Normal operators

1. Jul 30, 2009

evilpostingmong

1. The problem statement, all variables and given/known data
Prove that a normal operator on a complex inner-product space
is self-adjoint if and only if all its eigenvalues are real.

2. Relevant equations

3. The attempt at a solution
Let c be an eigenvalue. Now since T=T*, we have
<TT*v, v>=<v, TT*v> if and only if TT*v=cv on both sides and not -cv (-c is the complex conjugate of c made possible
by c being a complex number) on one side and cv on the other side. Therefore c must be real.

Last edited: Jul 30, 2009
2. Jul 30, 2009

evilpostingmong

For the other direction, suppose Tv=av with a being real. Now
<Tv, v>=<av, v>=a<v, v>=<v, av>. Now since Tv=av, <v, av>=<v, Tv>. This shows that <Tv, v>=<v, Tv> or Tv=T*v=av.
Now <T*Tv, v>=<T*av, v>=<T*va, v>=<a2v, v>=<v, a2v>
=<v, T*Tv>. Therefore, T*T is self adjoint.

3. Jul 30, 2009

Dick

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?

4. Jul 30, 2009

evilpostingmong

Ok Tv=cv. T=T*. Therefore <Tv, v>=<v, T*v>. Now <cv, v>=<v, T*v> if and only if T*v=cv.
Therefore c must be real. If c were to be complex, then T*v=c*v and c*v=/=cv.

Last edited: Jul 30, 2009
5. Jul 30, 2009

Dick

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?

6. Jul 30, 2009

evilpostingmong

c* is not a complex conjugate. c* and c are the same. I got mixed up thinking that the little star above the scalar
was used to show that it is a complex conjugate.

7. Jul 30, 2009

Dick

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". The * does mean complex conjugate. What do YOU think complex conjugate means?

Last edited: Jul 30, 2009
8. Jul 30, 2009

evilpostingmong

:rofl: I know.
For the other direction, suppose Tv=av with a being real. Now
<Tv, v>=a<v, v>=allvll2. And <v, T*v>=<v, v>a*=a*llvll2
Since a*=a, allvll2=a*llvll2 Thus a*llvll2=a*<v, v>
=<a*v, v>=<av, v>=<Tv, v>.

Last edited: Jul 30, 2009
9. Jul 30, 2009

Dick

It's not that funny. What is that supposed to prove? I'm regretting, as I have before, even responding to this thread.

10. Jul 30, 2009

evilpostingmong

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?

Last edited: Jul 31, 2009
11. Jul 31, 2009

evilpostingmong

hey could someone tell me whether or not im right or wrong>

12. Jul 31, 2009

Dick

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???

13. Jul 31, 2009

evilpostingmong

ugh god your right, meant <v, Tv>=<Tv, v>. ok ill fix this. cant believe I missed
that. :yuck:

14. Jul 31, 2009

evilpostingmong

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* .

Last edited: Jul 31, 2009
15. Jul 31, 2009

Dick

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?

16. Jul 31, 2009

evilpostingmong

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*.

17. Jul 31, 2009

Dick

That's better. So Tv=(T*)v. The problem now is that not ALL vectors are eigenvectors.

18. Jul 31, 2009

evilpostingmong

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.

19. Jul 31, 2009

Dick

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.

20. Jul 31, 2009

evilpostingmong

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 of
the form a-bi (since T* is the complex conjugate of the transpose) along its diagonal. But in our case, b=0. So T's diagonal matrix=T*'s.

Last edited: Jul 31, 2009