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.

  • $\begingroup$ 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

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 $\textrm {rank}\left(t\right)\leq n^{k-1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.