Remarks on the Symmetric Rank of Symmetric Tensors

Abstract

We give sufficient conditions on a symmetric tensor $\mathcal{S}\in\mathrm{S}^d\mathbb{F}^n$ to satisfy the following equality: the symmetric rank of $\mathcal{S}$, denoted as $\mathrm{srank\;}\mathcal{S}$, is equal to the rank of $\mathcal{S}$, denoted as $\mathrm{rank\;}\mathcal{S}$. This is done by considering the rank of the unfolded $\mathcal{S}$ viewed as a matrix $A(\mathcal{S})$. The condition is $\mathrm{rank\;}\mathcal{S}\in\{\mathrm{rank\;}A(\mathcal{S}),\mathrm{rank\;}A(\mathcal{S})+1\}$. In particular, $\mathrm{srank\;}\mathcal{S}=\mathrm{rank\;}\mathcal{S}$ for $\mathcal{S}\in\mathrm{S}^d\mathbb{C}^n$ for the cases $(d,n)\in\{(3,2),(4,2),(3,3)\}$. We discuss the analogues of the above results for border rank and best approximations of symmetric tensors.