Normal, self-adjoint and positive definite operators

  • Thread starter steinmasta
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  • #1
I have two questions:

1. Let V be a finite dimensional inner product space. Show that if U: V--->V is self-adjoint and T: V---->V is positive definite, then UT and TU are diagonalizable operators with only real eigenvalues.

2. Suppose T and U are normal operators on a finite dimensional complex inner product space V such that TU = UT

a) Show that UT* = T*U
b) show that there is an orthonormal basis for V consisting of vectors that are eigenvectors for both T and U
 

Answers and Replies

  • #2
What have you tried? We cannot help you before you have shown your work.
 
  • #3
What do you know about positive, definite operators, and self-adjoint operators? Do you happen to know what field you're working with?
 
  • #4
Regarding 1, I have tried to show that UT is self-adjoint with respect to the inner product <x,y>' = <T(x),y> but I've had trouble unwinding definitions.

Regarding 2a, I tried to show that if U commutes with T, then U commutes with T*. This corollary to the spectral theorem is useful: If F = C, T is normal iff T* = g(T) for some polynomial g, but I can't seem to make it work.

Regarding 2b, I am not sure where even to begin.
 
  • #5
2
0
woah woah woah, don't forget the J in this if you don't know about that then you need a degree and if you've got one....
you need another one
 
  • #6
1. Let V be a finite dimensional inner product space. Show that if U: V--->V is self-adjoint and T: V---->V is positive definite, then UT and TU are diagonalizable operators with only real eigenvalues.

Well you know that U:V--->V is self-adjoint, and therefore has an orthonormal basis of eigenvectors... you know that T:V---->V is Positive, and therefore also self-adjoint, and must also have an orthonormal basis of eigenvectors.

I believe you could also show that UT and TU are Self-adjoint operators from Inner Product algebra.

<Uv,u>=<v,Uu>
<Tv,u>=<v,Tu> and <Tv,v> > 0 for all v in V.
<UTv,u>=<v,(UT)*u>
<UTv,u>=<v,T*U*u>
<UTv,u>=<v,TUu>

(UT)* = T*U*=TU
(TU)*= U*T* = UT

Since U and T are both self-adjoint, TU=UT (Chose the orthonormal basis of eigenvectors, it's the product of the diagonal entries of the respective matrices, and diagonal matrices commute)

I hope that helps. (I might be mistaken, so if anything seems fishy let me know. I'm a freshman and just finished this stuff like three weeks ago... so yeah ;p)
 

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