Exchange of Columns as Sequence of Other Elementary Column Operations
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.
$\blacksquare$