Discusses basic properties and classifications of operators.
We start with a simple but useful property.
Lemma 1 If
$A\in B(X,X)$ and
$\u2329x,,,A,x\u232a=0$ 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
$$\begin{array}{cc}\hfill 0& =\u2329x,+,a,y,,,A,(,x,+,a,y,)\u232a,\hfill \\ & =\u2329x,+,a,y,,,A,x,+,a,A,y\u232a,\hfill \\ & =\u2329x,,,A,x\u232a+\overline{a}\u2329x,,,A,y\u232a+a\u2329y,,,A,x\u232a+{\lefta\right}^{2}\u2329y,,,A,y\u232a.\hfill \end{array}$$
Since
$x,y\in X$ , we have that
$\u2329x,,,A,x\u232a=\u2329y,,,A,y\u232a=0$ ;
therefore,
$0=\overline{a}\u2329x,,,A,y\u232a+a\u2329y,,,A,x\u232a$ .
If we set
$a=1$ , then
$0=\u2329x,,,A,y\u232a+\u2329y,,,A,x\u232a$ . So in this case,
$\u2329x,,,A,y\u232a=\u2329y,,,A,x\u232a$ .
If we set
$a=i$ , then
$0=i\u2329x,,,A,y\u232a+i\u2329y,,,A,x\u232a$ . So in this case,
$\u2329x,,,A,y\u232a=\u2329y,,,A,x\u232a$ .
Thus,
$\u2329x,,,A,y\u232a=0$ 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:
 A unique solution exists;
 multiple solutions exist; or
 no solution exists.
We consider these cases separately below.

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
$\widehat{x}={A}^{1}y$ ;

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 closedform expression for the minimumnorm 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
$\widehat{x}={A}^{*}z={A}^{*}{\left(A{A}^{*}\right)}^{1}y$ .

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 yAx\parallel $ . This is the wellknown 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
$\left\rightxAy\left\right$ 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
$\left\rightyu\left\right$ 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 normpreservation properties.
Theorem 3
$A\in B(X,X)$ is unitary if and only if
$\mathcal{R}\left(A\right)=X$ and
$\left\rightAx\left\right=\left\rightx\left\right$ for all
$x\in X$ .
Let
$A$ be unitary, then for any
$x\in X$ ,
$$\begin{array}{cc}\hfill \left\rightx\left\right& =\sqrt{\u2329x,,,x\u232a}=\sqrt{\u2329x,,,A,{A}^{*},x\u232a}=\sqrt{\u2329A,x,,,A,x\u232a}=\left\rightAx\left\right.\hfill \end{array}$$
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
$A\in B(X,X)$ is unitary, then
$\mathcal{R}\left(A\right)=X$ and
$\left\rightAx\left\right=\left\rightx\left\right$ for all
$x\in X$ .
From
[link] we can find that
$$\begin{array}{cc}\hfill 0& ={\left\rightx\left\right}^{2}{\left\rightAx\left\right}^{2}=\u2329x,,,x\u232a\u2329A,x,,,A,x\u232a=\u2329x,,,x\u232a\u2329x,,,{A}^{*},A,x\u232a=\u2329x,,,x,,{A}^{*},A,x\u232a.\hfill \end{array}$$
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
$\left\rightAxAy\left\right=\left\rightA(xy)\left\right=\left\rightxy\left\right$ . So
$x=y$ if and only if
$Ax=Ay$ , implying that
$A$ is onetoone.
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
$\left\rightAx\left\right=\left\rightx\left\right$ for all
$x\in X$ , then
$A\in B(X,X)$ is unitary.
Corollary 4 If
$X$ is finitedimensional, then
$A$ is unitary if and only if
$\left\rightAx\left\right=\left\rightx\left\right$ for all
$x\in X$ .