ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13

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Determinant is additive in one selected column but not under simultaneous whole-matrix addition

Example

Over $Z$ , determinant is additive when one column varies and the other is fixed, but it is not additive as a function of the whole matrix.

Facts & Assumptions

Given: $u = (1, 0)^{T}$ , $v = (0, 1)^{T}$ , $w = (0, 1)^{T}$ , and $A = B = I_{2}$ .

[F1]

$Z$ is a commutative ring (The integers form a commutative ring) and the displayed columns and matrices belong to the corresponding matrix sets (Finite rectangular matrices over a commutative ring, their entries, rows and columns).

[L1]

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

[F2]

The $2 \times 2$ determinant is the two-term Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

Verification

technique · direct

1.1

Direct calculation gives $\det [u + v, w] = 1, \det [u, w] + \det [v, w] = 1 + 0 = 1.$

F1F2algebra

1.2

For simultaneous whole-matrix addition, $A + B = 2 I_{2}$ , so $\det (A + B) = 4$ , whereas $\det (A) + \det (B) = 1 + 1 = 2$ .

F1F2algebra

2.1

This is the instance $\det [u + v, w] = \det [u, w] + \det [v, w]$ of columnwise multilinearity [L1], with the second column fixed.

step 1.1L1

3.1

Steps 2.1 and 1.2 isolate the distinction: additivity holds in a single selected column, but fails for the whole matrix argument.

step 2.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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