\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \fancyhead[CE]{Zynab Izadi, Rahmat Soltani} \fancyhead[CO]{Maps Preserving the Difference of Minimum and Surjectivity Moduli of Self-adjoint Operators} %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewenvironment{proof}{{\bfseries \noindent Proof.}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\thm@headfont{\bfseries}} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm %\setcounter{page}{} \thispagestyle{plain} {\noindent Journal of Mathematical Extension \\ Vol. XX, No. XX, (2019), pp-pp (Will be inserted by layout editor)}\\ ISSN: 1735-8299\\ URL: http://www.ijmex.com\\ %‎\vspace*{9mm} ‎ \begin{center} {\Large \bf Maps Preserving the Difference of Minimum and Surjectivity Moduli of Self-adjoint Operators\\} %%{\bf Do You Have a Subtitle? \\ If so, Write It Here} \let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)} %%%%%%%%%%%%%%%%%%%%%%%%% % Subject class; see http://www.ams.org/mathscinet/msc/msc2010.html %%%%%%%%%%%%%%%%%%%%%%%%%%%% %\subjclass[ 2010 Mathematics subject Classification:]{Primary....;Secondary.....} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % keywords; Note that the number of keywords must be at least 3 items and at most 5 items. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\keywords{} %%%%%%%%%%%%%%%%%%%%%%% {\bf Zynab Izadi}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Payamenoor University\\ P. O. Box 19395-3697,Tehran, Iran\\$zynab_{-}izadi$@pnu.ac.ir} \vspace{2mm} {\bf Rahmat Soltani$^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Payamenoor University\\ P. O. Box 19395-3697,Tehran, Iran\\ $r_{-}soltani$@pnu.ac.ir} %\vspace{2mm} \end{center} \vspace{2mm} {\footnotesize \begin{quotation} {\noindent \bf Abstract.} Let $ \mathcal{H}$ be a separable infinite dimensional complex Hilbert space and $ \mathcal{SA}(\mathcal{H})$ be the real Jordan algebra of all bounded self-adjoint operators acting on $\mathcal{ H}$. In this paper, we study the general form of surjective non-linear maps $\xi : \mathcal{SA}( \mathcal{H}) \rightarrow \mathcal{SA}(\mathcal{H})$, that preserve the difference of minimum and surjectivity moduli of self-adjoint operators in both directions. It turns out that \begin{align*} \xi(P)=EPE^{*} + R , \hspace{2cm} (P, R\in\mathcal{SA}(\mathcal{H})) \end{align*} where $E: \mathcal{H} \rightarrow \mathcal{H}$, is either a bounded unitary or an anti-unitary operator.\\ \noindent{\bf AMS Subject Classification:} 47B49; 47B15. \noindent{\bf Keywords and Phrases:} Non-linear preserver problems, algebraic operators, algebraic singularity \end{quotation}} \newpage \section{Introduction} \label{intro} \hspace{.71cm}Recently non-linear preserver problems have been investigated by many authors, see for instance [10, 14, 15, 18, 20].\hspace{.1cm}In [5], authors characterized surjective maps preserving the spectral radius of the difference of matrices. In [19], Molnar studied maps preserving the spectrum of operator or matrix products. His results have been extended in several direction for uniform algebras and semisimple commutative Banach algebras, and a number of results is obtained on maps preserving several spectral and local spectral quantities of operator or matrix product, or Jordan product, or difference; see for instance [9, 13, 14, 15, 17] and the references therein. In [18], authors characterized maps of $\xi : \mathcal{B}( X) \rightarrow \mathcal{B}( X) $ that preserve difference minimum moduli of operators. They proved that this map has the form $\xi(P)=EP^{\sharp}F+R $ for $E$ ,$F$ isometry operators and $P^{\sharp}$ denotes $P$, or $ P^*$.\\ In this paper, we attempt to determine the general form of $\xi$ when it is restricted to the real Jordan algebra $\mathcal {SA}(\mathcal H)$. Throughout this paper, $ \mathcal{H}$ stands for an infinite dimensional separable complex Hilbert space, and $\mathcal{B}( \mathcal{H}) $, $\mathcal{SA}(\mathcal{H})$ and $\mathcal{F}(\mathcal{H})$, $\mathcal{FS}(\mathcal{H})$ denote the space of all bounded operators, self-adjoint bounded operators, finite rank operator on $ \mathcal{H}$, finite rank operator on $ \mathcal{SA}(\mathcal{H})$, and $ \mathcal{A}_s(\mathcal{H})$ the set of all algebraic operators on $\mathcal{SA}(\mathcal{H})$. In [15], Havlicek and Semrl showed a complete characterization of bijective maps $\xi$ on $\mathcal{B}( \mathcal{H}) $ satisfying the condition \\ $~~~~~~~~~~~~~~~~~~\xi(Q)-\xi(P)~ is ~invertible\Longleftrightarrow Q-P~ is ~invertible$. \begin{theorem} ([15] Theorem 1.2): Let $ \mathcal{H}$ be an infinite dimensional complex Hilbert space and $\mathcal{B}( \mathcal{H}) $ denotes the algebra of all bounded linear operators on $ \mathcal{H}$. Let $\xi:\mathcal{B}( \mathcal{H})\rightarrow \mathcal{B}( \mathcal{H}) $ be bijective map such that for every pair $M,W\in \mathcal{B}( \mathcal{H}) $ the operator $M-W$ is invertible if and only if $\xi(M)-\xi(W)$ is invertible. Then there exist $R \in \mathcal{B}( \mathcal{H})$ and invertible $P,Q\in \mathcal{B}( \mathcal{H})$ such that either \begin{equation}\label{Pythagoras} %\begin{align*} \xi(M)=PMQ+R \end{equation} for ~every ~~ $M\in \mathcal{B}( \mathcal{H}) $,or \begin{equation}\label{Pythagoras} \xi(M)=PM^{t}Q+R \end{equation} for~ every ~~$ M\in \mathcal{B}( \mathcal{H})$ ,or \begin{equation}\label{Pythagoras} \xi(M)=PM^*Q+R \end{equation} for ~every ~~$M\in \mathcal{B}( \mathcal{H}) $, or \begin{equation}\label{Pythagoras} \xi(M)=P(M^{t})^*Q+R \end{equation} for ~every ~~$M\in \mathcal{B}( \mathcal{H}).$ ~~\\\\ Here $M^t$ and $M^*$ denote the transpose with respect to an arbitrary but fixed orthonormal basis, and the usual adjoint of $M$ in the Hilbert space sense, respectively. \end{theorem} \begin{definition}([23]) The minimum modulus of an operator $P\in \mathcal{B}( \mathcal{H})$ denoted by $\rho(P)$, is defined by $\rho(P)=inf\{||Ph|| :~h\in \mathcal{H}~,~||h||=1\}$.\\ The surjectivity modulus of $P$ is defined by \\ $\tau(P)=sup\{\varepsilon\geqslant o:\varepsilon B(0,1)\subset P(B(0,1))\}$, where $B(0,1)=\{x\in H:||x||<1\}$.\hspace{.2cm}Maximum modulus of $P$ is defined by\\ $N(P)=max\{ \rho(P),\tau(P)\}.$ \end{definition} \begin{definition}([23]) A linear map $\xi:\mathcal{B}( \mathcal{H})\rightarrow \mathcal{B}( \mathcal{H}) $ preserves the minimum modulus(resp. surjectivity modulus), if for all $P\in \mathcal{B}( \mathcal{H}) $,\\ $~~~~~~\rho(\xi(P))=\rho(P)~~~,\hspace{2cm}~ (resp. ~ \tau(\xi(P))=\tau(P))$. \end{definition} Note that $\rho(P^*)=\tau(P) ~$ and $~ \tau(P^*)=\rho(P)~~ for ~all ~P\in \mathcal{B}( \mathcal{H})$. \begin{theorem}([23] Theorem 3.5): Let $\xi:\mathcal{B}( \mathcal{H})\rightarrow \mathcal{B}( \mathcal{H}) $ be a surjective linear map. The following are equivalent: \begin{align*} 1. ~~ &\rho(\xi(P))=\rho(P)~,~ for~all~ P\in \mathcal{B}( \mathcal{H}). \\ 2.~~ &\tau(\xi(P))=\tau(P)~,~for~all~ P\in \mathcal{B}( \mathcal{H}). \\ 3.~~&There ~exist ~two~ unitary~ operators~ E\in \mathcal{B}( \mathcal{H})~ and ~ F\in \mathcal{B}( \mathcal{H})~ such~ that\\ &\xi(P)=EPF ~ for~ every~ P\in \mathcal{B}( \mathcal{H}). \end{align*} \end{theorem} In this paper, we let $\nu(.)$ stand for either $\rho(.) ~or~\tau(.)~or~N(.)$. We stablish a similar result to Theorem 1.1 of characterizing maps from $\mathcal{SA}( \mathcal{H})$ onto $\mathcal{SA}( \mathcal{H})$ preserving any of the surjectivity, the injectivity, and the boundedness from below of the difference and sum of operators. We show the adjacency of operators in term of any of the previous mentioned spectral quantities and use such a description to show that if a map $\xi$ from $\mathcal{SA}( \mathcal{H})$ ~onto~$\mathcal{SA}( \mathcal{H})$ preserves operator pairs difference is invertible and thus Theorem 1.1 ensures that such a map $\xi$ takes either (1) or (3). Then we describe maps $\xi$ from $\mathcal{SA}( \mathcal{H})$ ~onto~$\mathcal{SA}( \mathcal{H})$ satisfying \begin{equation}\label{Pythagoras} \nu(\xi(M)-\xi(W))=\nu(M-W) \end{equation} \begin{equation}\label{Pythagoras} \nu(\xi(M)+\xi(W))=\nu(M+W) \end{equation} for all $M,W\in \mathcal{SA}(\mathcal{H})$.\\ For $g,h \in \mathcal{H}$, $ $ stands for the inner product of g and h.\hspace{.1cm}For every $P\in\mathcal{B}(\mathcal{H})$, we use the notations $rank(P), ~ ker(P), ~ ran(P)$ and $\sigma(P)$ for the rank, kernel, range and the spectrum of $P$, respectively.\hspace{.1cm}A conjugate linear bijective operator $E$ on $ \mathcal{H}$ is called anti-unitary, provided that $=$ for all $x,y\in \mathcal{H}$.\hspace{.1cm}The identity operator on $ \mathcal{H}$ will be denoted by $I$.\hspace{.1cm}Two operators $Q,P$ in $\mathcal {SA}(\mathcal{H})$ are called adjacent, provided that $Q-P$ is a rank one operator.\hspace{.1cm}It is said that a surjective map $\xi:\mathcal {SA}( \mathcal{H}) \longrightarrow \mathcal {SA}( \mathcal{H})$ preserves adjacency of operators in both directions, if it preserves adjacent operators in both directions. Every self-adjoint rank one operator on $\mathcal{H}$ is of the form $\lambda b\otimes b$ for some non-zero $b\in\mathcal{H}$ and $\lambda \in \mathbb{R}$. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \section{Main results} ~~~ In this part, we recall some important lemmas that will be used in the proof of our results. Recall that the spectral radius of an operator $P\in \mathcal{B}( \mathcal{H})$ is \\ $r(P) =lim _{n\rightarrow \infty}||P^n||^{1/n}$ , and coincides with the maximum modulus of $\sigma(P)$, the spectrum of P. \begin{lemma}([12]) Let $A\in \mathcal{B}( \mathcal{H})$. Then $A=0$ if and only if \\ $r(A+P)=0$, for all nilpotent operators $P\in \mathcal{B}( \mathcal{H})$ of rank at most one. \end{lemma} \begin{lemma}([23]) Let $C,D\in \mathcal{B}( \mathcal{H})$ be two invertible operators. The following are equivalent: %Let $P\in \mathcal{SA}(\mathcal{H})$. The following statements hold: \begin{itemize} \item [\rm{(1)}] $ \rho(CPD)=\rho(P)~~~~~for~all~P\in \mathcal{B}( \mathcal{H})$, \item [\rm{(2)}] $\tau(CPD)=\tau(P)~~~~~for~all~P\in \mathcal{B}( \mathcal{H})$, \item [\rm{(3)}] there are two unitary operators $E \in \mathcal{B}(\mathcal{H})~ and $ $F\in \mathcal{B}( \mathcal{H})$~ such that~ $C=\alpha E$~,~$D=\beta F~$ where $\alpha,\beta \in\mathbb{C}\setminus \{0\}$ with $|\alpha \beta|=1$. \end{itemize} \end{lemma} Let $M_n(\mathbb{C})$ be the algebra of all $n\times n$ complex matrices, and note that $\rho(P)=\tau(P)=N(P)$ for all matrices $P\in M_n( \mathbb{C})$. \begin{theorem}([15]) A surjective map $\xi$ on $M_n( \mathbb{C})$ satisfies \begin{align*} \rho(\xi(Q)-\xi(P))=\rho(Q-P),~~~~~~~(Q,P\in M_n( \mathbb{C})) \end{align*} if and only if there are $E,F,R\in M_n( \mathbb{C})$ with $E$ and $F$ unitary matrices such that ~~~~~~~ $\xi(P)=EP^{\sharp}F+R~,~~~~(P\in M_n( \mathbb{C}))$\\ where $P^{\sharp}$ uses for $P~or~P^{tr}~or~P^*~or\bar{P}$, the complex conjugation of $P$. \end{theorem} We describe maps $\xi$ from $\mathcal{SA}( \mathcal{H})$ ~onto~$\mathcal{SA}( \mathcal{H})$ satisfying (5) and (6) and prove some details needed for the proof of the main results. These results generalize published results on linear or additive maps preserving the minimum and surjectivity moduli of operators. See for instance [6, 8, 23]. \begin{theorem} Let $\xi:\mathcal{SA}( \mathcal{H})\rightarrow \mathcal{SA}( \mathcal{H})$ be a surjective map satisfying (5). Then following situation hold:\\ There are $R\in \mathcal{B}( \mathcal{H})$ and unitary or anti-unitary operator $ E:\mathcal{H}\rightarrow\mathcal{H}$ such that \begin{equation}\label{Pythagoras} \xi(P)=EPE^*+R~,~~~P\in \mathcal{SA}(\mathcal{H}). \end{equation} If $\xi$ satisfies $(7)$, then $(5)$ is satisfied. \end{theorem} For our aims, we show a similar result to Havlicek and Semrl [15] and Hou and Huang [16] by replacing the invertibility by surjectivity and boundedness below. This result plays an important role in the proof of the Theorem 2.4. Let ­$\Delta_{Inj}(\mathcal{H}), \Delta_{­Surj}(\mathcal{H}), \Delta_{ ­LB}(\mathcal{H}), \Delta_{ Inj-or-Surj}(\mathcal{H}) ~and~\\ \Delta_{­LB-or-Surj}(\mathcal{H}) $ be respectively the subsets of all injective, surjective, lower bounded, injective or surjective, and lower bounded or surjective operators on $\mathcal{SA}( \mathcal{H})$. Assume that $\Delta­(\mathcal{H})$ denotes any of these sets, and note that every operator $P\in \Delta(\mathcal{H})$ is either injective or surjective and that \begin{align*} ~~~~~~~P.\Delta(\mathcal{H})=\Delta(\mathcal{H}).P=\Delta(\mathcal{H}) \end{align*} for all invertible operators $P\in\mathcal{SA}(\mathcal{H})$. \begin{theorem} If a surjective map $\xi:\mathcal{SA}( \mathcal{H})\longrightarrow \mathcal{SA}( \mathcal{H})$ satisfies \begin{equation}\label{Pythagoras} \xi(Q)-\xi(P) \in \Delta(\mathcal{H}) \Longleftrightarrow Q-P\in\Delta(\mathcal{H}), \end{equation} there is an operator $R\in\mathcal{SA}( \mathcal{H})$ such that $\xi$ is of the form \begin{align*} ~\xi(P)=APA^*+R~~,~~P\in \mathcal{SA}( \mathcal{H}) \end{align*} for some bijective continuous mapping $A:\mathcal{H}\rightarrow \mathcal{H}$. \end{theorem} For proving of this theorem, we show that if $\xi$ satisfies (8), then $\xi$ is a bijective map that preserves the invertibility of the difference of operators. But to show this, we first specify the adjacency of operators in term of operators in $\Delta­(\mathcal{H})$, similar to following lemma. \begin{lemma} Let $Q, P$ be two different operators in $\mathcal{SA}(\mathcal{H})$. Then $Q,P$ are adjacent, if and only if there exists $R\in\mathcal{SA}(\mathcal{H})\setminus \{Q,P\}$ such that $R-P\in \Delta(\mathcal{H})$ and for every $Y\in\Delta(\mathcal{H}),Y-R,Y-P\in\Delta(\mathcal{H})$ imply $Y-Q\in\Delta(\mathcal{H})$. \end{lemma} \begin{proof} We can restrict ourselves to the case where $P=0$. Assume $Q$ is a rank one operator. Hence $Q=\lambda h\otimes h$, where $h\in \mathcal{H}$ is a unit vector and $\lambda$ is non-zero real scalar. Let $ R=-Q$. Then $R\in \Delta(\mathcal{H})\setminus \{Q,0\}$. Let $Y\in \Delta(\mathcal{H})$ be such that $Y-R\in\Delta(\mathcal{H})$. We claim $Y-Q$ is non-invertible(non-injective or non-surjective). For this, we consider two cases: if $ker(Y)\cap \{h\}^{\perp}\neq\{0\}$, then $ker(Y-Q)\neq \{0\}$ and consequently $Y-Q$ is non-invertible. Assume $ker(Y) \cap \{h\}^{\perp}=\{0\}$, then $Y+Q$ is non-injective, as $Y-R=Y+Q\in \Delta(\mathcal{H})$. Let $k\in ker (Y+Q) $ be a non-zero unit vector. Then $Yk=-\lambdah$. Hence $ k \notin \{h\}^{\perp}$. As $\mathcal{H}=\{h\}^{\perp} \oplus \mathbb{C} h$, it follows that $k=\beta h$, for some non-zero scalars $\beta \in \mathbb{C}$. Hence $Yh=-\lambda h$. Consequently, as $Y-Q=Y(I+h\otimes h)$, applying the facts that $Y$ is non-invertible and $I+h\otimes h$ is invertible, it follows that $Y-Q$ is not invertible. So $Y-Q\in \Delta(\mathcal{H})$.\\ For the inverse direction, it is assumed $dim \hspace{.06cm} ran(Q)\geq 2$.\hspace{.06cm}We claim that for every $R\in \Delta(\mathcal{H}) \setminus \{Q,0\} $, there exists $Y\in\Delta(\mathcal{H})$ such that $Y-R\in \Delta(\mathcal{H})$ and $Y-Q\notin \Delta(\mathcal{H})$. For this, let $R\in \Delta(\mathcal{H})\setminus \{Q,0\}$ be fixed.\hspace{.06cm}There are two cases: If $Q\notin\Delta(\mathcal{H}) $, then it is enough to consider $Y=0$.\hspace{.19cm}If $Q\in \Delta(\mathcal{H})$, then $Q$ is not injective and there exist some $h\in \mathcal{H}$ such that $(R-Q)h\neq 0$, as $R\neq Q$.\hspace{.19cm}Applying the fact that $dim ~ ran(Q)\geq 2$, it follows that there exist some $k\in\mathcal{H}$ such that the vectors $\{(R-Q)h,Qk\}$ are linearly independent.\hspace{.19cm}By replacing $k$ with $k+\theta$, for some $\theta\in ker(Q)$ if necessary, we may assume $\{h,k\}$ are linearly independent.\hspace{198cm}Let $V=span\{h,k,(R-Q)h,Qk\}$.\hspace{.17cm}Then we can represent the operators $Q$ and $ R$ with respect to the decomposition of $\mathcal{H}=V\oplus V^{\perp}$ as follows:\\\ \begin{equation*} Q= %\begin{equation*} \begin{bmatrix} Q_1& Q_2 \\ Q_2^*& Q_3 \end{bmatrix}, %\end{equation*}, R= \begin{bmatrix} R_1 & R_2\\ R_{2}^* & R_3 \end{bmatrix}. \end{equation*} Let $Y\in \mathcal{SA}(\mathcal{H}) $ be the operator that with respect to the decomposition $\mathcal{H}=V\oplus V^{\perp}$ is represented as \begin{equation*} Y= %\begin{equation*} \begin{bmatrix} P+Q_1& Q_2 \\ Q_2^*&cI \end{bmatrix}, \end{equation*} where $c\notin \sigma(Q_3)$ and $P\in \mathcal{SA}(V)$ is an invertible operator such that $Ph=(R_1-Q_1)h$ and $Pk= -Q_1k$. It follows that $ R,Y$ and $Y-R$ are algebraic operators.\hspace{.19cm}But as $ Yk=(Y-R)h=0$, hence $Y,Y-R \in \Delta(\mathcal{H})$. On the other hand, as \begin{equation*} Y-Q= %\begin{equation*} \begin{bmatrix} P& 0 \\ 0&cI-Q_3 \end{bmatrix}, \end{equation*} it follows that $Y-Q$ is invertible, thus $Y-Q\notin \Delta(\mathcal{H})$, which completes the proof. \end{proof} \begin{proposition} %\textbf{Proposition 2.1}. Let $Q,P\in \mathcal{SA}(\mathcal{H})$ and for every invertible $Y\in \mathcal{SA}(\mathcal{H})$ we have \begin{align*} Q-Y\in\Delta(\mathcal{H}) \Longleftrightarrow P-Y\in \Delta(\mathcal{H}). \end{align*} Then $Q=P$. \end{proposition} \begin{proof} Since $0\in \Delta(\mathcal{H})$ by setting $Y=Q$, it follows $P-Q\in\Delta(\mathcal{H}) $. As $\Delta(\mathcal{H})$ does not contain any invertible operator and $P-Q\in\Delta(\mathcal{H})$, for asserting that $P=Q$, it is enough to show that $P-Q$ is a scalar multiple of the identity. If this is not so, $P-Q\neq cI$ for every $c\in \mathbb{R}$, then there exists a unit vector $k\in \mathcal{H}~$ such that $k, (P-Q)k$ are linearly independent. Let $V=span\{k,(P-Q)k\}$.\hspace{.19cm}Then $ P-Q$ has the matrix representation \begin{equation*} P-Q= \begin{bmatrix} A& B\\ B^*&C \end{bmatrix} \end{equation*} \begin{equation*} where ~~\hspace{.3cm} A=\begin{bmatrix} 0& 1\\ 1&\beta \end{bmatrix}. \end{equation*} Let \begin{equation*} R= \begin{bmatrix} 0&- B\\ -B^*&I-C \end{bmatrix}. \end{equation*} Then $R\in \mathcal{SA}(\mathcal{H})$ and since $Rk=0$, it follows that $ R$ is not invertible.\hspace{.17cm}Since by assumption $P-Q\in\Delta(\mathcal{H})$, it follows that $ C$ and hence $R$ are algebraic. Consequently, if we set $Y=Q-R$, then $Q-Y=R\in \Delta(\mathcal{H})$.\hspace{.19cm}But since \begin{equation*} P-Y= \begin{bmatrix} A&0\\ 0&I \end{bmatrix}, \end{equation*} is invertible, so $P-Y\notin\Delta(\mathcal{H})$. we get a contradiction. \end{proof} Lemma 2.8 take from [21], and characterize non-linear maps that preserving finite rank operators and preserve the operator that difference of them is rank one. Recall that a map $ B: \mathcal{H}\rightarrow \mathcal{H}$ is called semilinear if it is additive and there is an automorphism $\sigma$ of $\mathbb{C}$ such that $B(\alpha x)=\sigma(\alpha)Bx$ for all $x\in \mathcal{H}~and~\alpha\in \mathbb{C}$. Such a map is sometimes said $\sigma$-semilinear when the automorphism $\sigma$ is specified. \begin{lemma}(Petek-Semrl [21]) Assume that $X$ and $Y$ are Banach spaces of dimensions at least 2, and let $\xi$ be a bijective map from $\mathcal{F}(X)$ into $\mathcal{F}(Y )$ such that whenever $C,D$ are operators in $\mathcal{F}(X)$ that satisfies\\ $~~~~~~~~~~~~~C-D~ has ~rank ~one ~\Longleftrightarrow ~\xi(C)-\xi(D) ~has~ rank~ one$,\\ then one of the following properties is satisfied: \begin{itemize} \item [\rm{(1)}] There is an automorphism $\sigma$ of $\mathbb{C}$, $R\in\mathcal{B}(Y)$, and bijective $\sigma$-semilinear maps ~$S:X\rightarrow Y$ and $T:X^*\rightarrow Y^*$ such that \\$D\mapsto \xi(D)-R$ is an additive map defined by $\xi(x\otimes f)-R=Sx\otimes Tf.~~~~~~~~~(x\in X,f\in X^*)$. \item [\rm{(2)}] There is an automorphism $\sigma$ of $\mathbb{C}$, $R\in\mathcal{B}(Y)$, and bijective $\sigma$-semilinear maps $S:X\rightarrow Y^*$ and $T:X^*\rightarrow Y$ such that $D\mapsto \xi(D)-R$ is an additive map defined by $\xi(x\otimes f)-R=Tf\otimes Sx.~~~~~~~~~(x\in X,f\in X^*)$. \end{itemize} \end{lemma} This lemma is true for $X=Y=\mathcal{H}$ and $\mathcal{SA}(\mathcal{H})$.\\ Let $\mathcal{AI}_{s}(\mathcal{H})$ be the set of all invertible operators in $\mathcal{SA}(\mathcal{H})$. \begin{proposition} %\textbf{Proposition 2.2}. Let $Q,P\in \mathcal{SA}(\mathcal{H})$ and for every $N\in \mathcal{AI}_{s}(\mathcal{H})$ we have \begin{align*} Q-N\in \mathcal{AI}_{s}(\mathcal{H}) \Longleftrightarrow P-N\in \mathcal{AI}_{s}(\mathcal{H}). \end{align*} Then $Q=P$. \end{proposition} \begin{proof} First we claim that $\sigma(Q)=\sigma(P)$. For this note that for every scalar $ \lambda\in \mathbb{R}$, $\lambda\in \sigma(Q)$ if and only if $Q-\lambda I$ is not invertible, but by assumption, this holds precisely when $P-\lambda I$ is not invertible or equivalently $\lambda\in \sigma(P)$. For the rest of proof, it is enough to show that $Q-P$ is a scalar operator, since from $Q=P+\lambda I$ and the fact that $\sigma(Q)=\sigma(P)$, it follows that $\lambda=0$. If $Q-P$ is not a scalar operator, then there exists $k\in \mathcal{H}$ such that the vectors $k$ and $(Q-P)k$ are linearly independent.\hspace{.15cm}There are two cases: either $\{k,Pk\}$ or $\{k,Qk\}$ is a linearly independent set. It is enough to consider the first case, since the other one is similar. Put $V=span\{k,Qk,Pk\}$.\hspace{.15cm}Then the operator $Q$ can be represented as \begin{equation*} Q= \begin{bmatrix} Q_1&Q_2\\ Q^*_2&Q_3 \end{bmatrix} \end{equation*} regarding the decomposition of $\mathcal{H}=V\bigoplus V^{\perp}$. Before proceeding further, we make a claim:\\ \textbf{Claim.}\hspace{.15cm}There exists an invertible $A\in \mathcal{SA}(V)$ such that $Ak=Pk$ and $Q_1-A$ is invertible.\\ For this we consider two cases: Case1. Assume $dim(V)=3$. Let \begin{equation*} Q_1= \begin{bmatrix} q_{11}&q_{12}&q_{13}\\ q^*_{12}&q_{22}&q_{23}\\ q^*_{13}&q^*_{23}&q_{33} \end{bmatrix}, A= \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a^*_{12}&a_{22}&a_{23}\\ a^*_{13}&a^*_{23}&a_{33} \end{bmatrix} \end{equation*} be the representations of $Q_1$ and $A$ regarding the decomposition of $V=\{k\}\bigoplus \{Pk\}\bigoplus \{Qk\}$. Since $Q_1k=Q_1 \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} $, it follows that $q_{11}=q_{12}=0$ and $q_{13}=1$. Similarly, since $Ak=A \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} $, it follows that $a_{11}=a_{13}=0$ and $a_{12}=1$.\hspace{.14cm}Hence, it is enough to consider \begin{equation*} A= \begin{bmatrix} 0&1&o\\ 1&q_{22}-1&q_{23}\\ 0&q^*_{23}&q_{33}-1 \end{bmatrix}. \end{equation*} Then $A$ is invertible and satisfies $Ak=Pk$. \\ Case2. Assume $dim(V)=2$. Let $Qk=\lambda k+\mu Pk$, for $\lambda,\mu \in \mathbb{R}$.\hspace{.14cm}Since $\{k,(P-Q)k\}$ is linearly independent, $\mu\neq 1$. Let \begin{equation*} Q_1= \begin{bmatrix} q_{11}&q_{12}\\ q^*_{12}&q_{22} \end{bmatrix}, A= \begin{bmatrix} a_{11}&a_{12}\\ a^*_{12}&a_{22} \end{bmatrix} \end{equation*} be the representations of $Q_1$ and $A$ regarding the decomposition of $V=\{k\}\bigoplus \{Pk\}$. Since $Q_1k=Q_1 \begin{bmatrix} 1\\ 0 \end{bmatrix}=\lambda k+\mu Pk= \begin{bmatrix} \lambda\\ \mu \end{bmatrix} $, it follows that $q_{11}=\lambda$ ans $q_{12}=\mu$. Similarly, since $Ak=A \begin{bmatrix} 1\\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 1 \end{bmatrix} $, it follows that $a_{11}=0$ and $a_{12}=1$. Hence, it is enough to consider \begin{equation*} A= \begin{bmatrix} 0&1\\ 1&q_{22} \end{bmatrix}. \end{equation*} Then since $\mu\neq 1$, $A$ is invertible and satisfies $Ak=Pk$. Consider the operator $N\in \mathcal{SA}(\mathcal{H})$ such that regarding the decomposition of $\mathcal{H}=V\bigoplus V^{\perp}$ has the matrix representation \begin{equation*} N= \begin{bmatrix} A&Q_2\\ Q^*_2&\lambda I \end{bmatrix}, \end{equation*} where $\lambda\in \mathbb{R} \setminus\sigma(Q_3)$. Since $A$ and $Q_1-A$ are invertible, it follows from lemma 2.6 that $N$ and $Q-N$ are algebraicly invertible operators.\hspace{.14cm}But since $(P-N)k=0$, we conclude that $P-N\notin \mathcal{AI}_{s}(\mathcal{H})$, which is a contradiction. \end{proof} Now, we can prove Theorem 2.5.\\ %\end{proof} \begin{proof}(Proof of Theorem 2.5) Assume that $\xi$ satisfies in following relation \\ $~~~~~~~~~\hspace{1cm}~~~~~~~~\xi(Q)-\xi(P) \in \Delta(\mathcal{H}) \Longleftrightarrow Q-P\in\Delta(\mathcal{H})~~~~~~$\\ and note that $\xi-R$ satisfies to this relation. Thus, after replacing $\xi_1$ by $\xi-R$, we may assume that $\xi_1(0)=0$. We do this through a few step.\\ \textbf{Step 1}. $\xi_1$ is bijective and preserves adjacency of operators in both directions. Let $Q_1,Q_2\in\mathcal{SA}(\mathcal{H})$ be such that $\xi_1(Q_1)=\xi_1(Q_2)$.\hspace{.14cm}Then by assumption, for every operator we have \begin{align*} Q_1-N\in\Delta(\mathcal{H}) &\Longleftrightarrow~~\xi_1(Q_1)-\xi_1(N) \in \Delta(\mathcal{H})&\\ &\Longleftrightarrow~~ \xi_1(Q_2)-\xi(N) \in \Delta(\mathcal{H})&\\ &\Longleftrightarrow~~ Q_2-N \in \Delta(\mathcal{H})& \end{align*} for every $N\in \mathcal{SA}(\mathcal{H})$.\hspace{.14cm}Therefore, it follows from proposition 2.9 that $Q_1=Q_2$ and consequently $\xi_1$ is injective. It is also bijective because it is supposed to be surjective.\\ Now, let $A,B\in\mathcal{SA}(\mathcal{H})$ be such that $A-B$ has rank one.\hspace{.14cm}Then by lemma 2.6 it follows that there exists $R\in \mathcal{SA}(\mathcal{H})\setminus \{A,B\} $ such that $R-B\in \Delta(\mathcal{H})$ and for every $P\in \mathcal{SA}(\mathcal{H})$, the relations $P-R\in \Delta(\mathcal{H})$ and $P-B\in \Delta(\mathcal{H})$ yield that $P-A\in \Delta(\mathcal{H})$.\hspace{.14cm}As $\xi_1$ is injective, we get $\xi_1(R)\in \mathcal{SA}(\mathcal{H})\setminus \{\xi_1(A),\xi_1(B)\}$.\hspace{.14cm}By assumption $\xi_1(R)-\xi_1(B) \in \Delta(\mathcal{H})$. Suppose $Q \in \mathcal{SA}(\mathcal{H})$ such that $Q-\xi_1(R) \in \Delta(\mathcal{H})$ and $Q-\xi_1(B) \in \Delta(\mathcal{H})$. There exists $P\in \mathcal{SA}(\mathcal{H})$ that $\xi_1(P)=Q$, as $\xi_1$ is surjective. Thus $\xi_1(P)-\xi_1(R) \in \Delta(\mathcal{H})$ and $\xi_1(P)-\xi_1(B) \in \Delta(\mathcal{H})$, which implies $P-R \in \Delta(\mathcal{H})$ and $P-B \in \Delta(\mathcal{H})$.\hspace{.14cm}Hence, we have $P-A \in \Delta(\mathcal{H})$ and consequently $Q-\xi_1(A) \in \Delta(\mathcal{H})$.\hspace{.14cm}Now applying lemma 2.6 it follows that $\xi_1(A)-\xi_1(B)$ has rank one. On the other hand, as $\xi_1$ is bijective and $\xi_1^{-1}$ satisfies the same properties as $\xi_1$, it follows that $\xi_1$ preserves adjacency in both directions.\\ %$\xi_1$ %maps the \textbf{Step 2}. $\xi_1$ maps rank one operators onto rank one operators and maps $\mathcal{FS}(\mathcal{H})$ onto itself. Let $F$ be a rank one operators in $ \mathcal{SA}(\mathcal{H})$.\hspace{.14cm}Then $F$ is adjacent to 0.\hspace{.14cm}By step 1, $\xi_1(F)$ is adjacent to $\xi_{1}(0)=0$.\hspace{.14cm}Hence $\xi_1(F)$ is a rank one operator.\hspace{.14cm}On the other hand as every rank 2 operator is adjacent to a rank one operator, hence, if $ rank(E)=2$, then $rank(\xi_1(E))<\infty$. Similarly, it follows that $\xi_1$ maps $\mathcal{FS}(\mathcal{H})$ onto itself.\\\\ %\item[\rm{Step3}] .\\ \textbf{Step 3.} $\xi_1$ preserves projections of rank one and there exists either a bijective linear or conjugate-linear operator $A: \mathcal{H}\rightarrow \mathcal{H}$ such that for every $P\in\mathcal{FS}(\mathcal{H})$ \begin{align*} \xi_1(P)=\lambda APA^*. \end{align*} As $\xi_1: \mathcal{FS}(\mathcal{H})\rightarrow \mathcal{FS}(\mathcal{H})$ preserves adjacency and satisfies $\xi_1(0)=0$, it follows [22, Theorem 2.1] that either \begin{itemize} \item[\rm{$\bullet$}] %\bf{$\bullet$} There exists a rank one operator $R\in \mathcal{SA}(\mathcal{H})$ such that the range of $\xi_1$ is contained in the linear span of $R$; or \item[\rm{$\bullet$}] There exists an injective linear or conjugate-linear map $A:\mathcal{H}\rightarrow \mathcal{H}$ such that \begin{align*} \xi_1(\sum_{j=1}^kn_jy_j\otimes y_j)=\sum_{j=1}^k n_jA(y_j\otimes y_j)A^* \end{align*} for every $\sum_{j=1}^kn_jy_j\otimes y_j \in \mathcal{FS}(\mathcal{H})$; or \item[\rm{$\bullet$}] %$.$ There exists an injective linear or conjugate-linear map $A:\mathcal{H}\rightarrow \mathcal{H}$ such that \begin{align*} \xi_1(\sum_{j=1}^kn_jy_j\otimes y_j)=-\sum_{j=1}^k n_jA(y_j\otimes y_j)A^* \end{align*} \end{itemize} for every $\sum_{j=1}^kn_jy_j\otimes y_j \in \mathcal{FS}(\mathcal{H})$.\\ As $\xi_1$ is bijective, the first case doesnot happen. Since both $\xi_1$ and $\xi_1^{-1}$ have the same properties, from above discussion it follows that there exists either an invertible linear or conjugate-linear operator $A:\mathcal{H}\rightarrow \mathcal{H}$ such that for every $P\in \mathcal{FS}(\mathcal{H})$ \begin{align*} \xi_1(P)=\lambda APA^*, \end{align*} where $\lambda \in\{-1,1\}$. Note that for an arbitrary unit vector $f\in \mathcal{H}$, $I-f\otimes f\in\Delta(\mathcal{H})$. Hence, by assumption we should have \begin{align*} \xi_1(I)-\xi_1(f\otimes f)=I-\lambda Af\otimes Af \in\Delta(\mathcal{H}). \end{align*} But this happens precisely when $\lambda=1$. Now, consider an arbitrary vector $b\in\mathcal{H}$. Then \begin{align*} =1 &\Longleftrightarrow ~~I-b\otimes b \in \Delta(\mathcal{H})\\ &\Longleftrightarrow~~ I-Ab\otimes bA^*\in \Delta(\mathcal{H})\\ &\Longleftrightarrow ~~=1. \end{align*} Hence $\xi_1$ preserves projections of rank one. %\hspace{.14cm}Furthermore as for every unit vector $b\in\mathcal{H}$, $||Ab||=\sqrt{}=1$. %that $A$ is either a unitary or an anti-unitary operator on $\mathcal{H}$.\\ By replacing $\xi_1$ with $\xi_2=A^*\xi_1A$, in the sequel we may assume $\xi_2(F)=F$, for every $F\in\mathcal{FS}(\mathcal{H})$.\\ \textbf{Step 4.} %\item[\rm{Step4}] . $\xi_2$ preserves the difference of $\mathcal{AI}_s(\mathcal{H})$ in both directions, that is, for every $Q,P\in \mathcal{SA}(\mathcal{H})$ \begin{align*} Q-P\in\mathcal{AI}_{s}(\mathcal{H}) \Longleftrightarrow \xi_2(Q)-\xi_2(P) \in \mathcal{AI}_{s}(\mathcal{H}). \end{align*} Let $Q,P\in\mathcal{SA}(\mathcal{H})$ be such that $Q-P\in\mathcal{AI}_{s}(\mathcal{H})$.\hspace{.14cm}Then for some unit vectors $b\in \mathcal{H}$, we have \begin{align*} =1. \end{align*} Set $F=b\otimes b$.\hspace{.14cm}It follows that $(Q-P)-F$ is not invertible.\hspace{.14cm}Hence $Q-(P+F)\in\Delta(\mathcal{H})$ which implies $\xi_2(Q)-\xi_2(P+F)\in\Delta(\mathcal{H})$. On the other hand, since $(P+F)-P$ is rank one then so is $\xi_2(P+F)-\xi_2(P)$.\hspace{.14cm}Therefore, since \begin{align*} \xi_2(Q)-\xi_2(P)=\xi_2(Q)-\xi_2(P+F)+(\xi_2(P+F)-\xi_2(P)), \end{align*} it follows that $\xi_2(Q)-\xi_2(P)\in\mathcal{A}_{s}(\mathcal{H})$. But since by assumption $Q-P$ is invertible, $Q-P\notin\Delta(\mathcal{H})$, which implies $\xi_2(Q)-\xi_2(P)\notin\Delta(\mathcal{H})$. So $\xi_2(Q)-\xi_2(P)\in\mathcal{AI}_s(\mathcal{H})$. Similarly, let $Q,P\in\mathcal{SA}(\mathcal{H})$ be such that $\xi_2(Q)-\xi_2(P)\in\mathcal{AI}_{s}(\mathcal{H})$. Since $\xi_2^{-1}$ satisfies the same properties as $\xi_2$, we conclude $Q-P\in\mathcal{AI}_{s}(\mathcal{H})$.\\ \textbf{Step 5}. $\xi_2(P)=P$ for every $P\in\mathcal{AI}_{s}(\mathcal{H})$. Let $P\in\mathcal{AI}_{s}(\mathcal{H})$. Since $P-0\in\mathcal{AI}_{s}(\mathcal{H})$, it follows from step 4 that $\xi_2(P)=\xi_2(P)-\xi_2(0)\in\mathcal{AI}_{s}(\mathcal{H})$. If $\xi_2(P)\neq P$, then there exists a unit vector $e\in\mathcal{H}$ such that $P^{-1}e\neq\xi_2(P)^{-1}e$, $=1$ while $\neq 1$. It shows that $P-e\otimes e\notin\mathcal{AI}_{s}(\mathcal{H})$ but $\xi_2(P)-e\otimes e=\xi_2(P)-\xi_2(e\otimes e)\in\mathcal{AI}_{s}(\mathcal{H})$, there appears a contradiction. This contradiction shows that $\xi_2(P)=P$.\\ \textbf{Step 6}. $\xi_2(P)=P$ for every $P\in\Delta(\mathcal{H})$. Let $P\in\Delta(\mathcal{H})$.\hspace{.14cm}Then $\xi_2(P)=\xi_2(P)-\xi_2(0)\in\Delta(\mathcal{H})$.\hspace{.14cm}For every $N\in\mathcal{AI}_{s}(\mathcal{H})$, from step 5 we have $\xi_2(N)=N$ and \begin{align*} P-N\in\mathcal{AI}_{s}(\mathcal{H}) \Longleftrightarrow \xi_2(P)-N \in \mathcal{AI}_{s}(\mathcal{H}). \end{align*} Hence, from proposition 2.9 it follows that $\xi_{2}(P)=P$.\\ \textbf{Step 7}. $\xi_2(P)=P$ for every $P\in\mathcal{SA}(\mathcal{H})$. %Since by lemma 2.1, $\mathcal{AS}_0(\mathcal{H})=\mathcal{S}(\mathcal{H})$, if we %show that $\xi_2\mid _{\Delta(\mathcal{H})}$ is additive, then the desired result follows from step 6.\\ Let $P_1,P_2\in \Delta(\mathcal{H})$ be fixed and consider the map $\xi:\mathcal{SA}(\mathcal{H})\rightarrow \mathcal{SA}(\mathcal{H})$ that for every $P\in\mathcal{SA}(\mathcal{H})$ is defined by \begin{align*} \xi(P):=\xi_2(P+P_2)-P_2. \end{align*} Then it follows from previous steps that $\xi$ is bijective, preserves the difference of $\Delta(\mathcal{H})$ in both directions, $\xi(I)=I$ and $\xi(F)=F$ for all finite rank operator $F\in\mathcal{SA}(\mathcal{H})$. Hence, for every $P\in \Delta(\mathcal{H})$, $\xi(P)=P$. In particular, we have \begin{align*} P_1=\xi(P_1)=\xi_2(P_1+P_2)-P_2, \end{align*} that follows $\xi_2(P_1+P_2)=P_1+P_2$. Hence $\xi_2\mid _{\Delta(\mathcal{H})}$ is additive.\\ From step 7, it follows that for every $P\in \mathcal{SA}(\mathcal{H})$, $\xi_2(P)=P$. But from this we get \begin{align*} P=\xi_2(P)=A^*\xi_1(P)A=A^*(\xi(P)-R)A. \end{align*} Hence for every $P\in \mathcal{SA}(\mathcal{H})$ \begin{align*} \xi(P)=APA^*+R, \end{align*} which is the desired result and finishes the proof. %\end{itemize} \end{proof} \begin{proof} (Proof of Theorem 2.4) Assume that $\xi$ is a map satisfying \begin{align*} ~~~\nu(\xi(Q)-\xi(P))=\nu(Q-P).\end{align*} for all $Q,P\in \mathcal{SA}(\mathcal{H})$, and note that $\xi$ is a bijective map satisfying $(8)$, so $\xi$ is of the form\\ $~~~~~~~~~~~~~~~~~~~~~\hspace{2cm}~~~\xi(P)=APA^*+R$,\\ %and note that \\ %$\frac{1}{||T||}=c(T^{-1})=c(\Psi(T^{-1})-\Psi(0))=c(AT^{-1}A^*)=\frac{1} %{||(A^{*})^{-1}TA^{-1}||}$,\\ for all operators $P\in\mathcal{SA}(\mathcal{H})$. So we have \begin{align*} \nu(AQA^*+R-APA^*-R)=\nu(Q-P) \Longrightarrow \nu(A(Q-P)A^*)=\nu(Q-P), \end{align*} and since $\nu$ is $\rho$ or $\tau$, By lemma 2.2 there is an unitary operator $E:\mathcal{H}\rightarrow \mathcal{H}$ and scalar $\lambda$ such that $A=\lambda E$ and $|\lambda \lambda^*|=1$. Thus\\ $~~~~~~~~~~~~~~\hspace{2cm}~~~~~~~~~\xi(P)=\lambda EP(\lambda E)^*+R,$\\ so we have\\ $~~~~~~~~~~~~\hspace{2cm}~~~~~~~~~~~~~~~\xi(P)=EPE^*+R.$\\ %that $|\lambda^2|=1$. ~~~~~~~~~~~~~~~~~~~~~\hspace{4cm}~~~~~~~~~~~\end{proof} %\vspace{.cm} % BibTeX users please use one of %\bibliographystyle{spbasic} % basic style, author-year citations %\bibliographystyle{spmpsci} % mathematics and physical sciences %\bibliographystyle{spphys} % APS-like style for physics %\bibliography{} % name your BibTeX data base % Non-BibTeX users please use \begin{center} \begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bibitem{RefJ1} B. Aupetit, A Primer on Spectral Theory, Springer-Verlag, New York, (1991). \bibitem{RefJ2} B. Aupetit, Propri\'{e}t\'{e}s spectrales des algebres de Banach, Lecture Notes in Mathematics, Springer-Verlag, (1979). \bibitem{RefJ3} B. 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Mashreghi, Nonlinear maps preserving the reduced minimum modulus of operators, Linear %Algebra and its Applications,493, (2016), 426-432. %\item [\rm{[25]}] B. Aupetit, Sur les transformations qui conservent le spectre, In %Banach algebras 97, pages %55–78, de Gruyter, Berlin, 1998. %\end{itemize} \end{thebibliography} \end{center} {\small \noindent{\bf Zynab Izadi} \noindent Department of Mathematics \noindent Phd student of Mathematics \noindent Payamenoor University \noindent P. O. Box 19395-3697, Tehran, Iran \noindent E-mail: $zynab_{-}izadi$@pnu.ac.ir}\\ {\small \noindent{\bf Rahmat Soltani } \noindent Department of Mathematics \noindent Assistant Professor of Mathematics \noindent Payamenoor University \noindent P. O. Box 19395-3697, Tehran, Iran \noindent E-mail: $r_{-}soltani$@pnu.ac.ir}\\ \end{document}