- #1

Yagoda

- 46

- 0

## Homework Statement

Let [itex]\textbf{A}[/itex] be an m x n matrix and [itex]\lambda = \max\{ |a_{ij}| : 1 \leq i \leq m, 1 \leq j \leq n \}[/itex].

Show that the norm of the matrix [itex]||\textbf{A}|| \leq \lambda \sqrt{mn}[/itex].

## Homework Equations

The definition I have of the norm is that [itex]||\textbf{A}||[/itex] is the smallest number such that [itex]|\textbf{Ax}| \leq ||\textbf{A}|| \, |\textbf{x}|[/itex] for all [itex]\textbf{x}[/itex] in ℝ

^{n}.

## The Attempt at a Solution

I let [itex]\textbf{y} = \textbf{Ax} [/itex] and so [itex]|\textbf{y}| = \sqrt{\sum_{i=1}^{m}y_i^2}[/itex].

I started by looking at a matrix [itex]\textbf{A}[/itex] with all the same entries so that any entry could be thought of as [itex]\lambda[/itex]. So when you calculate [itex]|\textbf{y}|[/itex] you will get [itex]m \lambda^2[/itex]'s under the radical along with sum of the [itex]x_1, \dots, x_n[/itex] squared. So in the more general case that not all the entires of [itex]\textbf{A}[/itex] are equal I can see how [itex]\lambda[/itex] would provide an upper bound.

But I'm having trouble seeing how the [itex]\sqrt{n}[/itex] comes into it.