Elementary Matrix corresponding to Elementary Row Operation/Scale Row

Theorem
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.

Let $\mathbf I$ denote the unit matrix of order $m$ in $\map \MM {m, n}$.

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

for $1 \le k \le m$.

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

$\mathbf E$ is of the form:

$E_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end{cases}$ 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 $k$ of $\mathbf I$ are to be multiplied by $\lambda$.

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

Thus in $\mathbf E$:
 * $E_{k k} = \lambda \cdot 1 = \lambda$

The elements in all the other rows of $\mathbf E$ are the same as the corresponding elements of $\mathbf I$.

Hence the result.