Conjugate Homogeneity for Self-Adjoint Operators: Proof and Explanation

[evilpostingmong]

(aT)∗ = $$\bar{a}$$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
$$\bar{a}$$=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>
= $$\bar{λ}$$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.

 

[tiny-tim]

Hi evilpostingmong!

(it's \bar{\lambda} )

… = <Tv,v>
= <v,Tv>
= <v,λv>

Shouldn't that be

= <v,Tv>*
= <v,λv>*
= λ*<v,v>* ? ​

 

[evilpostingmong]

Hi evilpostingmong!

(it's \bar{\lambda} )


Shouldn't that be

= <v,Tv>*
= <v,λv>*
= λ*<v,v>* ? ​

yes it should

 

[HallsofIvy]

But, of course, <v, v> is a positive real number (part of the definition of "inner product") so that is $\lambda<v,v>= \lambda^*<v,v>^*= \lambda^*<v,v>$.