Editing Singular value decomposition (section)

=== Bounded operators on Hilbert spaces ===
The factorization {{tmath|\mathbf M {{=}} \mathbf U \mathbf \Sigma \mathbf V^*}} can be extended to a [[bounded operator]] {{tmath|\mathbf M}} on a separable Hilbert space {{tmath|H.}} Namely, for any bounded operator {{tmath|\mathbf M,}} there exist a [[partial isometry]] {{tmath|\mathbf U,}} a unitary {{tmath|\mathbf V,}} a measure space {{tmath|(X, \mu),}} and a non-negative measurable {{tmath|f}} such that

<math display=block>
\mathbf{M} = \mathbf{U} T_f \mathbf{V}^*
</math>

where {{tmath|T_f}} is the [[multiplication operator|multiplication by {{tmath|f}}]] on {{tmath|L^2(X, \mu).}}

This can be shown by mimicking the linear algebraic argument for the matrix case above. {{tmath|\mathbf V T_f \mathbf V^*}} is the unique positive square root of {{tmath|\mathbf M^* \mathbf M,}} as given by the [[Borel functional calculus]] for [[self-adjoint operator]]s. The reason why {{tmath|\mathbf U}} need not be unitary is that, unlike the finite-dimensional case, given an isometry {{tmath|U_1}} with nontrivial kernel, a suitable {{tmath|U_2}} may not be found such that

<math display=block>
\begin{bmatrix} U_1 \\ U_2 \end{bmatrix}
</math>

is a unitary operator.

As for matrices, the singular value factorization is equivalent to the [[polar decomposition]] for operators: we can simply write

<math display=block>
\mathbf M = \mathbf U \mathbf V^* \cdot \mathbf V T_f \mathbf V^*
</math>

and notice that {{tmath|\mathbf U \mathbf V^*}} is still a partial isometry while {{tmath|\mathbf V T_f \mathbf V^*}} is positive.