TITLE: Operator norm vs spectral radius for positive matrices
QUESTION [5 upvotes]: I believe the following statement should be true but somehow I don't see an argument:
For every integer $d>1$ there exists a constant $C=C(d)>1$ such that whenever $A$ is a $d \times d$ matrix with all entries integers $>0$, then 
$$ \frac{1}{C} \|A\| \le \sigma(A) \le C \|A\|.$$
Here $\sigma(A)$ is the spectral radius of $A$ (which in this case is the Perron-Frobenius eigenvalue of $A$), and $\|A\|=\sup_{\|v\|=1} \|Av\|$ 
is the operator norm of $A$.
Does anyone see how to prove (or disprove) this statement?
Many thanks,
Ilya.

REPLY [9 votes]: In fact, one can improve a little bit Terry's bound to show that
$\|A\|\le(\sigma(A))^2$, and therefore
  $$ \sigma(A) \le \|A\| \le (\sigma(A))^2. $$
Indeed, this is an immediate consequence of the fact that for a
matrix $A=(a_{ij})_{1\le i,j\le n}$ with positive integer entries one has
$(\sigma(A))^2\ge\sum_{i,j=1}^n a_{ij}$, and the observation that if
$x=(x_1,\ldots,x_n)$ is a unit-length vector, then
  $$ \|Ax\|^2 = \sum_{i=1}^n \Big( \sum_{j=1}^n a_{ij} x_j \Big)^2
              \le \sum_{i=1}^n \Big( \sum_{j=1}^n a_{ij} \Big)^2
              \le \Big( \sum_{i,j=1}^n a_{ij} \Big)^2, $$
implying $\|A\|\le \sum_{i,j=1}^n a_{ij}$.
Notice that by Mike Jury's example above, with $\sigma(A)=\sqrt n+1$
and $\|A\|>n$, the estimate $\|A\|\le(\sigma(A))^2$ is
essentially best possible.
Also notice that the argument just presented actually works for any matrix $A$ with all entries greater than or equal to $1$; the assumption that the entries are
integer is not really used.