Given a symmetric matrix $M\in\Bbb R^{n\times n}$ so that
- all off-diagonal entries are non-negative,
- $M$ is singular (not invertible), and
- $M$ has a unique positive eigenvalue of multiplicit one,
are the row sums of $M$ necessarily positive?
I observed this in certain applications, such as with the Alexandrov matrix of a polytope, or the Colin de Verdiére matrix of certain graphs, and I wondered what structure is necessary to guarantee this.
If this turns out false, I wonder whether it help to restrict to irreducible matrices: if you interpret the matrix as an adjacency matrix of a graph, where $0$ gives non-edges and everything else gives an edge, then irreducibility means that this graph is connected. This is motivated from past applications, but also reminds of the conditions in the non-negative Perron-Frobenius theorem, which is often surprisingly helpful in such questions (though I don't know how it helps here).
No. Suppose $n\ge3$ and $M=ee^T-(n+1)e_1e_1^T$. Then $M$ is irreducible (because it is entrywise nonzero), singular (because its rank is at most $2$) and it has positive off-diagonal entries.
Since $M\preceq ee^T$, its second largest eigenvalue is less than or equal to the second largest eigenvalue of $ee^T$, which is zero. Thus $M$ has at most one positive eigenvalue. However, as $M$ is not negative semidefinite (because it does not have a non-positive diagonal), it has at least one positive eigenvalue. It follows that $M$ has exactly one positive eigenvalue.
Yet, the first row sum of $M$ is $-1$.
For a concrete counterexample, consider the case where $n=3$. The eigenvalues of $$ M=\pmatrix{-3&1&1\\ 1&1&1\\ 1&1&1} $$ are $\frac12(\sqrt{33}-1),\,-\frac12(\sqrt{33}+1)$ and $0$.