Let $A$ and $B$ be $n \times n$ non-negative matrices with spectral radii $\lambda_A$ and $\lambda_B$, respectively. Suppose that $A_{ij} \leq B_{ij}$ for all $i,j$. Then, $$\lambda_A \leq \lambda_B$$
I found this theorem in a paper, but the author did not provide proper reference on the proof. After searching in some texts, I only found the version for when $B$ is irreducible.
If this is true, could anyone kindly point me a reference on this theorem?
Please kindly suggest me a suitable book about matrix theory and spectral radius since I haven't read much about this topic.
Thank you in advance for the help.
If $0\leqslant A_{ij}\leqslant B_{ij}$ for all $i,j$ then $$\sum_{i=1}^n |A_{ij}| \leqslant \sum_{i=1}^n |B_{ij}| \text{ for all } j, $$ and hence $\|A\|_1\leqslant \|B\|_1$, where $$ \|A\|_1 = \max_{1\leqslant j\leqslant n}\sum_{i=1}^n |A_{ij}|. $$ By monotonicity of the maps $A\mapsto A^k$ and $x\mapsto x^{1/k}$ it follows that $$\|A^k\|_1^{1/k}\leqslant\|B^k\|_1^{1/k}$$ for all $k>0$. Hence by Gelfand's formula $$ \rho(A) = \lim_{k\to\infty}\|A^k\|_1^{1/k}\leqslant \lim_{k\to\infty}\|B^k\|_1^{1/k}=\rho(B), $$ where $\rho(\cdot)$ denotes spectral radius.