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Problem packetResearch packetR194

R194Sourced evidence

Exact determinant of the disjoint-binary-support matrix

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Authored summary

The determinant is zero at odd orders and \((-1)^{n/2}\) at even orders.

The record cites sources for its explanation.

Recorded status: established

Recorded scope: every positive matrix order n

Complete recorded scope and conditions
{
  "kind": "universal",
  "statement": "every positive matrix order n"
}

Originating problem: Determinants of the disjoint-binary-support matrix

Authored record and scope
Authored title
Exact determinant of the disjoint-binary-support matrix
Record type
claim
Stored status
established
Evidence grade
sourced
Recorded scope data
{ "kind": "universal", "statement": "every positive matrix order n" }

2Authored explanation

For \(1\leq i,j\leq n\), let \((A_n)_{ij}=1\) when \(i\mathbin{\&}j=0\). Then \[ \det A_n=\begin{cases}0,&n\text{ odd},\\(-1)^{n/2},&n\text{ even}.\end{cases} \] For even order \(n=2m\), place odd indices before even indices. The matrix becomes \[ \begin{pmatrix}0&C_m\\C_m^T&A_m\end{pmatrix}, \] where \(C_m=(\mathbf1_{i\mathbin{\&}j=0})_{0\leq i<m,\,1\leq j\leq m}\). Swapping the two column blocks gives a block triangular matrix and contributes sign \((-1)^{m^2}=(-1)^m\). Since \(C_m\) is unimodular, \(\det A_{2m}=(-1)^m\).

At odd order \(2m+1\), the same split has an \((m+1)\times m\) off-diagonal block. Its transpose has a nonzero kernel vector, which yields a kernel vector of \(A_{2m+1}\). Thus the odd determinant vanishes.

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3Evidence

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Verification source: arxiv.org ↗, Mereb, On determinants of matrices related to Pascal's triangle, Corollary 1, combined with the parity block decomposition recorded here

4How it connects

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  "ref": "R194",
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  "slug": "dbm-claim-exact-determinant",
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  "title": "Exact determinant of the disjoint-binary-support matrix",
  "summary": "The determinant is zero at odd orders and \\((-1)^{n/2}\\) at even orders.",
  "relevance": "This records a result, observation, or conjecture used to assess the target problem.",
  "relevance_source": "derived",
  "body": "For \\(1\\leq i,j\\leq n\\), let \\((A_n)_{ij}=1\\) when \\(i\\mathbin{\\&}j=0\\). Then\n\\[\n\\det A_n=\\begin{cases}0,&n\\text{ odd},\\\\(-1)^{n/2},&n\\text{ even}.\\end{cases}\n\\]\nFor even order \\(n=2m\\), place odd indices before even indices. The matrix becomes\n\\[\n\\begin{pmatrix}0&C_m\\\\C_m^T&A_m\\end{pmatrix},\n\\]\nwhere \\(C_m=(\\mathbf1_{i\\mathbin{\\&}j=0})_{0\\leq i<m,\\,1\\leq j\\leq m}\\). Swapping the two column blocks gives a block triangular matrix and contributes sign \\((-1)^{m^2}=(-1)^m\\). Since \\(C_m\\) is unimodular, \\(\\det A_{2m}=(-1)^m\\).\n\nAt odd order \\(2m+1\\), the same split has an \\((m+1)\\times m\\) off-diagonal block. Its transpose has a nonzero kernel vector, which yields a kernel vector of \\(A_{2m+1}\\). Thus the odd determinant vanishes.",
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  "scope": {
    "kind": "universal",
    "statement": "every positive matrix order n"
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    {
      "slug": "R195",
      "title": "The required Pascal blocks are unimodular",
      "object_type": "claim",
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    {
      "slug": "R196",
      "title": "Rank and Smith normal form follow from the determinant formula",
      "object_type": "claim",
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      "direction": "incoming"
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    {
      "slug": "R193",
      "title": "Exact determinant sweep through order 130",
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    {
      "slug": "R1277",
      "title": "Lean proof of the exact disjoint-support determinant formula",
      "object_type": "formalization",
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      "direction": "incoming"
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    {
      "slug": "disjoint-binary-support-determinant",
      "title": "disjoint binary support determinant",
      "object_type": "problem",
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6Provenance

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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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