- #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=T

^{2}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

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.