Bounded Operators On Hilbert Spaces
The factorization can be extended to a bounded operator M on a separable Hilbert space H. Namely, for any bounded operator M, there exist a partial isometry U, a unitary V, a measure space (X, μ), and a non-negative measurable f such that
where is the multiplication by f on L2(X, μ).
This can be shown by mimicking the linear algebraic argument for the matricial case above. VTf V* is the unique positive square root of M*M, as given by the Borel functional calculus for self adjoint operators. The reason why U need not be unitary is because, unlike the finite dimensional case, given an isometry U1 with nontrivial kernel, a suitable U2 may not be found such that
is a unitary operator.
As for matrices, the singular value factorization is equivalent to the polar decomposition for operators: we can simply write
and notice that U V* is still a partial isometry while VTf V* is positive.
Read more about this topic: Singular Value Decomposition
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