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 $a = Re z$ , $b = Im z$ .

1.1

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 $∥ (T - z I) x ∥^{2} = ∥ (T - a I) x ∥^{2} + b^{2} ∥ x ∥^{2} \geq b^{2} ∥ x ∥^{2}$ .

A2A4algebra

1.2

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 $\ker (T - \overline{z} I) = {0}$ .

A2A4algebra

2.1

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 $(T - z I) x = y$ .

step 1.1A3A7algebra

3.1

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 $ran (T - z I) = \overline{ran (T - z I)} = H$ .

step 2.1step 1.2A5A6

4.1

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 $σ (T) \subseteq R$ .

step 2.1step 3.1A1∎

Depends on

Used by

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, Lemma 5.49, printed pp.238–240 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–13 (standard reference, not scraped)