LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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$\frac{d}{d x} \det (I - x A) = - tr (adj (I - x A) A)$

Statement

Let $R$ be a commutative ring, let $p \geq 1$ , let $A \in M_{p} (R)$ , and put $M (x) = I_{p} - x A$ . Then, with the formal derivative,

$\frac{d}{d x} \det M (x) = - {tr}_{R [x]} (adj (M (x)) A) .$

The subscript names the coefficient ring the trace is taken over. Since $adj (M (x)) A$ has entries in $R [x]$ , not in $R$ , the trace here is the one belonging to the commutative ring $R [x]$ ; the defining formula $\sum_{i < p} a_{i i}$ is the same.

Facts & Assumptions

Given: A commutative ring $R$ , a positive size $p$ , a matrix $A \in M_{p} (R)$ , and $M = I_{p} - x A$ .

[L1]

The ring trace is the finite sum of the diagonal entries (The trace of a square matrix over a commutative ring).

[L2]

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

[L3]

The determinant is the Leibniz sum $\det (M) = \sum_{σ} sgn (σ) \prod_{i} M_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L4]

The formal derivative is coefficientwise and sends $x$ to $1$ and constants to $0$ (The formal derivative $D (\sum a_{n} x^{n}) = \sum_{n \geq 1} n a_{n} x^{n - 1}$ ).

[L5]

Formal differentiation is linear and satisfies the product rule (Formal differentiation is linear and satisfies product, power, quotient, chain, and coefficient-recovery laws).

Proof

technique · direct

1.1

Differentiate the finite Leibniz sum [L3]. By [L5], each product contributes one term for each selected matrix entry, and grouping all terms that differentiate $M_{i j}$ leaves its cofactor $C_{i j} (M)$ .

givenL3L5

1.2

By [L1] and matrix multiplication, ${tr}_{R} (adj (M) A) = \sum_{i} \sum_{j} adj (M)_{i j} A_{j i}$ . Using [L2] and renaming the finite indices gives $\sum_{i, j} C_{i j} (M) A_{i j}$ .

L1L2algebra

2.1

Thus $(\det M)^{'} = \sum_{i, j} C_{i j} (M) M_{i j}^{'}$ . Since $M^{'} = - A$ by [L4], this is $- \sum_{i, j} C_{i j} (M) A_{i j}$ .

step 1.1L4algebra

3.1

Comparing steps 2.1 and 1.2 gives the displayed derivative identity. Positive size supplies every cofactor used in [L2].

step 2.1step 1.2L2∎

Depends on

Used by

Closed walks have trace and logarithmic-derivative generating functions Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 14 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

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Corollary 4.7.3 (standard reference, not scraped)

ddxdet⁡(I−xA)=−tr⁡(adj⁡(I−xA)A)dxd​det(I−xA)=−tr(adj(I−xA)A)