Linear Associativity and Tensor Products (10.4.12)

Dummit and Foote Abstract Algebra, section 10.4, exercise 12:

Let $V$ be a vector space over a field $F$ and let $v,v'∈V$ be nonzero. Prove $v \otimes_F v' = v' \otimes_F v ⇔ v=av'$ for some $a∈F$.

Proof: ($\Leftarrow$) $$v' \otimes v = v' \otimes av' = av' \otimes v' = v \otimes v'$$ ($⇒$) We may assume $V$ has a basis $\{e_i\}_{i∈I}$.

Let $v=\sum_{j∈J} f_je_j$ and $v'=\sum_{j∈J} f_j'e_j$ for appropriate, finite $J \subseteq I$. Let $W = \sum_{j∈J}Fe_j$, so that $W≅F^n$ for some finite $n$.

Now, $v \otimes v'=v \otimes \sum_{j∈J} f_j'e_j = \sum_{j∈J} v \otimes f_j'e_j = \sum_{j∈J} f_j'v \otimes e_j$, and similarly $v' \otimes v = \sum_{j∈J} f_jv' \otimes e_j$. We have $v \otimes v',v' \otimes v ∈ W \otimes W≅F^n \otimes F^n$ and by 10.4.10(a) we have $f_j'v=f_jv'$ for all $j∈J$ and since $v,v'≠0$ we may say $v=(f_j'^{-1}f_j)v'$.