Consider the Segre embedding $\mathbb{P}^n\times \mathbb{P}^n\rightarrow \mathbb{P}^N$, and let $S\subset\mathbb{P}^N$ be its image. 


Then $rank(Z)\leq k$ implies that $Z\in Sec\_k(S)$. Moreover if $Z\in Sec\_k(S)$ is general then $rank(Z)\leq k$. Does this last statement hold for any $Z\in Sec\_k(S)$ and not just for $Z\in Sec\_k(S)$ general? What if we replace the Segre $S$ with the Veronese $V$ parametrizing rank $1$ symmetric matrices or the Grassmannian $\mathbb{G}(1,n)$ parametrizing rank $2$ anti-symmetric matrices?



Rank of matrices and secant varieties
Your question is:



> 
> Let $S \subset \mathbb{P}^N$ be the image of the Segre map
>  $\mathbb{P}^n \times \mathbb{P}^n \to \mathbb{P}^N$.
>  Let $Z \in \operatorname{Sec}\_k(S)$. Does $Z$ have rank at most $k$?
> 
> 
> 


Yes. This is *matrix rank*. The elements of $\mathbb{P}^N$ are $(n+1) \times (n+1)$ matrices (up to a scalar factor) and rank with respect to $S$ is ordinary matrix rank. Every element of $\operatorname{Sec}\_k(S)$ has rank at most $k$.


A modified question is:



> 
> Let $S \subset \mathbb{P}^N$ be the image of the Segre map
>  $\mathbb{P}^{n\_1} \times \dotsm \times \mathbb{P}^{n\_s} \to \mathbb{P}^N$.
>  Let $Z \in \operatorname{Sec}\_k(S)$. Does $Z$ have rank at most $k$?
> 
> 
> 


When $s \geq 3$, no. A tensor of border rank $k$ may have rank strictly greater than $k$ or $k+1$. An example is below.


In the symmetric case, the answer is yes for Veronese varieties of degree $2$ (corresponding to quadratic forms), and no for Veronese varieties of degree $d \geq 3$. For example, the homogeneous binary form $x y^{d-1}$ has rank $d$, but lies on $\operatorname{Sec}\_2(V)$:
$$
 x y ^{d-1} = \lim\_{t \to 0} \frac{(tx+y)^d - y^d}{dt},
$$
where each $(1/dt)((tx+y)^d - y^d) \in \operatorname{Sec}\_2(V)$.


In the antisymmetric case I believe the answer is again no.


Landsberg's book is a good reference. See also more recently <https://arxiv.org/abs/1811.12725> and <https://arxiv.org/abs/1812.10267>.


For an example of a tensor with border rank $k=2$ and rank greater than $2$ take
$$
 T = x y^{d-1} = x \otimes y \otimes \dotsm \otimes y + y \otimes x \otimes y \otimes \dotsm \otimes y + \dotsb + y \otimes \dotsm \otimes y \otimes x,
$$
where $x,y \in \Bbbk^2$ are a basis. It visibly has rank at most $d$ and you can prove that it has rank equal to $d$. It has border rank $2$, since $T = \lim\_{t \to 0} ((tx+y)^{\otimes d} - y^{\otimes d})/t$.


