the tensor product question Well the exercise is

Let $\{u_{1}, \cdots u_{k}\}$ be linearly independent vectors in $U$, and $\{v_{1} \cdots , v_{k}\}$ arbitrary vectors in $V$. Show that $\sum u_{i} \otimes v_{i} = 0$ in $U \otimes V$ implies that $v_{1} = \cdots = v_{k} = 0$.

Can I get some advice??
Thanks!

You can assume the $u_{i}$ span $U$.
Choose an index $k$, and consider the bilinear map
$$
U \times V \to V
$$
that sends all $(u_{j}, v)$ to zero, for $j \ne k$, and sends $(u_{k}, v)$ to $v$, for each $v \in V$.
By the universal property of the tensor product, this factorizes through the tensor product, and sends $\sum u_{i} \otimes v_{i}$ to $v_{k}$, which must thus be zero.
