Matrix reciprocal positive to prove $λ_{max⁡}=n$ 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 $λ_{max}$ 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 already found an example of $5\times 5$ reciprocal matrix  to show this condition:
\begin{bmatrix}1&1/2&1&1&1/4\\2&1&2&2&1/2\\1&1/2&1&1&1/4\\1&1/2&1&1&1/4\\4&2&4&4&1\end{bmatrix}
It has simple characteristic equation $λ^5-5λ^4=0$, so the eigenvalues of matrix $A$ are $λ=0$ and $λ=5$ and it's shown that $λ_{max}=n=5$.
Is there anyone can help me to give another positive reciprocal matrix example  that has simple characteristic equation and integer number as its eigenvalues ($4\times 4, 5\times 5$ or $6\times 6$ matrix)?
 A: $$\pmatrix{1&2&3&4\cr1/2&1&3/2&2\cr1/3&2/3&1&4/3\cr1/4&1/2&3/4&1\cr}$$
A: You get an endless supply of such matrices of any size from the following recipe.


*

*Let $B$ be an $n\times n$ matrix of all $1$.

*Let $D$ be a diagonal $n\times n$ matrix with some positive entries.

*Let $A=DBD^{-1}$.


Here the eigenvalues of $B$ are $\lambda_{max}=n$ (multiplicity one) and $\lambda=0$ (multiplicity $n-1$). Therefore the same applies to $A$. Furthermore, $B$ is trivially reciprocal, and so is $A$, because 
$a_{ij}=d_i/d_j$.
Your example matrix is gotten from this recipe with $n=5$ and
$D=diag(1,2,1,1,4)$.

Actually all the reciprocal matrices satisfying the consistency property are of this form. If $a_{ij}=1/a_{ji}$, then we can use $D=diag(a_{11},a_{21},\ldots,a_{n1})$. This works because the consistency property implies that
$$
a_{ij}=\frac{a_{i1}}{a_{j1}}
$$
for any pair $i,j$.
for all $i,j$
A: If $A$ is a reciprocal matrix, then $D^{-1}AD$ is also a reciprocal matrix for any positive diagonal matrix $D$. Now, suppose $\rho(A)=n$. By Perron-Frobenius theorem, $A$ possesses a positive eigenvector $u$ for the eigenvalue $n$. Denote by $D$ the diagonal matrix whose diagonal is $u$. Then $Au=nu$ is equivalent to $Be=ne$, where $B=D^{-1}AD$ is a reciprocal matrix and $e$ is the vector of ones.
It follows that $e^TBe-n^2=0$. That is, $\sum_{i<j}\left(\sqrt{b_{ij}}-\frac1{\sqrt{b_{ij}}}\right)^2=0$. Hence $b_{ij}=1$ for all $i,j$, or $B=ee^T$.
Therefore $A=Dee^TD^{-1}$, i.e. $A=uv^T$, where $v$ is the entrywise reciprocal of the positive vector $u$. It is straightforward to show that every such $A$ satisfies the condition that $a_{ij}a_{jk}=a_{ik}$.
