Proving the square root of a positive operator is unique

In summary, the problem asks to prove that the operator T defined by ##T(u_i)=\sqrt{\lambda_i}u_i## is positive and unique in satisfying ##T^2=P## within the given context. This is achieved by showing that T is the only positive operator that satisfies ##T^2=P##, making it the positive square root of P. The proof for uniqueness involves showing that any other positive operator satisfying ##T'^2=P## must be equivalent to T, thus proving T's uniqueness.
  • #1
Adgorn
130
18

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.
 
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  • #2
Adgorn said:

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!)
 
  • #3
Ray Vickson said:
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.
 

Related to Proving the square root of a positive operator is unique

1. What is a positive operator?

A positive operator is a mathematical term used in linear algebra and functional analysis. It is a linear operator that maps a vector space to itself and preserves the positive elements of that space.

2. How do you prove uniqueness of the square root of a positive operator?

To prove uniqueness, we need to show that there is only one square root of the positive operator. This can be done by assuming that there are two distinct square roots and then showing that they are equal. This can be done using various methods such as using the spectral theorem or the polar decomposition theorem.

3. Can a positive operator have more than one square root?

No, a positive operator can only have one square root. This is because if we assume there are two distinct square roots, then we can show that they are equal using the methods mentioned above. Therefore, there can only be one unique square root of a positive operator.

4. Are there any applications of proving the uniqueness of the square root of a positive operator?

Yes, proving the uniqueness of the square root of a positive operator has many applications in quantum mechanics, where positive operators are used to represent physical observables. It is also used in functional analysis to study the properties of operators on vector spaces.

5. Is the square root of a positive operator always a positive operator?

Yes, the square root of a positive operator is always a positive operator. This is because the square root operation preserves the positive elements of a vector space. Therefore, if the original operator was positive, its square root will also be positive.

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