Answers are grounded in this paper and cite their sources.
We give sufficient conditions on a symmetric tensor SᵈFⁿ to satisfy the following equality: the symmetric rank of S, denoted as srank\;S, is equal to the rank of S, denoted as rank\;S. This is done by considering the rank of the unfolded S viewed as a matrix A(S). The condition is rank\;S∈\;A(S),rank\;A(S)+1\. In particular, srank\;S=rank\;S for SᵈCⁿ for the cases (d,n)∈\(3,2),(4,2),(3,3)\. We discuss the analogues of the above results for border rank and best approximations of symmetric tensors.
No takes yet. Share an insight, caveat, or question.
Shmuel Friedland (2016) studied this question.
Synapse has enriched 5 related papers. Consider them for comparative context: