Let $A$ and $B$ be two square real positive (all entries are positive) matrices that differ only in the first row. Let $\lambda\_A$ and $\lambda\_B$ be the maximal real eigenvalues of $A$ and $B$, respectively. Let $\lambda\_\*$ be the maximal real eigenvalue of the matrix $(A+B)/2$.
It is easy to see that $\lambda\_\* \leq \max\{\lambda\_A,\lambda\_B\}$.
I would like to know whether $\lambda\_\* \geq \min\{\lambda\_A,\lambda\_B\}$.



The maximal eigenvalue of average of positive matrices
Let $u$ be an eigenvector of $M = (A+B)/2$ for $\lambda\_\*$. By 
Perron-Frobenius we can choose $u \ge 0$. Now if $e\_j$ is the
$j$'th standard unit vector, $e\_j^T A = e\_j^T B$ for $j > 1$.
Thus for $j > 1$, $e\_j^T A u = e\_j^T B u = \lambda\_\* u\_j$. On the
other hand, $e\_1^T M u = \lambda\_\* u\_1$ implies that $\max(e\_1^T A u, e\_1^T B u) \ge \lambda\_\* u\_1$ and $\min(e\_1^T A u, e\_1^T B u) \le \lambda\_\* u\_1$. WLOG $e\_1^T A u \ge \lambda\_\* u\_1$
and $e\_1^T B u \le \lambda\_\* u\_1$. Thus $v^T A u \ge \lambda\_\* v^T u$ and $v^T B u \le \lambda\_\* v^T u$ for any nonnegative vector $v$. In particular, this is true for the Perron eigenvector $v\_A$ of $A^T$ and the Perron eigenvector $v\_B$ of $B^T$. Thus $\lambda\_A v\_A^T u \ge \lambda\_\* v\_A^T u$ and $\lambda\_B v\_B^T u \le \lambda\_\* v\_B^T u$. Since the matrices have strictly positive entries, so do the Perron eigenvectors, and thus $\lambda\_A \ge \lambda\_\*$ and $\lambda\_B \le \lambda\_\*$.


