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Eigenspaces of a self adjoint operator are orthogonal

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) and let $T \in B (H)$ be self-adjoint (Self-adjoint, positive, unitary and normal operators). Then:

every eigenvalue of $T$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism) is a real number: if $T x = λ x$ with $x \neq 0$ , then $λ \in R$ ;
eigenspaces belonging to distinct eigenvalues are orthogonal: if $λ \neq μ$ and $x \in E_{λ} (T)$ , $y \in E_{μ} (T)$ , then $⟨ x, y ⟩ = 0$ (Orthogonality and the orthogonal complement).

Facts & Assumptions

Given: A real or complex Hilbert space $H$ and a self-adjoint bounded operator $T$ on $H$ .

[A1]

Self-adjointness. $T^{*} = T$ , so $⟨ T x, y ⟩ = ⟨ x, T y ⟩ = \overline{⟨ T y, x ⟩}$ for all $x, y$ (Self-adjoint, positive, unitary and normal operators, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[A2]

Inner-product algebra. The pairing is linear in the first argument, conjugate-linear in the second, conjugate symmetric and positive definite, and $∥ x ∥^{2} = ⟨ x, x ⟩$ (Real and complex inner-product spaces and their induced length); in particular $⟨ T x, x ⟩ = ⟨ x, T x ⟩$ , so this number is its own conjugate and lies in $R$ .

[A3]

Eigen-data. $x \in E_{λ} (T) = \ker (T - λ I)$ means $T x = λ x$ , so $λ$ is an eigenvalue with eigenvector $x \neq 0$ for the nonzero members of that kernel, that is $T x = λ x$ , and then $∥ x ∥^{2} > 0$ by positive definiteness (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, Real and complex inner-product spaces and their induced length).

[A4]

Scalars. For $λ \in F$ one has $\overline{λ} = λ$ only for $λ \in R$ , and conjugation fixes every real scalar; if $⟨ x, x ⟩ \neq 0$ is real and $λ ⟨ x, x ⟩ = \overline{λ} ⟨ x, x ⟩$ , then $λ = \overline{λ}$ (Real and complex inner-product spaces and their induced length).

[A5]

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, a self-adjoint $T$ , and eigen-data as in the statement.

1.1A1A2A3algebra

Eigenvalues are real. Let $T x = λ x$ with $x \neq 0$ . By conjugate symmetry, [A1] applied to the pair $(T x, x)$ and conjugate-linearity in the second argument, $λ ⟨ x, x ⟩ = ⟨ T x, x ⟩ = ⟨ x, T x ⟩ = \overline{⟨ T x, x ⟩} = \overline{λ} ⟨ x, x ⟩$ ; since $⟨ x, x ⟩ = ∥ x ∥^{2} > 0$ is a nonzero real number, [A4] gives $λ = \overline{λ}$ , that is $λ \in R$ .

2.1step 1.1A1A2algebra

Distinct eigenvalues force orthogonality. Let $T x = λ x$ and $T y = μ y$ with $x, y \neq 0$ and $λ \neq μ$ . By [A1] and conjugate-linearity in the second argument, $λ ⟨ x, y ⟩ = ⟨ T x, y ⟩ = ⟨ x, T y ⟩ = \overline{μ} ⟨ x, y ⟩$ ; by [step 1.1] both $λ$ and $μ$ are real, so $\overline{μ} = μ$ and hence $(λ - μ) ⟨ x, y ⟩ = 0$ ; since $λ - μ \neq 0$ , it follows that $⟨ x, y ⟩ = 0$ .

3.1step 1.1step 2.1A3A5∎

Conclusion. Claim 1 is [step 1.1]. For claim 2 let $x \in E_{λ} (T)$ and $y \in E_{μ} (T)$ with $λ \neq μ$ : if both are nonzero then [step 2.1] gives $⟨ x, y ⟩ = 0$ , while if $x = 0$ or $y = 0$ then $⟨ x, y ⟩ = 0$ as well because the pairing is additive and homogeneous in the first argument and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

Eigenspaces of a self adjoint operator are orthogonal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources