Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Adjoint question

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    V: complex inner product space with adjoint T*

    Suppose that < T( x ) , x > = 0 for all x in V, then T is the zero transformation.


    3. The attempt at a solution

    < T( x ) , x > = < x, T*(x ) > = 0
    0 = < x, 0 > = < 0, x >
    < x, T*(x ) > = 0 = < x, 0 >

    if < x , y > = < x, z> , then y = z
    so T*(x ) = 0 for any x, which means T* is the zero transformation, which implies that T is the zero transformation..

    is this okay? thanks
     
  2. jcsd
  3. Jul 11, 2010 #2

    hunt_mat

    User Avatar
    Homework Helper

    Doesn't look quite right. Look at applying T to the basis elements [tex]<Te_{i},e_{j}>=0[/tex] and you should get that T is the zero operator. Another hint is write
    [tex]
    Te_{i}=\sum_{k=1}^{N}\alpha_{ik}e_{k}
    [/tex]
     
  4. Jul 11, 2010 #3
    thanks, I'm less interested in the right answer, but I'd like to know why the steps I posted above were wrong? thanks
     
  5. Jul 11, 2010 #4
    <x , y> = < x , z> does not by itself imply y=z. Consider, for example, the Euclidean inner product. The vector x can be normal to two different vectors y and z, and so <x,y>=<x,z>=0, but y [tex] \neq [/tex] z. I hope this helps.
     
  6. Jul 12, 2010 #5
    you're right, the implication in full is : if < x , y > = < x, z > for all x in V, then y = z
    in this case, x may be arbitrarily chosen, but 'y' ( a.k.a T*( x ) ) is fixed for each x , so we cannot consider < x , y > = < x, z > for any x, since y is never the same vector
    thanks for reading
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook