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Properties of Adjoint Operators 📂Hilbert Space

Properties of Adjoint Operators

Theorem1

Let’s call H,KH,K a Hilbert space. For a bounded linear operator T:K→HT : K \to H, T∗:H→KT^{\ast} : H \to K is called the adjoint operator of TT if it satisfies the following.

⟨Tv,w⟩H=⟨v,T∗w⟩K,∀v∈K \left\langle T \textbf{v} , \textbf{w} \right\rangle_{H} = \left\langle \textbf{v} , T^{\ast} \textbf{w} \right\rangle_{K},\quad \forall \textbf{v} \in K The adjoint operator has the following properties.

(a) T∗T^{\ast} is linear and bounded.

(b) (T∗)∗=T\left( T^{\ast} \right)^{\ast} = T

(c) ∥T∗∥=∥T∥\left\| T^{\ast} \right\| = \left\| T \right\|

Proof

(a)

  • Part 1. T∗T^{\ast} is linear

    According to the definition,

    T∗(αw+βu)=αT∗(w)+βT∗(u) T^{\ast}( \alpha \mathbf{w}+\beta \mathbf{u} )=\alpha T^{\ast}(\mathbf{w})+\beta T^{\ast}(\mathbf{u})

    we need to show that. By the definition of T∗T^{\ast} and the inner product, for v∈K\mathbf{v}\in K, the following holds.

    ⟨v,T∗(αw+βu)⟩K= ⟨Tv,αw+βu⟩H= α‾⟨Tv,w⟩H+β‾⟨Tv,u⟩H= α‾⟨v,T∗w⟩K+β‾⟨v,T∗u⟩K= ⟨v,αT∗w+βT∗u⟩K \begin{align*} \left\langle \mathbf{v}, T^{\ast}(\alpha \mathbf{w}+ \beta \mathbf{u}) \right\rangle_{K} =&\ \left\langle T\mathbf{v},\alpha \mathbf{w} + \beta \mathbf{u} \right\rangle_{H} \\ =&\ \overline{\alpha}\left\langle T\mathbf{v},\mathbf{w} \right\rangle_{H} +\overline{\beta} \left\langle T\mathbf{v}, \mathbf{u} \right\rangle_{H} \\ =&\ \overline{\alpha}\left\langle \mathbf{v}, T^{\ast}\mathbf{w} \right\rangle_{K} +\overline{\beta} \left\langle \mathbf{v}, T^{\ast} \mathbf{u} \right\rangle_{K} \\ =&\ \left\langle \mathbf{v}, \alpha T^{\ast} \mathbf{w}+\beta T^{\ast} \mathbf{u} \right\rangle_{K} \end{align*}

    For all v\mathbf{v}, if ⟨v,u⟩=⟨v,w⟩\left\langle \mathbf{v},\mathbf{u} \right\rangle=\left\langle \mathbf{v},\mathbf{w}\right\rangle , then u=w\mathbf{u}=\mathbf{w} holds,

    T∗(αw+βu)=αT∗(w)+βT∗(u) T^{\ast}( \alpha \mathbf{w}+\beta \mathbf{u} )=\alpha T^{\ast}(\mathbf{w})+\beta T^{\ast}(\mathbf{u})

  • Part 2. T∗T^{\ast} is bounded

    According to the definition,

    ∥T∗w∥≤C∥w∥,∀w∈H \left\| T^{\ast}\mathbf{w} \right\| \le C \left\| \mathbf{w} \right\|,\quad \forall \mathbf{w}\in H

    we need to show that there exists a constant CC that satisfies it. By the relationship between the inner product and the norm, the following holds.

    ∥T∗w∥=sup⁡v∈K∥v∥=1∣⟨v,T∗w⟩K∣ \left\| T^{\ast}\mathbf{w} \right\| = \sup \limits_{\substack{\mathbf{v}\in K \\ \left\| \mathbf{v} \right\|=1 }} \left| \left\langle \mathbf{v},T^{\ast}\mathbf{w} \right\rangle_{K} \right|

    Thus, by the definition of T∗T^{\ast} and the Cauchy-Schwarz inequality, the following holds.

    ∥T∗w∥= sup⁡v∈K∥v∥=1∣⟨v,T∗w⟩K∣= sup⁡v∈K∥v∥=1∣⟨Tv,w⟩H∣≤sup⁡v∈K∥v∥=1∥Tv∥∥w∥ \begin{align*} \left\| T^{\ast}\mathbf{w} \right\| =&\ \sup \limits_{\substack{\mathbf{v}\in K \\ \left\| \mathbf{v} \right\|=1 }} \left| \left\langle \mathbf{v},T^{\ast}\mathbf{w} \right\rangle_{K} \right| \\ =&\ \sup \limits_{\substack{\mathbf{v}\in K \\ \left\| \mathbf{v} \right\|=1 }} \left| \left\langle T\mathbf{v},\mathbf{w} \right\rangle_{H} \right| \\ \le& \sup \limits_{\substack{\mathbf{v}\in K \\ \left\| \mathbf{v} \right\|=1 }} \left\| T\mathbf{v} \right\| \left\| \mathbf{w} \right\| \end{align*}

    Since TT is bounded, ∥Tv∥≤∥T∥∥v∥\left\| T\mathbf{v} \right\|\le \left\| T \right\| \left\| \mathbf{v} \right\| holds. Therefore,

    ∥T∗w∥≤sup⁡v∈K∥v∥=1∥Tv∥∥w∥≤sup⁡v∈K∥v∥=1∥T∥∥v∥∥w∥≤∥T∥∥w∥ \begin{align*} \left\| T^{\ast}\mathbf{w} \right\| \le& \sup \limits_{\substack{\mathbf{v}\in K \\ \left\| \mathbf{v} \right\|=1 }} \left\| T\mathbf{v} \right\| \left\| \mathbf{w} \right\| \\ \le& \sup \limits_{\substack{\mathbf{v}\in K \\ \left\| \mathbf{v} \right\|=1 }} \left\| T \right\| \left\| \mathbf{v} \right\| \left\| \mathbf{w} \right\| \\ \le& \left\| T \right\| \left\| \mathbf{w} \right\| \end{align*}

    T∗T^{\ast} is bounded.

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(b)

It can be simply shown by the definition of T∗T^{\ast} and the inner product.

⟨Tv,w⟩H= ⟨v,T∗w⟩K= ⟨T∗w,v⟩K‾= ⟨w,(T∗)∗v⟩K‾= ⟨(T∗)∗v,w⟩K \begin{align*} \left\langle T\mathbf{v},\mathbf{w} \right\rangle_{H} =&\ \left\langle \mathbf{v},T^{\ast}\mathbf{w} \right\rangle_{K} \\ =&\ \overline{\left\langle T^{\ast}\mathbf{w},\mathbf{v} \right\rangle_{K}} \\ =&\ \overline{\left\langle \mathbf{w},(T^{\ast})^{\ast}\mathbf{v} \right\rangle_{K}} \\ =&\ \left\langle (T^{\ast})^{\ast}\mathbf{v},\mathbf{w} \right\rangle_{K} \end{align*}

As this holds for all w\mathbf{w}, the logic from Part 1. applies here as well.

Tv=(T∗)∗v  ⟹  T=(T∗)∗ T\mathbf{v}=(T^{\ast})^{\ast}\mathbf{v}\implies T=(T^{\ast})^{\ast}

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(c)

From the proof of (a), we obtained ∥T∗∥≤∥T∥\left\| T^{\ast} \right\| \le \left\| T \right\|. In the same way, we can obtain the inequality in the opposite direction.

∥Tv∥= sup⁡w∈H∥w∥=1∣⟨Tv,w⟩H∣= sup⁡w∈H∥w∥=1∣⟨v,T∗w⟩K∣≤sup⁡w∈H∥w∥=1∥v∥∥T∗w∥≤sup⁡w∈H∥w∥=1∥v∥∥T∗∥∥w∥≤∥T∗∥∥v∥ \begin{align*} \left\| T\mathbf{v} \right\| =&\ \sup \limits_{\substack{\mathbf{w}\in H \\ \left\| \mathbf{w} \right\|=1 }} \left| \left\langle T\mathbf{v},\mathbf{w} \right\rangle_{H} \right| \\ =&\ \sup \limits_{\substack{\mathbf{w}\in H \\ \left\| \mathbf{w} \right\|=1 }} \left| \left\langle \mathbf{v},T^{\ast}\mathbf{w} \right\rangle_{K} \right| \\ \le& \sup \limits_{\substack{\mathbf{w}\in H \\ \left\| \mathbf{w} \right\|=1 }} \left\| \mathbf{v} \right\| \left\| T^{\ast}\mathbf{w} \right\| \\ \le& \sup \limits_{\substack{\mathbf{w}\in H \\ \left\| \mathbf{w} \right\|=1 }} \left\| \mathbf{v} \right\| \left\| T^{\ast} \right\| \left\| \mathbf{w} \right\| \\ \le& \left\| T^{\ast} \right\| \left\| \mathbf{v} \right\| \end{align*}

Therefore, since ∥T∥≤∥T∗∥\left\| T^{} \right\|\le \left\| T^{\ast} \right\| holds,

∥T∥=∥T∗∥ \left\| T \right\| = \left\| T^{\ast} \right\|

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  1. Ole Christensen, Functions, Spaces, and Expansions: Mathematical Tools in Physics and Engineering (2010), p72 ↩︎