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11.7: Singular-value decomposition

( \newcommand{\kernel}{\mathrm{null}\,}\)

The singular-value decomposition generalizes the notion of diagonalization. To unitarily diagonalize TL(V) means to find an orthonormal basis e such that T is diagonal with respect to this basis, i.e.,

M(T;e,e)=[T]e=[λ100λn],

where the notation M(T;e,e) indicates that the basis e is used both for the domain and codomain of T. The Spectral Theorem tells us that unitary diagonalization can only be done for normal operators. In general, we can find two orthonormal bases e and f such that

M(T;e,f)=[s100sn],

which means that Tei=sifi even if T is not normal. The scalars si are called singular values of T. If T is diagonalizable, then these are the absolute values of the eigenvalues.

Theorem 11.7.1

All TL(V) have a singular-value decomposition. That is, there exist orthonormal bases e=(e1,,en) and f=(f1,,fn) such that

Tv=s1v,e1f1++snv,enfn,

where si are the singular values of T.

Proof

Since |T|0, it is also also self-adjoint. Thus, by the Spectral Theorem, there is an orthonormal basis e=(e1,,en) for V such that |T|ei=siei. Let U be the unitary matrix in the polar decomposition of T. Since e is orthonormal, we can write any vector vV as

v=v,e1e1++v,enen,

and hence

Tv=U|T|v=s1v,e1Ue1++snv,enUen.

Now set fi=Uei for all 1in. Since U is unitary, (f1,,fn) is also an orthonormal basis, proving the theorem.


This page titled 11.7: Singular-value decomposition is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

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