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Another positive operator proof

  1. Aug 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose that T is a positive operator on V. Prove that T is invertible
    if and only if <Tv,v > is >0 for every v ∈ V \ {0}.


    2. Relevant equations



    3. The attempt at a solution
    If T is invertible, then TT-1=I.Now let v=v1+...+vn and let Tv=a1v1+...+anvn. Now <Tv, v>=<a1v1, v>+...+<anvn, v>. Applying T-1 we get <T-1(a1v1), v>+...+<T-1(anvn),v> =<v1, v>+...+<vn, v>=<Iv, v>=<v, v>. Since v[tex]\notin[/tex]{0},<v, v> is >0. And since T is invertible,
    <Tv, v>=<v, v> if T=I, or <Tv, v> > <v, v> if T=/=I. Therefore <Tv, v> is >=<v, v>.
    And since <v, v> is >0, <Tv, v> is >0.
     
  2. jcsd
  3. Aug 10, 2009 #2
    I don't know how to handle the other direction. All we know is that <Tv, v> is >0.
    T doesn't have to be invertible. It can be a projection and still be >0.
     
  4. Aug 10, 2009 #3

    Dick

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    You know more than <Tv,v> >=0 for a positive operator. You also know T is self-adjoint. You stated the definition in a previous thread. If it weren't the theorem wouldn't be true. Take T to be rotation by 90 degrees in R^2. Then <Tv,v>=0. But T is invertible. But it's not self-adjoint.
     
  5. Aug 11, 2009 #4
    Alright for the first part the only thing that I will change is the definition of <Tv, v>.
    Since we know T is self adjoint, and postive, T has a positive square root. So
    <S^2v, v>=<v, S^2v>.
    I have no idea how to do the other direction.
     
    Last edited: Aug 11, 2009
  6. Aug 11, 2009 #5

    Dick

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    The reason you know T has a square root is because you know there is an orthogonal basis of the vector space all of which are eigenvectors of T. What can you say about the values of the eigenvalues?
     
  7. Aug 11, 2009 #6
    All of the eigenvalues should be >=0 so it follows that each eigenvalue has a positive square root. I take it that this is
    important because if an eigenvalue is negative, then the square root would be imaginary. So
    we'd have 0-i with its conjugate 0+i. This wouldn't allow self adjointness.
     
    Last edited: Aug 11, 2009
  8. Aug 11, 2009 #7

    Dick

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    Zero isn't positive. Who cares about a "square root"?? I thought you were trying to prove something about when T is invertible.
     
  9. Aug 11, 2009 #8
    what should I do now?
     
  10. Aug 11, 2009 #9

    Dick

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    Ack! Think about it! If T has a zero eigenvalue, is it invertible??
     
  11. Aug 11, 2009 #10
    no. If Tv=a1v1+...+0*vn, you're not going to get v back . If Tv=a1v1+...+anvn (no eigenvalue is 0) then you can
    get v back by dividing each eigenvalue by itself. Sorry for being lazy, its a hot day. I gotta pull my weight here.
     
    Last edited: Aug 11, 2009
  12. Aug 11, 2009 #11

    Dick

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    That's pretty incomprehensible.
     
  13. Aug 11, 2009 #12
    Its like this: Say the domain has a basis {(1 0 0), (0 1 0), (0 0 1)}.
    Now say T maps from here to a range with basis {(1 0 0) (0 1 0)}. So the
    matrix is 3x3 with a 0 row at the bottom. This implies that there is a 0 eigenvalue.
    Now lets say that the vector we map is v=a(1 1 1) (a=/=0). T(1 0 0 )+T(0 1 0 ) would
    never equate to a(1 1 1) no matter what T is since a(1 1 1) is not within the basis for the range.
    This is what I mean by "not being able to get v back". This applies to all vectors with three
    nonzero entries.
     
  14. Aug 11, 2009 #13

    Dick

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    You know if Tv=0*v for v not zero, then v is ker(T). If ker(T) is not {0} then T is not invertible. You KNOW that. You don't have to try to reprove that fact with awkward bad examples everytime you need it.
     
  15. Aug 11, 2009 #14
    The problem is that when given <Tv, v> is >0, T doesn't have to be invertible.
    T can be an orthogonal projection with all eigenvalues >0 (except for say, one eigenvalue being 0)
    and still be >0. I mean we know that <Tv, v> is >0 but we don't know whether or not it is invertible.
    But it is still possible for <Tv, v> to be >0 when T is invertible.
     
  16. Aug 11, 2009 #15

    Dick

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    That's a pretty poor example of a mapping where <Tv,v> >0. <Tv,v>=0 for the vectors that are projected out. What's your point?
     
  17. Aug 11, 2009 #16
    Lets say v=(1 1 1). Now Tv=(1 1 0 ) with (0 0 1) being in nullT.
    Now this is an orthogonal projection since the inner product between (1 1 0) and
    (0 0 1) is 0. But the inner product between (1 1 0 ) and (1 1 1) or <Tv, v> is
    greater than 0. Orthogonal projections are not invertible. But the
    inner product <Tv, v> is >0. The problem is using <Tv, v> >0 to prove that
    T is invertible.
     
  18. Aug 11, 2009 #17

    Dick

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    Look. The problem says prove T is invertible if <Tv,v> >0 for EVERY v not equal to zero (and T self-adjoint). A projection with a nontrivial kernel DOES NOT have that property. Period. End of discussion. Read the problem statement again, several times.
     
  19. Aug 12, 2009 #18
    Ok let <Tv, v> be >0. Now let v=e1+...+en where ek is an eigenvector of T from
    an orthonormal basis on V. Since T is positive and self adjoint, let T be a diagonal matrix with one positive eigenvalue for each ek. Should I assume this? wait nevermind, the eigenvalues should be nonnegative.
     
    Last edited: Aug 12, 2009
  20. Aug 12, 2009 #19

    Dick

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    State clearly what part of the proof you are trying to do. Say, I'm trying to prove that if ___ then ___. Fill in the blanks.
     
  21. Aug 12, 2009 #20
    <Tv, v> is >0, v is not a zero vector, and T is a positive operator, then T is invertible.
     
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