$n \times n$ positive matrix with $a_{ij} a_{ji} = 1$ has an eigenvalue not less than $n$

Question

$A$ is a real $n \times n$ matrix with positive elements $\{a_{ij}\}$. For all pairs $(i, j), a_{ij} a_{ji}=1$. Prove that $A$ has an eigenvalue not less than $n$.

Answer 1

Proof 1. By Perron-Frobenius theorem, $Av=\rho(A)v$ for some positive eigenvector $v$. Let $D=\operatorname{diag}(v)$ (the diagonal matrix whose diagonal is $v$), $e=(1,\ldots,1)^T$ and $B=D^{-1}AD$. Then $Be=\rho(A)e$. Since $B$ is also a positive matrix with $b_{ij}b_{ji}=1$ for all $i,j$, and $2\le b+\frac1b$ for every positive real number $b$, we have $n^2\le e^TBe=e^T\left(\rho(A)e\right)=n\rho(A)$ and the result follows.

Proof 2. For any (entrywise) nonnegative square matrix, we have (cf. Horn and Johnson, Topics in Matrix Analysis, 1/e, p.363, corollary 5.7.11) $$ \rho\left[A^{(1/2)}\circ(A^T)^{(1/2)}\right]\le\rho(A),\tag{1} $$ where the square roots in the above are taken entrywise. In our case, $A\circ A^T=E$, the matrix with all entries equal to one. Hence $(1)$ gives $\rho(A)\ge\rho(E)=n$. As $\rho(A)$ is an eigenvalue of $A$ (Perron-Frobenius theorem), the result follows.