LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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 complement of an eigenspace is invariant

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real or complex Hilbert space (Hilbert space), let $T \in B (H)$ be self-adjoint (Self-adjoint, positive, unitary and normal operators) and let $λ$ be an eigenvalue of $T$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism) with eigenspace $E_{λ} = E_{λ} (T) = \ker (T - λ I)$ . Then:

$E_{λ}$ is a closed linear subspace of $H$ and $T (E_{λ}) \subseteq E_{λ}$ ;
$E_{λ}^{⊥}$ (Orthogonality and the orthogonal complement) is a closed linear subspace of $H$ and $T (E_{λ}^{⊥}) \subseteq E_{λ}^{⊥}$ ;
the restrictions $T ∣_{E_{λ}}$ and $T ∣_{E_{λ}^{⊥}}$ satisfy the self-adjoint identity $⟨ T u, v ⟩ = ⟨ u, T v ⟩$ for all $u, v$ in the respective subspace.

Facts & Assumptions

Given: A Hilbert space $H$ , a self-adjoint bounded $T$ , an eigenvalue $λ$ , and $E_{λ} = \ker (T - λ I)$ .

[A1]

Eigenspace and self-adjointness. $E_{λ} = \ker (T - λ I) = {v : T v = λ v}$ is a linear subspace of $H$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, Kernel and image of a linear map, Linear subspace of a vector space); $T$ is self-adjoint, so $⟨ T x, y ⟩ = ⟨ x, T y ⟩$ for all $x, y$ (Self-adjoint, positive, unitary and normal operators, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[A2]

Continuity and limits. The bounded operator $T - λ I$ is continuous, so $x_{n} \to x$ implies $(T - λ I) x_{n} \to (T - λ I) x$ , and limits of convergent sequences in a metric space are unique (A bounded linear operator between normed spaces, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , A sequence in a metric space has at most one limit).

[A3]

Complements and closedness. For every subset $S$ of an inner-product space the orthogonal complement $S^{⊥}$ is a closed linear subspace (Orthogonal complements are closed, Orthogonality and the orthogonal complement); $v \in S^{⊥}$ means $⟨ v, s ⟩ = 0$ for every $s \in S$ , and the pairing is linear in the first argument and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

[A4]

Countable Choice is the standing hypothesis of this pair's Hilbert-space interface (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: Countable Choice and the data above.

1.1A1A2

$E_{λ}$ is a closed linear subspace. Let $x_{n} \in E_{λ}$ with $x_{n} \to x$ ; then $(T - λ I) x_{n} = 0$ for all $n$ , so by continuity [A2] both $(T - λ I) x_{n} \to (T - λ I) x$ and $(T - λ I) x_{n} \to 0$ , and uniqueness of limits gives $(T - λ I) x = 0$ , i.e. $x \in E_{λ}$ ; with the linear-subspace statement of [A1] this proves closedness.

1.2A1

$E_{λ}$ is $T$ -invariant. If $v \in E_{λ}$ then $T v = λ v \in E_{λ}$ by [A1] and linearity of the subspace.

1.3A3

$E_{λ}^{⊥}$ is closed. The general closedness of orthogonal complements [A3] applied to the subset $E_{λ}$ gives that $E_{λ}^{⊥}$ is a closed linear subspace of $H$ .

1.4A1A3

$E_{λ}^{⊥}$ is $T$ -invariant. Let $x \in E_{λ}^{⊥}$ and $y \in E_{λ}$ . By self-adjointness and $T y = λ y$ , $⟨ T x, y ⟩ = ⟨ x, T y ⟩ = ⟨ x, λ y ⟩ = \overline{λ} ⟨ x, y ⟩ = 0$ , since $⟨ x, y ⟩ = 0$ by definition of the orthogonal complement; as $y \in E_{λ}$ was arbitrary, $⟨ T x, y ⟩ = 0$ for all $y \in E_{λ}$ , that is $T x \in E_{λ}^{⊥}$ .

2.1step 1.1step 1.2step 1.3step 1.4A1A4∎

The restrictions are self-adjoint. If $u, v$ both lie in $E_{λ}$ , or both lie in $E_{λ}^{⊥}$ , then $u, v \in H$ and [A1] gives $⟨ T u, v ⟩ = ⟨ u, T v ⟩$ in $H$ , which is exactly the defining identity of self-adjointness for the restricted operator on that subspace; [step 1.2] and [step 1.4] show that each restriction maps its subspace into itself, and [step 1.1] and [step 1.3] give the closedness statements of claims 1 and 2.

Orthogonal complement of an eigenspace is invariant

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources