Dick said:You are going to have to do a lot better than that before I'll even start reading it. Who cares whether T*T is self-adjoint? That has nothing to do with the problem. Try and make a linear progression between what you are assuming and what you are trying to prove. Just this once, for my sake, ok? Start with Tv=cv. Assume T=T*. Show me c is real. And nothing else. Write the complex conjugate of c as (c*), not -c. And do it using a minimal number of digressions. Please, please?
Dick said:I appreciate the attempt at brevity, thanks. And I do appreciate it, really. But you are still somehow missing the point. Try using <Tv,v>=<v,(T*)v>=<v,Tv> and that <v,cv>=c<v,v> and <cv,v>=(c*)<v,v>. You know that <ax,x>=(a*)<x,x>, right? If c=(c*) what does that mean about c?
Dick said:You have a really odd way of expressing yourself. I know what you mean by that, but probably few other people in the world would. Here's what you meant to say: "if c=(c*) then the imaginary part of c is zero, so c is real".
Dick said:It's not that funny. What is that supposed to prove? I'm regretting, as I have before, even responding to this thread.
evilpostingmong said:its supposed to prove the other direction that if all eigenvalues
are real, then T=T*.
For the other direction, suppose Tv=av with a being real. Now
<Tv, v>=<av, v>=a<v, v>. And <v, T*v>=<v, a*v>=a*<v, v>.
Since a*=a, a<v, v>=a*<v, v>. Thus <v, T*v>=<v, a*v>=a*<v, v> =<a*v, v>=<av, v>=<Tv, v>.
is this right?
Dick said:It's wrong (in that it abuses '*' - for a general complex number c, <cv,v>=(c*)<v,v>). Since you are assuming a=(a*) you can, I suppose, move the '*'s around. The big problem is that it's just factoring 'a' in and out of a product over and over again. How is that supposed to prove T=T*? Your 'thus' concludes <v,T*v>=<Tv,v>. That doesn't prove T=T*. How can you not see that? What might a proof that T=T* look like?
evilpostingmong said:Ok assume the eigenvalue a is real. And let Tv=av for v is an eigenvector of T.
Now we have <Tv, v>=<v, T*v>. Or <av, v>=<v, a*v>. Since a is real, a=a*.
Thus <v, a*v>=<v, av>=<v, Tv>. And since <Tv, v>=<v, T*v>, <Tv, v>=<v, Tv>.
no factoring a or a* .
Dick said:It's still vacuous. Wouldn't a proof that T=T* involve showing T(x)=T*(x) for any vector x? I can't even see where you clearly stated that T(v)=T*(v) for your eigenvector v. And even if you had, how could T(v)=T*(v) for a single vector v tell you T=T* for all vectors?
evilpostingmong said:well ok Let v be any vector. Tv=av and T*v=a*v. Thus Tv-T*v=av-a*v
Or (T-T*)v=av-a*v. Since a=a*, (T-T*)v=av-av=0. Therefore (T-T*)v=0
and since (T-T*)v=0 , T must=T*.
Dick said:That's better. So Tv=(T*)v. The problem now is that not ALL vectors are eigenvectors.
evilpostingmong said:Ok let x be any vector in V. Now <Tx, x>=<x, T*x>. We represent T with a
a matrix. Since T "has" eigenvectors, T must be diagonizable. Therefore (after reducing the matrix to a diagonal one, note that TT represents T*) TT=T since all eigenvalues along T's diagonal are real and all real diagonal matrices
equal their transposes.
oh my bad...we're dealing with a complex space. So T's diagonal matrix has scalars of the form a+bi along its diagonal and T*'s diagonal matrix has scalars ofDick said:Ok. Except T isn't diagonalizable because it "has" eigenvectors. T is diagonalizable because it's normal. T* isn't the transpose of T, it's the complex conjugate of the transpose. So for any diagonal element of T, T_kk, T_kk=(T_kk)*. That tells you T_kk is real.
A normal operator is considered self-adjoint if it is equal to its own adjoint, meaning that it has the same matrix representation as its Hermitian conjugate. In other words, its eigenvalues and eigenvectors are real and orthogonal, respectively.
Proving that a normal operator is self-adjoint is important because it guarantees that the operator has real eigenvalues and orthogonal eigenvectors. This allows for a simpler and more efficient analysis of the operator's properties and behavior.
To prove that a normal operator is self-adjoint, we must show that it is equal to its adjoint. This can be done by using the definition of the adjoint and the properties of normal operators, such as commutativity and the spectral theorem.
No, a non-normal operator cannot be self-adjoint. This is because the defining property of a self-adjoint operator is that it is equal to its adjoint, and a non-normal operator does not have this property.
Yes, there are many real-world applications of self-adjoint normal operators. They are commonly used in quantum mechanics, signal processing, and differential equations, among other fields. They also have important implications in areas such as quantum computing and linear algebra.