Results 1 to 1 of 1

Thread: Euclidean Norm

  1. #1
    Junior Member
    Sep 2009
    Johannesburg, South Africa

    Euclidean Norm

    We have the Euclidean norm (or l_2-norm) for a vector on the vector space R^m ; \parallel{v}\parallel_2 = \sqrt{\sum_{j=1}^m\mid{v_j}\mid^2}.

    Given a vector norm ||\cdot|| on R^m the induced matrix norm for m x m matrices A is defined by
    ||A||\ =\ max_{v\neq{0}}\frac{||Av||}{||v||}
    That is , ||A|| is the smallest number \alpha such that  ||Av||\leq\alpha||v||\ \forall\ v \in R^m

    So given the Euclidian-norm for a vector the induced matrix norm ||A||_2=\sqrt{max.\ eigenvalue A^TA}
    Can anyone proof/explain why?
    In my book it is given but I donīt see it directly, i think it has something to do with the fact that A^TA is symmetric.
    Last edited by bram kierkels; Oct 4th 2009 at 09:02 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Find vectors using the euclidean norm and inner product.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 9th 2011, 06:07 AM
  2. Replies: 3
    Last Post: Jul 13th 2010, 07:37 PM
  3. Replies: 2
    Last Post: Nov 7th 2009, 01:13 PM
  4. Euclidean Norm and Maximum Norm
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 7th 2009, 05:26 AM
  5. Euclidean norm of U*x when norm of x is 1!
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Oct 2nd 2009, 01:45 AM

Search Tags

/mathhelpforum @mathhelpforum