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 $\frac{1}{3}n^2$.
Best Answer
The lower bound can be improved slightly to $n^3/(3n-2)$ by noting that in $x\otimes y\otimes z$ one can assume $|y|=|z|=1$. See also Chapter 20, "Typical Tensorial Rank", in the book Algebraic Complexity Theory, by Peter Bürgisser, Michael Clausen, and Mohammad Amin Shokrollahi.
An upper bound of $n^2−n−1$ is shown in http://arxiv.org/abs/0909.4262v4 so that is probably the best known upper bound.