Spectral Radius of Perturbed Matrix Let $A$ be an n-dimensional square matrix such that each entry of $A$ lies in $[0,1]$, i.e., $a_{ij} \in [0,1]$ for all $1\le i,j\le n$. Let $\widetilde{A}$ be a perturbed version of $A$ where some entries of $\widetilde{A}$ might be larger than those in $A$ and the rest equal, i.e.,
\begin{align*}
1 \ge \widetilde{a}_{ij} \ge a_{ij} \text{ for all } i,j \\
\widetilde{a}_{ij} > a_{ij} \text{ for some } i,j.
\end{align*}
I am working on a problem where it would help me to have the spectral radius $\widetilde{\rho}$ of $\widetilde{A}$ to be at least the spectral radius $\rho$ of $A$ and wondering if this would be true?
Any help would be appreciated on the intuition of why or why not it may hold. With thanks.
 A: For $A \in \mathbb{R}^{n\times n}$ and any matrix norm $\|\cdot\|$ it is known that $\lim_{n \to \infty} \|A^n\|^{1/n} = \rho(A)$. Now let $A=(a_{ij}),B=(b_{ij})$ be nonnegative matrices. If $A \le B$, that is $a_{ij} \le b_{ij}$ $(i,j=1,\dots,n)$, check that then $A^n \le B^n$ $(n \in \mathbb{N})$. Next, consider the row sum norm $\|\cdot\|_\infty$ on $\mathbb{R}^{n\times n}$ (or any other matrix norm which is nondecreasing on the set of nonnegative matrices). Now $\|A^n\|_\infty \le \|B^n\|_\infty$ $(n \in \mathbb{N})$, hence $\|A^n\|_\infty^{1/n} \le \|B^n\|_\infty^{1/n}$ $(n \in \mathbb{N})$. As $n \to \infty$ this yields $\rho(A) \le \rho(B)$. Summing up, if $\mathbb{R}^{n\times n}$ is ordered by the coordinatewise ordering then the spectral radius is nondecreasing on the set of nonnegative matrices.
Finally, a strict inequality in one entry in general does not imply a strict inequality for the spectral radius: For
$$
A=\left(\begin{array}{cc}
1 & 1 \\
0 & 1 \\
\end{array} \right), \quad
B=\left(\begin{array}{cc}
1 & 2 \\
0 & 1 \\
\end{array} \right)
$$
we have $\rho(A)=\rho(B)=1$.
