Abstract Algebra Theory and Applications - Computer Science ...

16 CHAPTER 0 PRELIMINARIESandB =then A ∼ B since P AP −1 = B forP =( ) −18 33,−11 20( ) 2 5.1 3Let I be the 2 × 2 identity matrix; that is,( ) 1 0I = .0 1Then IAI −1 = IAI = A; therefore, the relation is reflexive. To showsymmetry, suppose that A ∼ B. Then there exists an invertible matrix Psuch that P AP −1 = B. SoA = P −1 BP = P −1 B(P −1 ) −1 .Finally, suppose that A ∼ B and B ∼ C. Then there exist invertiblematrices P and Q such that P AP −1 = B and QBQ −1 = C. SinceC = QBQ −1 = QP AP −1 Q −1 = (QP )A(QP ) −1 ,the relation is transitive. Two matrices that are equivalent in this mannerare said to be similar.A partition P of a set X is a collection of nonempty sets X 1 , X 2 , . . .such that X i ∩ X j = ∅ for i ≠ j and ⋃ k X k = X. Let ∼ be an equivalencerelation on a set X and let x ∈ X. Then [x] = {y ∈ X : y ∼ x} is called theequivalence class of x. We will see that an equivalence relation gives riseto a partition via equivalence classes. Also, whenever a partition of a setexists, there is some natural underlying equivalence relation, as the followingtheorem demonstrates.Theorem 0.5 Given an equivalence relation ∼ on a set X, the equivalenceclasses of X form a partition of X. Conversely, if P = {X i } is a partition ofa set X, then there is an equivalence relation on X with equivalence classesX i .Proof. Suppose there exists an equivalence relation ∼ on the set X. Forany x ∈ X, the reflexive property shows that x ∈ [x] and so [x] is nonempty.Clearly X = ⋃ x∈X[x]. Now let x, y ∈ X. We need to show that either