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Or put it simply, for any $n$-by-$n$ doubly stochastic matrix $A$, you can always find $n$ non-zero entries in $A$, none of them lies in the same row or column. Why is that?

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I had an "answer" to this that some criticized; I'm going to remove it. Basically I did some gyrations on A to get it into the format with a 1 at upper left, such that the result is still doubly stochastic. But this has no bearing on the question, since there is no relation between the choices of nonzero entries of the original A and where nonzero entries might be in the "gyrated A". – coffeemath Oct 16 '12 at 23:58

According to the Birkhoff-von Neumann theorem the set of doubly stochastic $(n\times n)$ matrices is the convex hull of the set of $(n\times n)$ permutation matrices. For a simple proof see here (there are many proofs in the literature):

Given that, for any doubly stochastic $(n\times n)$ matrix $D$ by Caratheodory's theorem there are $<n^2$ permutation matrices $P^{(\ell)}$ and numbers $\lambda_\ell>0$ summing to $1$ such that $$D=\sum_\ell\lambda_\ell P^{(\ell)}\ .$$ It follows that $d_{ik}>0$ whenever $P_{ik}^{(1)}>0$, and the latter happens exactly once in each row and each column.

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