PropositionStatement: AI-generatedProof: AI-generated

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.

The constant projection and centering decomposition

Statement

Let $V$ be a finite nonempty set, $n = ∣ V ∣$ , and let $C^{V}$ have the inner product $⟨ f, g ⟩ = \sum_{x \in V} \overline{f (x)} g (x)$ . Write $1$ for the function identically equal to $1$ , $I$ for the identity operator, $J = 11^{*}$ , $P = J / n$ , and $Q = I - P$ . Here $1^{*} f = \sum_{x \in V} f (x)$ . The constant subspace is $C = {c 1 : c \in C}$ , and $1^{⊥} = {f : ⟨ 1, f ⟩ = 0}$ .

Then $P$ is the orthogonal projection onto $C$ and $Q$ is the orthogonal projection onto $1^{⊥}$ . For every $f \in C^{V}$ , $f = P f + Q f$ and $f^{*} J f = ∣ \sum_{x \in V} f (x) ∣^{2}$ .

If a linear operator $A$ is self-adjoint and satisfies $A 1 = d 1$ for some $d \in C$ , both $C$ and $1^{⊥}$ are invariant under $A$ , meaning that $A$ maps each subspace into itself, and $A - d P$ vanishes on $C$ . For a nonregular operator, meaning here one for which $A 1$ is not a scalar multiple of $1$ , such invariance is not assumed.

Statement references

Finite vertex Hilbert space in the article convention

The orthogonal projection P_Wv is the W-component in V=W⊕ W^⊥

Facts & Assumptions

[F1]

On $C^{V}$ , the stated inner product is conjugate-linear in its first argument and linear in its second; the conjugate transpose gives the adjoint, so a self-adjoint $A$ satisfies $⟨ u, A v ⟩ = ⟨ A u, v ⟩$ (Finite vertex Hilbert space in the article convention).

[F2]

In a finite-dimensional inner product space, the orthogonal projection onto a subspace $W$ assigns to each vector its unique $W$ component in a decomposition with the other component in $W^{⊥}$ (The orthogonal projection P_Wv is the W-component in V=W⊕ W^⊥).

Proof

Given: A finite nonempty $V$ , the operators and subspaces in the statement, and, for the final assertion, a self-adjoint $A$ with $A 1 = d 1$ .

Goal: Establish the projection, quadratic-form and invariance assertions.

1.1F1F2givenalgebra

For $f \in C^{V}$ , put $s = \sum_{x \in V} f (x)$ . Matrix multiplication gives $J f = s 1$ and $P f = (s / n) 1 \in C$ . Since $⟨ 1, 1 ⟩ = n$ , we have $⟨ 1, f - P f ⟩ = s - (s / n) n = 0$ . Thus $Q f = f - P f \in 1^{⊥}$ and $f = P f + Q f$ . Moreover $C^{⊥} = 1^{⊥}$ , since $⟨ c 1, g ⟩ = \overline{c} ⟨ 1, g ⟩$ . The decomposition therefore identifies $P$ as the orthogonal projection onto $C$ .

2.1F1F2step 1.1algebra

The same decomposition identifies $Q$ as the orthogonal projection onto $1^{⊥}$ : its remainder $f - Q f = P f$ lies in $C$ , which is orthogonal to $1^{⊥}$ . Also $f^{*} J f = ⟨ f, J f ⟩ = \overline{s} s = ∣ s ∣^{2}$ .

3.1F1step 1.1givenalgebra∎

If $c \in C$ , then $A (c 1) = c d 1 \in C$ . If $f \in 1^{⊥}$ , self-adjointness gives $⟨ 1, A f ⟩ = ⟨ A 1, f ⟩ = \overline{d} ⟨ 1, f ⟩ = 0$ ; hence $A f \in 1^{⊥}$ . Finally $P (c 1) = c 1$ , so $(A - d P) (c 1) = 0$ . These deductions use the stated eigenvector and self-adjoint hypotheses; no invariance is asserted for nonregular operators.

Depends on

Used by · 4

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.