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I read this sentence.

Suppose that the matrix $A_{ij}$ of dimension $n_i \times n_j$ has rank $k$ to precision $\epsilon$, then there exists a factorization of $A_{ij}$ of the form: $A_{ij} = L_i S_{ij} R_j + \text{O}(\epsilon)$.

I wonder what does matrix rank $k$ to precision $\epsilon$ mean?

Thank you.

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    Where did you read this?2011-10-25
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    It must mean that there's a rank-$k$ matrix within a distance of $\epsilon$ from $A$, for some appropriate (but unidentified) norm on the space of $n_i\times n_j$ matrices. I wonder what $k$ has to do with the conclusion of the claim, though.2011-10-25
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    @Henning: the "unidentified" norm is often the 2-norm, especially in the case of diagnosing badly-behaved least-squares problems and other problems that necessitate the use of orthogonal matrices for decompositions.2011-10-25
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    @ChrisEagle This is a restatement of Theorem 3 in ON THE COMPRESSION OF LOW RANK MATRICES by H.Cheng. You could get it through Google Scholar search.2011-10-25

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