Let $F$ be a field, $V=\oplus_{i=1}^m Fv_i$, $W=\oplus_{j=1}^n Fw_j$ be $F$-vector spaces, $\otimes: V\times W\to V\otimes_F W$ be a balanced product. How to prove that $\sum_{i=1}^m\sum_{j=1}^n a_{ij}v_i\otimes w_j$ $\in\text{im}(\otimes)$ if and only if $\text{rank}(A)\le 1$, where $A=(a_{ij})$?
Thanks in advance.
Let $v \in V$ and $w \in W$, so that $v = r_1v_1 + \dots + r_mv_m$ and $w = s_1w_1 + \dots + s_nw_n$ for some $r_1, \dots, r_m, s_1, \dots, s_n \in F$. Then
$$v\otimes w = \sum_{i=1}^m\sum_{j=1}^nr_is_jv_i\otimes w_j.$$
So the claim is equivalent to: $a_{ij} = r_is_j$ if and only if $A = [a_{ij}]$ has rank at most one.
If $a_{ij} = r_is_j$, then
$$A = \left[\begin{array}{cccc} r_1s_1 & r_1s_2 & \dots &r_1s_n\\ r_2s_1 & r_2s_2 & \dots & r_2s_n\\ \vdots & \vdots & \ddots&\vdots\\ r_ms_1 & r_ms_2 & \dots & r_ms_n \end{array}\right].$$
Now note that
\begin{align*} \operatorname{rank}(A) &= \dim\operatorname{Col}A\\ &= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1s_1\\ \vdots\\ r_ms_1\end{array}\right], \dots, \left[\begin{array}{c} r_1s_n\\ \vdots\\ r_ms_n\end{array}\right]\right\}\\ &= \dim\operatorname{span}\left\{s_1\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right], \dots, s_n\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\ &\leq \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\ &\leq 1. \end{align*}
Conversely, if $A$ has rank at most one, then $\operatorname{Col}(A) = \operatorname{span}\{r\}$ for some vector $r = \left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]$.
As $\operatorname{Col}(A)$ is spanned by the columns of $A$, there are $s_1, \dots, s_n \in F$ such that the $j^{\text{th}}$ column of $A$ is of the form $s_jr$ and therefore $a_{ij} = s_jr_i$.
Summarising, if $r \in F^m$ and $s \in F^n$, then the $m\times n$ matrix $rs^T$ (called the outer product of $r$ and $s$) has rank one, unless $r = 0$ or $s = 0$ in which case it has rank zero. Conversely, if an $m\times n$ matrix $A$ has rank one or zero, then it is equal to the outer product of some $r \in F^m$ and $s \in F^n$. More generally, the rank of $A$ can be characterised as the smallest $k$ for which $A$ can be written as the sum of $k$ outer products.