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Let $\left\{ e_1 , e_2 \right\}$ be the standard basis for $\mathbb{R}^2$. Is it true that $e_1 \otimes e_1 + e_2 \otimes e_2$ can be expressed as $v \otimes w$ for some $v,w \in \mathbb{R}^2$?

I believe this statement is false. However, I'm having trouble formulating something to go with my intuition.

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Suppose it can then $e_1\otimes e_1+e_2\otimes e_2=v\otimes w$ for some $v,w\in \mathbb{R}^2$. Now there exist $\alpha,\beta,\lambda,\mu\in \mathbb{R}$ such that $v=\alpha e_1+\beta e_2$ and $w=\lambda e_1+\mu e_2$. Hence

\begin{eqnarray}v\otimes w &=& (\alpha e_1+\beta e_2)\otimes (\lambda e_1+\mu e_2)\\ &=& \alpha\lambda e_1\otimes e_1 + \beta\lambda e_2\otimes e_1 + \alpha\mu e_1\otimes e_2+\beta\mu e_2\otimes e_2. \end{eqnarray}

Thus $$e_1\otimes e_1+e_2\otimes e_2=\alpha\lambda e_1\otimes e_1 + \beta\lambda e_2\otimes e_1 + \alpha\mu e_1\otimes e_2+\beta\mu e_2\otimes e_2$$ or equivalently $$(\alpha\lambda-1) e_1\otimes e_1 + \beta\lambda e_2\otimes e_1 + \alpha\mu e_1\otimes e_2+(\beta\mu-1) e_2\otimes e_2=0.$$

Since $\left\{e_i\otimes e_j\right\}_{1\leq i,j \leq 2}$ forms a basis of $\mathbb{R}^2\otimes \mathbb{R}^2$, they are linearly independent. Hence \begin{eqnarray} \alpha\lambda&=&1,\\ \beta\lambda &=&0,\\ \alpha\mu &=&0,\\ \beta\mu&=&1. \end{eqnarray} It follows that $\alpha,\beta,\lambda$ and $\mu$ are all non-zero but some of them are zero, a contradiction!

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  • $\begingroup$ Thank you for the detailed answer!!! $\endgroup$ – Dragonite Apr 13 '17 at 15:15
  • $\begingroup$ The other answer does exactly the same, but in more conceptual way actually. For finite-dimensional vector spaces though you need not worry about universal properties. $\endgroup$ – Mathematician 42 Apr 13 '17 at 15:17
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Consider the map $f: \Bbb R^2 \times \Bbb R^2 \to \Bbb R^4$ given by $(ae_1 + be_2, ce_1+de_2) \mapsto (ac, ad, bc, bd)$. This can be checked to be bilinear, which by the universal property of the tensor product reduces the given problem to showing that $f(e_1, e_1) + f(e_2, e_2)$ is not of the form $f(v,w)$ for any $v,w$.

(Here the intution is that the induced map $\Bbb R^2 \otimes \Bbb R^2 \to \Bbb R^4$ is actually the isomorphism which identifies the standard basis of $\Bbb R^4$ with the usual basis of $\Bbb R^2 \otimes \Bbb R^2 $)

Now note that $f(e_1, e_1) = (1,0,0,0)$ and $f(e_2, e_2) = (0,0,0,1)$. If we set $v = ae_1 + be_2, w = ce_1+de_2$, then $f(v,w) = (ac, ad, bc, bd) = (1,0,0,1)$ has no solutions, since if $ad=0$ then either $a=0$ or $d=0$, which contradict $ac = 1, bd=1$, respectively.

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  • $\begingroup$ You were a bit faster and actually I prefer your formulation. +1 $\endgroup$ – Mathematician 42 Apr 13 '17 at 15:15

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