# Column Equivalence is Equivalence Relation

## Proof

In the following, $\mathbf A$, $\mathbf B$ and $\mathbf C$ denote arbitrary matrices in a given matrix space $\map \MM {m, n}$ for $m, n \in \Z{>0}$.

We check in turn each of the conditions for equivalence:

### Reflexive

Let $\kappa_i$ denote an arbitrary column of $\mathbf A$.

Let $e$ denote the elementary column operation $\kappa_i \to 1 \kappa_i$ applied to $\mathbf A$.

Then trivially:

$\map e {\mathbf A} = \mathbf A$

and so $\mathbf A$ is trivially column equivalent to itself.

So column equivalence has been shown to be reflexive.

$\Box$

### Symmetric

Let $\mathbf A$ be column equivalent to $\mathbf B$.

Let $\Gamma$ be the column operation that transforms $\mathbf A$ into $\mathbf B$.

From Column Operation has Inverse there exists a column operation $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$.

Thus $\mathbf B$ is column equivalent to $\mathbf A$.

So column equivalence has been shown to be symmetric.

$\Box$

### Transitive

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

Let $\Gamma_1$ be the column operation that transforms $\mathbf A$ into $\mathbf B$.

Let $\Gamma_2$ be the column operation that transforms $\mathbf B$ into $\mathbf C$.

From Sequence of Column Operations is Column Operation, the application of $\mathbf C$ is column equivalent to $\mathbf A$.

So column equivalence has been shown to be transitive.

$\Box$

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

Hence by definition it is an equivalence relation.

$\blacksquare$