Page 168 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
156
7.46
Singular-Value Decomposition:
gular values s 1 ,...,s n . Then there exist orthonormal bases (e 1 ,...,e n )
and (f 1 ,...,f n ) of V such that Suppose T ∈L(V) has sin-
7.47 Tv = s 1 v, e 1 f 1 + ··· + s n v, e n f n
for every v ∈ V.
√
Proof: By the spectral theorem (also see 7.14) applied to T T,
∗
√
there is an orthonormal basis (e 1 ,...,e n ) of V such that T Te j = s j e j
∗
for j = 1,...,n. We have
v = v, e 1 e 1 + ··· + v, e n e n
√
for every v ∈ V (see 6.17). Apply T T to both sides of this equation,
∗
getting
√
∗
T Tv = s 1 v, e 1 e 1 +· · ·+ s n v, e n e n
This proof illustrates for every v ∈ V. By the polar decomposition (see 7.41), there is an
√
the usefulness of the isometry S ∈L(V) such that T = S T T. Apply S to both sides of the
∗
polar decomposition. equation above, getting
Tv = s 1 v, e 1 Se 1 + ··· + s n v, e n Se n
for every v ∈ V. For each j, let f j = Se j . Because S is an isometry,
(f 1 ,...,f n ) is an orthonormal basis of V (see 7.36). The equation above
now becomes

Tv = s 1 v, e 1 f 1 + ··· + s n v, e n f n

for every v ∈ V, completing the proof.

When we worked with linear maps from one vector space to a second
vector space, we considered the matrix of a linear map with respect
to a basis for the first vector space and a basis for the second vector
space. When dealing with operators, which are linear maps from a
vector space to itself, we almost always use only one basis, making it
play both roles.
The singular-value decomposition allows us a rare opportunity to
use two different bases for the matrix of an operator. To do this, sup-
pose T ∈L(V). Let s 1 ,...,s n denote the singular values of T, and let
(e 1 ,...,e n ) and (f 1 ,...,f n ) be orthonormal bases of V such that the
singular-value decomposition 7.47 holds. Then clearly