MHB Problem Of The Week # 323 - Sep 14, 2018

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The problem of the week discusses proving that for any basis vectors v_1 and v_2 in R^2, the expression v_1 ⊗ v_2 + v_2 ⊗ v_1 in R^2 ⊗_R R^2 cannot be simplified to a simple tensor. The solution was successfully provided by castor28, demonstrating the mathematical reasoning behind the assertion. The discussion emphasizes the properties of tensor products and their implications in linear algebra. Participants are encouraged to engage with the solution and explore the underlying concepts further. This problem highlights the complexities of tensor operations in vector spaces.
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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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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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