Elementary Column Operations as Matrix Multiplications

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Theorem

Let $e$ be an elementary column operation.

Let $\mathbf E$ be the elementary column matrix of order $n$ defined as:

$\mathbf E = e \paren {\mathbf I}$

where $\mathbf I$ is the unit matrix.


Then for every $m \times n$ matrix $\mathbf A$:

$e \paren {\mathbf A} = \mathbf A \mathbf E$

where $\mathbf A \mathbf E$ denotes the conventional matrix product.


Corollary

Let $\mathbf X$ and $\mathbf Y$ be two $m \times n$ matrices that differ by exactly one elementary column operation.

Then there exists an elementary column matrix $\mathbf E$ of order $n$ such that:

$\mathbf X \mathbf E = \mathbf Y$


Proof

Let $s, t \in \closedint 1 m$ such that $s \ne t$.


Case $1$

Let $e$ be the elementary column operation $\kappa_s \to \lambda \kappa_s$:

$E_{k i} = \begin{cases} \delta_{k i} & : i \ne s \\ \lambda \delta_{k i} & : i = s \end{cases}$

where $\delta$ denotes the Kronecker delta.

Then:

\(\ds \sqbrk {A E}_{i j}\) \(=\) \(\ds \sum_{k \mathop = 1}^m A_{j k} E_{k i}\)
\(\ds \) \(=\) \(\ds \begin {cases} A_{j i} & : i \ne r \\ \lambda A_{j i} & : i = r \end{cases}\)
\(\ds \leadsto \ \ \) \(\ds \mathbf {A E}\) \(=\) \(\ds e \paren {\mathbf A}\)

$\Box$


Case $2$

Let $e$ be the elementary column operation $\kappa_s \to \kappa_s + \lambda \kappa_t$:

$E_{k i} = \begin {cases} \delta_{k i} & : i \ne s \\ \delta_{k s} + \lambda \delta_{k t} & : i = s \end {cases}$

where $\delta$ denotes the Kronecker delta.

Then:

\(\ds \sqbrk {A E}_{j i}\) \(=\) \(\ds \sum_{k \mathop = 1}^m A_{j k} E_{k i}\)
\(\ds \) \(=\) \(\ds \begin {cases} A_{j i} & : j \ne s \\ A_{i j} + \lambda A_{j t} & : i = s \end {cases}\)
\(\ds \leadsto \ \ \) \(\ds \mathbf {A E}\) \(=\) \(\ds e \paren {\mathbf A}\)

$\Box$


Case $3$

Let $e$ be the elementary column operation $\kappa_s \leftrightarrow \kappa_t$:

By Exchange of Columns as Sequence of Other Elementary Column Operations, this elementary column operation can be expressed as:

$\paren {e_1 e_2 e_3 e_4 \mathbf A} = e \paren {\mathbf A}$

where the $e_i$ are elementary column operation of the other two types.

For each $e_i$, let $\mathbf E_i = e_i \paren {\mathbf I}$.

Then:

\(\ds e \paren {\mathbf A}\) \(=\) \(\ds e_1 e_2 e_3 e_4 \paren {\mathbf A}\) Definition of $e$
\(\ds \) \(=\) \(\ds \mathbf A \mathbf E_4 \mathbf E_3 \mathbf E_2 \mathbf E_1\) Cases $1$ and $2$
\(\ds \) \(=\) \(\ds \mathbf A e_4 \paren {\mathbf I} \mathbf E_3 \mathbf E_2 \mathbf E_1\)
\(\ds \) \(=\) \(\ds \mathbf A e_3 e_4 \paren {\mathbf I} \mathbf E_2 \mathbf E_1\)
\(\ds \) \(=\) \(\ds \mathbf A e_2 e_3 e_4 \paren {\mathbf I} \mathbf E_1\)
\(\ds \) \(=\) \(\ds \mathbf A e_1 e_2 e_3 e_4 \paren {\mathbf I}\)
\(\ds \) \(=\) \(\ds \mathbf A e \paren {\mathbf I}\)
\(\ds \) \(=\) \(\ds \mathbf A \mathbf E\)

$\blacksquare$

Also see


Sources

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