Abstract-This note studies the problem of nonsymmetric rank-one matrix completion. We show that in every instance where the problem has a unique solution, one can recover the original matrix through the second round of the sum-ofsquares/Lasserre hierarchy with minimization of the trace of the moments matrix. Our proof system is based on iteratively building a sum of N − 1 linearly independent squares, where N is the number of monomials of degree at most two, corresponding to the canonical basis (z α − z α 0 ) 2 . Those squares are constructed from the ideal I generated by the constraints and the monomials provided by the minimization of the trace.
I. NON SYMMETRIC MATRIX COMPLETIONThis paper introduces a deterministic recovery result for non symmetric rank-1 matrix completion by using the Lasserre hierarchy of semidefinite programming relaxations. To our knowledge, the closest result in the current literature is [1] where the authors use the same hierarchy and certify recovery in the case of tensor decomposition. Our paper also shares its deterministic nature with [2] where the authors derive recovery from the spectral properties of a graph Laplacian. Finally, this paper can also be related to [3] in which the authors study noisy tensor completion and use the sixth round of the Lasserre hierarchy to derive probabilistic recovery guarantees.We will use M(r; m × n) to denote the set of matrices of rank r. This set is an algebraic determinantal variety that can be completely characterized through the vanishing of the (r + 1)-minors. This determinantal variety has dimension (m+n−r)r.The general nonsymmetric rank-1 matrix completion problem consists in recovering an unknown matrix X ∈ M(1; m× n) such that X = xy T , given a fixed subset of its entries [4], find X subject to rank(X) = 1As a slight abuse, we also speak of constraints X ij = A ij as belonging to Ω. In relation to problem (1), we introduce the mapping R Ω : R m×n → R |Ω| that corresponds to extracting the observed entries of the matrix. We let R 1 Ω denote the restriction of R Ω to matrices of rank-1, i.e R 1 Ω : M(1; m × n) → R |Ω| . Invertibility of this restriction R 1 Ω is a natural question. In other words, when can one uniquely recover the matrix X from the knowledge of R Ω (X) and the fact that X has rank 1 ?In particular, this paper considers the completion problem on M * (1, m×n), where M * (1, m×n) denotes the restriction of M(1; m × n) to matrices for which none of the entries are zero. The reason for this is that if a rank-1 matrix has a zero element, then the corresponding row or column will be zero, and it is easy to see that the completion problem will generically lack injectivity.Respectively denote by V 1 , V 2 the row and column indices of X. We consider the bipartite graph G(V 1 , V 2 , E) associated to problem (1), where the set of edges in the graph is defined by (i, j) ∈ E iff (i, j) ∈ Ω. The conditions for the recovery of the matrix X from the set Ω are related to the properties of this bipartite graph. In particular, we have the f...