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Proving the square root of a positive operator is unique

  1. Jun 27, 2017 #1
    1. The problem statement, all variables and given/known data
    The problem relates to a proof of a previous statement, so I shall present it first:
    "Suppose P is a self-adjoint operator on an inner product space V and ##\langle P(u),u \rangle## ≥ 0 for every u ∈ V, prove P=T2 for some self-adjoint operator T.

    Because P is self-adjoint, there exists an orthonormal basis{ ##u_1,...,u_n##} of V consisting of eigenvectors of P; say, ##P(u_i)=\lambda_iu_i##. Since P is self-adjoint, the ##\lambda_i## are real, also, since ##0≤ \langle P(u_i),u_i \rangle = \lambda_i\langle u_i,u_i \rangle## and ##\langle u_i,u_i \rangle ≥ 0##, ##\lambda_i## ≥ 0 for all ##\lambda_i##."
    The rest of the proof defines ##T(u_i)=\sqrt{\lambda_i}u_i## for i=1,...,n and shows how ##T^2=P##

    Now for the question itself: Consider the operator T defined by ##T(u_i)=\sqrt{\lambda_i}u_i##, ##i=1,...,n## in the above proof. Show that T is positive and that it is the only positive operator for which ##T^2=P##.

    2. Relevant equations

    N/A

    3. The attempt at a solution
    I have already proven that ##T## is positive with relative ease, my problem is proving its uniqueness. I have tried to define a second positive operator ##T'## such that ##T'^2=P## and show that ##T'=T##. I've tried showing they agree on a basis, using their properties as self-adjoint operators by messing around with inner products and so on but with no success.
    Help would be appreciated.
     
  2. jcsd
  3. Jun 27, 2017 #2

    Ray Vickson

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    Why do you need uniqueness? The question (if accurately reproduced) does not claim that the ##T## you get is unique. (In fact, it cannot be unique, since both ##T## and ##-T## satisfy the theorem!)
     
  4. Jun 27, 2017 #3
    The original proof does not say claim T is unique but a remark below it says "The above operator T is the unique positive operator such that ##P=T^2##, it is called the positive square root of P". In any case even if the theorem does not claim it, the question asks me to prove it.
    Also while ##-T## might satisfy the theorem it does not contradict the question since the questions asks to prove T is the only positive operator for which ##T^2=p##, since ##-T## is clearly not positive it is not relevant to the question.
     
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