Showing a tensor is not a simple tensor Let $(e_1,e_2)$ be the standard basis of $\mathbb{R}^2$. Why can the element $e_1\otimes e_2 + e_2\otimes e_1$ in $ \mathbb{R}^2\otimes_{\mathbb{R}} \mathbb{R}^2 $ not be written as a simple tensor $v\otimes w$ for $v,w\in \mathbb{R}^2$?
If we suppose $e_1\otimes e_2 + e_2\otimes e_1=v\otimes w$, then can we conclude that $e_1=\lambda e_2$ for some $\lambda \in \mathbb{R}$? This would give a contradiction since $e_1$ and $e_2$ are linearly independent. However, I am not sure if the argument is valid. 

Hint In this case we can (and it is instructive to) proceed naively:
If $e_1 \otimes e_2 + e_2 \otimes e_1$ is simple, there are $v, w \in \Bbb R^2$ such that $$e_1 \otimes e_2 + e_2 \otimes e_1 = v \otimes w .$$ Writing $v = v^1 e_1 + v^2 e_2, w = w^1 e_1 + w^2 e_2$, substituting in the above equation, expanding, and comparing like coefficients gives a quadratic system. Show that it is inconsistent.

The system is$$\left\{\begin{array}{rcl}0 &=& v^1 w^1 \\1 &=& v^1 w^2 \\1 &=& w^1 v^2 \\0 &=& v^2 w^2 \end{array} \right. .$$

Hint for alternate solution Show that, when regarded as bilinear forms on $(\Bbb R^2)^*$, $e_1 \otimes e_2 + e_2 \otimes e_1$ is nondegenerate but any simple tensor $v \otimes w$ is degenerate.

The determinant of matrix representation of $e_1 \otimes e_2 + e_2 \otimes e_1$ with respect to the standard basis is $-1$, so it is nondegenerate. On the other hand, for any $\lambda \in \langle v\rangle^{\perp} \subset (\Bbb R^2)^*$, $(v \otimes w)(\lambda, \,\cdot\,) = \lambda(v) w = 0$, so $v \otimes w$ is degenerate.

