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Let's say I have an image representing a sampled function. It just so happens that I know this function can be represented as a sum of individual outer products along with some noise. So I might have an image defined as:

$I(x,y) = a_0b_0^T + a_1b_1^T + ... +\ \Omega$

Where the a and b vectors are the individual vectors forming the outer products, and $\Omega$ is a matrix of IID gaussian random variables.

If I decompose $I(x,y)$ using the SVD, I similarly get a sum of outer products:

$I(x,y) = U \Sigma V^T = s_0u_0v_0^T + s_1u_1v_1^T + ... s_Nu_Nv_N^t$

I would like to be able to extract the original image components (neglecting the noise), given these "singular images". Is there anything that I can confidently say about the relationship between my source inner products and output inner products?

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The SVD is usually employed to perform denoising on an image, that is the problem that you stated: given $\tilde{I} = I + \Omega$ (where $\Omega$=noise), try to approximate $I$.

One of the possible approaches to the denoising problem is to consider a truncated SVD of the given image $\tilde{I}$, i.e.:

  1. Compute $\tilde{I} = \sum_{i=1}^n \sigma_i u_iv_i^T$
  2. Consider just the first $k$ terms $I^{(k)} = \sum_{i=1}^k \sigma_i u_iv_i^T$.

Then you have $I^{(k)} \approx I$ for appropriate choices of the truncation index $k$, that depends both on the noise magnitude $\varepsilon = \|\Omega\|$ and on the decay rate of the $\sigma_i$'s.

Informally speaking, this process works mainly because the last singular vectors are the most noisy ones. I don't know a formal explanation of this fact, and I guess that it could be truly involved.

An apparently good reference I found by googling "svd + denoising" is this one.

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  • $\begingroup$ I think that's not quite what I want. I certainly want $I$ but I want it broken out like the original outer products. That's the thing that's tripping me up is if there's necessarily a relationship there. $\endgroup$ – gct Mar 8 '14 at 2:39
  • $\begingroup$ Sometimes there can be quite a lot of information in the last components of the SVD, see stats.stackexchange.com/questions/177102/… $\endgroup$ – kjetil b halvorsen Nov 3 '15 at 20:02

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