Row Equivalence is Equivalence Relation

Theorem
Row equivalence is an equivalence relation.

Proof
Checking in turn each of the critera for equivalence:

Reflexive
Any matrix is trivially row equivalent to itself: for example, multiply any row by $1$ by $r_i \to 1 r_i$.

So row equivalence has been shown to be reflexive.

Symmetric
Each elementary row operation is reversible.


 * The inverse of $r_i \to a r_i$ is $r_i \to \left({a^{-1}}\right) r_i$.
 * The inverse of $r_i \to r_i + a r_j$ is $r_i \to r_i - a r_j$.
 * The inverse of $r_1 \leftrightarrow r_j$ is $r_1 \leftrightarrow r_j$ (i.e. swap them back again).

So let $\mathbf A$ be row equivalent to $\mathbf B$.

Let $\Gamma$ be the sequence of elementary row operations that turn $\mathbf A$ into $\mathbf B$.

Then for each elementary row operation in $\Gamma$, the inverses of those operations can be applied in the reverse order on $\mathbf B$ to get $\mathbf A$.

Thus if $\mathbf A$ is row equivalent to $\mathbf B$, then $\mathbf B$ is row equivalent to $\mathbf A$.

So row equivalence has been shown to be symmetric.

Transitive
Let $\mathbf A, \mathbf B, \mathbf C$ be $m \times n$ matrices.

Let $\mathbf A$ be row equivalent to $\mathbf B$, and let $\mathbf B$ be row equivalent to $\mathbf C$.

Let $\Gamma_1$ be the sequence of elementary row operations that turn $\mathbf A$ into $\mathbf B$.

Let $\Gamma_2$ be the sequence of elementary row operations that turn $\mathbf B$ into $\mathbf C$.

It follows that, as $\Gamma_1$ and $\Gamma_2$ are finite sequences of operations, you can apply the operations in $\Gamma_1$ to $\mathbf A$ and follow them immediately by the operations in $\Gamma_2$ to get $\mathbf C$

Thus $\mathbf A$ is row equivalent to $\mathbf C$, and so row equivalence has been shown to be transitive.

Row equivalence has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.