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Factorized and composite linear inversion — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

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 $F^{- 1} G^{- 1} = (G F)^{- 1} = A^{- 1}$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Rank-deficient noncommuting factors break pseudoinverse composition

Statement refuted

For every pair of bounded linear maps $F : X \to Y$ and $G : Y \to Z$ between real or complex Hilbert spaces, the Moore–Penrose inverse of their composite satisfies $(G \circ F)^{†} = F^{†} G^{†}$ .

Statement references

Composite operator order and absence of a general product identity

Brandt and Horn, §2.1, pp. 2–3

Setup

Take $X = Y = Z = R^{2}$ with the Euclidean inner product and let $F = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $G = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ . Both maps have rank one, they do not commute, and for $A = G \circ F$ the product $F^{†} G^{†}$ is not $A^{†}$ .

Setup references

Finite-matrix Moore–Penrose pseudoinverse

Existence and uniqueness of the finite-matrix pseudoinverse

Facts & Assumptions

Given: The matrices $F = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $G = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ on Euclidean $R^{2}$ , with $A = G \circ F$ .

[F1]

Composite double regularization: The composite forward map has order $A = G \circ F$ ; a pseudoinverse product is not part of the composite definition.

[F2]

[F3]

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

Verification

1.1givenF1algebra

Direct multiplication gives $F G = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array})$ and $G F = A = \frac{1}{2} (\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array})$ , which are unequal. Each of $F, G, A$ is nonzero with a one-dimensional image, so both factors are rank deficient and noncommuting.

2.1givenstep 1.1F2F3algebra

Put $B = (\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array})$ . Multiplication gives $A B = G$ and $B A = F$ . Since $F$ and $G$ are self-adjoint idempotents, $A B A = (A B) A = G A = G^{2} F = A$ , $B A B = (B A) B = F B = B$ , and both $A B$ and $B A$ are self-adjoint. Thus $B$ satisfies all four equations in [F2], so $A^{+} = B$ by [F3].

3.1givenstep 2.1F2F3algebra∎

The same equations show $F^{+} = F$ and $G^{+} = G$ , because each is a self-adjoint idempotent and hence satisfies the four Penrose equations; uniqueness in [F3] identifies them. Therefore $F^{+} G^{+} = F G = \frac{1}{2} B \neq B = A^{+}$ . This refutes the universal identity.

Factorized and composite linear inversion — Examples and Counterexamples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

Commuting diagonal factors with an explicit composite estimator

Example

Example references

Facts & Assumptions

Verification

Rank-deficient noncommuting factors break pseudoinverse composition

Statement refuted

Statement references

Setup

Setup references

Facts & Assumptions

Verification

Sources