LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero

Statement

Let $m, n \geq 1$ , let $U \subseteq R^{m}$ be open, and let $r \in N$ . If the entries of $A : U \to M_{n} (R)$ are $C^{r}$ and $\det A$ never vanishes, then the entries of $A^{- 1}$ are $C^{r}$ .

Facts & Assumptions

Given: The matrix-valued map in the Statement. Cofactors have the meaning of Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring, and the reciprocal derivative rule is supplied by Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ .

[L1]

For an invertible square matrix, $A^{- 1} = \det (A)^{- 1} adj (A) .$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ )

[L2]

The function $\det : M_{n} (R) \to R$ is evaluation of the polynomial $\sum_{σ \in S_{n}} sgn (σ) x_{1, σ (1)} \dots x_{n, σ (n)} .$ (For every fixed finite size at least one, the determinant of a real square matrix is a polynomial in its matrix entries)

[L3]

Finite componentwise sums and products of $C^{k}$ Euclidean maps are $C^{k}$ , and a composite of composable $C^{k}$ Euclidean maps is $C^{k}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

Proof

technique · direct

1.1

Every cofactor and the determinant are polynomial expressions in the $C^{r}$ entries of $A$ , so they are $C^{r}$ by [L2] and [L3]; [L1] identifies the only remaining factor needed for the inverse.

L1L2L3

1.2

On $R ∖ {0}$ , repeated differentiation of $h (t) = t^{- 1}$ gives $h^{(j)} (t) = (- 1)^{j} j! t^{- j - 1}$ . This formula follows by induction from the reciprocal and product rules, and every derivative displayed is continuous on that domain.

givenalgebra

2.1

Since $\det A$ never vanishes, its image lies in the domain of step 1.2. Thus $(\det A)^{- 1}$ is $C^{r}$ by composition, and [L1] together with step 1.1 and [L3] makes every entry of $A^{- 1}$ $C^{r}$ .

step 1.1step 1.2L1L3∎

Depends on

Used by

Dependency tree · two levels

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Sources

J. Lebl, Basic Analysis II, §§8.5–8.6 (standard reference, not scraped)

Matrix inversion preserves CkCk regularity where the determinant is nonzero