Page 162 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
150

2
Sv = S v, e 1 e 1 +· · ·+Pv, e n e n
= v, e 1 Se 1 +· · ·+Pv, e n Se n
2 2 2 2
=| v, e 1 | + ··· + | v, e n |
2
= v ,
where the first and last equalities come from 6.17. Taking square roots,
we see that S is an isometry, proving that (e) implies (a).
Having shown that (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (e) ⇒ (a), we know at this
stage that (a) through (e) are all equivalent to each other. Replacing S
with S , we see that (f) through (j) are all equivalent to each other. Thus
∗
to complete the proof, we need only show that one of the conditions
in the group (a) through (e) is equivalent to one of the conditions in
the group (f) through (j). The easiest way to connect the two groups of
conditions is to show that (c) is equivalent to (h). In general, of course,
S need not commute with S . However, S S = I if and only if SS = I;
∗
∗
∗
this is a special case of Exercise 23 in Chapter 3. Thus (c) is equivalent
to (h), completing the proof.
The last theorem shows that every isometry is normal (see (a), (c),
and (h) of 7.36). Thus the characterizations of normal operators can
be used to give complete descriptions of isometries. We do this in the
next two theorems.

7.37 Theorem: Suppose V is a complex inner-product space and
S ∈L(V). Then S is an isometry if and only if there is an orthonormal
basis of V consisting of eigenvectors of S all of whose corresponding
eigenvalues have absolute value 1.

Proof: We already proved (see the first paragraph of this section)
that if there is an orthonormal basis of V consisting of eigenvectors of S
all of whose eigenvalues have absolute value 1, then S is an isometry.
To prove the other direction, suppose S is an isometry. By the com-
plex spectral theorem (7.9), there is an orthonormal basis (e 1 ,...,e n )
of V consisting of eigenvectors of S. For j ∈{1,...,n}, let λ j be the
eigenvalue corresponding to e j . Then

|λ j |= λ j e j = Se j = e j = 1.

Thus each eigenvalue of S has absolute value 1, completing the proof.