Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP Show that is an equivalence relation.

Answer:Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B. Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that [tex]A=J^{-1}AJ[/tex]. Thus, A↔A.Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that [tex]A=P^{-1}BP[/tex]. In this equality we can perform a right multiplication by [tex]P^{-1}[/tex] and obtain [tex]AP^{-1} =P^{-1}B[/tex]. Then, in the obtained equality we perform a left multiplication by P and get [tex]PAP^{-1} =B[/tex]. If we write [tex]Q=P^{-1}[/tex] and [tex]Q^{-1} = P[/tex] we have [tex]B = Q^{-1}AQ[/tex]. Thus, B↔A.Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have [tex]A=P^{-1}BP[/tex] and from B↔C we have [tex]B=Q^{-1}CQ[/tex]. Now, if we substitute the last equality into the first one we get[tex]A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)[/tex].Recall that if P and Q are invertible, then QP is invertible and [tex](QP)^{-1}=P^{-1}Q^{-1}[/tex]. So, if we denote R=QP we obtained that[tex]A=R^{-1}CR[/tex]. Hence, A↔C.Therefore, the relation is an equivalence relation.