Show that similarity is an equivalence relation on \(M_n(R)\), and *equivalence* is an equivalence relation on \(M_{m\times n}(R)\).
Solution
Recall that \(A,B\in M_n(R)\) are similar if \[ B=P^{-1}AP \] for some \(P\in\operatorname{GL}_n(R)\), and that \(A,B\in M_{m\times n}(R)\) are equivalent if \[ B=PAQ \] for some \(P\in\operatorname{GL}_m(R)\) and \(Q\in\operatorname{GL}_n(R)\).
1. Similarity is reflexive.
For every \(A\in M_n(R)\), \[ A=I_n^{-1}AI_n, \] so \(A\) is similar to itself.
2. Similarity is symmetric.
If \(B=P^{-1}AP\), then multiplying by \(P\) on the left and \(P^{-1}\) on the right gives \[ A=PBP^{-1}=(P^{-1})^{-1}B(P^{-1}). \] Since \(P^{-1}\) is invertible, \(A\) is similar to \(B\).
3. Similarity is transitive.
If \[ B=P^{-1}AP \qquad\text{and}\qquad C=Q^{-1}BQ, \] then \[ C=Q^{-1}P^{-1}APQ=(PQ)^{-1}A(PQ). \] Thus \(C\) is similar to \(A\).
4. Matrix equivalence is reflexive.
For every \(A\in M_{m\times n}(R)\), \[ A=I_mAI_n. \]
5. Matrix equivalence is symmetric.
If \(B=PAQ\) with \(P,Q\) invertible, then \[ A=P^{-1}BQ^{-1}. \] Hence \(A\) is equivalent to \(B\).
6. Matrix equivalence is transitive.
If \[ B=PAQ \qquad\text{and}\qquad C=UBV, \] with \(P,U\in\operatorname{GL}_m(R)\) and \(Q,V\in\operatorname{GL}_n(R)\), then \[ C=(UP)A(QV). \] Both \(UP\in\operatorname{GL}_m(R)\) and \(QV\in\operatorname{GL}_n(R)\), so \(C\) is equivalent to \(A\).
7. Therefore similarity and matrix equivalence are equivalence relations ontheir respective matrix sets.
Each relation is reflexive, symmetric, and transitive by the preceding steps.