Page 158 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
146
there is an orthonormal basis (e 1 ,...,e n ) of V consisting of eigen-
vectors of T. Let λ 1 ,...,λ n be the eigenvalues of T corresponding to
e 1 ,...,e n , so that each λ j is a nonnegative number. Define S ∈L(V)
by

Se j = λ j e j
for j = 1,...,n. Then S is a positive operator, as you should verify.
2
2
Furthermore, S e j = λ j e j = Te j for each j, which implies that S = T.
Thus S is a positive square root of T, and hence (c) holds.
Clearly (c) implies (d) (because, by definition, every positive operator
is self-adjoint).
Now suppose that (d) holds, meaning that there exists a self-adjoint
2
operator S on V such that T = S . Then T = S S (because S = S), and
∗
∗
hence (e) holds.
Finally, suppose that (e) holds. Let S ∈L(V) be such that T = S S.
∗
Then T ∗ = (S S) = S (S ) = S S = T, and hence T is self-adjoint.
∗
∗ ∗
∗
∗
∗
To complete the proof that (a) holds, note that
∗
Tv, v = S Sv, v
= Sv, Sv
≥0
for every v ∈ V. Thus T is positive.
Each nonnegative number has a unique nonnegative square root.
The next proposition shows that positive operators enjoy a similar
√
property. Because of this proposition, we can use the notation T
to denote the unique positive square root of a positive operator T, just
√
as λ denotes the unique nonnegative square root of a nonnegative
number λ.

A positive operator can 7.28 Proposition: Every positive operator on V has a unique positive
have infinitely many square root.
square roots (though
only one of them can Proof: Suppose T ∈L(V) is positive. Let λ 1 ,...,λ m denote the
be positive). For distinct eigenvalues of T; because T is positive, all these numbers are
example, the identity nonnegative (by 7.27). Because T is self-adjoint, we have
operator on V has
7.29 V = null(T − λ 1 I) ⊕· · · null(T − λ m I);
infinitely many square
roots if dim V> 1.
see 7.14.