If $u\otimes v_1 = v_2\otimes u$; $v_1,v_2,u \in V-\{0\}$ then $v_1 = v_2$? On the beginning I though that it was false. But, making several trials for counter-example I have failed in find one. Anyone has any suggestions to prove or disprove?
 A: Let $v_1 = \alpha u + \alpha_ie_i$, $v_2 = \beta u + \beta_ie_i$.
$$v_1\otimes u = \alpha u\otimes u + \alpha_i e_i\otimes u $$
$$u\otimes v_2 = \beta u\otimes u + \beta_i u\otimes e_i.$$
Then, $$\alpha u\otimes u + \alpha_i e_i\otimes u = \beta u\otimes u + \beta_i u\otimes e_i.$$
Then $\alpha = \beta$ necessarily.
$$ \alpha_i e_i\otimes u -\beta_i u\otimes e_i = 0 .$$
But then, $\alpha_i = \beta_i = 0$.
And then, $v_1 = \alpha u$, $v_2 = \alpha u.$
A: Hint: Choose a basis for $V$ which contains $u$ as one of the basis vectors, and write $u\otimes v_1$ and $v_2\otimes u$ in terms of the corresponding basis for $V\otimes V$.
A: Another solution:
It suffices to show that $u, v_1$ are linearly dependent. In fact, if for some $\lambda \neq 0$ we have  $v_1 \otimes \lambda v_1 = \lambda v_1 \otimes v_2 $, then  $v_1\otimes(\lambda v_1 - \lambda v_2 )= 0 $ implies $v_2 = v_1.$
Thus, we are left to show that  $u, v_1$ cannot be linearly independent. Let $\Gamma: V\otimes V \rightarrow \mbox{Hom}(V^*,V)$  be the canonical map satisfying $\Gamma(w_1\otimes w_2)(f) = f(w_1)w_2$ for every $f\in V^*$ and $w_1,w_2\in V$.
Take any $f\in V^*$ such that $f(u) =1 , f(v_1) =0$. Such a $f$ exists, since $u$ and $v_1$ are suppposed to be linearly independent. Thus,
$$   v_2 = f(u)v_2= \Gamma(u\otimes v_2)(f) = \Gamma (v_1\otimes u ) = f(v_1)u = 0$$
is a contradiction.
