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Math Help - Operators over real vector spaces are self adjoint right?

  1. #1
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    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?
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    Re: Operators over real vector spaces are self adjoint right?

    No responses? I thought this was a really simple question
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    Re: Operators over real vector spaces are self adjoint right?

    Consider T:\mathbb{R}^2 \rightarrow \mathbb{R}^2 with the usual inner product with T(x_1,x_2)=(x_2,-x_1) then \langle x,Tx \rangle =0 for all x\in \mathbb{R} ^2 but T is not self-adjoint because \langle y,Tx \rangle = y_1x_2-x_1y_2 \neq y_2x_1-y_1x_2 = \langle Ty,x \rangle (the equality holds only when y=\lambda x for some \lambda)
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Operators over real vector spaces are self adjoint right?

    In general, if T\in\textrm{End}(E) with (E,<,>) finite dimensional euclidean space, B=\{u_1,\ldots,u_n\} is basis of E, G is the Gram matrix with respect to B and A is the matrix of T with respect to B, is easy to prove that T is self adjoint iff GA=A^tG. Choosing for example B orthonormal we have G=I so, any non symmetric matrix represents a non self adjoint operator.
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    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?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Operators over real vector spaces are self adjoint right?

    Quote Originally Posted by durrrrrrrr View Post
    Now I have a different question, how do I show TT* is self adjoint?
    One way: use (T^{*})^*=T and (R\circ S)^*=S^*\circ R^*.
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