Question

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 1

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
`A=J^(-1)AJ`. 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
`A=P^(-1)BP`. In this equality we can perform a right multiplication by
`P^(-1)` and obtain
`AP^(-1) =P^(-1)B`. Then, in the obtained equality we perform a left multiplication by P and get
`PAP^(-1) =B`. If we write
`Q=P^(-1)` and
`Q^(-1) = P` we have
`B = Q^(-1)AQ`. Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have
`A=P^(-1)BP` and from B↔C we have
`B=Q^(-1)CQ`. Now, if we substitute the last equality into the first one we get

`A=P^(-1)Q^(-1)CQP = (P^(-1)Q^(-1))C(QP)`.

Recall that if P and Q are invertible, then QP is invertible and
`(QP)^(-1)=P^(-1)Q^(-1)`. So, if we denote R=QP we obtained that

`A=R^(-1)CR`. Hence, A↔C.

Therefore, the relation is an equivalence relation.