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Matrix norms 1,2 equivalence

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    wiki page on matrix norms


    3. 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| ?
     
    Last edited: Aug 18, 2012
  2. jcsd
  3. Aug 21, 2012 #2
    Noone?

    One another norm problem, i'm given.If you could verify / correct:
    [tex]
    A\in\mathbb{R}^{n,n} , x\in \mathbb{R}^n. \|Ax\|\ge \frac{\|x\|}{10}
    [/tex]
    and ask to show that [itex] \|A^{-1}\| \le 10[/itex]
    where [itex]||\cdot||[/itex] a norm and the corresponding matrix norm derived by it.

    [tex]
    \|Ax\|\ge \frac{\|x\|}{10} \iff \|A\| \|x\| \ge\|Ax\|\ge \frac{\|x\|}{10} \Rightarrow
    \|A\| \ge \frac{1}{10} (*)
    [/tex]
    Only way i managed was to assert a value and lead to contradiction,but i suspect there must be a more elegant way.Let [itex] \|A^{-1}\| > 10[/itex]
    [tex]
    \|A^{-1}\| > 10 \iff ||A|| \|A^{-1}\| > 10 \|A\| \\ but\\
    1 = \|A A^{-1}\| \le \|A\| \|A^{-1}\| \\
    (*) \Rightarrow 10 \|A\| \ge 1
    [/tex]
    , which is a contradiction, so [itex]\|A^{-1}\| > 10[/itex] is false.Thus [itex]\|A^{-1}\| \le 10[/itex]
     
    Last edited: Aug 21, 2012
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