 Letting 119902 = 119902

1 then 119876119875 = 119902119868

119899and 119891(119860) =

119902119875119860120575

119875minus1

This completes the proof of Theorem

3 Induced Map PreservingInverses of Matrices

Theorem 7 Let F be a field and let 119898 119899 ge 3 be positiveintegers Suppose that 119891 is a map on 119872

119899(F) induced by 119891

119894119895

such that 119891(0) = 0 Then 119891 preserves inverses of matrices ifand only if there exist an invertible and diagonal 119875 isin 119872

119899(F)

119888 isin minus1 1 and an injective endomorphism 120593 of F such that

119891 (119860) = 119888119875119860120593

119875minus1

forall119860 isin 119872119899(F) (27)

Proof The sufficiency is obviousWe will prove the necessarypart For any 119894 = 119895 isin [1 119899] 119886 isin F and 119887 isin Flowast since

[119864119894119894minus 119886119864119894119895+ 119864119895119894+ (119887minus1

minus 119886) 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= (1 minus 119886119887) 119864119894119894+ 119886119887119864

119894119895minus 119887119864119895119894+ 119887119864119895119895+ Σ119896 = 119894119895

119864119896119896

(28)

by 119891 preserving inverses of matrices we have

[

119891119894119894(1) 119891

119894119895(minus119886)

119891119895119894(1) 119891

119895119895(119887minus1

minus 119886)

] [

119891119894119894(1 minus 119886119887) 119891

119894119895(119886119887)

119891119895119894(minus119887) 119891

119895119895(119887)] = 1198682 (29)

so that

119891119894119894(1) 119891119894119895(119886119887) + 119891

119894119895(minus119886) 119891

119895119895(119887) = 0 (30)

119891119894119894(1) 119891119894119894(1 minus 119886119887) + 119891

119894119895(minus119886) 119891

119895119894(minus119887) = 1 (31)

Let 119887 = 1 then (30) turns into

119891119894119894(1) 119891119894119895(119886) + 119891

119894119895(minus119886) 119891

119895119895(1) = 0 (32)

Replacing 119886 by 119886119887 then the above turns into

119891119894119894(1) 119891119894119895(119886119887) + 119891

119894119895(minus119886119887) 119891

119895119895(1) = 0 (33)

It follows from (30) and (33) that

119891119894119895(minus119886119887) 119891

119895119895(1) = 119891

119894119895(minus119886) 119891

119895119895(119887) (34)

Replacing minus119886 by 119886 then the above turns into

119891119894119895(119886119887) 119891

119895119895(1) = 119891

119894119895(119886) 119891119895119895(119887) (35)

By [119886119864119894119894+ Σ119896 = 119894119864119896119896]minus1

= 119886minus1

119864119894119894+ Σ119896 = 119894119864119896119896 we have

119891119894119894(119886) 119891119894119894(119886minus1

) = 1 (36)

In particular

119891119894119894(1)2

= 1 (37)

From [119886119864119894119895+ 119886119864119895119894+ Σ119896 = 119894119895

119864119896119896]minus1

= 119886minus1

119864119894119895+ 119886minus1

119864119895119894+

Σ119896 = 119894119895

119864119896119896 we have

119891119894119895(119886) 119891119895119894(119886minus1

) = 1 (38)

In particular

119891119894119895(1) 119891119895119894(1) = 1 (39)

Multiplying119891119894119895(minus119887minus1

) by (31) we obtain by using (38) that

119891119894119894(1) 119891119894119894(1 minus 119886119887) 119891

119894119895(minus119887minus1

) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (40)

It follows from (35) and (38) that

119891119894119894(1) 119891119894119895((1 minus 119886119887) (minus119887

minus1

)) = 119891119894119894(1 minus 119886119887) 119891

119894119895(minus119887minus1

) (41)

This together with (40) implies that

119891119894119894(1) 119891119894119894(1) 119891119894119895(minus119887minus1

+ 119886) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (42)

Hence it follows from 119891119894119894(1)2

= 1 that

119891119894119895(minus119887minus1

+ 119886) + 119891119894119895(minus119886) = 119891

119894119895(minus119887minus1

) (43)

4 Abstract and Applied Analysis

Let 119909 = minus119887minus1 and 119910 = minus119886 we have

119891119894119895(119909 minus 119910) = 119891

119894119895(119909) minus 119891

119894119895(119910) forall119909 isin F

lowast

119910 isin F (44)

From

[119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896

(45)

we have

[

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] [

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] = 1198682 (46)

so that

119891119894119894(1) 119891119894119895(1) = minus119891

119894119895(1) 119891119895119895(minus1) (47)

This together with (39) implies

119891119894119894(1) = minus119891

119895119895(minus1) (48)

Similarly 119891119896119896(1) = minus119891

119895119895(minus1) hence

119891119894119894(1) = 119891

119896119896(1) forall119894 119896 isin [1 119899] (49)

It follows from

[119864119894119894+ 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896 forall119909 isin F

(50)

that

[

119891119894119894(1) 119891

119894119895(119909)

0 119891119895119895(1)] [

119891119894119894(1) 119891

119894119895(minus119909)

0 119891119895119895(1)] = 1198682 (51)

Hence

119891119894119894(1) 119891119894119895(minus119909) = minus119891

119894119895(119909) 119891119895119895(1) (52)

This together with (39) and (49) implies

119891119894119895(minus119909) = minus119891

119894119895(119909) forall119909 isin F (53)

Since

[119886119864119894119894+ 119864119894119895minus 119864119895119894+ Σ119896 = 119894119895

119864119896119896]

minus1

= minus119864119894119895+ 119864119895119894+ 119886119864119895119895+ Σ119896 = 119894119895

119864119896119896

(54)

we have

[

119891119894119894(119886) 119891

119894119895(1)

119891119895119894(minus1) 0

] [

0 119891119894119895(minus1)

119891119895119894(1) 119891

119895119895(119886)] = 1198682 (55)

so that

119891119894119894(119886) 119891119894119895(minus1) = minus119891

119894119895(1) 119891119895119895(119886) (56)

Hence

119891119894119894(119886) = 119891

119895119895(119886) (57)

For distinct 119894 119895 119896 isin [1 119899] and 119909 119910 isin F since

[119864119894119894+ 119909119864119894119895+ 119909119864119894119896+ 119864119895119895+ 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895minus 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897

(58)

we have

[

[

119891119894119894(1) 119891

119894119895(119909) 119891

119894119896(119909119910)

0 119891119895119895(1) 119891

119895119896(119910)

0 0 119891119896119896(1)

]

]

[

[

119891119894119894(1) 119891

119894119895(minus119909) 0

0 119891119895119895(1) 119891

119895119896(minus119910)

0 0 119891119896119896(1)

]

]

= 1198683

(59)

so that

119891119894119895(119909) 119891119895119896(minus119910) + 119891

119894119896(119909119910) 119891

119896119896(1) = 0 (60)

This together with (53) implies that

119891119894119895(119909) = 119891

119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

(61)

It follows from (35) and (61) that

119891119894119895(119909)

= 119891119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

= 119891119894119896(119909) 11989111(1) 119891119895119896(1)minus1

= (119891119896119896(1)minus1

1198911198941(1) 11989111(119909))

times 11989111(1) (119891

119896119896(1)minus1

1198911198951(1) 11989111(1))

minus1

= 1198911198941(1) 11989111(119909) 1198911198951(1)minus1

forall119894 = 119895 isin [1 119899]

(62)

Using this together with (57) we obtain

119891 (119860) = 119875 [11989111(119886119894119895)] 119875minus1

(63)

where 119875 = diag(11989111(1) 119891

1198991(1)) Let 119888 = 119891

11(1) isin minus1 1

120593(119909) = 11989111(1)11989111(119909) and then

119891 (119860) = 119888119875 [120593 (119886119894119895)] 119875minus1

= 119888119875119860120593

119875minus1

(64)

Let 120601(119860) = [120593(119886119894119895)] since 119891 preserves inverses of matrices

one can see that 120601 also preserves inverses of matrices By120593(1) = 119891

11(1)2

= 1 we obtain by using similar method to(35) (37) (44) and (53) that for any 119886 119887 isin F

120593 (119886119887) = 120593 (1) 120593 (119886119887) = 120593 (119886) 120593 (119887)

120593 (119886 + 119887) = 120593 (119886) + 120593 (119887)

(65)

This completes the proof

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 5

Acknowledgments

The authors show great thanks to the referee for hishercareful reading of the paper and valuable comments whichgreatly improved the readability of the paper Li Yang issupported by Vocational education institute in HeilongjiangProvince ldquo12th five-year development planrdquo The GuidingFunction and Practice Research of Mathematical modelingin Advanced Mathematics Teaching of New Rise FinancialInstitutionsrsquo (Grant no GG0666) Chongguang Cao is sup-ported by National Natural Science Foundation Grants ofChina (Grant no 11371109)

References

[1] J Kalinowski ldquoOn rank equivalence and rank preservingoperatorsrdquo Novi Sad Journal of Mathematics vol 32 no 1 pp133ndash139 2002

[2] J Kalinowski ldquoOn functions preserving rank of matricesrdquoMathematical Notes A Publication of the University of Miskolcvol 4 no 1 pp 35ndash37 2003

[3] S-W Liu and G-D Zhang ldquoMaps preserving rank 1 matricesover fieldsrdquo Journal ofNatural Science ofHeilongjiangUniversityvol 23 no 1 pp 138ndash140 2006

[4] S Du J Hou and Z Bai ldquoNonlinearmaps preserving similarityon 119861(119867)rdquo Linear Algebra and its Applications vol 422 no 2-3pp 506ndash516 2007

[5] T Oikhberg and A M Peralta ldquoAutomatic continuity oforthogonality preservers on a non-commutative 119871119901(120591) spacerdquoJournal of Functional Analysis vol 264 no 8 pp 1848ndash18722013

[6] M A Chebotar W-F Ke P-H Lee and N-C Wong ldquoMap-pings preserving zero productsrdquo Studia Mathematica vol 155no 1 pp 77ndash94 2003

[7] C-W Leung C-K Ng and N-C Wong ldquoLinear orthogonalitypreservers of Hilbert Clowast-modules over Clowast-algebras with realrank zerordquo Proceedings of the American Mathematical Societyvol 140 no 9 pp 3151ndash3160 2012

[8] I V Konnov and J C Yao ldquoOn the generalized vector vari-ational inequality problemrdquo Journal of Mathematical Analysisand Applications vol 206 no 1 pp 42ndash58 1997

[9] D Huo B Zheng and H Liu ldquoCharacterizations of nonlinearLie derivations of 119861(119883)rdquo Abstract and Applied Analysis vol2013 Article ID 245452 7 pages 2013

[10] H Yao and B Zheng ldquoZero triple product determined matrixalgebrasrdquo Journal of Applied Mathematics vol 2012 Article ID925092 18 pages 2012

[11] J Xu B Zheng and H Yao ldquoLinear transformations betweenmultipartite quantum systems that map the set of tensorproduct of idempotent matrices into idempotent matrix setrdquoJournal of Function Spaces and Applications vol 2013 ArticleID 182569 7 pages 2013

[12] J Xu B Zheng and L Yang ldquoThe automorphism group of theLie ring of real skew-symmetric matricesrdquo Abstract and AppliedAnalysis vol 2013 Article ID 638230 8 pages 2013

[13] X Song C Cao and B Zheng ldquoA note on 119896-potence preserverson matrix spaces over complex fieldrdquo Abstract and AppliedAnalysis vol 2013 Article ID 581683 8 pages 2013

[14] B Zheng J Xu and A Fosner ldquoLinear maps preserving rankof tensor products of matricesrdquo Linear and Multilinear Algebra2014

[15] L Yang W Zhang and J L Xu ldquoLinear maps on upper trian-gular matrices spaces preserving idempotent tensor productsrdquoAbstract and Applied Analysis vol 2014 Article ID 148321 8pages 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Let 119909 = minus119887minus1 and 119910 = minus119886 we have

119891119894119895(119909 minus 119910) = 119891

119894119895(119909) minus 119891

119894119895(119910) forall119909 isin F

lowast

119910 isin F (44)

From

[119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894+ 119864119894119895minus 119864119895119895+ Σ119896 = 119894119895

119864119896119896

(45)

we have

[

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] [

119891119894119894(1) 119891

119894119895(1)

0 119891119895119895(minus1)

] = 1198682 (46)

so that

119891119894119894(1) 119891119894119895(1) = minus119891

119894119895(1) 119891119895119895(minus1) (47)

This together with (39) implies

119891119894119894(1) = minus119891

119895119895(minus1) (48)

Similarly 119891119896119896(1) = minus119891

119895119895(minus1) hence

119891119894119894(1) = 119891

119896119896(1) forall119894 119896 isin [1 119899] (49)

It follows from

[119864119894119894+ 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895+ Σ119896 = 119894119895

119864119896119896 forall119909 isin F

(50)

that

[

119891119894119894(1) 119891

119894119895(119909)

0 119891119895119895(1)] [

119891119894119894(1) 119891

119894119895(minus119909)

0 119891119895119895(1)] = 1198682 (51)

Hence

119891119894119894(1) 119891119894119895(minus119909) = minus119891

119894119895(119909) 119891119895119895(1) (52)

This together with (39) and (49) implies

119891119894119895(minus119909) = minus119891

119894119895(119909) forall119909 isin F (53)

Since

[119886119864119894119894+ 119864119894119895minus 119864119895119894+ Σ119896 = 119894119895

119864119896119896]

minus1

= minus119864119894119895+ 119864119895119894+ 119886119864119895119895+ Σ119896 = 119894119895

119864119896119896

(54)

we have

[

119891119894119894(119886) 119891

119894119895(1)

119891119895119894(minus1) 0

] [

0 119891119894119895(minus1)

119891119895119894(1) 119891

119895119895(119886)] = 1198682 (55)

so that

119891119894119894(119886) 119891119894119895(minus1) = minus119891

119894119895(1) 119891119895119895(119886) (56)

Hence

119891119894119894(119886) = 119891

119895119895(119886) (57)

For distinct 119894 119895 119896 isin [1 119899] and 119909 119910 isin F since

[119864119894119894+ 119909119864119894119895+ 119909119864119894119896+ 119864119895119895+ 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897]

minus1

= 119864119894119894minus 119909119864119894119895+ 119864119895119895minus 119864119895119896+ 119864119896119896+ Σ119897 = 119894119895119896

119864119897119897

(58)

we have

[

[

119891119894119894(1) 119891

119894119895(119909) 119891

119894119896(119909119910)

0 119891119895119895(1) 119891

119895119896(119910)

0 0 119891119896119896(1)

]

]

[

[

119891119894119894(1) 119891

119894119895(minus119909) 0

0 119891119895119895(1) 119891

119895119896(minus119910)

0 0 119891119896119896(1)

]

]

= 1198683

(59)

so that

119891119894119895(119909) 119891119895119896(minus119910) + 119891

119894119896(119909119910) 119891

119896119896(1) = 0 (60)

This together with (53) implies that

119891119894119895(119909) = 119891

119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

(61)

It follows from (35) and (61) that

119891119894119895(119909)

= 119891119894119896(119909) 119891119896119896(1) 119891119895119896(1)minus1

= 119891119894119896(119909) 11989111(1) 119891119895119896(1)minus1

= (119891119896119896(1)minus1

1198911198941(1) 11989111(119909))

times 11989111(1) (119891

119896119896(1)minus1

1198911198951(1) 11989111(1))

minus1

= 1198911198941(1) 11989111(119909) 1198911198951(1)minus1

forall119894 = 119895 isin [1 119899]

(62)

Using this together with (57) we obtain

119891 (119860) = 119875 [11989111(119886119894119895)] 119875minus1

(63)

where 119875 = diag(11989111(1) 119891

1198991(1)) Let 119888 = 119891

11(1) isin minus1 1

120593(119909) = 11989111(1)11989111(119909) and then

119891 (119860) = 119888119875 [120593 (119886119894119895)] 119875minus1

= 119888119875119860120593

119875minus1

(64)

Let 120601(119860) = [120593(119886119894119895)] since 119891 preserves inverses of matrices

one can see that 120601 also preserves inverses of matrices By120593(1) = 119891

11(1)2

= 1 we obtain by using similar method to(35) (37) (44) and (53) that for any 119886 119887 isin F

120593 (119886119887) = 120593 (1) 120593 (119886119887) = 120593 (119886) 120593 (119887)

120593 (119886 + 119887) = 120593 (119886) + 120593 (119887)

(65)

This completes the proof

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Abstract and Applied Analysis 5

Acknowledgments

The authors show great thanks to the referee for hishercareful reading of the paper and valuable comments whichgreatly improved the readability of the paper Li Yang issupported by Vocational education institute in HeilongjiangProvince ldquo12th five-year development planrdquo The GuidingFunction and Practice Research of Mathematical modelingin Advanced Mathematics Teaching of New Rise FinancialInstitutionsrsquo (Grant no GG0666) Chongguang Cao is sup-ported by National Natural Science Foundation Grants ofChina (Grant no 11371109)

References

[1] J Kalinowski ldquoOn rank equivalence and rank preservingoperatorsrdquo Novi Sad Journal of Mathematics vol 32 no 1 pp133ndash139 2002

[2] J Kalinowski ldquoOn functions preserving rank of matricesrdquoMathematical Notes A Publication of the University of Miskolcvol 4 no 1 pp 35ndash37 2003

[3] S-W Liu and G-D Zhang ldquoMaps preserving rank 1 matricesover fieldsrdquo Journal ofNatural Science ofHeilongjiangUniversityvol 23 no 1 pp 138ndash140 2006

[4] S Du J Hou and Z Bai ldquoNonlinearmaps preserving similarityon 119861(119867)rdquo Linear Algebra and its Applications vol 422 no 2-3pp 506ndash516 2007

[5] T Oikhberg and A M Peralta ldquoAutomatic continuity oforthogonality preservers on a non-commutative 119871119901(120591) spacerdquoJournal of Functional Analysis vol 264 no 8 pp 1848ndash18722013

[6] M A Chebotar W-F Ke P-H Lee and N-C Wong ldquoMap-pings preserving zero productsrdquo Studia Mathematica vol 155no 1 pp 77ndash94 2003

[7] C-W Leung C-K Ng and N-C Wong ldquoLinear orthogonalitypreservers of Hilbert Clowast-modules over Clowast-algebras with realrank zerordquo Proceedings of the American Mathematical Societyvol 140 no 9 pp 3151ndash3160 2012

[8] I V Konnov and J C Yao ldquoOn the generalized vector vari-ational inequality problemrdquo Journal of Mathematical Analysisand Applications vol 206 no 1 pp 42ndash58 1997

[9] D Huo B Zheng and H Liu ldquoCharacterizations of nonlinearLie derivations of 119861(119883)rdquo Abstract and Applied Analysis vol2013 Article ID 245452 7 pages 2013

[10] H Yao and B Zheng ldquoZero triple product determined matrixalgebrasrdquo Journal of Applied Mathematics vol 2012 Article ID925092 18 pages 2012

[11] J Xu B Zheng and H Yao ldquoLinear transformations betweenmultipartite quantum systems that map the set of tensorproduct of idempotent matrices into idempotent matrix setrdquoJournal of Function Spaces and Applications vol 2013 ArticleID 182569 7 pages 2013

[12] J Xu B Zheng and L Yang ldquoThe automorphism group of theLie ring of real skew-symmetric matricesrdquo Abstract and AppliedAnalysis vol 2013 Article ID 638230 8 pages 2013

[13] X Song C Cao and B Zheng ldquoA note on 119896-potence preserverson matrix spaces over complex fieldrdquo Abstract and AppliedAnalysis vol 2013 Article ID 581683 8 pages 2013

[14] B Zheng J Xu and A Fosner ldquoLinear maps preserving rankof tensor products of matricesrdquo Linear and Multilinear Algebra2014

[15] L Yang W Zhang and J L Xu ldquoLinear maps on upper trian-gular matrices spaces preserving idempotent tensor productsrdquoAbstract and Applied Analysis vol 2014 Article ID 148321 8pages 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Acknowledgments

The authors show great thanks to the referee for hishercareful reading of the paper and valuable comments whichgreatly improved the readability of the paper Li Yang issupported by Vocational education institute in HeilongjiangProvince ldquo12th five-year development planrdquo The GuidingFunction and Practice Research of Mathematical modelingin Advanced Mathematics Teaching of New Rise FinancialInstitutionsrsquo (Grant no GG0666) Chongguang Cao is sup-ported by National Natural Science Foundation Grants ofChina (Grant no 11371109)

References

[1] J Kalinowski ldquoOn rank equivalence and rank preservingoperatorsrdquo Novi Sad Journal of Mathematics vol 32 no 1 pp133ndash139 2002

[2] J Kalinowski ldquoOn functions preserving rank of matricesrdquoMathematical Notes A Publication of the University of Miskolcvol 4 no 1 pp 35ndash37 2003

[3] S-W Liu and G-D Zhang ldquoMaps preserving rank 1 matricesover fieldsrdquo Journal ofNatural Science ofHeilongjiangUniversityvol 23 no 1 pp 138ndash140 2006

[4] S Du J Hou and Z Bai ldquoNonlinearmaps preserving similarityon 119861(119867)rdquo Linear Algebra and its Applications vol 422 no 2-3pp 506ndash516 2007

[5] T Oikhberg and A M Peralta ldquoAutomatic continuity oforthogonality preservers on a non-commutative 119871119901(120591) spacerdquoJournal of Functional Analysis vol 264 no 8 pp 1848ndash18722013

[6] M A Chebotar W-F Ke P-H Lee and N-C Wong ldquoMap-pings preserving zero productsrdquo Studia Mathematica vol 155no 1 pp 77ndash94 2003

[7] C-W Leung C-K Ng and N-C Wong ldquoLinear orthogonalitypreservers of Hilbert Clowast-modules over Clowast-algebras with realrank zerordquo Proceedings of the American Mathematical Societyvol 140 no 9 pp 3151ndash3160 2012

[8] I V Konnov and J C Yao ldquoOn the generalized vector vari-ational inequality problemrdquo Journal of Mathematical Analysisand Applications vol 206 no 1 pp 42ndash58 1997

[9] D Huo B Zheng and H Liu ldquoCharacterizations of nonlinearLie derivations of 119861(119883)rdquo Abstract and Applied Analysis vol2013 Article ID 245452 7 pages 2013

[10] H Yao and B Zheng ldquoZero triple product determined matrixalgebrasrdquo Journal of Applied Mathematics vol 2012 Article ID925092 18 pages 2012

[11] J Xu B Zheng and H Yao ldquoLinear transformations betweenmultipartite quantum systems that map the set of tensorproduct of idempotent matrices into idempotent matrix setrdquoJournal of Function Spaces and Applications vol 2013 ArticleID 182569 7 pages 2013

[12] J Xu B Zheng and L Yang ldquoThe automorphism group of theLie ring of real skew-symmetric matricesrdquo Abstract and AppliedAnalysis vol 2013 Article ID 638230 8 pages 2013

[13] X Song C Cao and B Zheng ldquoA note on 119896-potence preserverson matrix spaces over complex fieldrdquo Abstract and AppliedAnalysis vol 2013 Article ID 581683 8 pages 2013

[14] B Zheng J Xu and A Fosner ldquoLinear maps preserving rankof tensor products of matricesrdquo Linear and Multilinear Algebra2014

[15] L Yang W Zhang and J L Xu ldquoLinear maps on upper trian-gular matrices spaces preserving idempotent tensor productsrdquoAbstract and Applied Analysis vol 2014 Article ID 148321 8pages 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of