Is there a rote algorithm to tell whether a tensor (of two vectors) can be reduced to an elementary tensor? Given vector spaces $V$ and $W$ over a field $F$, is there an algorithm that can decide whether an expression
$$ v_{1}\otimes w_{1} + \cdots + v_{n}\otimes w_{n} $$
can be reduced to an expression of the form
$$ v\otimes w $$
(where $v_{i}, v\in V$ and $w_{i}, w\in W$)?
If this is possible to decide, is there an algorithm to make such a reduction?
 A: If $V$ and $W$ are finite-dimensional, then the answer is yes. Let $e_1, \dots, e_k$ be a basis of $V$, let $f_1, \dots, f_m$ be a basis of $W$. Then $\{e_i \otimes f_j \mid i \in \{1, \dots, k\}, j \in \{1, \dots, m\}\}$ is a basis of $V \otimes W$.
Step 1. Determine the decomposition of $v_1 \otimes w_1 + \dots + v_n \otimes w_n$ in terms of the basis vectors, i.e. calculate, for each $i$ and $j$, $a_{i,j}$ so that $\sum_{i,j} a_{i,j} e_i \otimes f_j = v_1 \otimes w_1 + \dots + v_n \otimes w_n$. This can be done in the obvious way - by decomposing each addend in terms of the basis vectors and then summing the results.
Step 2. What is left to check is whether the matrix $A$ determined by the entries $a_{i,j}$ has rank one. Assume, without loss of generality, that $a_{1,1} \neq 0$. Check if there is a nonzero row $a_i$ which has $a_{i,1} = 0$ - if this is the case, then the matrix is not rank-1. For each nonzero row $a_i$, divide it by $a_{i,1}$ and check if all such rows are equal to each other. If they are, the matrix is indeed rank-1, otherwise it's not.
I don't know how to formulate the problem in maximum generality when $V$ or $W$ is infinite-dimensional.
A: Suppose that $W=U^*$ (the linear dual of $U$). Then $V\otimes W=\mathrm{Hom}(U,V)$. Pure tensors in $V\otimes W$ are precisely rank one operators. Indeed, if $T\in \mathrm{Hom}(U,V)$ is a rank one operator, let $v$ span its image. Let $w\in W=U^*$ be defined by $w(u)v=T(v)$. Then $T=v\otimes w$.
So here is your algorithm: think of $v_i$ and $w_i$ as column vectors with respect to some bases of $V$ and $W$. $v\otimes w$ is a pure tensor if and only if the matrix $A=\sum_{i=1}^n v_i w_i^T$ is of rank one. If it is of rank one, then we can easily write $A$ in the form $vw^T$. We will have:
$$
\sum_{i=1}^n v_i\otimes w_i = v\otimes w.
$$
