Exchange of Columns as Sequence of Other Elementary Column Operations

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Theorem

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

Let $i, j \in \closedint 1 m: i \ne j$

Let $\kappa_k$ denote the $k$th column of $\mathbf A$ for $1 \le k \le n$:

$\kappa_k = \begin {pmatrix} a_{1 k} \\ a_{2 k} \\ \vdots \\ a_{m k} \end {pmatrix}$


Let $e$ be the elementary column operation acting on $\mathbf A$ as:

\((\text {ERO} 3)\)   $:$   \(\ds \kappa_i \leftrightarrow \kappa_j \)    Interchange columns $i$ and $j$             


Then $e$ can be expressed as a finite sequence of exactly $4$ instances of the other two elementary column operations.

\((\text {ERO} 1)\)   $:$   \(\ds \kappa_i \to \lambda \kappa_i \)    For some $\lambda \in K_{\ne 0}$, multiply column $i$ by $\lambda$             
\((\text {ERO} 2)\)   $:$   \(\ds \kappa_i \to \kappa_i + \lambda \kappa_j \)    For some $\lambda \in K$, add $\lambda$ times column $j$ to column $i$             


Proof

In the below:

$\kappa_i$ denotes the initial state of column $i$
$\kappa_j$ denotes the initial state of column $j$
$\kappa_i'$ denotes the state of column $i$ after having had the latest elementary column operation applied
$\kappa_j'$ denotes the state of column $j$ after having had the latest elementary column operation applied.


$(1)$: Apply $\text {ECO} 2$ to column $j$ for $\lambda = 1$:

$\kappa_j \to \kappa_j + \kappa_i$

After this operation:

\(\ds \kappa_i'\) \(=\) \(\ds \kappa_i\)
\(\ds \kappa_j'\) \(=\) \(\ds \kappa_i + \kappa_j\)

$\Box$


$(2)$: Apply $\text {ECO} 2$ to column $i$ for $\lambda = -1$:

$\kappa_i \to \kappa_i + \paren {-\kappa_j}$

After this operation:

\(\ds \kappa_i'\) \(=\) \(\ds \kappa_i - \paren {\kappa_i + \kappa_j}\)
\(\ds \) \(=\) \(\ds -\kappa_j\)
\(\ds \kappa_j'\) \(=\) \(\ds \kappa_i + \kappa_j\)

$\Box$


$(3)$: Apply $\text {ECO} 2$ to column $j$ for $\lambda = 1$:

$\kappa_j \to \kappa_j + \kappa_i$

After this operation:

\(\ds \kappa_i'\) \(=\) \(\ds -\kappa_j\)
\(\ds \kappa_j'\) \(=\) \(\ds \kappa_i + \kappa_j - \kappa_j\)
\(\ds \) \(=\) \(\ds \kappa_i\)

$\Box$


$(4)$: Apply $\text {ECO} 1$ to column $i$ for $\lambda = -1$:

$\kappa_i \to -\kappa_i$

After this operation:

\(\ds \kappa_i'\) \(=\) \(\ds -\paren {-\kappa_j}\)
\(\ds \) \(=\) \(\ds \kappa_j\)
\(\ds \kappa_j'\) \(=\) \(\ds \kappa_i\)

$\Box$


Thus, after all the $4$ elementary column operations have been applied, we have:

\(\ds \kappa_i'\) \(=\) \(\ds \kappa_j\)
\(\ds \kappa_j'\) \(=\) \(\ds \kappa_i\)

Hence the result.

$\blacksquare$


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