If $v\otimes w=0$ then $v=0$ or $w=0$.

Question

Let $v \otimes w \in V \otimes W$. If $v \otimes w = 0$ then $v=0$ or $w=0$.

I am having a difficult time proving "basic" things like this involving tensor products. I know that $v\otimes w$ is the decomposable bilinear form $V^*\times W^*\to \mathbb F$ given by $(\alpha,\beta)\mapsto\langle\alpha,v\rangle \langle\beta,w\rangle$. However, the notions of tensors still confuse me, any help would be much appreciated.

Answer 1

I assume you mean vector spaces over a field, otherwise it won't be true (e.g. for $\mathbb Z$-modules $\mathbb Z/5 \otimes \mathbb Z/3 = 0$). Use the definition of the tensor product: a linear map $V\otimes W \to U$ for any vector space $U$ is the same as a bilinear map from the pair $V,W$ to $U$. Now a tensor $\sum_i v_i \otimes w_i$ is nonzero if and only if there exists a bilinear map $f$ such that $\sum_i f(v_i, w_i) \ne 0$. I will leave to you to find such a map in your case.

Answer 2

Assume $v \neq 0$ and $w \neq 0$. Then we have $f,g \in V^*$ such that $f(v) \neq 0$ and $g(w) \neq 0$ (why?). This implies that $(v\otimes w)(f,g) = f(v)g(w) \neq 0$, and hence $v \otimes w \neq 0$.