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Normal, self-adjoint and positive definite operators

  1. Dec 15, 2008 #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
     
  2. jcsd
  3. Dec 16, 2008 #2
    What have you tried? We cannot help you before you have shown your work.
     
  4. Dec 16, 2008 #3
    What do you know about positive, definite operators, and self-adjoint operators? Do you happen to know what field you're working with?
     
  5. Dec 16, 2008 #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.
     
  6. Dec 16, 2008 #5
    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
     
  7. Dec 17, 2008 #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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