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Kernel of the adjoint of a linear operator

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Let V be an inner product space and T:V->V a linear operator. Prove that if T is normal, then T and T* have the same kernel (T* is the adjoint of T).

    2. Relevant equations



    3. The attempt at a solution
    Let us assume x is in the kernel of T. Then, TT*x =T*Tx = T*0= 0. So T*x is in the kernel of T but this doesn't mean T*x = 0 and I get stuck here. Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 29, 2009 #2

    Dick

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    You aren't using that T* is the adjoint. Look at x*(T*T)x and x*(TT*)x.
     
  4. Apr 29, 2009 #3
    I think I found out how to do this. Are my steps right?
    Let x be an element in the kernel of T. Since TT*x = T*Tx = T*0=0, (TT*x,x) = 0 where (,) indicates the inner product.
    Then, (TT*x,x) = (T*x,T*x) = ||T*x||^2 = 0 so T*x = 0 and x is in the kernel of T*. The other way can be done similarly.

    BTW, what does "x*" indicate in x*(T*T)x? I don't understand...
     
  5. Apr 29, 2009 #4

    Dick

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    x* is the adjoint of the vector x. So x*(TT*)x is (x,(TT*)x). Yes, that's it. Show (Tx,Tx)=(T*x,T*x)=||Tx||^2=||T*x||^2.
     
  6. Apr 29, 2009 #5
    O.K., thank you.
     
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