Canonical Polyadic Decomposition of Third-Order Tensors: Reduction to Generalized Eigenvalue Decomposition

Now, the statement (i) follows from (S.1.3) by setting y = x. (ii) Since the vectors ci1 , . . . , ciK−1 are linearly independent in R , it follows that there exists a vector y such that det [ ci1 . . . ciK−1 y ] 6= 0. Hence, by (S.1.3), the (i1, . . . , iK−1)-th column of B(C) is nonzero. (iii) follows from (S.1.3) and the fact that det [ ci1 . . . ciK−1 y ] = 0 if and only if y ∈ span{ci1 , . . . , ciK−1}. (iv) Assume that the (i1, . . . , iK−1)-th and the (j1, . . . , jK−1)-th column of B(C) are proportional to a nonzero vector t. Set E := span{ci1 , . . . , ciK−1 , cj1 , . . . , cjK−1}. Since kC = K, it follows that E = R . On the other hand, by (iii), the nonzero vector t is orthogonal to E, which is a contradiction. Lemma S.1.2. Let C ∈ RK×R, K ≤ R, and kC = K. Let also B(C) be defined by (S.1.1). Then (i) every column of the R×CK−1 R matrix CB(C) has exactly K−1 zero entries. Namely, if d = [ d1 . . . dR ] is the (j1, . . . , jK−1)-th column of C B(C), then dr = 0 if and only if r ∈ {j1, . . . , jK−1};