Square Matrix with Duplicate Columns has Zero Determinant

Theorem

If two columns of a square matrix over a commutative ring $\struct {R, +, \circ}$ are identical, then its determinant is zero.

If a square matrix has a zero column, its determinant is zero.

Let $\mathbf A$ be a square matrix over $R$ with two identical columns.

Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.

Then $\mathbf A^\intercal$ has two identical rows.

Then:

$\ds \map \det {\mathbf A}$

$=$

$\ds \map \det {\mathbf A^\intercal}$

$\ds $

$=$

$\ds 0$

$\blacksquare$

1994: Robert Messer: Linear Algebra: Gateway to Mathematics: $\S 7.2$
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 $1.13$
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): determinant (4)
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): determinant (4)
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): determinant (i)