Let $M\in\Bbb R^{n\times n}$ be a symmetric matrix, all whose off-diagonal entries are non-positive (problem variant: negative). Can the following be true at the same time?
- $M$ has a unique negative eigenvalue $\theta<0$ (of multiplicity 1).
- $M$ has at least one row whose entries sum up to a positive (problem variant: non-negative) number.
By Perron-Frobenius I know that $\theta$ has an eigenvector $v\in\Bbb R^n$ with positive entries (we might assume that the matrix is irreducible in the sense required by Perron-Frobenius). If $v$ were $(1,...1)$, this would answer the question in the negative, but in general we don't have this.
Consider the matrix $$M = \begin{bmatrix}2 & -1 \\ -1 & 0\end{bmatrix}.$$ Clearly, $M$ is square and symmetric, and all of the off-diagonal entries of $M$ are negative. The sum of the elements in the first row of $M$ is positive. Also, the eigenvalues of $M$ are $1+\sqrt{2} > 0$ and $1-\sqrt{2} < 0$, so $M$ has a unique negative eigenvalue of multiplicity $1$..