# 1.8 Properties of linear operators

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Discusses basic properties and classifications of operators.

Lemma 1 If $A\in B\left(X,X\right)$ 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

$\begin{array}{cc}\hfill 0& =〈x,+,a,y,,,A,\left(,x,+,a,y,\right)〉,\hfill \\ & =〈x,+,a,y,,,A,x,+,a,A,y〉,\hfill \\ & =〈x,,,A,x〉+\overline{a}〈x,,,A,y〉+a〈y,,,A,x〉+{|a|}^{2}〈y,,,A,y〉.\hfill \end{array}$

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 $A\in B\left(X,Y\right)$ is a bounded linear operator. Now pick $y\in Y$ . Then we pose the question: does a solution $\stackrel{^}{x}\in X$ 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 $A\left(x-{x}_{1}\right)=0$ . 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 $\stackrel{^}{x}={A}^{-1}y$ ;
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 $A\in B\left(X,Y\right)$ . Then $\stackrel{^}{x}$ is the solution to $Ax=y$ if and only if $\stackrel{^}{x}={A}^{*}z$ , 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 $\stackrel{^}{x}={x}_{1}-u$ , where $u\in \mathcal{N}\left(A\right)$ . We therefore search for the solution that achieves the minimum value of $||{x}_{1}-u||$ over $u\in \mathcal{N}\left(A\right)$ , i.e., the closest point to ${x}_{1}$ in $\mathcal{N}\left(A\right)$ . Assume $\stackrel{^}{u}$ is such a point; then, ${x}_{1}-\stackrel{^}{u}\perp \mathcal{N}\left(A\right)$ . Now, recall that ${\left(\mathcal{N}\left(A\right)\right)}^{\perp }=\mathcal{R}\left({A}^{*}\right)$ , so $\stackrel{^}{x}={x}_{1}-\stackrel{^}{u}\in \mathcal{R}\left({A}^{*}\right)$ . Thus $\stackrel{^}{x}={A}^{*}z$ for some $z\in Y$ , and $y=A\stackrel{^}{x}=A{A}^{*}z$ . Note that if $A{A}^{*}$ is invertible, then $\stackrel{^}{x}={A}^{*}z={A}^{*}{\left(A{A}^{*}\right)}^{-1}y$ .
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 $A\in B\left(X,Y\right)$ . The vector $\stackrel{^}{x}$ minimizes $||x-Ay||$ if and only if ${A}^{*}A\stackrel{^}{x}={A}^{*}y$ . Denote $u=Ax$ so that $u\in \mathcal{R}\left(A\right)$ . We need to find the minimum value of $||y-u||$ over $u\in \mathcal{R}\left(A\right)$ . Assume $\stackrel{^}{u}$ is the closest point to $y$ in $\mathcal{R}\left(A\right)$ , then $y-\stackrel{^}{u}\perp \mathcal{R}\left(A\right)$ , which means $y-\stackrel{^}{u}\in {\left(\mathcal{R}\left(A\right)\right)}^{\perp }$ . Recall that ${\left(\mathcal{R}\left(A\right)\right)}^{\perp }=\mathcal{N}\left({A}^{*}\right)$ , which implies that $y-\stackrel{^}{u}\in \mathcal{N}\left({A}^{*}\right)$ . So ${A}^{*}\left(y-\stackrel{^}{u}\right)=0$ . Thus ${A}^{*}y={A}^{*}\stackrel{^}{u}$ . Now, denoting $\stackrel{^}{u}=A\stackrel{^}{x}$ , we have that ${A}^{*}y={A}^{*}A\stackrel{^}{x}$ . Note that if ${A}^{*}A$ is invertible, then we have $\stackrel{^}{x}={\left({A}^{*}A\right)}^{-1}{A}^{*}y$ .

## Unitary operator

Definition 1 An operator $A\in B\left(X,X\right)$ 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 $A\in B\left(X,X\right)$ is unitary if and only if $\mathcal{R}\left(A\right)=X$ and $||Ax||=||x||$ for all $x\in X$ .

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

$\begin{array}{cc}\hfill ||x||& =\sqrt{〈x,,,x〉}=\sqrt{〈x,,,A,{A}^{*},x〉}=\sqrt{〈A,x,,,A,x〉}=||Ax||.\hfill \end{array}$

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

$\begin{array}{cc}\hfill 0& ={||x||}^{2}-{||Ax||}^{2}=〈x,,,x〉-〈A,x,,,A,x〉=〈x,,,x〉-〈x,,,{A}^{*},A,x〉=〈x,,,x,-,{A}^{*},A,x〉.\hfill \end{array}$

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 $||Ax-Ay||=||A\left(x-y\right)||=||x-y||$ . 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 $||Ax||=||x||$ for all $x\in X$ , then $A\in B\left(X,X\right)$ is unitary.

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

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