- #1

garrus

- 17

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## Homework Statement

Show that [itex]||A||_1 \le \sqrt{n} ||A||_2 , ||A||_2 \le \sqrt{n} ||A||_1[/itex] , where

[tex]||A||_1 = \max_{1\le j\le n}\sum_{i=1}^n |a_{ij}| \\

||A||_2 = (p(A^TA))^\frac{1}{2} \\

p(B) = \max|\lambda_B|

[/tex]

with [itex]A,B\in \mathbb{R}^{n,n}, i,j\in[1...n] , \lambda_A[/itex]the eigenvalues of matrix A

## Homework Equations

wiki page on matrix norms

## The Attempt at a Solution

I figured i could go the same way in proving that [itex]||x||_1 \le \sqrt{n} ||x||_2 , x\in \mathbb{R}^n[/itex] via the Cauchy Schwartz inequality.But since the max operator is not linear, isn't it a mistake to write

[tex]

||A||_1 = \max_{1\le j\le n} \sum_{i=1}^{n}|a_{ij}| =

\max_{1\le j\le n}\sum_{i=1}^{n}|a_{ij}| * 1 \le

\max_{1\le j\le n} (\sum_{i=1}^{n}|a_{ij}|^2 * \sum_{i=1}^{n}1^2)^{\frac{1}{2}}

[/tex] ?

Any hints?

edit:

Slightly off topic question: In triangle inequality, it holds that

|a - b| >= |a| - |b| . can we also bound it from below, by |a-b| = |a +(-b)| <=|a|+|b| ?

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