# Maximal rank of tensors in $F^n\otimes …\otimes F^n$.

What is the largest possible rank of a tensor in the space $$F^n\otimes ...\otimes F^n$$ where we have $$k$$ copies of $$F^n$$? It is quite easy to see that it is at most $$n^{k-1}$$ (I have commented the proof for this). For $$k=2$$ this bound is in fact tight. What about for $$k\geq2$$?

In particular, I'm looking for a tighter bound, or an example of a tensor with rank $$n^{k-1}$$ in these spaces. If this is too much, the spaces with $$k=3,4$$ would suffice.

• The rank of $t\in F^{n_{1} }\otimes ... F^{n_{k}}$ is the minimal value $r$ such that we can write $$t=\sum_{i=1}^r u_{1,i}\otimes ... u_{k,i}$$ with $u_{i,j}\in F^{n_i}$. – Joshua Tilley Oct 30 '18 at 13:01
• What notion of rank do you have? – Joshua Tilley Oct 30 '18 at 13:03
• The proof does not work the other way around: the decomposition one would get in the last step would be of the wrong form. – Joshua Tilley Oct 30 '18 at 13:06
• I failed to mention in the proof that the $u_i$ have to be rank one themselves for $r$ to equal the rank. Doing the same with the $v_i$ would not yield rank one $u_i$s. – Joshua Tilley Oct 30 '18 at 13:08
• Ah right, sorry. I see what you mean now. – Arnaud D. Oct 30 '18 at 13:13

Note that to get the bound $$\textrm {rank}\left(t\right)\leq n^{k-1}$$ for $$t\in F^n\otimes ...\otimes F^n$$, suppose that $$t=\sum _{i=1}^r u_i \otimes v_i$$ with $$u_1,...,u_r\in F^n\otimes ...\otimes F^n$$ themselves rank-one with $$k-1$$ copies of $$F^n$$, and $$v_1,...,v_r\in F^n$$. Suppose that $$r$$ is the minimal number for such a decomposition. We will show that $$u_1,...,u_r$$ is linearly independent, and hence that $$r\leq n^{k-1}$$.
Suppose for contradiction that $$u_1,...,u_r$$ is linearly dependent $$\sum _{i=1}^r a_iu_i=0$$ without loss of generality, assume that $$a_r\neq 0$$ and hence we may assume $$a_r= -1$$. We have $$t=\sum _{i=1}^r u_i \otimes v_i$$ $$=\sum _{i=1}^{r-1} u_i \otimes v_i+u_r \otimes v_r$$ $$=\sum _{i=1}^{r-1} u_i \otimes v_i+\sum _{i=1}^{r-1} a_iu_i \otimes v_r$$ $$=\sum _{i=1}^{r-1} u_i \otimes \left(v_i+a_iv_r\right)$$ but this contradicts the minimality of $$r$$. Hence, $$u_1,...,u_r$$ is linearly independent, thus $$\textrm {rank}\left(t\right)\leq n^{k-1}$$.