LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Eigenspaces of the diagonal operator

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space), let $(e_{n})_{n \in N}$ be a complete orthonormal family of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), let $(λ_{n})_{n \in N} \subseteq R$ , and let $A$ be the self-adjoint diagonal operator of Diagonal adjoint and graph-core lemma: $D (A) = {x \in H : \sum_{n} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty}, A x = \sum_{n} λ_{n} ⟨ x, e_{n} ⟩ e_{n} .$ Then:

(a) for every $λ \in R$ the eigenspace of $A$ at $λ$ is the closed linear span of the basis vectors carrying the eigenvalue $λ$ : $\ker (A - λ I) = \overset{}{span} {e_{n} : λ_{n} = λ};$ in particular $λ$ is an eigenvalue of $A$ exactly when $λ = λ_{n}$ for some $n$ ;

(b) if $λ_{m} \neq λ_{n}$ whenever $m \neq n$ , then $\ker (A - λ_{n} I) = C e_{n}$ for every $n$ : every eigenspace is one-dimensional with eigenvector $e_{n}$ , the eigenspaces are pairwise orthogonal, and every nonzero eigenvector of $A$ is a nonzero multiple of some $e_{n}$ ;

(c) in the situation of (b) the linear span of all eigenspaces of $A$ is $span {e_{n} : n \in N}$ , which is dense in $H$ .

Facts & Assumptions

Given: a complex Hilbert space $H$ ; a complete orthonormal family $(e_{n})_{n \in N}$ of $H$ ; real numbers $λ_{n}$ ; the diagonal operator $A$ displayed in the Statement; a real number $λ$ .

[F1]

$D (A)$ and $A$ are as displayed, $A$ is self-adjoint, $A e_{n} = λ_{n} e_{n}$ for every $n$ , and the algebraic span $D_{0} = span {e_{n} : n \in N}$ is a core for $A$ (Diagonal adjoint and graph-core lemma clauses (b), (d), (e)).

[F2]

$⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ ; a series $\sum_{n} a_{n} e_{n}$ converges in $H$ exactly when $\sum_{n} ∣ a_{n} ∣^{2} < \infty$ , and then $∥ \sum_{n} a_{n} e_{n} ∥^{2} = \sum_{n} ∣ a_{n} ∣^{2}$ ; every $x \in H$ satisfies $x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n}$ and $\sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2} = ∥ x ∥^{2}$ ; and the coefficients of a convergent expansion in the family are unique (Orthonormal families, complete orthonormal systems and Hilbert bases, Parseval equivalences for an orthonormal family).

Proof

Given: the Hilbert space $H$ , the complete orthonormal family $(e_{n})$ , the real numbers $λ_{n}$ , the diagonal operator $A$ of [F1], and a real number $λ$ .

1.1givenF1F2

[One inclusion] Let $x \in D (A)$ with coefficients $c_{n} = ⟨ x, e_{n} ⟩$ satisfy $A x = λ x$ . By [F1] both series $\sum_{n} λ_{n} c_{n} e_{n}$ and $\sum_{n} λ c_{n} e_{n}$ converge, so their difference $\sum_{n} (λ_{n} - λ) c_{n} e_{n} = 0$ converges; its coefficients in the orthonormal family are $(λ_{n} - λ) c_{n}$ , and by the uniqueness of coefficients in [F2] all of them vanish. Hence $c_{n} = 0$ whenever $λ_{n} \neq λ$ , so $x = \sum_{n : λ_{n} = λ} c_{n} e_{n}$ lies in the closed linear span of ${e_{n} : λ_{n} = λ}$ .

1.2givenF1F2algebra

[The converse inclusion] Conversely let $I : = {n : λ_{n} = λ}$ and let $x = \sum_{n \in I} c_{n} e_{n}$ be an element of the closed linear span of ${e_{n} : n \in I}$ , so that $\sum_{n \in I} ∣ c_{n} ∣^{2} < \infty$ and the series converges by [F2]. Then $\sum_{n} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} = \sum_{n \in I} λ^{2} ∣ c_{n} ∣^{2} < \infty$ , so $x \in D (A)$ by [F1], and $A x = \sum_{n \in I} λ c_{n} e_{n} = λ x$ .

2.1step 1.1step 1.2given

[Clause (a)] Steps 1.1 and 1.2 give $\ker (A - λ I) = \overset{}{span} {e_{n} : λ_{n} = λ}$ for every real $λ$ , and this span is nonzero exactly when $λ \in {λ_{n} : n \in N}$ because $e_{n} \neq 0$ .

3.1step 2.1F1F2

[Clause (b)] Assume the $λ_{n}$ are pairwise distinct and let $λ = λ_{k}$ . Then ${n : λ_{n} = λ} = {k}$ , so step 2.1 gives $\ker (A - λ_{k} I) = C e_{k}$ ; every such eigenspace is one-dimensional with eigenvector $e_{k} \neq 0$ , and distinct eigenspaces are orthogonal because $⟨ e_{m}, e_{k} ⟩ = 0$ for $m \neq k$ by [F2]. If $x \neq 0$ satisfies $A x = λ x$ , then $λ = λ_{k}$ for some $k$ by step 2.1 and $x \in C e_{k}$ ; conversely every nonzero multiple of $e_{k}$ is an eigenvector by [F1].

4.1step 3.1F1∎

[Clause (c)] The span of the union of the eigenspaces contains every $e_{n}$ , hence contains $D_{0} = span {e_{n}}$ , which is dense in $H$ by [F1]; therefore the span of the eigenspaces is dense.

Eigenspaces of the diagonal operator

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources