ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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 singular local LLE covariance still has a unique minimum-norm weight vector

Example

Take $x_{1} = 0$ , $x_{2} = 1$ , $x_{3} = - 1$ , and $x_{4} = 0$ in $R$ , with $i = 1$ and $k = 3$ . The neighbor order is $N_{1} = (4, 2, 3)$ . Define the local difference matrix $Z_{1} = [x_{1} - x_{j}]_{j \in N_{1}}$ and its local covariance matrix $C_{1} = Z_{1}^{T} Z_{1}$ ; then $C_{1}$ has rank one and is singular. The affine reconstruction minimizers are $w_{4} = 1 - 2 t$ , $w_{2} = w_{3} = t$ for real $t$ , and the unique minimum-norm choice is $w_{4} = w_{2} = w_{3} = 1 / 3$ .

Example references

Hastie, Tibshirani, and Friedman, ESL II, §14.9, printed pp. 573–575

Cameron, Linear and Nonlinear Dimensionality Reduction, §7, PDF pp. 14–16

Facts & Assumptions

Given: The four real points, the subject $i = 1$ , neighbor count $k = 3$ , and the specified tie rule.

[F1]

The local LLE weights minimize squared reconstruction error under weights summing to $1$ and then minimize Euclidean norm among the minimizers (Minimum-norm affine reconstruction weights for a finite LLE neighborhood).

[F2]

For every finite nonempty neighborhood, that affine least-squares problem has a unique minimum-norm minimizer, including singular local covariance cases (Finite LLE affine least-squares weights have a unique minimum-norm minimizer).

Verification

1.1givenF1algebra

The distances from $x_{1} = 0$ to $x_{4} = 0$ , $x_{2} = 1$ , and $x_{3} = - 1$ are $0, 1, 1$ , so the deterministic list is $N_{1} = (4, 2, 3)$ . The corresponding row matrix is $Z_{1} = (0, - 1, 1)$ and $C_{1} = Z_{1}^{T} Z_{1} = (\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & - 1 & 1 \end{array})$ , which is nonzero of rank one and hence singular.

2.1step 1.1F1algebra

Write the weights in the order $(w_{4}, w_{2}, w_{3})$ . The constraint is $w_{4} + w_{2} + w_{3} = 1$ , and the reconstructed point is $w_{2} - w_{3}$ . The minimum residual is zero exactly when $w_{2} = w_{3} = t$ , giving $w_{4} = 1 - 2 t$ .

3.1step 2.1F2algebra∎

Along this affine family, the squared norm is $(1 - 2 t)^{2} + 2 t^{2} = 6 (t - 1 / 3)^{2} + 1 / 3$ , uniquely minimized at $t = 1 / 3$ . Thus the minimum-norm weights are $(1 / 3, 1 / 3, 1 / 3)$ in the ordered list, and [F2] guarantees the selection is well-defined despite singularity.

Depends on

Used by

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Sources

The Elements of Statistical Learning, second edition, corrected 12th printing (standard reference, not scraped)
Maria Cameron, Linear and Nonlinear Dimensionality Reduction, lecture notes (standard reference, not scraped)