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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
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
The converse is also true, although it is not so simple to prove. Indeed,
Theorem - $T$ is positive if and only if $\langle Tv, v \rangle \geq 0 \;\;\;\;\forall_{v \in H}$
The above notion can be generalized to elements in an arbitrary $C^*$ -algebra.
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
for some element $a \in \mathcal{A}$ .
$Remark -$ Positive elements are clearly self-adjoint.
It can be shown that the positive elements of $\mathcal{A}$ are precisely the normal elements of $\mathcal{A}$ with a positive spectrum. We state 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 normal and $\sigma(x)\subset \mathbb{R}_{0}^+$ .
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 even weaker assumptions): If $x$ admits a self-adjoint square root then $x$ is positive, since
Another interesting fact about positive elements is that they form a proper convex cone 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
- $a+b$ is also positive
- $\lambda a$ is also positive for every $\lambda \geq 0$
- If both $a$ and $-a$ are positive then $a=0$ .
Theorem - The set of positive elements in $\mathcal{A}$ is norm closed.
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