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. ∎
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.
Amazing videos and book. Thanks alot!
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.
∎
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.