13.12. ROOTS OF POSITIVE LINEAR MAPS 363

13.12 Roots of Positive Linear MapsIn this section, H will be a Hilbert space, real or complex, and T will denote an operatorwhich satisfies the following definition. This will be a more general result than the abovebecause it will hold for infinite dimensional spaces.

Definition 13.12.1 Let T satisfy T = T ∗ (Hermitian) and for all x ∈ H,

(T x,x)≥ 0 (13.21)

Such an operator is referred to as positive and self adjoint. It is probably better to refer tosuch an operator as “nonnegative” since the possibility that T x = 0 for some x ̸= 0 is notbeing excluded. Instead of “self adjoint” you can also use the term, Hermitian. To saveon notation, write T ≥ 0 to mean T is positive, satisfying 13.21. When we say A ≤ B thismeans B−A≥ 0.

A useful theorem about the existence of roots of positive self adjoint operators is pre-sented. This proof is very elementary. I found it in [28] for square roots.

13.12.1 The Product of Positive Self Adjoint OperatorsWith the above definition here is a fundamental result about positive self adjoint operators.

Proposition 13.12.2 Let S,T be positive and self adjoint such that ST = T S. Then ST isalso positive and self adjoint.

Proof: It is obvious that ST is self adjoint.

(ST x,y) = (T Sx,y) = (Sx,Ty) = (x,STy)

The only problem is to show that ST is positive. The idea is to write S = Sn+1 +∑nk=0 S2

kwhere S0 = S and the operators Sk are self adjoint. This is because if you have

(T S2x,x

),

where everything commutes, this equals (ST Sx,x) = (T Sx,Sx)≥ 0. Thus it will be possibleto deal with the terms of the sum which are squared. First assume (Sx,x)≤ (x,x) so S≤ I.

Define a sequence recursively as follows.

Sn+1 = Sn−S2n, S≡ S0 (13.22)

Then ∑nk=0 S2

k = ∑nk=0 (Sk−Sk+1) = S−Sn+1, S = Sn+1+∑

nk=0 S2

k . Now S0 ≥ 0 by assump-tion. Assume Sn ≥ 0. Then

Sn+1 = Sn−S2n = (I−Sn)Sn (Sn +(I−Sn)) = S2

n (I−Sn)+(I−Sn)2 Sn

It follows that Sn+1 ≥ 0 because clearly those two terms on the end are positive. Therefore,

(Sx,x) = (Sn+1x,x)+n

∑k=0

(S2

kx,x)≥

n

∑k=0∥Skx∥2 , (Sx,x)≥

∞

∑k=0∥Skx∥2

also and so limk→∞ ∥Skx∥= 0.

T Sx = T Sn+1x+n

∑k=0

T S2kx