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positive element (Definition)

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

$\displaystyle T=A^*A $
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

$\displaystyle \langle T v , v \rangle = \langle A^*A v , v \rangle = \langle A v , Av \rangle = \Vert Av\Vert^2 \geq 0 $

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}$

Generalization to $ C^*$-algebras

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

$\displaystyle x=a^*a $
for some element $ a \in \mathcal{A}$.

$ Remark -$ Positive elements are clearly self-adjoint.

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 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}^+$.

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 even weaker assumptions): If $ x$ admits a self-adjoint square root then $ x$ is positive, since

$\displaystyle x=b^2=bb=b^*b $

The positive cone

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$.

Norm closure

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



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Other names:  positive
Also defines:  positive operator, positive cone, square root of positive element
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Cross-references: closed, norm, square root, spectrum, normal elements, converse, adjoint, bounded operator, Hilbert space, complex
There are 26 references to this entry.

This is version 5 of positive element, born on 2007-08-27, modified 2007-11-10.
Object id is 9898, canonical name is PositiveElement3.
Accessed 1368 times total.

Classification:
AMS MSC47A05 (Operator theory :: General theory of linear operators :: General )
 47L07 (Operator theory :: Linear spaces and algebras of operators :: Convex sets and cones of operators)
 46L05 (Functional analysis :: Selfadjoint operator algebras :: General theory of $C^*$-algebras)

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The converse by FunctorSalad on 2007-10-06 18:37:58
Hello,
in:
"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}$"

do you assume that T is self-adjoint in the first place, so that <Tv,v> is real (and can therefore be compared to 0)? Or do you mean that [<Tv,v> is real and nonnegative] implies positivity for arbitrary T? (the latter sounds rather strong)
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