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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.

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  • $\begingroup$ 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 $u_{i,j}\in F^{n_i}$. $\endgroup$ – Joshua Tilley Oct 30 '18 at 13:01
  • $\begingroup$ What notion of rank do you have? $\endgroup$ – Joshua Tilley Oct 30 '18 at 13:03
  • $\begingroup$ The proof does not work the other way around: the decomposition one would get in the last step would be of the wrong form. $\endgroup$ – Joshua Tilley Oct 30 '18 at 13:06
  • $\begingroup$ 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. $\endgroup$ – Joshua Tilley Oct 30 '18 at 13:08
  • $\begingroup$ Ah right, sorry. I see what you mean now. $\endgroup$ – Arnaud D. Oct 30 '18 at 13:13
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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 $\textrm {rank}\left(t\right)\leq n^{k-1}$.

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