Proof that the set of doubly-stochastic matrices forms a convex polytope? Does the set of all doubly-stochastic matrices form a convex polytope?  In general, I wonder how the proofs of convexity and geometry can be established for sets of matrices of this kind? Anything to do with the birkhoff's theorem!? Am not sure, if I am thinking in the right direction and hence the post.
 A: I only write this answer to summarize the comments.
We identify the space of matrix $\mathbb{R}^{n\times n}$ with $\mathbb{R}^{n^2}$.
A matrix $(a_{i,j})_{i,j\leq n}$ is doubly stochastic if


*

*$0\leq a_{i,j}\leq 1\ \forall i,j$

*$\sum_i a_{i,j}=1 \ \forall j$

*$\sum_j a_{i,j}=1 \ \forall i$


It is direct to check that if  $(a_{i,j})_{i,j\leq n}$ and $(b_{i,j})_{i,j\leq n}$ are doubly stochastic then $(t a_{i,j}+ (1-t) b_{i,j})_{i,j\leq n}$ is doubly stochastic for any $0\leq t\leq1$ (the conditions above are clearly fulfilled). This gives the convexity.
To observe that the set of doubly stochastic matrices is a polytope in $\mathbb{R}^{n^2}$ we just need to have in mind that a polytope can be defined as a finite intersection of half-spaces.
The conditions above are actually $2n^2+4$ inequalities (have in mind that $a=b\iff (a\leq b \text{ and } b\leq a)$ ).
I think it is usual to consider that a polytope is bounded, here this is clear because of condition $1$.
Edit: As you suggested in your question, this is a direct corollary of Birkhoff-Von Neumann theorem which is much stronger. The Birkhoff–von Neumann theorem states that the set of doubly stochastic matrices is the convex hull of the set of permutation matrices. Their is a finite number of permutations and the convex hull of a finite number of points is a convex body, so the corollary is immediate. 
But if you just want to see that this is a convex polytope without beeing more precise you don't need to use this theorem, the elementary arguments used above are sufficients.
