Remarks on the symmetric rank of symmetric tensors

TL;DR

This paper establishes conditions under which the symmetric rank of a tensor equals its rank, focusing on symmetric tensors and their unfolded matrix representations, with specific cases and implications for border rank.

Contribution

It provides sufficient conditions linking symmetric rank and rank of symmetric tensors via their unfolded matrix, including specific cases and discussions on border rank.

Findings

srank(S) = rank(S) when rank(S) is in {rank(A(S)), rank(A(S))+1}

Equality holds for certain tensor dimensions, e.g., (3,2), (4,2), (3,3)

Discusses extensions to border rank and tensor approximations

Abstract

We give sufficient conditions on a symmetric tensor S in S^dF^n to satisfy the 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) is in {rank(A(S)),rank (A(S))+1}. In particular, srank(S)=rank(S) for S in S^dC^n for the cases (d,n) in {(3,2),(4,2),(3,3)}. We discuss the analogs of the above results for border rank and best approximations of symmetric tensors.