# $v \otimes v' = v' \otimes v$ 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!

• Wrong. $r_ie_i \otimes s_je_j \ne s_je_j\otimes r_i e_i$ – Kenny Lau Dec 18 '17 at 19:17
• A generalization of this result is Theorem 5.15 of kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod2.pdf, describing when two elementary tensors in a tensor product of vector spaces are equal. – KCd Jan 18 at 5:08

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.

• This answer is same as Eric's. He mentions this at the end of his answer. – Vishal Gupta Dec 18 '17 at 19:37

I suppose it is nice to remark that in general, if $V$ is a f.d. vector space and if $n\in\mathbb N$, there is a map $\xi:V^{\otimes n}\to V^{\wedge n}\subseteq V^{\otimes n}$ such that $$v_1\otimes\cdots\otimes v_n\longmapsto v_1\wedge\cdots \wedge v_n :=\sum(-1)^\sigma v_{\sigma 1}\otimes\cdots\otimes v_{\sigma n}$$

where the sum runs through all permutations of $n$ elements and $(-1)^\sigma$ is the sign of $\sigma$. In your example, $n=2$ so this map sends $v\otimes v'$ to $v\otimes v'-v'\otimes v$.

The antisymmetrization map $\xi$ detects linear dependence, in the sense that a set of vectors $\{v_1,\ldots,v_n\}$ is linearly dependent if and only if $\xi(v_1\otimes \cdots\otimes v_n)$ is zero. Indeed, unraveling the definitions one sees that the coefficients of $e_{i_1}\wedge \cdots \wedge e_{i_n}$ appearing on the right are the maximal minors of the corresponding matrix of the $v_i$, and this matrix is of full rank if and only if the vectors are independent.

In your example, if $v = \sum a_i e_i$ and $w= \sum b_ie_i$, the coefficient of $e_i\wedge e_j = e_i\otimes e_j- e_j\otimes e_i$ is the minor $a_ib_j-a_jb_i$ of columns $i$ and $j$ of the matrix $$\begin{pmatrix} a_1 &\cdots & a_n \\ b_1 &\cdots &b_n \end{pmatrix},$$ and this is zero for every $i,j$ if and only if $\{v,v'\}$ is linearly dependent.