Is a doubly stochastic matrix positive semidefinite if the maximum of each row occurs on the diagonal? Suppose that $A$ is doubly stochastic, symmetric with nonnegative entries. Suppose that the diagonal entries are the largest in each row (or column). Does this imply that $A$ is positive semidefinite?

The answer should be affirmative when the size of $A$ is at most $3$, otherwise it's negative in general. Counterexample: $$ \frac15\pmatrix{2&2&1&0\\\ 2&2&0&1\\\ 1&0&2&2\\\ 0&1&2&2} \pmatrix{-1\\\ 1\\\ 1\\\ -1}=-\frac15\pmatrix{-1\\\ 1\\\ 1\\\ -1}. $$ **Edit.** Here is another counterexample. Pick any $t>0$. Then $$ \frac1{t+2}\pmatrix{1&0&1&t\\\ 0&1&t&1\\\ 1&t&1&0\\\ t&1&0&1} \pmatrix{1\\\ 1\\\ -1\\\ -1}=\frac{-t}{t+2}\pmatrix{1\\\ 1\\\ -1\\\ -1}. $$ Note that when $t=0$, the matrix is diagonally dominant, but when you perturb the anti-diagonal by a small $t$, the matrix immediately becomes indefinite.