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I am given that $\{u_i\}$ and $\{v_i\}$ are bases of $U$ and $V$ then $u_1\otimes v_2+u_2\otimes v_1$ is not a pure tensor.

I don't see how to prove this. Any hints?

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  • $\begingroup$ A little bit of clarification on definitions such as "pure tensor" and what you know about the problem or have tried so far would be be immensely helpful, and would result in more answers for you. Just a friendly FYI. $\endgroup$ – The Count Dec 8 '16 at 21:09
  • $\begingroup$ What facts about $U\otimes V$ are you familiar with? For instance, do you know that $\{u_i\otimes v_j\}_{i,j}$ is a basis? $\endgroup$ – user228113 Dec 8 '16 at 21:12
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Assume there exists $u \in U$ and $v \in V$ such that

$$ u \otimes v = u_1 \otimes v_2 + u_2 \otimes v_1. $$

Writing $u = \sum_{i \in I} a_i u_i$ and $v = \sum_{j \in I} b_j v_j$ and plugging in the equation, we get

$$ \left( \sum_{i \in I} a_i u_i \right) \otimes \left( \sum_{j \in I} b_j v_j \right) = \sum_{i,j \in I} (a_i b_j) (u_i \otimes v_j) = u_1 \otimes v_2 + u_2 \otimes v_1. $$

Since $(u_i \otimes v_j)$ is a basis for $U \otimes V$, we must have $a_1 b_2 = a_2 b_1 = 1$ and $a_1 b_1 = a_2 b_2 = 0$ which is impossible.

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