Conjugate Homogeneity for Self-Adjoint Operators: Proof and Explanation

In summary, this conversation discusses the self-adjointness of operators in linear algebra and proves the proposition that every eigenvalue of a self-adjoint operator is a real number. The proof involves using the definition of inner product and the properties of self-adjointness. The key concept is that for a self-adjoint operator, the matrix for T should not be a conjugate transpose of the matrix for T*, which makes sense since the eigenvalues should be real numbers.
  • #1
evilpostingmong
339
0
(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.
 
Last edited:
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  • #2
Hi evilpostingmong! :smile:

(it's \bar{\lambda} :wink:)
evilpostingmong said:
… = <Tv,v>
= <v,Tv>
= <v,λv>

Shouldn't that be

= <v,Tv>*
= <v,λv>*
= λ*<v,v>* ? :redface:
 
  • #3
tiny-tim said:
Hi evilpostingmong! :smile:

(it's \bar{\lambda} :wink:)


Shouldn't that be

= <v,Tv>*
= <v,λv>*
= λ*<v,v>* ? :redface:

yes it should
 
  • #4
But, of course, <v, v> is a positive real number (part of the definition of "inner product") so that is [itex]\lambda<v,v>= \lambda^*<v,v>^*= \lambda^*<v,v>[/itex].
 

1. What is conjugate homogeneity?

Conjugate homogeneity is a mathematical property that describes how a function's value changes when its input is multiplied by a constant. Specifically, a function is said to be conjugately homogeneous if f(ax) = a^k * f(x) for some constant k and for all values of x and a.

2. How is conjugate homogeneity related to homogeneity?

Conjugate homogeneity is a specific type of homogeneity, which is a more general mathematical concept. A function is homogeneous if f(ax) = a^k * f(x) for all values of x and a, but k can be any constant. In contrast, a function is conjugately homogeneous if k is a specific constant for all values of x and a.

3. Can you give an example of a conjugately homogeneous function?

Yes, the function f(x) = x^2 is conjugately homogeneous with k = 2. This can be seen by plugging in values for a and x: f(ax) = (ax)^2 = a^2 * x^2 = a^2 * f(x). This holds true for all values of a and x, satisfying the definition of conjugate homogeneity.

4. How is conjugate homogeneity used in scientific research?

Conjugate homogeneity is a useful property in many areas of science, including physics, economics, and statistics. It is used to model relationships between variables and to simplify equations in mathematical models. In physics, for example, conjugate homogeneity is used to describe the behavior of certain physical systems and to make predictions about their behavior.

5. What are the limitations of conjugate homogeneity?

Conjugate homogeneity is a useful property, but it does have some limitations. For example, it only applies to functions that have a specific form, and not all functions exhibit conjugate homogeneity. Additionally, it may not accurately describe the behavior of a system if other factors are at play, such as non-linear relationships or interactions between variables.

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