Determinant of Elementary Column Matrix/Exchange Columns
Theorem
Let $e_3$ be the elementary column operation $\text {ECO} 3$:
\((\text {ECO} 3)\) | $:$ | \(\ds \kappa_i \leftrightarrow \kappa_j \) | Exchange columns $i$ and $j$ |
which is to operate on some arbitrary matrix space.
Let $\mathbf E_3$ be the elementary column matrix corresponding to $e_3$.
The determinant of $\mathbf E_3$ is:
- $\map \det {\mathbf E_3} = -1$
Proof
Let $\mathbf I$ denote the unit matrix of arbitrary order $n$.
By Determinant of Unit Matrix:
- $\map \det {\mathbf I} = 1$
Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$.
From Parity of K-Cycle, $\map \sgn \rho = -1$.
By definition we have that $\mathbf E_3$ is $\mathbf I$ with columns $i$ and $j$ transposed.
By the definition of a determinant:
- $\ds \map \det {\mathbf I} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }$
By Permutation of Determinant Indices:
- $\ds \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$
We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
\(\ds \map \det {\mathbf E_3}\) | \(=\) | \(\ds \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\map \det {\mathbf I}\) |
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 $1.13$