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The answer to question 1 is affirmative. There are several lower bounds in various papers. I'll take the idea from https://arxiv.org/abs/1503.08253 (Buczyński and myself, "Some examples of forms of hi …

The analogous conjecture for border ranks, that the border rank and border symmetric rank (border Waring rank) of a symmetric tensor are equal, seems to be still open (as far as I know, as of September …

This map, that sends a tensor to its $i_2$'th $2$-slice, is surjective: any matrix arises as the $2$-slice of a tensor given by sticking that matrix into the appropriate entries, and filling the rest with …

A tensor of border rank $k$ may have rank strictly greater than $k$ or $k+1$. An example is below. … 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 …

If $T$ is a symmetric tensor of tensor rank $m$, does $T$ have a symmetric decomposition with $m$ terms? Probably not. … Suppose that $T$ is a symmetric tensor with both tensor rank and symmetric tensor rank equal to $m$. In this case, is it possible to find, in some sense, a symmetric decomposition with $m$ terms? …

Let's write $T \in \mathcal{A}^* \otimes \mathcal{B}^* \otimes \mathcal{C}$ for the tensor, and $L_T : \mathcal{A} \otimes \mathcal{B} \to \mathcal{C}$ for the linear map given by $A \otimes B \mapsto … You gave an expression for $T$ as a sum of four simple tensors, so the tensor rank satisfies $\operatorname{rank}(T) \leq 4$. …

For tensors in $\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3$ or in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3$, the maximum rank is $5$. See Bremner, Hu, On Kruskal's theorem t …

If $T$ is a decomposable tensor (simple tensor, rank one tensor) then every flattening of $T$ has rank one. … It is also true for symmetric tensors with (symmetric) border rank, either by the same argument, or by an inequality between symmetric border rank and (ordinary, non-symmetric) tensor border rank. …