LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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For $A \in M_{n} (R)$ and columns $u, v$ over a commutative ring, $\det (A + u v^{T}) = \det (A) + v^{T} adj (A) u$

Statement

Let $R$ be a commutative ring, $n \geq 1$ , $A \in M_{n} (R)$ , and $u, v \in M_{n \times 1} (R)$ . Then

$\det (A + u v^{T}) = \det (A) + v^{T} adj (A) u .$

Facts & Assumptions

Given: $R, n, A, u, v$ as in the statement.

[L1]

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

[L2]

Expansion along column $j$ is $\det (B) = \sum_{i} b_{i j} C_{i j} (B)$ (Laplace expansion computes the determinant along every row and every column over a commutative ring).

[F1]

$adj (A)_{j i} = C_{i j} (A)$ (Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring).

[F2]

Matrix products and transposes are given by their entry formulas (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L3]

Matrix addition and multiplication obey the usual distributive laws (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · direct

1.1

Column $j$ of $A + u v^{T}$ is $A_{j} + v_{j} u$ . Expanding the determinant by column multilinearity gives one term for each subset of columns chosen from $u v^{T}$ .

F2L1L3

2.1

The empty subset contributes $\det (A)$ . Every term choosing at least two columns from $u v^{T}$ vanishes, since those chosen columns are scalar multiples of the same column $u$ and alternation makes the determinant zero.

step 1.1L1

2.2

For the singleton subset ${j}$ , pull out $v_{j}$ and expand the determinant of $A$ with column $j$ replaced by $u$ along that column. Deleting that replaced column leaves exactly the same minors as deleting column $j$ from $A$ , so its contribution is $v_{j} \sum_{i} u_{i} C_{i j} (A)$ .

step 1.1L1L2

3.1

Summing step 2.2 over $j$ and using [F1] gives $\sum_{i, j} v_{j} C_{i j} (A) u_{i} = v^{T} adj (A) u$ . Together with step 2.1, this is the claimed identity.

step 2.1step 2.2F1L3∎

Depends on

Used by

Over a commutative ring, adj(P⁻¹AP)=P⁻¹adj(A)P for every invertible P Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Jaroslav Vrabel, A note on the matrix determinant lemma (standard reference, not scraped)
András Pál, Introduction to Commutative Algebra (standard reference, not scraped)

For A∈Mn(R)A∈Mn​(R) and columns u,vu,v over a commutative ring, det⁡(A+uvT)=det⁡(A)+vTadj⁡(A)udet(A+uvT)=det(A)+vTadj(A)u