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positive element

positive operator, positive cone, square root of positive element
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Mathematics Subject Classification

46L05 no label found47L07 no label found47A05 no label found


"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)

I mean the latter: [<Tv,v> is real and nonnegative] implies positivity for arbitrary T.

Self-adjointness comes as a consequence. Notice that

<(T-T*)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 T-T*= 0 , i.e. T is self-adjoint.

- Self-adjointness is really more than a consequence. In order to prove positivity we need to prove self-adjointness 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 non-negative spectrum].

Hope this helps..

P.S. - I would dare to say that all of this is false for real Hilbert spaces.

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).

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