R = {(A, B) | B = P–1 AP} A = I–1AI ⇒ (A, A) ∈ R ⇒ R is reflexive Let (A, B) ∈ R , B = P–1AP PB = AP ⇒ PBP–1 = A ⇒ A = (P–1)–1 B(P–1) ⇒ (B, A) ∈R, ⇒ R is symmetric Let (A, B) ∈ R, (B, C) ∈ R A = P–1BP and B = Q–1 CQ A = P–1Q–1C QP = (QP)–1 C(QP) ⇒ (A, C) ∈ R