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Math Help - Show that rho(A) <= ||A||

  1. #1
    Member Mollier's Avatar
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    Show that rho(A) <= ||A||

    I have been trying to understand matrix norms for a while and I do not find it easy.
    If someone has a good writeup on the subject for us slow ones, please share.

    Anyway, here's the problem I'm currently working on:

    problem:
    Let ||*|| denote any norm on \mathbb{C}^m and also the induced matrix norm on \mathbb{C}^{m\times m}.
    Show that \rho(A)\leq ||A||, where \rho(A) is the spectral radius of A, i.e., the largest absolute value |\lambda| of an eigenvalue of A.

    attempt:
    Ax=\lambda x
    - I take the norm on both sides. (I don't really know if this is a valid step)
    ||Ax||=||\lambda x||=|\lambda| ||x||

    I now use the fact that ||A|| is the smallest number for which the inequality (1) holds for any x\in\mathbb{C}^n

    (1) ||Ax||\leq C||x||

    ||Ax||=||\lambda x||=|\lambda| ||x||\leq C||x||

    and so
    <br />
|\lambda|\leq C.
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  2. #2
    Senior Member roninpro's Avatar
    Joined
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    Looks fine to me!

    (By the way, Wikipedia has some thoughts on this problem: Spectral radius - Wikipedia, the free encyclopedia.)
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