Problem Of The Week # 323 - Sep 14, 2018

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Conversation:Person 1: "Have you watched the new Marvel movie yet?"Person 2: "Yes, I did! It was amazing. The action scenes were top-notch."In summary, Person 2 has watched the new Marvel movie and thought it was amazing, particularly praising the action scenes.
  • #1
Euge
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MHB
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Here is this week's POTW:

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Show that, for any basis $v_1, v_2 \in \Bbb R^2$, the sum $v_1 \otimes v_2 + v_2 \otimes v_1$ in $\Bbb R^2 \otimes_{\Bbb R} \Bbb R^2$ cannot be reduced to a simple tensor.
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  • #2
This week's problem was solved correctly by castor28. You can read his solution below.
Assume by contradiction that $v_1\otimes v_2 + v_2\otimes v_1=(\alpha v_1+\beta v_2)\otimes(\gamma v_1+\delta v_2)$.
We have:
$$
(\alpha v_1+\beta v_2)\otimes(\gamma v_1+\delta v_2)=\alpha\gamma (v_1\otimes v_1)+\alpha\delta (v_1\otimes v_2) + \beta\gamma (v_2\otimes v_1) + \beta\delta (v_2\otimes v_2)
$$
As the $(v_i\otimes v_j)$ constitute a basis for $\mathbb{R}\otimes\mathbb{R}$, we must have:
\begin{align*}
\alpha\gamma &= 0\\
\alpha\delta &= 1\\
\beta\gamma &= 1 \\
\beta\delta &= 0
\end{align*}
As $\alpha\delta=1$, we have $\alpha\ne0$. As $\alpha\gamma = 0$, we have $\gamma=0$, and this contradicts the fact that $\beta\gamma=1$.
 

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