1.8 Properties of linear operators

We start with a simple but useful property.

Lemma 1 If $A\in B(X,X)$ and $〈x,,,A,x〉=0$ for all $x\in X$ , then $A=0$ .

Let $a\in \mathbb{C}$ , then for any $x,y\in X$ , we have that $x+ay\in X$ . Therefore, we obtain

Since $x,y\in X$ , we have that $〈x,,,A,x〉=〈y,,,A,y〉=0$ ; therefore, $0=\overline{a}〈x,,,A,y〉+a〈y,,,A,x〉$ .

If we set $a=1$ , then $0=〈x,,,A,y〉+〈y,,,A,x〉$ . So in this case, $〈x,,,A,y〉=-〈y,,,A,x〉$ .

If we set $a=i$ , then $0=-i〈x,,,A,y〉+i〈y,,,A,x〉$ . So in this case, $〈x,,,A,y〉=〈y,,,A,x〉$ .

Thus, $〈x,,,A,y〉=0$ for all $x,y\in X$ , which means $Ay=0$ for all $y\in X$ . So we can come to the conclusion that $A=0$ .

Solutions to operator equations

Assume $X$ and $Y$ are two normed linear spaces and $A\in B(X,Y)$ is a bounded linear operator. Now pick $y\in Y$ . Then we pose the question: does a solution $\widehat{x}\in X$ to the equation $Ax=y$ exist?

There are three possibilities:

1. A unique solution exists;
2. multiple solutions exist; or
3. no solution exists.

We consider these cases separately below.

1. Unique solution : Assume $x$ and $x_1$ are two solutions to the equation. In this case we have $Ax=A{x}_1$ . So $A(x-x_1)=0$ . Therefore, $x-x_1\in \mathcal{N}\left(A\right)$ . If the solution $x$ is unique then we must have $x=x_1$ and $x-x_1=0$ . Therefore, $\mathcal{N}\left(A\right)=\left\{0\right\}$ . Since the operator has a trivial null space, then $A^{-1}$ exists. Thus, the solution to the equation is given by $\widehat{x}=A^{-1}y$ ;
2. Multiple solutions : In this case, we may prefer to pick a particular solution. Often, our goal is to find the solution with smallest norm (for example, to reduce power in a communication problem). Additionally, there is a closed-form expression for the minimum-norm solution to the equation $Ax=y$ . Theorem 1 Let $X$ , $Y$ be Hilbert spaces, $y\in Y$ , and $A\in B(X,Y)$ . Then $\widehat{x}$ is the solution to $Ax=y$ if and only if $\widehat{x}=A^{*}z$ , where $z\in Y$ is the solution of $A{A}^{*}z=y$ . Let $x_1$ be a solution to $Ax=y$ . Then all other solutions can be written as $\widehat{x}=x_1-u$ , where $u\in \mathcal{N}\left(A\right)$ . We therefore search for the solution that achieves the minimum value of $\left|\right|x_1-u\left|\right|$ over $u\in \mathcal{N}\left(A\right)$ , i.e., the closest point to $x_1$ in $\mathcal{N}\left(A\right)$ . Assume $\widehat{u}$ is such a point; then, $x_1-\widehat{u}\perp \mathcal{N}\left(A\right)$ . Now, recall that ${\left(\mathcal{N}\left(A\right)\right)}^{\perp }=\mathcal{R}\left(A^{*}\right)$ , so $\widehat{x}=x_1-\widehat{u}\in \mathcal{R}\left(A^{*}\right)$ . Thus $\widehat{x}=A^{*}z$ for some $z\in Y$ , and $y=A\widehat{x}=A{A}^{*}z$ . Note that if $A{A}^{*}$ is invertible, then $\widehat{x}=A^{*}z=A^{*}{\left(A{A}^{*}\right)}^{-1}y$ .
3. No solution : In this case, we may aim to find a solution that minimizes the mismatch between the two sides of the equation, i.e., $\parallel y-Ax\parallel$ . This is the well-known projection problem of $y$ into $\mathcal{R}\left(A\right)$ . Theorem 2 Let $X$ , $Y$ be Hilbert spaces, $y\in Y$ , and $A\in B(X,Y)$ . The vector $\widehat{x}$ minimizes $\left|\right|x-Ay\left|\right|$ if and only if $A^{*}A\widehat{x}=A^{*}y$ . Denote $u=Ax$ so that $u\in \mathcal{R}\left(A\right)$ . We need to find the minimum value of $\left|\right|y-u\left|\right|$ over $u\in \mathcal{R}\left(A\right)$ . Assume $\widehat{u}$ is the closest point to $y$ in $\mathcal{R}\left(A\right)$ , then $y-\widehat{u}\perp \mathcal{R}\left(A\right)$ , which means $y-\widehat{u}\in {\left(\mathcal{R}\left(A\right)\right)}^{\perp }$ . Recall that ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }=\mathcal{N}\left(A^{*}\right)$ , which implies that $y-\widehat{u}\in \mathcal{N}\left(A^{*}\right)$ . So $A^{*}(y-\widehat{u})=0$ . Thus $A^{*}y=A^{*}\widehat{u}$ . Now, denoting $\widehat{u}=A\widehat{x}$ , we have that $A^{*}y=A^{*}A\widehat{x}$ . Note that if $A^{*}A$ is invertible, then we have $\widehat{x}={\left(A^{*}A\right)}^{-1}{A}^{*}y$ .

Unitary operator

Definition 1 An operator $A\in B(X,X)$ is said to be unitary if $A{A}^{*}=A^{*}A=I$ .

This implies that $A^{*}=A^{-1}$ . Unitary operators have norm-preservation properties.

Theorem 3 $A\in B(X,X)$ is unitary if and only if $\mathcal{R}\left(A\right)=X$ and $\left|\right|Ax\left|\right|=\left|\right|x\left|\right|$ for all $x\in X$ .

Let $A$ be unitary, then for any $x\in X$ ,

Since $I:X\to X$ and $A{A}^{*}:\mathcal{R}\left(A\right)\to \mathcal{R}\left(A\right)$ , then $X=\mathcal{R}\left(A\right)$ . So if $A\in B(X,X)$ is unitary, then $\mathcal{R}\left(A\right)=X$ and $\left|\right|Ax\left|\right|=\left|\right|x\left|\right|$ for all $x\in X$ . From [link] we can find that

Since this is true for all $x\in X$ we have that $x-A^{*}Ax=0$ for all $x\in X$ , which means that $x=A^{*}Ax$ for all $x\in X$ . Therfore, we must have $A^{*}A=I$ . Additionally, since the operator is unitary, we find that for any $x,y\in X$ we have that $\left|\right|Ax-Ay\left|\right|=\left|\right|A(x-y)\left|\right|=\left|\right|x-y\left|\right|$ . So $x=y$ if and only if $Ax=Ay$ , implying that $A$ is one-to-one. Since $\mathcal{R}\left(A\right)=X$ , then $A$ is onto as well. Thus, $A$ is invertible, which means $A^{-1}A=I$ . So $A^{-1}A=I=A^{*}A$ , and therefore $A^{-1}=A^{*}$ . The result is that $A$ is unitary. We have shown that if $\mathcal{R}\left(A\right)=X$ and $\left|\right|Ax\left|\right|=\left|\right|x\left|\right|$ for all $x\in X$ , then $A\in B(X,X)$ is unitary.

Corollary 4 If $X$ is finite-dimensional, then $A$ is unitary if and only if $\left|\right|Ax\left|\right|=\left|\right|x\left|\right|$ for all $x\in X$ .