PositiveElement3 2007-08-27 20:55:59 2007-11-10 19:39:36 Definition positive operator positive cone square root of positive element % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $H$ be a complex Hilbert space. Let $T:H \longrightarrow H$ be a bounded operator in $H$. {\bf Definition -} $T$ is said to be a {\bf positive operator} if there exists a bounded operator $A: H \longrightarrow H$ such that \begin{displaymath} T=A^*A \end{displaymath} where $A^*$ denotes the adjoint of $A$. Every positive operator $T$ satisfies the very strong condition $\langle T v , v \rangle \geq 0$ for every $v \in H$ since \begin{displaymath} \langle T v , v \rangle = \langle A^*A v , v \rangle = \langle A v , Av \rangle = \|Av\|^2 \geq 0 \end{displaymath} The converse is also true, although it is not so \PMlinkescapetext{simple} to prove. Indeed, {\bf Theorem -} $T$ is positive if and only if $\langle Tv, v \rangle \geq 0 \;\;\;\;\forall_{v \in H}$ \subsection{Generalization to $C^*$-algebras} The above notion can be generalized to elements in an arbitrary \PMlinkname{$C^*$-algebra}{CAlgebra}. In what follows $\mathcal{A}$ denotes a $C^*$-algebra. {\bf Definition -} An element $x \in \mathcal{A}$ is said to be {\bf positive} (and denoted $0 \leq x$) if \begin{displaymath} x=a^*a \end{displaymath} for some element $a \in \mathcal{A}$. $Remark -$ Positive elements are clearly \PMlinkname{self-adjoint}{InvolutaryRing}. \subsection{Positive spectrum} It can be shown that the positive elements of $\mathcal{A}$ are precisely the normal elements of $\mathcal{A}$ with a positive spectrum. We \PMlinkescapetext{state} it here as a theorem: {\bf Theorem -} Let $x \in \mathcal{A}$ and $\sigma(x)$ denote its spectrum. Then $x$ is positive if and only if $x$ is \PMlinkescapetext{normal} and $\sigma(x)\subset \mathbb{R}_{0}^+$. \subsection{Square roots} Positive elements admit a unique positive square root. {\bf Theorem -} Let $x$ be a positive element in $\mathcal{A}$. There is a unique $b \in \mathcal{A}$ such that \begin{itemize} \item $b$ is positive \item $x=b^2$. \end{itemize} The converse is also true (with \PMlinkescapetext{even weaker} assumptions): If $x$ admits a \PMlinkescapetext{self-adjoint} square root then $x$ is positive, since \begin{displaymath} x=b^2=bb=b^*b \end{displaymath} \subsection{The positive cone} Another interesting fact about positive elements is that they form a \PMlinkname{proper convex cone}{Cone5} in $\mathcal{A}$, usually called the {\bf positive cone} of $\mathcal{A}$. That is stated in following theorem: {\bf Theorem -} Let $a, b$ be positive elements in $\mathcal{A}$. Then \begin{itemize} \item $a+b$ is also positive \item $\lambda a$ is also positive for every $\lambda \geq 0$ \item If both $a$ and $-a$ are positive then $a=0$. \end{itemize} \subsection{Norm closure} {\bf Theorem -} The set of positive elements in $\mathcal{A}$ is norm closed.