Elementary Matrix corresponding to Elementary Row Operation/Scale Row and Add

Theorem
Let $\mathbf I$ denote the unit matrix of order $m$ over a field $K$.

Let $e$ be the elementary row operation acting on $\mathbf I$ as:

for $1 \le i \le m$, $1 \le j \le m$.

Let $\mathbf E$ be the elementary row matrix defined as:
 * $\mathbf E = e \paren {\mathbf I}$

$\mathbf E$ is the square matrix of order $m$ of the form:
 * $E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$

where:
 * $E_{a b}$ denotes the element of $\mathbf E$ whose indices are $\tuple {a, b}$
 * $\delta_{a b}$ is the Kronecker delta:
 * $\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$

Proof
By definition of the unit matrix:
 * $I_{a b} = \delta_{a b}$

where:
 * $I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$.

By definition, $\mathbf E$ is the square matrix of order $m$ formed by applying $e$ to the unit matrix $\mathbf I$.

That is, all elements of row $i$ of $\mathbf I$ are to have the corresponding elements of row $j$ added to them after the latter have been multiplied by $\lambda$.

By definition of unit matrix:
 * all elements of row $i$ are $0$ except for element $I_{i i}$, which is $1$.
 * all elements of row $j$ are $0$ except for element $I_{j j}$, which is $1$.

Thus in $\mathbf E$:
 * where $a \ne i$, $E_{a b} = \delta_{a b}$
 * where $a = i$:
 * $E_{a b} = \delta_{a b}$ where $b \ne j$
 * $E_{a b} = \delta_{a b} + \lambda \cdot 1$ where $b = j$

That is:
 * $E_{a b} = \delta_{a b}$ for all elements of $\mathbf E$

except where $a = i$ and $b = j$, at which element:
 * $E_{a b} = \delta_{a b} + \lambda$

That is:
 * $E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$

Hence the result.

Also presented as
This can also be seen presented as:


 * $E_{a b} = \begin{cases} \delta_{a b} & : a \ne i \\ \delta_{i b} + \lambda \delta_{j b} & : a = i \end{cases}$

when it is considered desirable to make the nature of the rows more easily understood.