Kruskal's uniqueness inequality is sharp
Let \(F\) be a field and \(V\) an \(F\)-vector space. For a subset \(A\subset V\), the Kruskal rank \(k(A)\) is the largest integer such that \(A\) has at least \(k(A)\) elements and every subset of \(A\) with \(k(A)\) elements is linearly independent. For \(m\) vector spaces \(V_1,V_2,\dots,V_m\) over \(F\), \textit{J. B. Kruskal} [Linear Algebra Appl. 18, 95--138 (1977; Zbl 0364.15021)] proved that a tensor in \(V_1\otimes V_2\otimes \dots\otimes V_m\) of rank \(r\) has a unique decomposition as a sum of \(r\) pure tensors if a certain inequality is satisfied. The author of the paper shows that the uniqueness fails if the inequality is weakened.
scientific article; zbMATH DE number 6125886