# Determinant of Elementary Column Matrix/Scale Column and Add

## Theorem

Let $e_2$ be the elementary column operation $\text {ECO} 2$:

\((\text {ECO} 2)\) | $:$ | \(\displaystyle \kappa_i \to \kappa_i + \lambda \kappa_j \) | For some $\lambda$, add $\lambda$ times column $j$ to column $i$ |

which is to operate on some arbitrary matrix space.

Let $\mathbf E_2$ be the elementary column matrix corresponding to $e_2$.

The determinant of $\mathbf E_2$ is:

- $\map \det {\mathbf E_2} = 1$

## Proof

By Elementary Matrix corresponding to Elementary Column Operation: Scale Column and Add, $\mathbf E_2$ is of the form:

- $E_{a b} = \delta_{a b} + \lambda \cdot \delta_{b i} \cdot \delta_{j a}$

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}$

Because $i \ne j$ it follows that:

- if $a = i$ and $b = j$ then $a \ne b$

Hence when $a = b$ we have that:

- $\delta_{b i} \cdot \delta_{j a} = 0$

Hence the diagonal elements of $\mathbf E_2$ are all equal to $1$.

We also have that $\delta_{b i} \cdot \delta_{j a} = 1$ if and only if $a = i$ and $b = j$.

Hence, all elements of $\mathbf E_2$ apart from the diagonal elements and $a_{i j}$ are equal to $0$.

Thus $\mathbf E_2$ is a triangular matrix (either upper or lower).

From Determinant of Triangular Matrix, $\map \det {\mathbf E_2}$ is equal to the product of all the diagonal elements of $\mathbf E_2$.

But as we have seen, these are all equal to $1$.

Hence the result.

$\blacksquare$

## Sources

- 1998: Richard Kaye and Robert Wilson:
*Linear Algebra*... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$