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Inner product question.

  1. Jun 23, 2007 #1
    let V be a vector space with inner product, and T:V->V linear trans.
    then for V on R, prove that for every v in V, <v,T(v)>=0 iff T*=-T.

    now i got so far that: from <v,T(v)>=0 we have <v,(T+T*)(v)>=0 for every v, here im stuck, i guess if it's for every v, if i were to write (T+T*)(v)=av for some scalar a, then i would get: a<v,v>=0 for every v, so a=0, and we get what we wanted, but im not sure that T and T* are eigen functions i.e in the form ive given.
    any pointers?
     
  2. jcsd
  3. Jun 23, 2007 #2

    StatusX

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    The adjoint is defined by the requirement:

    <v,Tw>=<T*v,w>
     
  4. Jun 23, 2007 #3
    yes i know, and i got to this equation by getting <T*(v),v>=0 and it's symmetric bacuase it's on R, but still i don't see how your remark helps me here?
     
  5. Jun 23, 2007 #4

    StatusX

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    Um... then use T*=-T, and then there's one more step. I'm really doing nothing but copying definitions for you.
     
  6. Jun 23, 2007 #5
    but T* doesnt necessarily equals -T, or am i missing something here?
    T* is defined as a linear trans between V to V that satisifes <v,T(w)>=<T*(w),v> how from this definition you conclude that T*=-T?

    and i need to prove that if <v,Tv>=0 for every v in V then T*=-T, if i were using the antecendent that i werent proving were i?
     
  7. Jun 23, 2007 #6

    Hurkyl

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    One very useful fact about quadratic forums Q, such as
    Q(v) := <v, T(v)>,​
    is that you can construct a bilinear form by considering
    Q(v+w).​

    Maybe this is helpful?
     
  8. Jun 23, 2007 #7
    doh, i shouldv'e know that it has something to do with this.
    anyway, if Q(v)=<v,Tv>=0
    then
    <(T*+T)(v),(T*+T)(v)>=1/2(Q((T+T*)v)-Q((T+T*)(v))-Q((T*+T)(v)))=0
    so (T*+T)(v)= for every v in V.
    is this good enough?

    thanks hurkyl.
     
  9. Jun 23, 2007 #8

    StatusX

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    Sorry, I didn't notice the second f in iff. I guess you already did the direction I was thinking of. I'm not sure what you're doing in your last post though. You have:

    <(T*+T)(v),(T*+T)(v)>=1/2(Q((T+T*)v)-Q((T+T*)(v))-Q((T*+T)(v)))=0

    which, given your definition of Q, translates to:

    <(T*+T)(v),(T*+T)(v)>=1/2(<(T+T*)v,T(T+T*)v>-<(T+T*)v,T(T+T*)v>-<(T+T*)v,T(T+T*)v>)

    I don't think that's right. I'm pretty sure Hurkyl wants you to use a v and a w in order to get a relation between <v,Tw> and <Tv,w>
     
  10. Jun 24, 2007 #9
    but q(v) is defined as q(v)=f(v,v) for some symmetric bilinear form f.
    so if i define: q(v):=<v,Tv>
    so: q(v+w)=<v+w,Tv+Tw>=<v,Tv>+<w,Tw>+<v,Tw>+<w,Tv>=
    =<v+w,T(w+v)>=0
    so from this i can <(T+T*)v,(T+T*)v>=1/2(q(2(T+T*)(v))-2q((T+T*)(v))
    q((T+T*)(v))=0 and so is the first term.
    isn't this correct?
     
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