text:
Are you or your partner age 65 or over and on a low income?
If so you may be affected by some changes to benefits rules starting this May which mean you could be up to £7,000 a year worse off. Currently couples where one person is over pension age and one person is under pension age can claim Pension Credit, but from 15 May the rules are changing.
After this date these 'mixed age' couples will need to claim Universal Credit instead and only couples who are BOTH over pension age will be able to claim Pension Credit.
Pension Credit is a benefit for older people who are on a low income to top up their weekly income. Over a million people who are entitled don't claim it and around 50,000 of these people are estimated to be in mixed age couples.
Universal Credit is a new benefit for working age people that replaces a number of existing benefits and tax credits. It is designed to support people who have a low (or no) income with their basic living expenses and housing costs.
As well as all of the conditions attached to Universal Credit – such as the five week wait for your first payment, monthly rather than weekly or fortnightly payments and the need to sign a claimant commitment that may mean you must look for a set number of hours work – the amount of money you could be eligible for are very different.
The amounts will vary based on your own circumstances but for working age couples the maximum level of personal support from Universal Credit is £114.85 a week but from April this year Pension Credit pays £255.25 a week, a difference of up to £140 a week. And this will apply every week until both you and your partner reach pension age.
There could also be further loses if you receive help with Council Tax or have more bedrooms than the government allow you to claim help for.
The change means that all couples where one person is over 65 should check their benefit entitlement before 15 May. If you are entitled to Pension Credit and claim it before 15 May you won't be affected but if you try to claim after the deadline you will be too late and will have to claim Universal Credit instead.
If you have any questions contact one of our Financial Support Officers today.
*Costs per minute up to 10p from landlines and between 3p and 55p from mobile phones. Calls are free from your mobile if you have free minutes in your package."
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text:
Q: unable to get this class working in java on client code - no compilation errors but runtime I have pasted my code here. below the code I give descriptin of what I am trying to do
 import java.util.*;
    import java.io.*;

//given a string, gives letters counts etc
public class LetterDatabase{
    private String word;
    private int[] charCounts;
    public static final int TOTAL_LETTERS=26;

    //constructor
    public LetterDatabase(String word) {
        this.word=word;
        int[] charCounts= new int[TOTAL_LETTERS];
        //this function fillup charCounts with no of occurances
        //more below
        fillupcharArray();
    }

    /*Returns a count of letters.
    public int get(char letter) {
    return charCounts[Character.getNumericValue(letter)-10];
}

/*Sets the count for the given letter to the given value.
public void set(char letter,int value) {
    int index=Character.getNumericValue(letter-'a');
    charCounts[index]=value;
 }

/* converts string to Array of chars containing only alphabets
private char[] convertToArray() {
    String str = word.replaceAll("[^a-zA-Z]+", "");
    char[] charArr=new char[str.length()];
    for (int i = 0; i < str.length(); i++) {
            charArr[i] = str.charAt(i);
        }
    return charArr;
}

private void fillupcharArray(){
    char[] charArr=convertToArray();
    for(int i=0;i<charArr.length;i++) {
       for(int j=0;j<26;j++) {
            if (Character.toLowerCase(charArr[i])==Character.toLowerCase((char)('a'+j))) {
           charCounts[j]+=1;
            }
        }
    }
}

}
my client code that I used to test is below
import java.util.*;
import java.io.*;
public class Testdatabase{
 public static void main(String args[]) {
    String str="my name is Dummy!!!:..,";
    LetterDatabase first= new LetterInventory(str);
    System.out.println(first.get('a'));
    System.out.println(first.set('a'));
    System.out.println(first.get('a'));
    }
 }

Explanation: I define a LetterDatabase class which computes the count of letters--only alphabets (type char) in given string. I have a get method, that returns the occurance of particular letter and a set method, that sets the value of letter to a set value.
in my constructor, I am calling a function that fills the array(charCounts), so that I can easily lookup for occurrence of a given char. Firstly my constructor does not work. my class code compiles and my client code above compiles. when I run the clientcode, commenting out the getter and setter calls, i get the following error.
Exception in thread "main" java.lang.NullPointerException
    at LetterDatabase.fillupcharArray(LetterInventory.java:55)
    at LetterDatabase.<init>(LetterInventory.java:17)
    at Testdatabase.main(hw1test.java:7)

I am unable to figure what is going wrong. the fillupcharArray seems to work fine when I test it individually. I am not pasting that here.
secondly, the way I define get and set method in my class is not very good. it would be nice if I dont have to use Character.getNumericValue
I am open to hear on any other improvements. Thanks for your time

A: In your constructor, you've defined a local variable charCounts:
int[] charCounts= new int[TOTAL_LETTERS];

This shadows the instance variable charCounts, to which I think you meant to assign it.
Assign it to the instance variable charCounts like this:
charCounts = new int[TOTAL_LETTERS];

This means that, in your code, the instance variable charCounts remains null until you access and a NullPointerException results.

A: You are shadowing the array charCounts in the constructor of LetterDatabase. Replace
int[] charCounts = new int[TOTAL_LETTERS];

with
charCounts = new int[TOTAL_LETTERS];

Inside a class method, when a local variable has the same name as an instance variable, the local variable shadows the instance variable inside the method block. In this case the local variable charCounts is shadowing the instance variable charCounts.
From wikipedia

In computer programming, variable shadowing occurs when a variable declared within a certain scope (decision block, method, or inner class) has the same name as a variable declared in an outer scope

redpajama_set_name: RedPajamaStackExchange


text:
Feature Release: Boater beware: Vessel documentation fraud lurks online
U.S. Coast Guard sent this bulletin at 10/16/2017 07:45 PM EDT
Feature Release
U.S. Coast Guard 17th District PA Detachment Anchorage
Contact: Coast Guard PA Detachment Anchorage
Editors' Note: Click on images to download high resolution version.
Story by Petty Officer 1st Class Bill Colclough
A new scam is targeting boat owners looking to save a little time online, but it's costing them hundreds of dollars: Websites offering documentation renewal services for a fee.
These websites lure boaters with the appearance and convenience of an official government website, but, just because a site looks like the Coast Guard and works like the Coast Guard, that does not mean it's the Coast Guard. Boaters using these websites can end up spending three times the standard fee, and Coast Guard boarding officers will not accept their vessel's documentation as valid.
Why? The U.S. Coast Guard's National Vessel Documentation Center, located in Falling Waters, West Virginia, is the only authorized entity to issue Certificates of Documentation required for vessels engaged in commercial trade and optional for vessels weighing five or more net tons engaged in recreational use and activities.
The NVDC is aware there are commercial entities that offer to manage the certification and renewal process on behalf of vessel owners for a fee. The Coast Guard does not endorse any of these companies, and the companies do not operate on behalf of the Coast Guard in any way. The services they provide are legal, but the certificates issued are not deemed in compliance.
Any fees charged beyond the $26 renewal fee or other agreements offered by such companies are in no way associated with the NVDC certification process. In addition, these companies are not authorized to issue any form of documentation, including travel letters and/or permits that authorize operation of any vessel.
"These are legitimate companies similar to the DMV [Department of Motor Vehicles] that have satellite offices open on the weekends and after hours. They'll give you your tags, but it's going to cost you twice as much because you're paying for their service," said Russell Hazlett, commercial fishing vessel examiner for Coast Guard Sector Anchorage. "These companies are not issuing the certificate, but rather, they are the middleman who charges a fee for processing the paperwork on your behalf."
According to Hazlett, many fishermen in Alaska sign up with a company to handle their annual documentation and end up paying hundreds of dollars. They often realize too late they are not dealing directly with the Coast Guard.
During dockside safety examinations, Hazlett provides operators with a copy of the marine safety advisory bulletin sent by the NVDC May 16, 2017, which advises vessel operators to be aware of third party service providers.
"Just this year alone I have come across probably eight people who have fallen victim to it," said Hazlett. "I have warned them: When you go to renew your documentation, just be aware there is another company that provides that service."
Hazlett noted that many third party websites resemble a Coast Guard website complete with an anchor-themed logo and color scheme that closely matches the Department of Homeland Security seal.
A librarian in Naknek, Alaska, filed for her husband after she received an official-looking letter marked "reinstatement courtesy notice" from a website offering documentation services. The letter even listed the boat's hull number and looked every bit like an official Coast Guard document. The website bore all the markers for an official domain except one: the non-official site ended in ".us" instead of ".mil."
"It looked like a Coast Guard site, but when you scrolled all the way to the bottom, the site claimed in the tiniest font they are not affiliated with the U.S. Coast Guard," said the librarian.
The librarian cancelled her credit card and proceeded to dispute the charge. It was when she entered "coast guard renewal scam" in an online search engine that she realized she was not alone. It turns out there are a number of forums online of fishermen in Alaska and across the rest of the country, who share their stories and seek information on how to get their money back from third-party providers.
"I was worried we got scammed, which we did. I ended up paying about $150 when it only really costs $26. I was really worried we wouldn't get the certificate in time for my husband to go fishing," said the librarian.
The wife of a fisherman in Whittier, Alaska, paid twice the standard fee of $26 after her husband received an email that stated his permit could be renewed with a fee.
"I thought it was free, so I told him I'll renew it, because the website looked so official. And that's why I was probably duped," she said.
The charge on her credit card statement coupled with unreturned calls from the third party provider tipped her off that the transaction was not legitimate.
"We thought because of budget cuts there was now a charge, but I thought to myself, hmmm, it smells like rotten fish," she said.
She disputed the charge with her credit card company on the grounds of fraud; however, the representative told her she was responsible for the charge because she agreed to it.
"I said: No. You'll need to look up Webster for the definition of fraud. Fraud means that someone misrepresented their intention," she said.
She eventually received a refund after talking to a supervisor, but not every boat owner is so lucky.
When it comes time to renew the documentation, always remember the Coast Guard website offering documentation uses a .mil domain name. Web domains using .us, .com and .org are not authorized. Get the Coast Guard-approved certificate that helps save money just like a Coast Guard-approved life jacket helps save lives.
For more information on Certificates of Documentation call the Coast Guard's NVDC at 800-799-8362 or visit http://www.uscg.mil/nvdc/.

redpajama_set_name: RedPajamaCommonCrawl


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For sale in aliwal road, all property types.


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text:
\section{Introduction }

 Let $H$ and $K$ be  two complex Hilbert spaces  and  $\mathcal{B}(H, K)$ be the Banach space of all bounded linear operators from $H$ into $K$. In the case
  $K=H$, $\mathcal{B}(H,H)$  is simply denoted by  $\mathcal{B}(H) $ and is a  Banach algebra.

  For an arbitrary operator $T\in \mathcal{B}(H, K)$, we denote by $\mathcal{R}(T)$, $\mathcal{N}(T)$ and  $T^*$  the range,  the null subspace and   the operator adjoint  of $T$ respectively.
  For $T\in \mathcal{B}(H)$, the spectrum of $T$ is denoted by $\sigma(T)$.

 An operator $T\in \mathcal{B}(H,K)$ is  a {\it partial isometry} when $T^*T$ is an orthogonal projection (or, equivalently $TT^*T = T$).
  In particular $T$ is an {\it isometry} if $T^*T=I_{H}$, and {\it unitary} if $T$ is a surjective isometry.

As usually, for $T\in \mathcal{B}(H)$ we denote the module of $T$ by $|T|=(T^*T)^{1/2}$ and we shall always write, without further
mention, $T=V|T|$ to be the unique polar decomposition  of $T$, where $V$ is the appropriate partial isometry satisfying  $\mathcal{N}(V)=\mathcal{N}(T)$.
The Aluthge transform was   introduced in \cite{alu}  as
 $$\Delta(T)=|T|^{\frac{1}{2}}V|T|^{\frac{1}{2}}, \quad  T\in \mathcal{B}(H),$$
   to  extend some properties  of hyponormal  operators.
Later, in \cite{oku},    Okubo introduced a more general notion called  $\lambda-$Aluthge
transform which has also been studied in detail.

 For $\lambda \in [ 0, 1]$, the $\lambda-$Aluthge transform is  defined by,
 $$
 \Delta_{\lambda}(T)=|T|^{\lambda}V|T|^{1-\lambda}, \quad T\in \mathcal{B}(H).
 $$
  Notice  that $\Delta_0(T) =V|T|=T$, and $\Delta_1(T) = |T|V$ which is known as Duggal's
  transform.  It has since been studied in many different contexts and considered by a
  number of authors (see for instance,  \cite{ alu, ay, ams1, ams2,  kp1, kp2,  kp3} and some
  of the references there).  One of the interests  of the Aluthge transform lies in the fact that
   it respects many   properties of the original operator. For example,
\begin{equation}
   \sigma_*(\Delta_{\lambda}(T)) = \sigma_*(T),   \mbox{ for every } \; \; T \in \mathcal{B}(H),
\end{equation}
 where $  \sigma_*$ runs over a large family of spectra. See \cite[Theorems 1.3, 1.5]{kp3}.
 Another important property is that $Lat(T)$, the lattice of $T$-invariant subspaces of
 $H$, is nontrivial if and only if $Lat(\Delta(T))$ is nontrivial
(see \cite[Theorem 1.15]{kp3}).

   In \cite{bmg},   the authors described  the   linear  bijective mappings on von Neumann algebras which commute with the $\lambda$-Aluthge transform
      $$\Delta_{\lambda}(\Phi (T))=\Phi (\Delta_{\lambda}(T )) \mbox{ for   every }  T \in \mathcal{B}(H). $$

 In \cite{chf} the first author gives a complete description of the bijective maps  $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$ which satisfies  following condition,
  \begin{equation}
\Delta_{\lambda}(\Phi (A)\Phi (B))=\Phi (\Delta_{\lambda}(A B))   \mbox{ for   every }   A, B\in \mathcal{B}(H),
 \end{equation}
for some  $\lambda \in]0,1[$.

 \bigskip

  In this paper we will be concerned with the  Jordan product  commuting maps  with the  $\lambda$-Aluthge transform  in the following sense,
  \begin{equation}\label{c1}
\Delta_{\lambda}(\Phi(A)\circ \Phi(B))=\Phi(\Delta_{\lambda}(A \circ B)) \: \mbox{ for  all}\;  A, B\in \mathcal{B}(H),
 \end{equation}
 where $A\circ B=\frac{1}{2}(AB+BA)$ is the Jordan product of  $A$ and $B$.

Our main result gives a complete description of the bijective maps $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$ which satisfies Condition (\ref{c1}), the precise theorem being as follows

  \begin{thm} \label{th1}
 Let $H$ and $K$ be  two  complex Hilbert space, such that $H$ is of dimension greater than $2$. Let   $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map. Then
  $\Phi$ satisfies  (\ref{c1}) for some  $\lambda \in]0,1[\:$   if and only if  there exists  a unitary operator $U : H\to K $ such that
 $$\Phi(A)=UAU^*\;\;\;\;\text{ for every} \; A\in\mathcal{B}(H).$$
 \end{thm}

 The paper is organized as follows. In the second section, we establish some useful results on the Aluthge transform.
  These results are needed for proving our main theorem in the  section 3.

\section{  Some properties of the  Aluthge transform}

In this section we establish some results,  properties of the  Aluthge transform. These results are necessary for the proof of the main theorem.

We first recall some basic notions that are used in the sequel. An operator $T\in\mathcal{B}(H)$  is  normal if $T^*T=TT^*$,  and  is  quasi-normal, if  it commutes with $T^*T$ ( i.e. $TT^*T=T^*T^2$), or equivalently  $|T|$ and $V$ commutes ($T = V\vert T\vert$  a  polar decomposition of $T$). In finite dimensional spaces  every quasi-normal operator  is normal.  It is easy to see that if $T$ is quasi-normal, then $T^2$ is also quasi-normal, but the converse is false as shown by nonzero  nilpotent operators .

Also, it is well known that  the quasi-normal operators  are exactly the fixed points of  $\Delta_{\lambda}$ (see \cite[Proposition 1.10]{kp3}). That is,
   \begin{equation}\label{qn}
 T \; \; \mbox{ quasi-normal} \; \;  \iff  \Delta_{\lambda}(T) = T.
 \end{equation}

  For nonzero $x,y \in H$ , we denote by  $x\otimes y$ the rank one operator defined by
 $$(x\otimes y)u=<u,y>x \: \mbox{   for } \:  u\in H.$$
  It is easy to show that every rank one   operator has  the previous form and that   $x\otimes y$ is an orthogonal projection, if and only if $x=y$ and $\|x\|=1$.

\bigskip

 We start with  the following proposition.  It is found in \cite{chf}.  To facilitate the reading we include  the proof.

  \begin{pro}\label{p1} (\cite{chf})  Let $x,y \in H$ be nonzero vectors. We have
 $$\Delta_{\lambda}(x\otimes y)=\dfrac{<x,y>}{\|y\|^2}(y\otimes y)\: \:  \mbox{ for every }  \: \lambda \in ]0,1[.$$
\end{pro}
\begin{proof}
Denote $T=x\otimes y$, then
$$T^*T=|T|^2=\|x\|^2(y\otimes y)=\big(\frac{\|x\|}{\|y\|}(y\otimes y)\big)^2\;\;\text{and}\;\;|T| = \frac{\|x\|}{\|y\|}(y\otimes y).$$
It follows that
$$|T|^2=\|x\|\, \|y\| |T|\;\; \text{and}\;\; |T|^\gamma =(\|x\|\, \|y\|)^{\gamma-1}|T| \: \text{ for every}  \; \gamma>0.$$

Now,  let  $T=V|T|$ be the polar decomposition of $T$. We have
\begin{align*}
\Delta_{\lambda}(x\otimes y)=\Delta_{\lambda}(T)&=|T|^{\lambda}V|T|^{1-\lambda}\\
&=(\|x\|\, \|y\|)^{{\lambda-1}}(\|x\|\, \|y\|)^{{-\lambda}}|T|V|T|\\
&=\dfrac{1}{\|x\|\, \|y\|}|T|T =\frac{1}{\|y\|^2}(y\otimes y)\circ (x\otimes y)\\
&=\dfrac{<x,y>}{\|y\|^2}(y\otimes y).
\end{align*}
Thus , $\Delta_{\lambda}(x\otimes y)=\dfrac{<x,y>}{\|y\|^2}(y\otimes y)\: $ as desired
\end{proof}

 \begin{pro}\label{l1} Let $A\in \mathcal{B}(H)$ and  $P=x\otimes x$ be a rank one projection on $H$. Then
$$\Delta_{\lambda}(A \circ P) = P  \;\;\text{if and only if}\;\; PA=P.$$
\end{pro}
   \begin{proof} The "if" part  follows directly from the previous proposition. We show the "only if" part.  Suppose that $\Delta_{\lambda}(A \circ P) = P$.
   First, it is easy to see the following
       $$ PA = P\:  \iff \: A^*x = x.$$
       So,  to complete the proof of the proposition, it suffices to show that $A^*x = x$.

  Put  $T = A \circ P $. Then  $T = \frac{1}{2}(Ax\otimes x+x\otimes A^*x)$,  and thus
 \begin{equation}\label{eq-}
T^2=\frac{1}{4}(<Ax,x>Ax\otimes x + Ax\otimes A^*x+<A^2x,x> x\otimes x+<Ax,x> x\otimes A^*x).
 \end{equation}
  In the other hand, we have
 $$
  \sigma(T)=\sigma(\Delta_{\lambda}(T))=\sigma(P)=\{0,1\}  \; \; \mbox{ and thus} \;  \sigma(T^2)=\{0,1\}.
   $$
  Since the rank of $T$ is at most $2$,  $tr(T^2)=tr(T)=1$.  We also have $tr(T)=<Ax,x>$ and $tr(T^2)=\frac{1}{2}(<A^2x,x>+<Ax,x>)$. Therefore
  $$<Ax,x> =  <A^2x,x> =1.$$
  It follows that
  $Tx=\frac{1}{2}(Ax+x)$ and $T^*x=\frac{1}{2}(A^*x+x)$. Hence
  $$<Tx,x> = 1 \: \mbox{and} \: <T^*x,x>=1.$$

  From (\ref{eq-}) we get
\begin{equation} \label{eq1}
 T^2=\frac{1}{4}(Ax+x)\otimes (A^*x+x)=Tx\otimes T^*x.
\end{equation}

  Let  $T=U|T|$ be the polar decomposition of $T$.  It holds
  \begin{eqnarray*}
  T^2 &=& U|T|U|T|= U|T|^{1-\lambda}(|T|^{\lambda}U|T|^{1-\lambda})|T|^{\lambda}\\
  &=& U|T|^{1-\lambda} \Delta_{\lambda}(T)|T|^{\lambda} =  U|T|^{1-\lambda}(x \otimes  x) |T|^{\lambda}   \\
  & = &    U|T|^{1-\lambda}x \otimes |T|^{\lambda} x.
 \end{eqnarray*}
   From  (\ref{eq1}) we get
   \begin{equation}\label{eq+}
Tx\otimes T^*x = U|T|^{1-\lambda}x \otimes |T|^{\lambda}x.
   \end{equation}
 Since $<T^*x,x> =1$, we get that
  $$Tx = <x,|T|^{\lambda}x> U|T|^{1-\lambda}x.$$
 Thus
 $$ |T|^{\lambda}|T|x = |T|^{\lambda}U^*Tx =  <x,|T|^{\lambda}x>U^* U|T|x = <x,|T|^{\lambda}x>|T|x.$$
  By function calculus, we obtain
   $$|T||T|x=( <x,|T|^{\lambda}x>)^{1/\lambda}|T|x.$$
    It follows that
  $$<|T|^2x,x>=\||T|x\|^2=( <x,|T|^{\lambda}x>)^{1/\lambda}<|T|x,x>.$$

   Now, by Holder-Mc Carthy inequality (see \cite{fur} page 123), we deduce

 $$\big{\|}|T|x\big{\|}^2=( <x,|T|^{\lambda}x>)^{1/\lambda}<|T|x,x>\leq (<|T|x,x>)^2\leq \big{\|}|T|x\big{\|}^2.$$
  Hence $\big{\|}|T|x\big{\|}=<|T|x,x>$. Therefore  there exists $\alpha \in \mathbb{C}$ such that $|T|x=\alpha x$.\\
  The last equality and Equation  (\ref{eq+}) imply
  $$Tx\otimes T^*x=\alpha Ux \otimes x.$$
Thus, $T^*x = \beta x$ for some $\beta \in \mathbb{C}$.  Since,  $<T^*x, x> = 1, \;$ it holds $ \; \beta = 1$.  On the other hand $T^*x=\frac{1}{2}(A^*x+x)$. Then
$ \frac{1}{2}(A^*x+x) = x$. Finally we conclude that  $A^*x=x$ and hence  $PA=P$. The proof is completed.
   \end{proof}

  \bigskip

   The following lemma gives a property  of  rank one projections

    \begin{pro}\label{l2} Let $A\in \mathcal{B}(H)$ and  $P=x\otimes x$ be a rank one projection on $H$. Then
 $$\Delta_{\lambda}(A\circ P)=A \;\text{ if and only if} \;\; A=\alpha P \;\;\text{for some}\; \alpha \in \mathbb{C}.$$
 \end{pro}

 \begin{proof}
   The "if" part  is obvious. We show the "only if" part. Put $T= A\circ P$. Then
   $$T =\frac{1}{2}(Ax\otimes x+x\otimes A^*x) \; \;  \mbox{and} \; \;   T^* =\frac{1}{2}(x\otimes Ax + A^*x\otimes x).$$
   By assumption,  $A = \Delta_{\lambda}(T)$.
  Since  $\mathcal{R}( \Delta_{\lambda}(T)), \mathcal{R}( \Delta_{\lambda}(T)^*) \subseteq \mathcal{R}(T^*)$,  we have

 $$\mathcal{R}(A), \mathcal{R}(A^*)\subseteq \mathcal{R}(T^*)\subseteq H_0 := span\{x,A^*x\}.$$
 Note that  $H_0$ is an invariant subspace of $A$ and $A^*$.

  {\bf  Claim:}   there exists $\delta\in\mathbb{C}$  such that $A^*x=\delta x$.
  In this case, $A$ and $A^*$ are rank one operators. Moreover, their ranges  $\mathcal{R}(A)$ and  $\mathcal{R}(A^*)$ are generated by $x$. It is easy to deduce from this, that
   $A=\overline{\delta}P.$

  We prove the claim by contradiction.
Assume on contrary that $A^*x$ and $x$ are linearly independent. Let $\{x, e\}$ be an orthonormal basis of $H_0$.  We can choose $e \in H_0$  such that  $<e,Ax>\geq 0$.
In this basis $A$ has the following form
 \begin{equation}\label{eq3}
A=x\otimes A^*x +e\otimes A^*e.
 \end{equation}
 Let us consider the following cases :

 \textbf{Case 1 :}  $A^*e=0$. In this case,  we show $A^*x$ and $x$ are linearly dependent, which is a contradiction.

From (\ref{eq3}), we have $A=x\otimes A^*x$. Hence $$T=P\circ A= 1/2 ( x\otimes A^*x+<Ax,x>x\otimes x)=1/2 x\otimes\big(A^*x+<A^*x,x>x\big).$$
Since $\mathcal{R}(A) \subseteq \mathcal{R}(T^*)$, then  $x$ and $A^*x+<A^*x,x>x$
are linearly dependent it follows that $x$ and $A^*x$ are also linearly dependent,  which is a contradiction.

\textbf{Case 2 :}  $A^*e\ne 0$. In this case,  we show that $A=2T$ which is a contradiction  with the fact that
$\Delta_{\lambda}(T)=\Delta_{\lambda}(A/2)=A,$ and
$$\|A\|=\|\Delta_{\lambda}(A/2)\|\leq \frac{\|A\|}{2}.$$
Thus  $A=0$,  which is not possible.

  Note that since $A=\Delta_{\lambda}(T)$ it holds
   $$\|A\|=\|\Delta_{\lambda}(T)\|\leq  \|T\|  =   \|\frac{1}{2}(Ax\otimes x+x\otimes A^*x)\|\leq\frac{1}{2}(\|Ax\|+\|A^*x\|)\leq \|A\|.$$
  Hence
   \begin{equation}\label{eq2}
  \|A\|=\|Ax\|=\|A^*x\|.
  \end{equation}

Now, by (\ref{eq3}), we have

$$
A=2T \Leftrightarrow e\otimes A^*e=Ax\otimes x \Leftrightarrow
\left\{
      \begin{aligned}
        Ax= a e ;\\
        A^*e= \bar{a} x,\\
      \end{aligned}
    \right. \;\;  \mbox { for some } a \in \mathbb{C}.
  $$
So, we just need to prove  the equivalence.
First,  we show that
$$A^*e=\|A^*e\| x.$$

 Indeed by assumption, we get
$$tr(A)=tr(\Delta_{\lambda}(T)) = tr(T) = <Ax,x>.$$
  By (\ref{eq3}), we get $$tr(A)=<Ax,x> + <Ae,e>.$$
  Thus $<Ae,e>=0$.  Since $A^*e \in H_0$ and $<e,Ax>\geq 0$, it follows that
 \begin{equation}\label{eq4}
A^*e=<A^*e,x>x=\|A^*e\|x.
 \end{equation}

Second, we show $$Ax=\|A^*e\| e.$$ Indeed
 From (\ref{eq3}), we get
 \begin{equation*}
  AA^*=x\otimes AA^*x +e\otimes AA^*e.
 \end{equation*}
 Thus
$$AA^*x=\|A^*x\|^2 x+<x,AA^*e>e,$$
moreover, by (\ref{eq2})  it follows that
  $$\|AA^*x\|^2=\|A^*x\|^4 +|<x,AA^*e>|^2\leq \|AA^*\|^2=\|A\|^4=\|A^*x\|^4.$$
 Therefore
 $$<x,AA^*e>=0.$$
 This, together  with the fact $A^*e=\|A^*e\| x$,  show that
   $$ AA^*x=\|A\|^2 x \;\;\mbox{ and}\;\;  AA^*e=\|A^*e\|^2e=\|A^*e\| Ax.$$
 It follows that
\begin{equation} \label{eq4'}
 Ax=\|A^*e\|e.
\end{equation}

   This completes the proof.
  \end{proof}

 \begin{rem}  If $T = V \vert T\vert \in \mathcal{B}(H)$ is the polar decomposition of $T$,  then $V$ is unitary if and only if  $T$ and $T^*$ are one-to-one.

 \end{rem}

   \begin{lem}\label{l33} Let $T\in\mathcal{B}(H)$ and $\lambda \in]0,1[$. Suppose that $T$ and $T^*$ are one-to-one. Then,
  $$\Delta_{\lambda}(T^2)=T \;\;  \Longrightarrow  \;\; T^2 =  T^*.$$
  \end{lem}
  \begin{proof}  Let us consider $T^2=U|T^2|$ the polar decomposition of $T^2$. Since $T$ and $T^*$ are  injective, then $T^2$ and $(T^2)^*$ are also injective.
   Thus $U$ is unitary operator.  We have
  \begin{align*}
   \Delta_{\lambda}(T^2) =  |T^2|^{\lambda}U|T^2|^{1-\lambda} = T & \Longrightarrow   |T^2|^{\lambda}U|T^2|^{1-\lambda}   |T^2|^{\lambda} = T  |T^2|^{\lambda}\\
    & \Longrightarrow   |T^2|^{\lambda}U |T^2| = T  |T^2|^{\lambda}\\
     & \Longrightarrow   |T^2|^{\lambda}T^2 = T  |T^2|^{\lambda}.
  \end{align*}
  On the other hand, $T \Delta_{\lambda}(T^2)=T^2$, hence
    \begin{align*}  T |T^2|^{\lambda}U|T^2|^{1-\lambda} = T^2 = U|T^2| = U  |T^2|^{\lambda}|T^2|^{1-\lambda}
   \end{align*}
   Since  $T^2$ is injective, we get
   $$ T |T^2|^{\lambda}U = U  |T^2|^{\lambda}.$$
   Thus,
   $$  |T^2|^{\lambda} T^2 U = U  |T^2|^{\lambda} \; \; \text{and} \; \;   |T^2|^{\lambda}  T^2= U  |T^2|^{\lambda} U^* \geq 0.$$
 It follows that
   \begin{align*}
    |T^2|^{\lambda}  T^2 = T^{*2}  |T^2|^{\lambda} & =  |T^2| U^*  |T^2|^{\lambda} \\
     & =  |T^2|^{\lambda}   |T^2|^{1 - \lambda} U^*  |T^2|^{\lambda} \\
     & =  |T^2|^{\lambda}  \Delta_{\lambda}(T^2) ^* \\
     & =  |T^2|^{\lambda} T^{*}.
  \end{align*}
  Finally we obtain  $ T^2 = T^*$, as claimed.
    \end{proof}

     \begin{lem}\label{l22} Let $S\in\mathcal{B}(H)$ and $\lambda \in]0,1[$. Suppose that $S$ and $S^*$ are one-to-one. Then,
  $$\Delta_{\lambda}(S)=S^* \;\;  \Longrightarrow  \;\; S =  S^*.$$
  \end{lem}
  \begin{proof}  Let us consider $S=U|S|$ the polar decomposition of $S$. Since $S$ and $S^*$ are  injective,  $U$ is unitary operator.  We have,
 $$\Delta_{\lambda}(S) \Delta_{\lambda}(S)^* = S^*S = |S|^2 =  |S|^{\lambda}   |S|^{1 - \lambda}  |S|^{1 - \lambda}   |S|^{\lambda}.$$
 By a simple calculation, we deduce
 $$ U  |S|^{2(1 -\lambda)}  = |S|^{2(1 -\lambda)}U$$
 By the continuous functional calculus, we obtain $S$ quasi-normal. \\
 Consequently, $ S = \Delta_{\lambda}(S) = S^*$, as desired.
  \end{proof}

 \begin{rem}
  It is well known that  $T$ is quasi-normal  does not imply that his adjoint $T^*$  is always quasi-normal.

  For example :  take the right Shift operator $S$ on a separable Hilbert  space $H$,  with orthonormal bas is  $(e_n)_n$.  We have
  $$S^*S=I \: \text{ and} \: SS^*=I-P_1$$
   where  $P_1$ is the projection onto  the span of the first vector $e_1$.  Hence
   $$S^*SS^*=S^* \: \text{and} \:  S(S^*)^2=S^*-P_1S^*\neq S^*.$$
       This shows that $S^*$ is not quasi-normal.

       \bigskip

        Now, we replace the condition $S$ and $S^*$ are injective in preceding lemma by $S$ is quasi-normal. We have the following result

   \end{rem}

 \begin{lem}\label{l22}
  Let $S\in\mathcal{B}(H)$ and $\lambda \in]0,1[$. Suppose that $S$ is quasi-normal. Then
  $$\Delta_{\lambda}(S^*)=S \;\;  \Longrightarrow  \;\; S =  S^*.$$
 \end{lem}
  \begin{proof}
  Let us consider     the polar decomposition of $S$ and $S^*$,       $S=U|S|$ and $S^*=U^*|S^*|$  respectively.
It is  well known that  $|S^*|^p=U|S|^pU^*$ for all $p > 0$.

Now, since $S$ is quasi-normal,  $U|S|^p=|S|^pU$ and  $U^*|S|^p=|S|^pU^*$ for every  $p>0$. Hence,   we have the following equalities
 \begin{eqnarray*}
S =  \Delta_{\lambda}(S^*)&=&|S^*|^{\lambda}U^*|S^*|^{1-\lambda}\\&=&U|S|^{\lambda}U^*U^*U|S|^{1-\lambda}U^*\\&=& U(U^*)^2|S|.
 \end{eqnarray*}
It follows that
 \begin{equation}
 U(U^*)^2|S|=U|S|.
 \end{equation}
 After multiplying this equality by $U^*$, we get
  $$(U^*)^2|S|=|S|.$$
  Hence
 $$(U^*)^2|S|=|S|U^2=U|S|U=|S|.$$ Thus
 $$S= U|S| = |S|U=U^*U|S|U=U^*|S|=|S|U^*=S^*.$$
 Thus $ S = S^*$ and the lemma is proved.
 \end{proof}

  Combining  Lemmas \ref{l22} and \ref{l33}, we obtain the following corollary which plays an important role in the proof of $\Phi(I) = I$.

   \begin{cor}\label{l3}
  Let $T\in\mathcal{B}(H)$ and $\lambda \in]0,1[$. Suppose that $T$ and $T^*$ are one-to-one. Then,
  $$\Delta_{\lambda}(T^2)=T  \;\;\text{if and only if }\;\; T=I.$$
  \end{cor}

  \bigskip

  The next lemma,  was established by S. Garcia in the case $\lambda = \frac{1}{2}$, see \cite{gar}. It  identifies the kernel of the $\lambda$-Aluthge transform as the set of all operators which are nilpotent of order two. We need this lemma later.

  \begin{lem}\label{00}
Let $\lambda \in ]0,1]$ and $T\in \mathcal{B}(H)$. Then
$$ {\Delta_{ \lambda}}(T)=0 \: \text{ if and only if} \: T^2 = 0.$$
\end{lem}

\begin{proof}
 Let $T=V|T|$ be the polar decomposition of $T$.

If $\;  {\Delta_{ \lambda}}(T) = 0$, then we have
$$T^2=V|T|V|T|=V|T|^{1-\lambda} {\Delta_{ \lambda}}(T)|T|^{\lambda}=0,  \: \text{and thus} \;   T^2 = 0.$$

Conversely, suppose that  $T^2=0$. Then
$V|T|V|T|= T^2 = 0$.  Hence $V^*V|T|V|T|=0$, which  implies $|T|V|T|=0$,  since $V^*V$ is the projection onto $\overline{{\mathcal{R}(|T|)}}$.
Whence $|T|V$ vanishes on $\overline{{\mathcal{R}(|T|)}}$. Furthermore $\mathcal{N}(V) = \mathcal{N}(|T|)$, from which one concludes $|T|V$ vanishes on $\mathcal{N}(|T|)$.
Consequently, $|T|V=0$. Thus  ${\Delta_{ \lambda}}(T) = |T|^{\lambda}V|T|^{1-\lambda} = 0$.
\end{proof}

  \section{ Proof of the main theorem}

   An  idempotent self adjoint  operator $P\in\mathcal{B}(H)$  is said to be an orthogonal projection.
 Clearly  quasi-normal idempotents  are orthogonal projections. \\
Two  projections $P,Q\in\mathcal{B}(H)$ are said to be orthogonal if
$$PQ=QP=0,$$
 in this case, we denote $P\perp Q$.
  A partial ordering between orthogonal projections is defined as follows,
  $$Q\leq P\:  \;  \mbox{  if  } \:  \; PQ = QP = Q.$$

\begin{rem}  We have
$$ P\perp Q \: \iff \: P+Q \; \text{orthogonal projection},$$
and
   $$ Q \leq P \:  \iff PQ + QP = 2Q \: \iff P\circ Q = Q .$$
   \end{rem}

 \bigskip

\begin{pro}\label{cor1} Let $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map satisfying  (\ref{c1}). Then
 $$\Phi(0)=0 \quad \text{ and} \quad  \Phi(I)=I.$$
\end{pro}
 \begin{proof}
  Since $\Phi$ is onto, let $A\in \mathcal{B}(H)$ such that $\Phi(A)=0$. By  (\ref{c1}) we have
 $$\Phi(0)=\Delta_{\lambda}(\Phi(0)\circ \Phi(A)) = \Delta_{\lambda}(0) = 0.$$
 This proves $\Phi(0) = 0$.

 New, we prove that $\Phi(I) = I$. For the sake of simplicity,   write  $T=\Phi(I)$. First, we show that $T$ and $T^*$ are injective. Indeed, let $y\in K$ such that $Ty=0$. Since $\Phi$ is onto, there exists $B\in\mathcal{B}(H)$ such that $\Phi(B)=y\otimes y$. By  (\ref{c1}) we get
  $$\Delta_{\lambda}((y\otimes y)\circ T) =  \Delta_{\lambda}(\Phi(B)\circ T)=\Delta_{\lambda}(\Phi(B)\circ \Phi(I))=\Phi(\Delta_{\lambda}(B)).$$
 This can be rewritten in the other therms,
  $$\frac{1}{2}\Delta_{\lambda}(Ty\otimes y+y\otimes T^*y)= \frac{1}{2}\Delta_{\lambda}(y\otimes T^*y)=\Phi(\Delta_{\lambda}(B)).$$
 In the other hand  $(y\otimes T^*y)^2=<Ty,y>y\otimes T^*y=0$, since $Ty = 0$. By Lemma \ref{00},  $\Delta_{\lambda} (y\otimes T^*y)=0$ and thus $\Phi(\Delta_{\lambda}(B))=0$. Therefore $\Delta_{\lambda}(B)=0$, because $\Phi$ is bijective  and $\Phi(0) = 0$. Again, by  Lemma \ref{00}, $B^2=0$.

 Now, using  Condition (\ref{c1}),  we get
  $$\|y\|^2 y\otimes y=\Phi(B)^2=\Delta_{\lambda}(\Phi(B)\circ \Phi(B))=\Phi(\Delta_{\lambda}(B^2))=0,$$
   which implies that $y=0$ and whence $T$ is injective. With a similar argument we prove that $T^*$ is  injective.

Again from (\ref{c1}), we get
$$\Delta_{\lambda}(T^2)=\Delta_{\lambda}(\Phi(I)\circ \Phi(I))=\Phi(\Delta_{\lambda}(I))=\Phi(I)=T.$$
From Corollary \ref{l3}, we deduce that $T=\Phi(I)=I$, and completes the proof.
\end{proof}

 \begin{thm}\label{cor2} Let $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map satisfying  (\ref{c1}). Then
\begin{enumerate}[(i)]
\item$\Delta_{\lambda}(\Phi(A))=\Phi(\Delta_{\lambda}(A))$, for all $A\in\mathcal{B}(H)$. In particular $\Phi$ preserves the set of quasi-normal operators in  both directions.
\item $\Phi (A^2)=(\Phi (A))^2$ for all  $A$ quasi-normal.
\item $\Phi $ preserves the set of  orthogonal projections.
\item $\Phi $ preserves the orthogonality between the projections ; $$P\perp Q \Leftrightarrow \Phi (P)\perp\Phi (Q).$$
\item $\Phi $ preserves the order relation on the set of orthogonal projections in  both directions ;
 $$Q\leq P \Leftrightarrow \Phi (Q) \leq \Phi  (P).$$
\item  $\Phi (P+Q)=\Phi (P)+\Phi (Q)$ for all orthogonal projections $P,Q$ such that $P\perp Q$.
\item  $\Phi $ preserves the set of  rank one  orthogonal projections in  both directions.
\end{enumerate}
 \end{thm}
 \begin{proof}
 (i)  Since, from Proposition \ref{cor1}, $\Phi(I) = I$,  for  $B=I$ in (\ref{c1})  we obtain  $\Delta_{\lambda}(\Phi(A))=\Phi(\Delta_{\lambda}(A))$, for all $A\in\mathcal{B}(H)$.
  Now, by  $(2.1)$ we get that
$\Phi$ preserves the set of quasi-normal operators in  both directions.

 (ii)  Let $A\in \mathcal{B}(H)$ be a quasi-normal operator. Since $\Phi$ preserves the set of quasi-normal operators,
  $\Phi(A), \Phi(A^2), (\Phi(A))^2$  are quasi-normal. By (\ref{c1}), we get  $\Delta_{\lambda}((\Phi(A))^2)=\Phi(\Delta_{\lambda} (A^2))$.
  Hence  $(\Phi(A))^2=\Phi(A^2)$ follows from $(2.1)$.

 (iii) It is an immediate consequence of (i) and (ii).

 \bigskip

 In the rest of the proof $P, Q$ are two orthogonal  projections.

 (iv) Assume that $P,Q$ are orthogonal ($P\perp Q)$. Since $\Phi$ preserves the set of orthogonal  projections,  $\Phi(P)\circ \Phi(Q)$ is self-adjoint operator.
 Hence by (\ref{c1}) and $(2.1)$, we get
 $$\Phi(P)\circ \Phi(Q)=\Delta_{\lambda}(\Phi(P)\circ\Phi(Q))=\Phi(\Delta_{\lambda}(P\circ Q))=0.$$
 Thus
 \begin{equation}\label{eq7}
 \Phi(P)\Phi(Q)+\Phi(Q)\Phi(P)=0
 \end{equation}
 Multiplying  (\ref{eq7}) by $\Phi(P)$ on  both sides,  we get
 \begin{equation}\label{eq8}
 \Phi(P)\Phi(Q)+\Phi(P)\Phi(Q)\Phi(P)=0 \;\;\text{ and }\;\;
\Phi(P)\Phi(Q)\Phi(P)+\Phi(Q)\Phi(P)=0.
  \end{equation}
 It follows that $\: \Phi(P)\Phi(Q)=\Phi(Q)\Phi(P)=0$.

 (v) Suppose that $Q \leq P$.  Using Remark 3.1 and   (\ref{c1}), we get

$$\Phi(P)\circ \Phi(Q) = \Delta_{\lambda}(\Phi(P)\circ \Phi(Q)) = \Phi(\Delta_{\lambda}(P\circ Q)) = \Phi(Q).$$
Hence,
 $$\Phi(Q)\leq \Phi(P).$$
  (vi) Since  $P\perp Q$,  $P+Q$ is an orthogonal  projection and $P,Q\leq P+Q$. By (v),   $\Phi(P),\Phi(Q) \leq \Phi(P+Q)$. Thus $\Phi(P)+\Phi(Q)\leq \Phi(P+Q)$.
  Since $\Phi^{-1}$ satisfies the same assumptions as $\Phi$, it follows that $\Phi(P)+\Phi(Q)=\Phi(P+Q)$.

 (vii)  Let $P=x\otimes x$ be a rank one projection. Then $\Phi(P)$ is a non zero projection. Let $y\in K$ be an unit vector such that
 $y\otimes y \leq \Phi(P)$.  Thus $\Phi^{-1}(y\otimes y) \leq P$.  Since $P$ is a minimal projection and $\Phi^{-1}(y\otimes y)$ is a non zero projection,
  $\Phi^{-1}(y\otimes y)= P$. Therefore  $\Phi(P)=y\otimes y$ is a rank one projection.
 \end{proof}

\bigskip

 In the following proposition we introduce a function  $h:\mathbb{C}\to\mathbb{C}$, which will be used later to prove the linearity of $\Phi$.

 \begin{pro}\label{cor3}   Let $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map satisfying  (\ref{c1}). Then there exists a bijective function $h:\mathbb{C}\to\mathbb{C}$ such that
$$h(0)=0\;\; \text{and}\;\;h(1)=1, $$
 and it  satisfies the following properties :
 \begin{enumerate}[(i)]
 \item $\Phi(\alpha I)=h(\alpha)I$ for all $\alpha \in \mathbb{C}$.
 \item $h(\alpha\beta)=h(\alpha)h(\beta)$ for all $\alpha, \beta \in \mathbb{C}$.
 \item $h(-\alpha)=-h(\alpha)$ for all $\alpha \in \mathbb{C}$.
 \item $\Phi(\alpha A)=h(\alpha)\Phi(A)$ for all $A$ quasi-normal and $\alpha \in \mathbb{C}$.
 \end{enumerate}
 \end{pro}
  \begin{proof}
  First, we show that there exists a function
  $$h : \mathbb{C} \to  \mathbb{C} \: \text{such that}  \: \Phi(\alpha I)=h(\alpha)I\: \text{ for all} \: \alpha \in\mathbb{C}.$$

 When  $\alpha\in\{0,1\}$ the function $h$ is defined by $h(\alpha)=\alpha$ since $\Phi(\alpha I)=\alpha I$.

  Now, we suppose that $\alpha \ne 0$. Let  $y\in K$ an arbitrary unit vector.  By Theorem 3.1 (iii), there exists $P = x\otimes x \in \mathcal{B}(H)$ with $\Vert x\Vert = 1$,  such that  $\Phi(P)=y \otimes y$. \\
   By (\ref{c1}) with $A=P$ and $B=\alpha P$ we get
  $$\Delta_{\lambda}\Big( \Phi(\alpha P) \circ \Phi(P)\Big) =\Phi(\Delta_{\lambda}(\alpha P))=\Phi(\alpha P).$$
 From  Proposition \ref{l2}, there exists $h_P(\alpha) \in \mathbb{C}$ such that
 $$\Phi(\alpha P)=h_P (\alpha) P.$$
 Since $\Phi$ is bijective and $\alpha \not=0$,   $h_P(\alpha)\not=0$.

 Again using  (\ref{c1}) with $A=\alpha I$ and $B=P$, we get
 $$\Delta_{\lambda}\Big(\Phi(\alpha I) \circ \Phi(P)\Big)=\Phi(\alpha P)=h_p(\alpha) \Phi(P).$$
 From Proposition \ref{l1},  we get that
  \begin{eqnarray*}
 \Phi(P)\Phi(\alpha I) = h_P(\alpha)\Phi(P) = y\otimes \Phi(\alpha I)^*y=h_P(\alpha)y\otimes y
  \end{eqnarray*}
 Then
  $\Phi(\alpha I)^* y=\overline{h_P(\alpha)}y$ for every $y\in K$.
  It follows that $\Phi(\alpha I)=h_P(\alpha)I$. Thus $h := h_P$ does not depend on $P$.

  On the other hand, since $\Phi$ is bijective and $\Phi^{-1}$ satisfies the same properties as $\Phi$,  we conclude that  the function $h:\mathbb{C}\to\mathbb{C}$ is well defined,  bijective   and   $\Phi(\alpha I)=h(\alpha)I$.

(ii)  Let $\alpha, \beta \in \mathbb{C}$,  taking  $A=\alpha I$ and $B=\beta I$ in (\ref{c1}),  we get
 $$h(\alpha)h(\beta)I=\Delta_{\lambda}(\Phi(\alpha I)\circ \Phi(\beta I))=\Phi(\Delta_{\lambda}(\alpha I\circ \beta I)=h(\alpha\beta) I.$$
 Therefore $h$ satisfies the point (ii).

 From (ii),  we have $(h(-1))^2=1$.  Hence $h(-1)=-1$,  since $h(1)=1$ and $h$ is bijective. Hence
 $$h(-\alpha)=h(-1)h(\alpha)=-h(\alpha) \: \text{ for all} \: \alpha\in\mathbb{C}.$$
 Thus (iii) is satisfied.

 To prove (iv), let us consider $A$ quasi-normal and $\alpha \in \mathbb{C}$. By (\ref{c1}), we have $$h(\alpha)\Phi(A)=\Delta_{\lambda}(h(\alpha)\Phi(A))=\Delta_{\lambda}(\Phi(A)\circ \Phi(\alpha I))=\Phi(\Delta_{\lambda}(\alpha A))=\Phi(\alpha A).$$
 Thus  (iv) holds.
 \end{proof}

  \begin{lem}\label{l4}
 Let $P=x\otimes x $, $P'=x'\otimes x'$ be two rank one orthogonal projections such that $P\perp P'$ (that is $<x,x'> = 0$). Then
$$\Phi(\alpha P+\beta P')=h(\alpha)\Phi(P)+h(\beta)\Phi(P'),  \: \text{for every}  \:   \alpha ,\beta \in \mathbb{C}$$
 \end{lem}
 \begin{proof} If $\alpha =0$ or $\beta=0$ the result follows from preceding Proposition.

 Suppose that $\alpha\ne 0$ and $\beta \ne 0$.   Note that $(\alpha P+\beta P')\circ (P+P')=\alpha P+\beta P'$. \\
 By  (\ref{c1}),  applied to  $B=\alpha P+\beta P'$ and $A=P+P'$, we obtain
 \begin{eqnarray*}
\Phi(\alpha P+\beta P')&=&\Phi(\Delta_{\lambda}(\alpha P+\beta P'))\\&=&\Phi(\Delta_{\lambda}((\alpha P+\beta P')\circ(P+P')))\\&=&\Delta_{\lambda}(\Phi(\alpha P+\beta P')\circ\Phi(P+P'))\\&=&\Delta_{\lambda}(\Phi(\alpha P+\beta P')\circ(\Phi(P)+\Phi(P'))).
\end{eqnarray*}
It follows that
\begin{equation}\label{eq9'}
\Phi(\alpha P+\beta P')=\Delta_{\lambda}\Big(\Phi(\alpha P+\beta P')\circ\Phi(P)+\Phi(\alpha P+\beta P')\circ\Phi(P')\Big)
\end{equation}
Again by  (\ref{c1}) with $B=\alpha P+\beta P'$ and $ A=P$,  we get that
\begin{eqnarray*}
\Delta_{\lambda}\Big(\Phi(\alpha P+\beta P')\circ\Phi(P)\Big)&=&\Delta_{\lambda}(\Phi((\alpha P+\beta P')\circ P)))\\
&=&\Phi(\Delta_{\lambda}(\alpha P)) \quad \quad \text{(since $(\alpha P+\beta P')\circ P=\alpha P$)}\\
&=&\Phi(\alpha P)=h(\alpha)\Phi(P).
  \end{eqnarray*}
  By Proposition \ref{l1} and  $h(\alpha)\ne 0$, it follows that
  \begin{equation}\label{eq10}
 \Phi(P)\Phi(\alpha P+\beta P')=h(\alpha)\Phi(P).
  \end{equation}
  In the same manner, we also have
   \begin{equation}\label{eq11}
 \Phi(P')\Phi(\alpha P+\beta P')=h(\beta)\Phi(P').
  \end{equation}
  Now, we denote by $T=\Phi(\alpha P+\beta P')$, $\Phi(P)=y\otimes y$ and $\Phi(P')=y'\otimes y'$. \\
  Since $P$ and $P'$ are orthogonal, then $y$ and $y'$ are orthogonal too.\\
    By (\ref{eq10}) and (\ref{eq11}), we get that
  $$T^*y=\overline{h(\alpha)} y \quad \text{and} \quad T^*y'=\overline{h(\beta)}y',$$
   and also
  \begin{equation}\label{eq12}
\Phi(\alpha P+\beta P')\circ\Phi(P)=\frac{1}{2}(h(\alpha)\Phi(P)+\Phi(\alpha P+\beta P')\Phi(P))=\frac{1}{2}\big((h(\alpha)y+Ty)\otimes y\big),
  \end{equation}
  and
   \begin{equation}\label{eq13}
   \Phi(\alpha P+\beta P')\circ\Phi(P')=\frac{1}{2}(h(\beta)\Phi(P')+\Phi(\alpha P+\beta P')\Phi(P'))=\frac{1}{2}\big((h(\alpha)y'+Ty')\otimes y'\big).
     \end{equation}
 By (\ref{eq13}), it follows that
  \begin{equation}\label{eq14}
  T=\frac{1}{2}\Delta_{\lambda}\Big(\big(h(\alpha)y+Ty\big)\otimes y+\big(h(\alpha)y'+Ty'\big)\otimes y'\Big).
  \end{equation}
 This implies that  the rank of $T$ and $T^*$ are less than two. Moreover $\mathcal{R}(T), \mathcal{R}(T^*)\subseteq span\{y,y'\}$.  Since $T^*y=\overline{h(\alpha )}y$ and $T^*y'=\overline{h(\beta )}y'$, then $T= h(\alpha )y\otimes y+h(\beta )y'\otimes y'$. Therefore
 $$T=\Phi(\alpha P+\beta P')=h(\alpha)\Phi(P)+h(\beta)\Phi(P'),$$
 which is the desired equality.
 \end{proof}

 \bigskip

 If  $\Phi$ satisfies  (\ref{c1}), then $\Phi$ preserves the set of rank one projection and  preserves also  the orthogonality between projections (see Theorem \ref{cor2} (iii) and (iv)).
 Then  for two vectors $x,x' \in H$ such that $\|x\|=\|x'\|=1$ and $<x,x'>=0$, there exist two vectors $y,y' \in K$ such that
 $$\| y\|=\|y'\|=1, \: <y, y'>=0, \: \: \text{and} \:  \: \Phi(x\otimes x)=y\otimes y, \: \Phi(x'\otimes x')=y'\otimes y'.$$
  With the preceding notations, we have the following lemma

 \begin{lem}\label{l5}  Let $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map satisfying  (\ref{c1}).
Then there exists $\mu \in\mathbb{C}$ such that $|\mu|=1$ and
$$\Phi(x\otimes x'+x'\otimes x)=\mu y\otimes y'+ \bar{\mu}y'\otimes y).$$
In addition, we obtain
$$h(2)=2 \;\;  \text{and }\;\;h(\frac{1}{2}) = \frac{1}{2}.$$
\end{lem}

\begin{proof} Let us denote by $A=x\otimes x'+x'\otimes x$. First, note that  $A$ is self adjoint operator  of rank  two, and we have
$$A^2=x\otimes x+x'\otimes x',$$
which is an non-trivial orthogonal projection. Therefore, the spectrum of $A$ is $\sigma(A)=\{-1,0,1\}$. Hence we can find $f_1, f_2$ two unit and orthogonal vectors from
$H$ ~~($\|f_1\|=\|f_2\|=1$ and $<f_1,f_2>=0$) \: such that
\begin{equation}\label{eq3.9}
A= f_1 \otimes f_1-f_2 \otimes f_2.
\end{equation}

 Let us denote  $T=\Phi(A)$. By Lemma \ref{l4}, we have
$$T=\Phi(f_1 \otimes f_1-f_2 \otimes f_2)=\Phi(f_1 \otimes f_1)+h(-1)\Phi(f_2 \otimes f_2)=\Phi(f_1 \otimes f_1)-\Phi(f_2 \otimes f_2),$$
 Now, since $\Phi(f_1 \otimes f_1)$ and $\Phi(f_2 \otimes f_2)$ are two orthogonal projections, then we deduce that  $T$ is a self adjoint operator of rank two, and
$$tr(T)=tr(\Phi(f_1 \otimes f_1))-tr(\Phi(f_2 \otimes f_2))=0.$$

In the other hand, since $A$  is self adjoint, by   Theorem \ref{cor1} (ii) and (vi) , we get
 $$T^2= \Phi(A)^2=\Phi(A^2)=\Phi(x\otimes x+x'\otimes x') = y\otimes y + y'\otimes y'.$$
Moreover,
\begin{equation}\label{eq15}
T^2=y\otimes y+y'\otimes y'.
\end{equation}
It follows that $y,y'\in \mathcal{R}(T)$, and thus
$$\mathcal{R}(T)=span\{y,y'\}.$$

Clearly,  $A=2 A \circ (x\otimes x)$.  By (\ref{c1}), we get
\begin{eqnarray*}
T=\Delta_{\lambda}(T)&=&\Delta_{\lambda}(\Phi(2 A \circ (x\otimes x))\\&=&h(2)\Phi(\Delta_{\lambda}(A\circ (x\otimes x)))\\&=&h(2)\Delta_{\lambda}(\Phi(A)\circ \Phi(x\otimes x)))\\&=&h(2)\Delta_{\lambda}(T\circ (y\otimes y)).
\end{eqnarray*}
Therefore
\begin{equation}\label{eq16}
T=\frac{h(2)}{2}\Delta_{\lambda}(Ty\otimes y+y\otimes T^*y)=\frac{h(2)}{2}(Ty\otimes y+y\otimes Ty).
\end{equation}
If follows that
$$tr(T)=h(2)<Ty,y>=0.$$
 Hence $<Ty,y>=0$. Since $\{y,y'\}$ is an orthonormal basis of $\mathcal{R}(T)$ then there exists $\mu \in\mathbb{C}$ such that $Ty=\mu y'$.
  From (\ref{eq16}) we get
 \begin{equation}
 T=\frac{h(2)}{2}(\mu y'\otimes y+ \bar{\mu}y\otimes y').
 \end{equation}

  In particular, we also have $Ty=\frac{h(2)}{2} \mu y'=\mu y'$, it follows that
  $$h(2)=2 \quad \text{and thus} \quad h(\frac{1}{2}) = \frac{1}{2} \; \:  \text{(since $h$ multiplicative and $h(1) = 1$)} .$$
Hence
$$T=\mu y'\otimes y+ \bar{\mu }y\otimes y',$$
and we have
$$T^2=|\mu|^2 (y\otimes y+y'\otimes y')=y\otimes y+y'\otimes y'.$$
 Hence $|\mu|=1$. This completes the proof.
\end{proof}

As a direct consequence from the preceding lemma, we have the following corollary.
 \begin{cor}
The function $h:\mathbb{C}\to \mathbb{C}$ defined in Proposition \ref{cor3} is additive.
\end{cor}
\begin{proof}
Let $x,x'\in H$ be two unit and orthogonal vectors for $H$ and $\alpha, \beta \in \mathbb{C}$.
Denote
$$P=x\otimes x,  \; P'=x'\otimes x', \: \:  Q=\Phi(P)=y\otimes y, \; Q'=\Phi(P')=y'\otimes y'.$$
Put  $\; A=x\otimes x'+x'\otimes x$,   and $B=\alpha P + \beta P' $.
Observe that
$$A\circ P=\frac{1}{2} A\; \; \text{and}\;\; A\circ P'=\frac{1}{2}A,$$
thus
$$A\circ B=\dfrac{\alpha+\beta}{2} A.$$
 By Lemma \ref{l5},  there exists $\mu \in\mathbb{C}$ such that $|\mu|=1$ and
 $$\Phi(A)=\mu y\otimes y'+ \bar{\mu} y'\otimes y.$$
 Then, by a simple  calculation, we get
 $$ \Phi(P)\circ\Phi( A)= Q\circ\Phi( A)=\frac{1}{2}\Phi( A).$$
  Hence, using (\ref{c1}), we immediately obtain the following
\begin{eqnarray*}
h(\frac{\alpha+\beta}{2})\Phi(A)&=& \Phi(\frac{\alpha+\beta}{2}A) = \Phi(\Delta_{\lambda}(\dfrac{\alpha+\beta}{2}(A)))
\\&=& \Phi(\Delta_{\lambda}( B\circ A)) = \Delta_{\lambda}(\Phi(B)\circ\Phi( A))\\&=&
\Delta_{\lambda}\Big(\Phi(\alpha P + \beta P')\circ\Phi( A)\Big)\\&=&\Delta_{\lambda}\Big(\Phi(\alpha P)\circ\Phi( A) + \Phi(\beta P')\circ\Phi( A)\Big)\\&=&
\Delta_{\lambda}\Big(h(\alpha) \Phi(P)\circ\Phi( A) + h(\beta) \Phi(P')\circ\Phi( A)\Big)
 \\&=&\frac{1}{2} ( h(\alpha)+h(\beta))\Phi(A).
\end{eqnarray*}
Since $h(1/2)=1/2$, we have
$$\frac{1}{2} h(\alpha +\beta)=h(\dfrac{\alpha+\beta}{2})=\frac{1}{2}(h(\alpha)+h(\beta)).$$
Thus $h(\alpha+\beta)=h(\alpha)+h(\beta)$. It follows  that $h$ is additive.
\end{proof}

\bigskip

Now, we are in position to prove our main Theorem. \\

{\it   \bf Proof of Theorem \ref{th1}.}
\begin{proof} The "if" part is  immediate.

We show the "only if" part. Assume that $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$ is bijective and  satisfies  (\ref{c1}).   The proof of theorem is organized in several steps.

\bigskip

\begin{enumerate}[\bf Step. 1]
\item For every  $A\in\mathcal{B}(H)$, for all $x,y \, \text{such that} \; \Vert x\Vert = \Vert y\Vert = 1 \, \text{and} \,  \Phi (x\otimes x)=y\otimes y$, we have
 \begin{equation}\label{eq}
<\Phi (A)y,y>=h(<Ax,x>),
  \end{equation}

  Put  $\; T=  2(A\circ (x\otimes x)) = Ax\otimes x+x\otimes A^*x$.
  Let  $\alpha_1, \alpha_2$ be the non zero  eigenvalues  of $T$. By the Schur decomposition  of $T$, there exist two unit and orthogonal vectors $e_1,e_2$ such that
 $$T=\alpha_1 e_1\otimes e_1+\alpha_2 e_2\otimes e_2+ \beta e_1\otimes e_2.$$
First, we show that
  \begin{equation}\label{eq'}
tr(\Delta_{\lambda}(\Phi(T)))=2h(<Ax,x>).
  \end{equation}
 If $\alpha_1=\alpha_2=0$, we have $T=\beta e_1\otimes e_2$,  thus $T^2 = 0$ and $\Delta_{\lambda}(T) = 0$, by Lemma 2.4.
 Hence $\Delta_{\lambda}(\Phi(T))=\Phi(\Delta_{\lambda}(T))=0$, since $\Phi$ commutes with $\Delta_{\lambda}$. Consequently,
 $$tr(T)=2<Ax,x> = 0 = tr(\Delta_{\lambda}(\Phi(T))),$$
  and, in this case,   (\ref{eq'}) is satisfied.

 Now suppose that $(\alpha_1,\alpha_2)\not=(0,0)$.

  First, note that
 $$T\circ (e_1\otimes e_1)=\frac{1}{2}(e_1\otimes (\bar{\beta} e_2+2\bar{\alpha_1} e_1))=\frac{1}{2}(e_1\otimes v), $$

 with  $v=\bar{\beta} e_2+2\bar{\alpha_1} e_1 \;\;\text{and }\;\; <e_1,v>=2\alpha_1.$
From Proposition 2.1, it follows that
$$\Delta_{\lambda}(T\circ (e_1\otimes e_1))=\frac{1}{2}\Delta_{\lambda}((e_1\otimes v)=\alpha_1\frac{1}{\|v\|^2}(v\otimes v) .$$
By  (\ref{c1}), we get
\begin{eqnarray*}
h(\alpha_1)\Phi(\frac{1}{\|v\|^2}(v\otimes v))&=& \Phi(\Delta_{\lambda}(T\circ (e_1\otimes e_1)))\\&=&\Delta_{\lambda} (\Phi(T)\circ \Phi(e_1\otimes e_1)).
\end{eqnarray*}
 We have also
$$T\circ ( e_2\otimes e_2)=\frac{1}{2}(\beta e_1+2\alpha_2 e_2)\otimes e_2. $$
Hence
$$\Delta_{\lambda}(T\circ ( e_2\otimes e_2))=\alpha_2 e_2\otimes e_2.$$
Again (\ref{c1}) implies that
\begin{eqnarray*}
h(\alpha_2)\Phi(e_2\otimes e_2)&=&\Phi(\Delta_{\lambda}(T\circ ( e_2\otimes e_2)))\\&=&\Delta_{\lambda} (\Phi(T)\circ \Phi(e_2\otimes e_2))
\end{eqnarray*}
Now, observe  that $T\circ (e_1\otimes e_1 +e_2\otimes e_2)=T$.  We apply again (\ref{c1}), then
\begin{eqnarray*}
\Delta_{\lambda}(\Phi(T))&=&\Phi(\Delta_{\lambda}(T\circ (e_1\otimes e_1 +e_2\otimes e_2)))\\&=&
\Delta_{\lambda}(\Phi(T)\circ \Phi(e_1\otimes e_1 +e_2\otimes e_2))\\&=&
\Delta_{\lambda}(\Phi(T)\circ \Phi(e_1\otimes e_1) +\Phi(T)\circ\Phi(e_2\otimes e_2)).
\end{eqnarray*}
Since $\Phi(\frac{1}{\|v\|^2}(v\otimes v))$ and $\Phi(e_2\otimes e_2)$ are orthogonal projections,   then
\begin{eqnarray*}
tr(\Delta_{\lambda}(\Phi(T)))&=&tr(\Delta_{\lambda}(\Phi(T)\circ \Phi(e_1\otimes e_1) +\Phi(T)\circ\Phi(e_2\otimes e_2)))\\&=&tr(\Phi(T)\circ \Phi(e_1\otimes e_1) +\Phi(T)\circ\Phi(e_2\otimes e_2))\\&=&tr(\Phi(T)\circ \Phi(e_1\otimes e_1))+tr( \Phi(T)\circ\Phi(e_2\otimes e_2))\\&=&tr(h(\alpha_1)\Phi(\frac{1}{\|v\|^2}(v\otimes v))+tr(h(\alpha_2)\Phi(e_2\otimes e_2))\\&=&h(\alpha_1)+h(\alpha_2)\\&=&h(\alpha_1+\alpha_2)=h(tr(T))\\&=&2h(<Ax,x>).
\end{eqnarray*}
Now, we show that
\begin{equation}\label{eq''}
tr(\Delta_{\lambda}(\Phi(T))=2<\Phi(A)y,y>.
\end{equation}
By (\ref{c1}) and  $h(2)=2$,  we get
\begin{eqnarray*}
\Delta_{\lambda}(\Phi(T))&=&\Delta_{\lambda}(\Phi(2A\circ (x\otimes x))=\Delta_{\lambda}(h(2)\Phi(A\circ (x\otimes x))\\&=&h(2)\Phi(\Delta_{\lambda}(A\circ (x\otimes x)))\\&=&2\Delta_{\lambda}( \Phi(A)\circ \Phi(x\otimes x))\\&=&2\Delta_{\lambda}( \Phi(A)\circ (y\otimes y)).
\end{eqnarray*}
Thus
\begin{eqnarray*}
 tr(\Delta_{\lambda}(\Phi(T)) &=& 2tr(\Delta_{\lambda}( \Phi(A)\circ (y\otimes y)))=2tr( \Phi(A)\circ (y\otimes y))\\
 &=& 2<\Phi(A)y,y>.
\end{eqnarray*}
From (\ref{eq'}) and (\ref{eq''}), it  follows  that
$$ <\Phi(A)y,y>=h(<Ax,x>),$$
 for all $A\in\mathcal{B}(H)$ and for all $x\in H$ and $y\in H$  such $\Phi(x\otimes x)=y\otimes y$, which completes the proof of (3.13).

\bigskip

 \item The function $h$ is continuous.

  Let   $\mathcal{E}$ be a bounded subset  in $\mathbb{C}$ and  $A\in\mathcal{B}(H)$  such that  $\mathcal{E}\subset W(A)$, where $W(A)$ is the numerical range of A.
By   (\ref{eq}),
 \begin{equation*}
h(\mathcal{E})\subset h(W(A))   = W(\Phi (A))
 \end{equation*}
  Since  $W(\Phi (A))$ is bounded,    $h$ is bounded on the bounded subset, which implies that $h$ is continuous,  since it  is additive (see Corollary 3.1).
   We then have, using Proposition 3.2 (ii) and Corollary 3.1,  that $h$ is an automorphism continuous   over the complex field $\mathbb{C}$.
    It follows that $h$ is the identity or the  complex conjugation map   (see, for example, \cite{ks}).

\bigskip

\item $\Phi $ is linear or anti-linear.

 Let  $y\in K$ and $x\in H$ be unital  vectors such that $y \otimes y=\Phi ( x\otimes x)$. Let $\alpha\in \mathbb{C}$ and $A ,B\in \mathcal{B}(H)$ be arbitrary. Using (\ref{eq}), we get
  \begin{eqnarray*}
 <\Phi (A+B)y,y>&=&h(<(A+B)x,x>)\\&=&h(<Ax,x>+<Bx,x>)\\&=&h(<Ax,x>)+h(<Bx,x>)
 \\&=&<\Phi (A)y,y>+<\Phi (B)y,y>\\&=&<(\Phi (A)+\Phi (B))y,y>,
  \end{eqnarray*}
and
\begin{eqnarray*}
 <\Phi (\alpha A)y,y>&=&h(<\alpha A x,x>)\\
 &=& h(\alpha)h(<Ax,x>) \\
 &=& h(\alpha)<\Phi (A)y,y>.
\end{eqnarray*}

  Hence, we immediately obtain the following
   $$\forall A,B\in\mathcal{B}(H), \forall  \alpha \in \mathbb{C}, \; \;   \Phi (A+B)=\Phi (A)+\Phi (B)  \: \text{and} \: \Phi (\alpha A)=h(\alpha)\Phi (A).$$
   Therefore  $\Phi $ is linear or anti-linear,  since $h$ is the identity or the complex conjugation.

   \bigskip

\item $\Phi (A)=UAU^*$  every $A\in \mathcal{B}(H)$, for some  unitary  operator  $U\in \mathcal{B}(H,K)$.
   By  Step.3, we have  $\Phi$ or  $\Phi^*$ is linear, where $\Phi^*$  is defined by $\Phi^*(A)=(\Phi(A))^*$.  From Theorem 3.1 (i),  $\Phi$  commute
    with $\Delta_{\lambda}$. Now, by \cite[Theorem 1]{bmg},   there exists a unitary operator $V: H\to K$, such that  $\Phi$ take one of the following forms
   \begin{equation}\label{f1}
   \Phi(A)=VAV^* \;\; \text{for all}\;\; A\in\mathcal{B}(H),
   \end{equation}
 either
   \begin{equation}\label{f2}
   \Phi(A)=VA^*V^* \;\; \text{for all}\;\; A\in\mathcal{B}(H).
   \end{equation}
   In order to complete the proof we have to show $\Phi $ can not take the form (\ref{f2}).

   Seeking a contradiction,   suppose that (\ref{f2}) holds. Then  for every $\; A\in\mathcal{B}(H)$,
   \begin{eqnarray*}
   {\Delta_{\lambda }}(A^{*})&=&  {\Delta_{\lambda }}(V^*\Phi(A)V) \\
   &=& V^*{\Delta_{\lambda }}(\Phi(A)) V\\
   &=&  V^*\Phi({\Delta_{\lambda }}(A))V\\
   &=& ({\Delta_{\lambda }}(A))^*.
   \end{eqnarray*}

Therefore
\begin{equation}\label{ee}
{\Delta_{\lambda }}(A^{*})= ({\Delta_{\lambda}}(A))^{*},\quad \mbox{ for every} \; A\in\mathcal{B}(H).
\end{equation}
Now, let us consider  $A=x\otimes x'$ with $x, x'$  are  unit,  independent and non-orthogonal vectors  in $H$. Then $A^*=x'\otimes x$ and  by Proposition \ref{p1}, we have
 $$ ({\Delta_{\lambda}}(A))^*=<x', x>(x'\otimes x') \; \;  \mbox{and} \; \; {\Delta_{\lambda}}(A^{*})=<x', x>(x\otimes x),$$
which contradicts  (\ref{ee}).  So if we take $U = V$,  $\; \Phi$  gets the  form
$$ \Phi(A)=UAU^*,  \;\; \text{for all}\;\; A\in\mathcal{B}(H).$$
\end{enumerate}
 This completes the proof of our main theorem.
\end{proof}

\section{Star Jordan Product Commuting  Maps  }

We end this paper by  characterizing  the  bijective maps   $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$,  which verify
  \begin{equation}\label{c2}
\Delta_{\lambda}(\Phi(A)\circ (\Phi(B))^*)=\Phi(\Delta_{\lambda}(A \circ B^*)) \quad \text{for all} \; \;  A, B\in \mathcal{B}(H).
 \end{equation}

  \begin{thm} \label{th3}
 Let $H$ and $K$ two  complex Hilbert space, such that $H$ is of dimension greater than $2$. Let   $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map.
 Then $\Phi$ satisfies  (\ref{c2}), if and only if, there exists  an unitary operator $U : H\to K $ such that
 $$\Phi(A)=UAU^*\;\;\;\;\text{ for all $A\in\mathcal{B}(H)$.}$$
 \end{thm}

�\begin{rem}  Based on the arguments and the methods  developed in the proof of Theorem 1.1,  it is not difficult to proof Theorem 4.1. For example,
 If $\Phi :\mathcal{B}(H)\to\mathcal{B}(K)$ is a bijective map, satisfying (\ref{c2}), then
 \begin{enumerate}[(i)]
 \item $\Phi(0)=0.$,
 \item $\Phi(I)=I.$,
 \item $\Delta_{\lambda}(\Phi(A))=\Phi(\Delta_{\lambda}(A))$ and $\Delta_{\lambda}((\Phi(A))^*)=\Phi(\Delta_{\lambda}(A^*))$ for all $A\in\mathcal{B}(H)$,
 \item $\Phi$ preserve the set of quasi-normal operator in both directions,
 \item $\Phi$ preserves the set of self adjoint operators in both directions.
 \end{enumerate}

\begin{proof}
 (i) Since $\Phi$ is onto, then there exists  $B\in \mathcal{B}(H)$ such that $\Phi(B)=0$. Hence, by (\ref{c2}) $\Phi(0)=\Delta_{\lambda}(\Phi(0)\circ \Phi(B)^*)=0$.

 (ii) Let us denote $T:=\Phi(I)$. First we show that $T$ is one-to-one. Let $y\in K$ such that $Ty=0$. Now since $\Phi$ is onto, there exists  $B\in\mathcal{B}(H)$ such that
 $\Phi(B)=y\otimes y$. By (\ref{c2}), we have $\Delta_{\lambda}(T\circ y\otimes y)=y\otimes y$.  Thus $y\otimes y=0$ and  it follows that $y=0$.

 Again from  (\ref{c2}) it  follows that $T\circ T^*=T$. Hence $T$ is self adjoint and $T^2=T\circ T^*=T$. Thus $T=I$ since $T$ is one-to-one.

 (iii) is immediate, and (iv) is deduced directly from (iii).

 (v) Let $B$ be a self adjoint operator, and $S:=\Phi(B)$.  By (iii) $S$ is quasi-normal operator, and by (\ref{c2}) with $A=I$ we get, that $\Delta(S^*)=S$.
 By Lemma \ref{l22}, $\; S=S^*$ and thus $\Phi$ preserves the self adjoint operators.
  \end{proof}

The rest of the proof of Theorem 4.1 is similar to that of Theorem 1.1. We do not give details of those arguments.

\end{rem}

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\emph{}{ \textit {The iterated Aluthge transforms of a matrix converge}},  Adv. Math.  226 (2011), 1591-1620.

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redpajama_set_name: RedPajamaArXiv


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text:
Q: Error Printing Foreign Key ID. Symfony2 Doctrine I have an Entity called Keyword, and it has two rows:
id  comment_id  foo_id  text
2       1         1     Jajajaja :) Hola.
3       2         1     Chao

foo_id and comment_id are foreign keys and integer values. Foo_id from table Foo and comment_id from table Comments.
I'm trying to print comment_id and foo_id in a twig template.
{% for k in keywords %}
    {{ k.id}} , {{ k.text}}, {{ k.comment }}, {{ k.foo}}
{% else %}

but it gives me this error:
An exception has been thrown during the rendering of a template ("Catchable Fatal Error: Object of class
Proxies\PruebasRBundleEntityCommentsProxy could not be
converted to string in       C:\wamp\www\sym\app\cache\dev\twig\52\c9\3138bf2dc905760b186f2d006484.php
line 74") in PruebasRBundle:Default:keywords.html.twig at line 24.

So, I tried printing the values that are not Foreign keys
I'm trying to print comment_id and foo_id in a twig template.
{% for k in keywords %}
    {{ k.id}} , {{ k.text}} <!--No k.comment no k.foo -->
{% else %}

it works this way, but I can't print foreign keys values and I need them.
It makes me think that there's something with the relationships :(.

A: That's because Doctrine automatically detects the k.comment foreign key, and returns you an object representing the foreign comment row.
This object is a PruebasRBundle\Entity\Comments instance. (Doctrine2 will wrap it in a proxy to ease lazy loading).
Your problem is that you try to cast this object as a string.
Etiher modify your twig like this:
{% for k in keywords %}
    {{ k.id}} , {{ k.text}}, {{ k.comment.id }}, {{ k.foo}}
{% else %}

Or implement the __toString method in your Comment class.

A: If you want to print the numeric values try this way:
{% for k in keywords %}
    {{ k.id}} , {{ k.text}}, {{ k.comment_id }}, {{ k.foo_id}}
{% else %}

If you want to print some value from related entities try creating a magic method __toString() for entities Foo and Comments. Then:
{% for k in keywords %}
    {{ k.id}} , {{ k.text}}, {{ k.comments }}, {{ k.foo}}
{% else %}


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text:
Q: Compilation Error when change .Net framework from 4.5.2 to 4.5 in VS 2015 Community edition After changing my .Net framework from 4.5.2 to 4.5 in VS 2015 Community edition. I am getting following error. Did anyone know what is the issue and how to solve this. Look like compile version is different, but I have already modify my web.config to 4.5

Server Error in '/' Application.
Compilation Error    Description: An error occurred during the
  compilation of a resource required to service this request. Please
  review the following specific error details and modify your source
  code appropriately.
Compiler Error Message: CS1617: Invalid option '6' for /langversion;
  must be ISO-1, ISO-2, 3, 4, 5 or Default
Source Error:
[No relevant source lines]
Source File:    Line: 0
Show Detailed Compiler Output:
C:\Program Files (x86)\IIS Express>
  "C:\Windows\Microsoft.NET\Framework\v4.0.30319\csc.exe" /t:library
  /utf8output /R:"C:\Users\admin\AppData\Local\Temp\Temporary ASP.NET
  Files\root\479d40ee\add72695\assembly\dl3\31bb6ea7\6a8168a8_dae8d001\Microsoft.CodeDom.Providers.DotNetCompilerPlatform.dll"
  /R:"C:\WINDOWS\Microsoft.Net\assembly\GAC_MSIL\System.Configuration\v4.0_4.0.0.0__b03f5f7f11d50a3a\System.Configuration.dll"
  /R:"C:\WINDOWS...................
Microsoft (R) Visual C# Compiler version 4.6.0079.0
for C# 5 Copyright (C) Microsoft Corporation. All rights reserved.
This compiler is provided as part of the Microsoft (R) .NET Framework,
  but only supports language versions up to C# 5, which is no longer the
  latest version. For compilers that support newer versions of the C#
  programming language, see
  http://go.microsoft.com/fwlink/?LinkID=533240
error CS1617: Invalid option '6' for /langversion; must be ISO-1,
  ISO-2, 3, 4, 5 or Default

A: C# in Visual Studio 2015 is C# 6.0, however the error message you're getting is from the ASP.NET Compiler (not the Visual Studio compiler) informing you that C# 6.0 is not available, only C# up-to version 5, so not version 6.
I suspect this error message is coming from your webhosting provider, which means they haven't installed the very latest version of the .NET Framework on their systems, you'll need to downgrade your C# language compilation options in your web.config back to what they were before and refrain from using any C# 6.0 language features in runtime-compiled code. I don't know what effect running C# 6.0 language-features in a .NET 4.5.0 environment will have if they rely on any new CLR instructions.
Update:
Alternatively, ensure you're using the latest version of the Microsoft.Net.Compilers NuGet package, which causes the latest version of csc to be bundled with your project when deployed, so it will have support for C# 6, 7, 8, etc.

A: I am answering my question just to let other user know what's the solution. After some googling and personal research. So I precisely did two thing.

*

*Go to  each project property -> Build -> Advance and change the
language version to C# 5.0

*Open web.config for the startup project (MVC in my case) and remove
the codedom section that contain the custom compile information

This fix my issue.

A: I was facing the same building with MSBuild and the following helped me:
In Visual Studio > Select your project > Right Click > Properties > Library > Targeting > Change > .Net Framework 4.6
This link Helped me

A: This solution also worked for me.
Window 10
Visual studio 2013
.Net framework 4.5
Go to each project property -> Build -> Advance and change the language version to C# 5.0
Open web.config for the startup project (MVC in my case) and remove the codedom section that contain the custom compile information
This fix my issue.


The food industry is turning to IoT to bring its ecosystems into the modern technological era. By facilitating improved prevention and improving response and traceability, IoT is well positioned to improve food safety. Blockchain, wireless sensors, smartphones and the cloud are reducing food contamination and tracking contaminants to the source.
When appropriate, it's smart to support two competing, IoT low-power wide-area network (LPWAN) standards like NB-IoT and LTE-M, but maybe now we need four-in-one. Hedge your bets when developing IoT solutions so that they can survive long term.
Combing high-end processing hardware with low-power operation is difficult enough, but for IoT solution providers, it's also important to ensure the wireless interface is optimized. Online RF calculation tools can help.
Water utilities are lagging behind electric utilities in terms of IoT and data analytics. Only about 20 percent of U.S. drinking water utilities count themselves as smart. The number for electric utilities is much higher at 60 percent. Ready to make up the difference, water utilities are starting to embrace smart meters, IoT and data analytics.
Retail stores are experiencing a harsh reality of declining foot traffic. However, the sales numbers for in-store retail purchases are not as bleak as you might think. According to multinational accounting firm EY, 93 percent of purchases are still made in shops. But buying patterns are different, and shops have to keep pace. Solution providers need to help retailers accelerate their pick-up portals.
The Internet of Things is now bringing intelligence to every nook and cranny in the home, including to beds, bookcases, desks and chairs. A robotics start-up company, Ori, has created a voice-controlled modular furniture system that can automatically glide about the room, allowing home dwellers to maximize spaces in their tiny smart places.
IoT directed at boosting New Zealand's prosperity holds huge potential for both the country and the tech sector, according to new IoT research study. Ranking among the leading countries in the world on their readiness for IoT deployment, New Zealand is unusually well positioned to embrace—and reap the significant benefits of—the deployment of IoT technologies.


text:
Q: How to auto fill other websites which I donot own I need to fill a website which I donot own many times. I want to write a script or software to let my information fill the website's form. I know how script like "document.getElementByID( "zipCode" ).value = "98105";" works. But I donot have access to the server side of the website. How can I fill these forms automatically by simply access the  client side. Thank you.

A: There are firefox addons for the same:
https://addons.mozilla.org/en-US/firefox/addon/autofill-forms/reviews/
https://addons.mozilla.org/en-US/firefox/addon/autofill-262804/


Today's 5 day forecast: HOT!
This is the kind of forecast we have been hedging against all season, and this is what we thought an early harvest may bring. It is not just that it is hot, but that it is hot in early September versus early October. What that means is longer days, higher sun angles, and heat for a longer period of time. And what that provides is greater dehydration pressure for the same high temperature.
Thankfully we have no Pinot or Chardonnay left to worry about. The Syrah is off, and several core Cabernet lots were brought in with healthy ripeness over the last several days. With a few exceptions what remains are blocks of Cabernet Franc and Cabernet Sauvignon that are at a point of their ripening trajectory where this heat is less likely to do much harm. The exceptions are being harvest today, tomorrow, and Saturday. We moved our favorite Merlot block from the top of the vineyard (block 26C) forward in the schedule to bring it in Saturday. Tomorrow we process Petit Verdot, Malbec, and Cabernet Franc.
As noted in the last email we believed the second round of Pinot picks would bring more power and concentration. This is proving to be true and provides us a nice set of components to work with. The Cabernet aromas continue to intoxicate the cellar with beautiful blackberry, cassis, sage, and crushed rock.
At this stage nearly all the Drum Canyon Pinot Noir has been put to barrel save 5 small fermenters, much of the Santa Maria Pinot will be put to barrel by week's end. All the Chardonnay's are finished with primary fermentation and we are evaluating which ought to go into secondary (ML) and which ought to be sulfured. Early next week we will have finished wines including Sauvignon blanc, Chardonnay, Rose and Pinot noir! I must note the Rose is killer!
It is easy to feel very satisfied with how things have flowed, particularly because of the heat that is coming. However it is premature to declare victory as we must employ patience and observe what truly emerges from vintage. Prettiness in Pinot, precision in Chardonnay, depth and supple Bordelaise? We seem headed in that direction.
I have attached a photo I took yesterday of some Cabernet set to be harvested Saturday. That cluster is about the size of my hand, maybe smaller. Beautiful.
Here is the harvest info since the last time point. We are up to over 550 tons total thus far for ourselves and over 900 tons including clients.
One final note. I am going to be very disappointed if the Cardinals do not make the playoffs in a vintage where I will likely be able to watch all the games live! A first for me! It is looking more and more likely we may be finished before Oct. 1st.


text:
How Much Is The New York Times Newspaper
How To See Niagara Falls From New York
What Is The Best Health Insurance In New York
What Is Time Square In New York
How To File An Llc In New York State
StateTrending
When Is The New York Biennial Statement Due
NYS LLC | How to file an LLC in New York (quick tips)
The filing period for your New York Biennial Statement is the month in which your original Articles of Organization were filed. Its due every other year.
When you sign up for Northwest, we send you reminders for your Biennial Statement due dates. Want one less thing to worry about? With our business renewal service, we can complete and submit your Biennial Statement for you for $100 plus the state fee.
Creating An Llc In Ny: Eligibility Requirements
The New York Department of State sets the eligibility requirements for LLCs in the state of New York. You need to meet one of two qualifications to create an LLC in NY:
You, the business owner, must be a New York State residentâor
The business is located in NY state
There are also industry, licensing and zoning rules to follow, many of which are specific to your city or municipality. For example, starting a daycare, a food service establishment, or a home-based business all require special permits. In NYC, even temporary situations or one-time events, such as a grand opening in a public park or on a city block, require express permission from the city.
Research the rules for location-specific requirements and get the necessary licenses and permits so you can legally start your LLC in NY. For guidance on legal obligations, consult the business resources your city offers. In NYC, you can use the step-by-step online business wizard to determine exactly how to start a business in the city.
Limited Liability Companies And Limited Liability Partnerships
A limited liability company is an unincorporated organization, with one or more members, each having limited liability for the contractual obligations and other liabilities of the company, other than a partnership or trust, that is formed for any lawful business purpose under the Limited Liability Company Law of New York State or under the laws of any other jurisdiction.
A limited liability partnership is a partnership whose partners are authorized to provide professional services and that has registered as a limited liability partnership under Article 8-B of the Partnership Law of New York State or under the laws of another jurisdiction.
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How To File Articles Of Incorporation In New York
Filing the Articles of Incorporation Electronically Use the New York State Department of Corporations online filing system. Select Domestic Business Corporation. Under the heading of Online Filings, you will select Domestic Business Corporation to begin creating your online Certificate of Incorporation.
How Can I Withdraw My Nys Dba
Contact your county clerk for county-specific instructions for withdrawing your DBA for your unincorporated business.
For incorporated businesses, you will need to file for discontinuance of your assumed name with the NY Department of State. You can find the Certificate of Discontinuance of Assumed Name form here.
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Obtain Business Licenses And Permits
The licenses and permits required for an LLC in can vary, depending on a number of variables:
If you would like to see more in-depth information on licenses and permits, please feel free to visit the content in our learning library that covers business licenses and permits.
How Do I Open A Bank Account For My New York Llc
To open a bank account for your New York LLC, you will need to bring the following with you to the bank:
A copy of the New York LLC Articles of Organization
The LLC operating agreement
The LLCs EIN
If there are multiple members in the LLC, you may also want to bring an LLC resolution to open a bank account that states that the person going to the bank is authorized by the members to open the account in the name of the LLC. Northwest can help with this as wellLLC bank resolutions are one of the many free legal forms we provide to ensure you can get your LLC started fast.
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Step : Create A New York Llc Operating Agreement
All New York LLCs are required to have an operating agreement. This may be a verbal or written agreement.
What is an operating agreement? An operating agreement is a legal document outlining the ownership and operating procedures of an LLC.
Why are operating agreements important? A comprehensive operating agreement ensures that all business owners are on the same page and reduces the risk of future conflict.
For more information on operating agreements, read our New York LLC operating agreement guide.
Recommended: Download a template or create a Free Operating Agreement using our tool.
FAQ: Creating a New York LLC Operating Agreement
Do I need to file my operating agreement with the State of New York?
No. The operating agreement is an internal document that you should keep on file for future reference. However, many states like New York do legally require LLCs to have an operating agreement in place.
Service And Membership Restrictions
HOW I FORMED AN LLC IN NEW YORK STATE 2020 *PAID ONLY THE FILING FEES*
A New York PLLC can only provide the professional services indicated in its articles of organization. The practice of multiple professional disciplines is not allowed for PLLCs in the professions of medicine, dentistry, veterinary medicine, licensed clinical social work, mental health counseling, psychoanalysis, creative arts therapy, and marriage and family therapy. PLLCs that combine design professions are allowed as long as there is at least one PLLC member licensed for each professional service provided by the PLLC. A New York PLLC generally can engage in any other, non-professional business activities permitted under the state's LLC laws. In general, only licensed professionals can be members of a New York PLLC. And, finally, remember that New York PLLCs and/or their members are subject to the regulation of the relevant state professional licensing authorities.
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How To Become A Corporation In New York
The New York template Certificate of Incorporation suggests reporting 200 shares, without par value. This is how most new corporations begin. You should meet with an attorney to discuss your planning and strategies related to issuing corporate shares. Name the Secretary of State for service of process.
Step : New York Llc Ein You Need One Heres How To Get It
What is an EIN? The Employer Identification Number , Federal Employer Identification Number , or Federal Tax Identification Number , is a nine-digit number issued by the Internal Revenue System an Employer ID Number is used to identify a business entity and keep track of a businesss tax reporting. It is essentially a Social Security number for the company.
Why do I need an EIN? An EIN number is required for the following:
To open a business bank account for the company
For Federal and State tax purposes
To hire employees for the company
Where do I get an EIN? An EIN is obtained from the IRS by the business owner after forming the company. This can be done online or by mail.
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I Have A New York State Llc How Do I File My Taxes
Clients will frequently tell me they have an LLC and expect that to explain everything. However, as an accountant, the LLC status does not tell me anything . In other words, LLC status is more a legal term rather than a tax-related term. What you want to know about your business is: are you a:
Single member LLC
S Corp
C Corp
Being an LLC means you can actually be filing under any of the above. If you have not specifically indicated to the IRS or any state that you are a partnership or corporation, then you are a single-member LLC and can file on your schedule C. This is actually beneficial for several reasons:
Schedule C tax returns are flow-through, so all losses incurred in the ordinary course of business will be deductible on your personal return
Schedule C tax returns are easier to complete and accounting fees are typically much lower for this sort of business as opposed to running a partnership or corporation return and then separately booking the income into your individual return
The drawbacks to filing your business as self-employment income include the following:
This schedule then feeds directly into your personal return. This is the easiest type of return to produce, and will eventually end up on your NY state IT-201 return. If you are a partnership or corporation, your taxes will go on a form 1065 or 1120/1120s return.
What Is A Foreign New York Llc
A foreign New York LLC is any LLC formed outside of New York but registered to do business in the state. For example, if you formed an LLC in Connecticut but wanted to conduct business in New York as well, you would register as a foreign LLC in New York by filing an Application for Authority with the New York Secretary of State. Like domestic LLCs, foreign LLCs in the state have to file New York Biennial Statements. Northwest can register your foreign New York LLC for you today!
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Complete The Process By Filing With The New York Department Of State
Once youve obtained our consent and prepared the Certificate of Dissolution, youre ready to file with the Department of State. Your filing must include:
one Certificate of Dissolution and
a check for $60 payable to the New York Department of State.
You may file in person or by mailing to:
NEW YORK DEPARTMENT OF STATEDIVISION OF CORPORATIONS99 WASHINGTON AVEALBANY NY 12231
You may also fax the completed Certificate of Dissolution, the consent of the New York State Department of Taxation and Finance, consent of the New York City Department of Finance, if required, and a Credit Card/Debit Card Authorization Form to the Division of Corporations at 474-1418.
Once the Department of State accepts the Certificate of Dissolution for filing, theyll issue your corporation a filing receipt. The filing receipt establishes the date that the corporations existence comes to an end.
Tax Treatment Of Llcs And Llps
The New York personal income tax and the corporate franchise tax conform to the federal income tax classification of LLCs and LLPs.
An LLC or LLP that is treated as a partnership for federal income tax purposes will be treated as a partnership for New York tax purposes.
An LLC or LLP that is treated as a corporation, including an S corporation, for federal income tax purposes will be treated as a corporation for New York tax purposes or as a New York S corporation if the New York S election is made .
A single-member LLC that is treated as a disregarded entity for federal income tax purposes will be treated as a disregarded entity for New York tax purposes.
If the SMLLC is disregarded and the single member is an individual, the SMLLC will be treated as a sole proprietorship for New York tax purposes.
If the SMLLC is disregarded and the single member is a corporation, including an S corporation, the SMLLC will be considered a division of the corporation for New York tax purposes.
If the SMLLC is disregarded and the single member is a partnership, the SMLLC will be considered a division of the partnership.
For information regarding the tax treatment of an LLC or LLP for purposes of the New York City Business Corporation Tax, New York City General Corporation Tax , and the New York City Unincorporated Business Tax , please visit the New York City Department of Finance Business webpage.
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What Is Membership Interest
Membership interest is your percentage of ownership of the LLC. Membership interest is normally proportionate to your investment. So how does this work?
Imagine your LLC has 5 members. 4 members each invest $1,000 in the business. One member invests $6,000. The total contribution of all members is $10,000. The 4 members each own 10% of the business. The moneybags member who shelled out $6,000 owns 60% of the business.
Typically, this also means that the 4 members would each get 10% of any profits, and moneybags would get 60%. Exactly how profits and losses are allocated, however, can be adjusted in the operating agreement as long as changes are in line with IRS requirements and New York laws.
Publish Notice Of Llc Formation
How to start an LLC in New York State (3 methods)
NY state law requires you to publish a copy of your articles of organization or a notice of your LLC formation in two newspapers for six consecutive weeks within 120 days of your LLC formation.
You have to publish your notice in newspapers approved by the NY county where your LLC is located. To find out which newspapers are acceptable, contact your county clerk. If you choose to publish a notice instead of your articles of organization, be sure the LLC name in the notice matches the Department of States records as set forth in your initial articles of organization.
To get your notice published, contact the newspapers yourself or hire a third-party service to handle it for you. The newspapers will provide you with an affidavit of publicationâtheyll also charge a publication fee. NYC-based business may have to pay several thousand dollars due to steep publishing rates for NYC newspapers. To save money, you might choose a registered agent outside NYC so you can use that address as your LLCs office address. That way you could run your publication notice in a newspaper in that county, which likely has lower publishing rates.
Once youve fulfilled the publication requirement, youll need to submit a Certificate of Publication, along with a $50 filing fee, to the New York Department of State.
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Create A Name For Your New York Llc
Before you register your LLC, you'll need to come up with a name. Your business name needs to be catchy for branding purposes and legitimate for legal purposes.
We go into depth on this topic in our LLC naming guide. Well go over the basics below.
Naming Your LLC
First, brainstorm some possibilities. Use our LLC name generator to get the ideas flowing.
Next, make sure your name meets New York guidelines:
it contains the words limited liability company, limited company, LLC or L.L.C.
it doesn't contain words like academy, bank, lawyer, mortgage, university, or doctor
it must be distinguishable from any other entity or trade name registered in New York
full list of New York naming guidelines
Next, do a New York LLC name search with the Secretary of State to find out if your name is available in New York.
If your business name is available and meets guidelines, youll be able to register it when you file your New York Articles of Organization with the Secretary of State.
Is the URL available? Before you commit 100% to a name, check to see if theres a good web domain available.
We recommend buying your domain right away because even if launching a business website isnt on your radar right now, it will be soon.
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How To Get Articles Of Organization New York State
The New York state articles of organization LLC is available to the public and you can get it via:
Online- by going to the state site, you can request the forms online after creating an account. This method works great for people who want to complete the process online i.e., filling the details, making the payment and submitting the paperwork
The state site the form is readily accessible on the New Yorks Business Express website and you do not even have to open an account. All you need to do is to download it, fill it and send it to the authorities.
Accessing the New York LLC articles of organization will not cost you anything. However, please note that you will need to pay a fee when filing the form with the state.
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What Is A Professional Service
Under the relevant New York law, professional service is defined only generally, as any type of personal service which may be lawfully rendered by a member of a profession within the purview of that profession. Other states' laws often provide itemized lists of professional services. These lists typically include, at a minimum, physicians, surgeons, dentists, lawyers, certified public accountants, professional engineers, architects, and veterinarians, but usually include other professions, as well.
While New York's LLC law does not provide an itemized list of professional services, it does have provisions that indicate that all of the followingwhen properly licensed by the stateconstitute professional services:
the practice of medicine
professional engineering, land surveying, architectural and/or landscape architectural services
clinical social work services
mental health counseling services
applied behavior analysis services, and
the practice of law.
If you're unsure whether your New York licensed or authorized profession is considered a professional service for the purpose of forming a PLLC, check with a local business attorney.
Domestic New York Llc Name Change Instructions
We recommend using the Certificate of Amendment Name Change Only as its easier than Form 1358. The instructions below are for Form 2120.
Insert Name of Domestic Limited Liability Company: On the first line, enter your existing/old LLC name. Make sure to include the designator, such as LLC or L.L.C..
FIRST: The name of the limited liability company is: Enter your existing/old LLC name again.
If the name of the limited liability company has been changed, the name under which it was organized is: For most filers, this will be left blank. Youd only have something here if this is the second time youre changing your LLC name.
SECOND: The date of filing of the articles of organization is: This is the date your LLC was approved by the New York Division of Corporations. You can find this date on your approved Articles of Organization. Or you can search your LLC name on the Business Entity Database page. Click on your LLC name and look for the Initial DOS Filing Date.
THIRD: The amendment effected by this certificate of amendment is as follows: The state has technically already filled this section in for you. It will read, Paragraph FIRST of the Articles of Organization relating to the name of the limited liability company is hereby amended to read as follows:
FIRST: The name of the limited liability company is: Enter your new LLC name here. Make sure to include the designator. As per section 204 of the NY LLC law, the allowable designators in New York State are:
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text:
Q: Skyrim missing stone texture near Winterhold In Skyrim, there is a weird stone that seems to be missing a texture on one side.
Pic:

Location:

Is this normal? And how do i fix it?
I tried restarting Skyrim, it didn't work.

A: There are many such graphical glitches in Skyrim.  The Unofficial Skyrim Patch addresses many of these issues, but its very hard to find all of them.  In case you are using the HD Textures, there is also an unofficial patch for that as well you should get.
In case this is simply a glitch for your machine, you should try to verify the integrity of your game cache.  I recommend doing this step first.