Writing $x\in U\otimes V$ as $x=\sum u_i \otimes v_i$

Question

Let $U, V$ be vector spaces. Let $u_i,\dots,u_n \in U $ be linearly independent. For $x \in U\otimes V$ show that there exist uniquely determined $v_1,\dots,v_n \in V$ such that $x = \sum u_i \otimes v_i$.

I am curious to know if we should use the Universal Property to prove this? And if not, what steps should we do to prove it?

Answer 1

Suppose that $U$ is a $k$-vector space, and for $(u,a) \in U \times k$ consider $u \otimes a := au$. Note that if $b : U \times k \to Z$ is bilinear, then $b(\_,1) : U \to Z$ satisfies $b = b(\_,1) \circ \otimes$, which means that $U \otimes k \cong U$. Now, it is not necessarily true that for every $x \in U$ there are unique scalars $a_1,\dots,a_n \in k$ such that $x = a_1u_1 + \cdots + a_nu_n$.