# Exchange of Columns as Sequence of Other Elementary Column Operations

Jump to navigation Jump to search

## 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.

$\blacksquare$