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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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    @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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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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    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.2011-10-25