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 ith 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.

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