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Thread: Euclidean Norm

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

    Euclidean Norm

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

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

    So given the Euclidian-norm for a vector the induced matrix norm $\displaystyle ||A||_2=\sqrt{max.\ eigenvalue A^TA}$
    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 $\displaystyle A^TA$ is symmetric.
    Last edited by bram kierkels; Oct 4th 2009 at 08:02 AM.
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