Problem packetLean verificationR864
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Originating problem: Determinants of the Fibonacci-sum matrix
Authored record and environment
- Authored title
- Almost-TU determinant and nonvanishing cofactors
- Authored summary
- Lean proves that a least-order bad sign minor has determinant plus or minus two and none of its cofactors vanish.
- Stored status
- draft
- Evidence grade
- unverified_formalization
- Lean world
- lean-4.33.0-rc1/mathlib4@4608056c77c52468b80773e8dcd585ef821c7c5e+theoremdb@d575c4e2ff28345440c4f8a42bf0178bcb3f6f41b703a45d9d1cbb709036f0dc
2Authored explanation
The proof uses the adjugate identities and minimality. Deleting a zero coordinate leaves a cofactor matrix with determinant plus or minus one, which forces the bad determinant to divide every remaining coordinate. An adjugate column rules out zero cofactors. A signed difference of two adjugate columns forces the determinant to divide two.
3Formal statement
theorem minimal_bad_square_minor_det_and_adjugate_nonzero {m n : Type*} [Fintype m] [DecidableEq m] [Fintype n] [DecidableEq n] (A : Matrix m n ℤ) (hentries : ∀ i j, A i j ∈ Set.range SignType.cast) (k : ℕ) (f : Fin k → m) (g : Fin k → n) (hf : f.Injective) (hg : g.Injective) (hbad : (A.submatrix f g).det ∉ Set.range SignType.cast) (hminimal : ∀ (l : ℕ), l < k → ∀ (f' : Fin l → m) (g' : Fin l → n), f'.Injective → g'.Injective → (A.submatrix f' g').det ∈ Set.range SignType.cast) (hk : 3 ≤ k) : ((A.submatrix f g).det = 2 ∨ (A.submatrix f g).det = -2) ∧ ∀ i j, (A.submatrix f g).adjugate i j ≠ 0 := by
exact ⟨minimal_bad_square_minor_det_eq_two_or_neg_two A k f g hf hg hbad hminimal hk, minimal_bad_square_minor_adjugate_nonzero A k f g hf hg hbad hminimal (by omega)⟩Continue this work
Replay material: partial
4Verification
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Verification source: mathoverflow.net ↗, formal/lean/TheoremDB/Matrix/Camion.lean
5What was measured
6How it connects
Depended on by
- formalization
Cite this record
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Machine-readable record
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{
"schema": "theoremdb-agent-record-v1",
"ref": "R864",
"content_hash": null,
"slug": "fib-formalization-camion-minimal-obstruction",
"type": "formalization",
"title": "Almost-TU determinant and nonvanishing cofactors",
"summary": "Lean proves that a least-order bad sign minor has determinant plus or minus two and none of its cofactors vanish.",
"relevance": "For fib problem determinant range; fib problem nonzero support, record fib-formalization-camion-minimal-obstruction (“Almost-TU determinant and nonvanishing cofactors”) states a machine-checkable theorem or proof obligation. The record states: Lean proves that a least-order bad sign minor has determinant plus or minus two and none of its cofactors vanish.",
"relevance_source": "recorded",
"body": "The proof uses the adjugate identities and minimality. Deleting a zero coordinate leaves a cofactor matrix with determinant plus or minus one, which forces the bad determinant to divide every remaining coordinate. An adjugate column rules out zero cofactors. A signed difference of two adjugate columns forces the determinant to divide two.",
"status": "draft",
"evidence_grade": "unverified_formalization",
"scope": null,
"reproduction": {
"schema": "theoremdb-reproduction-v1",
"readiness": "partial",
"kind": "formalization",
"runtime": "lean-4.33.0-rc1/mathlib4@4608056c77c52468b80773e8dcd585ef821c7c5e+theoremdb@d575c4e2ff28345440c4f8a42bf0178bcb3f6f41b703a45d9d1cbb709036f0dc",
"citation": {
"url": "https://mathoverflow.net/questions/513340/is-the-determinant-of-this-fibonacci-sum-indicator-matrix-always-1-0-or/513372",
"locator": "formal/lean/TheoremDB/Matrix/Camion.lean"
},
"missing": [
"source",
"command",
"expected_output"
]
},
"formal_statement": "theorem minimal_bad_square_minor_det_and_adjugate_nonzero {m n : Type*} [Fintype m] [DecidableEq m] [Fintype n] [DecidableEq n] (A : Matrix m n ℤ) (hentries : ∀ i j, A i j ∈ Set.range SignType.cast) (k : ℕ) (f : Fin k → m) (g : Fin k → n) (hf : f.Injective) (hg : g.Injective) (hbad : (A.submatrix f g).det ∉ Set.range SignType.cast) (hminimal : ∀ (l : ℕ), l < k → ∀ (f' : Fin l → m) (g' : Fin l → n), f'.Injective → g'.Injective → (A.submatrix f' g').det ∈ Set.range SignType.cast) (hk : 3 ≤ k) : ((A.submatrix f g).det = 2 ∨ (A.submatrix f g).det = -2) ∧ ∀ i j, (A.submatrix f g).adjugate i j ≠ 0 := by\n exact ⟨minimal_bad_square_minor_det_eq_two_or_neg_two A k f g hf hg hbad hminimal hk, minimal_bad_square_minor_adjugate_nonzero A k f g hf hg hbad hminimal (by omega)⟩",
"source": {
"url": "https://mathoverflow.net/questions/513340/is-the-determinant-of-this-fibonacci-sum-indicator-matrix-always-1-0-or/513372",
"locator": "formal/lean/TheoremDB/Matrix/Camion.lean"
},
"models": [],
"relations": [
{
"slug": "R860",
"title": "Camion's total-unimodularity criterion",
"object_type": "formalization",
"relation": "depends_on",
"direction": "incoming"
}
]
}8Provenance
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