# Is there any conclusions generalized Singular Value Decomposition into Hilbert Space

Spectrum decomposition can be regarded as the generalizations of the following fact that: Every Hermitian matrix $A$ can be decomposed into $A=U^{*}\Lambda U$,where $U$ is a unitary matrix

Singular vector decomposition can be expressed as Every Matrix $A_{mn}$ can be decomposed in to $A=U\Lambda V^{*}$, where $U$,and $V$ are unitary matrices. Does it can be extended in to decompostion of linear operators on Hilbert Space. ?

I searched the internet and several traditional books about the topic "Singular Vector Decomposition into Hilbert space", However, to my disappointment, I find no similar conclusion. Thanks for your help.

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It's worth mentioning that there's a concept in numerical linear algebra called the "Generalized Singular Value Decomposition (GSVD)" that still has to do with matrices and doesn't have anything to do with Hilbert space. – Brian Borchers Apr 9 '11 at 5:15
What is the relevance of the "r-matrix" tag? – Yemon Choi Apr 9 '11 at 6:04
To paraphrase the answer below by Brian Borchers: what you are looking for is part of the spectral theory of self-adjoint operators on Hilbert space. This is covered in functional analysis textbooks; and so I am voting to close this question. – Yemon Choi Apr 9 '11 at 6:06
Be that as it may, what you are after is (at least for compact operators) the Schmidt decomposition. Your original question is quite vague and of the form, "I have seen something for finite matrices, does something similar work for operators on infinite-dimensional Hilbert space". I still think this is not really a focused question suitable for MO. (For general, not necessarily self-adjoint, operators on Hilbert space, things are much more complicated than for matrices; you shouldn't expect everything you know for matrices to work for operators.) – Yemon Choi Apr 9 '11 at 8:40
By the way, one problem may be that "singular vector decomposition" is incorrect terminology, so if you search for those words you may not find what you want. The correct terminology is "singular value decomposition". – Yemon Choi Apr 9 '11 at 8:41

The simplest generalization is that a "compact self adjoint linear operator" on a Hilbert space can be diagonalized in terms of (an infinite number of) eigenvalues and eigenfunctions that are elements of the Hilbert space. This can be extended to non-compact but still self adjoint operators, but it's more complicated because you might not have a discrete spectrum. It can also be extended to a kind of singular value decomposition for compact but non self adjoint operators. All of this is covered in textbooks on (linear) functional analysis.

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Thanks for your answers, however, this question is more complicated than text books on linear function analysis. At least, Rudin and Conway's books just discuss the compact operators and adjoint operators. – yaoxiao Apr 9 '11 at 7:45
I first learned about this stuff in a course that used "Functional Analysis" by Bachman and Narici as the text. The spectral theorem for bounded normal operators is covered in chapter 28. This book is available in an inexpensive reprint edition from Dover. – Brian Borchers Apr 9 '11 at 20:09
(Just to clarify for anyone reading this answer: Brian's answer was in response to an older formulation of the question which did not make it clear that the OP wanted something for not-necessarily-normal operators.) – Yemon Choi Apr 10 '11 at 6:30

This should really be a comment, not an answer, but I decided it perhaps could do with extra visibility.

The current version of your question, asks, among other things, once we have corrected the terminology:

Do operators on Hilbert space have a decomposition $A=U\Lambda V^*$ where $U$ and $V$ are unitary and $\Lambda$ is diagonal?

Leaving aside the subtlety about different versions of "diagonal" for operators in infinite dimensions, let me just note that if an injective operator $A$ is of the form $U\Lambda V^*$ where $\Lambda$ is diagonal and $U$ and $V$ are invertible, then it must have dense range; this is an easy exercise. The forward shift on $\ell^2({\mathbb N})$ is a simple example of an injective operator on Hilbert space that does not have dense range.

So the answer to your question, at least in the general form you have posed, is "no".

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This suggests that if any form of SVD is valid in Hilbert space, then the word unitary should not be understood as unitary and invertible. I suggest to keep the terminology for $U$ such that $\|Ux\|=\|x\|$ (not necessarily surjective), and look for to unitary operators $U,V$ such that $AV=U\Lambda$ with $\Lambda$ `diagonal'. This must have been studied an the answer must exist somewhere. – Denis Serre Apr 9 '11 at 9:16
In other words, we allow (co)isometries? Well, isn't this more or less just the polar decomposition of an operator, then? – Yemon Choi Apr 9 '11 at 9:26
I am also not particularly keen to spend too much time thinking about what a better version of the original question might be. This is, to re-use my old favourite idiom, a case of providing the food that goes into someone else's stone soup. – Yemon Choi Apr 9 '11 at 9:34
Thanks for Yemon Choi and Dennis Serre's comment. – yaoxiao Apr 9 '11 at 10:27

I believe what you are looking for is here:

http://mathworld.wolfram.com/PolarDecomposition.html

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Oh，thank you. may be this theorem "Every bounded operator acting on a Hilbert space has a decomposition , where and is a partial isometry. This decomposition is called polar decomposition. If is invertible, then can be chosen to be unitary." given by mathworld is related to some deep operator theory. – yaoxiao Apr 9 '11 at 12:33