How to show this tensor is not decomposable Let $V$ a vector space over $\mathbb{F}$ with basis $B=\{e_1,\dots, e_n \}$ and $B'=\{e^1,\dots,e^n \}$ adjoint basis in $V^*$, Consider the following tensor:
$$t^{ij}=(i+j) $$
I need to see that is not decomposable. This is the proof of my professor:
$\delta=\sum_{i,j=1}^{n}(i+j)(e_i\otimes e_j), $ suppose $\delta=\left(\sum_{i=1}^{n}x_ie_i\right) \otimes \left(\sum_{i=1}^{n}y_j e_j \right)$. As $n\geq 2$, we have $x_1y_1=2$,$x_1y_2=3$, so $y_1/y_2=2/3$, but $x_2y_1=3$, $x_2y_2=4$ and $y_1/y_2=3/4$ that is a contradiction.
I am not sure why this numbers are related with the fact that is decomposable, tensors are pretty hard to me, can you help me to understand this?
 A: 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

            $\bullet$ $x_1y_1=1+1=2$                         $\bullet$ $x_2y_1=1+2=3$
            $\bullet$ $x_1y_2=2+1=3$                         $\bullet$ $x_2y_2=2+2=4$

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 $3$.)
