Proving the square root of a positive operator is unique

[Adgorn]

Homework Statement

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. Homework Equations
N/A

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.

 

[Ray Vickson]

Homework Statement

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. Homework Equations
N/A

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.

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!)

 

[Adgorn]

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!)

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.