I ask you for this *possibly not so simple* task:

**Explicitly**construct a 4th-order tensor $A \in \mathbb{C}^{n_1 \times \ldots \times n_4}$ that**does not**have (border) rank $r$, but for which each matricization**does**have rank $r$; that is $$\forall J \subset \{1,\ldots,4\},\ |J| \notin \{0,4\}\ :\ r_J := \mathrm{rank}(A^{[J]}) = r,$$ where each $A^{[J]} \in \mathbb{C}^{n_J \times n_{\{1,\ldots,4\} \setminus J}}$ for $n_S := \prod_{s \in S} n_s$ is a simple reshaping into a matrix*(see below)*.

For simplicity, let $n_j = \hat{n} \in \mathbb{N}$ for all $j = 1,\ldots,4$.

You may choose $\hat{n},r \in \mathbb{N}$ as you wish. Convincing numerical evidence is welcome as well.

*The following Matlab code displays all relevant matricization ranks:*

```
n = size(A);
r_1 = rank( reshape(A,n(1),[]) )
r_2 = rank( reshape(A,n(2),[]) )
r_3 = rank( reshape(A,n(3),[]) )
r_4 = rank( reshape(A,n(4),[]) )
r_12 = rank( reshape(A,n(1)*n(2),[]) )
r_13 = rank( reshape( permute(A,[1,3,2,4]) ,n(1)*n(3),[]) )
r_14 = rank( reshape( permute(A,[1,4,2,3]) ,n(1)*n(4),[]) )
```

*Note that $\mathrm{rank}(A^{[J]}) = \mathrm{rank}((A^{[J]})^T) = \mathrm{rank}(A^{[\{1,\ldots,d\} \setminus J]})$.*