Matrix Equivalence is Equivalence Relation

Theorem
Matrix equivalence is an equivalence relation.

Proof
Checking in turn each of the critera for equivalence:

Reflexive
Let $\mathbf {I_m}$ and $\mathbf {I_n}$ denote the identity matrices of order $m$ and $n$ respectively.

Let $\mathbf A$ be an arbitrary $m \times n$ matrix.

Then:
 * $\mathbf A = \mathbf {I_m}^{-1} \mathbf A \mathbf {I_n}$

trivially.

Thus reflexivity holds.

Symmetric
Let $\mathbf A$ and $\mathbf B$ be arbitrary $m \times n$ matrices such that $\mathbf A \equiv \mathbf B$.

Then by definition there exist invertible matrices $\mathbf P$ and $\mathbf Q$ such that:
 * $\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$

That is:
 * $\mathbf B \equiv \mathbf A$

Thus symmetry holds.

Transitive
Let $\mathbf A$, $\mathbf B$ and $\mathbf B$ be arbitrary $m \times n$ matrices such that:
 * $\mathbf A \equiv \mathbf B$
 * $\mathbf B \equiv \mathbf C$

Then by definition there exist invertible matrices $\mathbf P_1$, $\mathbf P_2$, $\mathbf Q_1$ and $\mathbf Q_2$ such that:


 * $\mathbf B = \mathbf Q_1^{-1} \mathbf A \mathbf P_1$
 * $\mathbf C = \mathbf Q_2^{-1} \mathbf B \mathbf P_2$

Then:
 * $\mathbf C = \mathbf Q_2^{-1} \mathbf Q_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$

From Inverse of Matrix Product:
 * $\mathbf Q_2^{-1} \mathbf Q_1^{-1} = \paren {\mathbf Q_1 \mathbf Q_2}^{-1}$

By Product of Matrices is Invertible iff Matrices are Invertible, both $\mathbf Q_1 \mathbf Q_2$ and $\mathbf P_1 \mathbf P_2$ are invertible.

Hence:
 * $\mathbf A \equiv \mathbf C$

Thus transitivity holds.

Hence the result by definition of equivalence relation.