What Happens When the V Matrix is Zero in SVD?

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Discussion Overview

The discussion revolves around the implications of obtaining a zero matrix for the V component in the Singular Value Decomposition (SVD) of a given matrix. Participants explore the definitions and properties of SVD, particularly in the context of a specific problem from a practice final exam. The conversation touches on theoretical aspects, mathematical reasoning, and potential misunderstandings related to the SVD process.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about obtaining a zero matrix for V in the SVD, questioning the validity of this outcome.
  • Another participant asserts that U and V cannot be zero, referencing the definition of SVD which states they are orthonormal matrices.
  • A participant mentions that if the eigenvalues of O*O are all zero, then Σ would also be a zero matrix, suggesting that SVD is unique only to Σ, allowing for arbitrary orthonormal bases for U and V.
  • There is a challenge to the uniqueness of U and V, with a participant explaining that in cases of repeated eigenvalues, V is not uniquely defined, except for scaling factors.
  • One participant provides a specific example involving the zero transformation in R², arguing that multiple linearly independent vectors can serve as eigenvectors, highlighting the relevance of the zero matrix case.
  • Another participant emphasizes that the original poster's (OP's) claim of a zero V matrix is likely incorrect, noting that the matrix in question does not have repeated singular values.
  • There is agreement that U and V are not unique when singular values are repeated, indicating a potential area of confusion in the OP's understanding.

Areas of Agreement / Disagreement

Participants generally disagree on the possibility of obtaining a zero V matrix in SVD, with some asserting it is impossible while others acknowledge the conditions under which it could occur. The discussion remains unresolved regarding the OP's specific situation and the implications of the zero matrix.

Contextual Notes

Participants note that the uniqueness of U and V matrices can depend on the presence of repeated singular values, and the discussion highlights the importance of understanding eigenvalues and eigenvectors in the context of SVD.

SELFMADE
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Semester is over but still want to figure this out

Prob 8 on here:

http://www.math.uic.edu/~akers/310PracticeFinal.pdf

When trying to SVD that matrix one of the U or V matrix turns out to be zero but the answer key has just the general formula for SVD

can anyone explain thanks
 
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SELFMADE said:
Semester is over but still want to figure this out

Prob 8 on here:

http://www.math.uic.edu/~akers/310PracticeFinal.pdf

When trying to SVD that matrix one of the U or V matrix turns out to be zero but the answer key has just the general formula for SVD

can anyone explain thanks

U and V can't be zero. The the definition of the SVD says they are orthonormal matrices.

How are you doing this by hand? One way is to finding the eigenvalues and vectors of A^T.A and A.A^T . The U and V matrices are the eigenvectors of those matrices, so it doesn't make any sense for them to be zero.
 
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Since the eigenvalues of O*O are all 0, Σ would be the same exact zero matrix as the given. SVD is unique only to Σ in general, so you can pick any orthonormal bases for your domain and codomain and you have U and V.
 
zcd said:
SVD is unique only to Σ in general, so you can pick any orthonormal bases for your domain and codomain and you have U and V.

Are you sure about that?

A = U S VT

AT A = V S UT U S VT = V S2 VT

AT A V = V S2 VT V = V S2

So S2 and V are the eigenvalues and vectors of AT A

Except in the special case where there are repeated eigenvalues, V is uniquely defined (apart from scaling factors of +/-1).

Starting with A AT, the same is true of U.
 
But in the case of T:R^2 -> R^2 where T is the zero transformation, any two linearly independent vectors from R^2 can be eigenvectors of T. Even with the restriction that the two eigenvectors we pick from R^2 are orthogonal, there's more than one way we can pick the vectors aside from ordering. Since his matrix in question is specifically the zero matrix, the pathological example is actually relevant.
 
I think the OP's problem was that he got a zero V matrix, but that is wrong. Since we don't know what he/she actually did to get V = 0, it's impossible to say what the mistake was.

If you look at the question in the OP's link, the matrix does not have any repeated singular values.

I agree that U and V are not unique if there are repeated singular values.
 

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