Exercise 6.3: Nonnegative Matrices and Operator Norms
(a)
Using that $0 \le A \le B$ element-wise, \begin{equation} [A^2]_{ij} = \sum_{k=1}^d A_{ik} A_{kj} \le \sum_{k=1}^d B_{ik} B_{kj} = [B^2]_{ij}, \end{equation} and deduce $0 \le A^m \le B^m$ by induction.
(b)
By (a), we have $0 \le (A^m)^2 \le (B^m)^2$. Therefore, summing all elements, $\norm{A^m}_F \le \norm{B^m}_F$, so $\norm{A^m}_F^{1/m} \le \norm{B^m}_F^{1/m}$. Then take $m \to \infty$ to find $\lambda_1(A) \le \lambda_1(B)$.
(c)
Like (a), \begin{equation} [C^2]_{ij} = \sum_{k=1}^d C_{ik} C_{kj} \le \sum_{k=1}^d |C_{ik}| |C_{kj}| = [|C|^2]_{ij}, \end{equation} and deduce $(C^m)^2 \le (|C|^m)^2$ by induction. Therefore, like (b), summing all elements, $\norm{C^m}_F \le \norm{|C|^m}_F$, so $\norm{C^m}_F^{1/m} \le \norm{|C|^m}_F^{1/m}$, and take $m \to \infty$ to find $\lambda_1(C) \le \lambda_1(|C|)$.