TITLE: A question about symmetric matrix
QUESTION [6 upvotes]: Let $A= (a_{ij})_{ij}, 1 \leq i, j \leq n$ be a symmetric $n \times n$ matrix. Suppose 
(1) $a_{ij} \geq 0$ are real numbers;
(2) The sum of each row $\sum_{j=1}^{n} a_{ij} = 1$ for $1 \leq i \leq n$.
Then I want to show the following: there must exists a nonzero $\prod_{i=1}^n a_{i, \sigma(i)}$, where $\sigma \in S_n$ is an element of the symmetric group $S_n$. In other words, there must exists a nonzero summand in the expression of $\det A$.

REPLY [2 votes]: An other solution (but Darij's is enough). It is classical that if all these products are zero, then $A$ contains a submatrix $k\times\ell$ of zeros, with $k+\ell>n$ (see for instance Exercise 7, Chapter 3 of my book, 2nd edition). Then an elementary calculus shows that the sums of entries in the opposite block equals $n-k-\ell$. This is negative, contradicting the fact that the entries are non-negative.