Exchange of Rows as Sequence of Other Elementary Row 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 $r_k$ denote the $k$th row of $\mathbf A$ for $1 \le k \le m$:

$r_k = \begin {pmatrix} a_{k 1} & a_{k 2} & \cdots & a_{k n} \end {pmatrix}$


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

\((\text {ERO} 3)\)   $:$   \(\ds r_i \leftrightarrow r_j \)    Interchange rows $i$ and $j$             


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

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


Proof

In the below:

$r_i$ denotes the initial state of row $i$
$r_j$ denotes the initial state of row $j$
$r_i'$ denotes the state of row $i$ after having had the latest elementary row operation applied
$r_j'$ denotes the state of row $j$ after having had the latest elementary row operation applied.


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

$r_j \to r_j + r_i$

After this operation:

\(\ds r_i'\) \(=\) \(\ds r_i\)
\(\ds r_j'\) \(=\) \(\ds r_i + r_j\)

$\Box$


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

$r_i \to r_i + \paren {-r_j}$

After this operation:

\(\ds r_i'\) \(=\) \(\ds r_i - \paren {r_i + r_j}\)
\(\ds \) \(=\) \(\ds -r_j\)
\(\ds r_j'\) \(=\) \(\ds r_i + r_j\)

$\Box$


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

$r_j \to r_j + r_i$

After this operation:

\(\ds r_i'\) \(=\) \(\ds -r_j\)
\(\ds r_j'\) \(=\) \(\ds r_i + r_j - r_j\)
\(\ds \) \(=\) \(\ds r_i\)

$\Box$


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

$r_i \to -r_i$

After this operation:

\(\ds r_i'\) \(=\) \(\ds -\paren {-r_j}\)
\(\ds \) \(=\) \(\ds r_j\)
\(\ds r_j'\) \(=\) \(\ds r_i\)

$\Box$


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

\(\ds r_i'\) \(=\) \(\ds r_j\)
\(\ds r_j'\) \(=\) \(\ds r_i\)

Hence the result.

$\blacksquare$


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