Fix any non-negative matrix $M \in \mathbb{R}_{\geq 0}^{m \times n}$ that contains no zero-row and no zero-column. Further, fix any positive vector $r \in \mathbb{R}_{> 0}^m$. With $nz(M) := \{(i,j) \ | \ M_{i,j} > 0\}$ the index set of non-zero-entries in $M$, and $\mathbf{1}$ the all-ones-vector, define the set

$$S_{r, W} := \{A \in \mathbb{R}_{\geq 0}^{m \times n} \ | \ A \mathbf{1} = r, \ nz(A) \subseteq nz(M)\}$$

of all non-negative matrices that have row sum $r$ at that have at least the zero-entries of $M$, that is $M_{i,j} = 0 \Rightarrow A_{i,j} = 0$. This set is non-empty, because $diag(r) \cdot diag(M \mathbf{1})^{-1} M \in S_{r, W}$.

Now consider the following set:

$$A_{r, W} := \{A \in \mathbb{R}^{m \times n} \ | \ A \mathbf{1} = r, \ nz(A) \subseteq nz(M)\} \quad \supseteq \quad S_{r, W}$$ which in contrast to $S_{r, W}$ allows for arbitrary matrices under these constraints.

It is easy to see that every affine combination of elements from $S_{r, W}$ gives some element from $A_{r, W}$, hence, the affine hull satisfies $$aff(S_{r, W}) \subseteq A_{r, W}.$$

I suppose that it even holds that $aff(S_{r, W}) \supseteq A_{r, W}$.

I tried to prove this by constructing an affine combination of non-negative matrices explicitly for any element from $A_{r, W}$, but I failed on this approach.

So, my question is how to prove this (if it is true at all)?

Is there probably even a more elegant argument, for example from the theory of convex polytopes, or by some neat characterization of the affine hull?