TITLE: Relation between coefficients of a matrix and its eigenvalues
QUESTION [5 upvotes]: Let $A \in \mathbb{R}^{nxn}$ be a matrix and $\rho(A)$ its largest eigenvalue (or largest module of its eigenvalues). Let $a_{ij}$ be a typical entry of the matrix $A$ at the $i$-th row and $j$-th column such that $a_{ij} \in \{0,1\}$ and $a_{ii} \equiv 0$. If the following condition is satisfied for some constant $\alpha > 0$ $$ \alpha \rho(A) < 1 $$ can we deduce that $$\alpha a_{ij} < 1 $$ for all $i$ and $j$ in $\{1, \ldots n \}$ ? 
For example call the following matrix A, \begin{pmatrix} 
   0 & 0 & 1 \\
   1 & 0 & 1 \\
1 & 1 & 0 
\end{pmatrix}
has eigenvalues  $-0.6180$ and $1.6180$ and $-1.0000$. Since $\rho(A)= 1.618$ it seems to be true for this special case. Can anyone see an obvious counter example? 
We know from Gershgorin circle theorem that every eigen value of the square matrix $A$ lies in at least one of Gershgorin's disc $D(a_{ii} , R_i)$, where $D(a_{ii} , R_i)$ is a closed disc centered at $a_{ii}$ with radius $R_i = \sum_{ j \neq i } |a_{ij} |$. So we have an estimate of the range of the eigenvalues but it doesn't directly answer my question.

REPLY [2 votes]: Question: Let $A=(a_{i,j})\in\{0,1\}^{n\times n}$, find $I\subset [0,\infty)$ such that $$\alpha \in I\qquad \iff \qquad \begin{cases}\alpha\ \rho(A)<1 \\ \alpha\ a_{i,j}<1 & \forall i,j=1,\ldots,n\end{cases}$$
Answer:
   - If $\rho(A)>0$, then $I=\big[0,\min\{1,\rho(A)^{-1}\}\big).$
   - If $\rho(A)=0$ and $A\neq 0$, then $I= [0,1)$
   - If $A=0$, then $I =[0,\infty)$.

Note that if $\rho(A)> 0$
$$ \alpha \ \rho(A)< 1 \quad\iff\quad \alpha < \rho(A)^{-1} \qquad \text{and}\qquad \alpha \ a_{i,j}<1\quad \forall i,j\quad \iff\quad \alpha <1.$$
Where we have used that $\rho(A)>0$ implies $A\neq 0$ and thus there is $i,j$ such that $a_{i,j}=1$.
If $\rho(A)=0$ and $A\neq 0$, then
$$ 0=\alpha \ \rho(A)< 1 \qquad \text{and}\qquad \alpha \ a_{i,j}<1\quad \forall i,j\quad \iff\quad \alpha <1$$
If $A=0$ then the results is obvious.