CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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A square matrix with a zero column or two equal columns has determinant zero

Statement

Let $n \geq 1$ . If a matrix in $M_{n} (R)$ over a commutative ring has a zero column or has two equal columns, then its determinant is zero.

Facts & Assumptions

Given: A square matrix $A$ over a commutative ring.

[L1]

Determinant is column-multilinear and alternating (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

Proof

technique · direct

1.1L1algebra

If one column is zero, multilinearity gives $\det (A) = \det (A) + \det (A)$ by writing that column as $0 + 0$ ; cancellation in the additive group gives $\det (A) = 0$ .

2.1step 1.1L1∎

If two columns are equal, alternation in [L1] gives $\det (A) = 0$ directly, regardless of whether the columns are adjacent.

Depends on

The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring

Dependency tree · 2 levels

Sources

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Theorem 4.19