364 CHAPTER 13. MATRICES AND THE INNER PRODUCT

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

∑k=0

(T S2

kx,x)= (Sn+1x,T x)+

n

∑k=0

(T Skx,Skx)

so passing to a limit as n→ ∞,(T Sx,x) = 0+ limsupn→∞ ∑nk=0 (T Skx,Skx)≥ 0.

Thus if S≤ I, the theorem is proved. If S is general, S∥S∥ ≤ I and in this case, it follows

that(

T S∥S∥x,x

)=(

S∥S∥T x,x

)≥ 0 and so (ST x,x)≥ 0. ■

The proposition is like the familiar statement about real numbers which says that whenyou multiply two nonnegative real numbers the result is a nonnegative real number.

13.12.2 Roots of Positive Self Adjoint OperatorsWith this preparation, it is time to give the theorem about roots.

Theorem 13.12.3 Let T ∈ L (H,H) be a positive self adjoint linear operator. Then form ∈N, there exists a unique mth root A with the following properties. Am = T,A is positiveand self adjoint, A commutes with every operator which commutes with T .

Proof: Define the following sequence of operators:

A0 ≡ 0, An+1 ≡ An +1m(T −Am

n )

Say T ≤ I.Claim 1: An ≤ I.Proof of Claim 1: True if n = 0. Assume true for n. Then

I−An+1 = I−An +1m(Am

n −T )≥ I−An +1m(Am

n − I)

= I−An−1m(I−Am

n )

= (I−Am)−1m(I−Am)

(I + · · ·+Am−1

n)

Now, since An ≤ I, I + · · ·+Am−1n ≤ mI, it follows that

= (I−Am)

(I− 1

m

(I + · · ·+Am−1

n))≥ (I−Am)(I− I) = 0

so by induction, An ≤ I.Claim 2: An ≤ An+1.Proof of Claim 2: From the definition of An, this is true if n = 0 because

A1 = T ≥ 0 = A0.

Suppose true for n. Then from Claim 1,

An+2−An+1 = An+1 +1m

(T −Am

n+1)−[

An +1m(T −Am

n )

]= An+1−An +

1m

(Am

n −Amn+1)