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Normal proof

  1. Jul 20, 2013 #1

    Zondrina

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    Normal Operator Proof

    1. The problem statement, all variables and given/known data

    Prove an operator ##T \in L(V)## is normal ##⇔ ||T(v)|| = ||T^*(v)||##.

    2. Relevant equations

    (1) ##T \in L(V)## is normal if ##TT^*= T^*T##.

    (2) If T is a self-adjoint operator on V such that ##<T(v), v> = 0, \space \forall v \in V##, then ##T=0##.

    3. The attempt at a solution

    ##"\Rightarrow"## Assume ##T## is normal (1) :

    ##TT^*= T^*T##
    ##TT^* - T^*T = 0##

    Now using (2) we can write :

    ##<(TT^* - T^*T)(v), v> = 0, \space \forall v \in V##

    Using some inner product rules yields :

    ##<T^*T(v), v> = <TT^*(v), v>, \space \forall v \in V##
    ##||T(v)||^2 = ||T^*(v)||^2, \space \forall v \in V##
    ##||T(v)|| = ||T^*(v)||, \space \forall v \in V##


    ##"\Leftarrow"## : The proof will be exactly as above, except I start at ##||T(v)|| = ||T^*(v)||## and I finish at ##TT^*= T^*T## I believe?
     
    Last edited: Jul 20, 2013
  2. jcsd
  3. Jul 22, 2013 #2
    You cannot use (2) here since T=0 is not the assumption rather the result.
     
  4. Jul 22, 2013 #3

    Zondrina

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    This confused me a bit, I used (2) because ##TT^* - T^*T## is a self adjoint operator.

    Using the fact it is self adjoint, and that ##TT^* - T^*T = 0##, I could choose any vector in ##V## and still get a result of zero.
     
  5. Jul 22, 2013 #4
    The result is right, what I'm pointing out is the reason for why the result holds. From what I understand, you didn't use (2) to deduce that ##<(TT^* - T^*T)(v), v> = 0, \space \forall v \in V##
     
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