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

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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]

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.

[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.

Depends on

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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)
Roger Penrose, A generalized inverse for matrices (standard reference, not scraped)