- #1
evilpostingmong
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(aT)∗ = [tex]\bar{a}[/tex]T∗ for all a ∈ C and T ∈ L(V,W);
This doesn't make much sense to me. Isn't a supposed to be=x+iy and
[tex]\bar{a}[/tex]=x-iy? Not a fan of complex numbers. And
this proof also confuses me.7.1 Proposition: Every eigenvalue of a self-adjoint operator is real.
Proof: Suppose T is a self-adjoint operator on V. Let λ be an
eigenvalue of T, and let v be a nonzero vector in V such that Tv = λv.
Then
λllvll2 = <λv,v>
= <Tv,v>
= <v,Tv>
= <v,λv>
= [tex]\bar{λ}[/tex]llvll2 that 955 is lambda
so I am guessing that for <Tv, v> to=<v, Tv>, the eigenvalue
should not have a conjugate of the form x-iy but just a real number.
I mean that makes sense I guess. In other words, the matrix for T* should not be a conjugate
transpose since the matrix of T must=the matrix of T* for self adjointness.
This doesn't make much sense to me. Isn't a supposed to be=x+iy and
[tex]\bar{a}[/tex]=x-iy? Not a fan of complex numbers. And
this proof also confuses me.7.1 Proposition: Every eigenvalue of a self-adjoint operator is real.
Proof: Suppose T is a self-adjoint operator on V. Let λ be an
eigenvalue of T, and let v be a nonzero vector in V such that Tv = λv.
Then
λllvll2 = <λv,v>
= <Tv,v>
= <v,Tv>
= <v,λv>
= [tex]\bar{λ}[/tex]llvll2 that 955 is lambda
so I am guessing that for <Tv, v> to=<v, Tv>, the eigenvalue
should not have a conjugate of the form x-iy but just a real number.
I mean that makes sense I guess. In other words, the matrix for T* should not be a conjugate
transpose since the matrix of T must=the matrix of T* for self adjointness.
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