Let a matrix $M\in\mathbb R^{m\times n}$ be given. Define $\vec 1 = [1, \ldots ,1]^T$ to be a vector of all ones (I don't specify its length since it will be clear from the context.)
Under what conditions on $M$ can we find $v\in\mathbb R^n$ and $w\in\mathbb R^m$ such that $(\ref{e})$ holds?
$$ M=w\vec 1 ^T + \vec 1v^T \label{e}\tag{1} $$
Under what conditions on $M$ will such $v$ and $w$ be unique?
My attempt: I tried assuming that some $x \in \text{ker}(w^T)$, then we'd have $$x^T M = x^T \vec 1 v^T$$ which doesn't seem to bring us any further.
Let's write $u$ instead of $1$. So you want $$ M = wu^t + u v^t $$ Well, let $s_i = e_1 - e_i$. Then $u^t s_i = 0$. So $$ \begin{align} Ms_i &= wu^ts_i + u v^t s_i\\ &= u v^t s_i \end{align} $$ Since $Ms_i$ is just the $i$th column of $M$ minus the first column of $M$, the left side's easy to understand.
On the right, $v^t s_i$ is just $q = v_i - v_1$. You multiply that by $u$ and you get a column of $q$s.
In short:
If there's a solution, then the difference between the $i$th column of $M$ and the first column should be a column of constants. (Same thing applies to rows).
Now you've got all the differences $v_i - v_1$ and $w_i - w_1$; how can you find the actual $v$ and $w$ vectors (if they exist)? That's a little messy. Once you know $v_1$, everything else is simple, for $v_2$ is just $(v_2 - v_1)$, which you've computed, added to $v_1$; the same goes for all the other entries of $v$. And $m_{11} = v_1 + w_1$, so you can find $w_1 = v_1 - m_11$.
The problem is that if $v = au$ and $w = bu$, then $v' = (a+b)u, w' = 0$ is another solution to the problem, i.e., in certain cases, there may not be a single solution, but rather infinitely many solutions.
Wait...in general, you can add $au$ to $v$ and subtract $au$ from $w$ to get a new solution! So if there's a solution, there are infinitely many solutions. And one of them has $v_1 = 0$.
So here goes, in simple form:
If all column-differences and all row-differences are constant vectors, there's a solution. If not, there isn't.
In the first case, you can compute a solution $v, w$ as follows:
$$ v_1 = 0\\ v_2 = m_{12}\\ \ldots \\ v_n = m_{1,n} $$ That gets you $v$. Now $$ w_1 = m_{11}\\ w_2 = m_{21}-m_{11}\\ \ldots w_n = m_{n,1} - m_{11} $$ And I think that's it!