# 1.8 Properties of linear operators

 Page 1 / 1
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 $x\in X$ .

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
why its coecients must have a power-law rate of decay with q > 1/p. ?