Uniqueness of Tensor Decompositions (Aren't Matrix Decompositions a Special Case?)

Question

It seems that higher-order tensors (of order 3 or higher) generally have unique decompositions under relatively mild conditions. For example, Kruskal proved that if an order-3 Tensor $T$ can be decomposed as the outer product of matrices $A, B, C$, with Kruskal ranks $k_A, k_B, k_C$ respectively, then as long as the rank $R$ of $T$ satisfies

$$R \le \frac{k_A + k_B + k_C - 2}{2},$$

then the decomposition is unique.

My question is: aren't matrix decompositions a special case of tensor decompositions? In other words, I can think of a matrix $M$ as a tensor whose third dimension is of size 1. If I was to apply the same result to matrix decompositions, I would get that that the matrix decomposition is unique as long as its rank is:

$$R \le \frac{k_A + k_B + 1 - 2}{2} = \frac{k_A + k_B - 1}{2},$$

(where we have assumed that $C$ is just a matrix of ones), but we know that this is generally not true. You can take a matrix, and rotate its component matrices, and get a different decomposition (the so-called "rotation problem", so matrix decompositions are not unique.

How do we reconcile these facts?

Answer 1

Kruskal rank can not be higher than the canonical rank, so you can never have:

                         2*R <= 2*R - 1

which means that the decomposition is never unique in the general case.