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 
 $e_1⊗e_2+e_2⊗e_1 = (e_1+e_2⊗e_1+e_2)  - e_1⊗e_1 - e_2⊗e_2$ 
 A: 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 $u_1,u_2,v_1,v_2 \in \Bbb R$.
A: Assume that $e_1 \otimes e_2 + e_2 \otimes e_1 = u \otimes v$ and write
$$ u = u_1 e_1 + u_2 e_2, \,\,\, v = v_1 e_1 + v_2 e_2. $$
Expanding $u \otimes v$, we have
$$ u \otimes v = (u_1 e_1 + u_2 e_2) \otimes (v_1 e_1 + v_2 e_2) = 
\\
 (u_1 v_1) (e_1 \otimes e_1) + (u_1 v_2) (e_1 \otimes e_2) + (u_2 v_1) (e_2 \otimes e_1) + (u_2 v_2) (e_2 \otimes e_2) = \\
1 \cdot (e_1 \otimes e_2) + 1 \cdot (e_2 \otimes e_1). $$
Since $e_i \otimes e_j$ is a basis of $\mathbb{R}^2 \otimes \mathbb{R}^2$, we must have
$$ u_2 v_2 = u_1 v_1 = 0, \,\,\, u_1 v_2 = u_2 v_1 = 1. $$
Show that this leads to a contradiction.
A: Imagine that $e_1 \otimes e_2 + e_1 \otimes e_2 = u \otimes v$. If $u = u_1 e_1 + u_2 e_2$ and $v = v_1 e_1 + v_2 e_2$, then
$$e_1 \otimes e_2 + e_1 \otimes e_2 = (u_1 e_1 + u_2 e_2) \otimes (v_1 e_1 + v_2 e_2) = \\
u_1 v_1 \ e_1 \otimes e_1 + u_1 v_2 \ e_1 \otimes e_2 + u_2 v_1 \ e_2 \otimes e_1 + u_2 v_2 \ e_2 \otimes e_2$$
which means
$$0 = u_1 v_1, \quad 1 = u_1 v_2, \quad 1 = u_2 v_1, \quad 0 = u_2 v_2 .$$
The fact that $0 = u_1 v_1$ implies that either $u_1 = 0$, or $v_1 = 0$. If $u_1 = 0$ then the equality $1 = u_1 v_2$ becomes impossible; if $v_1 = 0$ then the equality $1 = u_2 v_1$ becomes impossible. In either case one obtains a contradiction, therefore there exist no $u$ and $v$ such that $e_1 \otimes e_2 + e_1 \otimes e_2 = u \otimes v$.
