Operators over real vector spaces are self adjoint right?

From Linear Algebra Done Right:

"Positive Operators

An operator T ∈ L(V ) is called positive if T is self-adjoint and ⟨T v , v ⟩ ≥ 0

for all v ∈ V. Note that if V is a complex vector space, then the condition that T be self-adjoint can be dropped from this definition (by 7.3)."

If V is a real vector space then <Tv,v>=<v,Tv> so T is automatically self-adjoint. Doesn't this mean that the condition that T is self-adjoint can be dropped whatever the situation?

Re: Operators over real vector spaces are self adjoint right?

No responses? I thought this was a really simple question

Re: Operators over real vector spaces are self adjoint right?

Consider with the usual inner product with then for all but is not self-adjoint because (the equality holds only when for some )

Re: Operators over real vector spaces are self adjoint right?

In general, if with finite dimensional euclidean space, is basis of , is the Gram matrix with respect to and is the matrix of with respect to , is easy to prove that is self adjoint iff . Choosing for example orthonormal we have so, any non symmetric matrix represents a non self adjoint operator.

Re: Operators over real vector spaces are self adjoint right?

Ok I see my mistake:

<Tv,w>=<v,T*w> which doesn't necessarily equal <v,Tw>

Now I have a different question, how do I show TT* is self adjoint?

Re: Operators over real vector spaces are self adjoint right?

Quote:

Originally Posted by

**durrrrrrrr** Now I have a different question, how do I show TT* is self adjoint?

One way: use and .