3
$\begingroup$

Im my algorithm, I am working with Singular Value Decomposition (SVD).

I have an input matrix $A_{in} \in \{0,1\}^{(m * n)} $, made by $n$ rows and $m$ colums. All the entries are 0 or 1.

I decompose it in $A = U * \Sigma * V^{T}$

After choosing a proper truncation level $k$, I construct an output matrix $A_{out} \in \mathbb{R}$, this way:

$ U_{k} * \Sigma_{k} * \; V^{T}_{k} = A_{out}$

All the $A_{out}$ entries are real valued.

What is the meaning of each real value?

Is it the likelihood that the element is near to 1.0 in the reconstruction?

Or are the real values just to create an order, from the less likely to the most likely?

Improve this question
$\endgroup$
2
  • $\begingroup$ Could you clarify the sense in which you use "likely" here, given that you offer no probability model nor is there anything in this setting that can be interpreted naturally as a probability? You do use the term "reconstruction": that says the output matrix is an approximation to the input matrix. Isn't that explicitly the "meaning" of its coefficients? (Note that the coefficients of $A_\text{out}$ can lie outside the range $[0,1]$, so it makes no sense in general to think of them as "likelihoods.") $\endgroup$
    – whuber ♦
    Aug 23, 2013 at 15:18
  • $\begingroup$ @whuber Yes, you're right: the output matrix is an approximation of the input matrix. The output matrix elements approximate the input matrix elements. I am wondering: beyond this approximation, are there other eventual meaning to this values? For example, if we order them from the highest to the lowest, can we see an hidden sense in this ranking? Thanks $\endgroup$
    – DavideChicco.it
    Aug 23, 2013 at 16:09

1 Answer 1

Reset to default
5
$\begingroup$

The individual real values don't necessarily represent any particular ordering. Aout is the closest rank-k approximation to A, in the sense that the overall sum of the squares of the differences between the entries of the two matrices has been minimized. That doesn't say much, however, about any individual value of Aout. With your original matrix restricted to 0/1 values, consider whether the SVD is the best approach to your problem.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.