A Technical proofs Lemma 1 (Kruskal [15]) Suppose A,B,C, ¯ A, ¯ B, ¯ C are six matrices with R columns. There exist integers I$_{0}$ , J$_{0}$ , and K$_{0}$ such that I$_{0}$ + J$_{0}$ + K$_{0}$ ≥ 2 R + 2 . In addition, every I$_{0}$ columns of A are linearly independent, every J$_{0}$ columns of B are linearly independent, and every K$_{0}$ columns of C are linearly independent. Define a triple product to be a three-way array [ A,B,C ] = ( d$_{ijk}$ ) where d$_{ijk}$ = ∑ R r $_{=1}$a$_{ir}$ b$_{ir}$c$_{kr}$ . Suppose that the following two triple products are equal [ A,B,C ] = [ ¯ A, ¯ B, ¯ C ] . Then, there exists a column permutation matrix P , we have ¯ A = AP Λ , ¯ B = BPM, ¯ C = CPN , where Λ ,M,N are diagonal matrices such that Λ MN = identity. Column permutation matrix is a matrix acts on the righthand side of another matrix and permutes the columns of that matrix. Proof of Theorem 1. For each item j , let p$_{j}$ α = P ( Y$_{j}$ = 1 | α ) . that takes two possible values. Let $^{p$_{j}$}$− or p$_{j}$$_{+}$ be these two values. According to Lemma 1, it is sufficient to show that the T -matrices corresponding to the three subsets of items T$_{I}$$_{1}$ , T$_{I}$$_{2}$ , and T$_{I}$$_{3}$ are all of full column rank. Suppose that there are n$_{i}$ items in I$_{i}$ . For each item j ∈ I$_{i}$ , define We further define which is a 2 n$_{i}$ by 2 n$_{i}$ matrix. Because $^{p$_{j}$}$− = p$_{j}$$_{+}$ , each P$_{j}$ is a full-rank matrix and is of rank 2. Thus, P c is rank 2 n$_{i}$ matrix and is a full-rank matrix. Each column of T$_{I}$$_{i}$ is precisely one of the column vector in P $_{i}$. In addition, there is no identical columns in T$_{I}$$_{i}$ , thus its columns vectors are linearly independent. Thus, T$_{I}$$_{i}$ is of full column rank. We construct three groups of items ˜ I$_{1}$ = I$_{1}$, ˜ I$_{2}$ = I$_{2}$ and ˜ I$_{3}$ = { 1 ,... ,J }\ ( I$_{1}$ ⋃ I$_{2}$ ). These three groups are non-overlapping and I$_{3}$ ⊂ ˜ I$_{3}$ . Notice that T$_{˜}$ I$_{1}$ = T$_{I}$$_{1}$,T$_{˜}$ I$_{2}$ = T$_{I}$$_{2}$ and T$_{I}$$_{3}$ is a submatrix of 35 P$_{j}$ =    $^{p$_{j}$}$− p$_{j}$$_{+}$ 1 − $^{p$_{j}$}$− 1 − p$_{j}$$_{+}$    . P i = ⊗ $^{j}$∈I$_{i}$ P$_{j}$