This question is a generalization of this one.
For the sake of this question, a tensor product of $𝑉$ and $𝑊$ is a couple $(𝑇,ℎ)$ where $𝑇$ is a vector space and $ℎ:𝑉×𝑊→𝑇$ is a bilinear map satisfying the the universal property : for every vector space $𝑍$ and bilinear map $𝑓:𝑉×𝑊→𝑍$, there exist an unique linear map $𝑓̃ :𝑇→𝑍$ such that $𝑓=\tilde f ∘ℎ$.
Suppose that
$\sum_{ij}a_{ij}v_i\otimes w_j=\sum_{kl}b_{kl}a_k\otimes b_l$.
I want to know if, up to some permutations of the indexes, there exists $\lambda_i$ and $\mu_j$ such that $v_i=\lambda_i a_i$ and $w_j=\mu_jb_j$.
In other words, I'd like to know, $\omega\in V\otimes W$ being given, how unique a decomposition $\omega=\sum_{ij}u_i\otimes v_j$ is.
Best Answer
This is false even with the additional hypothesis that the four sets of vectors $\{ v_i \}, \{ w_j \}, \{ a_k \}, \{ b_k \}$ are each individually linearly independent and there are easy counterexamples like
$$(v_1 + v_2) \otimes w_1 = v_1 \otimes w_1 + v_2 \otimes w_1$$
(where we take $a_1 = v_1 + v_2, b_1 = w_1$ and $v_1, v_2$ are linearly independent).
In a positive direction it is true that if $\{ v_i \}$ and $\{ w_j \}$ are linearly independent then so is the family $\{ v_i \otimes w_j \}$ in the tensor product.