Suppose a square matrix $A$ has spectral radius $\rho(A) < 1$. Fixing the last row and scaling other entries by $r \in (0,1)$, will $\rho(A)<1$?

Question

Suppose $A \in M_n(\mathbb R)$ has spectral radius small than $1$, i.e., $\rho(A) < 1$. Denote $A = \pmatrix{a_1^T \\ \vdots \\ a_n^T}$, where $a_j^T$ denotes the $j^{th}$ row of $A$.Putting $B=\pmatrix{a_1^T \\ \vdots \\ a_{n-1}^T}$, i.e., the first $n-1$ rows of $A$, if $r \in (0, 1)$ what should the spectral radius of $(A)_r = \pmatrix{rB \\ a_n^T}$ be? That is, $(A)_r$ is the matrix we fix the last row of $A$ but scale other rows by a factor $r$. I feel that $\rho( (A)_r)$ should be smaller than $1$ too but could not prove it or give a counterexample.

Answer 1

The answer is no. Let $A$ be any nilpotent matrix such that $|a_{nn}|>1$, such as $A=\alpha\pmatrix{-1&-1\\ 1&1}$ for any $\alpha>1$. Then $\rho(A)=0$ but $\rho(A_r)\to|a_{nn}|>1$ when $r\to0$.

However, the answer is yes if $A$ is (entrywise) nonnegative. In this case, as $0\le A_r^k\le A^k$ for all positive integer $k$, by using Gelfand's formula with Frobenius norm, we have $$ \rho(A_r)=\lim_{k\to\infty}\|A_r^k\|^{1/k}\le\lim_{k\to\infty}\|A^k\|^{1/k}=\rho(A)<1. $$