Tensor product. Simple tensors. I have the following problem in my assignment:
Prove that every element of $V \otimes W$ can be represented as $\sum \limits_{i = 1}^{k} v_i \otimes w_i$, where $v_i$ are linearly independent and $w_i$ are linearly independent.
I thought that we can do the following:
Every tensor can be written as a linear combination of basis vectors $e_i \otimes f_j$, so we can represent this tensor as sum: $\sum \limits_{i = 1}^{n} e_i \otimes w_i$ (using tensor and summation properties). So, I have similar representation, but second components might be linearly dependent. My question is am I thinking in the right direction, and if so how I can transform this representation to make vectors in second component linearly independent?
 A: Given an element $\alpha \in V \otimes W$, we know it can be written as $\sum_{i=1}^k v_i \otimes w_i$ for some $v_i \in V, w_i \in W$ and $n \in \mathbb{N}$. Consider such a presentation of $\alpha$ in which $k$ is minimal. Let us show that in such a presentation, the vectors $v_1,\dots,v_k$ must be linearly independent and similarly the vectors $w_1,\dots,w_k$ must be linearly independent. 
If $v_1,\dots,v_k$ are linearly dependent then without loss of generality (by reordering the $(v_i,w_i)$) we can assume that $v_1 = \sum_{i=2}^k a_i v_i$ for some $a_2,\dots,a_k$. But then
$$ \alpha = \sum_{i=1}^k v_i \otimes w_i = v_1 \otimes w_1 + \sum_{i=2}^k v_i \otimes w_i = \left( \sum_{i=2}^k a_i v_i \right) \otimes w_1 + \sum_{i=2}^k v_i \otimes w_i \\
= \sum_{i=2}^k a_i (v_i \otimes w_1) + \sum_{i=2}^k v_i \otimes w_i = \sum_{i=2}^k v_i \otimes (a_i w_1) + \sum_{i=2}^k v_i \otimes w_i = \sum_{i=2}^k v_i \otimes (w_i + a_i w_1)$$
contradicting the minimality of $k$. The argument for the independent of the $(w_i)_{i=1}^k$ is exactly the same.
