ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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Commuting diagonal factors with an explicit composite estimator

Example

Let $X = Y = Z = R^{2}$ with the Euclidean inner product, $F = diag (2, 1)$ , $G = diag (3, 4)$ , and $A = G \circ F$ . For $α, γ > 0$ , use the Tikhonov maps $R_{α}^{G} = (G^{*} G + α I)^{- 1} G^{*}$ and $R_{γ}^{F} = (F^{*} F + γ I)^{- 1} F^{*}$ . Compute the two-stage estimate $R_{γ}^{F} R_{α}^{G} z$ for arbitrary $z = (z_{1}, z_{2})$ and its limit as $α, γ ↓ 0$ .

Example references

Composite estimator order

Tikhonov spectral filter

Brandt and Horn, §2.1

Clason, Chapter 6

Facts & Assumptions

Given: $X = Y = Z = R^{2}$ , $F = diag (2, 1)$ , $G = diag (3, 4)$ , $A = G \circ F$ , $α, γ > 0$ , and $z = (z_{1}, z_{2})$ .

[F1]

Composite double regularization: The forward map is $A = G \circ F$ , and the composite estimator applies the $G$ -regularizer first and the $F$ -regularizer second.

[F2]

Compact spectral filters, source conditions, and qualification: Tikhonov regularization for a compact operator $H$ is $R_{η} = (H^{*} H + η I)^{- 1} H^{*}$ .

[F3]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: For a matrix $M$ , its Moore–Penrose pseudoinverse $M^{+}$ satisfies $M M^{+} M = M$ , $M^{+} M M^{+} = M^{+}$ , and the square products $M M^{+}$ and $M^{+} M$ are self-adjoint.

[F4]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: Every finite real or complex matrix has a unique matrix satisfying those four Penrose equations.

Verification

1.1givenF2algebra

Since $G^{*} G = diag (9, 16)$ and $F^{*} F = diag (4, 1)$ , the specified Tikhonov maps are $R_{α}^{G} = diag (3 / (9 + α), 4 / (16 + α))$ and $R_{γ}^{F} = diag (2 / (4 + γ), 1 / (1 + γ))$ .

2.1givenF1step 1.1algebra

Multiplying in the required order gives $A = diag (6, 4)$ and $R_{γ}^{F} R_{α}^{G} z = diag (6 / ((4 + γ) (9 + α)), 4 / ((1 + γ) (16 + α))) z$ .

3.1givenstep 2.1F3F4algebra∎

As $α, γ ↓ 0$ , the diagonal coefficients tend to $1 / 6$ and $1 / 4$ , so the estimate tends to $A^{- 1} z$ for $A = diag (6, 4)$ . Since $A A^{- 1} = I = A^{- 1} A$ , we have $A A^{- 1} A = A$ , $A^{- 1} A A^{- 1} = A^{- 1}$ , and both square products are the self-adjoint identity; [F3]–[F4] therefore give $A^{+} = A^{- 1}$ . The same argument gives $F^{+} = F^{- 1}$ and $G^{+} = G^{- 1}$ , and $F^{- 1} G^{- 1} = (G F)^{- 1} = A^{- 1}$ .

Depends on

Used by

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Sources

Christina Brandt and Patrick Horn, Regularisation and Parameter Choice Rules for Composite Linear Inverse Problems, §2.1 (standard reference, not scraped)
Christian Clason, Regularization of Inverse Problems, full lecture notes, Chapter 6 (standard reference, not scraped)