Prove that $\lambda_{\max}\geq n$ for a positive reciprocal ($a_{ij}=\frac{1}{a_{ji}}$ for all $i,j\in\{1,2,\dots,n\}$) square matrix $A$, where $n$ is the number of rows of $A$ and $\lambda_{\max}$ is its dominating eigenvalue. Also prove that $\lambda_{\max}=n$ if and only if $A$ is consistent ($a_{ij}a_{jk}=a_{ik}$ for all $1\leq i,j,k\leq n.$)
If anyone knows the proof or can give a link to it, it would be very much appreciated as I cannot find it anywhere.
I assume positive reciprocal matrix means all components are positive, with off diagonal elements obeying the reciprocal vs transpose, and diagonal elements necessarily being one.
I have left the equality conditions as an open item for OP. The proof of the inequality is given below.
The heart of this problem is
(1) the fact that for $x \in (0,\infty)$ we have
$2 \leq x+x^{-1}$ or $1 = x^\frac{1}{2}(x^{-1})^\frac{1}{2} \leq \frac{1}{2}\big(x+x^{-1}\big)$ by $\text{GM}\leq \text{AM}$
(2) since all components are positive, there is some nice structure afforded by Perron Theory.
(3) For any positive diagonal matrix $D$, if $A$ is 'positive reciprocal' then so is the similar matrix $\big(D^{-1}AD\big)$
nice case
suppose that the Perron vector $\mathbf v$ is both a left and right eigenvector of $A$, i.e. $A^T\mathbf v = \lambda_1 \mathbf v$ and $A\mathbf v = \lambda_1 \mathbf v$. This implies that $\lambda_{max} = \lambda_1 = \sigma_1$. Where singular values are in the usual ordering of $\sigma_1\geq \sigma_2\geq ... \geq \sigma_n$. To confirm this, check the Perron vector and hence the maximal eigenvalue of positive matrix $\big(A^TA\big)$.
using the the Operator 2 norm we know
$\lambda_{Perron}\Big(\frac{1}{2}\big(A+A^T\big)\Big)\leq \Big\Vert\frac{1}{2}\big(A+A^T\big)\Big\Vert_2 \leq\Big\Vert \frac{1}{2}A^T \Big\Vert_2 + \Big\Vert \frac{1}{2}A \Big\Vert_2 = \sigma_{max}\Big(A\Big)=\sigma_1 = \lambda_1$
(it's easy to check that these are in fact both met with equality)
By Perron Theory, the Perron Root is bounded below by the minimal row sum of any positive matrix (and the inequality is strict unless the minimal row sum equals the maximal row sum). So consider minimal row sum occurring in row k and we have
$\lambda_{Perron}\Big(\frac{1}{2}\big(A+A^T\big)\Big)$
$\geq \mathbf e_k^T\Big(\frac{1}{2}\big(A+A^T\big)\Big)\mathbf 1$
$= \frac{1}{2}\Big(\big(\mathbf e_k^T A\mathbf 1\big)+ \big(\mathbf e_k^T A^T\mathbf 1\big)\Big)$
$= \frac{1}{2}\Big(\big(\mathbf e_k^T A\mathbf 1\big)+ \big(\mathbf 1^T A\mathbf e_k\big)\Big)$
$= \frac{1}{2}\Big(\sum_{j=1}^n (a_{k,j}+a_{k,j}^{-1})\Big) $
$\geq \frac{1}{2}\Big(\sum_{j=1}^n 2\Big) $
$=n$
where the second inequality is (1)
Thus we've proven for the nice case that
$\lambda_1\geq n$ i.e. The Perron root of $A$ is bounded below by $n$
more general case
$A\mathbf v = \lambda_1 \mathbf v$, but $ A^T\mathbf v \neq \lambda_1 \mathbf v$
We can find a similar matrix $C$ that has $\mathbf z$ as both its left and right Perron vector. Since it is similar to $A$ it has the same spectrum. Furthermore we'll only use positive diagonal matrices to effect this similarity transform, so by (3) we know $C$ is positive reciprocal and by the 'nice case' we have
$\lambda_1 = \lambda_{max}\big(A\big)= \lambda_{max}\big(C\big)\geq n$.
Thus it remains to construct $C$.
step 1: homogenize the row sums of $A$
recalling that the Perron vector has strictly positive components,
define diagonal matrix $D := \text{diag}\big(\mathbf v\big)$
$B:=\big(D^{-1}AD\big)$
$ B\mathbf 1 = \big(D^{-1}AD\big)\mathbf 1=D^{-1}A\big(D\mathbf 1\big) = D^{-1}A\mathbf v = \lambda_1 D^{-1}\mathbf v = \lambda_1 \mathbf 1$
so the one's vector is the Perron vector for $B$.
step 2: symmetrize Perron vector
now $B^T$ has a Perron vector $\mathbf w$. And define
$\mathbf z:= \mathbf w^\frac{1}{2}\cdot \frac{1}{\big \Vert \mathbf w^\frac{1}{2}\big \Vert_2}$
(where the square root is understood to be taken component-wise)
$\Gamma:= \text{diag}\big(\mathbf z\big)$
$C:= \Gamma B \Gamma^{-1}$
has $\mathbf z$ as the left and right Perron vector, which completes the proof.
