What is the spectral radius of a non-diagonal matrix?

Question

This is my first question in Math StackExchange.

Assume that I know the spectral radius of matrix $A$. The matrix $\bar{A}$ is created from $A$ by removing all the $A$'s diagonal entries (i.e., $\bar{a}_{ij} = {a}_{ij}$ if $i \neq j$, $0$ otherwise). Can I find the value, or at least the upper bound, of spectral radius of $\bar{A}$?

Thank you in advance.

Answer 1

An attempt:

An upper bound for the spectral radius of $\overline{A}$.

First note that $\overline{A}=A-D$ where $D=[a_{11},a_{22},\ldots,a_{nn}]$, the diagonal matrix with diagonal entries $a_{ii}$. Therefore $\text{spectral of radius}(\overline{A})= \text{spectral of radius}(A-D)$.

Therefore we will search for the best upper bound for the spectral radius of $A-D$

Also if $A$ is a matrix with zero diagonal, then $\text{spectral of radius}(\overline{A})= \text{spectral of radius}(A)+\text{spectral of radius}(D)$, since $\overline{A}=A$ and $D=0$.