Page 159 - Linear Algebra Done Right

Isometries
an eigenvalue of S.If v ∈ null(S − αI), then Sv = αv, which implies
that Now suppose S ∈L(V) is a positive square root of T. Suppose α is 147
2
2
7.30 Tv = S v = α v,
2
2
so v ∈ null(T − α I). Thus α is an eigenvalue of T, which means

2
that α must equal some λ j . In other words, α = λ j for some j.
Furthermore, 7.30 implies that

7.31 null(S − λ j I) ⊂ null(T − λ j I).
In the paragraph above, we showed that the only possible eigenval-

ues for S are λ 1 ,..., λ m . Because S is self-adjoint, this implies that

7.32 V = null(S − λ 1 I) ⊕· · · null(S − λ m I);
see 7.14. Now 7.29, 7.32, and 7.31 imply that

null(S − λ j I) = null(T − λ j I)

for each j. In other words, on null(T − λ j I), the operator S is just

multiplication by λ j . Thus S, the positive square root of T, is uniquely
determined by T.

Isometries

An operator S ∈L(V) is called an isometry if The Greek word isos
means equal; the Greek
Sv = v word metron means
measure. Thus
for all v ∈ V. In other words, an operator is an isometry if it preserves
norms. For example, λI is an isometry whenever λ ∈ F satisfies |λ|= 1. isometry literally
means equal measure.
More generally, suppose λ 1 ,...,λ n are scalars with absolute value 1 and
S ∈L(V) satisfies S(e j ) = λ j e j for some orthonormal basis (e 1 ,...,e n )
of V. Suppose v ∈ V. Then

7.33 v = v, e 1 e 1 + ··· + v, e n e n
and

2
2
2
7.34 v =| v, e 1 | +· · ·+| v, e n | ,