A tensor rank of tree dimentional matrix $M[i,j,k], i,j,k\in [1,\ldots,n]$ is a minimal number of vectors $x_i,y_i,z_i$, such that $M=\sum_{i=1}^d x_i\otimes y_i\otimes z_i$. From dimension argument it easily follows that there exists a matrix of tensor rank at least $\frac{1}{3}n^2$. One can also easily show that every matrix is of tensor rank at most $n^2$.

So I know that maximal tensor rank is between $\frac{1}{3}n^2$ and $n^2$. Does any one knows what is the maximal tensor rank.

p.s. As far as I understand maximal border rank is $\frac{1}{3}n^2$.

The tensor rank of a three dimensional array $M[i,j,k], i,j,k\in [1,\ldots,n]$ is the minimal number of vectors $x_i,y_i,z_i$, such that $M=\sum_{i=1}^d x_i\otimes y_i\otimes z_i$. From dimension argument it easily follows that there exists a tensor of tensor rank at least $\frac{1}{3}n^2$. One can also easily show that every tensor is of tensor rank at most $n^2$.

So I know that the maximal tensor rank is between $\frac{1}{3}n^2$ and $n^2$. Does any one know what is the maximal tensor rank?

p.s. As far as I understand maximal border rank is $\frac{1}{3}n^2$.