Positive Operators

Positive operators. Square roots of operators. Characterization of positive operators. Each positive operator has a unique positive square root.

[speedbird]

Hi professor,
Your lectures are really awesome! complete and short.
Would you please explain more about the analogy of real numbers and the self adjoint operators.
Thank you.

[spiderjerusalem]

Alternative proof to the uniqueness of the positive square root of each positive operator (that is different from the 2nd, 3rd, and 4th edition) (13:48) :

Let 𝑇 be a positive operator, and R be a positive square root of 𝑇.

Suppose x is an eigenvector of 𝑇 (whose existence is guaranteed since 𝑇 is self-adjoint).
There is then a non-negative λ such that Tx=λx.
We shall show that Rx=√λx.


We will divide this into 2 cases.

*• Case 1: λ=0*
x∈null 𝑇 = null R²
⇔ Rx∈null R ∩ range R
= null R* ∩ range R
= (range R)ᗮ ∩ range R
= {0}
⇔ Rx = 0
⇔ x∈null R

In other words, null R=null 𝑇 .
which proves Rx=0=√λx.


*• Case 2: λ>0*
λ>0 ⇒ √λ>0.
Since the eigenvalues of positive operators are nonnegative, R+√λ𝐼 is therefore injective.

Thus
(R²–λ𝐼 )x = 0
⇔ (R+√λ𝐼 )(R–√λ𝐼 )x =  0
⇒ (R–√λ𝐼 )x = 0
⇔ Rx = √λx

As was to be demonstrated.


By spectral theorem, there is an orthonormal basis {e₁, ..., eₙ} of 𝑽 such that
𝑇eₖ=λₖeₖ
for each k=1, ..., n, where {λ₁, ..., λₙ} is the set of all eigenvalues of 𝑇.
Each of the λₖ's is non-negative since 𝑇 is a positive operator.

Therefore,
Reₖ = √λₖeₖ for each k=1, ..., n.

In other words, the positive square root of 𝑇 is uniquely determined.
∎