Bounds on the tensor rank

Abstract

We give a sufficient criterion for a lower bound of the cactus rank of a tensor. Then we refine that criterion in order to be able to give an explicit sufficient condition for a non-redundant decomposition of a tensor to be minimal and unique.

Tensor decomposition for infectious disease incidence data

“…Characterizing the rank of a tensor is a complex mathematical problem without a simple solution (Alexeev, Forbes, & Tsimerman, 2011; Ballico, Bernardi, Chiantini, & Guardo, 2018; Kolda & Bader, 2009; Stegeman & Friedland, 2017). Therefore, we attempt tensor decomposition beginning by selecting a single component, and increasing the number of components until the algorithm consistently converges.…”

Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties

“…If z := r X (q) < r, say q = S with S ⊂ X and #S = z = 1, q is in the linear span of all S ∪ {o}, o ∈ S \ {o}, but none of these sets irredundantly spans q. For tensors a general tensor of each rank s is irredundantly spanned by sets of t points if s < t ≤ s in infinitely many ways ( [38], Theorem 3.8). The following example with z any positive integer shows that sometimes above its rank z := r X (q) a point q may be irredundantly spanned by a unique set of cardinality z + 1.…”

Ranks and symmetric ranks of cubic surfaces

“…The conjecture has been proved in many special cases: when the symmetric rank is at most two [10], when the rank is less than or equal to the order [27], and when the rank is at most the flattening rank plus one [12]. Furthermore, the conjecture has been proved to generically hold in certain families of tensors [2]. On the other hand, a counter-example to Comon's conjecture for complex rank has been found by Shitov [22], a tensor of size 800 × 800 × 800.…”