I was reading one of the theorem in Roger A. Horn's Matrix Analysis and yet failed to understand how to prove it.
Let $A=[a_{ij}] \in M_n$ be nonnegative and $\rho(A)$ is spectral radius of $A$. Then $$\min_{1 \le i \le n} \sum_{j=1}^n a_{ij} \le \rho(A) \le \max_{1 \le i \le n} \sum_{j=1}^n a_{ij}$$ and $$\min_{1 \le j \le n} \sum_{i=1}^n a_{ij} \le \rho(A) \le \max_{1 \le j \le n} \sum_{i=1}^n a_{ij}$$ Can anyone help me to give me brief explaination and detailed proof of this theorem?
It suffices to show the result when $A$ is a positive matrix (the $(a_{i,j})$ are $>0$); indeed we conclude, using the continuity of $\rho(A)$ wrt the $(a_{i,j})$.
When $A$ is positive, $\rho(A)\in spectrum(A)$ and $A$ admits an eigenvector $v=(v_i)$ s.t. $v_i>0$ and $Av=\rho(A)v$.
That is, for every $i$, $(*)$ $\sum_ja_{i,j}v_j=\rho(A) v_i$.
Let $v_k=\sup_j v_j$. Then (take $i=k$ in $(*)$) $\rho(A)\leq \sum_j a_{k,j}\leq \max_i \sum_j a_{i,j}$ .
Let $v_l=\inf_j v_j$. Then (take $i=l$ in $(*)$) $\rho(A)\geq \sum_j a_{l,j}\geq \min_i \sum_j a_{i,j}$.
For the last two inequalities, consider $A^T$. Note that $A^T$ is positive and has same spectral radius as $A$.