# Determinant of Elementary Column Matrix/Scale Column

## Theorem

Let $e_1$ be the elementary column operation $\text {ECO} 1$:

 $(\text {ECO} 1)$ $:$ $\displaystyle \kappa_k \to \lambda \kappa_k$ For some $\lambda \ne 0$, multiply column $k$ by $\lambda$

which is to operate on some arbitrary matrix space.

Let $\mathbf E_1$ be the elementary column matrix corresponding to $e_1$.

The determinant of $\mathbf E_1$ is:

$\map \det {\mathbf E_1} = \lambda$

## Proof

By Elementary Matrix corresponding to Elementary Column Operation: Scale Column, the elementary column matrix corresponding to $e_1$ 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 $\mathbf E_1$ 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$} \end {cases}$

Thus when $a \ne b$, $E_{a b} = 0$.

This means that $\mathbf E_1$ is a diagonal matrix.

 $\ds \displaystyle \map \det {\mathbf E_1}$ $=$ $\ds \prod_i E_{i i}$ Determinant of Diagonal Matrix where the index variable $i$ ranges over the order of $\mathbf E_1$ $\ds$ $=$ $\ds \prod_i \paren {\begin {cases} 1 & : i \ne k \\ \lambda & : a = k \end{cases} }$ $\ds$ $=$ $\ds \prod_{i \mathop \ne k} 1 \times \prod_{i \mathop = k} \lambda$ $\ds$ $=$ $\ds 1 \times \lambda$ $\ds$ $=$ $\ds \lambda$

$\blacksquare$