# Matrix Factorization in Recommender Systems: Uniqueness of SVD?

I was studying the collaborative filtering approach about recommender system and I read about matrix factorization approach. In SVD version, I have not figured out how the non-uniqueness of the decomposition is not a problem for the recommendation process. By varying the decomposition recommendations have to change. Where theorically am I doing wrong ?

• Non-uniqueness is not catastrophic, it does not imply that a solution is grossly biased or anything like that. (to paraphrase a great) Approximate answers to exact questions can be extremely useful. :) Apr 27, 2016 at 21:25
• Hello, thanks for the answer, then I try to rephrase the question. What guarantees that two different approximations , are close to each other in the latent factors space ? Apr 28, 2016 at 6:12
• It is a comment not an answer. :) Welcome to the community by the way. I would suggest you edit our original post as well as its title to reflect your new take on the question. It will attract more attention (hopefully). Apr 28, 2016 at 7:22
• The singular vectors in SVD need not be unique, and it's ok. Jun 7, 2016 at 5:20
• We have no guarantee about the closeness in latent space. Two different decompositions can be arbitrarily far from one another (depending how you measure). What's important is that the ratings themselves stay the same. See the answer below.
– Amir
Feb 16, 2017 at 12:25

Here is what you are probably missing: while the decomposition might not be unique, the reconstruction is unique. In other words, let's assume that $(U,V)$ and $(\tilde U,\tilde V)$ are two decompositions, which differ, e.g. by permuting the rows of both matrices (this corresponds to a permutation among the top $k$ of the singular values). While the decompositions are indeed different, the reconstruction matrices will be the same, i.e.
$$UV^T = \tilde U \tilde V^T.$$