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Sign and positive negative parts of a self adjoint operator

Example

Assume AC. Let $T$ be a bounded self-adjoint operator on a nonzero complex Hilbert space $H$ , so that $σ (T) \subseteq R$ (Spectrum of a self adjoint operator is real), and let $E$ be its spectral projection valued measure. Write $∣ T ∣ : = ∣ λ ∣ (T), T_{+} : = \max (λ, 0) (T), T_{-} : = \max (- λ, 0) (T), sgn (T) : = s (λ) (T),$ where $∣ λ ∣$ , $\max (λ, 0)$ , $\max (- λ, 0)$ are continuous on $σ (T)$ and $s (t) : = 1$ for $t > 0$ , $s (t) : = - 1$ for $t < 0$ , $s (0) : = 0$ is bounded Borel on $σ (T)$ , so all four operators are given by the continuous, respectively bounded Borel, functional calculus (Borel functional calculus for a bounded normal operator). Then

$T = T_{+} - T_{-}, ∣ T ∣ = T_{+} + T_{-}, T_{+} T_{-} = 0, T_{\pm} = \frac{1}{2} (∣ T ∣ \pm T), sgn (T)^{2} = I - E ({0}),$

and $∣ T ∣$ agrees with the absolute value $(T^{*} T)^{1 / 2}$ of Absolute value of a bounded operator; moreover $I - E ({0}) = E (σ (T) ∖ {0})$ is the orthogonal projection onto $(\ker T)^{⊥}$ .

Facts & Assumptions

[A1]

A bounded self-adjoint operator has $σ (T) \subseteq R$ , and its Borel calculus is a unital $*$ -homomorphism: $(f g) (T) = f (T) g (T)$ , $\overline{f} (T) = f (T)^{*}$ and $1_{B} (T) = E (B)$ for every Borel $B \subseteq σ (T)$ ; it extends the continuous calculus on continuous $f$ (Spectrum of a self adjoint operator is real, Borel functional calculus for bounded normal operators, Borel functional calculus for a bounded normal operator).

[A2]

Scalar identities for real $λ$ : $λ = \max (λ, 0) - \max (- λ, 0)$ , $∣ λ ∣ = \max (λ, 0) + \max (- λ, 0)$ , $\max (λ, 0) \max (- λ, 0) = 0$ , $s (λ)^{2} = 1_{R ∖ {0}} (λ)$ , $∣ λ ∣^{2} = λ^{2}$ and $∣ λ ∣ \geq 0$ ; the functions $\max (λ, 0)$ , $\max (- λ, 0)$ and $∣ λ ∣$ are continuous on the compact real spectrum and $s$ is Borel and bounded by $1$ . [algebra]

[A3]

$E ({0}) H = \ker T$ and $E ({0}) = I - E (σ (T) ∖ {0})$ , so $I - E ({0})$ is the orthogonal projection onto $(\ker T)^{⊥}$ (Spectral projections and resolution of the identity).

[A4]

For self-adjoint $T$ one has $T^{*} T = T^{2}$ , and a bounded positive operator has a unique positive square root; the calculus value of a nonnegative continuous function is positive, and the calculus is isometric (Absolute value of a bounded operator, Positive square root, Self-adjoint, positive, unitary and normal operators, Continuous functional calculus properties, Projection valued measure).

[A5]

The order on bounded self-adjoint operators is the quadratic-form order, and $S \geq 0$ means $⟨ S x, x ⟩ \geq 0$ for all $x$ (Order on bounded self adjoint operators).

[A6]

AC is the declared choice hypothesis of this page from the construction item onward (The Axiom of Choice).

Verification

technique · direct

Given: A bounded self-adjoint $T$ on a nonzero complex Hilbert space, its spectral PVM $E$ and Borel calculus, and the functions $∣ λ ∣$ , $\max (λ, 0)$ , $\max (- λ, 0)$ , $s (λ)$ on $σ (T) \subseteq R$ .

1.1

The three continuity identities pass to the calculus: $T = Φ_{E} (λ) = Φ_{E} (\max (λ, 0)) - Φ_{E} (\max (- λ, 0)) = T_{+} - T_{-}$ and $∣ T ∣ = Φ_{E} (∣ λ ∣) = T_{+} + T_{-}$ , by linearity of the Borel calculus applied to the pointwise scalar identities, since the involved functions are continuous on the compact spectrum.

A1A2

1.2

Orthogonality of the parts: $T_{+} T_{-} = Φ_{E} (\max (λ, 0) \max (- λ, 0)) = Φ_{E} (0) = 0$ by multiplicativity.

A1A2

1.3

Sign: $s (λ)^{2} = 1_{σ (T) ∖ {0}} (λ)$ , hence $sgn (T)^{2} = Φ_{E} (1_{σ (T) ∖ {0}}) = E (σ (T) ∖ {0}) = I - E ({0})$ , and $E ({0})$ is the orthogonal projection onto $\ker T$ , so $I - E ({0})$ is the orthogonal projection onto $(\ker T)^{⊥}$ .

A1A2A3

1.4

The absolute value agrees with the earlier definition: $∣ T ∣ = Φ_{E} (∣ λ ∣)$ is positive because $∣ λ ∣ \geq 0$ , and $∣ T ∣^{2} = Φ_{E} (∣ λ ∣^{2}) = Φ_{E} (λ^{2}) = T^{2} = T^{*} T$ ; by the uniqueness of the positive square root of $T^{*} T$ , $∣ T ∣ = (T^{*} T)^{1 / 2}$ .

A1A2A4A5

2.1

The operators $T_{\pm}$ are the half-sum and half-difference: from the two identities of step 1.1, $∣ T ∣ + T = 2 T_{+}$ and $∣ T ∣ - T = 2 T_{-}$ , so $T_{\pm} = \frac{1}{2} (∣ T ∣ \pm T)$ .

step 1.1

3.1

Therefore $T = T_{+} - T_{-}$ , $∣ T ∣ = T_{+} + T_{-}$ , $T_{+} T_{-} = 0$ , $T_{\pm} = \frac{1}{2} (∣ T ∣ \pm T)$ , $sgn (T)^{2} = I - E ({0})$ with $I - E ({0})$ the projection onto $(\ker T)^{⊥}$ , and $∣ T ∣$ coincides with $(T^{*} T)^{1 / 2}$ .

step 1.2step 2.1step 1.3step 1.4A6∎

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3–5.7, printed pp.244–296 (standard reference, not scraped)
Gerald Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., §4.1, printed pp.113–115 (standard reference, not scraped)