Existence of Inverse Elementary Column Operation

Theorem
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.

Let $\mathbf A \in \map \MM {m, n}$ be a matrix.

Let $\map e {\mathbf A}$ be an elementary column operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.

Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.

Then $e'$ is an elementary column operation which always exists and is unique.

Proof
Let us take each type of elementary column operation in turn.

For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new matrix $\mathbf A'' \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.

In the below, let:
 * $\kappa_k$ denote column $k$ of $\mathbf A$
 * $\kappa'_k$ denote column $k$ of $\mathbf A'$
 * $\kappa_k$ denote column $k$ of $\mathbf A$

for arbitrary $k$ such that $1 \le k \le m$.

By definition of elementary column operation:
 * only the column or columns directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$

and similarly:
 * only the column or columns directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A''$.

Hence it is understood that in the following, only those columns directly affected will be under consideration when showing that $\mathbf A = \mathbf A''$.

$\text {ERO} 1$: Scalar Product of Column
Let $\map e {\mathbf A}$ be the elementary column operation:


 * $e := \kappa_k \to \lambda \kappa_k$

where $\lambda \ne 0$.

Then $\kappa'_k$ is such that:
 * $\forall a'_{k i} \in \kappa'_k: a'_{k i} = \lambda a_{k i}$

Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A''$:
 * $e' := \kappa_k \to \dfrac 1 \lambda \kappa_k$

Because it is stipulated in the definition of an elementary column operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists.

Hence $e'$ is defined.

So applying $e'$ to $\mathbf A'$ we get:

It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.

$\text {ERO} 2$: Add Scalar Product of Column to Another
Let $\map e {\mathbf A}$ be the elementary column operation:


 * $e := \kappa_k \to \kappa_k + \lambda r_l$

Then $\kappa'_k$ is such that:
 * $\forall a'_{i k} \in \kappa'_k: a'_{i k} = a_{i k} + \lambda a_{i l}$

Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A''$:
 * $e' := \kappa'_k \to \kappa'_k - \lambda \kappa'_l$

Applying $e'$ to $\mathbf A'$ we get:

It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.

$\text {ERO} 3$: Exchange Columns
Let $\map e {\mathbf A}$ be the elementary column operation:


 * $e := \kappa_k \leftrightarrow \kappa_l$

Thus we have:

Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A''$:
 * $e' := \kappa'_k \leftrightarrow \kappa'_l$

Applying $e'$ to $\mathbf A'$ we get:

It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.

Thus in all cases, for each elementary column operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible elementary column operation which transforms $\mathbf A'$ to $\mathbf A$.

Hence the result.