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

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Spectrum of a positive operator is nonnegative

Statement

Assume Countable Choice. If $T$ is a bounded positive operator on a nonzero complex Hilbert space, then $σ (T) \subseteq [0, + \infty)$ .

Facts & Assumptions

[A1]

$T$ is positive when $⟨ T x, x ⟩$ is a real number in $[0, + \infty)$ for every $x$ ; positivity is a condition on the values of the quadratic form and does not presuppose self-adjointness (Self-adjoint, positive, unitary and normal operators).

[A2]

$⟨ T^{*} x, y ⟩ = ⟨ x, T y ⟩$ , and for a fixed $w$ the expansion of $⟨ T x, x ⟩$ at $x + t y$ uses the linear/conjugate-linear inner-product conventions (The Hilbert-space adjoint of a bounded operator, Real and complex inner-product spaces and their induced length). The adjoint algebra laws give $(T - z I)^{*} = T^{*} - \overline{z} I$ (Hilbert-adjoint identities).

[A3]

$∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ for all vectors $u, v$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A4]

$λ \in ρ (T)$ 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).

[A5]

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

[A6]

$S^{⊥} = {v : ⟨ v, s ⟩ = 0 \forall s \in S}$ and ${0}^{⊥} = H$ (Orthogonality and the orthogonal complement).

[A7]

A Hilbert space is complete for its induced norm; $z = Re z + i Im z$ and $z \notin [0, + \infty)$ means $Im z \neq 0$ or $Re z < 0$ (Hilbert space, Real and imaginary parts, complex conjugation, and modulus, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A8]

Countable Choice is the hypothesis of the adjoint and orthogonality suppliers used below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ , a bounded positive operator $T \in B (H)$ and a scalar $z$ outside $[0, + \infty)$ .

1.1A1A7algebra

For $x = 0$ the lower bounds below are immediate. For $x \neq 0$ the number $r : = ⟨ T x, x ⟩ / ∥ x ∥^{2}$ is real and nonnegative, so writing $z = a + b i$ one has $∣ ⟨ (T - z I) x, x ⟩ ∣ = ∣ r - z ∣ ∥ x ∥^{2}$ with $∣ r - z ∣ \geq ∣ b ∣$ always and $∣ r - z ∣ = r + ∣ a ∣ \geq ∣ a ∣$ when $b = 0$ and $a < 0$ .

1.2A1A2A7algebra

If $(T^{*} - z I) x = 0$ with $x \neq 0$ , then testing against $x$ and using the adjoint identity gives $⟨ T x, x ⟩ = ⟨ x, T^{*} x ⟩ = \overline{z} ∥ x ∥^{2}$ ; the left side is a nonnegative real number while $z \notin [0, + \infty)$ makes $\overline{z} ∥ x ∥^{2}$ non-real or negative, so $\ker (T^{*} - z I) = {0}$ .

2.1step 1.1A3A7algebra

If $z \notin R$ then $∥ (T - z I) x ∥ \geq ∣ Im z ∣ ∥ x ∥$ by the estimate and Cauchy–Schwarz, and the same lower bound with $∣ Re z ∣$ holds when $z$ is real and negative; in either case there is $c > 0$ with $∥ (T - z I) x ∥ \geq c ∥ x ∥$ , so $T - z I$ is injective. Its range is closed: if $(T - z I) x_{n} \to u$ , the inequality makes $(x_{n})$ Cauchy, completeness gives $x_{n} \to x$ , and boundedness gives $(T - z I) x = u$ .

3.1step 2.1step 1.2A2A5A6A8

For such $z$ the orthogonal complement of $ran (T - z I)$ is $\ker (T^{*} - \overline{z} I) = {0}$ , the vanishing being step 1.2 applied to the scalar $\overline{z}$ , which also lies outside $[0, + \infty)$ ; the closed range equals its closure, so $ran (T - z I) = H$ .

4.1step 2.1step 3.1A4A7∎

Hence every $z \notin [0, + \infty)$ lies in $ρ (T)$ : $T - z I$ is bijective with bounded inverse, and its inverse has norm at most $1 / c$ by step 2.1; changing sign gives the bounded inverse of $z I - T$ . Thus $σ (T) \subseteq [0, + \infty)$ .

Spectrum of a positive operator is nonnegative

Statement

Facts & Assumptions

Proof

Depends on

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