Find a way to express $u:= e_1\otimes e_1 + e_2\otimes e_2 $ as $u:= v\otimes v' + w\otimes w'$ with $v,v',w,w'\in \mathbb{R^2}$ 
Let $ U:= \mathbb{R^2}\otimes_{K}\mathbb{R^2}$ be a tensor product with $e_1,e_2$  as the standard basis of  $\mathbb{R^2}$.
  Let $u:= e_1\otimes e_1 + e_2\otimes e_2 \in U$
Show that $u$ cannot be written as  $ v\otimes v'$ with $ v,v' \in\mathbb{R^2}$ and find $v,v',w,w'\in \mathbb{R^2}$ such that the expression $u:= v\otimes v' + w\otimes w'$ is valid and neither $v\otimes v'$ nor $w\otimes w'$ is a multiple of $e_1\otimes e_1$  or $e_2\otimes e_2$.

I'm really having trouble solving this one here. Maybe you guys can help me out a bit . Any help is greatly appreciated. 
 A: I'll give a partial answer, since I'm in a bit of a hurry here, but for the first part, assume that $u = v \otimes v'$, and let's get a contradiction. Write $v = ae_1+be_2$ and $v' = ce_1+de_2$, so that $$e_1 \otimes e_1 + e_2 \otimes e_2 = ac e_1 \otimes e_1 + bc e_2 \otimes e_1 + ad e_1 \otimes e_2 + bd e_2\otimes e_2.$$By linearly independence we get $$ 1 = ac, \quad 0 = bc, \quad 0 = ad, \quad 1 = bd,$$which immediately leads to a contradiction.
If you don't manage to do the second one thinking of what I did above, I'll give you a boost later.
A: Assume $K=\mathbb{R}$. Note that $e_i\otimes e_j$ is a basis of $\mathbb{R}^2\otimes \mathbb{R}^2$.
Let $v=ae_1+be_2$ and $v^{\prime}=a^{\prime}e_1+b^{\prime}e_2$. Assume that $v\otimes v^{\prime} = e_1\otimes e_1 + e_2\otimes e_2$. Since $v\otimes v^{\prime} = aa'e_1\otimes e_1 + ab'e_1\otimes e_2 + ba'e_2\otimes e_1 + bb'e_2\otimes e_2$ we have $$aa'=1, \; ab' = 0, \; a'b = 0, \; bb'=1, $$ which is not possible. 
For $v = e_1$, $v' = e_1+e_2$, $w=e_2-e_1$, $w'=e_2$ we have
\begin{eqnarray}
v\otimes v'+w\otimes w' & = & e_1\otimes (e_1+e_2) + (e_2-e_1)\otimes e_2\\
& = & e_1\otimes e_1 +e_1\otimes e_2 + e_2\otimes e_2 - e_1\otimes e_2 \\
& = & e_1\otimes e_1 + e_2\otimes e_2 = u,
\end{eqnarray}
and $v \otimes v' = e_1\otimes (e_1+e_2) = e_1\otimes e_1 + e_1\otimes e_2 \notin \langle e_1\otimes e_1,e_2\otimes  e_2\rangle$. Analogoulsy for $w,w'$.
