PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge 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.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A constraint removes kernel ambiguity exactly when its differences avoid the kernel

Statement

Let $X$ and $Y$ be real vector spaces, let $A : X \to Y$ be linear, and let $C \subseteq X$ be nonempty. The restriction $A ∣_{C} : C \to Y$ is injective if and only if

$(C - C) \cap \ker A = {0} .$

This condition characterizes exact uniqueness on $C$ ; it gives no stability bound for perturbed data.

Statement references

Admissible constraint classes and restricted identifiability

Linearity of the forward map

Kernel of the forward map

Christian Clason, Regularization of Inverse Problems, Chapter 3 §§3.1–3.2

Tristan van Leeuwen and Christoph Brune, 10 Lectures on Inverse Problems and Imaging, Lecture 1 §1.2

Facts & Assumptions

Given: Real vector spaces $X, Y$ , a linear map $A : X \to Y$ , and a nonempty constraint class $C \subseteq X$ .

[F1]

Admissible constraint classes and restricted identifiability with $U = X$ and the discrete metrics on $X, Y$ just specified there (used only to instantiate the model; the argument is algebraic): $C$ is nonempty, $C - C = {x - x^{'} : x, x^{'} \in C}$ , and restricted identifiability means $A x = A x^{'}$ for $x, x^{'} \in C$ implies $x = x^{'}$ .

[F2]

Linear map between vector spaces over the same field: linearity gives $A (x - x^{'}) = A x - A x^{'}$ .

[F3]

Kernel and image of a linear map: $\ker A = {x \in X : A x = 0_{Y}}$ .

Proof

1.1givenF1F2F3choosealgebra

Suppose $A ∣_{C}$ is injective and take $v \in (C - C) \cap \ker A$ . By [F1], write $v = x - x^{'}$ with $x, x^{'} \in C$ . By [F2] and [F3], $A x - A x^{'} = A v = 0_{Y}$ , so $A x = A x^{'}$ . Injectivity gives $x = x^{'}$ , hence $v = 0$ . Since $C$ is nonempty, choose $x \in C$ ; then $0 = x - x \in C - C$ . Linearity also gives $A (0_{X}) = A (0_{X} + 0_{X}) = A (0_{X}) + A (0_{X})$ , hence $A (0_{X}) = 0_{Y}$ and $0_{X} \in \ker A$ . Therefore $(C - C) \cap \ker A = {0}$ .

2.1givenF1F2F3algebra∎

Conversely, suppose $(C - C) \cap \ker A = {0}$ and take $x, x^{'} \in C$ with $A x = A x^{'}$ . Then $x - x^{'} \in C - C$ by [F1], and $A (x - x^{'}) = A x - A x^{'} = 0_{Y}$ by [F2], so $x - x^{'} \in \ker A$ by [F3]. Thus $x - x^{'} = 0$ , hence $x = x^{'}$ . Therefore $A ∣_{C}$ is injective.

Depends on

Used by

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Sources

Christian Clason, Regularization of Inverse Problems, Chapter 3 §§3.1–3.2 (standard reference, not scraped)
Tristan van Leeuwen and Christoph Brune, 10 Lectures on Inverse Problems and Imaging, Lecture 1 §1.2 (standard reference, not scraped)