Yes. It's exactly the collection of non-negative matrices subject to:
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The entries sum to 1;
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The diagonal entries are 0;
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For each $i$, the $i$th column and $i$th row have equal sums. 

The proof is by induction on the number of non-zero entries.
Suppose that matrices satisfying the above conditions with at most $k$ non-zero entries are in the convex hull. Then given a matrix $A$ with $k+1$ non-zero entries, pick an entry $(i,j)$ of the matrix so that $A_{ij}$ is non-zero. Let $i_0=i$ and $i_1=j$. Since the $i_1$st row has a non-zero entry, the $i_1$st column must also have a non-zero entry. Continue in this way until $i_l=i_m$ for some $l<m$. Then there is a cycle $i_l,i_{l+1},\ldots,i_m,i_l$ where all the entries corresponding to an edge are non-zero. Take the minimum of the edge-weights $w$ and let $B$ be the matrix with weights $1/(m-l)$ on the $m-l$ edges forming the cycle. Then $A=w(m-l)B+(1-w(m-1))C$ for a matrix $C$ satisfying the constraints and having at least one less non-zero entry.