## In a tensor square over a field, a simple tensor is equal to its reverse if and only if its components are linearly dependent

Let $F$ be a field, let $V$ be an $F$-vector space, and consider the tensor square $V \otimes_F V$. Prove that for all nonzero $v,w \in V$, $v \otimes w = w \otimes v$ if and only if $v = aw$ for some $a \in F$.

Let $B = \{e_i\}_I$ be a basis for $V$. Note then that every simple tensor (hence every element) of $V \otimes_F V$ can be written in the form $\sum v_i \otimes e_i$. For each $i \in I$, we have an $F$-module injection $\psi_i : V \rightarrow V \otimes_F V$ given by $v \mapsto v \otimes e_i$. By the universal property of direct sums, we have an $F$-module homomorphism $\Psi : \bigoplus_I V \rightarrow V \otimes_F V$ such that $\Psi(v_i) = \sum v_i \otimes e_i$. Now define $\varphi : V \times V \rightarrow \bigoplus_I V$ by $\varphi(v, \sum a_ie_i) = (va_i)$. This mapping is well defined since every element of $V$ is a finite linear combination of the $e_i$. Moreover, $\varphi$ is clearly bilinear. Thus we have an $F$-module homomorphism $\Phi : V \otimes_F V \rightarrow \bigoplus_I V$ such that $\Phi(v \otimes e_i) = \iota_i(v)$, where $\iota_i$ denotes the $i$th natural injection $V \rightarrow \bigoplus_I V$.

We claim that $\Phi$ and $\Psi$ are mutual inverses. To see this, note that $(\Phi \circ \Psi)(v_i) = \Phi(\sum v_i \otimes e_i) = (v_i)$ and $(\Psi \circ \Phi)(v \otimes (\sum a_ie_i)) = \Psi(va_i)$ $= \sum va_i \otimes e_i = v \otimes \sum a_ie_i$.

Suppose now that $v \otimes w = w \otimes v$. Let $v = \sum a_ie_i$ and $w = \sum b_ie_i$. Then $\Phi(v \otimes w) = \Phi(w \otimes v)$, so that $(vb_i) = (wa_i)$ for all $i \in I$. If $(vb_i) = 0$, then $vb_i = 0$ for all $i$, and since $v \neq 0$, we have $b_i = 0$ for all $i$. Thus $w = 0$, a contradiction. Thus $vb_i = wa_i \neq 0$ for some $i$, and we have $v = b_i^{-1}a_iw$. Thus $v$ and $w$ are linearly dependent.

Conversely, suppose $w = av$. Then $v \otimes w = v \otimes av$ $= va \otimes v$ $= w \otimes v$.