Square Matrix with Duplicate Columns has Zero Determinant
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Theorem
If two columns of a square matrix over a commutative ring $\struct {R, +, \circ}$ are identical, then its determinant is zero.
Corollary
If a square matrix has a zero column, its determinant is zero.
Proof
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:
\(\ds \map \det {\mathbf A}\) | \(=\) | \(\ds \map \det {\mathbf A^\intercal}\) | Determinant of Transpose | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Square Matrix with Duplicate Rows has Zero Determinant |
$\blacksquare$
Also see
Sources
- 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 $1.13$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): determinant (i)