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Problem packetResearch packetR676

R676Sourced evidence

The binary order-n singular probability equals the sign-matrix order-(n+1) singular probability

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Authored summary

Bordering and sign normalization identify the binary problem with the standard Bernoulli sign model one order higher.

The record cites sources for its explanation.

Recorded status: established

Recorded scope: every n by n binary matrix and the corresponding normalized (n+1) by (n+1) sign matrix, for every positive integer n

Complete recorded scope and conditions
{
  "kind": "universal",
  "statement": "every n by n binary matrix and the corresponding normalized (n+1) by (n+1) sign matrix, for every positive integer n"
}

Originating problem: Number of singular ten by ten binary matrices over the reals

Authored record and scope
Authored title
The binary order-n singular probability equals the sign-matrix order-(n+1) singular probability
Record type
claim
Stored status
established
Evidence grade
sourced
Recorded scope data
{ "kind": "universal", "statement": "every n by n binary matrix and the corresponding normalized (n+1) by (n+1) sign matrix, for every positive integer n" }

2Authored explanation

For a binary n by n matrix A, define the normalized sign matrix \[ H(A)=\begin{pmatrix}1&\mathbf1^{\mathsf T}\\ \mathbf1&J-2A\end{pmatrix}. \] Subtracting the first row from every other row gives \[ \det H(A)=(-2)^n\det A. \] Every sign matrix can be normalized to first row and first column equal to one by row and column sign changes. Each normalized matrix has the same number 2^(2n+1) of sign matrices in its normalization fiber. Consequently, the singular probability for n by n uniform binary matrices equals the singular probability for (n+1) by (n+1) uniform sign matrices.

Tikhomirov proved that the latter probability is (1/2+o(1))^(n+1). This asymptotic theorem explains the scale of the problem. Its constants do not give a numerical order-eleven bound, so the finite interval in rsbm10-claim-certified-count-interval uses exact elementary counts.

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3Evidence

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Verification source: doi.org ↗, Tikhomirov, Singularity of random Bernoulli matrices, Annals of Mathematics 191 (2020), 593-634; the bordering identity is checked algebraically in this record

4What was measured

5How it connects

Recorded for

Machine-readable record

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json
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  "ref": "R676",
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  "slug": "rsbm10-claim-bernoulli-normalization",
  "type": "claim",
  "title": "The binary order-n singular probability equals the sign-matrix order-(n+1) singular probability",
  "summary": "Bordering and sign normalization identify the binary problem with the standard Bernoulli sign model one order higher.",
  "relevance": "For Number of singular ten by ten binary matrices over the reals, record rsbm10-claim-bernoulli-normalization (“The binary order-n singular probability equals the sign-matrix order-(n+1) singular probability”) records a bound, answer, status fact, or structural consequence. The record states: Bordering and sign normalization identify the binary problem with the standard Bernoulli sign model one order higher.",
  "relevance_source": "recorded",
  "body": "For a binary n by n matrix A, define the normalized sign matrix\n\\[\nH(A)=\\begin{pmatrix}1&\\mathbf1^{\\mathsf T}\\\\ \\mathbf1&J-2A\\end{pmatrix}.\n\\]\nSubtracting the first row from every other row gives\n\\[\n\\det H(A)=(-2)^n\\det A.\n\\]\nEvery sign matrix can be normalized to first row and first column equal to one by row and column sign changes. Each normalized matrix has the same number 2^(2n+1) of sign matrices in its normalization fiber. Consequently, the singular probability for n by n uniform binary matrices equals the singular probability for (n+1) by (n+1) uniform sign matrices.\n\nTikhomirov proved that the latter probability is (1/2+o(1))^(n+1). This asymptotic theorem explains the scale of the problem. Its constants do not give a numerical order-eleven bound, so the finite interval in rsbm10-claim-certified-count-interval uses exact elementary counts.",
  "status": "established",
  "evidence_grade": "sourced",
  "scope": {
    "kind": "universal",
    "statement": "every n by n binary matrix and the corresponding normalized (n+1) by (n+1) sign matrix, for every positive integer n"
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  "reproduction": {
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    "citation": {
      "url": "https://doi.org/10.4007/annals.2020.191.2.6",
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  "source": {
    "url": "https://doi.org/10.4007/annals.2020.191.2.6",
    "locator": "Tikhomirov, Singularity of random Bernoulli matrices, Annals of Mathematics 191 (2020), 593-634; the bordering identity is checked algebraically in this record"
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    {
      "slug": "R677",
      "title": "The singular count is between 126,174,821,830,345,268,667,240,568,576 and 901,210,462,928,281,273,073,900,978,176",
      "object_type": "claim",
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    {
      "slug": "real-singular-binary-matrices-ten",
      "title": "real singular binary matrices ten",
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7Provenance

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