Elementary Matrix corresponding to Elementary Column Operation/Scale Column
Theorem
Let $\mathbf I$ denote the unit matrix of order $n$ over a field $K$.
Let $e$ be the elementary column operation acting on $\mathbf I$ as:
\((\text {ECO} 1)\) | $:$ | \(\ds \kappa_k \to \lambda \kappa_k \) | For some $\lambda \in K_{\ne 0}$, multiply column $k$ of $\mathbf I$ by $\lambda$ |
for $1 \le k \le n$.
Let $\mathbf E$ be the elementary column matrix defined as:
- $\mathbf E = e \paren {\mathbf I}$
$\mathbf E$ is the square matrix of order $n$ 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:
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 column $k$ of $\mathbf I$ are to be multiplied by $\lambda$.
By definition of unit matrix, all elements of column $k$ 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 columns of $\mathbf E$ are the same as the corresponding elements of $\mathbf I$.
Hence the result.
$\blacksquare$