Tensor Products and a Basis of $\Bbb R^2 ⊗ \Bbb R^2$ Let ${e_1, e_2}$ be the standard basis of $\Bbb R^2$.
 Show that $e_1⊗e_2+e_2⊗e_1$ cannot be written in the form $u ⊗ v$ with $u,v \in \Bbb R^2$.
I am just being introduced to tensor spaces and I know that 
$e_1⊗e_1,e_1⊗e_2,e_2⊗e_1,e_2⊗e_2$ is a basis of $\Bbb R^2 ⊗ \Bbb R^2$ but I am not sure how to show a contradiction. Any hints appreciated.
Edit: I also know that 
 $e_1⊗e_2+e_2⊗e_1 = (e_1+e_2⊗e_1+e_2)  - e_1⊗e_1 - e_2⊗e_2$ 

It is helpful to note that the map $u \otimes v \mapsto uv^T$ is an isomorphism of vector spaces.
Now, the question becomes as follows: why do there exist no $u$ and $v$ such that
$$
uv^T = \pmatrix{0&1\\1&0}?
$$
think in terms of the rank of a matrix.

Or, if you prefer anomaly's approach: note that
$$
u \otimes v = (u_1v_1) e_1 \otimes e_1 + (u_1v_2) e_1 \otimes e_2 + (u_2v_1) e_2 \otimes e_1 + (u_2v_2) e_2 \otimes e_2
$$
Now, it suffices to show that the system
$$
u_1v_1 = 0\\
u_1v_2 = 1\\
u_2 v_1 = 1\\
u_2 v_2 = 0
$$
has no solutions for $u_1,u_2,v_1,v_2 \in \Bbb R$.
