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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? – Chris Eagle Oct 25 '11 at 14:17
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. – Henning Makholm Oct 25 '11 at 14:19
@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. – J. M. Oct 25 '11 at 14:37
@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. – Yao Jin Oct 25 '11 at 16:59

The rank of a matrix is the number of its nonzero singular values.

Rank to precision $\epsilon$ means that in computing the rank of the matrix, we consider every singular value of the matrix that is less than $\epsilon$ as zero.

This is also known as "numerical rank": the number of singular values greater than $\epsilon$.

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I see. Thank you. – Yao Jin Oct 25 '11 at 14:28
Put another way: the SVD is a useful tool for diagnosing the numerical rank of a matrix. If the matrix has any singular values less than machine epsilon (or more often, the norm of the matrix times machine epsilon), you have what is called an ill-conditioned matrix. – J. M. Oct 25 '11 at 14:34

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