LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Spectrum of a self adjoint operator is real

Statement

Assume Countable Choice. If $T$ is a bounded self-adjoint operator on a nonzero complex Hilbert space, then $σ (T) \subseteq R$ , and $∥ (T - z I) x ∥ \geq ∣ Im z ∣ ∥ x ∥$ for every $z \notin R$ and every $x \in H$ .

Facts & Assumptions

[A1]

A scalar $λ$ lies in the resolvent set $ρ (T)$ exactly when $λ I - T$ is bijective with bounded inverse; $σ (T)$ is the complement of $ρ (T)$ (Spectrum and resolvent of a bounded operator).

[A2]

For a self-adjoint $T$ one has $⟨ T x, y ⟩ = ⟨ x, T y ⟩$ for all $x, y$ , and consequently $⟨ T x, x ⟩$ is real; the adjoint is conjugate-linear, so $(T - z I)^{*} = T - \overline{z} I$ (Self-adjoint, positive, unitary and normal operators, Hilbert-adjoint identities).

[A3]

A Hilbert space is complete for its induced norm (Hilbert space).

[A4]

For $z \in C$ the numbers $Re z$ and $Im z$ are real with $z = Re z + i Im z$ , $\overline{z} = Re z - i Im z$ , and $z \notin R$ exactly when $Im z \neq 0$ (Real and imaginary parts, complex conjugation, and modulus).

[A5]

For every bounded $S$ one has $(ran S)^{⊥} = \ker S^{*}$ and $\overline{ran S} = (\ker S^{*})^{⊥}$ (Kernel–range orthogonality for Hilbert adjoints).

[A6]

$S^{⊥} = {v : ⟨ v, s ⟩ = 0 for all s \in S}$ and ${0}^{⊥} = H$ ; a vector orthogonal to every vector of a set spanning a dense subspace is zero (Orthogonality and the orthogonal complement).

[A7]

Countable Choice is the hypothesis under which the adjoint, orthogonality and completeness suppliers are stated, and $∥ S ∥ \leq C$ means $∥ S x ∥ \leq C ∥ x ∥$ for every $x$ (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ and a bounded self-adjoint $T \in B (H)$ , and a scalar $z = a + b i$ with $a = Re z$ , $b = Im z$ .

1.1A2A4algebra

Since $T^{*} = T$ , $z I$ is normal and $(T - z I)^{*} (T - z I) = (T - a I)^{2} + b^{2} I$ , so for every $x$ the expansion of $⟨ (T - z I)^{*} (T - z I) x, x ⟩$ gives $∥ (T - z I) x ∥^{2} = ∥ (T - a I) x ∥^{2} + b^{2} ∥ x ∥^{2} \geq b^{2} ∥ x ∥^{2}$ .

1.2A2A4algebra

If $b \neq 0$ and $(T - \overline{z} I) x = 0$ for some $x \neq 0$ , then testing against $x$ gives $⟨ T x, x ⟩ = \overline{z} ∥ x ∥^{2}$ ; the left side is real by self-adjointness while $\overline{z} \notin R$ because $b \neq 0$ , so no such $x$ exists and $\ker (T - \overline{z} I) = {0}$ .

2.1step 1.1A3A7algebra

If $b \neq 0$ then $∥ (T - z I) x ∥ \geq ∣ b ∣ ∥ x ∥$ for every $x$ , so $T - z I$ is injective and its range is closed: from $(T - z I) x_{n} \to y$ the estimate makes $(x_{n})$ Cauchy, hence convergent to some $x$ with $(T - z I) x = y$ .

3.1step 2.1step 1.2A5A6

If $b \neq 0$ then $(ran (T - z I))^{⊥} = \ker (T - z I)^{*} = \ker (T - \overline{z} I) = {0}$ , and since the range is closed it equals its own closure, so $ran (T - z I) = \overline{ran (T - z I)} = H$ .

4.1step 2.1step 3.1A1∎

For $z \notin R$ the operator $T - z I$ is therefore bijective, and for $y = (T - z I) x$ the lower bound gives $∥ (T - z I)^{- 1} y ∥ = ∥ x ∥ \leq ∣ b ∣^{- 1} ∥ y ∥$ , so the inverse is bounded and $z \in ρ (T)$ ; hence $σ (T) \subseteq R$ .

Spectrum of a self adjoint operator is real

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources