Measure, Integration & Real Analysis, 2021a

332 Chapter 10 Linear Maps on Hilbert Spaces
Singular Value Decomposition
The next result provides an important generalization of 10.106(b) to arbitrary compact
operators that need not be self-adjoint or normal. This generalization requires two
orthonormal families, as compared to the single orthonormal family in 10.106(b).
10.113 Singular Value Decomposition
Suppose T is a compact operator on a Hilbert space V. Then there exist a
countable set Ω, orthonormal families {e k } k∈Ω and {h k } k∈Ω in V, and a family
{s k } k∈Ω of positive numbers such that
10.114 Tf = ∑ s k 〈 f , e k 〉h k
k∈Ω
for every f ∈ V.
Proof
If α is an eigenvalue of T ∗ T, then (T ∗ T) f = α f for some f ̸= 0 and
α‖ f ‖ 2 = 〈α f , f 〉 = 〈T ∗ Tf, f 〉 = 〈Tf, Tf〉 = ‖Tf‖ 2 .
Thus α ≥ 0. Hence all eigenvalues of T ∗ T are nonnegative.
Apply 10.106(b) and the conclusion of the paragraph above to the self-adjoint
compact operator T ∗ T, getting a countable set Ω, an orthonormal family {e k } k∈Ω in
V, and a family {s k } k∈Ω of positive numbers (take s k = √ α k ) such that
10.115 (T ∗ T) f = ∑ s 2 k 〈 f , e k 〉e k
k∈Ω
for every f ∈ V. The equation above implies that (T ∗ T)e j = s 2 j e j for each j ∈ Ω.
For k ∈ Ω, let
h k = Te k
.
s k
For j, k ∈ Ω,wehave
〈h j , h k 〉 = 1
s j s k
〈Te j , Te k 〉 = 1
s j s k
〈T ∗ Te j , e k 〉 = s j
s k
〈e j , e k 〉.
The equation above implies that {h k } k∈Ω is an orthonormal family in V.
If f ∈ span {e k } k∈Ω , then
)
Tf = T(
∑ 〈 f , e k 〉e k = ∑ 〈 f , e k 〉Te k = ∑ s k 〈 f , e k 〉h k ,
k∈Ω
k∈Ω
k∈Ω
showing that 10.114 holds for such f .
If f ∈ ( span {e k } k∈Ω
) ⊥, then 10.115 shows that (T ∗ T) f = 0, which implies
that Tf = 0 (because 0 = 〈T ∗ Tf, f 〉 = ‖Tf‖ 2 ); thus both sides of 10.114 are 0.
Hence the two sides of 10.114 agree for f in a closed subspace of V and for f in
the orthogonal complement of that closed subspace, which by linearity implies that
the two sides of 10.114 agree for all f ∈ V.
Measure, Integration & Real Analysis, by Sheldon Axler