This question is a generalization of this one.
For the sake of this question, a tensor product of $$ and $$ is a couple $(,ℎ)$ where $$ is a vector space and $ℎ:×→$ is a bilinear map satisfying the the universal property : for every vector space $$ and bilinear map $:×→$, there exist an unique linear map $̃ :→$ such that $=\tilde f ∘ℎ$.
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.
In general, if $\dim(V)=n$, $\dim(W)=m$ and $\{v_i:1\leq i\leq n\}$ is a basis of $V$ and $\{w_1,\ldots,w_m\}$ is a basis of $W$ then $\{v_i\otimes w_j: 1\leq i\leq n, 1\leq j\leq m\}$ is a basis of $V\otimes W$, so that any tensor can be written as a linear combination of the $n.m$ tensors $\{v_i\otimes w_j\}$.
However, an arbitrary tensor is given as a sum of pure tensors $\eta = \sum_{i=1}^k a_i\otimes b_i$, so it seems more natural to ask to what extent such a representation is unique. First it should be noted that $k$ need never exceed $\min \{m,n\}$: indeed we may always write $\eta = \sum_{i,j}a_{ij} v_i\otimes w_j$, and hence $$ \eta = \sum_{i=1}^n v_i\otimes (\sum_{j=1}^m a_{ij}w_j) = \sum_{j=1}^m\big(\sum_{i=1}^n a_{ij}v_i\big) \otimes w_j $$
The above captures essentially all that one can say, but it is possible to be a bit more explicit:
Update: I have added proofs of the Facts, and slightly refined the statement of Fact 1.
Fact 1: If $\eta\in V\otimes W$ and $\{v_1,\ldots,v_k\}\subseteq V$ is linearly independent in $V$, then there is at most one representation of $\eta$ of the form $\sum_{i=1}^k v_i\otimes w_i$ (i.e. it is unique when it exists). Moreover, if $\text{span}\{v_1',\ldots,v_k'\}= \text{span}\{v_1,\ldots,v_k\}$ then $\eta$ can be written in the form $\sum_{i=1}^k v_i\otimes w_i$ if and only if it can be also be written in the form $\sum_{i=1}^k v_i'\otimes w_i'$.
Proof: Given a linearly independent set $\{v_1,\ldots,v_k\}\subseteq V$, we may extend it to a basis $\{v_1,\ldots,v_n\}$ of $V$, and pick any basis $\{b_1,\ldots,b_m\}$ of $W$. Then $\eta \in V\otimes W$ can be written uniquely as a sum $\sum_{i,j} \lambda_{i,j}v_i\otimes b_j$, and thus there is an expression for $\eta$ of the form $\sum_{i=1}^k v_i\otimes w_i$ if and only if $\lambda_{i,j}=0$ for all $i \notin \{1,2,\ldots,k\}$. Indeed if such an expression exists, then writing $w_i = \sum_{r=1}^m \alpha_{i,j}b_j$ we see that $\eta = \sum_{1\leq i \leq k,j}\alpha_{i,j} v_i\otimes b_j$, and hence the coordinates of $\eta$ are nonzero only for $i\in\{1,2,\ldots,k\}$. For the converse one simply notes that $\sum_{i=1}^k \lambda_{i,j}v_i\otimes b_j = \sum_{i=1}^k v_i\otimes w_i$ where $w_i =\sum_{j=1}^m \lambda_{i,j}b_j$.
For the final sentence of Fact 1, if $\eta$ can be expressed in terms of $\{v_1,\ldots,v_k\}$ then in order to express it as in terms of the $\{v_1',\ldots,v_k'\}$ one simply expresses the $v_i$ in terms of the $v_i'$s and "collects terms" in the $W$ factor in the same fashion we used above.
Fact 2: If $m$ is the smallest integer for which there is an expression $\eta = \sum_{i=1}^m v_i\otimes w_i$, then the subspaces $\text{span}\{v_i: 1\leq i\leq m\} = V(\eta)$ and $\text{span}\{w_i:1\leq i\leq m\}= W(\eta)$ do not depend on the choice of representative $\sum_{i=1}^m v_i\otimes w_i$ and are both $m$-dimensional.
Proof: Suppose that $\eta = \sum_{i=1}^m v_i\otimes w_i$ for $v_i \in V, w_i\in W$ ($1\leq i \leq m$) is any representation of $\eta$. Then the set $\{v_1,\ldots,v_m\}$ contains a maximal linearly independent subset, which by reordering if necessary we may assume is $\{v_1,\ldots,v_k\}$. Thus for $j>k$ we have $v_i = \sum_{i=1}^k c_{i j} v_j$, and hence $$ \sum_{i=1}^m v_i\otimes w_i = \sum_{i=1}^k v_i\otimes w_i + \sum_{j>k}\sum_{i=1}^k c_{ij} v_i\otimes w_j = \sum_{i=1}^k v_i\otimes \left(w_i+\sum_{j>k} c_{ij}w_j\right) $$ and hence setting $u_i = w_i +\sum_{j>k} c_{ij}w_j$, we have $\eta = \sum_{i=1}^k v_i\otimes u_i$. Since we will need it in a moment, note that if $\{w_1,\ldots,w_k\}$ are linearly independent, then so are $\{u_1,\ldots,u_k\}$.
But now by symmetry, if $\eta$ can be written as $\sum_{i=1}^k v_i\otimes u_i$ where $\{v_i:1\leq i \leq k\}$ is a linearly independent set, we may apply the same technique to replace $\{u_1,\ldots,u_k\}$ with a linearly independent subset, by combining some of the $v_i$s into linear combinations as we did with the $w_j$s above. Thus we obtain an representation of $\eta$ of the form $\sum_{i=1}^l t_i\otimes u_i$ where both $\{u_1\ldots, u_l\}$ and $\{t_1,\ldots,t_l\}$ are linearly independent.
It follows that if $\eta = \sum_{i=1}^k a_i \otimes b_i$ and $k$ is the smallest integer for which $\eta $ can be written as a sum of $k$ pure tensors, then both the sets $\{a_i:1\leq i \leq k\}$ and $\{b_j: 1\leq j \leq k\}$ are linearly independent.
Now suppose that $\eta = \sum_{i=1}^k a_i\otimes b_i$ is a representation of $\eta$ for which $\{a_i:1\leq i \leq k\}$ and $\{b_i: 1\leq i \leq k\}$ are linearly independent, and that $\eta = \sum_{j=1}^l c_j\otimes d_j$ is another such representation, with $\{c_i: 1\leq j \leq l\}$ and $\{d_j: 1\leq j \leq l\}$ also linearly independent. Then let $S_1 = \mathrm{span}\{a_1,\ldots,a_k\}$ and $S_2 = \mathrm{span}\{c_1,\ldots,c_l\}$. Picking a basis of $\{s_1,\ldots,s_r\}$ of $S_1\cap S_2$, we may extend it to a basis $B_1=\{s_1,\ldots,s_r,t_1,\ldots, t_{k-r}\}$ of $V_1$ and a basis $B_2=\{s_1,\ldots,s_r,u_1,\ldots, u_{l-r}\}$ of $V_2$. But then $\eta$ must have a unique representation in terms of the basis $B_1$ and another in terms of $B_2$. But since $B_1\cup B_2$ is linearly independent, these representations must be the same, and hence we must have $V_1=V_2$. It follows that $V(\eta)$ is well-defined (and hence, by symmetry it follows that $W(\eta)$ is also well-defined).