TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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The determinant differential is $D \det (A) [H] = tr (adj (A) H)$ at every matrix, and Jacobi's formula holds on the invertible locus

Statement

Let $A$ be an $n \times n$ matrix and let $H$ be a direction matrix. Then

$D \det (A) [H] = tr (adj (A) H) .$

If $A$ is invertible, then

$D \det (A) [H] = \det (A) tr (A^{- 1} H) .$

Facts & Assumptions

Given: An $n \times n$ matrix $A$ and a perturbation direction $H$ .

[L1]

On the invertible locus, the derivative of inversion is $D inv (A) [H] = - A^{- 1} H A^{- 1}$ (On the invertible locus, $D inv (A) [H] = - A^{- 1} H A^{- 1}$ ).

Proof

technique · direct

1.1givenalgebra

The determinant is multilinear in the columns. In $\det (A + t H)$ , the coefficient of $t$ comes from replacing exactly one column of $A$ by the corresponding column of $H$ . Those cofactors are precisely the entries of $adj (A)$ , so $D \det (A) [H] = tr (adj (A) H)$ .

2.1L1step 1.1algebra∎

If $A$ is invertible, then $adj (A) = \det (A) A^{- 1}$ . Substituting this identity into step 1.1 gives $D \det (A) [H] = \det (A) tr (A^{- 1} H)$ , which is Jacobi's formula on the invertible locus. The domain restriction matters because $A^{- 1}$ appears explicitly there, unlike the adjugate formula of step 1.1.

Depends on

Used by

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alan Edelman and Steven G. Johnson, Matrix Calculus for Machine Learning and Beyond (standard reference, not scraped)

The determinant differential is Ddet⁡(A)[H]=tr⁡(adj⁡(A)H)Ddet(A)[H]=tr(adj(A)H) at every matrix, and Jacobi's formula holds on the invertible locus