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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
$\langle Tv,v\rangle=\langle A^{*}Av,v\rangle=\langle Av,Av\rangle=\Av\^{2}\geq 0$ 
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.
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 selfadjoint.
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 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 $\sigma(x)\subset\mathbb{R}_{{0}}^{+}$.
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 even weaker assumptions): If $x$ admits a selfadjoint 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 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 $a=0$.
0.5 Norm closure
Mathematics Subject Classification
46L05 no label found47L07 no label found47A05 no label found Forums
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The converse
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 selfadjoint 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)
Re: The converse
Hi,
I mean the latter: [<Tv,v> is real and nonnegative] implies positivity for arbitrary T.
Selfadjointness comes as a consequence. Notice that
<(TT*)v,v> = <Tv,v>  <T*v,v> = <Tv,v>  <v,Tv> =
= <Tv,v>  conjugate{<Tv,v>} = <Tv,v>  <Tv,v> = 0
for all vectors v in H.
Since we are assuming that the Hilbert space is complex, we can conclude that TT*= 0 , i.e. T is selfadjoint.
 Selfadjointness is really more than a consequence. In order to prove positivity we need to prove selfadjointness first (at least in the proofs I'm thinking about), so your question makes every sense.
To prove positivity you can either use the spectral theorem for normal operators (general version) or the theorem (stated in the entry) about the positive spectrum: T is positive iff [T is normal and has nonnegative spectrum].
Hope this helps..
P.S.  I would dare to say that all of this is false for real Hilbert spaces.
Re: The converse
Thanks for your answer; that is a nice simple proof.
Yes, for real Hilbert spaces it is false, the standard counterexample being R^2 with T a rotation by pi/2. <Tv,v> = 0 for all v but T is not self adjoint (T=T^* even).