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LA - Trace - Is this an elegant solution or an incorrect one? (Breif glance needed)

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose V is an inner product space and [tex]T\in L(V)[/tex]. Prove that if [tex]\left\|T^{*}v\right\|[/tex][tex]\leq[/tex][tex]\left\|Tv\right\|[/tex] then T is Normal.

    Sorry for being bothersome; this is the first year I've ever written proofs so I'm a bit sketchy about them.


    2. Relevant equations



    3. The attempt at a solution Define an Inner Product on V [tex]\left\langle Sv, Tv\right\rangle[/tex] = [tex]\ Trace(ST^{*})[/tex].
    Then [tex]\left\|T^{*}v\right\|[/tex][tex]\leq[/tex][tex]\left\|Tv\right\| = Trace (TT^{*}) \leq Trace( T^{*}T)[/tex].
    However, [tex]\ Trace (TT^{*}) = Trace (T^{*}T)[/tex]. Which implies [tex]\left\|T^{*}v\right\|[/tex] = [tex]\left\|Tv\right\|[/tex], and therefore T is normal

    I worked up another solution, which I'm almost sure is correct, but if this one does indeed check out I'd prefer to use it ;p. It just seems wrong to me for some reason; because it implies that basically all linear maps are normal; but I can't place what I did incorrectly.
     
    Last edited: Nov 16, 2008
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  3. Nov 16, 2008 #2

    HallsofIvy

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    Re: LA - Trace - Is this an elegant solution or an incorrect one? (Breif glance neede

    That is not an inner product on V, it is an inner product on L(V).

    What justification do you have for [tex]\left\|T^{*}v\right\|\leq\left\|Tv\right\|[/tex]?

     
  4. Nov 16, 2008 #3
    Re: LA - Trace - Is this an elegant solution or an incorrect one? (Breif glance neede

    We assumed that T*v was less than or equal to Tv; in order to show that T had to be normal.

    An inner product on L(V), right. I think that destroys my "proof." Since it doesn't make sense to take the "Trace" of a vector.

    I have another breif question; If all the eigenvalues of a particular operator are equal to 0, does that imply the operator is equal to 0? or just Nilpotent? The other proof I worked on shows that TT*-T*T is positive, and since Trace of that operator is 0, It must have all zeros along the diagonal. Then I sort of hand-wavily assumed I could say TT*-T*T = 0, and therefore TT* = T*T and T is a normal operator. Is there any solid justification for that?

    EDIT: Nevermind I think I figured it out. T is Positive, so it's also self-adjoint; which means V has a basis of eigenvectors, but since all the eigenvalues are 0, TT*-T*T must be the zero map.

    Hurray!
     
    Last edited: Nov 16, 2008
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