$\newcommand\P{\mathbb{P}}$Let $V$ and $W$ be complex vector spaces, and let $\P(V)$, $\P(W)$ and $\P(V\otimes W)$ be the projective spaces attached to $V$, $W$ and $V\otimes W$, respectively. If $v\in V$ is non-zero, I'll denote by $[v]$ the point of $\P(V)$ corresponding to it; it is the equivalence class of $v$ in $V\setminus0$ for the equivalence relation of linear dependence.
Since decomposability of a tensor does not change when we multiply it by a non-zero scalar, we can talk about the indecomposable elements of $\P(V\otimes W)$. Your question is therefore more or less equivalent to
how can we describe the set of indecomposable elements of $\P(V\otimes W)$?
Now, there is a map $f:\P(V)\times\P(W)\to\P(V\otimes W)$ which maps $([v],[w])$ to $[v\otimes w]$. This is a map of projective varieties (in the sense of algebraic geometry) and its image is precisely the set of indecomposable tensors. The image is in fact a subvariety of $\P(V\otimes W)$, which means that it is the common zero set of a finite set of polynomials. Finding these polynomials is a classical problem solved long ago; see Segre embedding for more information (most introductions to algebraic geometry will say something as well).
In the special case where $\dim V=3$ and $\dim W=2$, with bases $\{x_1,x_2,x_3\}$ and $\{y_1,y_2\}$, we want the coefficients of a tensor $$\sum_{\substack{1\leq i\leq 3\\1\leq j\leq2}}f_{i,j}x_i\otimes y_j$$
to be equal to a product $$\Bigl(\sum_{1\leq i\leq 3}v_ix_i\Bigr)\otimes\Bigl(\sum_{1\leq j\leq 2}w_iy_i\Bigr).$$
It is easy to see that we must have $$f_{i,j}f_{k,l}=f_{k,l}f_{i,l}$$ for all $i,k\in\{1,2,3\}$ and all $j,l\in\{1,2\}$ for that to happen, and some work will show that these conditions are in fact sufficient. We can express all these conditions by saying that the matrix $$\begin{pmatrix}f_{1,1}&f_{1,2}\\f_{2,1}&f_{2,2}\\f_{3,1}&f_{3,2}\end{pmatrix}$$ has rank $1$. Proving this is «just» linear algebra.
The answer in the general case where the dimensions are arbitrary is of the same spirit.
N.B.: it is interesting to know that the question «which tensors have rank $k$?» when $k\geq2$ and there are more than two factors is much, much harder, and very important—I think this is unsolved in general. Someone who knows algebraic geometry might be able to tell us.
Let the columns of $3 \times 3$ matrix $\rm V$ be the given vectors, i.e.,
$$\rm V := \begin{bmatrix}1 & 0 & 0\\ 0 & 3 & 2\\ -1 & -6 & -4\end{bmatrix}$$
Computing the Gram matrix, we obtain
$$\rm V^{\top} V = \begin{bmatrix} 2 & 6 & 4\\ 6 & 45 & 30\\ 4 & 30 & 20\end{bmatrix}$$
What can we conclude, then?
Best Answer
Let $\delta=\sum_{i,j=1}^n(i+j)(e_i\otimes e_j)$, as you have. By definition, a decomposable element of $V\otimes V$ is one of the form $v\otimes w$, for some $v,w\in V$. To show that $\delta$ is not decomposable, we therefore need to show that $\delta\neq v\otimes w$, for any $v,w\in V$. On the other hand, since $B$ is a basis of $V$, every element of $V$ is of the $\sum_{i=1}^nx_ie_i$, for some $x_i\in \mathbb{F}$. Thus it suffices to show that $\delta\neq\left(\sum_{i=1}^nx_ie_i\right)\otimes\left(\sum_{j=1}^ny_je_j\right)$ for any $x_i,y_i\in\mathbb{F}$.
Assume for contradiction therefore that $\delta$ is of that form, for some $x_i$ and $y_i$. We can expand the second expression as follows, using the bilinear and $\mathbb{F}$-balanced properties of the tensor product: \begin{align} \left(\sum_{i=1}^nx_ie_i\right)&\otimes\left(\sum_{j=1}^ny_je_j\right)=\sum_{i=1}^n\left(x_ie_i\otimes \sum_{j=1}^ny_je_j\right) \\ &=\sum_{i,j=1}^n(x_ie_i\otimes y_je_j)=\sum_{i,j=1}^n(x_iy_j)(e_i\otimes e_j). \end{align} If this equals $\delta=\sum_{i,j=1}^n(i+j)(e_i\otimes e_j)$, then – because $\{e_i\otimes e_j\}_{i,j\leqslant n}$ is a basis for $V\otimes V$ – we must have $x_iy_j=i+j$ for every $i,j\leqslant n$.
In particular, we have
This gives that $$\frac{y_1}{y_2}=\frac{x_1y_1}{x_1y_2}=\frac{2}{3}\ \ \ \ \ \ \ \ \text{ and }\ \ \ \ \ \ \ \ \frac{y_1}{y_2}=\frac{x_2y_1}{x_2y_2}=\frac{3}{4},$$ which gives the desired contradiction.
(In dividing by $3$ and $4$, we are assuming here that $\mathbb{F}$ does not have characteristic $3$ or $2$. I suspect you are probably assuming that $\mathbb{F}$ has characteristic $0$, so I assumed this too. However, do correct me if I'm wrong; the result still holds no matter what $\operatorname{char}(\mathbb{F})$ is, we just need a tiny bit of extra casework in the case where it equals $2$ or $3$.)