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.

  • $\begingroup$ Thanks. Do you know of any results that are related to my question. I know that diagonally dominant matrices will be positive semidefniite, but I am trying to determine if all of the above conditions allow you to weaken this so that the diagonal entries are largest but perhaps the matrix is not diagonally dominant. $\endgroup$ – user3281410 Feb 28 '18 at 13:08
  • $\begingroup$ @user3281410 Sorry, I don't know, but this might be hard. See my new edit. $\endgroup$ – user1551 Feb 28 '18 at 14:51
  • $\begingroup$ Thank you again for this new example. This has been very helpful. $\endgroup$ – user3281410 Mar 1 '18 at 16:07

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