Exchange of Rows as Sequence of Other Elementary Row Operations
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 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.
$\blacksquare$