positive element

Let $H$ be a complex Hilbert space. Let $T:H\longrightarrow H$ be a bounded operator in $H$ .

Definition - $T$ is said to be a positive operator if there exists a bounded operator $A:H\longrightarrow H$ such that

T=A^{*}A

where $A^{*}$ denotes the adjoint of $A$ .

Every positive operator $T$ satisfies the very strong condition $\langle Tv,v\rangle\geq 0$ for every $v\in H$ since

\langle Tv,v\rangle=\langle A^{*}Av,v\rangle=\langle Av,Av\rangle=\|Av\|^{2}\geq 0

The converse is also true, although it is not so to prove. Indeed,

Theorem - $T$ is positive if and only if $\langle Tv,v\rangle\geq 0\;\;\;\;\forall_{v\in H}$

0.1 Generalization to $C^{*}$ -algebras

The above notion can be generalized to elements in an arbitrary $C^{*}$ -algebra (http://planetmath.org/CAlgebra).

In what follows $\mathcal{A}$ denotes a $C^{*}$ -algebra.

Definition - An element $x\in\mathcal{A}$ is said to be positive (and denoted $0\leq x$ ) if

x=a^{*}a

for some element $a\in\mathcal{A}$ .

$Remark-$ Positive elements are clearly self-adjoint (http://planetmath.org/InvolutaryRing).

0.2 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 it here as a theorem:

Theorem - Let $x\in\mathcal{A}$ and $\sigma(x)$ denote its spectrum. Then $x$ is positive if and only if $x$ is and $\sigma(x)\subset\mathbb{R}_{0}^{+}$ .

0.3 Square roots

Positive elements admit a unique positive square root.

Theorem - Let $x$ be a positive element in $\mathcal{A}$ . There is a unique $b\in\mathcal{A}$ such that

•

$b$ is positive
•

$x=b^{2}$ .

The converse is also true (with assumptions): If $x$ admits a square root then $x$ is positive, since

x=b^{2}=bb=b^{*}b

0.4 The positive cone

Another interesting fact about positive elements is that they form a proper convex cone (http://planetmath.org/Cone5) in $\mathcal{A}$ , usually called the positive cone of $\mathcal{A}$ . That is stated in following theorem:

Theorem - Let $a, b$ be positive elements in $\mathcal{A}$ . Then

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$a+b$ is also positive
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$\lambda a$ is also positive for every $\lambda\geq 0$
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If both $a$ and $-a$ are positive then $a=0$ .

0.5 Norm closure

Theorem - The set of positive elements in $\mathcal{A}$ is norm closed.