unicity of tensor product decomposition

Question

For the sake of this question, a tensor product of two vector spaces $V$ and $W$ over a field $K$ is a couple $(T,h)$ where $T$ is a vector space over $K$ and $h:V\times W\to T$ is a bilinear map satisfying the universal property: For every vector space $Z$ over $K$ and bilinear map $f:V\times W\to Z$ there exist an unique linear map $\tilde f:T \to Z$ such that $f=\tilde f\circ h$.

In this context, $u\otimes v$ is a notation for $h(u,v)$.

I want to prove that, if $u\otimes v=a\otimes b$, then there exist $\lambda,\mu \in K$ such that $u=\lambda a$ and $v=\mu b$.

QUESTIONS:

1. Is it true?
2. How to prove it?

My first try to prove it is to use the universal property taking, as $f$, the projection $V\times W\to V$. But $f$ is not bilinear.

EDIT: The vector spaces are finite dimensional real (or complex if possible) vector spaces.

Answer 1

It is not true when the vectors are zero, consider $u \otimes 0 = 0 \otimes 0$ for example. But when all the vectors are non-zero, it is true.

For two linear maps $f : V \to K$, $g : W \to K$, the map $f * g : V \times W \to K$ defined by $(v,w) \mapsto f(v) \cdot g(w)$ is bilinear, and hence lifts to a linear map $V \otimes W \to K$. We can make use of this.

Let $v_1,v_2 \in V$ and $w_1,w_2 \in W$ be non-zero vectors with $$v_1 \otimes w_1 = v_2 \otimes w_2.$$ Choose basis $\{b_i\}$ of $V$ and a basis $\{c_j\}$ of $W$. (We need to work with bases since the claim is not true for general modules, as mentioned by Qiaochu.) Let $f_i : V \to K$ denote the $i$-th projection (meaning $f_i(b_s) = \delta_{i,s})$ and similarly $g_j : W \to K$. Then we get $(f_i * g_j)(v_1 \otimes w_1) = (f_i * g_j)(v_2 \otimes w_2)$, hence $$f_i(v_1) \cdot g_j(w_1) = f_i(v_2) \cdot g_j(w_2)$$ holds in $K$. Choose some $i_0$ with $f_{i_0}(v_1) \neq 0$. Since $g_j(w_1) \neq 0$ for some $j$, we then also have $f_{i_0}(v_2) \neq 0$. Thus, $\lambda = f_{i_0}(v_2) / f_{i_0}(v_1) \in K$ is well-defined and non-zero.

But $\lambda$ does not really depend on $i_0$: We have $\lambda = g_j(w_1) / g_j(w_2)$ whenever $g_j(w_2) \neq 0$. From this we can easily deduce $f_i(v_2) = \lambda f_i(v_1)$ for all $i$, and hence $v_2 = \lambda v_1$. A similar argument shows $w_1 = \lambda w_2$.