If $\lambda_{max} = n$, then $n\times n$ positive, reciprocal matrix A is consistent At the end of chapter 3. 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?
 A: Let $A$ be an $n\times n$ positive and reciprocal matrix whose maximum eigenvalue is $n$. 
We show that $A$ is consistent by showing that there exists a vector $\mathbf{w}=(w_1,\ldots,w_n)$ such that $(A)_{ij}=\frac{w_i}{w_j}$ for all $i,j$.
The vector that we will see works is any eigenvector $\mathbf{w}=(w_1,\ldots,w_n)$ for the largest eigenvalue $\lambda_\max$ of $A$. By the Perron-Frobenius Theorem, $\mathbf{w}$ is always positive, i.e., each entry is positive.
Let $W$ be the matrix with $(W)_{ij}=\frac{w_i}{w_j}$. Let $E$ be the matrix defined by $(E)_{ij}=(A)_{ij}\frac{w_j}{w_i}$. Then $E$ is also positive and reciprocal and $W\circ E = A$, where $\circ$ denotes the Hadamard product of $W$ and $E$ (i.e., the entry-wise product).
Now let's add up the entries in row $i$ of $E$. Since $\mathbf{w}$ is an eigenvector with eigenvalue $\lambda_\max$ we have \begin{eqnarray*} \sum_{j=1}^n (E)_{ij} &=& \sum_{j=1}^n (A)_{ij}\frac{w_j}{w_i} \\ &=& \frac{1}{w_i} \sum_{j=1}^n (A)_{ij}w_j \\ &=& \frac{1}{w_i} \lambda_\max w_i \\ &=& \lambda_\max.\end{eqnarray*}
Therefore the sum of all entries of $E$ is equal to $n\lambda_\max$. 
On the other hand, since $E$ is a positive reciprocal matrix, the diagonal entries are equal to $1$. Hence the sum of all entries of $E$ is equal to \begin{eqnarray*} \sum_{i,j=1}^n (E)_{i,j} &=& \sum_{i=1}^n (E)_{i,i} + \sum_{1\leq i< j \leq n} \left ((E)_{i,j} + (E)_{i,j}^{-1}\right)\\ &\geq& n+{n\choose 2} 2,\end{eqnarray*}
where the inequality follows from the fact that $x+x^{-1}\geq 2$ for all positive $x$ with equality if and only if $x=1$.
So we have $n+{n\choose 2} 2 \leq n\lambda_\max$ with equality if and only if each $(E)_{i,j}=1$. 
But if we assume that $\lambda_\max =n$, then $n+{n\choose 2} 2 = n^2$ and so we do have equality. Therefore $(E)_{i,j}=1$ for all $i,j$ from which we conclude that $A=W$ and so $A$ is consistent.
