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.

Rescaling can change covariance-PCA directions

Example

Let $(U, V)$ be uniform on ${- 1, 1}^{2}$ and set $X = (3 U, U + \sqrt{3} V)^{T}$ . Then $Cov (X) = (\begin{array}{cc} 9 & 3 \\ 3 & 4 \end{array}), R = D^{- 1} Cov (X) D^{- 1} = (\begin{array}{cc} 1 & 1 / 2 \\ 1 / 2 & 1 \end{array}) .$ The leading correlation-PCA line is $span (1, 1)$ , while this line is not an eigenline of the covariance matrix. The two PCA conventions therefore give different leading directions for the same data after standardization.

Facts & Assumptions

Given: A four-point probability space $Ω = {- 1, 1}^{2}$ with probability $1 / 4$ at each point, $U (u, v) = u$ , $V (u, v) = v$ , and $X = (3 U, U + \sqrt{3} V)^{T}$ .

[F1]

Covariance matrix of a finite-dimensional random vector: covariance entries are expectations of centered coordinate products.

[F2]

Covariance and correlation principal component analysis: correlation-PCA divides centered coordinates by their positive standard deviations.

[F3]

The leading PCA eigenspace maximizes projected variance: the leading eigenspace is the maximizing line for projected variance.

Verification

technique · direct

1.1F1givenalgebra

On this four-point space, $E U = E V = 0$ , $E U^{2} = E V^{2} = 1$ , and $E [U V] = 0$ because two points have $U V = 1$ and two have $U V = - 1$ . Thus $E X = 0$ , $Var (3 U) = 9$ , $Var (U + \sqrt{3} V) = 1 + 3 = 4$ , and $Cov (3 U, U + \sqrt{3} V) = 3$ . This gives the stated covariance matrix.

2.1F2step 1.1algebra

The positive standard-deviation matrix is $D = diag (3, 2)$ , so [F2] gives $D^{- 1} Cov (X) D^{- 1} = (\begin{array}{cc} 1 & 1 / 2 \\ 1 / 2 & 1 \end{array})$ . This matrix sends $(1, 1)^{T}$ to $\frac{3}{2} (1, 1)^{T}$ and $(1, - 1)^{T}$ to $\frac{1}{2} (1, - 1)^{T}$ ; hence its leading line is $span (1, 1)$ .

3.1F1F3step 1.1step 2.1algebra∎

The covariance matrix has characteristic roots $(13 \pm \sqrt{61}) / 2$ , so its leading eigenspace is a line. It sends $(1, 1)^{T}$ to $(12, 7)^{T}$ , which is not a scalar multiple of $(1, 1)^{T}$ . Thus the correlation-PCA leading line is not the covariance-PCA leading line, and [F3] identifies each as the maximizing line for its own projected variance.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Online Bayesian Data Analysis, Lecture Notes 8: Learning Signal Representations (11 April 2016) (standard reference, not scraped)
Soumendu Sundar Mukherjee, Multivariate Analysis, Lecture 8: Principal Component Analysis (standard reference, not scraped)