Elementary Row Operations as Matrix Multiplications

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Theorem

Let $e$ be an elementary row operation.

Let $\mathbf E$ be the elementary row matrix of order $m$ defined as:

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

where $\mathbf I$ is the unit matrix.


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

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

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


Corollary

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

Then there exists an elementary row matrix of order $m$ such that:

$\mathbf {E X} = \mathbf Y$


Proof

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


Case $1$

Let $e$ be the elementary row operation $r_s \to \lambda r_s$:

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

where $\delta$ denotes the Kronecker delta.

Then:

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

$\Box$


Case $2$

Let $e$ be the elementary row operation $r_s \to r_s + \lambda r_t$:

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

where $\delta$ denotes the Kronecker delta.

Then:

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

$\Box$


Case $3$

Let $e$ be the elementary row operation $r_s \leftrightarrow r_t$:

By Exchange of Rows as Sequence of Other Elementary Row Operations, this elementary row operation can be expressed as:

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

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

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

Then:

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

$\blacksquare$


Also see


Sources