Rank of tensors in terms of ranks of associated linear maps Let $V$ be a vector space over a field $k$, let $w \in \otimes^l V$ be a tensors. We call $w$ a simple tensor if it can be written as
$$
w=w_1 \otimes w_2 \otimes \ldots \otimes w_l,
$$
where $w_i \in V$ for $i=1,\dots, l$. Simple tensors are also called rank $1$ tensors, and a tensor $w$ is called rank $r$ tensor if $r$ is the minimum number of terms in a presentation of $w$ as a sum of simple tensors. 
With any tensor $w$ we can associate a finite sequence of linear maps
$$
L_i : \otimes^i V^* \to \otimes^{l-i} V,
$$
where $1 \leq i \leq l-1$, $V^*$ is the dual space, and the maps $L_i$ acts on $\otimes^i V^*$ by evaluating tensor product of vectors on a tensor product of forms.
Is it true that if $\operatorname{rk}(L_i)=1$ for $1 \leq i \leq l-1$ then tensor $w$ is simple? In other direction it is clearly true. Thus, it could provide a criterion for a tensor to be simple in terms of rank of linear maps.
If the answer to previous question is positive, is it true for higher rank tensors?
 A: The answer for your first question is yes. The answer for your second is no.
The next proposition is well known. It was proved here.

Proposition: Let $V,W$ be vector spaces over $k$. Let $w=\sum_{i=1}^ra_i\otimes b_i=\sum_{i=1}^sv_i\otimes w_i\in V\otimes W$.


*

*If $\{a_1,\ldots, a_r\}$ is linear independent then $\text{span
    }\{b_1,\ldots,b_r\}\subset\text{span }\{w_1,\ldots,w_s\}$.

*If $\{b_1,\ldots, b_r\}$ is linear independent then $\text{span }\{a_1,\ldots,a_r\}\subset\text{span }\{v_1,\ldots,v_s\}$.

*If $\{a_1,\ldots, a_r\}$, $\{b_1,\ldots, b_r\}$ are linear independent then $s\geq r$ and the tensor rank of $w$ is $r$.



Answer for problem 1:
Now, let $w\in\bigotimes^lV$. Thus we can consider $w\in V\otimes W$, where $W=\bigotimes^{l-1}V$.
Assume each $L_i$ has rank $1$. Let $w=\sum_{i=1}^ra_i\otimes b_i$, where $a_i\in V$ and $b_i\in W$ and $\{a_1,\ldots, a_r\}$, $\{b_1,\ldots, b_r\}$ are linear independent. We can find $f_i\in \bigotimes^{l-1}V^*$, for $1\leq i\leq r$, such that $f_i(b_j)=\delta_{ij}$, for $1\leq i,j\leq r$.
Thus, $Id\otimes f_i(w)=a_i$, for $1\leq i\leq r$. Thus, one of your maps, $L_{l-1}$, has rank $r$. Therefore, $r=1$ and $w=a_1\otimes b_1$, where $a_1\in V$ and $b_1\in \bigotimes^{l-1}V$. By induction on $l$, we can prove that $b_1=v_2\otimes\ldots\otimes v_l$ and we are done.
Answer for problem 2:
Now, let us prove that this idea does not work for rank 2.
Now, let $a,b\in V$ be linear independent. Consider $w=a\otimes a\otimes a+a\otimes b\otimes b+b\otimes a\otimes b$.
Notice that any map $L_i$ has rank 2. Please check it.
Now, assume $w= c\otimes d\otimes e+f\otimes g\otimes h$, for some $c,d,e,f,g,h\in V$.
Notice that $w$ as a tensor in $V\otimes W$, where $W=\bigotimes^2V$, has tensor rank 2, because $w=a\otimes(a\otimes a+b\otimes b)+b\otimes (a\otimes b)$ and by item 3 of proposition above.  
Thus, $\{c,f\}$ must be linear independent. 
Analogously, we can prove that $\{d,g\}$ is linear independent and $\{e,h\}$ is too. Using  this proposition again we have $\text{span}\{a,b\}=\text{span}\{c,f\}=\text{span}\{d,g\}=\text{span}\{e,h\}$.
Let $\delta_a,\delta_b\in V^*$ be such that $\delta_a(a)=
\delta_b(b)=1$ and $\delta_a(b)=\delta_b(a)=0$.
Thus, $Id\otimes Id\otimes \delta_a(w)=a\otimes a= c\otimes d\delta_a(e)+f\otimes g \delta_a(h)$. 
Since $\{c,f\}$ and $\{d,g\}$ are linear independent then, by item 3 of the proposition, we must have or $\delta_a(e)=0$ or $\delta_a(h)=0$. Assume $\delta_a(e)\neq0$ and $\delta_a(h)=0$.
Now remind that $e=\delta_a(e)a+\delta_b(e)b$ and $h=\delta_b(h)b$, because $\text{span}\{a,b\}=\text{span}\{e,h\}$. 
Thus, $c\otimes d\otimes e=c\otimes d\otimes (\delta_a(e)a+\delta_b(e)b)=a\otimes a\otimes a+a\otimes a\otimes \dfrac{\delta_b(e)}{\delta_a(e)}b$.
Therefore $a\otimes b\otimes b+b\otimes a\otimes b=a\otimes a\otimes \dfrac{\delta_b(e)}{\delta_a(e)}b+f\otimes g\otimes \delta_b(h)b$
Thus, $m=a\otimes (b-\dfrac{\delta_b(e)}{\delta_a(e)}a)\otimes b+b\otimes a\otimes b=f\otimes g\otimes \delta_b(h)b$. 
Finally, $Id\otimes Id\otimes \delta_b(m)=a\otimes (b-\dfrac{\delta_b(e)}{\delta_a(e)}a)+b\otimes a=f\otimes g\delta_b(h)$, which is impossible. The left side of the equation has tensor rank 2 and the right side has tensor rank 1, by item 3 of proposition.
