Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

Orthogonal-projector power error inequality

Statement

Let A∈Fm×n, let P be an orthogonal projector on Fm, and let q≥0 be an integer. Then ∥(I−P)A∥2≤∥(I−P)(AA∗)qA∥21/(2q+1).

Facts & Assumptions

Given: A finite real or complex matrix A, an orthogonal projector P, and an integer q≥0.

[F4]

Jensen's inequality holds for finite probability distributions (Jensen's inequality for expectation).

Proof

1.1F1F2F3algebra

Choose an SVD A=UΣV∗ and put D=ΣΣ∗ and R=U∗(I−P)U. Then D is diagonal positive semidefinite and R is an orthogonal projector. By [F1]–[F3], ∥(I−P)A∥22=∥RDR∥2 and ∥(I−P)(AA∗)qA∥22=∥RD2q+1R∥2.

2.1step 1.1F2F3F4choosealgebra

If λ=∥RDR∥2=0, [step 1.1] gives ∥(I−P)A∥2=0, so the asserted inequality follows. If λ>0, choose a unit eigenvector x for the top eigenvalue of RDR; then Rx=x and λ=x∗Dx=∑idi∣xi∣2, with di≥0. The weights ∣xi∣2 sum to 1, so [F4] for t↦t2q+1 gives λ2q+1≤∑idi2q+1∣xi∣2=x∗D2q+1x≤∥RD2q+1R∥2.

3.1step 1.1step 2.1algebra∎

Substitute λ=∥(I−P)A∥22 from step 1.1 into step 2.1 and take the nonnegative (2q+1)st root. This yields the asserted inequality, also for q=0.

Depends on

Used by

Dependency tree · two levels

17 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