[Paper Review] Nonnegative rank depends on the field II

Yaroslav Shitov|arXiv (Cornell University)|May 23, 2016

graph theory and CDMA systems13 references3 citations

TL;DR

This paper resolves a long-standing problem in nonnegative matrix factorization by constructing a 21×21 nonnegative integer matrix that admits a nonnegative rank-19 factorization over the reals but not over the rationals. The proof uses structured matrix blocks with irrational parameters (e.g., √2, 1+√0.5) to force irrationality in any minimal rational factorization, demonstrating that nonnegative rank depends on the underlying field, as conjectured by Cohen and Rothblum in 1993.

ABSTRACT

We provide an example of a $21 imes 21$ matrix with nonnegative integer entries which can be written as a sum of $19$ nonnegative rank-one matrices but not as a sum of $19$ rational nonnegative rank-one matrices. This gives a solution for a problem posed by Cohen and Rothblum in 1993.

Motivation & Objective

• To resolve Problem 1 posed by Cohen and Rothblum (1993), asking whether nonnegative rank can differ over ℚ and ℝ for rational matrices.
• To construct a concrete example of a rational matrix where the nonnegative rank over ℝ differs from that over ℚ.
• To demonstrate that the minimal nonnegative rank factorization can require irrational entries even when all matrix entries are rational integers.
• To provide a definitive counterexample showing that nonnegative rank is not invariant under field extension, particularly from ℚ to ℝ.
• To clarify the role of irrational numbers in nonnegative matrix factorization, with implications for computational complexity and polyhedral extended formulations.

Proposed method

• Construct a 21×21 matrix 𝒜 composed of structured submatrices: 𝒞(α,α,α,α,√2), 𝒷(2−α,2−α), 𝒷(2−α), and 𝒷(2−√2), with α = 1+√0.5.
• Use known rank bounds on submatrices: Claim 5 shows the 4×4 block V has nonnegative rank 4; Claim 7 gives exact conditions for rank 3 in 𝒞.
• Prove 𝒜 has nonnegative rank ≤19 over ℝ by decomposing it into sum of 19 rank-one matrices using the irrational parameters.
• Assume a rational factorization with 19 rational rank-one matrices and derive a contradiction via color-based decomposition of entries into red, blue, yellow, magenta, and uncolored components.
• Use Claim 4 to show that if the parameters in a 𝒷 block are unequal, its nonnegative rank is at least 5, which forces irrationality in any rational factorization.
• Apply Claim 2 to show that any rational factorization of the uncolored part U must have rank at most 3, leading to a contradiction when combined with the irrationality of required entries in the red submatrix.

Experimental results

Research questions

• RQ1 Does there exist a rational matrix A such that Rank₊(A,ℚ) ≠ Rank₊(A)?
• RQ2 Can nonnegative rank be strictly smaller over ℝ than over ℚ for a rational matrix?
• RQ3 Is the minimal nonnegative rank factorization of a rational matrix necessarily rational?
• RQ4 What structural properties force irrational entries in minimal nonnegative factorizations?
• RQ5 How do field extensions affect the nonnegative rank of rational matrices?

Key findings

• The 21×21 matrix 𝒜 has nonnegative rank 19 over ℝ, as shown by constructing a factorization using irrational parameters α = 1+√0.5 and d = √2.
• The same matrix 𝒜 has nonnegative rank at least 20 over ℚ, proven by contradiction assuming a rational factorization with 19 rational rank-one matrices.
• The irrational parameters in the matrix blocks (e.g., √2, 1+√0.5) are essential for achieving rank 19 over ℝ, and their absence in rational factorizations leads to structural contradictions.
• The red submatrix in the factorization must have equal entries at positions (5,4) and (5,5), which is only possible with irrational parameters, violating rationality.
• The uncolored part of any rational factorization must have nonnegative rank at most 3, but this leads to a contradiction due to the required structure of the red submatrix.
• The result confirms that nonnegative rank is not invariant under field extension, providing a negative answer to Cohen and Rothblum’s 1993 problem.