$u \otimes v = u’ \otimes v’ \Leftrightarrow u’=\lambda u \textrm{ and } v’= \lambda^{-1} v$ Let $U$ and $V$ be $K$-vector spaces , and let $u, u’ \in U$ and $v,v’ \in V$ such that $u \otimes v \neq 0$. Prove that:
$$u \otimes v = u’ \otimes v’ \iff u’=\lambda u \textrm{ and } v’= \lambda^{-1} v$$
The backward implication is easy, I don’t know how to prove the “only if”.
 A: We want to show that given $u \otimes v \neq 0$, $u \otimes v = u' \otimes v'$ implies that $u' = \lambda u$ and $v' = \lambda^{-1}v$ for some $\lambda \in K$.
First, note that because $u \otimes v \neq 0$, we must have $u \neq 0$ and $v \neq 0$.
From there, one approach is to give a proof by contrapositive using the universal definition of the tensor product. In particular, if $u'$ is not a multiple of $u$, then $u,u'$ are linearly independent. We note (for instance, using the existence of a basis that extends the linearly independent set $\{u,u'\}$) that there exists a linear map $\phi:U \to K$ for which $\phi(u) = 1$ and $\phi(u') = 0$. By a similar kind of argument, there exists a linear map $\psi:V \to K$ such that $\psi(v) \neq 0$.
Now, let $\Phi:U \times V \to K$ denote the bilinear map $\Phi(u,v) = \phi(u)\psi(v)$. By the universal definition of the tensor product, there is an induced map $\tilde \Phi : U \otimes V \to K$. We see that $$
\tilde \Phi(u \otimes v) = \Phi(u,v) \neq 0, \quad \tilde \Phi(u' \otimes v') = \Phi(u',v') = 0.
$$
So, $u \otimes v$ and $u' \otimes v'$ must be distinct elements of $U \otimes V$.
We similarly conclude that if $v'$ is not a multiple of $v$, $u \otimes v \neq u' \otimes v'$.
