ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Sample covariance is singular when dimension exceeds sample size

Example

Let $X_{1}, \dots, X_{n} \in R^{p}$ be any observations and let ${\hat{Σ}}^{*}$ be their sample-centered covariance from Sample covariance estimators and squared operator loss.

Facts & Assumptions

Given: Integers $n \geq 1$ , $p > n$ , and observations $X_{1}, \dots, X_{n} \in R^{p}$ .

[F1]

The sample-centered covariance is $n^{- 1} \sum_{i} (X_{i} - \overset{ˉ}{X}) (X_{i} - \overset{ˉ}{X})^{T}$ , with $\overset{ˉ}{X} = n^{- 1} \sum_{i} X_{i}$ (Sample covariance estimators and squared operator loss).

[F2]

For an $m \times p$ matrix $A$ , $rank (A) + \dim N (A) = p$ (For an $m \times n$ matrix $A$ , $rank (A) + \dim N (A) = n$ ).

[F3]

Matrix rank is the dimension of the image and the nullspace is the set mapped to zero (Row space, column space, nullspace, row rank, column rank and matrix rank).

Proof

1.1F1F2F3givenalgebra

Let $C$ be the $n \times p$ matrix with rows $(X_{i} - \overset{ˉ}{X})^{T}$ . Since $\sum_{i} (X_{i} - \overset{ˉ}{X}) = 0$ , every column of $C$ lies in $\ker (1_{n}^{T})$ . The functional $1_{n}^{T} : R^{n} \to R$ is nonzero because $n \geq 1$ , so its rank is one and [F2] gives kernel dimension $n - 1$ . Thus $rank (C) \leq n - 1$ .

2.1F1F2F3step 1.1algebra∎

By [F1], ${\hat{Σ}}^{*} = C^{T} C / n$ . If $v \in N (C)$ , then $C^{T} C v = 0$ , so $N (C) \subseteq N (C^{T} C)$ . The matrices have the same $p$ -dimensional domain; [F2] and this nullspace inclusion give $rank (C^{T} C) \leq rank (C) \leq n - 1 < p$ . Multiplication by $1 / n \neq 0$ does not change rank, so ${\hat{Σ}}^{*}$ is singular.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Philippe Rigollet, 18.S997 High-Dimensional Statistics, Chapter 4, Section 4.3 (standard reference, not scraped)
Philippe Rigollet and Jan-Christian Hütter, High-Dimensional Statistics: Lecture Notes, Chapter 4 (standard reference, not scraped)