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Let $X$ be a vector space and $x \otimes y \in X \otimes X$. Under which conditions is $x \otimes y = y \otimes x$? Does it nessecarily follow that $x = \lambda y$ for some $\lambda$ in the underlying field?

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    $\begingroup$ Given that $X$ is a finite-dimensional $k$-vector space, the tensor product $x \otimes y$ of the vectors $x, y$ in $X$ is the matrix whose $j$th column is the column vector $(x_i y_j),$ where $x_i$ and $y_j$ are the coefficients of $x$ and $y$ with respect to some basis. $\endgroup$ – Carlo May 19 at 16:26
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    $\begingroup$ For a simple example, consider the $\mathbb R$-vector space $\mathbb R^2$ with vectors $x = (1, 2)$ and $y = (1, 1).$ We have that $x \otimes y = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}.$ $\endgroup$ – Carlo May 19 at 16:28
  • $\begingroup$ @EricWofsey Yes! $\endgroup$ – Jannik Pitt May 19 at 16:51
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You're not entirely correct (consider $y=0, x\neq0$). But yes, $x\otimes y=y\otimes x$ means that $x$ and $y$ are collinear / linearly dependent.

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  • $\begingroup$ I see, how can one prove this? This should be a simple algebraic manipulation but I can't figure it out. $\endgroup$ – Jannik Pitt May 19 at 16:24
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    $\begingroup$ Chose some suitable linear forms $\alpha$ and $\beta$ and evaluate $\alpha\otimes \beta$ on $x\otimes y$ $\endgroup$ – DIdier_ May 19 at 16:27

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