Show two nonzero vectors $v$ and $w$ have the property $v\otimes w=w\otimes v$ iff there is $\lambda$ such that $v=\lambda w$

Question

I am struggling with the following problem.

Let $V$ be a $F$-free module on a finite set over a field $F$. Prove that in $V\otimes_F V$ two nonzero vectors $v$ and $w$ have the property $v\otimes w=w\otimes v$ if and only if there is $\lambda$ such that $v=\lambda w$.

I have the converse argument to be the following:

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($\Leftarrow)$ Let $v$ and $w$ be two nonzero vectors. Suppose there exists $\lambda$ such that $v=\lambda w$. We want to show that $v\otimes w=w\otimes v$. Consider \begin{eqnarray*} v\otimes w &=& \lambda w \otimes w \text{ because $v=\lambda w$}\\ &=& w \otimes \lambda w \text{ by a relation of the tensor product}\\ &=& w\otimes v\text{ because $v=\lambda w$} \end{eqnarray*}

Thus $v\otimes w=w\otimes v$ as desired.

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I feel good about my proof of the converse. The forward direction is giving me trouble however. Here is my attempt:

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($\Rightarrow)$ Let $v$ and $w$ are two nonzero vectors. Suppose we have the property $v\otimes w=w\otimes v$. We want to show that there exists a $\lambda$ such that $v=\lambda w$. Notice $V$ has a finite basis, call it $B$. By definition of a free module, for every nonzero $v,w\in V$ there exist unique nonzero $r_i,s_i\in F$ and unique $b_i\in B$ where $i\in\{1,\ldots,n\}$ such that $$ v=\sum_{i=1}^n r_ib_i\hspace{3mm}\text{ and }\hspace{3mm} w=\sum_{i=1}^n s_ib_i $$ Because $B$ is finite, we have $V=Fb_1\oplus \cdots \oplus Fb_n\cong F^n$.

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This is where I get struck. I thought about trying to apply the universal property of a tensor product, but I am not sure how to go about this exactly.

Questions:

(1) Is a good approach to prove $(\Rightarrow)$ or is there a better way?

(2) Is what I have done so far correct?

Thanks in advance.

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Revision after help below:

Let $v$ and $w$ are two nonzero vectors. We will prove this problem using the contrapositive. Suppose $v\neq \lambda w$. This means $v$ and $w$ are linearly independent. We want to show $v\otimes w\neq w\otimes v$.

Notice $V$ has a finite basis, call it $B$. By definition of a free module, for every nonzero $v,w\in V$ there exist unique nonzero $r_i,s_i\in F$ and unique $b_i\in B$ where $i\in\{1,\ldots,n\}$ such that $$ v=\sum_{i=1}^n r_ib_i\hspace{3mm}\text{ and }\hspace{3mm} w=\sum_{i=1}^n s_ib_i $$ Because $B$ is finite, we have $V=Fb_1\oplus \cdots \oplus Fb_n\cong F^n$ (pg. 354). By corollary 19 (pg. 374), we have $V\otimes_F V$ to be a free $F$-module of rank $n^2$ with basis $b_i\otimes b_j$ such that $i\in\{1,\ldots,n\}$. Therefore \begin{eqnarray} F^n \otimes_F F^n\cong F^{n^2}=F^n\times F^n. \end{eqnarray} Claim $\{v,w\}$ is contained in some basis of $V$. Clearly $v,w$ are linearly independent by our assumption. Additionally $v,w\in\text{span}(B)$ by the definition of $V$ as a $F$-free module. Therefore we have $v$ and $w$ are in some basis of $V$. This means $v\otimes w$ and $w\otimes v$ are in some basis of $V\otimes_F V$. Then $v\otimes w=v\times w$ and $w\otimes v=w\times v$. Notice $v\times w= w\times v$ only if $v=w$, which is false because they are linearly independent. Thus $v\otimes w\neq w\otimes v$, proving the contrapositive.

Hence if $v\otimes w=w\otimes v$, then $v=\lambda w$ for some $\lambda\in F$.

Answer 1

For the forward direction, suppose $v$ and $w$ are linearly independent and find a basis of $V$ containing $\{v,w\}$. The corresponding basis of $V \otimes V$ contains $v \otimes w$ and $w \otimes v$. In particular, $v \otimes w \not = w \otimes v$.

Your proof of the converse direction is good.

Answer 2

I just found a way with the universal property that needs not you trust that $e_i \otimes e_j$ form a basis. (As Olivier showed, if you trust that, then the statement is trivial since it is just focuses on the independence of a particular pair ${e_i \otimes e_j, e_j \otimes e_i}$ out of the entire set whose independence you already trust.)
I'll try to explain my personal thought process which lead me to this result because I'm still a beginner in category language and I hope it'd be helpful to others, too.

Thought process

In loose terms, the mapping $\otimes : V\times V\to V\otimes V$ is (the) most general bilinear mapping out of $V\times V$ in the sense that it

1. "forgets" no information from $V\times V$, and
2. adds no more than necessary.

These two are captured by the universal property which makes sure any other bilinear function out of $V\times V$ is 1. recoverable from $\otimes$ (no data loss!) in 2. a unique way (no excess data!).

With $u\otimes v=v\otimes u$ (or, $\otimes(u,v)=\otimes(v,u)$ to stress that $\otimes$ is a map out of $V\times V$), we kind of lose the information that $(u,v)$ and $(v,u)$ were different before being processed by $\otimes$. It seems $\otimes$ violates point 1. of the universallity.

How can we show this? Point 1. is about recovability of bilinear functions out of $V\times V$, so we can hope to find a bilinear map which cannot be constructed by composition with $\otimes$.

And this is where I was stuck for a few days because I didn't know how to construct bilinear functions until I found the answer in the body of this question: for a fixed basis $\{v_i\}$ of V, a bilinear map $B:V\times V\to W$ is determined uniquely by (linear extension of) the equations $B(v_i,v_j)=w_{ij}$ for any choices of $w_{ij}\in W$. In this simple case it suffices to use the underlying field for $W$.

Answer

1. Construct a bilinear map $B:V\times V\to F$ that tells the difference between $u$ and $v$ with $B(u,v)=1$ and $B(v,u)=0$. This is possible if $u,v$ are independent because 1) they can be completed to a basis and 2) a bilinear map can be determined on a basis this way.
2. Apply the universal property of $\otimes$ for $B$: since $\otimes$ is universal, then $B$ should be recoverable from it by some linear $h:V\otimes V\to F$: $$B=h\circ\otimes.$$
3. Evaluate this equation on both $(u,v)$ and $(v,u)$: $$1=B(u,v)=(h\circ\otimes)(u,v)=h(u\otimes v)\\0=B(v,u)=(h\circ\otimes)(v,u)=h(v\otimes u).$$ Obviously, if $u\otimes v=v\otimes u$, we'd get $1=0$, i.e. the independence of $u,v$ is inconsistent with the $u\otimes v=v\otimes u$.

Moral

If $\otimes$ forgot the difference between $(u,v)$ and $(v,u)$, then we wouldn't be able to reconstruct from it any other bilinear function which does tell the difference. Since $\otimes$ is universal, that's impossible, i.e. such bilinear functions shouldn't exists. But that can only happen if $u,v$ are linearly dependent.

For me, that last piece was missing - how these mappings are constructed. Always know your maps!