$v \otimes v' = v' \otimes v$ implies $v = av'$

In Dummit and Foote problem 12 of section 10.4, I need to show that if $V$ is some vector space over a field $F$ and $v,v'$ are nonzero elements of $V$, supposing we have that $v \otimes v' = v' \otimes v$ then this implies that $v = av'$ for some $a \in F$. I was trying to consider some basis $\mathcal{B} = \{e_i\}_{i\in I}$ for the vector space $V$, show that we can write $v$ and $v'$ as linear combinations of this basis elements and then argue that since we have $v \otimes v' = \sum_{i \in I}r_i e_i \otimes \sum_{i\in I} s_i e_i = \sum_{i\in I} s_i e_i \otimes \sum_{i \in I}r_i e_i = v' \otimes v$ then if we look at a particular element of the sum on LHS and RHS, namely $r_ie_i \otimes s_je_j = s_je_j\otimes r_i e_i$ then we can send the $r_i$ to the other side or do some manipulation of that sort and obtain the desired result.

I know this argument is not rigorous at all, so I am having a hard time doing it more rigorous (if this is a good approach) or actually attack this problem efficiently. I was wondering if this would be the right approach, like working with basis elements and so on or if there is some better way to look at this problem. Perhaps some use of universal property for modules would be helpful? Any help or suggestion with this problem is highly appreciated! Thanks so much!

Your approach is good, but your last step is wrong: you can't conclude from the equality $$\sum_{i \in I}r_i e_i \otimes \sum_{j\in I} s_j e_j = \sum_{j\in I} s_j e_j \otimes \sum_{i \in I}r_i e_i$$ that the $(i,j)$ term on one side is equal to the $(i,j)$ term on the other side. Instead, if you write the right side as $\sum_{i\in I} s_i e_i \otimes \sum_{j \in I}r_j e_j$, you can conclude the $(i,j)$ terms on each side are equal, since the elements $e_i\otimes e_j$ form a basis for $V\otimes V$, and so the coefficients of $e_i\otimes e_j$ on each side must be the same. This gives $r_is_j=s_ir_j$ for each $(i,j)$.

From here you could argue that these equations (and the assumption that $v,v'\neq 0$) imply that there is some $a$ such that $r_i=as_i$ for all $i$. However, there's actually a trick you can use to make it a lot easier. Everything we've done is valid for any basis, so let's choose a nice basis. In particular, assuming that $v$ is not a scalar multiple of $v'$ (so they're linearly independent), let's choose a basis with $e_1=v$ and $e_2=v'$. Then in this basis we have $v\otimes v'=e_1\otimes e_2$ and $v'\otimes v=e_2\otimes e_1$. Since these are distinct elements of our basis for $V\otimes V$, they cannot be equal.

If u and v are not linearly dependent, then you can construct a basis which contains them both, and from that a basis of the tensor product. That basis then contains those two elementary tensors that appear in your question, and they are therefore different.