# Proving the square root of a positive operator is unique

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1. Jun 27, 2017

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. Jun 27, 2017

### Ray Vickson

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!)

3. Jun 27, 2017

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.