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Adjoint operators

  1. Jul 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that if V is a real inner-product space, then the set
    of self-adjoint operators on V is a subspace of L(V).

    2. Relevant equations



    3. The attempt at a solution
    Let M be the matrix representing T. Since we are dealing with real numbers, and T is self-adjoint, T=T* so M=MT.
    Let u and w be vectors. Now <Mu, w>=<u, MTw>. Since M=MT, <u, MTw>=<u, Mw>.
    If M=MT, M is an nxn matrix. And since M and MT are nxn, M and MT map a vector from an n-dimensional space V back to an n-dimensional space V. Since nxn matrices can only multiply into n-tall vectors, u and w must be n-tall vectors. Therefore u and w must be in V. Since M and MT are nxn matrices and can map an n-tall vector from V to an n-dimensional space (V), the set of self adjoint operators (which M and M^T are a part of)are within the set of L(V).
     
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  3. Jul 22, 2009 #2

    Office_Shredder

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    The set of self adjoint operators is obviously contained within L(V) by definition of self adjoint... you don't need to do that matrix mumbo jumbo to prove it. What the question is asking is to show it's a subspace which you haven't done
     
  4. Jul 22, 2009 #3

    Office_Shredder

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    Look back at your post. Did you prove:
    1) 0 is self adjoint
    2) If T and S are self adjoint, T+S is self adjoint
    3) If T is self adjoint, a real, then aT is self adjoint.

    It looks like you tried to prove T* is linear or something. That's not what you need to do. L(V) is a vector space of linear functions from V to V. Show the set of self adjoint linear functions is a vector subspace
     
  5. Jul 22, 2009 #4
    Ok Let w and k be vectors in V. To show 0 is a self adjoint transformation, <0w, k>=<w, 0k>=<0,k>=<w, 0>=0. Since <w, 0k>=<0w, k>, 0 is self adjoint.

    Now let T and S be self adjoint so that <w, Tk>=<Tw, k> and <w, Sk>=<Sw, k>.
    Now <(S+T)w, k>=<Sw+Tw, k>. If S*=S and T*=T, S*+T*=S+T therefore
    S+T=(S+T)*. So S+T is self adjoint

    I'll do the third, just want to know if I have it here.
    It kinda gets me how you can span make a span out of operators. I mean unless you
    represent operators as matrices and put those in a span.
     
  6. Jul 22, 2009 #5

    Office_Shredder

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    Have you proven that S*+T*=(S+T)*? Likely what's expected is something more like

    <(S+T)w,k> = <Sk+Tk,w> = <Sk,w> + <Tk,w> = <k,Sw> + <k,Tw> = <k,Sw+Tw> = <k,(S+T)w>

    but assuming you have S*+T* = (S+T)* then you're correct
     
  7. Jul 22, 2009 #6
    Well yeah I figured that since S is self adjoint and T is self adjoint, and were not
    dealing with the complex space, S*=S and T*=T. So S*+T*=S+T because
    S=S* and T=T*. And (S+T)* =S*+T* is an identity, I apologize, as I didn't know I
    actually had to prove (S+T)*=S*+T* since I didn't think that it the main aim of the proof.
    I'm doing the third one.
     
  8. Jul 22, 2009 #7

    Office_Shredder

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    Even thought S*+T* = S+T, without knowing that S*+T*=(S+T)* you can't conclude (S+T)*=S+T
     
  9. Jul 22, 2009 #8
    3) If T is self adjoint, a real, then aT is self adjoint.
    Now using w and k in V, we have <Tw, k>=<w, Tk>=<T*w, k>=<w, T*k>.
    Now <aw, k>=<w, ak>=a<w,k>. If <Tw, k>=<w,Tk> then a<Tw, k>=a<w, Tk>.
    Or <aTw,k>=<w,aTk>.
     
    Last edited: Jul 22, 2009
  10. Jul 23, 2009 #9
    am i right?:cool:
     
  11. Jul 23, 2009 #10

    Office_Shredder

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    Looks good to me
     
  12. Jul 23, 2009 #11
    Thanks! But here's part two. This one is similar to the first.

    Show that if V is a complex inner-product space, then the
    set of self-adjoint operators on V is not a subspace of
    L(V).

    I think that this has to do with the fact that say A is a complex matrix rep. T
    and A* is a complex matrix rep. T*. A* is a complex conjugate=/=A.
    But the problem is that the set needs to contain operators that are self-adjoint.
    So I guess that A is a real matrix. But since the set of real adjoint operators
    cannot be within L(V), since V is a complex space, the set cannot be a
    subspace of L(V).

    After I get this out of the way, I'll prove this statement.
     
    Last edited: Jul 23, 2009
  13. Jul 23, 2009 #12

    Office_Shredder

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    Don't fall back on matrix representation arguments. Try to complete the same steps as you did for the real case. One of them won't work, and it will be because you're in a complex space instead of a real space.
     
  14. Jul 23, 2009 #13
    lol I need to go to matriholics anonymous. alright that was the worst joke ever posted
    in the history of Physics Forums

    Ok
    prove 0 is self adjoint

    let k+wi and j+hi be vectors in V. Now <0(k+wi), j+hi>=<k+wi, 0(k+wi)>=<0,j+hi>
    =<k+wi, 0>=0.
    works

    prove T+S is self adjoint where T is a real transformation and S is a complex transformation.

    Using the same vectors k+wi and j+hi we have <T(k+wi), j+hi>=<k+wi, T*(j+hi)>
    and <S(k+wi), j+hi>=<k+wi, S*(j+hi)>. Now since T is a real
    transformation, it is self adjoint so it is within the set of self adjoint transformations
    in L(V). But since L(V) is complex, S is also in L(V). And S*(k+wi) is the complex conjugate
    of S(k+wi). Therefore S=/=S*. Since S=/=S*, T+S=T*+S but T+S=/=(T+S)*
    To show this consider <(T(k+wi), j+hi>+<S(k+wi), j+hi>=<k+wi, T*(j+hi)>+<k+wi, S*(j+hi)>. Since T*=T, but S*=/=S, we have <(T(k+wi), j+hi>+<S(k+wi), j+hi>=<k+wi, T(j+hi)>+<k+wi, S*(j+hi)> or <(T+S)(k+wi), j+hi>=<k+wi, (T+S*)(j+hi)>. This shows that (T+S)*=T*+S*=T+S* but (T+S)*=/=T+S.
     
  15. Jul 23, 2009 #14

    Office_Shredder

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    Ummm... close. You almost hit upon the right point, but your proof if far more complicated than it needs to be. If you have (T+S)*=T*+S*, and you assumed T and S are self adjoint, then (T+S)*=T+S./ You got yourself confused trying to introduce complex numbers, but those are dealt with in the proof for closure under scalar multiplication. Start there
     
  16. Jul 23, 2009 #15
    Ok I'm pretty sure the closure under scalar multiplication would
    be far simpler.

    Prove aT is self adjoint.

    Let k+wi and j+hi be vectors in V. Let T be a self adjoint operator. Now <T(k+wi), j+hi>=<k+wi, T(j+hi)>. Let a=x+yi a scalar. Now a<T(k+wi), j+hi>=<aT(k+wi), j+hi>. Now we have <(aT)(k+wi), j+hi>. Since aT is complex, a*T* is the complex conjugate of aT, so aT=/=a*T*. Thus <(aT)(k+wi), j+hi>=<k+wi, (a*T*)(j+hi)>=/=<k+wi, (aT)(j+hi)> since
    a=x+yi and a*=x-yi. Thus a<T(k+wi), j+hi> is not self adjoint which disproves closure.
     
    Last edited: Jul 23, 2009
  17. Jul 23, 2009 #16

    Office_Shredder

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    What is this k+wi stuff? Why can't you just pick a vector?

    If T is self adjoint then

    <aTu,w> = a<Tu,w> = <u,Tw> = <u,a*Tw> where a* is the conjugate of a. See where the problem is?
     
  18. Jul 23, 2009 #17
    k+wi is a complex vector. That proof you have is pretty much the same as mine though I
    admit I've done some unnecessary things. I determined that aT=/=(aT)* since a*=/=a
    so a*T*=/=aT There is a complex scalar a that when multiplied by T produces a complex
    transformation (aT). So the fact that a exists prevents the set of self adjoints to be closed. But again,
    this isn't the most elegant way to do it.
     
    Last edited: Jul 23, 2009
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