Student’s Name: Student’s ID No.:College Name:The study of Equivalence RelationsAbstractAccording to some relative definitions and properties, to proof that if B can be obtained from A by performing elementary row operations on A, ~ is an equivalence relation, and to find the properties that are shared by all the elements in the same equivalence class. To proof that if B is can be obtained from A by performing elementary operations, Matrix S A ∈ is said to be equivalent to matrix S B ∈, and ~A B means that matrix S A ∈ is similar to S B ∈, if let S be the set of m m ⨯ real matrices.IntroductionThe equivalence relations are used in the matrix theory in a very wide field. An equivalence relation on a set S divides S into equivalence classes. Equivalence classes are pair-wise disjoint subsets of S . a ~ b if and only if a and b are in the same equivalence class.This paper will introduce some definitions and properties of equivalence relations and proof some discussions.Main ResultsAnswers of Q1(a) The process of the proof is as following,obviously IA=A,therefore ~ is reflexive;we know B can be obtained from A by performing elementary row operations on A,we assume P is a matrix which denote a series of elementary row operations on A.Then ,we have PA=B,(A~B),and P is inverse,obviously we have A=P -1B,(B~A).So ~ is symmetric.We have another matrix Q which denote a series of elementary row operations on B,and the result is C,so we have QB=C.And we can obtain QB=Q(PA)=QPA=C,so A~C.Therefore,~ is transitive. Hence, ~ is an equivalence relation on S .(b) The properties that are shared by all the elements in the same equivalence class are as followings: firstly,the rank is the same;secondly,the relation of column is not changed;thirdly,two random matrices are row equivalent;fourthly,all of the matricescan be raduced as ⎥⎦⎤⎢⎣⎡00X E n. (c) ⎥⎦⎤⎢⎣⎡0000 is a representative element for each equivalence class corresponding to rank 0.⎥⎦⎤⎢⎣⎡00y x (x and y can not be zero at the same time)is a representative element for each equivalence class corresponding to rank 1.⎥⎦⎤⎢⎣⎡1001 is a representative element for each equivalence class corresponding to rank 2.Answers of Q2(a)The process of the proof is as following,,obviously IAI=A,therefore ~ is reflexive;we know B can be obtained from A by performing elementary operations on A,we assume P and Q is a matrix which denote a series of elementary operations onA.Then ,we have PAQ=B,(A~B),and P is inverse,obviously we have A=P -1BQ -1,(B~A).So ~ is symmetric.We have other matrices M and N which denote a series of elementary operations on B,and the result is C,so we have MBN=C.And we can obtain MBN=M(PAQ)N=MPAQN=C,so A~C.Therefore,~ is transitive. Hence, ~ is an equivalence relation on S .(b)Obviously,the rank haven ’t changed.Secondly,all of the matrices can be raduced as ⎥⎦⎤⎢⎣⎡000n E (c) ⎥⎦⎤⎢⎣⎡0000 is a representative element for each equivalence class corresponding to rank 0.⎥⎦⎤⎢⎣⎡0001 is a representative element for each equivalence class corresponding to rank 1.⎥⎦⎤⎢⎣⎡1001 is a representative element for each equivalence class corresponding to rank 2.Answers of Q3(a)The process of the proof is as following,obviously IAI -1=A,therefore ~ is reflexive;we assume P and Q are inverse matrices.Then,if we have PAP -1=B,(A~B),obviously we have A=P -1BP,(B~A).So~is symmetric.If We have QBQ -1=C.And we can obtain QBQ -1=Q(PAP -1)Q -1=(QP)A(P -1Q -1)=C,so A~C.Therefore,~ is transitive. Hence, ~ is an equivalence relation on S .(b)The properties that are shared by all the elements in the same equivalence class are as followings: firstly,the rank is the same;secondly,they have the same determinant;thirdly,all the matrices have the same characteristic equation and the same eigenvalues;the last but not the least,if the matrices are inverse and similar,then,the inverse of both are similar,that ’s to say,if A~B,then,A -1~B -1.(c) ⎥⎦⎤⎢⎣⎡0000 is a representative element for each equivalence class corresponding to rank 0.We assume the format of the 2⨯2 nonzero matrix is ⎥⎦⎤⎢⎣⎡d c b a .So,the characteristic equation is =-A I λbc ad d a bc d a d c b a -++-=---=⎥⎦⎤⎢⎣⎡----λλλλλλ)())((2 If 0=-bc ad ,then eigenvalues are ;,0d a +==λλIf a+d=0,we have a=-d,and 0=+=d a λ;Hence some matrices of rank 1 can be represent as ⎥⎦⎤⎢⎣⎡-a c b a ,otherwise,a+d ≠0,the matrix have eigenvalues ;,0d a +==λλSo,some matrices of rank 1 can be represent as ⎥⎦⎤⎢⎣⎡d c b a (a+d ≠0,0=-bc ad ) Else if 0≠-bc ad ,then from the format of characteristic equation we obtain )(4)(2bc ad d a --+=∆;If 0)(4)(2<--+=∆bc ad d a ,then some matrices of rank 2 can be represented as ⎥⎦⎤⎢⎣⎡d c b a (0)(4)(2<--+bc ad d a ,0≠-bc ad ). If 0)(4)(2=--+=∆bc ad d a ,then some matrices of rank 2 can be represented as ⎥⎦⎤⎢⎣⎡d c b a ()(4)(2bc ad d a -=+,0≠-bc ad ). If 0)(4)(2>--+=∆bc ad d a ,the matrix have two different eigenvalues,the matrix can be diagonalized to a diagonalizable matrix,then some matrices of rank 2 can berepresented as ⎥⎦⎤⎢⎣⎡2100λλ()21λλ≠.Answers of Q4(a)The process of the proof is as following,obviously I -1AI=I T AI=A,therefore ~ is reflexive;we assume P and Q are inverse and orthogonal matrices .Then,if we have P -1AP=P T AP=B,(A~B),obviously we haveA=PBP -1=PBP T ,(B~A).So~is symmetric.If We have Q -1BQ=Q T BQ=C.And we canobtain Q -1BQ=Q T (PAP -1)Q=(Q T P)A(P -1Q)=C,so A~C.Therefore,~ is transitive. Hence, ~ is an equivalence relation on S .(b)Firstly,the rank is the same;secondly,they have the same determinant;thirdly,all the matrices have the same characteristic equation and the same eigenvalues;fourthly,the matrices can be diagonalized; the last but not the least,if the matrices are inverse and similar,then,the inverse of both are similar,that ’s to say,if A~B,then,A -1~B -1.(c) We assume the format of the 2⨯2 real symmetric matrix is ⎥⎦⎤⎢⎣⎡c b b a .So,the characteristic equation is=-A I λ222)())((b ac c a b c a c b b a -++-=---=⎥⎦⎤⎢⎣⎡----λλλλλλ If 02=-b ac ,then eigenvalues are ;,0c a +==λλHence some matrices of rank 1can be represent as ⎥⎦⎤⎢⎣⎡+c a 000, Else if 02≠-b ac ,then from the format of characteristic equation we obtain 0)(4)(4)(2222≥-+=--+=∆c a b b ac c a ,the matrix have two nonzero eigenvalues 2)(4221c a b c a -+++=λand 2)(4222c a b c a -+-+=λ,the matrix can be diagonalized to a diagonalizable matrix,then some matrices of rank 2 can berepresented as ⎥⎦⎤⎢⎣⎡2100λλ()21λλ≠.Answers of Q5(a)The process of the proof is as following,obviously,A is isomorphic to A,owing to they have the same dimension,therefore ~ is reflexive;we assume A is isomorphic toB.From Theorem 3.5.2,we obtain A is isomorphic to B.So,A and B areisomorphic.So~is symmetric.If We have A is isomorphic to B, B is isomorphic toC.So,Dimension(A)=Dimension(B),Dimension(B)=Dimension(C),soDimension(A)=Dimension(C),then A is isomorphic to C.so A~C.Therefore,~ is transitive. Hence, ~ is an equivalence relation on S .(b)The properties that are shared by all the elements in the same equivalence class are listed as followings.Firstly,they have the same dimension;Secondly,they are all one-to-one linear mapping.(c)12~V V if there exists an isomorphism from 1V onto 2V ,W V =)(δ is a linearone-to-one mapping.So,we can obtain infinite vector space.For example,if we define a mapping δ:R R →+ by )()(x In x =δ.It is easy to verity that the mapping δ is a one-to-one mapping.And δis a linear mappingsince ).(ln )ln()(),()(ln ln )ln()ln()(x x x x y x y x xy y x y x αδααδδδδ===+=+==⊕=⊕∂Therefore,δis an isomorphism from ,R R →++R is isomorphic to R .Conclusion and AcknowledgementIn this project, we know that the matrices through elementary row operations or performed elementary operations, the similar or orthogonally matrices and the isomorphic matrices have the equivalence relation in their equivalence class. In our study, we must learn to summarize some common properties, look for the law, only in this way, we can learn more and better.References[1] 曹荣美. Lecture Notes On Matrix Theory.南京航空航天大学出版社，2012.[2] 戴华.矩阵论.北京：科学出版社，2001.。