2.5 Adjoint operators

Adjoint operators allow us to translate inner products in the destination space to inner products in the source space.

Definition 1 Let $X$ and $Y$ be inner product spaces and $A\in B(X,Y)$ . The adjoint operator $A^{*}$ : $Y\to X$ is defined by the equation ${⟨Ax,y⟩}_Y={⟨x,A^{*}y⟩}_X$ for all $x\in X$ , $y\in Y$ .

The subindices in the inner product notation clarify the inner product space we refer to, but it is often implicit from the inputs.

Example 1 Pick $X={\mathbb{R}}^n$ , $Y={\mathbb{R}}^n$ , and define the operator $A\in B(X,Y)$ by an $m\times n$ matrix; we want to find its adjoint $A^{*}$ . We appeal to the definition:

${⟨Ax,y⟩}_Y=y^T\left(Ax\right)=y^{T}Ax=\left(y^{T}A\right)x={\left(A^{T}y\right)}^{T}x={⟨x,A^{T}y⟩}_X$

so according to the definition, we have that $A^{*}y=A^{T}y$ , resulting in $A^{*}=A^T$ .

Theorem 1 If $A\in B(X,Y)$ then the adjoint operator $A^{*}\in B(Y,X)$ and $\parallel A^{*}\parallel =\parallel A\parallel$ .

To see $A^{*}$ is linear, note that

so $A^{*}(a_1{y}_1+a_2{y}_2)=a_1{A}^{*}{y}_1+a_2{A}^{*}{y}_2$ . We will next show that $\parallel A^{*}\parallel \le \parallel A\parallel$ and $\parallel A\parallel \le \parallel A^{*}\parallel$ , which implies that $\parallel A^{*}\parallel =\parallel A\parallel$ . First, note that

since this is true for all $x\in X$ , $\parallel A^{*}{\parallel }^2\le \parallel A\parallel \parallel A^{*}\parallel$ , and so $\parallel A^{*}\parallel \le \parallel A\parallel$ .

Next, pick $x_0\notin \mathcal{N}\left(A\right)$ and set $y_0=\frac{A{x}_0}{\parallel A{x}_0\parallel }$ . Note that if $\mathcal{N}\left(A\right)=X$ , then $Ax=0$ for all $x$ and the adjoint $A^{*}y=0$ for all $y$ satisfies the theorem. For such a choice of $x_0$ and $y_0$ , we note that

so $\parallel A{x}_0\parallel \le \parallel x_0\parallel \parallel A^{*}\parallel$ , and $\frac{\parallel A{x}_0\parallel }{\parallel x_0\parallel }\le \parallel A^{*}\parallel$ ; thus, $\parallel A\parallel \le \parallel A^{*}\parallel$ . Therefore, $\parallel A\parallel =\parallel A^{*}\parallel$ .

Fact 1 Some quick facts on adjoint operators:

1. $I^{*}=I$ .
2. ${(A_1+A_2)}^{*}=A_1^{*}+A_2^{*}$ .
3. ${\left(\alpha A\right)}^{*}=\alpha A^{*}$ .
4. ${\left(A_1{A}_2\right)}^{*}=A_2^{*}{A}_1^{*}$ .
5. If $A^{-1}$ exists, then ${\left(A^{-1}\right)}^{*}={\left(A^{*}\right)}^{-1}$ .

Definition 2 An operator $A\in B(X,X)$ is said to be self-adjoint if $A^{*}=A$ .

Definition 3 Let $X$ be a Hilbert space. A self-adjoint linear operator $A\in B(X,X)$ is said to be positive semidefinite if $⟨x,Ax⟩\ge 0$ for all $x\in X$ .

Example 2 Let $X=L_2[0,1]$ and define an operator $A\in B(X,X)$ by

where ${\int }_0^1{\int }_0^1{\left|K(t,s)\right|}^{2}dsdt<\infty$ . What is its adjoint $A^{*}$ ?

To find the adjoint, we use its definition $⟨Ax,w⟩=⟨x,A^{*}w⟩$ :

where $v\left(s\right)=\left(A^{*}w\right)\left(s\right)={\int }_0^{1}K(t,s)w\left(t\right)dt$ ; changing variables, $\left(A^{*}w\right)\left(t\right)={\int }_0^{1}K(s,t)w\left(s\right)ds$ .

Example 3 Let $X=L_2[0,1]$ and define an operator $A\in B(X,X)$ by

What is its adjoint $A^{*}$ ? Once again, from the definition,

We have that $\left(A^{*}w\right)\left(s\right)={\int }_s^{1}K(t,s)w\left(t\right)dt$ ; changing variables, we obtain the adjoint $\left(A^{*}w\right)\left(t\right)={\int }_t^{1}K(s,t)w\left(s\right)ds$ .

Example 4 Let $X=L_2[0,1]$ , $Y={\mathbb{R}}^N$ , and define the operator $A\in B(X,Y)$ as

What is its adjoint $A^{*}$ ? Once again, from the definition of adjoint,

where the subscript denotes the corresponding space for the sake of clarity. Then,

where $v\left(t\right)=\left(A^{*}w\right)\left(t\right)$ . We can successfully match these two inner products by using delta functions to define $v$ . Recall the properties of delta functions: ( $i$ ) $x\left(t\right)\delta (t-t_0)=x\left(t_0\right)\delta (t-t_0)$ , ( $ii$ ) ${\int }_0^1\delta (t-t_0)dt=1$ . Therefore, we may set $v\left(t\right)={\sum }_{i=1}^N{w}_i\delta (t-t_i)$ , which then provides

This confirms that $v\left(t\right)=A^{*}w\left(t\right)={\sum }_{i=1}^N{w}_i\delta (t-t_i)$ .

Theorem 2 Let $X,Y$ be normed spaces and $A\in B(X,Y)$ , then ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }=\mathcal{N}\left(A^{*}\right)$ .

We will show the double inclusion ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }\subseteq \mathcal{N}\left(A^{*}\right)$ and $\mathcal{N}\left(A^{*}\right)\subseteq {\left(\mathcal{R}\left(A\right)\right)}^{\perp }$ . Let $y\in \mathcal{N}\left(A^{*}\right)$ , i.e., $A^{*}y=0$ . Then for all $x\in X$ , we have

and so $y\in {\left(\mathcal{R}\left(A\right)\right)}^{\perp }$ , which implies that $\mathcal{N}\left(A^{*}\right)\subseteq {\left(\mathcal{R}\left(A\right)\right)}^{\perp }$ . Now, pick $y\in {\left(\mathcal{R}\left(A\right)\right)}^{\perp }$ , i.e., for all $x\in X$ we have that $⟨Ax,y⟩=0$ . Then, $⟨x,A^{*}y⟩=0$ , and since this is true for all $x\in X$ , then $A^{*}y=0$ and $y\in \mathcal{N}\left(A^{*}\right)$ . Therefore, ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }\subseteq \mathcal{N}\left(A^{*}\right)$ . The two inclusions imply that ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }=\mathcal{N}\left(A^{*}\right)$ . The following statements can be proved in a similar fashion.

Theorem 3 Let $X,Y$ be normed spaces and $A\in B(X,Y)$ , then

1. $\overline{\mathcal{R}\left(A\right)}={\left[\mathcal{N}\left(A^{*}\right)\right]}^{\perp }$
2. ${\left[\mathcal{R}\left(A^{*}\right)\right]}^{\perp }=\mathcal{N}\left(A\right)$
3. $\overline{\mathcal{R}\left(A^{*}\right)}={\left[\mathcal{N}\left(A\right)\right]}^{\perp }$

Example 5 Let $A=\left[\begin{array}{cc}2& 1\\ 0& 2\\ 2& 3\end{array}\right]$ , then $\mathcal{R}\left(A\right)=\mathrm{span}\left\{\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right],,,\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right\}$ , $A^{*}=\left[\begin{array}{ccc}2& 0& 2\\ 1& 2& 3\end{array}\right]$ , and $\mathcal{N}\left(A^{*}\right)=\mathrm{span}\left\{\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right]\right\}$ . It is easy to check that ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }=\mathcal{N}\left(A^{*}\right)$ .