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.
At the end of . the author states that  Suppose we have $n \times n$ matrix $A$ having only positive elements and satisfying the property $a_{ij}=1/a_{ji}$ (a matrix satisfying this property is called a reciprocal matrix).  If its largest eigenvalue $\lambda_{max}$ is equal to $n$, then the matrix $A$ satisfies the property (consistency property) $a_{ij} a_{jk} = a_{ik}$ where $i,j,k=1,2,...,n$.  I'm not convinced this theorem is true. Could anyone help?
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