Exercise 6.3: Nonnegative Matrices and Operator Norms

(a)

Using that $0 \leq A \leq B$ element-wise, $\begin{matrix} (1) & [A^{2}]_{i j} = \sum_{k = 1}^{d} A_{i k} A_{k j} \leq \sum_{k = 1}^{d} B_{i k} B_{k j} = [B^{2}]_{i j}, \end{matrix}$ and deduce $0 \leq A^{m} \leq B^{m}$ by induction.

(b)

By (a), we have $0 \leq (A^{m})^{2} \leq (B^{m})^{2}$ . Therefore, summing all elements, $‖ A^{m} ‖_{F} \leq ‖ B^{m} ‖_{F}$ , so $‖ A^{m} ‖_{F}^{1 / m} \leq ‖ B^{m} ‖_{F}^{1 / m}$ . Then take $m \to \infty$ to find $λ_{1} (A) \leq λ_{1} (B)$ .

(c)

Like (a), $\begin{matrix} (2) & [C^{2}]_{i j} = \sum_{k = 1}^{d} C_{i k} C_{k j} \leq \sum_{k = 1}^{d} | C_{i k} | | C_{k j} | = [| C |^{2}]_{i j}, \end{matrix}$ and deduce $(C^{m})^{2} \leq (| C |^{m})^{2}$ by induction. Therefore, like (b), summing all elements, $‖ C^{m} ‖_{F} \leq ‖ | C |^{m} ‖_{F}$ , so $‖ C^{m} ‖_{F}^{1 / m} \leq ‖ | C |^{m} ‖_{F}^{1 / m}$ , and take $m \to \infty$ to find $λ_{1} (C) \leq λ_{1} (| C |)$ .

Published on 9 April 2021.