3

Recall the definition of border-rank of a tensor T:

border-rank(T) = the minimum r such that $\forall \epsilon > 0$ there exists an approximate tensor $T' = \sum^r_{i=1} u_i \otimes v_i \otimes w_i $, such that $ \|T - T' \| \leq \epsilon $

Intuitively, what is the smallest r, so that when I restrict myself to rank r tensors, I can get arbitrarily close.

Recall the definition of rank of a tensor T is:

rank(T) = the minimum number r, such that we can write T as a sum of r rank one tensors, i.e. minimum r such that $T = \sum^r_{i=1} u_i \otimes v_i \otimes w_i$.

For order 2 tensors (i.e. matrices) there is an intrinsic limit to how well one can approximate any order 2 tensor. And its governed by the SVD of the matrix.

However, when the order of the tensor is 3 things change. What is so special about order 2 tensors that their border rank and rank is the same? When one increases the order of the tensor, this changes completely and the border rank exists for higher order tensors. What is the intuition (or precise/rigorous explanation) for why order 3 tensors have this special behavior wrt to ranks? Or in fact, why increasing the order makes this possible.

Charlie Parker
  • 6,546
  • 2
    One important distinction: higher order tensors cannot be decomposed as the direct sum of symmetric and antisymmetric tensors. There are other irreducible forms. I'm not sure if this is the reason for your question, hence this is a comment. More than this, I'd like to see a lot more on the MSE about rank of a tensor in general. I have much to learn :) – James S. Cook Mar 24 '15 at 21:56
  • 3
    The following paper proves some theorems regarding border rank. See for example thm 1.1: http://arxiv.org/pdf/math/0607647v2.pdf – Daniel Mar 24 '15 at 22:16
  • See https://math.stackexchange.com/questions/1182746/proof-that-border-rank-and-the-rank-of-a-matrix-order-2-tensors-are-equivalent – Zach Teitler Sep 16 '19 at 15:42

0 Answers0