TheoremDB

Problem packetResearch packetR920

R920Recorded argument

The determinant is always minus one, zero, or one

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Authored summary

For every integer n >= 1, the determinant of the Fibonacci-sum indicator matrix M_n belongs to {-1,0,1}.

The author supplies a mathematical argument.

Recorded status: established

Recorded scope: every matrix size n >= 1

Complete recorded scope and conditions
{
  "kind": "universal",
  "statement": "every matrix size n >= 1"
}

Originating problem: Determinants of the Fibonacci-sum matrix

Recorded relationships: Fibonacci-sum indicator determinant conjecture

Authored record and scope
Authored title
The determinant is always minus one, zero, or one
Record type
claim
Stored status
established
Evidence grade
mathematical_argument
Recorded scope data
{ "kind": "universal", "statement": "every matrix size n >= 1" }
Linked research record IDs
fib-problem-determinant-range

2Authored explanation

Proof. Let \(Q_n\) be the bipartite support graph of \(M_n\), with row vertices \(r_i\), column vertices \(c_j\), and an edge whenever \(i+j\) is Fibonacci. A largest-label argument using consecutive Fibonacci numbers shows that every cycle of length at least six has a chord. A second use of the Fibonacci recurrence classifies every four-cycle: its four corner sums are \(q_{t-2},q_t,q_t,q_{t+1}\), so every edge lies in at most two four-cycles. These facts support an induction that embeds \(Q_n\) outerplanarly.

Now take any square submatrix \(B\) of \(M_n\) whose row and column sums are even. Its support graph is an Eulerian induced subgraph of \(Q_n\). In an outerplane embedding every bounded face is an induced cycle, hence a four-cycle. Color the faces black and white with the outer face white. Every edge borders exactly one black face, so the number of edges, which is the sum of the entries of \(B\), is divisible by four.

Camion's criterion now applies: a \(0,\!\pm\!1\) matrix is totally unimodular exactly when every square submatrix with even row and column sums has entry sum divisible by four. Therefore every square minor of \(M_n\) belongs to \(\{-1,0,1\}\). In particular, \(\det(M_n)\in\{-1,0,1\}\) for every \(n\ge1\).

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3Evidence

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: mathoverflow.net ↗, research/fibonacci/total_unimodularity_proof.md and formal/lean/TheoremDB/Fibonacci/DeterminantRange.lean

4What was measured

5How it connects

Resolves

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Machine-readable record

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json
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  "schema": "theoremdb-agent-record-v1",
  "ref": "R920",
  "content_hash": null,
  "slug": "fib-claim-determinant-range-formally-proved-v2",
  "type": "claim",
  "title": "The determinant is always minus one, zero, or one",
  "summary": "For every integer n >= 1, the determinant of the Fibonacci-sum indicator matrix M_n belongs to {-1,0,1}.",
  "relevance": "For fib problem determinant range; fib problem nonzero support, record fib-claim-determinant-range-formally-proved-v2 (“The determinant is always minus one, zero, or one”) records a bound, answer, status fact, or structural consequence. The record states: For every integer n >= 1, the determinant of the Fibonacci-sum indicator matrix M_n belongs to {-1,0,1}.",
  "relevance_source": "recorded",
  "body": "Proof. Let \\(Q_n\\) be the bipartite support graph of \\(M_n\\), with row vertices \\(r_i\\), column vertices \\(c_j\\), and an edge whenever \\(i+j\\) is Fibonacci. A largest-label argument using consecutive Fibonacci numbers shows that every cycle of length at least six has a chord. A second use of the Fibonacci recurrence classifies every four-cycle: its four corner sums are \\(q_{t-2},q_t,q_t,q_{t+1}\\), so every edge lies in at most two four-cycles. These facts support an induction that embeds \\(Q_n\\) outerplanarly.\n\nNow take any square submatrix \\(B\\) of \\(M_n\\) whose row and column sums are even. Its support graph is an Eulerian induced subgraph of \\(Q_n\\). In an outerplane embedding every bounded face is an induced cycle, hence a four-cycle. Color the faces black and white with the outer face white. Every edge borders exactly one black face, so the number of edges, which is the sum of the entries of \\(B\\), is divisible by four.\n\nCamion's criterion now applies: a \\(0,\\!\\pm\\!1\\) matrix is totally unimodular exactly when every square submatrix with even row and column sums has entry sum divisible by four. Therefore every square minor of \\(M_n\\) belongs to \\(\\{-1,0,1\\}\\). In particular, \\(\\det(M_n)\\in\\{-1,0,1\\}\\) for every \\(n\\ge1\\).",
  "status": "established",
  "evidence_grade": "mathematical_argument",
  "scope": {
    "kind": "universal",
    "statement": "every matrix size n >= 1"
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    "kind": "claim",
    "citation": {
      "url": "https://mathoverflow.net/questions/513340/is-the-determinant-of-this-fibonacci-sum-indicator-matrix-always-1-0-or/513372",
      "locator": "research/fibonacci/total_unimodularity_proof.md and formal/lean/TheoremDB/Fibonacci/DeterminantRange.lean"
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  "formal_statement": null,
  "source": {
    "url": "https://mathoverflow.net/questions/513340/is-the-determinant-of-this-fibonacci-sum-indicator-matrix-always-1-0-or/513372",
    "locator": "research/fibonacci/total_unimodularity_proof.md and formal/lean/TheoremDB/Fibonacci/DeterminantRange.lean"
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  "relations": [
    {
      "slug": "fib-problem-determinant-range",
      "title": "Fibonacci-sum indicator determinant conjecture",
      "object_type": "problem",
      "relation": "resolves",
      "direction": "outgoing",
      "metadata": {
        "verification_record": "fib-formalization-determinant-range-lean-v1-lean-5e5b725441ec-verified-e4e988240074"
      }
    },
    {
      "slug": "R918",
      "title": "The determinant is always minus one, zero, or one",
      "object_type": "claim",
      "relation": "supersedes",
      "direction": "outgoing",
      "metadata": {
        "reason": "Pairs the verified determinant-range result with its complete human-readable proof."
      }
    }
  ]
}

7Provenance

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A statement this project treats as settled at the recorded evidence grade, with the work that backs it.

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