Let $A$ be a non-negative $N \times N$ square matrix with $a\_{i,i}=0, 1 \leq i \leq N$. Also, let $r\_i$ be the $i$-th row sum of $A$. 


I know that $\rho(A)$, the spectral radius of $A$, is bounded as follows:


$min(r\_i) \leq \rho(A) \leq max(r\_i); 1 \leq i \leq N$


If I increase the row sum of an arbitrary row of $A$, then is $\rho(A) \leq \rho(A')$ true? 


$A'$ is the resulting matrix where a random row sum has been increased. Note that increasing the row sum may involve decreasing the value of an element in that row but increasing other element(s) of the row by a larger amount (i.e., $a\_{i,j}\leq a'\_{i,j}$ is not necessarily true for all $i,j$). 


Let's assume there will never be a negative element and the elements in the principal diagonal are not changed. 


\*This question is related to one of my previous questions but I find this formulation more general and probably/hopefully more people may find it useful. 



Spectral radius's relation with row sum
No. For example, the spectral radii of 
$$ A = \pmatrix{0 & 1 & 1\cr 1 & 0 & 1\cr 0 & 0 & 0\cr},\ A' = \pmatrix{0 & 0 & 3\cr 1 & 0 & 1\cr 0 & 0 & 0\cr}$$
are $1$ and $0$ respectively.


