Theorem 1   Let $V_1,V_2$ be two vector spaces over a field $F$ . If $x_1,\ldots,x_n\in V_1$ are linearly independent and $y_1,\ldots,y_n\in V_2$ then $$ \sum x_i \otimes y_i =0 \Rightarrow y_i=0,\mbox{ for all } i$$

Proof. Take the dual vectors $x_i^{*}$ to the vectors $x_i$ , i.e. $x_i^{*}(x_j)=\delta_{i,j}$ . Given arbitrary linear functionals $f_i:V_2 \rightarrow F$ , define a bilinear form $f:V_1\times V_2\rightarrow F$ by $$ f(x,y)=\sum_{j=1}^n x_{j}^{*}(x)f_j(y)$$ By the definition of tensor product there exists a unique linear functional $\phi:V_1\otimes V_2\rightarrow F$ such that $\phi \circ \iota =f$ . Therefore \begin{eqnarray*} 0&=&\phi\left(\sum_{i} x_i\otimes y_i\right) \\ &=&\sum_{i} \phi\circ \iota (x_i,y_i) \\ &=&\sum_{i} f(x_i,y_i) \\ &=&\sum_i \sum_j x_{j}^{*}(x_i)f_j(y_i)\\ & =&\sum_{i} f_i(y_i). \end{eqnarray*} Since the $f_i$ are arbitrary, it follows that $y_i=0$ for all $i$ .